Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Use Integration by Parts to prove that $\int x^{n}\ln{x}\ dx=\frac{x^{n+1}}{(n+1)^{2}}\left[-1+(n+1)\ln{x}\right]+c$ I've gotten most of the way, but I can't see how I can transform my answer into the form in the assignment:
Use Integration by Parts to prove that $\displaystyle\int x^{n}\ln{x}\ dx=\frac{x^{n+1}}{(n+1)^... | Your work is fine. We just need to algebraically "manipulate" the result a bit to get the answer to match the given equality (as stated in your title):
Starting with your second-to-last line, we find a common denominator, $(n+1)^2$, and then factor out the common factor, which is the term you list in your last line, b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/319921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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How to find last items of order to get its sum Find the sum of order:
$$\sum_{n=1}^{∞}\left(\frac{\frac{3}{2}}{2n+3}-\frac{\frac{3}{2}}{2n-1}\right)$$
There is how they count it in book:
$$s_{n} = \left(\frac{3}{10}-\frac{3}{2}\right)+\left(\frac{3}{14}-\frac{1}{2}\right)+\left(\frac{1}{6}-\frac{3}{10}\right)+\left(\fr... | I hope i didn't missunterstood your question, this sum is a so called telescoping sum, the other terms are canceled so you only have those left which are mentioned.
Those are
\begin{align*}
\frac{3}{4n-10}&= \frac{3}{4(n-1)-6}\\
\frac{3}{4n-6}&= \frac{3}{4(n-1)-2} \\
\end{align*}
And $$\frac{3}{4n-2}$$ is canceled from... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/321319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Factorization of three variables Prove that : $(a+b+c)^3-(b+c)^3-(c+a)^3-(a+b)^3+a^3+b^3+c^3=6abc$
Since $(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a) $
Therefore the equation becomes : $2(a^3+b^3+c^3)+3(a+b)(b+c)(c+a) - [(c+a)^3 +(a+b)^3 +(b+c)^3]$
Putting $A:=(c+a)$ ; $B:=(a+b)$ ; $C:=(b+c)$
$[(c+a)^3 +(a+b)^3 +(b+c)^3] $ ... | If you simply expand everything and collect terms, everything will cancel out and reduce down to 6abc. However, if you want to factorize it, think about the other terms you have -
As you said, $(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a)$. So then what does $(a+b)^3$ equal? If you expand it out following the same format as... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/321402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Finding the critical point and local extreme value? How would I find the critical point of the following function.
$f(x)=x^{2/3}+2x^{-1/3}$
This is what I did.
$\frac{2}{3}x^{-1/3}-\frac{2}{3}x{^\frac{-4}{3}}$
$\frac{2}{3\sqrt[3]{x}}-\frac{2}{3\sqrt[3]{x^4}}$
but how do I simplify it and get the critical point.
| $$
\begin{align}
f'(x) = \frac{2}{3}x^{-1/3}-\frac{2}{3}x^{-4/3}
& = \frac 23\left(x^{-1/3} - x^{-4/3}\right) \\ \\
& = \frac 23\left(x^{3/3}x^{-4/3} - x^{-4/3}\right) \\ \\
& = \frac 23 x^{-4/3}\left(x^{3/3} - 1\right) \\ \\
& = \frac 23 x^{-4/3}(x - 1) \\ \\
& = \frac{2(x-1)}{3x^{4/3}} \\ \\
& = \frac{2(x-1)}{3\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/321657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate these $4$ integrals?
\begin{align}
(1) & \int \frac{\sqrt{x^4 + \frac{1}{x^4} + 2}}{x^3}dx\\
(2) & \int \sqrt{1 - \sin(2x)}dx & \forall x \in [-\pi/4, \pi/4]\\
(3) & \int \cos^2(2x) \cos(x)dx\\
(4) & \int \frac{dx}{\sqrt{x +1}(x+5)}
\end{align}
I'd be very grateful for any help and hints!
| $1$. $\sqrt{x^4 + \dfrac1{x^4} + 2} = x^2 + \dfrac1{x^2}$.
$2$. $\sqrt{1-\sin(2x)} = \sqrt{\cos^2(x) + \sin^2(x) -2 \sin(x) \cos(x)} = \cos(x) - \sin(x)\,\,\, \forall x \in [-\pi/4,\pi/4]$.
$3$. $\cos^2(2x) \cos(x) = \dfrac{1+\cos(4x)}2 \cdot \cos(x) = \dfrac{\cos(x)}2 + \dfrac{\cos(5x) - \cos(3x)}4$.
$4$. $\dfrac{dx}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/322918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Closed form for the sum of even fibonacci numbers? I recently took a look a project euler, and I am trying to think of a smart way to do number 2. I looked at the sequence, and I saw that the question is basically asking for
$$
\sum_{i=1}^n F_{3i}
$$
For whatever n is gives me the nth even under a million. So, I did so... | $$F_k=\frac{\left(1+\sqrt{5}\right)^k}{2^k\sqrt{5}}-\frac{\left(1-\sqrt{5}\right)^k}{2^k\sqrt{5}}$$
$$\sum_{k=1}^nF_{3k}=\sum_{k=1}^n\frac{\left(1+\sqrt{5}\right)^{3k}}{2^{3k}\sqrt{5}}-\sum_{k=1}^n\frac{\left(1-\sqrt{5}\right)^{3k}}{2^{3k}\sqrt{5}}$$
$$=\frac{1}{\sqrt5}\sum_{k=1}^n\left(\frac{1+\sqrt{5}}{2}\right)^{3k}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/323058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Integrating $\int^0_\pi \frac{x \sin x}{1+\cos^2 x}$ Could someone help with the following integration:
$$\int^0_\pi \frac{x \sin x}{1+\cos^2 x}$$
So far I have done the following, but I am stuck:
I denoted $ y=-\cos x $ then:
$$\begin{align*}&\int^{1}_{-1} \frac{\arccos(-y) \sin x}{1+y^2}\frac{\mathrm dy}{\sin x}\\&... | Let $y=x-\frac{\pi}{2}$.
\begin{array}{l}\displaystyle \int_{0}^{\pi} \frac{x \sin x}{1+\cos ^{2} x} d x&=\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\left(y+\frac{\pi}{2}\right) \cos y}{1+\sin ^{2} y} d y\\&=\displaystyle \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{y \cos y}{1+\sin ^{2} y} d y +\frac{\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/323109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 3
} |
Using Hensel's Lemma to Factor a Polynomial over $\mathbb{Z}_4[x]$ We recently learned about codes over $\mathbb{Z}_4$, and Hensel's Lemma. The lemma is as follows:
Let $f(x) \in \mathbb{Z}_4[x]$. Suppose $\mu(f(x)) = h_1(x)h_2(x) \cdots h_k(x)$, where $h_1(x), h_2(x), \ldots , h_k(x)$ are pairwise coprime polynomials... | We are going to lift the $\mathbb{F}_2$-factorization $$\bar{f} = (\underbrace{x^3 + x + 1}_{=:g})(\underbrace{x^4 + x^2 + x + 1}_{=:h})$$ to $\mathbb{Z}_4$. That is, considering $g$ and $h$ as polynomials in $\mathbb{Z}_4[x]$, we look for lift-polynomials $a,b\in\mathbb{F}_2[x]$, $\deg(a)\leq 2$, $\deg(b)\leq 3$ such ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/323956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Consider the ordinary differential equation, $\frac{dy}{dx}=\frac{4+y^2}{1+x^2}$ Consider the ordinary differential equation, $\frac{dy}{dx}=\frac{4+y^2}{1+x^2}$
If $y(1)=2$, then find the value of $y(2)$
$\int \frac{dy}{4+y^2}=\int \frac{dx}{1+x^2} -- (|)$
$\frac{1}{2}\ tan^{-1}\frac{y}{2}=\tan^{-1}x+c$
$tan^{-1}\frac... | $$\frac12\arctan \frac y2=\arctan x+c$$
As $y(1)=2, \frac12\arctan 1=\arctan 1+c$
$\implies c=-\frac12\arctan 1=-\frac\pi8$
So, $$\frac12\arctan\frac y2=\arctan x-\frac\pi8\iff \arctan y=2\arctan x-\frac\pi4$$
$$\implies \frac y2=\tan\left(2\arctan x-\frac\pi4\right)$$
$$=\frac{\tan(2\arctan x)-\tan\frac\pi4}{1+\tan(2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/324829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Variable Exponents Question If $a,b$ and $c$ are different positive integers and $2^a\cdot2^b\cdot2^c =64$ then $2^a+2^b+2^c$=?
This is so far my work: I got $2^a\cdot2^b\cdot2^c=2^6$ then $abc=6$ is this so far in the right track?
| Here's the $log$ solution,
$log 2^{a+b+c}=log 64$
$(a+b+c).(0.3010)=6.(0.3010)$
$a+b+c=6$
$a \neq b \neq c \implies log 2^a\neq log 2^b \neq log 2^c$
$log 2^{a+b+c}=log2^a+log2^b+log2^c=6.log2$
$a,b,c$ are distinct, therefore $(a,b,c)=(1,2,3),(3,2,1),(1,3,2),(2,3,1),(2,1,3),(3,1,2)$
Therefore, $2^a+2^b+2^c= 2+ 4+ 8 = 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/325127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
A recurrence relation on two variables How to solve the recurrence relation for $n >=m$:
$$P_{n,m}=\frac{n}{n+m}P_{n-1, m} + \frac{m}{m+n}P_{n,m-1}$$
$$P_{11}=\frac{1}{2}; P_{i,0}=1 \forall i > 0; P_{i,j}=0 \forall i<j$$
| Hint: Have you tried calculating the first few numbers. There is a interesting pattern.
$$\begin{array}{c|cccc}
&0&1&2&3\\
\hline
1&\frac{2}{2} & \frac{3}{3} & \frac{4}{4}& \frac{5}{5}\\
2&\frac{1}{2} & \frac{2}{3} & \frac{3}{4} & \ddots\\
3&0 & \frac{1}{3} & \frac{2}{4} & \ddots\\
4&0 & 0 & \frac{1}{4} & \ddots\\
\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/325437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Norm $N$ on $\Bbb Z[\sqrt d]$ is multiplicative, $N(x)$ prime $\Rightarrow x$ irreducible For the ring $\Bbb{Z}[\sqrt{d}]=\{a+b\sqrt{d}\mid a,b \in \Bbb{Z}\}$ where $d$ is not divisible by the square of a prime, prove that the norm $N(a+b\sqrt{d})= |a^2-db^2|$ satisfies the assertions: $N(x)=0$ if and only if $x=0$; $N... | We define $\mathbb Z[\sqrt d] = \{a + b \sqrt d : a, b \in \mathbb Z\}$ where $d$ is a square-free integer.
Claim: $N(x) = 0$ if and only if $x = 0$.
Proof: If $N(x) = N(a + b \sqrt d) = 0$, then $N(x) = \vert a^2 - db^2 \vert = 0$ which means $a^2 = \vert d \vert b^2$. If either $a$ or $b$ are not zero, the other one... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/325562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Solve using the definition of Big-$\mathcal{O}$ By using the definition of Big-$\mathcal{O}$, show that
\begin{align} 2^{(n+2)} + 3^{(n+1)} & \text{ is } \mathcal{O}(3^n)\\ \sqrt{10n^2+7n+3} & \text{ is } \mathcal{O}(n)\end{align}
I'm not sure where to start with this, especially with the last one. Can someone help p... | By the definition of the Big-$O$ function:
$$ \left| 2^{n+2} + 3^{n+1} \right| \le A \cdot \left| 3^n \right| $$
$$ \left| 4 \cdot 2^n + 3 \cdot 3^n \right| \le A \left| 3^n \right| $$
By the triangular inequality:
$$ \left| 4 \cdot 2^n + 3 \cdot 3^n \right| \le \left| 4 \cdot 2^n \right| + \left| 3 \cdot 3^n \right| ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/326850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Linear algebra proof, dependent or independent Suppose that $p_1=4-3x+6x^2+2x^3$, $p_2=1+8x+3x^2+x^3$, and $p_3=3-2x-x^2$ are vectors in $P_3$. Determine if $p_1$,$p_2$,and $p_3$ are linearly independent or dependent. Justify your answer.
So far, what I did was say:
Suppose $k_1$, $k_2$ and $k_3$ are constants.
Then I ... | That absolutely justifies that $p_1,p_2,p_3$ are linearly independent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/328722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Approximating a sequence with funny recurrence Consider the sequence $a_n$ defined as $a_1=a_2=1, a_{n+1}(1+a_{n})=n+1$.
This sequence describes the average number of fixed points of an involution on an $n$-set, and one can approach the problem of approximating it using complex analysis, but I want to avoid it. Can so... | After you have guessed the bounds, it is now straightforward to prove them using induction.
We shall induct on $n$ to prove the statement $-\frac{1}{2} + \sqrt{n+\frac{1}{4}} \leq a_n \leq -\frac{1}{2} + \sqrt{n+\frac{5}{4}}$.
When $n=1$, we have $-\frac{1}{2}+\sqrt{1+\frac{1}{4}} \leq 1=a_1=-\frac{1}{2}+\sqrt{1+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/328891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Prove $((n+1)!)^n < 2!\cdot4!\cdots(2n)!$ so I know I need to prove this via induction, but I am somewhat stuck. Here is what I have does so far.
Let $p(n) = (n+1)!^n \le 2!\cdot4!\cdot\ldots\cdot(2n)!$
*
*$p(2) = 3!^2\le 2!\cdot4!$
*Assume $p(n)$ is true.
*Prove $p(n+1)$
*$p(n+1) = (n+2)!^{n+1}\le 2!\cdot4!\cdot... | Since you have reached $ (n+2)^n.(n+2)! < (2n+2)! $ , I will lead you from here.
From this we can observe that there are 'n' number of $(n+2)$ terms on the LHS.
we can easily conclude that:
$(n+2) < (n+3)$ ---> $1st (n+2)$
$(n+2) < (n+4)$ ---> $2nd (n+2)$
.......and so on till
$(n+2) < (n+n+2)$ ---> $nth (n+2)$
So, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/329039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Finding $a_n$ for $a_{n+1}=\frac{1}{2}\left(a_n+\frac{1}{a_n}\right)(n=1,2,3,\cdots),~a_1=2$ I would appreciate if somebody could help me with the following problem:
Q: find $a_n=?$
$$a_{n+1}=\frac{1}{2}\left(a_n+\frac{1}{a_n}\right)(n=1,2,3,\cdots),~a_1=2$$
There are quite a few posts on the main site concerning conv... | Rewrite $a_n$ as $\frac{p_n}{q_n}$ for two sequences $p_n, q_n$ that need to be determined.
We have:
$$a_{n+1} = \frac12 ( a_n + \frac{1}{a_n} ) \iff \frac{p_{n+1}}{q_{n+1}} = \frac{p_n^2+q_n^2}{2p_nq_n}$$
We can choose $p_{n+1} = p_n^2 + q_n^2$ and $q_{n+1} = 2p_nq_n$ and get:
$$ p_{n+1} \pm q_{n+1} = ( p_n \pm q_n )... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/329240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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The sum of infinite series $\sum_{k=1}^{\infty}2\sin\left(\frac{1}{k}-\frac{1}{k+1}\right)\cos\left(\frac{1}{k}+\frac{1}{k+1}\right)$
Determine the sum of the infinite series
$$\sum_{k=1}^{\infty}2\sin\left(\frac{1}{k}-\frac{1}{k+1}\right)\cos\left(\frac{1}{k}+\frac{1}{k+1}\right).$$
| This turns out to be a telescoping sum:
$$\sum_{k=1}^{\infty} \left ( \sin{\frac{2}{k}} - \sin{\frac{2}{k+1}}\right ) = \sin{2} - \sin{1} + \sin{1} - \sin{\frac{2}{3}} + \ldots = \sin{2}$$
This sum converges because, as $k \rightarrow \infty$, the summand approaches
$$\frac{2}{k (k+1)} \sim \frac{2}{k^2}$$
so the sum c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/330639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Solve for $X$ in a simple equation system. I cannot really understand how to read this question, so please help me out here.
$$\left[ \begin{array}{ccc} 0 & -6 & 4\\ 1 & 2 & 7\end{array} \right] = 4X + 5 \left[ \begin{array}{ccc} 3 & 4 & 3\\ 8 & -4 & 8\end{array} \right]$$
First, how should i read this? Secondly how do... | Two matrices are the same when all of their values are the same.
Lets call
$$X=\begin{pmatrix} x_{11} & x_{12} & x_{13} \\ x_{21} & x_{22} & x_{23} \end{pmatrix}$$
So your equations are
\begin{align*}
0&= 4 x_{11} +5 \cdot 3\\
-6&= 4 x_{12} + 5 \cdot 4\\
4&=4 x_{13} + 5 \cdot 3\\
1&= 4 x_{21} + 5\cdot 8\\
2&= 4x_{22} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/331014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Number of permutations of a specific cycle decomposition
Let $X(n)$ denote number of all permutations of $\left\{1,\ldots,n \right\}$ that have only cycles of even length. Let $Y(n)$ denote number of all permutations of the same set that have only cycles of odd length. Count $X(2n)-Y(2n)$.
I don't know even how to st... | The exponential generating function for permutations containing only even cycles is
$$ G_1(z) =
\exp\left( \sum_{k\ge 1} \frac{z^{2k}}{2k} \right) =
\sqrt{ \frac{1}{1-z^2}} = \frac{\sqrt{1-z^2}}{1-z^2}$$
and the one for odd cycles is
$$ G_2(z) = \exp\left( \sum_{k\ge 0} \frac{z^{2k+1}}{2k+1} \right).$$
We are lookin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/331504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Partial Fraction Decomposition Problem... half answered... $$\int \frac{5x^3+19x^2+27x-3}{(x+3)^2(x^2+3)}dx$$
I know I will be using partial fraction decomposition on this problem, at least it seems that way. so far, what I have is this:
$$\frac{5x^3+19x^2+27x-3}{(x+3)^2(x^2+3)}=\frac{A}{x+3}{}+\frac{B}{(x+3)^2}+\frac{... | There are many options:
*
*Differentiate and substitute $x=-3$.
*Substitute simple values such as $x=0$ and $x=1$ and $x=-1$ to get three equations in three unknowns.
*The method @Shu mentions in the comments.
*Substitute $x=\sqrt{-3}$ and $x=-\sqrt{-3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/333234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Partial Fractions - combinatorics - trouble Having a very hard time with this question:
Q: Use partial fractions to find the power series expansion of $$\frac{1+5x}{1-2x-3x^2}$$
| The partial fraction decomposition is
$$-\frac{1}{1+x} - \frac{2}{-1+3x}$$
As the geometric series is
$$\frac{1}{1-x}=\sum_{k=0}^\infty x^k $$
So we have
$$\frac{1+5x}{1-2x-3x^2}=-\frac{1}{1+x}+\frac{2}{1-3x}=-\sum_{k=0}^\infty
(-1)^k x^k + 2\cdot \sum_{k=0}^\infty 3^k x^k=\sum_{k=0}^\infty (2 \cdot 3^k -(-1)^k) x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/333565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding real roots of $ P(x)=x^8 - x^7 +x^2 -x +15$ Let $ P(x)=x^8 - x^7 +x^2 -x +15 $, Descartes' Rule of Signs tells us that the polynomial has 4 positive real roots , but if we group the terms as $$ P(x)= x(x-1)(x^6+1) +15 $$ we find that $ P(x) $ does not have any positive roots.
Where did I err ?
| we have to calculate no. of ral roots of $x^8-x^7+x^2-x+15=0$
$\bf{Solution::}$ Let $f(x) = x^8-x^7+x^2-x+15$, we check real solution in $\bf{x\in \mathbb{R}}$
$\bullet$ If $x\leq 0$, Then $f(x) = x^8-x^7+x^2-x+15>0$
$\bullet$ If $0<x<1$, Then $f(x) = x^8+x^2(1-x^5)+(1-x)+14>0$
$\bullet$ If $x\geq 1$, Then $f(x)= x^7(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/335653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 1
} |
Establish convergence of the series: $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-...$ Establish convergence of the series: $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-...$
The number of signs increases by one in each "block".
I have an idea. Group the series like this: $1-(\frac{1}... | We can see that the general term of this series is
$$a_n=(-1)^n\sum_{k=\frac{n(n+1)}{2}+1}^{\frac{n(n+1)}{2}+n+1}\frac{1}{k}$$
and we have
$$H_n=\sum_{k=1}^n\frac{1}{k}=\log n+\gamma+\frac{1}{2n}+O(\frac{1}{n^2})$$
so
\begin{align}|a_n|=H_{\frac{n(n+1)}{2}+n+1}-H_{\frac{n(n+1)}{2}+1}&=\log(1+\frac{2}{n})+\frac{1}{(n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/336035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 6,
"answer_id": 3
} |
Why $\sum_{k=1}^{\infty} \frac{k}{2^k} = 2$? Can you please explain why
$$
\sum_{k=1}^{\infty} \dfrac{k}{2^k} =
\dfrac{1}{2} +\dfrac{ 2}{4} + \dfrac{3}{8}+ \dfrac{4}{16} +\dfrac{5}{32} + \dots =
2
$$
I know $1 + 2 + 3 + ... + n = \dfrac{n(n+1)}{2}$
| Consider the function
$$
f(z) = \sum_{i=0}^{\infty}\frac{z^i}{2^i} = \frac z{2-z}.
$$
Its radius of convergence is $2$.
Its derivative is
$$
f'(z) = \sum_{i=1}^{\infty}\frac{i}{2^i}z^{i-1} = \frac 2{(2-z)^2}.
$$
Since $1<2$, it is allowed to evaluate the derivative at $z=1$ with the written formula and obtain
$$
\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/337937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 12,
"answer_id": 10
} |
Simplify : $\sqrt{\frac{a}{2}}+\sqrt{\frac{a}{2}}\ge \sqrt{a}$? Just out of curiosity, is $$\sqrt{\frac{a}{2}}+\sqrt{\frac{a}{2}}\ge \sqrt{a},a \gt0\quad?$$
Thanks
| $$\sqrt{\frac{a}{2}} + \sqrt{\frac{a}{2}} = 2\sqrt{\frac{a}{2}} = \frac{2}{\sqrt{2}}\sqrt{a}\geq \sqrt{a}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/338127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Rank of the elliptic curve $y^2=x^3+px$ I need to prove that the rank of the curve $y^2=x^3+px$ is $0$, if $p\equiv 7 \pmod {16}$ is a prime.
Using the standard technique, we need to show that none of the following two equations admits an integer solution in M, N and e (with M, N and e pairwise coprime; M, e non-zero):... | For $2M^4-2pe^4=N^2$, we see that $N=2n$ for some integer $n$. Substituting in the equation and reducing by 2 we get $2n^2=M^4-pe^4$. Now if $p\equiv 7(mod 16)$ then we get
$n^2\equiv -3 (mod 16)$ so $n^2\equiv -3 (mod 8)$ which is a contradiction. Because
$x^2\equiv 0,1,4(mod 8) $ for integer $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/338657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
factorisation of algebra homework question that asks to perform the multiplication and division and simplification.
$$\frac{x^2+7x+10}{x^2+5x+4} \times \frac{x^2+3x+2}{x^2+4x+4} =$$
$$\frac{(x+5)(x+2)}{(x+4)(x+1)} \times \frac{(x+1)(x+2)}{(x+2)(x+2)} =$$
$$ = \frac{(x+5)}{(x+4)}$$
is my working out and answer correct?
... | Yes, it is correct. The $x+2$'s and $x+1$ terms do cancel. If there was no multiplication, the second part would be $\frac{x+1}{x+2}$, and the left side would be as is. However, you're OK.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/342606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Can't find solutions for $\tan{2x} = \tan{x}$ Solving $\tan{2x}=\tan{x}$
Reducing the left side: $$\frac{\sin{2x}}{\cos{2x}} = \frac{2\sin{x}\cos{x}}{2\cos^{2}(x)-1}.$$
Reducing the right side: $$\tan{x} = \frac{\sin{x}}{\cos{x}}.$$
therefore:
$$\frac{2\sin x\cos x}{2\cos^2x-1} = \frac{\sin x}{\cos x}$$
$$\frac{2\cos x... | We know $\tan A=\tan B\implies A=n\pi+B$ where $n$ is any integer (Proof Below )
So, $\tan2x=\tan x\implies 2x=n\pi+x\implies x=n\pi$ where $n$ is any integer
[
Proof:
$\tan A=\tan B\implies \sin A\cos B=\cos A\sin B$
$\implies \sin(A-B)=0\implies A-B=n\pi$ where $n$ is any integer
]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/344015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Lagrange multipliers - maximum and minimum values given constraint Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer does not exist, enter DNE.)
$ \ \ f(x, y, z) = xyz \ ; \ \ x^2 + 2y^2 + 3z^2 = 96$
What I have gotten to:
$\Delta f = \ <yz... | I do not see where you used Lagrange multipliers. Any critical points will satisfy the Lagrange multipliers equation
$$
\begin{bmatrix}
yz & xz & xy
\end{bmatrix} = \lambda \begin{bmatrix}
2x & 4y & 6z
\end{bmatrix} \ .
$$
This gives you the system of equations
$$
\begin{align}
yz &= 2\lambda x & (1) \\
xz &= 4\lambda ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/344214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Proving $\gcd(n^2(n^2+1),2n+1)=\gcd(2n+1,5)$ We suppose $\forall n \in \mathbb {N}\setminus{0}$.
How can I prove that $\gcd(n^2(n^2+1),2n+1)=\gcd(2n+1,5)$?
| \begin{align}
(2n+1)^4 - 16n^2(n^2+1) & = 32n^3 + 8n^2 + 8n+1\\
4(2n+1)^3 - (32n^3 + 8n^2 + 8n+1) & = 40n^2+16n+3\\
10(2n+1)^2 - (40n^2+16n+3) & = 24n+7\\
12(2n+1) - (24n+7) &= 5
\end{align}
Hence,
\begin{align}
\gcd(n^2(n^2+1),2n+1) & = \gcd(32n^3 + 8n^2 + 8n+1,2n+1)\\
& = \gcd(40n^2+16n+3,2n+1)\\
& = \gcd(24n+7,2n+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/345154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Sum of two quadratic forms Suppose I have two quadratic forms $Q_i(x)=(x-a_i)^T A_i(x-a_i)+c_i$, $i=1,2$ where $x,a_i \in \Bbb{R}^n$ and $A_i$ are positive-definite $n\times n$ matrices.
Then $Q(x)=Q_1(x)+Q_2(x)$ is also a quadratic form, $Q(x)=(x-a)^T A(x-a)+c$, with $A=A_1+A_2$ (easy to see by considering just the qu... | Let's turn to homogenous coordinates. For example, for $n=2$ you get
\begin{align*}
Q(x) &= (x-a)^T\cdot A\cdot (x-a) + c
\\ &=
\begin{pmatrix}x_1\\x_2\\1\end{pmatrix}^T\cdot\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
-a_1 & -a_2 & 1
\end{pmatrix}\cdot\begin{pmatrix}
A_{1,1} & A_{1,2} & 0 \\
A_{2,1} & A_{2,2} & 0 \\
0 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/345789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Help evaluating $\lim_{\left|x\right| \to \infty}y$, given $\frac{y^2}{x^2}=\frac{b^2}{a^2}-\frac{b^2}{x^2}$ I'm trying to understand this step in a derivation of the standard equation of a hyperbola. We have constants $a$ and $b$, and variables $x$ and $y$. We've gotten to a point where we have
$$\frac{y^2}{x^2}=\frac... | Since $$\left(\frac yx\right)^2=\frac{y^2}{x^2}\to\frac{b^2}{a^2}=\left(\frac ba\right)^2$$ as $|x|$ grows large, then $$\left(\frac yx-\frac ba\right)\left(\frac yx+\frac ba\right)=\left(\frac yx\right)^2-\left(\frac ba\right)^2\to 0$$ as $|x|$ grows large, so $$\frac yx-\frac ba\to 0\quad\text{or}\quad\frac yx+\frac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/345924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $a$ and $b$ are integers such that $9$ divides $a^2 + ab + b^2$ then $3$ divides both $a$ and $b$. If $a$ and $b$ are integers such that $9$ divides $a^2 + ab + b^2$ then show that $3$
divides both $a$ and $b$.
can anyone tell me please how to solve these types of problem oe which formula is required
| The divisibility is equivalent to $$3^2\mid 4(x^2+xy+y^2)=(2x+y)^2+3y^2$$. Since $3\mid 3y^2$ then $3\mid (2x+y)^2$. But $(2x+y)^2$ is a square so $3^2\mid (2x+y)^2$, implying that $3^2\mid 3y^2$, i.e. $3\mid y$. Together with $3\mid 2x+y$ we get also $3\mid x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/348964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Find the Laplace transform of $f(t) = \begin{cases} 0, & \text{if $t<5$} \\ t^2−10t+31, & \text{if $t\ge 5$} \\ \end{cases} $ Find the Laplace transform of
$$f(t) =
\begin{cases}
0, & \text{if $t<5$} \\
t^2−10t+31, & \text{if $t\ge 5$} \\
\end{cases}
$$
$F(s)=$ __________?
Here is my work. I went wrong somewhere. Can... | Your simplification is wrong. Note that
$(t-5)^2+1 = t^2 -10t + 26$ and you should have $t^2-10t+31$, so you need to use $(t-5)^2+6$ instead.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/350490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Trigonometrical limit $\lim\limits_{ x\to 0 } \frac{\sin x - x\cos x}{x^3}?$ Can you help me solve this without using de l'Hôpital's rule (just using Standard rules):
$$ \lim_{ x\to 0 } \frac{\sin x - x\cos x}{x^3}? $$
| Use that
$$\sin(x)=x-\frac{x^3}{3!} + \frac{x^5}{5!}\mp \cdots$$
and that
$$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{4!}\pm\dots$$
Hence
$$\frac{\sin(x)-x\cos(x)}{x^3}=\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}- x+\frac{x^3}{2}- \frac
{x^5}{4!} \pm \dots}{x^3}=\frac{1}{3} +\frac{x^2}{5!}-\frac{x^2}{4!}\pm \dots$$
Hence the Limi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/351038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Finding Determinants Recursively From the MIT OCW Linear Algebra (18.06) final exam, question 9:
For square matrices with 3's on the diagonal, 2s on the diagonal
above, and 1s on the diagonal below:
$$A_1=\begin{pmatrix} 3 \end{pmatrix}, A_2=\begin{pmatrix} 3 & 2 \\ 1
& 3\\ \end{pmatrix}, A_3=\begin{pmatrix} 3&2&0\... | It should
$$\begin{pmatrix} D_2\\ D_1\end{pmatrix} = \begin{pmatrix} 7\\ 3\end{pmatrix}$$
and not
$$\begin{pmatrix} D_2\\ D_1\end{pmatrix} = \begin{pmatrix} 3\\ 7\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/353368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
3 complex-variable equation
Moderator Note: This is a current contest question on Brilliant.org.
$x,y,z$ are complex numbers satisfying
$$
\begin{align}
x+y+z & =1\\
x^2+y^2+z^2 & =2\\
x^3+y^3+z^3 & =3
\end{align}
$$
The value of $x^4+y^4+z^4$ can be expressed as $\dfrac ab$, where $a$ and $b$ are positive co-prime ... | HINT:
$(x+y+z)^2=(x^2+y^2+z^2)+2(xy+yz+zx)$
$\implies 2(xy+yz+zx)=(x+y+z)^2-(x^2+y^2+z^2)=1^2-2=-1$
Again, $x^3+y^3+z^3-3xyz=(x+y+z)\{(x+y+z)^3-3(xy+yz+zx)\}$
So, we have found $x+y+z,xy+yz+zx, xyz$
So, we can form the cubic equation whose roots are $x,y,z$
Can you take it from here?
Alternatively,
$$x^4+y^4+z^4=(x^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/353900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Evaluate $\int \frac{x^2 + x+3}{x^2+2x+5}\ dx$ How can we evaluate $$\displaystyle\int \frac{x^2 + x+3}{x^2+2x+5} dx$$
To be honest, I'm embarrassed. I decomposed it and know what the answer should be but
I can't get the right answer.
| Hint Use the decomposition
$$\frac{x^2 + x+3}{x^2+2x+5}=1-\frac{ x+2}{x^2+2x+5}=1-\frac{1}{2}\frac{ 2x+2}{x^2+2x+5}-\frac{ 1}{x^2+2x+5}$$
and
$$\frac{ 1}{x^2+2x+5}=\frac{ 1}{(x+1)^2+4}=\frac{1}{4}\frac{ 1}{(\frac{x+1}{2})^2+1}$$
the first fraction is on the form $\frac{f'}{f}$ and the second have the form $\frac{1}{u^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/355296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Find the volume of the region contained above $z=1$ and below $x^{2}+y^{2}+z^{2}=4$ Why doesn't this work?
Find the volume of the region contained above $z=1$ and below $x^{2}+y^{2}+z^{2}=4$
going to cylindrical this should be easy. $z=(4-r^{2})^{\frac {1}{2}}$ and $z=1$
Clearly $z=1$ yields $x^{2}+y^{2}=3 \to r=(3)... | $$\displaystyle{\int _0^{2\pi }\int _0^{\sqrt{3}}\int _1^{\sqrt{4-r^2}}rdzdrd\theta} $$
$$\displaystyle{2\pi \int_0^{\sqrt{3}} \left(r\sqrt{4-r^2}-r\right) \, dr} $$
The first part you solve it by changing variables:
$$u=4-r^2 \qquad du=-2rdr \qquad \frac{-du}{2}=rdr$$
After solving that, the volume is $\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/355531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Find the point or points on C closest to the origin.
A curve $C$ in space is defined implicitly on the cylinder $x^2 + y^2 = 1$ by the additional equation $x^2 - xy + y^2 - z^2 = 1$. Find the point or points on $C$ closest to the origin.
This is an optimization problem. I tried constraining the distance $d = (x^2 + y... | The square of the distance from the origin is $x^2+y^2+z^2=1+z^2$ for any point on the cylinder. Any point on the curve satisfies $xy+z^2=0$, or $z^2=-xy$, so the square of the distance to the origin is $1-xy$, where $x^2+y^2=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/355607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove the following ceiling and floor identities? Could someone help me prove these identities? I'm completely lost:
$$\begin{align*}
&(1)\quad \left\lceil \frac{\left\lceil \frac{x}{a} \right\rceil} {b}\right\rceil = \left\lceil {\frac{x}{ab}}\right\rceil\\\\
&(2)\quad\left\lfloor \frac{\left\lfloor \frac{x}{a} \right... | You can work directly from the definitions. I’ll do $(2)$ and $(3)$ to illustrate.
$(2)$ First note that you want $a$ and $b$ to be positive. If not, you might have $x=\frac12$ and $a=b=-1$, in which case the lefthand side is $$\left\lfloor-\left\lfloor-\frac12\right\rfloor\right\rfloor=\lfloor-(-1)\rfloor=1\;,$$ and t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/356168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 0
} |
On comparing fractions , fraction with smaller difference between numerator and denominator is greater than the other A text book proposed that "when comparing fractions ,if the compared fractions's are such that numerator is smaller than denominator ,then fraction with more difference(absolute) between numerator and... | I have never seen this claim in any textbook; in any case it's wrong. The claim seems to be that if you have two fractions $\frac{a}{b}$ and $\frac{c}{d}$ with $a < b$ and $c < d$, then $|a - b| < |c - d| \iff \frac{a}{b} < \frac{c}{d}$.
This is false. We can write $\frac{a}{b} = 1 - \frac{b-a}{b}$ and $\frac{c}{d} = 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/358704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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How can I determine the Jordan Form of a matrix? How can I go about proving that the characteristic polynomial, minimal polynomial, and the dim(eigenspace) is enough to determine the Jordan Form of a matrix for n<7?
| I realize this is 5 years old, but the only answer is wrong, given that the two matrices in DonAntonio's answer do not have matching minimal polynomials. So in case anyone comes across it, here is an answer:
If Dimension=7, then you could have two matrices such as
$$
A=\begin{pmatrix}
0 & 1 \\
& 0 & 1 \\
& & 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/358777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Calculate $\sum_{n=2}^\infty ({n^4+2n^3-3n^2-8n-3\over(n+2)!})$ Calculate $\sum_{n=2}^\infty ({n^4+2n^3-3n^2-8n-3\over(n+2)!})$
I thought about maybe breaking the polynomial in two different fractions in order to make the sum more manageable and reduce it to something similar to $\lim_{n\to\infty}(1+{1\over1!}+{1\over2... | First step, we find the Taylor series of $x^4+2x^3-3x^2-8x-3$ at the point $x=-2$ and then use it to write
$$ n^4+2n^3-3n^2-8n-3 = 1-4\, \left( n+2 \right) +9\, \left( n+2 \right)^{2}-6\, \left( n+2
\right) ^{3}+ \left( n+2 \right) ^{4}.$$
Using the above expansion and shifting the index of summation ($n \longleftrigh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/360293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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How to compare two ratios of gamma functions? How would I show that for $x \ge 2$:
$$\frac{\Gamma(x-1)}{[\Gamma(\frac{x}{2})]^2} \le \frac{\Gamma(x)}{[\Gamma(\frac{x+1}{2})]^2}$$
Any hints or suggestions are greatly appreciated.
| The answer is to show that $\frac{d}{dx}(\frac{\Gamma(x-1)}{\Gamma([\frac{x}{2}]^2)}) > 0$
This will be established if I show that $\frac{d}{dx}[\ln{\Gamma(x-1)}- 2\ln\Gamma(\frac{x}{2})] > 0$.
I will use this series ψ:
$$\frac{d}{dx}(\ln\Gamma(x)) = \frac{\psi(x)}{dx} = -\gamma + \sum_{k=0}^\infty(\frac{1}{k+1} - \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/361898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Minimum value of inverse trigo function (cubic) The minimum value of $(\sin^{-1}x)^3+(\cos^{-1}x)^3$ is equal to ( following options)
a) $\displaystyle \frac{\pi^3}{32}$
b) $\displaystyle\frac{5\pi^3}{32}$
c) $\displaystyle\frac{9\pi^3}{32}$
d) $\displaystyle\frac{11\pi^3}{32}$
Can we go like this :
$ -\frac{\pi}{2} \... | As $\sin^{-1}x+\cos^{-1}x=\frac\pi2$
$$(\sin^{-1}x)^3+(\cos^{-1}x)^3$$
$$=(\sin^{-1}x)^3+\left(\frac\pi2-\sin^{-1}x\right)^3$$
$$=\left(\frac\pi2\right)^3-3\left(\frac\pi2\right)^2\sin^{-1}x+3\left(\frac\pi2\right)(\sin^{-1}x)^2$$
$$=\left(\frac\pi2\right)^3+3\left(\frac\pi2\right) \{(\sin^{-1}x)^2-\frac\pi2\sin^{-1}x\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/362185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
how to prove that $(1 + \frac{1}{n})^{n+1}$ is decreasing? Please, help me to prove that $$x_n=\left(1+\frac{1}{n}\right)^{n+1}$$decreases.
I know I must to prove that that $$\frac{x_n}{x_{n+1}}> 1$$
What to do next?
| Write out
$$
\frac{x_{n+1}}{x_n} = \frac{\left(1+\frac{2}{n}\right)^2}{\left(1+\frac{1}{n}\right)^3} \left( \frac{1+\frac{2}{n}}{\left(1+\frac{1}{n} \right)^2} \right)^n =
\frac{\left(1+\frac{2}{n}\right)^2}{\left(1+\frac{1}{n}\right)^3} \left( 1 - \frac{\frac{1}{n^2}}{\left(1+\frac{1}{n} \right)^2} \right)^n =
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/365680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Calculate the number of real roots of $x^8-x^5+x^2-x+1 = 0$ Calculate the number of real roots of $x^8-x^5+x^2-x+1 = 0$
My try: $$\left(x^4-\frac{x}{2}\right)^2+\frac{3}{4}x^2-x+1 = \left(x^4-\frac{x}{2}\right)^2+\frac{3}{4}\left(x^2-\frac{4}{3}x+\frac{4}{3}\right)$$
$$\implies \left(x^4-\frac{x}{2}\right)^2+\frac{3}{4... | AM-GM Inequality gives us,
$$\large\frac{1}{2} \left(x^{8}+x^{2}\right)\geq x^{5}$$
$$\large\frac{1}{2} \left(x^{2}+1\right)\geq x$$
$$\large\therefore x^{8}+x^{2}+1>\frac{1}{2} \left(x^{8}+x^{2}\right)+\frac{1}{2} \left(x^{2}+1\right)\geq x^{5}+x$$
$$\large\implies x^{8}-x^{5}+x^{2}-x+1> 0$$
Hence there cannot exist a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/366093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
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Confusion over a limit. Different ways of solving give different answers? Qn: If it is given that
$$
\lim_{x\to\infty} \frac{x^2 - x - 2}{x + 1} - ax - b = 1
$$
then a and b must be?
Now, I tried doing this by 2 methods.
Method 1:
$$ \frac{x^2 - x - 2}{x + 1} - ax - b $$
$$ = (x - 2) - ax - b $$
Since the limit is fini... | Method $1$ is correct
In Method $2$, you cannot apply limit to a part of the fraction. So
$$ = \frac{x^2(1 - \frac 1 x - \frac2 {x^2})}{x(1 + \frac1x)} - ax - b $$
is not equal to
$$ = x - ax - b$$
In the above step you have applied limit to just a part of that fraction leaving behind $\dfrac{x^2}{x}$.
hope the answer... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/367080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Show $x\sqrt{n} - n \ln\left(1+\frac{x}{\sqrt{n}}\right) \to \frac{x^2}{2}$ With $x > 0$, show
$$
L=\lim_{n \to \infty} x\sqrt{n} - n \ln\left(1+\frac{x}{\sqrt{n}}\right) = \frac{x^2}{2}.
$$
I tried to write
$$
x\sqrt{n}=\ln \left( e^{x\sqrt{n}} \right),
$$
so that
$$
L = \lim_{n\to \infty}
\frac{e^{x\sqrt{n}}}{\lef... | $$\begin{aligned}x\sqrt{n}-n\ln\left(1+\frac{x}{\sqrt{n}}\right)&=x\sqrt{n}-n\left(\frac{x}{\sqrt{n}}-\frac{x^2}{2n}+\frac{x^3}{3n\sqrt{n}}-\cdots\right)\\&=\frac{x^2}{2}-\frac{x^3}{3\sqrt{n}}+\cdots\to\frac{x^2}{2}\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/368630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Solving two simultaneous recurrence relations If we have the two recurrence relations $$a_n = 3a_{n-1} + 2b_{n-1}$$ $$b_n = a_{n-1} + 2b_{n-1}$$ with $a_0 = 1$ and $b_0 = 2$.
My solution is that we first add two equations and assume that $f_n = a_n + b_n$. The result is $f_n = 4f_{n-1}$. This can be solved easily and... | Use generating functions, $A(z) = \sum_{n \ge 0} a_n z^n$ and similarly $B(z)$.
Write:
$$
\begin{align*}
a_{n + 1} &= 3 a_n + 2 b_n \\
b_{n + 1} &= a_n + 2 b_n
\end{align*}
$$
By properties of generating functions:
$$
\begin{align*}
\frac{A(z) - a_0}{z} &= 3 A(z) + 2 B(z) \\
\frac{B(z) - b_0}{z} &= A(z) + 2 B(z)
\end{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/369377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
} |
Finding eigenvectors with square root eigenvalues I have a matrix
$$\begin{bmatrix}1 &-1 &2\\2 &-2 &4\\0 &1 &1\end{bmatrix}$$
Its eigenvalues are $0$, $\sqrt{5}$ and $-\sqrt{5}$
(These are checked in MATLAB to be correct). I have found its eigenvectors for $0$, and also don't seem to have a problem in similar situation... | If you want results to compare against, you should get for $A = \begin{bmatrix}1 &-1 &2\\2 &-2 &4\\0 &1 &1\end{bmatrix}$, the following characteristic polynomial:
$$5 \lambda - \lambda^3 = 0$$
This CP yields the following eigenvectors and eigenvalues:
$\displaystyle \lambda_1 = -\sqrt{5}, v_1 = \left(\frac{1}{2} (-1-\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/369507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
What's the Maclaurin series for $\arcsin(x)$? I solved the problem by using a known series: $\frac{1}{\sqrt{1-x^2}}$, but the solution I got is wrong. Also, I'm not sure what to do with the constant of integration $C$. Where is my mistake?
$$ \frac{1}{\sqrt{1-x^2}} = 1 + \frac{x^2}{2} + \frac{3x^4}{8} + \frac{5x^6}{16}... | You did a great job:
Just small mistakes.
*
*To find constant of integration, substitute known value of $\arcsin(x)$, $x=0$ is a good choice. You need to have a well defined interval while dealing with inverse trigonometric functions.
*The formula to integrate is $$\int x^n \, dx=\frac{x^{n+1}}{n+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/371611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
} |
What's the value of $ y^{(n)}$when $ y=\frac{x^n}{(x+1)^2(x+2)^2}$ What's the value of $\displaystyle y^{(n)}$when $\displaystyle y=\frac{x^n}{(x+1)^2(x+2)^2}$?
My Try:Let $\displaystyle y_n=\frac{x^n}{(x+1)^2(x+2)^2}$,so $\displaystyle y_n=xy_{n-1}$.According to Leibniz's formula,$$y_n^{(n)}=ny_{n-1}^{(n-1)}+xy_{n-1... | Assertion:
$$y_n^{(n)} = \frac{A_nx+B_n}{(x+1)^{n+2}}+\frac{C_nx+D_n}{(x+2)^{n+2}}$$
Now, we are going to need
$$
xy_{n}^{(n+1)}=\frac{-(n+1)A_nx^2+(A_n-B_n(n+2))x}{(x+1)^{n+3}}+\frac{-(n+1)C_nx^2+(2C_n-(n+2)D_n)x}{(x+2)^{n+3}}
$$
and
$$
(n+1)y_{n}^{(n)}=(n+1)\frac{(A_nx+B_n)(x+1)}{(x+1)^{n+3}}+(n+1)\frac{(C_nx+D_n)(x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/372426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Simplifying $y=2^{2/3} + 2^{-1/3}$ I am working on a calculus problem where I have to find the local minimum. The value I got was $$y=2^{2/3} + 2^{-1/3}.$$ I simplified it and got this:
$$ y=2^{2/3} + \frac{1}{2^{1/3}}$$
$$y=\frac{2^{2/3}2^{1/3}}{2^{1/3}} + \frac{1}{2^{1/3}}\frac{2+1}{2^{1/3}}=\frac{3}{2^{1/3}}$$
Accor... | Your simplification is entirely correct. Using your original evaluation of $y$, $$y=e^{2\frac{\ln2}{3}}+e^{-\frac{\ln2}{3}}=2^{2/3} + 2^{-1/3} = \dfrac{3}{2^{1/3}}$$
Are you sure your value for $y$ is a correct mimimum?
EDIT:
Original function: $$y(x)=e^{2x}+e^{-x}$$
Critical point (minimum) at $x = \dfrac{-\ln 2}{3} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/373115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Closed form for $(a_n)$ such that $a_{n+2} = \frac{a_{n+1}a_n}{6a_n - 9a_{n+1}}$ with $a_1=1$, $a_2=9$ $$a_1 = 1; a_2 = 9; a_{n+2} = \frac{a_{n+1}a_n}{6a_n - 9a_{n+1}}$$
I need to find non-recurring formula for $a_n$. Is there any good way to do this? The only one comes to mind is to guess the formula and then prove it... | As in Did's answer, but simpler to handle:
$$
b_{n + 2} = 6 b_{n + 1} - 9 b_n \qquad b_1 = 1, b_2 = 1/9
$$
Define $B(z) = \sum_{n \ge 0} b_{n + 1} z^n$. By the properties of generating functions:
$$
\frac{B(z) - b_1 - b_2 z}{z^2} = 6 \frac{B(z) - b_1}{z} - 9 B(z)
$$
From here:
$$
B(z) = \frac{9 - 53 z}{9 - 54 z + 81 z^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/375215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
} |
How to count permutations with restrictions on how items are grouped I am trying to solve the following problem:
A town contains $4$ people who repair televisions. If $4$ sets break down, what is the probability that exactly $i$ of the repairers are called? Solve the problem for $i=1,2,3,4$.
For $i=1$, there are ${}_... | *
*Probability that exactly 2 repairers are called.
There are two ways how two repairers are called:
*
*Both gets call for two broken sets each
First repairer $A$ can be anyone. Hence, will have probability $\frac{4}{4}=1$. The same repairer will be called for any one of 2nd, 3rd or 4th broken TV set. [Hence the to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/377270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Integral Question - $\int\frac{1}{x^2-6x}\,\mathrm dx$ How I can evaluate the indefinite integral? :
$$\int\frac{1}{x^2-6x}\,\mathrm dx$$
Do I need to bring it to this format? : $\displaystyle \int\frac{1}{x^2-a^2}\,\mathrm dx\;$?
Thanks!
| $$\int\frac{1}{x^2-6x}\,\mathrm dx = \int\frac{1}{x(x-6)}\,\mathrm dx$$
Now use partial fraction decomposition to obtain an integral of the form $$\int\frac{1}{x(x-6)}\,\mathrm dx =\int \left(\frac{A}{x} + \frac{B}{x-6}\right)\,\mathrm dx$$
Now all you need to do is to determine the (constant) values of $A$ and of $B$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/377410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
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Small inequality on unit open disc For $|u|,|z|<1$, $u,z$ complex numbers, how to show the inequality:
$|\frac{u-z}{1-\bar uz}|<1$?
| Let $u=a+ib, z=c+id$
So, $|u-z|=\sqrt{(a-c)^2+(b-d)^2}=\sqrt{a^2+c^2-2ca+b^2+d^2-2bd}$
$1-\bar uz=1-(a-ib)(c+id)=1-ac-bc+i(bc-ad)$
So, $|1-\bar uz|=\sqrt{(1-ac-bc)^2+(bc-ad)^2}$
$|u-v|<|1-\bar uz|$
$\implies \sqrt{a^2+c^2-2ca+b^2+d^2-2bd}<\sqrt{(1-ac-bc)^2+(bc-ad)^2}$
$\implies a^2+c^2-2ca+b^2+d^2-2bd <(1-ac-bc)^2+(bc-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/378074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Find the limit of $ x_n = \prod_{j=2}^{n} \left(1 - \frac{2}{j(j+1)}\right)^2$ I am stuck on the following problem:
Let $x_n=(1-\frac{1}{3})^2(1-\frac{1}{6})^2(1-\frac{1}{10})^2 \ldots...(1-\frac{1}{n(n+1)/2})^2, \text{where} \space n \geq 2$. Then $\lim_{n \to \infty}x_n=?$
I see that $x_n^{\frac{1}{2}}=\frac{2}{3... | Note you really have a telescoping product. To see this intuitively, write out a few terms. You start with $\frac{1 \cdot 4}{2 \cdot 3}$
$$\frac{1 \cdot 4}{2 \cdot 3} \times \frac{2 \cdot 5}{3 \cdot 4} = \frac{1 \cdot 5}{3 \cdot 3}$$
and adding another term
$$\frac{1 \cdot 5}{3 \cdot 3} \times \frac{3 \cdot 6}{4 \cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/379511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
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Why is $\;n^2-\frac{n^2}{2} =\frac{n^2}{2}\;$? Could someone please expand on how to get from $\;\displaystyle\left( n^2-\frac{n^2}{2}\right)\;$ to $\;\left(\dfrac{n^2}{2}\right)\;?\;$
I can't seem to wrap my head around that.
| Simple arithmetic:
\begin{equation}
n^2 - \dfrac{n^2}{2} = n^2(1 - \dfrac{1}{2}) = n^2(\dfrac{1}{2}) = \dfrac{1}{2}n^2 = \dfrac{n^2}{2}
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/379713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 15,
"answer_id": 5
} |
Simplifying $\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}$ How do I simplify $\sqrt{(4+2\sqrt{3})}+\sqrt{(4-2\sqrt{3})}$?
I've tried to make it $x$ and square both sides but I got something extremely complicated and it didn't look right.
I got $2\sqrt{3}$ on wolframalpha, but I'm not sure how is it possible?
Help would be a... | Since $(4+2\sqrt3)(4-2\sqrt3)=16-12=4$, try squaring:
$$
\begin{align}
\left(\sqrt{4+2\sqrt3}+\sqrt{4-2\sqrt3}\right)^2
&=(4+2\sqrt3)+(4-2\sqrt3)+2\sqrt{(4+2\sqrt3)(4-2\sqrt3)}\\
&=8+2\sqrt{16-12}\\[6pt]
&=12
\end{align}
$$
Therefore, $\sqrt{4+2\sqrt3}+\sqrt{4-2\sqrt3}=2\sqrt3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/383975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 7,
"answer_id": 0
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Can both $x^2 + y+2$ and $y^2+4x$ be squares? Prove that there exist no positive integers $x$ and $y$ such that both $x^2+y+2$ and $y^2+4x$ are perfect squares.
I thought I could perhaps solve this by square bounding but I couldn't get anywhere with it.
Thanks in advance for any help.
| Thanks to Erick Wong for setting me on the right track.
Assume for sake of contradiction that $x^2+y+2$ and $y^2+4x$ are both perfect squares.
Then as $y$ is a positive integer, $x^2+y+2 \geq (x+1)^2=x^2 +2x+1$
$$\implies y+2\geq 2x+1$$
$$y+1 \geq 2x \; \; \; \; \; (1)$$
Following a similar argument, as $x$ is a positi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/384100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Integral of $\int\frac{(x^4+1)\,dx}{x^3+4x}$ I followed the steps to solve this integral and want to know if I did it right and if $C=0? $
$$\int\frac{(x^4+1)\,dx}{x^3+4x} = \int\frac{(x^4+1)\,dx}{x(x^2+4)} = \frac{A}{x}+\frac{Bx+C}{x^2+4}$$
$$(x^2+4)A+x(Bx+C)=x^4+1$$
$$x=0 => 4A=1 => A=\frac{1}{4}$$
$$Ax^2+4A+Bx^2... | Whatever $A$, $B$, $C$ may be, the numerator of $\frac{A}{x} + \frac{B x+C}{x^2+4}$ is no greater than 3. The proper ansatz is
$$
\frac{x^4+1}{x^3+4x} = x + \frac{1-4x^2}{x(x^2+4)} = x + \frac{A}{x} + \frac{B x+ C}{x^2+4}
$$
and you should get $A = \frac{1}{4}$ and $C=0$, $B = -\frac{17}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/384496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How do I prove the arithmetic-geometric mean inequality? I am following along with this bare-bones proof of the arithmetic-geometric mean inequality with two real numbers. I'm having difficulty understanding the logic behind this step:
$$
\frac{a}{2}+\frac{a}{2}< \frac{b}{2}+\frac{a}{2}\Rightarrow a< \frac{b+a}{2}
$$
H... | Assume $0< a < b$.
The first thing we want to show for this proof is that given $$a<b,$$ we want $$a<\frac{a+b}{2} <b.$$ To do this, we show the left and right inequalities separately. Now, it is a rule of elementary algebra that addition preserves ordering, i.e. $$a+k < b+k$$ for all $k\in \mathbb{R}$. If we choose $k... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to prove this trigonometric expression? How would you go about proving the following?
$${1- \cos A \over \sin A } + { \sin A \over 1- \cos A} = 2 \operatorname{cosec} A $$
This is what I've done so far:
$$LHS = {1+\cos^2 A -2\cos A + 1 - \cos^2A \over \sin A(1-\cos A)}$$
....no idea how to proceed .... X_X
| Hint:
$1-\cos A=1-(1-2 \sin^2\dfrac{A}{2})=2\sin^2 \dfrac{A}{2}$
$\dfrac{2 \sin^2 \dfrac{A}{2}}{2 \sin \dfrac{A}{2} \cos \dfrac{A}{2}}=\tan \dfrac{A}{2}$
The other expression will be $\cot \dfrac{A}{2}$
$(\tan^2 \dfrac{A}{2}+1) /\tan\dfrac{A}{2}= \dfrac{\sec^2 A \cos \dfrac{A}{2}}{\sin \dfrac{A}{2}}= \dfrac{1}{2 \sin A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/385537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
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An easier way to find the integral of: $\int {x\sqrt {2 + x} {\rm{ }}dx} $, where ${u^2} = 2 + x$ My attempt at the question:
$\eqalign{
& \int {x\sqrt {2 + x} {\rm{ }}dx} \cr
& {u^2} = 2 + x \cr
& 2u{{du} \over {dx}} = 1 \cr
& {{du} \over {dx}} = {1 \over {2u}} \cr
& u = \sqrt {2 + x} \cr
& x =... | If $u=2+x \implies dx=du$ then the integral becomes
$$\int (u-2)\sqrt u\,du$$
which can now be integrated easily.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/386161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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How to derive $\cos\frac{n\pi}{3}=\frac{1+3(-1)^{[\frac{n+1}{3}]}}{4}$ Consider the following formula to calculate a trigonometric function:
$$\cos\frac{n\pi}{3}=\frac{1+3(-1)^{[\frac{n+1}{3}]}}{4}$$ $[x]$ denotes the integer part of $x$. The formula is valid for $n=0,2,4,6,...$
I'm curious how this formula is de... | HINT:
The formula can be written as $$\cos\frac{2m\pi}3=\frac{1+3(-1)^{\left[\frac{2m+1}3\right]}}4$$ for all integer $m\ge0$
Now, the value of $(-1)^{\left[\frac{2m+1}3\right]}$ can be $+1,-1$ based on whether $\left[\frac{2m+1}3\right]$ is even or is odd
Observe that the fractional part of $\frac{2m+1}3$ can be $\fr... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Proof of the identity $2^n = \sum\limits_{k=0}^n 2^{-k} \binom{n+k}{k}$ I just found this identity but without any proof, could you just give me an hint how I could prove it?
$$2^n = \sum\limits_{k=0}^n 2^{-k} \cdot \binom{n+k}{k}$$
I know that $$2^n = \sum\limits_{k=0}^n \binom{n}{k}$$ but that didn't help me
| Suppose we seek to verify that
$$S_n = \sum_{k=0}^n 2^{-k} {n+k\choose k} = 2^n.$$
We introduce the Iverson bracket
$$[[k\le n]] =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-k+1}} \frac{1}{1-z}
\; dz$$
so we may extend $k$ to infinity, getting
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} \frac{1}{1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/389099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 8,
"answer_id": 2
} |
Integrate $\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx$ Please help me to solve this integral:
$$\int_0^\pi\frac{3\cos x+\sqrt{8+\cos^2 x}}{\sin x}x\ \mathrm dx.$$
I managed to calculate an indefinite integral of the left part:
$$\int\frac{\cos x}{\sin x}x\ \mathrm dx=\ x\log(2\sin x)+\frac{1}{2} \I... | Begin with $u$-substitution using
$$\begin{align}
u & =x\\
dv & = \frac{3\cos(x)+\sqrt{8+\cos^2(x)}}{\sin(x)}\,dx\end{align}$$
so that $du=dx$, and my CAS tells me (which I suppose could be verified through differentiation and identities) that
$$\begin{align}
v & = \sinh^{-1}\left(\frac{\cos(x)}{\sqrt{8}}\right)+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/390957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 3,
"answer_id": 0
} |
Finding asymptotes to general functions I stumbled upon this problem and didn't quite understand why the solution worked. I have two questions:
*
*Why is it that the taylor series is enough to find the oblique asymptote? Specifically, why do the first two terms suffice? It seems like that may not work for all functi... | In general, the Taylor Series is the last thing you should be looking at.
In fact, the Taylor Series was not looked at; they calculated the Laurent Series. The point is that:
$$\operatorname{e}^y = 1 + y + \frac{1}{2}y^2 + \frac{1}{3!}y^3 + \cdots $$
Replacing $y$ with $2/x$ gives:
\begin{array}{ccc}
\operatorname{e}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/392622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Further explanation needed :Finding all $z$ such that the modulus of $f(z)=e^{(z+1)/(z-1)}$ is equal to/at most $1$ I was solving a previous exam paper and there I got stuck on the following problem:
Let $f(z)=e^{\frac{z+1}{z-1}}$. Then find all $z \in \Bbb C$ for which
*
*$|f(z)|=1$,
*$|f(z)|\le ... | Let $ z = a + bi, (a, b) \in \mathbb{R}^2 $. Hence $$\begin{align*} \left|e^{\frac{z+1}{z-1}}\right| &= \left|e^{\frac{(a + 1 + bi)(a - 1 - bi)}{(a - 1 + bi)(a - 1 - bi)}} \right| \\&= \left|\ e^{\frac{a^2 - 1 + b^2 -2bi}{(a - 1)^2 + b^2}}\ \ \right| \\ &= \left|\ e^{ \frac{a^2 -1 + b^2}{(a - 1)^2 + b^2}}\ \ \ \ \ \rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/394490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Proving that $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{100}}<20$ How do I prove that:
$$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\dots+\frac{1}{\sqrt{100}}<20$$
Do I use induction?
| Prove the following claim using induction on $n$:
$$\sum_{k=1}^n \dfrac1{\sqrt{k}} < 2 \sqrt{n}$$
In the induction, you will essentially need to show that
$$2\sqrt{n} +\dfrac1{\sqrt{n+1}} < 2 \sqrt{n+1} \tag{$\star$}$$
To prove $(\star)$, note that
$$\sqrt{n} < \sqrt{n+1} \implies \sqrt{n} + \sqrt{n+1} <2 \sqrt{n+1} \i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/395979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
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Laurent Series Expansion Problems Expand in a laurent Series :
1- $f_{1} (z) = \frac{z^{2} - 2z +5 }{(z^{2}+1) (z-2)}$
in the ring : $1 < |z| < 2 $
2- $ f_{2} (z) = \frac{1 }{(z-3) (z+2)}$
In :
$i. 2 < |z| < 3
\\ ii. 0 < |z+2| < 5$
I managed to solve the second one but not sure if it is correct
For i. $2 < |z| < 3$... | In case 1, note that $1\lt|z|\lt2$ means that $1/|z|\lt1$ and $|z|/2\lt1$, hence the well known expansion
$1/(1-u)=\sum\limits_{n\geqslant0}u^n$, valid for every $|u|\lt1$, applied twice, yields
$$
\frac1{z^2+1}=\frac1{z^2(1+1/z^2)}=\sum\limits_{n\geqslant0}\frac{(-1)^n}{z^{2n+2}},$$
and
$$
\frac1{z-2}=-\frac12\frac1{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/401851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Telescoping sum of powers
$$
\begin{array}{rclll}
n^3-(n-1)^3 &= &3n^2 &-3n &+1\\
(n-1)^3-(n-2)^3 &= &3(n-1)^2 &-3(n-1) &+1\\
(n-2)^3-(n-3)^3 &= &3(n-2)^2 &-3(n-2) &+1\\
\vdots &=& &\vdots & \\
3^3-2^3 &= &3(3^2) &-3(3) &+1\\
2^3-1^3 &= &3(2^2) &-3(2) &+1\\
1^3-0^3 &= &3(1^2) &-3(1) &+1\\
\underline{\hphantom{(... | An idea: since
$$A^3-B^3=(A-B)(A^2+AB+B^2)$$
we get then
$$(n-1)^3-(n-2)^3=\left[(n-1)-(n-2)\right]\left[(n-1)^2+(n-1)(n-2)+(n-2)^2\right]\ldots\ldots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/402445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
A tricky logarithms problem? $ \log_{4n} 40 \sqrt{3} \ = \ \log_{3n} 45$. Find $n^3$.
Any hints? Thanks!
| By elementary arithmetic operations (after / describing next action):
$$\log_{4n}40\sqrt{3}=\log_{3n}45\ \ \ \mbox{ / definition of logarithm}$$
$$(4n)^{\log_{3n}45}=40\sqrt{3}\ \ \ \mbox { / } 4=\frac{4}{3}\cdot 3$$
$$\left({4\over 3}\cdot 3n\right)^{\log_{3n}45}=40\sqrt{3}\ \ \ \mbox { / }(ab)^c=a^cb^c$$
$$\left({4\o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/404389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to calculate $\sum_{n=1}^\infty\frac{(-1)^n}n H_n^2$? I need to calculate the sum $\displaystyle S=\sum_{n=1}^\infty\frac{(-1)^n}n H_n^2$,
where $\displaystyle H_n=\sum\limits_{m=1}^n\frac1m$.
Using a CAS I found that $S=\lim\limits_{k\to\infty}s_k$ where $s_k$ satisfies the recurrence relation
\begin{align}
& s_{... | Here is a solution using simple tools
We have
$$\sum_{n=1}^\infty x^nH_n=-\frac{\ln(1-x)}{1-x}$$
Replace $x$ with $-x$ then multiply both sides by $-\frac{\ln(1-x)}{x}$ and use the fact that $-\int_0^1 x^{n-1}\ln(1-x)\ dx=\frac{H_n}{n}$
$$\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n}=\int_0^1\frac{\ln(1-x)\ln(1+x)}{x(1+x)}\ d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/405356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 5,
"answer_id": 3
} |
Determine the general solution for $2\cos^2x-5\cos x+2=0$ Determine the general solution for $ 2\cos^{2}x-5\cos x+2=0$
My attempt:
$2u^2 - 5u + 2 = 0$
$(2u - 1)(u - 2) = 0$
$u = \frac{1}{2}$ or $u = 2$
$\cos(x) = \frac{1}{2}$ or $\cos(x) = 2$
The answers on the worksheet are as follows but my answer does not match...
... | [\begin{array}{l}
\\
\\
{\rm{ }}2{\cos ^2}x - 5\cos x + 2 = 0\\
2\cos x(\cos x - 2) - 1(\cos x - 2) = 0\\
{\rm{ }}(2\cos x - 1)(cosx - 2) = 0\\
2\cos x - 1 = 0{\rm{ OR }}\cos x - 2 = 0\\
\\
{\rm{Case 1}}\\
2\cos x - 1 = 0\\
{\rm{ }}\cos x = \frac{1}{2}\\
{\rm{ }}\cos x{\rm{ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/406368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Given that $\;\sin^3x\sin3x = \sum^n_{m=0}C_m\cos mx\,,\; C_n \neq 0\;$ is an identity . Find the value of n. Problem :
Given that $\sin^3 x \sin 3x = \sum^n_{m=0}C_m \cos mx, C_n \neq 0 $ is an identity. Find the value of n.
I tried : $\sin3x = 3\sin x - 4\sin^3 x$ but unable to reach to any point.... Please suggest ... | HINT:
As you have noticed, $4\sin^3x=3\sin x-\sin3x,$
$$\text{So, }\sin^3x\sin3x=\frac{(3\sin x-\sin3x)\sin3x}4=\frac{3(2\sin x\sin3x)-2\sin^23x}8$$
As $2\sin A\sin B=\cos(A-B)-\cos(A+B),$
$\implies 2\sin x\sin3x=\cos(3x-x)-\cos(3x+x)=\cos2x-\cos4x$
and as $\cos2y=1-2\sin^2y\implies 2\sin^2y=1-\cos2y,$
$\implies 2\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/410343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Equation in polar coordinates to rectangular coordinates The equation in polar coordinates is $r=2\sin(\theta)\tan(\theta)$.
Show that the equation in rectangular coordinates is $y^2=\frac{x^3}{2-x}$ x cannot equal 2.
$r=2\sin(\theta)\tan(\theta)$
1.) I multiplied both sides by r so. $r^2=2r\sin(\theta)\tan(\theta)$
2.... | Good start! You want to gather the $y^2$ terms on one side, though, so instead, from $$x^2+y^2=\frac{2y^2}x,$$ subtract $y^2$ from both sides and factor, yielding $$x^2=\frac{2y^2}x-y^2=\left(\frac2x-1\right)y^2=\frac{2-x}xy^2.$$ Can you finish it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/412520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
help understanding step in derivation of correlation coefficient I'm looking to understand the starred step in the derivation below (also, if someone could help with the LaTex alignment, I'd appreciate it).
The regression line is $y= b_0 + b_1 x$, where $b_0$ and $b_1$ can be found by:
1) taking the difference between ... | For an alternative method for the same derivation:
$$
\begin{align}
\text{solving for } b_1\\
\frac{ \partial }{\partial_{b_1} }SS &= \sum (-2Y_iX_i + 2b_0 X_i + 2 b_1 X_i^2) &\text{ partial derivative wrt } b_1\\
0 &= 2\sum (-Y_iX_i + b_0 X_i + b_1 X_i^2) \\
0 &= \sum(X_iY_i - X_i\bar{Y} + b_1X_i\bar{X}-b_1X_i^2) &\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/413396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Dividing polynomial by binomial How can I divide this using long division?
$$\frac{ax^3-a^2x^2-bx^2+b^2}{ax-b}$$
Edit
Sorry guys I wrote it wrong... Fixed it now.
| $$
\begin{array}{rccccccccccccc}
& & x^2 & - & ax & - & b \\[12pt]
ax-b & ) & ax^3 & - & (a^2+b)x^2 & + & b^2 \\
& & ax^3 & - & bx^2 \\[12pt]
& & & & -a^2x^2 & + & b^2 \\
& & & & -a^2x^2 & + & abx \\[12pt]
& & & & & & -abx & + & b^2 \\
& & & & & & -abx & + & b^2 \\[12pt]
& & & & & & & & 0
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/413982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Interval of convergence of $\sum\limits_{n\geq0} \binom{2n}{n} x^n$ We consider the power series $\displaystyle{\sum_{n\geq0} {2n \choose n} x^n}$. By Ratio Test, the radius of convergence is easily shown to be $R=\frac{1}{4}$.
For $x=\frac{1}{4}$, Stirling equivalent and Ratio Test imply that the series is divergent. ... | Let $a_n=\frac{1}{4^n}\binom{2n}{n}$. Then
\begin{eqnarray}
a_n&=&\frac{(2n)\cdot(2n-1)\cdot\dots\cdot 2 \cdot 1}{(2n)^2 \cdot (2n-2)^2 \cdot \dots \cdot 2} \\
&=&\frac{(2n-1) \cdot (2n-3) \cdot \dots \cdot 3 \cdot 1}{(2n) \cdot (2n-2) \cdot \dots \cdot 4 \cdot 2} \\
&=& \prod_{k=1}^n\left(1-\frac{1}{2k}\right)\\
&<& ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/414420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
How can I evaluate $\lim_{x\to7}\frac{\sqrt{x}-\sqrt{7}}{\sqrt{x+7}-\sqrt{14}}$? I need to evaluate the following limit, if it exists:
$$\lim_{x\to7}\frac{\sqrt{x}-\sqrt{7}}{\sqrt{x+7}-\sqrt{14}}$$
How can I solve it without using differentiation or L'Hôpital?
| We can get a nice function whose limit will not be indeterminate, by multiplying numerator and denominator each by the conjugate of $\sqrt{x + 7} - \sqrt{14}$:
$$\lim_{x\to7}\frac{\sqrt{x}-\sqrt{7}}{\sqrt{x+7}-\sqrt{14}}\cdot\dfrac{(\sqrt{x+7} + \sqrt{14})}{(\sqrt{x+7} + \sqrt{14})}$$
In the denominator, we obtain a d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/415903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
$1=2$ | Continued fraction fallacy It's easy to check that for any natural $n$
$$\frac{n+1}{n}=\cfrac{1}{2-\cfrac{n+2}{n+1}}.$$
Now,
$$1=\frac{1}{2-1}=\frac{1}{2-\cfrac{1}{2-1}}=\frac{1}{2-\cfrac{1}{2-\cfrac{1}{2-1}}}=\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{1}{2-1}}}}=\ldots
=\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{... | Another example where dots are misleading: $$1= \frac{ 1 \cdot \color{blue}{2} \cdot \color{green}{3} \cdot \color{red}{4} \cdots}{ 2 \cdot \color{blue}{3} \cdot \color{green}{4} \cdot \color{red}{5} \cdots} \leq \frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/417280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "89",
"answer_count": 7,
"answer_id": 0
} |
Showing that $\int_0^{\pi/3}\frac{1}{1-\sin x}\,\mathrm dx=1+\sqrt{3}$
Show that $$\int_0^{\pi/3}\frac{1}{1-\sin x}\,\mathrm dx=1+\sqrt{3}$$
Using the substitution $t=\tan\frac{1}{2}x$
$\frac{\mathrm dt}{\mathrm dx}=\frac{1}{2}\sec^2\frac{1}{2}x$
$\mathrm dx=2\cos^2\frac{1}{2}x\,\mathrm dt$
$=(2-2\sin^2\frac{1}{2}x)\... | The Weierstrass substitution sets $t = \tan \frac{x}{2}$ and it's possible that's what you mean to do. In that case, $$\frac{\mathrm{d}t}{\mathrm{d}x} = \frac{1}{2} \mathrm{sec}^2 \frac{x}{2} = \frac{1}{2} (1 + \tan^2 \frac{x}{2}) = \frac{1 + t^2}{2}$$ which gives you $$\mathrm{d}x = \frac{2}{1+t^2} \mathrm{d}t.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/417877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Proving that $\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{100}} < 20$ How am I suppose to prove that:
$$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{100}} < 20$$
Do I use the way like how we count $1+2+ \cdots+100$ to estimate?
So $1/5050 \lt 20$, implying that it is indeed le... | We can estimate the sum by integration: because the inverse square root function is strictly descreasing, hence for $n\in \mathbb{N}$
$$
\frac{1}{\sqrt{n}}< \frac{1}{\sqrt{x}} \quad \text{ for } x\in (n-1,n).
$$
Integrating both sides on this interval:
$$
\frac{1}{\sqrt{n}}= \int^n_{n-1}\frac{1}{\sqrt{n}}\,dx< \int^{n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/420733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 6,
"answer_id": 0
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Finding $\sin^6 x+\cos^6 x$, what am I doing wrong here? I have $\sin 2x=\frac 23$ , and I'm supposed to express $\sin^6 x+\cos^6 x$ as $\frac ab$ where $a, b$ are co-prime positive integers. This is what I did:
First, notice that $(\sin x +\cos x)^2=\sin^2 x+\cos^2 x+\sin 2x=1+ \frac 23=\frac53$ .
Now, from what was... | $(\sin^2 x + \cos^2 x)^3=\sin^6 x + \cos^6 x + 3\sin^2 x \cos^2 x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/425664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
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Given: $f(n)=2n^2-3$ and $g(n)=3n+4$, find $k(n)=(fgg)(n)$? Given: $f(n)=2n^2-3$ and $g(n)=3n+4$, find $k(n)=(fgg)(n)$?
solution:
\begin{align*}
fgg(n) &= f(g(g(n)))\\
&= f(g(3n+4)))\\
&= f(3(3n + 4) + 4)\\
&= f(9n + 16)\\
&= 2(9n + 16)^2 - 3\\
&= 2(81n^2 + 288n + 256) - 3 \\
&= 162n^2 + 576n + 509
\end{align*}
Check: ... | It looks to me as though you need to multiply functions $(f\cdot g \cdot g),$ given the solutions. Here, you are doing function composition, which given your earlier problem, seems you denote as $f\circ g\circ g$. Composing the functions, you did that correctly, no mistakes (I ran through it myself.) So solution to com... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/425831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Showing that $1^k+2^k + \dots + n^k$ is divisible by $n(n+1)\over 2$
For any odd positive integer $k\geq1$, the sum $1^k+2^k + \dots + n^k$ is divisible by $n(n+1)\over 2$.
I used induction principle for the solution but cannot prove it.
I took $P(k) = 1^k+2^k+\dots+n^k$.
For $P(1)$ it is true.
For $P(n)$ let it be t... | For odd $k$ we have that
$$
a^k+b^k=(a+b)(a^{k-1}-a^{k-2}b+a^{k-3}b^2-\dots+b^{k-1})
$$
Thus, each column of
$$
\begin{align}
&0^k+\hphantom{(n-\ )}1^k+\hphantom{(n-\,\,)}2^k+\dots+n^k\\
&n^k+(n-1)^k+(n-2)^k+\cdots+0^k
\end{align}
$$
is divisible by $n$ and each column of
$$
\begin{align}
&1^k+\hphantom{(n-\ )}2^k+\hph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/427744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 4,
"answer_id": 2
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Find number of solutions of $|2x^2-5x+3|+x-1=0$
Problem: Find number of solutions of $|2x^2-5x+3|+x-1=0$
Solution:
Case 1: When $2x^2-5x+3 \geq 0$
Then we get, $2x^2-5x+3+x-1=0$
x=1,1
Case 2: When $2x^2-5x+3 < 0$
Then we get, $-2x^2+5x-3+x-1=0$
x=1,2
In both cases, common value of x is 1
Hence solution is x=1
Am I do... | $$|2x^2-5x+3|+x-1=
\begin{cases}
2x^2-5x+3+x-1 & x\ge\frac32 \quad \text{and} \quad x\le1 \\
-(2x^2-5x+3)+x-1 & 1\lt x\lt\frac32\\
\end{cases}\Rightarrow\begin{cases}
2x^2-4x+2 & x\ge\frac32 \quad \text{and} \quad x\le1 \\
-2x^2+6x-4 & 1\lt x\lt\frac32\\
\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/428303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Determine the determinant and the inverse to the matrix A. I am currently working on determining the determinant and the inverse to the matrix A.
$A =\begin{bmatrix} 1 & 0 & 0 & 0 & 0\\0 & 1 & 0 & 0 & 0\\0 & 0 & 1 & 0 & 0\\0 & 0 & 0 & a & b\\0 & 0 & 0 & c & d\\\end{bmatrix}$
The determinant is $ad-bc$. But how do I fin... | Since the OP asked in several comments whether this can be done by Gauss elimination/row operations, I wil also provide this computation. I will only do this for the right lower block.
$$
\left(\begin{array}{cc|cc}
a & b & 1 & 0\\
c & d & 0 & 1
\end{array}\right)\sim
\left(\begin{array}{cc|cc}
ac & bc & c & 0\\
ac & ad... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/431398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Evaluating $\int^1_0 \frac2{\sqrt{2-x^2}} dx$ $$\int^1_0 \frac2{\sqrt{2-x^2}} dx$$
using substitution $x=\sqrt 2 \sin \theta$
$$\int^{\pi/4}_0 \frac{2\cos \theta d\theta}{\sqrt{2-2\sin^2 \theta}} = \int^{\pi/4}_0 \frac{2\cos\theta d\theta}{\sqrt2 \cos\theta} = \int^{\pi/4}_0 \frac{2d\theta}{\sqrt2} = \int^{\pi/4}_0 \f... | HINT:
Putting $x=\sqrt2\sin\theta, dx=\sqrt2\cos\theta d\theta$ (you missed this $\sqrt2$)
$$\int^1_0 \frac2{\sqrt{2-x^2}}dx=2\int_0^{\frac\pi4}\frac{\sqrt2\cos\theta d\theta}{\sqrt2\cos\theta}=2\int_0^{\frac\pi4}d\theta$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/432533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Which of the following sets is a complete set of representatives modulo 7? Which of the following sets is a complete set of representatives modulo 7?
1) (1, 8, 27, 64, 125, 216, 343)
1 mod 7 = 1
8 mod 7 = 1
27 mod 7 = 6
64 mod 7 = 1
125 mod 7 = 6
216 mod = 6
343 mod = 0
2) (1, -3, 9, -27, 81, -243, 0)
1 mod 7 = 1
-3 mo... | $3)~~ (0, 1, -2, 4, -8, 16, -32)$ is only complete set of $\pmod 7$.
$$0 \pmod 7 = 0$$
$$1 \pmod 7 = 1$$
$$-2 \pmod 7 = 5$$
$$4 \pmod 7 = 4$$
$$-8 \pmod 7 = 6$$
$$16 \pmod 7 = 2$$
$$-32 \pmod 7 = 3$$
they all have a unique solution from $0$ to $6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/432676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Unintentional Negative Sign in Limit Evaluation I've been working on evaluating the following limit:
$$\lim_{x\to 0} \left(\csc(x^2)\cos(x)-\csc(x^2)\cos(3x) \right)$$
According to my calculator, the limit should end up being 4.
Though I've tried using the following process to find the limit, I continue to get -4:
$$\b... | HINT:
Using Prosthaphaeresis Formulas,
$$\cos x-\cos3x=2\sin 2x\sin x$$
$$\implies \dfrac{\cos x-\cos3x}{\sin (x^2)}=2\dfrac{\sin 2x\sin x}{\sin (x^2)}=2\cdot2\cdot\dfrac{\dfrac{\sin2x}{2x}\cdot\dfrac{\sin x}x}{\dfrac{\sin (x^2)}{x^2}}$$
Now use $\displaystyle\lim_{h\to0}\frac{\sin h}h=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/432762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\sum_{n=1}^{\infty} {(-1)^n \cdot \frac{2^{2n-1}}{(2n+1)\cdot 3^{2n-1}}}$
Calculate the summation $\sum_{n=1}^{\infty} {(-1)^n \cdot \frac{2^{2n-1}}{(2n+1)\cdot 3^{2n-1}}}$.
So I said:
Mark $x = \frac{2}{3}$. Therefore our summation is $\sum_{n=1}^{\infty} {(-1)^n \cdot \frac{x^{2n-1}}{(2n+1)}}$.
But h... | One can notice that
$$\frac{1}{x^2}\int x^{2n}\mathrm dx=\frac{x^{2n-1}}{(2n+1)}$$
So:
$$\sum_{n=1}^{\infty} {(-1)^n \frac{x^{2n-1}}{(2n+1)}}=\frac{1}{x^2}\sum_{n=1}^{\infty} {(-1)^n }\int x^{2n}\mathrm dx=\frac{1}{x^2}\int \left(\sum_{n=1}^{\infty} {(-1)^n }x^{2n}\right)\mathrm dx$$
And (keeping in mind that $x\leq 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/435956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Using trig substitution to evaluate $\int \frac{dt}{( t^2 + 9)^2}$ $$\int \frac{\mathrm{d}t}{( t^2 + 9)^2} = \frac {1}{81} \int \frac{\mathrm{d}t}{\left( \frac{t^2}{9} + 1\right)^2}$$
$t = 3\tan\theta\;\implies \; dt = 3 \sec^2 \theta \, \mathrm{d}\theta$
$$\frac {1}{81} \int \frac{3\sec^2\theta \mathrm{ d}\theta}{ \se... | The integral $$ \int \cos^2 \theta \, \mathrm d\theta=\frac{\theta}{2}+\frac{1}{2} \sin \theta \cos \theta+C. $$
(You had the coefficient wrong)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/436309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Prove that: $a^2+b^2+(1-a-b)^2\ge \frac {1}{3}$ Where $a$ and $b$ are any given real number. I have tried solving it using partial derivative.
$$
s=a^2+b^2+(1-a-b)^2$$
$$\frac{\partial s}{\partial a}=2a-2(1-a-b) \tag{1}$$
$$\frac{\partial s}{\partial b}=2b-2(1-a-b) \tag{2}$$
for maxima both (1) and (2) are 0..from here... | Taking it from where you left it:
$$a=1-a-b\implies a=\frac{1-b}2\\
b=1-a-b\implies b=\frac{1-a}2$$
So substituting the first eq. into the second one and using the symmetry of both:
$$b=\frac{1-\frac{1-b}2}2=\frac{1+b}4\implies b=\frac13\;\;\text{, and thus also}\;\;a=\frac13$$
Continuing with the partial derivatives
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/436587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
} |
Does $y_n=\frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n}$ converge or diverge? I have to show whether $y_n=\frac{1}{n+1}+\frac{1}{n+2}+..+\frac{1}{2n}$ is convergent or divergent. I tried using the squeeze theorem to prove it was convergent. So what I did was
bound ${y_n}$ in between $\frac{-1}{n}$ and $\frac{1}{n}$. Tha... | Note that
$\frac1{k+1}
\le \int_k^{k+1} \frac{dt}{t}
\le \frac1{k}
$.
Summing from $n+1$ to $n^2$,
$\sum_{k=n}^{n^2-1}\frac1{k+1}
\le \sum_{k=n}^{n^2-1}\int_k^{k+1} \frac{dt}{t}
\le \sum_{k=n}^{n^2-1}\frac1{k}
$,
or
$\sum_{k=n+1}^{n^2}\frac1{k}
\le \int_n^{n^2} \frac{dt}{t}
\le \sum_{k=n}^{n^2-1}\frac1{k}
$.
Usin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/437244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"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.