Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Solutions for $\frac{3}{x+1}\le\frac{2}{2x+5}$ Im in search of the solutions for:
$$\frac{3}{x+1}\le\frac{2}{2x+5}$$
So first i tried to combine the two sites:
$$\frac{6x + 15 - 2x + 2}{2x^2 +7x + 5}\le{0}$$
$$\frac{4x + 17}{2x^2 +7x +5}\le{0}$$
My problem is that now i have two solutions for the denominator and i dont... | If you assume assume $2x+5>0$ and $x+1>0$ you end up with :$$3(2x+5)\leq 2(x+1)$$
$$3(2x+5)\leq 2(x+1)\Rightarrow 6x+15\leq2x+2\Rightarrow 4x\leq -13\Rightarrow x\leq??$$
Suppose we assume $x+1$ is positive, then we would end up with above conclusion (Why??)
Suppose we assume $2x+5$ is positive but $x+1 <0$ then???
Ple... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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If $n\in\mathbb N$ and $4^n+2^n+1$ is prime, prove that there exists an $m\in\mathbb N\cup\{0\}$ such that $n=3^m$. If $n\in\mathbb N$ and $4^n+2^n+1$ is prime, prove that there exists an $m\in\mathbb N\cup\{0\}$ such that $n=3^m$.
I.e. if $4^n+2^n+1$ is prime, prove that $n=3^m$, where $m\in\mathbb N\cup\{0\}$.
I don'... | Lemma: $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$
Proof: Several proofs can be found here
$4^n+2^n+1 = \frac{8^n-1}{2^n-1}$. Let $n=3^t \times k$, where $k$ is not divisible by $3$. The numerator becomes $8^{3^t \times k}-1 = \left( {8^{3^t}} \right) ^k-1$, which has a factor $\left(8^{3^t}-1 \right)$.
The denomin... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving quadratic system If $a,b,c\in \mathbb{R}$ satisfy the system
$a^2+ab+b^2=9$;
$b^2+bc+c^2=16$;.
$c^2+ac+a^2=25$.
Find $ab+ac+bc$
| The following is a geometric solution. Let $P$ be a point, and consider three line segments $PA,PB$ and $PC$ making an angle of $120$ degrees with each other. Thus $\angle APB = \angle BPC = \angle CPA = 120^o.$ Here $|PA| = a, |PB| = b$ and $|PC| = c.$
Then by the law of cosines, triangle $ABC$ has sides $3,4$ and $5... | {
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How does mod multiplication work? For example, $10^{10} \equiv 4\pmod{6}$
If I used $\pmod{2}$ and $\pmod{3}$, how does the multiplication process work?
Since $10^{10} \equiv 0 \pmod{2}$ and $10^{10}\equiv 1\pmod{3}$,
$$
10^{10}\equiv (0,1) \pmod{(2,3)}
$$
how do we get the value $4$ at the end? do we list out the pos... | For integer $n\ge0$
$$10^{n+1}-10=10(10^n-1)\equiv0\pmod{10\cdot9}$$ as $10^n-1$ is divisible by $10-1=9$
$$\implies 10^n\equiv10\pmod{90}\equiv10\pmod6$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Why the integral $\int_0^3 \frac{1}{x^2-6x+5}dx$ doesn't exist? Why the integral $\int_0^3 \frac{1}{x^2-6x+5}dx$ doesn't exist?
$$
I=\frac{1}{2} \lim_{t\to 1^{-}} [\arctan(2-3/2) - \arctan(-3/2)] + \lim_{t\to 1^+}[\arctan(0/2) - \arctan(1-3/2)]
$$
I can find $\arctan(0)= \pi/2$
So why the answer in my book is doesn't e... | Note that $\frac{1}{(x-1)(x-5)}$ is undefined at $x = 1$ so
$$\int_0^3\frac{1}{(x - 1)(x-5)}dx = \lim_{s\to 1^-}\int_0^s\frac{1}{(x-1)(x-5)}dx + \lim_{t\to 1^+}\int_t^3\frac{1}{(x-1)(x-5)}dx.$$
To evaluate the two integrals on the right, we need to find an antiderivative. We do this by first applying partial fraction.... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finite sum related to Stirling numbers I am wondering if there is a closed form solution for the following sum:
$$
\sum _{k =0}^{n-1} \frac{(-1)^{k} (n-k)^{n+1} }{(k+1)(k+2)}\binom{n}{k}.
$$
If the the factors $(k+1)(k+2)$ in the denominator weren't there, the sum would equal $n!S(n+1,n)$, where $S(n,m)$ denotes a Stir... | We can get a better closed form using a different technique.
Suppose we seek to evaluate
$$\sum_{k=0}^{n-1} \frac{(-1)^k(n-k)^{n+1}}{(k+1)(k+2)}
{n\choose k}.$$
This is
$$\frac{1}{(n+2)(n+1)}
\sum_{k=0}^{n-1} (-1)^k (n-k)^{n+1}
{n+2\choose k+2}
\\ = \frac{1}{(n+2)(n+1)}
\sum_{k=0}^{n-1} (-1)^{k+2} (n+2-(k+2))^{n+1}
{n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/622840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Analytic hierarchy process (AHP), what is the significance of eigenvalues/eigenvectors? I was to a new years party today. Mathematics interested as I am I tried to discuss
eigenvalues with my friend there. He did not know what eigenvalues are but said
that he has heard about them in his research in something called
Ana... | The second link by Amzoti in the comment above is very clear about how the algorithm goes.
Illustration of Analytic hierarchy process
I have begun to study this by looking at the more arithmetic table $n/k$ as the pairwise comparison:
$$\begin{array}{llllll}
1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & ... | {
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Find the value of $n$ such that $(n-6)$ is divisible by $6$, $(n-7)$ is divisible by $7$ and $(n-8)$ is divisible by $8$. If $(n-6)$ is divisible by $6$, $(n-7)$ is divisible by $7$ and $(n-8)$ is divisible by $8$, then what is the value of $n $?
| If $n - 6$ is divisible by $6$, then so is $n$. Likewise for $7$ and $8$; that is, $n$ is divisible by $6$, $7$ and $8$. There are infinitely many such values; the smallest positive value is the least common multiple of $6$, $7$ and $8$.
| {
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"timestamp": "2023-03-29T00:00:00",
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manipulation of subtraction I am trying to solve an induction problem and got stuck at this part.
$$ 1 - \frac{n+2}{(n+2)!} + \frac{n+1}{(n+2)!} = 1 - \frac{(n+2) - (n+1)}{(n+2)!} $$
Shouldn't it be
$$ 1 - \frac{n+2}{(n+2)!} + \frac{n+1}{(n+2)!} = 1 - \frac{(n+2) + (n+1)}{(n+2)!} $$
How do you get the left expression t... | Note that
$$-\frac{A+B}{C}=(-1)\times \left(\frac{A+B}{C}\right)=(-1)\times\left(\frac{A}{C}+\frac{B}{C}\right)=-\frac{A}{C}-\frac{B}{C}=\frac{-A-B}{C}.$$
The minus sign in front of the first fraction does influence the plus sign in front of $B$ of the numerator of the first fraction.
| {
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prove by induction that $n(n+1)(n+2)(n+3)$ is an integer multiple of $24$ prove by induction that $n(n+1)(n+2)(n+3)$ is an integer multiple of $24$
Let $P(n)$ be the proposition we want to prov, ie: $P(n):=24 \mid(n)(n+1)(n+2)(n+3)$
For $P(1)$ we have: $24 \mid(1)(1+1)(1+2)(1+3)\implies6 \mid(1)(2)(3)(4)\implies24 \mi... | Continuing from where you left, we just need to prove that for $n=k+1$, $P(n)$ is an integer multiple of $24$.
$$P(k+1) = (k+1)(k+2)(k+3)(k+4)$$
$$P(k+1) = k(k+1)(k+2)(k+3)+4(k+1)(k+2)(k+3)$$
$1st$ term on right hand side is $P(k)$ which is an integer multiple of $24$ from your inductive hypothesis.
$2nd$ term on the r... | {
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"url": "https://math.stackexchange.com/questions/632245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How find this $\sum_{i=0}^{5}\frac{1}{2+\cos{\left(x+\frac{i\pi}{3}\right)}}\cdot \frac{1}{2+\cos{\left(x+\frac{(i+1)\pi}{3}\right)}}$ Find this follow function $f(x)$ range ,where $x\in R$,
$$f(x)=\sum_{i=0}^{5}\dfrac{1}{2+\cos{\left(x+\dfrac{i\pi}{3}\right)}}\cdot \dfrac{1}{2+\cos{\left(x+\dfrac{(i+1)\pi}{3}\right)}... | HINT
I should start simplifying the sum of the first and the sixth terms (result = A), then the sum of the second and fifth terms (result = B), then the sum of the third and fourth terms (result = C). Now, I should simplify A + B (result = D) and finally simplify C + D.
You will arrive to a surprizingly simple resu... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Searching for an explicit functional form Let $f:\mathbb N \to \mathbb R$ be a strictly decreasing function. Suppose $$\frac{f(x)}{f(x+1)}=2^x,\qquad \forall x\in\mathbb N.$$ Is it possible to find an explicit functional form of $f$? Any hint at finding a solution is appreciated.
| Hint: We compute a little to find out what's going on.
Let us rewrite the equation as $f(n+1)=\frac{f(n)}{2^{n}}$. Let $f(1)=a$.
We have $f(2)=\frac{a}{2^1}$, and therefore $f(3)=\frac{a}{2^{1}\cdot 2^2}$, and therefore $f(4)=\frac{a}{2^{1}\cdot 2^2\cdot 2^3}$, and therefore $f(5)=\frac{a}{2^{1}\cdot 2^2\cdot 2^3\cdo... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Determining y(2) knowing that $y(1)=1+e^3$ and $ty'+(2t^2)y=2t^2$ How do I determine y(2) knowing that $y(1)=1+e^3$ and $ty'+(2t^2)y=2t^2$ ?
| By dividing both sides by $t$ (to obtain the standard form), express the ordinary differential equation as
$$\begin{aligned}
t\dfrac{dy}{dt} + 2t^2y &= 2t^2\\
\dfrac{dy}{dt} + 2ty &= 2t
\end{aligned}$$
Since $p(t) = 2t$ by the standard form of differential equation, the integrating factor of this equation is
$$\begin{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/635237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sum of a Hyper-geometric series. (NBHM 2011)
How to find the sum of the following series
$$\frac{1}{5} - \frac{1\cdot 4}{5\cdot 10} + \frac{1\cdot 4\cdot 7}{5\cdot 10\cdot 15} - \dots\,.?$$
I have no idea. I have written the general term and tested its convergence by Gauss' test for convergence, but they are neither ... | From the Generalized Binomial Theorem (for $|x|<1$), $$\left(1+x\right)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\cdots$$
Method $1:$
$\displaystyle -\frac{1\cdot 4\cdot 7}{5\cdot 10\cdot 15}=\frac{-\frac13\left(-\frac13-1\right)\left(-\frac13-2\right)3^3}{3!\cdot 5^3}=\frac{-\frac13\left(-\frac13-1\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/635315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Partial sum of a geometric series Consider the geometric series $\sum_{n=0}^\infty ar^n$ where $a=1$
and $r=-\frac{1}{2}$. Since $|r| < 1$, the series converges to
$S = \sum_{n=0}^\infty ar^n = \frac{1}{1+\frac{1}{2}} = \frac{2}{3}$.
I would like to arrive at the same sum by computing
$\lim_{N \to \infty} S_N$ where... | Hint:
$$a + ar + ar^2 + \ldots + ar^{n-1} = \frac{ a( 1- r^n) } { (1-r) }. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/636308",
"timestamp": "2023-03-29T00:00:00",
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If $\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b}=\frac{1}{a+b}$, then show that $\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}=\frac{1}{(a+b)^2}$
If $\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b}=\frac{1}{a+b}$, then show that $\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}=\frac{1}{(a+b)^2}$
My work:
$(\frac{\sin^4 x}{a}+\frac{\cos^4 x}{... | One's first inclination might be to reduce powers by invoking the Half-Angle Identities, which can be written as
$$\sin^2 x = \frac{1}{2}( 1 - k ) \qquad \cos^2 x = \frac{1}{2}(1 + k) \qquad\text{, where}\qquad k := \cos 2x$$
Then,
$$\begin{align}
\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b} = \frac{1}{a+b} \quad
&\implies\q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/639223",
"timestamp": "2023-03-29T00:00:00",
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Limit $\lim_{n\to\infty}\left(1-\frac{1}{3}\right)^2\left(1-\frac{1}{6}\right)^2\cdots\left(1-\frac{1}{\frac{n(n+1)}{2}}\right)^2$ How can I find the following limit :$$\lim_{n\to\infty}\left(1-\frac{1}{3}\right)^2\left(1-\frac{1}{6}\right)^2\left(1-\frac{1}{10}\right)^2\cdots\left(1-\frac{1}{\frac{n(n+1)}{2}}\right)^2... | $$1-\frac{2}{n(n+1)}=\frac{(n-1)(n+2)}{n(n+1)}$$
In the following equation, you can simplify $(i+1)$ with the next $i$, except for the first (that is equals to $1$ anyway), and $(i+3)$ with the next $(i+2)$ except for the first one (that is equals to $3$)
$$\prod_{i=1}^{\infty}\left(1-\frac{2}{(i+1)(i+2)}\right)^2=\pro... | {
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"timestamp": "2023-03-29T00:00:00",
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Show that $\int_0^{\pi/3} \big((\sqrt{3}\cos x-\sin x)\sin x\big)^{1/2}\cos x \,dx =\frac{\pi\sqrt{3}}{8\sqrt{2}}. $ I have run a FORTRAN code and I have obtained strong evidence that
$$\int_0^{\pi/3} \!\!
\big((\sqrt{3}\cos\vartheta-\sin\vartheta)\sin\vartheta\big)^{\!1/2}\!\cos\vartheta \,d\vartheta
=\frac{\pi\sqrt{3... | OK, you can start by noting that
$$\sqrt{3} \cos{x} - \sin{x} = 2 \sin{\left ( \frac{\pi}{3}-x\right)}$$
which means that the integral is
$$ \sqrt{2} \int_0^{\pi/3} dx \, \cos{x} \sqrt{\sin{x} \sin{\left ( \frac{\pi}{3}-x\right)}} = \sqrt{2} I$$
You can make a substitution $x \mapsto \frac{\pi}{3}-x$ and see that
$$I =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/640585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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Calculus $T_1=\prod_{k=1}^{n-1} \cos\frac{k\pi}{2n}$ Calculus: $$T_1=\prod_{k=1}^{n-1} \cos\frac{k\pi}{2n}$$ and $$T_2=\prod_{k=1}^{n-1}\sin\frac{k\pi}{2n}$$
My tried:
I use Euler's formal: $$z_k=e^{i\frac{k\pi}{2n}}=\cos\frac{k\pi}{2n}+i\sin \frac{k\pi}{2n}$$
$$\to\left\{\begin{matrix}\cos\frac{k\pi}{2n}=\frac{1}{2}\... | First of all, $\displaystyle\cos\frac{k\pi}{2n}=\sin\left(\frac\pi2-\frac{k\pi}
{2n}\right)=\sin\left(\frac{(n-k)\pi}{2n}\right)$
So, $T_1=T_2$
Now, $\displaystyle\sin\frac{k\pi}{2n}=\sin\left(\pi-\frac{k\pi}{2n}\right)=\cos\frac{(2n-k)\pi}{2n}$
So, $\displaystyle T_2=\prod_{k=1}^{n-1}\sin\frac{k\pi}{2n}=\prod_{k=n+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/641941",
"timestamp": "2023-03-29T00:00:00",
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How to integrate fraction Hi can anybody give me some hint how to integrate this type of fractions
$$\int \frac{1}{(x^2 +a)^2} dx $$ where $ a \in \mathbb N $
Thanks
| Take the integral
$$\int \frac{dx}{x^2+a^2}$$
and we do an integration by parts:
$$\int 1\cdot\frac{dx}{x^2+a^2}=\frac{x}{x^2+a^2}+2\int\frac{x^2dx}{(x^2+a^2)^2}$$
for the last integral we add and subtract $1$ in the numerator and we find
$$\int\frac{dx}{(x^2+a^2)^2}=\frac 1 2\left(\int \frac{dx}{x^2+a^2}+\frac{x}{x^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/642690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Need assistance on geometry problem
Having a really hard time solving this problem. Given:
*
*a circle of radius $a$
*an ellipse with minor axis $g$ and major axis $f$
*the ellipse is oriented so that the major axis is parallel with the vector between the circle and ellipse
*lines which are tangent to both the e... | Here's a coordinate-based approach.
Set the center of the circle at the origin. Then the tangent line through $T(a \cos\phi, a \sin\phi)$ has $x$- and $y$-intercepts $a\sec\phi$ and $a\csc\phi$, respectively. (Note that the $x$-intercept is $d_1$ in your figure.) In intercept-intercept form, the line's equation is
$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/645770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Determine if $\sum_{n=1}^{\infty}\frac{(-1)^nn^2+n}{n^3+1}$ converges or diverges. Another series I found I'm struggling with.
Determine if the following series converges or diverges.$$\sum_{n=1}^{\infty}\frac{(-1)^nn^2+n}{n^3+1}$$
Ratio test and n-th root test are both inconclusive, Leibniz - criterion cannot be app... | The series
$$\sum_{n=1}^{\infty}(-1)^n\frac{n^2}{n^3+1}$$ converges by Leibniz criterion, in addition the series
$$\sum_{n=1}^{\infty}\frac{n}{n^3+1}$$ converges by the comparison test.
Hence
$$\sum_{n=1}^{\infty}(-1)^n\frac{n^2}{n^3+1} + \sum_{n=1}^{\infty}\frac{n}{n^3+1} = \sum_{n=1}^{\infty}\frac{(-1)^nn^2+n}{n^3+1}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding a least common multiple (LCM) My Algebra 2 book explains how to find a least common multiple:
Find the least common multiple of $4x^2 - 16$ and $6x^2 - 24x + 24$.
Solution
Step 1 Factor each polynomial. Write numerical factors as products of primes.
$4x^2 - 16 = 4(x^2 - 4) = (2^2)(x + 2)(x - 2)$
$6x^2 - ... | It's kind of a weird way of saying it. $$\mathrm{lcm}(P(x),Q(x)) = \frac{P(x)\cdot Q(x)}{\gcd(P(x),Q(x))}$$
Where $$\gcd(P(x),Q(x))$$ is just the product of the terms that appear in both factorizations.
For instance $$\mathrm{lcm}(((x+1)(x+1)(x-1)),((x+1)(x-1)(x-1)))=\frac{(x+1)^3(x-1)^3}{(x+1)(x-1)}=(x+1)^2(x-1)^2 $... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Use the disc method to find the volume of solid of revolution. Use the disc method to find the volume of solid of revolution obtained by revolving the the given region around the x-axis.
$R\quad is\quad bounded\quad by\quad the\quad graphs\quad of\quad k(x)\quad =\quad { x }^{ 2 }+x+2\quad and\quad m(x)\quad =\quad 2x+... | You may want to do these without the absolute value brackets, as I really think that is confusing the issue (and isn't even really correct...). The line is the "upper function", so the evaluation should be
$$ \pi \ \left[ \ ( - \frac{32}{5} - 8 - \frac{8}{3} + 24 + 24 ) \ - \ (\frac{1}{5} - \frac{1}{2} + \frac{1}{3} +... | {
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"timestamp": "2023-03-29T00:00:00",
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$AX=B$ solve for $X$ ....... in MATRIX $$ 2x - 3y + 4z = -19\\
6x + 4y - 2z =8 \\
x + 5y + 4z = 23
$$
what I have done so far is I put the nubmer and $x, y $ and $ z$ in matrix form:
$$
\begin{bmatrix}
2 & -3 & 4\\
6 & 4 &-2\\
1 & 5 & 4
\end{bmatrix}
\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}=
\begin{b... | You can also use Cramer's Rule:
$$
x=\frac{\left|\begin{array}{r}\color{#C00000}{-19}&-3&4\\\color{#C00000}{8}&4&-2\\\color{#C00000}{23}&5&4\end{array}\right|}{\left|\begin{array}{r}2&-3&4\\6&4&-2\\1&5&4\end{array}\right|}=-2
$$
$$
y=\frac{\left|\begin{array}{r}2&\color{#C00000}{-19}&4\\6&\color{#C00000}{8}&-2\\1&\colo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/649080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Computing $\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right) \, dx$
For $a\ge 0$ let's define $$I(a)=\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right)dx.$$ Find explicit formula for $I(a)$.
My attempt: Let
$$\begin{align*}
f_n(x)
&= \frac{\ln\left(1-2
\left(a+\frac{1}{n}\right)\cos x+\left(a+\frac{1}{n}\right)^2\right)-\ln\lef... | Here is an elementary way to compute the integral.
First, let us prove some initial results.
*
*Making the substitution $x \mapsto \pi - x$ yields
$I(a) = \int^\pi_0 \log \left (1 + 2a\cos x + a^2 \right ) \, dx = I(-a)$
so that
$$I(a) = I(-a). \tag{$\dagger$}$$
*Then, consider
$$\begin{align*}
I(a) + I(-a)
&= \int... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 7,
"answer_id": 3
} |
Solve $2000x^6+100x^5+10x^3+x-2=0$ One of the roots of the equation $2000x^6+100x^5+10x^3+x-2=0$ is of the form $\frac{m+\sqrt{n}}r$, where $m$ is a non-zero integer and $n$ and $r$ are relatively prime integers.Then the value of $m+n+r$ is?
Tried to use the fact that another root will be $\frac{m-\sqrt{n}}r$ as coeff... | The usual trick is to divide through by $x^3.$ Then notice that taking $w = 10x - \frac{1}{x}$ seems to allow writing the thing, and we get
$$ 2 w^3 + w^2 + 60 w + 30 = 0. $$
This has a rational root, namely $-1/2,$ and factors as
$$ (2w+1)(w^2 + 30) $$
Setting
$$ 10 x - \frac{1}{x} = -\frac{1}{2} $$
leads to
$$ 20 ... | {
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"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
Evaluating $\int_{\pi/3}^{\pi/2}\frac{1}{\sin\left(x\right)-\cos\left(x\right)+1} \operatorname dx$ $$\int \limits_{\pi/3}^{\pi/2}\dfrac{1}{\sin\left(x\right)-\cos\left(x\right)+1} \operatorname dx$$
If I try solving this integral with universal substitutions: $$\tan\left(\frac{x}{2}\right)=t;\:\sin\left(x\right)=\frac... | What you have got
$$\displaystyle \int \limits_{\frac{\sqrt{3}}{3}}^1\:\dfrac{dt}{t\left(t+1\right)}$$
is certainly correct. Now,
\begin{align}\displaystyle \int \limits_{\frac{\sqrt{3}}{3}}^1\:\dfrac{dt}{t\left(t+1\right)} & = \displaystyle \int \limits_{\frac{\sqrt{3}}{3}}^1\:dt\left(\frac{1}{t}-\frac{1}{t+1}\right)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/652582",
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"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
is there a nicer way to $\int e^{2x} \sin x\, dx$? I have to solve this:
$\int e^{2x} \sin x\, dx$
I managed to do it like this:
let $\space u_1 = \sin x \space$ and let $\space \dfrac{dv}{dx}_1 = e^{2x}$
$\therefore \dfrac{du}{dx}_1 = \cos x \space $ and $\space v_1 = \frac{1}{2}e^{2x}$
If I substitute these values in... | $$
\begin{align*}
\\ \int e^{2x}\sin x dx &= \Im \int e^{2x}(\cos x + i\sin x) dx
\\ &= \Im \int e^{(2 + i)x}dx
\\ &= \Im \frac{e^{(2 + i)x}}{2+i} + C
\\ &= \Im \frac15 e^{2x}(\cos x + i\sin x)(2-i) + C
\\ &= \frac15 e^{2x}(2\sin x - \cos x) +C
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/657389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Simplify the surd expression. Simplify the surd.
$(2\sqrt 3 + 3\sqrt 2)^2$
I know I should us this formula: $(a^2+2ab+b^2)$
But this gets complicated later. Please explain. :(
| We are given:
$$(2\sqrt 3 + 3\sqrt 2)^2$$
Expand:
$$(2\sqrt 3 + 3\sqrt 2)(2\sqrt 3 + 3\sqrt 2)$$
Multiply using techniques such as the FOIL method:
$$(2\sqrt 3 + 3\sqrt 2)(2\sqrt 3 + 3\sqrt 2)=4\sqrt9+6\sqrt6+6\sqrt6+9\sqrt4$$
Simplify:
$$4(3)+12\sqrt6+9(2)\\=12+18+12\sqrt6\\=30+12\sqrt6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/658736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Limit $\lim\limits_{x\to0}\frac1{\ln(x+1)}-\frac1x$ The limit is
$$\lim\limits_{x\to0}\frac1{\ln(x+1)}-\frac1x$$
The problem is I don't know if I can calculate it normally like with a change of variables or not. Keep in mind that I'm not allowed to use L'Hôpital's rule nor the $\mathcal O$-notation.
| It is bit difficult to avoid LHR or series expansions here. I present here a technique which is almost like using series expansion, but a bit simpler conceptually. For this purpose I need to use the standard definition of $\log x$ as $\int_{1}^{x}(1/t)\,dt$.
Let us assume that $0 < t < 1$. Then it can be checked using ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Prove that $\sum_{k=0}^{\infty} (k-1)/2^k = 0$ How to prove that this series converges, and that the limit is 0 ?
| $$
\sum_{k=0}^\infty \frac{k-1}{2^k} = -1 + \frac{1}{2}\sum_{k=1}^\infty \frac{k-1}{2^{k-1}}.
$$
If we call $n=k-1$, then
$$
\sum_{k=0}^\infty \frac{k-1}{2^k} = -1 + \frac{1}{2}\sum_{n=0}^\infty \frac{n}{2^n} = -1 + \frac{1}{2}\sum_{n=0}^\infty \frac{1}{2^n}+ \frac{1}{2}\sum_{n=0}^\infty \frac{{n-1}}{2^n}.
$$
Therefore... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Simplify the expression where $x>y>0$ Simplify $$\frac{\sqrt {x^2+y^2}+x}{y+\sqrt {x^2-y^2}}\cdot\frac{x-\sqrt {x^2+y^2}}{\sqrt {x^2-y^2}-y}$$
Help please, I tried but the answer doesn't match. I did it by multiplying of course and then normally simplifying.
I tried and what came is $\dfrac{-2xy -y^2}{2xy + x^2}$ and t... | $$\frac{\sqrt {x^2+y^2}+x}{y+\sqrt {x^2-y^2}}\cdot\frac{x-\sqrt {x^2+y^2}}{\sqrt {x^2-y^2}-y}= \frac{x + \sqrt {x^2+y^2}}{\sqrt {x^2-y^2}+y}\cdot\frac{x-\sqrt {x^2+y^2}}{\sqrt {x^2-y^2}-y}$$
Now you've got a function of the form $$\dfrac{(a+b)(a- b)}{(c+d)(c- d)} = \dfrac{a^2 - b^2}{c^2 - d^2}$$
The rest is then simpli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/661114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Computing this limit $$\lim_{n\to\infty}{\sqrt[3]{n^3+5n^2+6}-\sqrt{n^2+3n+4}}$$
At first glance, we see that it's an indeterminate form ($\infty-\infty$). Here are my tries:
I.) I tried to form $a^2-b^2$ in the numerator, where $\cdots$ represents something that $\to 0$:
$$\frac{n^2(\sqrt[3]{1+\cdots}-1)-3n-4}{n(\sqrt... | Using series:
\begin{align*}
\sqrt[3]{n^3+5n^2+6}-\sqrt{n^2+3n+4} &= n\left( \left(1+\frac{5}{n}+\frac{6}{n^3}\right)^{1/3} - \left(1+\frac{3}{n}+\frac{4}{n^2}\right)^{1/2} \right) \\
&= n\left( 1+\frac{5}{3n} + o\left(\frac{1}{n}\right) - \left(1+\frac{3}{2n}+o\left(\frac{1}{n}\right)\right) \right) \\
&= n\cdot\fra... | {
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"url": "https://math.stackexchange.com/questions/662060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\sum_{k=1}^{\infty} (2k+1)\,x^{2k}$,$\,$ for $|x| < 1$ Let $f(x)$ be $\sum_{k=1}^{\infty} x^{2k+1}$. This sum equals $$x + x^3 + x^5 + \dots - x= x(1 + x^2 + (x^2)^2 + \dots) - x = \frac{x}{1-x^2} - x = \frac{x^3}{1-x^2}$$
Now, okay, the original sum is the derivative of $f$, but I don't think it is the concl... | Yes you are allowed to differentiate, as $f(x)$ is differentiable for every $|x|<1$.
Then
$$
f'(x)=\frac{3x^2}{1-x^2}+\frac{2x^4}{(1-x^2)^2}.
$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is it true that $(2^n+n^2)(n^3+3^n)$ is $O(6^n)$? $(2^n+n^2)$ is $O(2^n)$ and $(n^3+3^n)$ is $O(3^n)$, therefore I conclude that $(2^n+n^2)(n^3+3^n)$ is $O(2^n*3^n)=O(6^n)$
| Your answer is correct, yet if you were not 100% sure you can always check using the definition of "big O":
$$\limsup_{n\rightarrow\infty}\frac{(2^n+n^2)(n^3+3^n)}{6^n}=\limsup_{n\rightarrow\infty}\frac{n^5+n^32^n+n^23^n+6^n}{6^n}=\limsup_{n\rightarrow\infty}(n^5(1/6)^n+n^3(1/3)^n+n^2(1/2)^n+1)=1$$
$$\Rightarrow (2^n+n... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Evaluate integral as a logarithm plus an arctangent. Evaluate the integral as a logarithm plus an arctangent.
$$ \int \frac{x}{3x^2-18x+45} \ dx $$
I just completed the square and couldn't continue.
$$ \int \frac{x}{3(x-3)^2+18} \ dx $$
Fixed a typo $18$ changed to $18 x$
| $$\begin{align}
\int \frac{x}{3(x-3)^2+18} \ dx & = \frac 1 3 \cdot \frac 1 2 \int\frac{2(x-3)}{(x-3)^2 + 6} + \frac 1 3 \int \frac{3}{(x-3)^2+6}\,dx.
\end{align}
$$
The first integral can be handled by the substitution $u=(x-3)^2+6$ and $du=2(x-3)\,dx$.
Then:
$$
\int \frac{dx}{(x-3)^2+6} = \frac 1 6 \int \frac{dx}{\fr... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Solving $\sqrt{x+2}+\sqrt{x-3}=\sqrt{3x+4}$ I try to solve this equation:
$$\sqrt{x+2}+\sqrt{x-3}=\sqrt{3x+4}$$
So what i did was:
$$x+2+2*\sqrt{x+2}*\sqrt{x-3}+x-3=3x+4$$
$$2*\sqrt{x+2}*\sqrt{x-3}=x+5$$
$$4*{(x+2)}*(x-3)=x^2+25+10x$$
$$4x^2-4x-24=x^2+25+10x$$
$$3x^2-14x-49$$
But this seems to be wrong! What did i wron... | first step on the right hand side should be $3x+5$ not $3x+4$
Then proceeding the same way:
$$
\begin{split}
4(x^2-x-6) &= x^2 + 12x + 36 \\
3x^2 - 16x - 60 &= 0
\end{split}
$$
Since the discriminant of that equation is $16^2 - 4 \cdot 3 \cdot 60 < 0$ there are no solutions.
UPDATE you changed the problem now, so you g... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
For which $x$ does the series $∑_{n=1}^∞(1+\frac{1}{2}+ \frac{1}{3}+⋯+\frac{1}{n})\, x^n$ converge? Determine for what value of $x$ the series converges
$$\sum_{n=1}^\infty \left(1+\frac{1}{2}+ \frac{1}{3}+⋯+\frac{1}{n}\right) x^n $$
Observe that
$∑_{n=1}^∞(1+\frac{1}{2}+ \frac{1}{3}+⋯+\frac{1}{n}) x^n =∑_{n=1}^∞ (∑_{... | Note that
$$
1\le 1+\frac{1}{2}+\cdots+\frac{1}{n}\le \log (n+1),
$$
and hence, using for example the comparison test, the radius of convergence of the series
$$\sum_{n=1}^\infty \left(1+\frac{1}{2}+ \frac{1}{3}+⋯+\frac{1}{n}\right) x^n $$
is $r=1$, since the radius of convergence of both series
$$
\sum_{n=0}^\infty... | {
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"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Partial fractions: why does $\int dt \implies t + C$ I am working on a partial fraction problem here, I understand everything in the problem except $t+C$, so I'd like to know where did the $t+C$ come from ?
I want to solve this integral
$$
\int \frac{dy}{(y+2)(1-y)} = \int dt
$$
$$\begin{align}
1 &= \frac{A}{y+2} + \... | Your original equation is $$\int \frac{dy}{(y+2)(1-y)} = \int \,dt$$
$t + C$ comes from integrating the righthand side of the original equation: $$\int \,dt = t + C$$
The $C$ is the constant of integration.
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How can I check the convergence of the sequence? Does it diverge? How can I check the convergence of the sequence $\frac{1}{\sqrt{n^2+1}}+\frac{2}{\sqrt{n^2+2}}+\cdots+\frac{n}{\sqrt{n^2+n}}$? I think that it diverges,because it is bounded below from $\frac{n(n+1)}{2\sqrt{n^2+n}} $ and above from $\frac{n(n+1)}{2\sqrt{... | HINT: you can use Squeeze Theorem.
$\frac{1}{\sqrt{n^2+n}}+\frac{2}{\sqrt{n^2+n}}+\cdots+\frac{n}{\sqrt{n^2+n}}\leq\frac{1}{\sqrt{n^2+1}}+\frac{2}{\sqrt{n^2+2}}+\cdots+\frac{n}{\sqrt{n^2+n}}\leq\frac{1}{\sqrt{n^2+1}}+\frac{2}{\sqrt{n^2+1}}+\cdots+\frac{n}{\sqrt{n^2+1}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/672710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Given positive real numbers $a, b, c$ with $aI am trying to prove the following:
$$\frac{a}{1+a} < \frac{b}{1+b} + \frac{c}{1+c}$$
given that $a, b, c > 0$ and $a < b+c$. I tried various rearrangements but can't seem to get anywhere with it.
| Function $f(x)=\frac{x}{1+x}$ is increasing (it is easy to check first derivative). Therefore $f(a)<f(b+c)$, for $a<b+c$. Because of that, we have $$\frac{a}{1+a}<\frac{b+c}{1+b+c}=\frac{b}{1+b+c}+\frac{c}{1+b+c},$$ from where we get (since $a,b,c>0$):
$$\frac{a}{1+a}<\frac{b}{1+b}+\frac{c}{1+c}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/672881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Generalization of "Sum of cube of any 3 consecutive integers is divisible by 3" I have this question posted by professor in graduate Number Theory class. First he asked for proof that the sum of cube of 3 consecutive integers is divisible by 3, which is very easy to prove, but then he continued by asking to prove its g... | Since this is a graduate level number theory class, I think it's safe to assume that you are familiar with modulo arithmetic?
Given any list of $n$ consecutive integers, $a, a+1, a+2, \dots, a+n-1$, modulo $n$ this list is equivalent to $0,1,2,3,\dots,n-1$ modulo $n$. (Note that I am not saying $a \equiv 0 \pmod{n}$).... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Find all the $a$ such $539|a3^{253}+5^{44}$ This is what i thought:
Given that $539|a3^{253}+5^{44}$ then $11|a3^{253}+5^{44}$ and $7^2|a3^{253}+5^{44}$
using congruences I get:
$$a3^{253}+5^{44} \equiv 0 \pmod{7^2}$$
and
$$a3^{253}+5^{44} \equiv 0 \pmod{11}$$
then as a condition, I get $(a , 11)\not=1$ and $(a, 49)\... | $539=7^2\cdot11$. Per Fermat's Little Theorem, we have $3^{10}\equiv1\text{ mod }11\iff3^{253}\equiv3^3\equiv5\text{ mod }11$ and $5^{10}\equiv1\text{ mod }11\iff5^{44}\equiv5^4\equiv9\equiv-2\text{ mod }11\iff5a-2\equiv0\text{ mod }11\iff5a\equiv2$ $\text{mod }11\iff a\equiv7\text{ mod }11$: The first condition.
By b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/675086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding the real solutions to $16^{x^{2} + y } + 16^{y^{2}+ x} = 1$ We have , $16^{x^{2} + y } + 16^{y^{2}+ x} = 1$ , then we have to find all the real values of $x$ and $y$.I tried this question but i am not able to proceed because I am not able to simplify this expression to an extent that it could be solved.
| $x^2 + y^2 + x + y = (x + 1/2)^2 + (y + 1/2)^2 - 1/2 \geq -1/2$ and equality occurs only when $x = y = - 1/2$.
Using AM-GM inequality $16^{x^2 + y} + 16^{y^2 + x} \geq 2\cdot\sqrt{16^{x^2+y^2+x+y}} \geq 2\cdot16^{-1/4} = 1$ and equality occurs only when $x=y=-1/2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/676358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 0
} |
Divisibility test for $4$
Claim: A number is divisible by $4$ if and only if the number formed by the last two digits is divisible by $4$.
Here's where I've gotten so far.
Let $x$ be an $(n+1)$-digit number. So $x= a_na_{n-1} \dots a_2a_1a_0$. If $a_1 = 0$ and $a_0 =0$, then $x$ is a multiple of $100$ and therefore c... | All integers can be written in the form:
$$a_{0}10^n+a_{1}10^{n-1}+...+a_{n-1}10^1+a_{n}$$
That can be rewritten as:
$$100(a_{0}10^{n-2}+a_{1}10^{n-3}+...+a_{n-3}10^1+a_{n-2})+a_{n-1}10^1+a_{n}$$
$$=4\times 25(a_{0}10^{n-2}+a_{1}10^{n-3}+...+a_{n-3}10^1+a_{n-2})+a_{n-1}10^1+a_{n}$$
Since the term $(a_{0}10^{n-2}+a_{1}1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/676431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How find this maximum $f(x)=\sqrt{\frac{1+x^2}{2}}+\sqrt{x}-x$ let $x\ge 0$,Find the follow maximum
$$f(x)=\sqrt{\dfrac{1+x^2}{2}}+\sqrt{x}-x$$
I find
$$f'(x)=\dfrac{1}{2}\left(\dfrac{\sqrt{2}x}{\sqrt{x^2+1}}+\dfrac{1}{\sqrt{x}}-2\right)$$
so
$$f'(x)=0\Longrightarrow x=1$$
so I think
$$f(x)_{max}=f(1)$$
But I this me... | Let us show that $f(x)+x\leqslant x+1$ for every $x$. This holds true if and only if
$$
\sqrt{\frac{x+\frac1x}2}+1\leqslant\sqrt{x}+\frac1{\sqrt{x}}.
$$
Let $u=\sqrt{x}+\frac1{\sqrt{x}}$, then $u\geqslant2$ for every $x\geqslant0$ and $x+\frac1x=u^2-2$ hence it is enough to show that $g(u)\leqslant0$ for every $u\geqsl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/677326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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Flexibility of form Prove that for all real $x,y,a,b$, one has
$$(5ax+ay+bx+3by)^2\leq(5a^2+2ab+3b^2)(5x^2+2xy+3y^2)$$
Here's my try but I can't continue.
$$(5ax+ay+bx+3by)^2=(\sqrt5a\cdot\sqrt5x+a\cdot y+b\cdot x+\sqrt3b\cdot\sqrt3y)^2\leq(5a^2+a^2+b^2+3b^2)(5x^2+y^2+x^2+3y^2)$$
| By C-S we obtain:
$$(5a^2+2ab+3b^2)(5x^2+2xy+3y^2)=$$
$$=\left(5a^2+2ab+\frac{1}{5}b^2+\frac{14}{5}b^2\right)\left(5x^2+2xy+\frac{1}{5}y^2+\frac{14}{5}y^2\right)=$$
$$=\left(\left(\sqrt5a+\frac{1}{\sqrt5}b\right)^2+\frac{14}{5}b^2\right)\left(\left(\sqrt5x+\frac{1}{\sqrt5}y\right)^2+\frac{14}{5}y^2\right)\geq$$
$$\geq\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/677446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Ratio and proportion problem
Question:
If $$(a+b):(b+c):(c+a)=6:7:8$$ and $a+b+c=14$, then find the value of $c$.
My solution:
*
*$$\frac{(a+b)(c+a)}{(b+c)}=\frac{(6)(8)}{7}$$$$\Rightarrow \frac{ac + a^2 + bc + ba}{b+c} = \frac{48}{7}$$$$\Rightarrow \frac{a(b+c)+a^2+bc}{b+c}=\frac{48}{7}$$$$\Rightarrow ????$$
*... | Let $$x=b+c,$$ $$y=a+c,$$ $$z=a+b.$$ Then, $x+y+z=2(a+b+c)=28$ and $z:x:y=6:7:8$.
By substituting $z=6k$, $x=7k$, $y=8k$ into $x+y+z=28$, we have $21 k=28$, $k=\frac{4}{3}$, from where we get: $$z=6k=8,$$ $$x=7k=\frac{28}{3},$$ $$y=8k=\frac{32}{3}.$$
Finally, $$a=a+b+c-x=\frac{14}{3},$$ $$b=a+b+c-y=\frac{10}{3},$$ $$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/677516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $x+\frac1x=5$ find $x^5+\frac1{x^5}$. If $x>0$ and $\,x+\dfrac{1}{x}=5,\,$ find $\,x^5+\dfrac{1}{x^5}$.
Is there any other way find it?
$$
\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)=23\cdot 110.
$$
Thanks
| \begin{align}
x+\frac{1}{x}&=5\\
x^2+1&=5x\\
x^2-5x+1&=0\\
x &=\frac{1}{2} \left( 5 +\sqrt{21}\right)\\
x^5+\frac{1}{x^5}&=\cdots
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/678650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 10,
"answer_id": 3
} |
how to form a quadratic equation If $\alpha$ and $\beta$ are the root of the equation $x^2 - 4x +6 =0$ , find the equation whose roots are $\alpha + 1/\beta$ and $\beta + 1/\alpha$.
| $x^2 - \frac{28}{6}x + \frac{49}{6}$.
$\alpha + \beta = 4$ and $\alpha\beta = 6$. Expand $(x - (\alpha + \frac{1}{\beta}))(x-(\beta + \frac{1}{\alpha}))$. This gives us $x^2 - (\alpha + \beta)(1 + \frac{1}{\alpha\beta})x + (\alpha\beta + 2 + \frac{1}{\alpha\beta})$ So, doing the necessary substitutions gives us: $x^2 -... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Change of variables in Double integral - explanation of solved example. I'm exercising double integrals and here's an solved example that I'm not understand something.
Calculate $\iint_D xy(x^2 + y^2) dx\,dy$ where $D$ in the first quadrant bounded by:
$1 \le xy \le 2$ and $5 \le x^2 - y^2 \le 9.$
Here's the solutio... | It was cancelled. There is an $x^2+y^2$ in the numerator and in the denominator.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
For which $n \in \mathbb{Z^+}$ is $(4n+9)/(2n^2+7n+6)$ a terminating decimal? I saw this problem here.
My approach:
Let $A = (4n+9)/(2n^2+7n+6) = \frac{6}{2n + 3} - \frac{1}{n + 2}$
If $\frac{6}{2n + 3}$ and $\frac{1}{n + 2}$ are terminating decimals, then so is $A$.
A number terminates if and only if it can be written... | The only prime that can simultaneously divide numerator and denominator is $3$. For if an odd prime $p$ divides $4n+9$ and $2n+3$, it must divide $(4n+9)-2(2n+3)=3$. And it is easy to see that $4n+9$ and $n+2$ are relatively prime.
We first examine the case where $3$ does not simultaneously divide numerator and denomin... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove that, $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\ge 4$ where we do not use AM-GM inequality on the given statement to prove it. Prove that, $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{d}+\dfrac{d}{a}\ge 4$ where we do not use AM-GM inequality on the given statement to prove it. Typically, I am actually looking for... | $$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=\left(\frac{a}{b}+\frac{c}{d}\right)+\left(\frac{b}{c}+\frac{d}{a}\right)=$$
$$=\left[\sqrt{\frac{a}{b}+\frac{c}{d}}-\sqrt{\frac{b}{c}+\frac{d}{a}}\right]^2+2\sqrt{\left(\frac{a}{b}+\frac{c}{d}\right)\cdot\left(\frac{b}{c}+\frac{d}{a}\right)}=$$
$$=\left[\sqrt{\frac{a}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/686791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Integral of $\cos(3x) \cos(4x) \cos(5x)$ The integral $$\int_0^{\pi/8}\cos(3x)\cos(4x)\cos(5x) \,dx$$ is equal to $k/24$. Find the constant $k$.
So far, I assume that the best way to solve this question is to solve the integral and compare the answer to find $k$.
I have thought about rewriting the integral by using t... | Using
\begin{align}
\cos(x) \cos(y) &= \frac{1}{2} \, (\cos(x+y) + \cos(x-y) ) \\
\sin\left(\frac{\pi}{4} \right) &= \sin\left(\frac{3 \pi}{4}\right) = \frac{1}{\sqrt{2}} \\
\sin\left(\frac{3 \pi}{2}\right) &= - \sin\left(\frac{\pi}{2}\right) = -1
\end{align}
then
\begin{align}
\cos(3 x) \cos(4 x) \cos(5 x) &= \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/687630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Looking for the recurrence relation for certain trigonometric integrals By assuming that:
$$ \int_{\pi/4}^{\pi/2} \frac{\cos^4(x)}{\sin^5(x)}\,dx = k,$$
what does the integral $$ \int_{\pi/4}^{\pi/2} \frac{\cos^6(x)}{\sin^7(x)}\,dx$$ equal in terms of k?
I have manipulated both integrals to get $\int_{\pi/4}^{\pi/2} (\... | By setting
$$ I_n = \int_{\pi/4}^{\pi/2}\frac{\cos^{2n}x}{\sin^{2n+1}x}dx$$
we have
$$ I_n = \int_{0}^{1}\frac{u^{2n}}{\sqrt{1+u^2}}du.\tag{1}$$
This follows from the variable changes $x=\arcsin t, t=\sqrt{s}, s=1/r, r=u+1$, or just $x=\arcsin\left(\frac{1}{\sqrt{1+u}}\right)$, for short.
Since:
$$\frac{d}{du}\operat... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Solve polynomial congruence using Hensel's lemma Solve $x^4+2x+46 \equiv 0$ $(\mod 4375 )$ for x.
.
My attempt:
$x^4+2x+46 \equiv 0$ $(\mod 5^47 )$
breaks down to
a Chinese Remainder Problem with the 2 following congruence's':
(1) $x^4+2x+46 \equiv 0$ $(\mod 7 )$ which by inspection is x $\equiv 2,6 (\mod 7)$
and
(2)... | The Hensel lemma states that if there exist integer $x$ such that:
$$f(x) \equiv 0 \pmod {p^k} \quad \quad \text{and} \quad \quad f'(x) \equiv 0 \pmod p$$
then there exist integer $s$ such that:
$$f(s) \equiv 0 \pmod {p^{k+m}} \quad \quad \text{and} \quad \quad s \equiv r \pmod {p^k}$$
Now it's obvious that $s = r + tp... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the maximum or minimum value of the quadratic function. Find the maximum or minimum value of the quadratic function by completing the squares. Also, state the value of $x$ at which the function is maximum or minimum.
$y=2x^2-4x+7$
$x^2$ has a coefficient of $2$, how should I complete the squares?
| $$y=2x^2-4x+7=2\left(x^2-2\cdot x\cdot1+1^2\right)+7-2\cdot1=2(x-1)^2+5$$
or $$2y=4x^2-8x+14=(2x)^2-2\cdot2x\cdot2+2^2+14-4=(2x-2)^2+10$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/690157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Making sure I'm not doing something wrong I'm solving a matrix equation: $$2(A-B+X) = 3(X-A)$$
Where $A = \left( \begin{matrix} 1&2 \\ 3&4 \end{matrix} \right)$ and $B = \left( \begin{matrix} -1&0 \\ 1&1 \end{matrix} \right) $
So far I've done: $$2(A-B+x) = 3(X - A)$$ $$\Rightarrow 2 \left( \begin{matrix} 1&2 \\ 3&4 \e... | It might be more clear if you don't write out the elements until necessary.
$$2(A-B+X) = 3(X-A) \\
2A-2B+2X = 3X-3A \\
5A-2B = X$$
Matrices add and subtract normally; that is, $A+A = 2A, 10Q-8Q = 2Q, 3X-2X = X,$ etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/690411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int_0^{\pi/2}\log\cos(x)\,\mathrm{d}x$ How can you evaluate $$\int\limits_0^{\pi/2}\log\cos(x)\,\mathrm{d}x\;?$$
| I first treat the integral as a derivative of a beta function
$$
I(a)=\int_0^{\frac{\pi}{2}} \cos ^a x d x=\frac{1}{2} B\left(\frac{a+1}{2}, \frac{1}{2}\right)
$$
$$
\begin{aligned}
I&\left.=\frac{\partial}{\partial a} I(a)\right|_{a=0} \left.=\left.\frac{1}{4} B\left(\frac{a+1}{2}, \frac{1}{2}\right)\left(\psi\left(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/690644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 9,
"answer_id": 8
} |
Proof by mathematical induction - Fibonacci numbers and matrices Using mathematical induction I am to prove:
$ \left( \begin{array}{ccc}
1 & 1 \\
1 & 0 \end{array} \right)^n $ =
$ \left( \begin{array}{ccc}
F_{n+1} & F_n \\
F_n & F_{n-1} \end{array} \right) $
where $F_k$ represents the $k^{th}$ Fibonacci number.
my ... | To prove it for $n=1$ you just need to verify that
$ \left( \begin{array}{ccc}
1 & 1 \\
1 & 0 \end{array} \right)^1 $ =
$ \left( \begin{array}{ccc}
F_2 & F_1 \\
F_1 & F_0 \end{array} \right) $
which is trivial.
After you established the base case, you only need to show that assuming it holds for $n$ it also holds f... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Solving a curve integral around part of an elipse I'm having trouble calculating a curve integral in a vector field:
$\int_C y (18x + 1)\ dx + 2y^2\ dy$
where $C$ is the curve along the ellipse
$9x^2 + y^2 = 64$
going counterclockwise from the point ( $-\frac{4}{3}\sqrt{3} $ , $4$) to the point (-$\frac{4}{3}$ , $... | As @dfan suggested, substitute $u=3x$ leading to a circular curve $C'$ defined by $$u^2+y^2=r^2$$ of radius $r=8$. Your integral becomes
$$J:=\int_C dx\,y(18x+1)+\int_C dy\,2y^2=\frac{1}{3}\int_{C'}du\,y(6u+1)+2\int_{C'}dy\,y^2.$$ Taking the total differential of the circle equation, we have $udu=-ydy$, so the second i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/694115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Infinite series for partial sums of square roots. Can you prove these infinite series for partial sums of square roots?
$$\sqrt{1}=\sum\limits_{n=1}^{\infty } \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)$$
$$\sqrt{1}+\sqrt{2}=\sum\limits_{n=1}^{\infty}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}-\frac{2}{\sqrt... | Let's see how we find the first sum which's known as telescoping sum and the other sums are almost the same: the idea is to change the index and then cancel most of the terms
$$\sum\limits_{n=1}^{\infty } \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)=\lim_{N\to\infty}\sum\limits_{n=1}^{N} \left(\frac{1}{\sqrt{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/695073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Do not understand formula... How does;
$x^{n} -y^{n}=(x-y)(x^{n-1} + x^{n-2}y+...+x y^{n-2}+ y^{n-1} )$
work on $x^{2} - y^{2}$
When I attempt to apply the formula on $x^{2} - y^{2}$
I get the following
$x^{2} - y^{2} =(x-y)( x^{1} + x^{0}y+...+x y^{0} + y^{1} )$
$x^{2} - y^{2} =(x-y)(2x+2y)$
which is obviously false... | $$x^2 - y^2 = (x -y)(x^1y^0 + x^0y^1) =(x-y)(x+y)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/695164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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How to evaluate $\int_0^\infty\operatorname{erfc}^n x\ \mathrm dx$? Let $\operatorname{erfc}x$ be the complementary error function.
I successfully evaluated these integrals:
$$\int_0^\infty\operatorname{erfc}x\ \mathrm dx=\frac1{\sqrt\pi}\tag1$$
$$\int_0^\infty\operatorname{erfc}^2x\ \mathrm dx=\frac{2-\sqrt2}{\sqrt\pi... | Integrating by parts twice, we get
$$ \begin{align} \int_{0}^{\infty} \text{erfc}^{3}(x) \, dx &= x \, \text{erfc}^{3}(x) \Bigg|^{\infty}_{0} + \frac{6} {\sqrt{\pi}} \int_{0}^{\infty}x \, \text{erfc}^{2}(x) e^{-x^{2}} \, dx \\ &= \frac{6} {\sqrt{\pi}} \int_{0}^{\infty}x \, \text{erfc}^{2}(x) e^{-x^{2}} \, dx \\ &= - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/696630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "53",
"answer_count": 2,
"answer_id": 0
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Determine the value of the integral $I=\int_{0}^{\infty} \frac{\ln\left(a^2+x^2\right)}{b^2+x^2}dx$ Determine the value of the integral $$I(a)=\int_{0}^{\infty} \frac{\ln\left(a^2+x^2\right)}{b^2+x^2}dx$$
My try:
$\to I'(a)=\int_{0}^{\infty}\frac{2a}{(a^2+x^2)(b^2+x^2)}dx=\frac{\pi}{b(a+b)}$
Hence $I(a)=\frac{\pi}{b}\l... | Firstly, we introduce a lemma:
\begin{equation}
\int_0^{\frac{\pi}{2}}\ln(\sin(x))dx=\int_0^{\frac{\pi}{2}}\ln(\cos(x))dx\\
\mbox{Then}\\
\int_0^{\frac{\pi}{2}}\ln(\tan(x))dx=0
\end{equation}
The proof see here.
Set $a=0$ and use the substitution that $x=b\tan(u)$, we have:
\begin{equation}
I(0)=2\int_0^\infty\frac{\ln... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Prove that $n^2+n+41$ is prime for $n<40$ Here's a problem that showed up on an exam I took, I'm interested in seeing if there are other ways to approach it.
Let $n\in\{0,1,...,39\}$. Prove that $n^2+n+41$ is prime.
I shall provide my own solution, though I am curious does anyone know how to do this without using PI... | Below is problem $\rm 6\ (B3)$ from IMO $1987$ and its solution (from AoPS, or Scholes)
Theorem. If $k^2+k+n$ is prime for all integers $k$ such that $0\le k\le \sqrt\frac{n}{3}$, then $k^2+k+n$ is prime for all integers $0\le k\le n-2$.
Proof. First observe that if $m$ is relatively prime to $b+1$, $b+2$, $\cdots$, $... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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"answer_id": 0
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Can $\frac {100-100}{100-100}=2$? \begin{align*}
\frac{0}{0} &= \frac{100-100}{100-100} \\
&= \frac{10^2-10^2}{10(10-10)} \\
&= \frac{(10+10)(10-10)}{10(10-10)} \\
&= \frac{10+10}{10} \\
&= \frac{20}{10} \\
&= 2
\end{align*}
I know that $0/0$ isn't equal to $2$, the what is wrong in this proof? I wasn't satisfied after... | You do some thing like this
$$\frac{0\cdot 5}{0\cdot 9}=\frac{5}{9}$$you haven't to divide over zero
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/710551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 9,
"answer_id": 2
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Arithmetic Problem Prime numbers Let $p\equiv 2\mod 3$ an odd prime number.
Prove that:
$p \mid (x^3+y^3) \implies p \mid (x+y)$ , for any integers $x,y$
$p\mid (x+y)(x^2-xy+y^2) \implies p\mid (x+y)$ or $p\mid (x^2-xy+y^2) $
if $p\mid (x+y)$ so problem solved
if $p\mid (x^2-xy+y^2)$ I can't find a contradiction... | We give two proofs, one by a Legendre symbol calculation, the second much simpler.
First Proof: If $p$ divides one of $x$ or $y$, then it must divide the other, and swe are finished. So suppose $p$ divides neither $x$ nor $y$. We show that $p$ cannot divide $x^2-xy+y^2$.
Suppose to the contrary that it does. Then $p$ ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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On last digit of 4 consecutive primes less than 10 apart $\def\mod{\mathrm{\;mod\;}}\def\pq{\{p_1,p_2,p_3,p_4\}}$$\def\dq{\{d_1,d_2,d_3,d_4\}}$
This question concerns quartets of consecutive primes $p_1 < p_2 < p_3 < p_4$ such that $p_4 - p_1 < 10$, and the associated quartets $\dq$ consisting of the last decimal digit... | If you consider the additions to $x$ modulo $3$ within
$\begin{array}{llllllll}
\{& x+3, & x+7, & x+9, & x+11 & & &\} \\
\{& & x+7, & x+9, & x+11, & x+13 & &\} \\
\{& & & x+9, & x+11, & x+13, & x+17 &\}
\end{array}$
You get
$\begin{array}{llllllll}
\{& 0, & 1, & 0, & 2 & & &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/714321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
if $f'''(x)$ is continuous everywhere and $\lim_{x \to 0}(1+x+ \frac{f(x)}{x})^{1/x}=e^3$ Compute $f''(0)$
if $f'''(x)$ is continuous everywhere and $$\lim_{x \to 0}(1+x+ \frac{f(x)}{x})^{1/x}=e^3$$ Compute $f''(0)$
The limit equals to $$\begin{align} \lim_{x \to 0} \frac{\log(1+x+ \frac{f(x)}{x})}{x}-3=0. \end{alig... | I will use the following formula for $f''(0)$. We have
$$
f(x)=f(0)+f'(0)x+f''(0)\,\frac{x^2}2+o(x^3),
$$
$$
f(-x)=f(0)-f'(0)x+f''(0)\,\frac{x^2}2-o(x^3).
$$
Adding and solving for $f''(0)$, we get
$$\tag{1}
f''(0)=\frac{f(x)+f(-x)-2f(0)}{x^2}+o(x).
$$
Starting from your
$$
1+x+\frac{f(x)}{x} = e^{3x+o(x)},
$$
we hav... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/714564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Infinite Geometric Series and its Value Calculate the value of $\sum_1^\infty (\frac{1}{3})^{2n}$.
I can clearly see that after listing a few partial sums this seems to tend toward $\frac{1}{8}$ and each partial sum is greater than the last.
$\sum_1^2 (\frac{1}{3})^{2n} = \frac{10}{81}$
$\sum_1^3 (\frac{1}{3})^{2n} = ... | The formula for the sum of a geometric series is $\dfrac{a}{1-r}$ where $a$ is the first term in the series and $r$ is the ratio you're multiplying by.
Here, $a = (\dfrac{1}{3})^{2\times1} = \dfrac{1}{9}$, and $r = \dfrac{1}{9}$.
So, we have $\dfrac{\dfrac{1}{9}}{1-\frac{1}{9}} = \dfrac{1}{9-1} = \dfrac{1}{8}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\int \cos^{-1}\left(\frac{x^2+a^2}{x^2-b^2}\right)x^2 dx$ How to evaluate the integral
$$\int \cos^{-1}\left(\frac{x^2+a^2}{x^2-b^2}\right)x^2 dx$$$a<b.$
I posted a similar question here.
Thanks in advance.
| Hint:
$\int\cos^{-1}\left(\dfrac{x^2+a^2}{x^2-b^2}\right)x^2~dx$
$=\int\cos^{-1}\left(\dfrac{x^2+a^2}{x^2-b^2}\right)d\left(\dfrac{x^3}{3}\right)$
$=\dfrac{x^3}{3}\cos^{-1}\left(\dfrac{x^2+a^2}{x^2-b^2}\right)-\int\dfrac{x^3}{3}d\left(\cos^{-1}\left(\dfrac{x^2+a^2}{x^2-b^2}\right)\right)$
$=\dfrac{x^3}{3}\cos^{-1}\left... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving a combintorical problem using generating function
In how many ways $n$ balls can be divided into groups where each group may contain only one ball or two.
Each ball can be a singleton, or coupled with one of the $n-1$ options. Hence,
$a_0 = 1$
$a_1 = 1$
$a_n = a_{n-1} + (n-1)a_{n-2}$
The correspond gener... | Though you did not state it in the question, from your recurrence I guess you are dealing with distinguishable (labelled) balls. Thus for instance, $\{\{1, 2\}, 3\}$ and $\{\{1, 3\}, 2\}$ are distinct groupings. So your recurrence $a_n = a_{n-1} + (n-1)a_{n-2}$ is correct: given $n$ balls labelled $1$ to $n$, you put t... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Limit involving logarithms I have to solve the limit,
$ \displaystyle \lim_{x \to 0} \frac{ \log ((k+x)^{b} - (k-x)^{b} )}{\log x}$
where $k \in (0,1)$ and $b \in (0,1)$ are constant . I have tried using Taylor expansion but it does not work. Thank you.
| You could use Newton's Binomial Theorem to say that
$$
\begin{align*}
(k+x)^b-(k-x)^b&=k^b\left[\left(1+\frac{x}{k}\right)^b-\left(1-\frac{x}{k}\right)^b\right]\\
&=k^b\sum_{n=0}^{\infty}\binom{b}{n}\left[\left(\frac{x}{k}\right)^n-\left(-\frac{x}{k}\right)^n\right]\\
&=k^b\sum_{m=0}^{\infty}\binom{b}{2m+1}2\left(\frac... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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finding bases for row space and null space of matrix. My problem is:
For the matrix
$$A = \begin{bmatrix}
1& 4& 5& 6& 9\\
3& −2& 1& 4& −1\\
−1& 0& −1& −2& −1\\
2& 3& 5& 7& 8\end{bmatrix}$$
(a) Find a basis for the row space of A.
(b) Find a basis for the null space of A.
(c) Find the ... | The row space is the span of the rows of $A$. (or the column space of $A^{T}$).
$$
A^{T} = \begin{bmatrix} 1 & 3 & -1 & 2\\
4 & -2 & 0 & 3 \\
5 & 1 & -1 & 5 \\
6 & 4 & -2 & 7\\
9 & -1 & -1 & 8 \end{bmatrix} \vec{x} = \begin{bmatrix} 1\\ 4 \\ 5 \\ 6 \\ 9\end{bmatrix}x_1 + \begin{bmatrix} 3\\ -2 \\ 1 \\ 4 \\ -1\end{bmat... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to solve system of two equation with two unknown and using substitution? $$\begin{align}
a^2 - b^2 = 3\\
a \cdot b = 2
\end{align}$$
In aforementioned equations, we can mentally find out the value of $a = 2, b = 1$. But what is the general way to solve this system algebraically?
I tried to use substitution but I g... | Your last equation, $b^2+3b^2-4=0$ is a quadratic in $b^2$. You can define $c=b^2$ and find $c^2+3c-4=0$. Maybe you can factor this, if not you can use the quadratic formula. Solve for $c$, then take the square root to get $b$. This works for quartics that only have the even power terms.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find complicated Taylor Series According to some software, the power series of the expression,
$$\frac{1}{2} \sqrt{-1+\sqrt{1+8 x}}$$
around $x=0$ is
$$\sqrt{x}-x^{3/2}+\mathcal{O}(x^{5/2}).$$
When I try to do it I find that I can't calculate Taylor because there are divisions by zero. Also I do not understand how Tayl... | Close to $x=0$,$$\frac{1}{2} \sqrt{-1+\sqrt{1+8 x}} \simeq \frac{1}{2} \sqrt{-1+ (1+4x)}\simeq \sqrt x$$ So, this must be the start of the development (in order that, locally, your expansion looks like the formula)
If you start using $$\sqrt{1+8 x} \simeq 1+4 x-8 x^2+32 x^3+O\left(x^4\right)$$ then $$\frac{1}{2} \sqrt... | {
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What's the easiest way to factor $5^{10} - 1$? What's the easiest way to factor $5^{10} - 1$?
I believe $5 - 1$ is a factor based off the binomial theorem.
From there I do not know.
We are using congruence's in this class.
| Note that
$$
5^{10} - 1 = (5^5+1)(5^5-1)
$$
And that
$$
5^5-1 = (5 - 1)(5^4 + 5^3 + 5^2 + 5 + 1)\\
5^5+1 = (5+1)(5^4 - 5^3 + 5^2 - 5 + 1)
$$
That should give you a fairly good start. We cannot factor this further using formal factorization alone.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the equation $(x-3)(x+9)(x+5)(x-7)=385$ I have one question please...I solved it in this way..so I am not sure..that is it right or not? and If I solved it in the wrong way..so would like to know about the correct way and method to solve it..so please :
Solve the equation (x-3) (x+9) (x+5) (x-7) = 385
and here i... | Let $y = x + 1$ (I got this by taking the average of the four). Then the problem becomes:
$$(y-4)(y+8)(y+4)(y-8) = 385$$
which can be simplified nicely (by multiplying out similar terms) into:
$$\begin{align}(y^2 - 16)(y^2 - 64) &= 385 \\&= 7\cdot55\\&=-7\cdot-55\end{align}$$
The difference between $7$ and $55$ is $48$... | {
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"timestamp": "2023-03-29T00:00:00",
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Last decimal digit of any perfect square must be $0,1,4,5,6$ or $9$ Last decimal digit of any perfect square must be $0,1,4,5,6$ or $9$
My Proof:
Ten cases exist, yielding the following equalities:
$$(1\mod{10})^2 = 1\mod{10}$$
$$(2\mod{10})^2 = 4\mod{10}$$
$$(3\mod{10})^2 = 9\mod{10}$$
$$(4\mod{10})^2 = 6\mod{10}$$
$$... | For a more "clever" solution (but yours looks perfect):
As $10=5\times 2$, let us consider
$$
x^2 = 10a + 5b+ r
\\ 0\le 5b+r< 10
\\ 0\le r< 5
$$
then reducing modulo 5:
$$
r = x^2 \in \{ 0,1,4
\}\mod 5\\
5b+r \in \{ 0,1,4,0+5=5,1+5=6,4+5=9
\}\mod 5\\
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$ \frac{x}{4 \ \sqrt{x^2+1}} \ = \ \frac{y}{5 \ \sqrt{y^2+1}} \ = \ \frac{z}{6 \ \sqrt{z^2+1}} $ and $ \ x+y+z \ = \ xyz \ $ Consider the system of equations in real numbers $ \ x,y,z \ $ satisfying
$$ \frac{x}{4 \ \sqrt{x^2+1}} \ = \ \frac{y}{5 \ \sqrt{y^2+1}} \ = \ \frac{z}{6 \ \sqrt{z^2+1}} $$
and $ \ x+y+z \ = \... | If you have $$15\sin(\theta_1)=12\sin(\theta_2)=10\sin(\theta_3)$$ as suggested by ruler501, you can extract $\theta_2$ and $\theta_3$ as a function of $\theta_1$. This leads to $$\theta_2=\sin ^{-1}\left(\frac{5 \sin (\theta_1)}{4}\right)$$ $$\theta_3=\sin ^{-1}\left(\frac{3 \sin (\theta_1)}{2}\right)$$ So, the equat... | {
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"source": "stackexchange",
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Find closed formula $f(n)$ from generating function I'm asked to find a closed formula for
$f(n)=6f(n-1)-9f(n-2)$ for $n>1$ with $f(0)=-1. f(1)=0$,
using the ordinary generating function $F(X)$.
I found $F(X)=-1/(1-3x)^2$ but from there I don't manage to get a satisfactory formula for $f(n)$. Can anyone give me a hint... | Write the recurrence as:
$$
f(n + 2) = 6 f(n + 1) - 9 f(n) \qquad f(0) = - 1, f(1) = 0
$$
Multiply by $z^n$ and sum over $n \ge 0$. Recognize:
\begin{align}
\sum_{n \ge 0} f(n + 1) z^n &= \frac{F(z) - f(0)}{z} \\
\sum_{n \ge 0} f(n + 2) z^n &= \frac{F(z) - f(0) - f(1) z}{z^2}
\end{align}
and get:
$$
\frac{F(z) + 1}{z^2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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About asymptotic behaviour of a divergent integral. I have the function $f(x) = x \tanh(\pi x) \log (x^2 +a^2)$ where $a$ is some positive real number. For the logarithm I am assuming a branch-cut along the positive imaginary axis starting at $x = ia$. So the branching effect on the "log" is that its imaginary part is ... | You're interested in the asymptotic behavior of
$$
\begin{align}
I_a(A) &= \int_0^A x \tanh(\pi x) \log(x^2+a^2)\,dx \\
&= \int_1^A x \tanh(\pi x) \log(x^2+a^2)\,dx + \int_0^1 x \tanh(\pi x) \log(x^2+a^2)\,dx \tag{1}
\end{align}
$$
as $A \to \infty$. Let's rewrite the integrand as
$$
\begin{align}
&x \tanh(\pi x) \log... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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modular arithmetic problem (when solving elliptic curves) Given E: (elliptical curve)
$y^2 = x_3+2x+2 \bmod 17$
Recall: $y^2 = x^3+ax+b$
point $P=(5,1)$
Compute:
$2P = P+P = (5,1)+(5,1)= (x_3,y_3)$
Now the formula used here is
slope $m = \dfrac{3x_1^2 +a}{2y_1}$
and what they got is $s = (2 · 1)^{−1} (3 · 5^2 + 2) = ... | $$2^{-1}=9\pmod{17}\;\;,\;\;3\cdot 5^2=3\cdot 8=7\pmod{17}\implies$$
$$(2\cdot 1)^{-1}(3\cdot 5^2+2)=9(7+2)=9^2=13\pmod{17}$$
BTW, this is exactly what you got, since
$$38.5=4.5=\frac92=9\cdot2^{-1}=9\cdot9=13\pmod{13}\ldots$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve $x^4-8x^3+24x^2-32x+16=0$ How can we solve this equation?
$x^4-8x^3+24x^2-32x+16=0.$
| As $x\ne0,$ dividing either sides by $x^2$
$$x^2+\left(\frac4x\right)^2-8\left(x+\frac4x\right)+24=0$$
Now as $\displaystyle x^2+\left(\frac4x\right)^2=\left(x+\frac4x\right)^2-2\cdot x\cdot\frac4x$
Setting $x+\dfrac4x=y,$ we get $\displaystyle y^2-8-8y+24=0\implies(y-4)^2=0\iff y=4$
So, we have $\displaystyle x+\frac... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Gaussian Quadrature -Deriving a Formula- The following is an exercise in the problem section of the Gaussian Quadrature chapter.
The theorem:
Derive a formula of the form $$\int_{a}^{b} f(x)dx \approx w_0f(x_0) + w_1f(x_1) + w_2f'(x_2) + w_3f'(x_3)$$
I don't really know how to even start this. I have the Gaussian Qua... | Let's look first at $\int_{-1}^{1} f(t) \ dt$.
We want the formula to evaluate $\int_{-1}^{1} dt, \int_{-1}^{1} t \ dt, \int_{-1}^{1} t^{2} \ dt, \int_{-1}^{1} t^{3} \ dt, \int_{-1}^{1} t^{4} \ dt, \int_{-1}^{1} t^{5} \ dt, \int_{-1}^{1} t^{6} \ dt $ and $\int_{-1}^{1} t^{7} \ dt$ exactly.
That leads to the following s... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof by induction that $ 169 \mid 3^{3n+3}-26n-27$ $ 169$ | $3^{3n+3}-26n-27$ ?
Fulfilled for $n=0$. Induction to $n+1$:
An integer $x$ exists so that
$ 169x= 3^{3n+6}-26n-27-26$
$ 169x= 27*3^{3n+3}-26n-27-26$
$ 169x= 26*3^{3n+3}+3^{3n+3}-26n-27-26$
An integer $m$ exists so that
$ 169x= 26*3^{3n+3}+169m-26$
($ 13x= 2... | The same, but using modules.
Let's prove by induction that $3^{3n+3} - 26n - 27 \equiv 0 \pmod {169}$
Working with module $169$:
*
*Base case. If $n = 1$, then $3^3 - 27 \equiv 0$.
*Induction. Let fix an integer $m \geq 0$ and supose that (induction hypothesis):
$$3^{3m+3}-26m -27 \equiv 0 \pmod{169}$$
or, with oth... | {
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"timestamp": "2023-03-29T00:00:00",
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Fibonacci Proof Using Induction
$$f(n) = \left\{\begin{matrix}
0 & n=1\\
1 & n=2\\
f_{n-1} + f_{n-2} &
n\geqslant 2\end{matrix}\right.$$
How can I prove by induction that $$f_{n} \leq \left ( \frac{1+\sqrt{5}}{2} \right )^{n-1}$$ for all$$ n\geq l_{a}$$, I have to find the smallest value for $$l_{a}$$
| $(1)$ $f_1 \leq (\frac{1+\sqrt{5}}{2})^{1-1} \Rightarrow 0 \leq 1,f_2 = 1 \leq \frac{1+ \sqrt{5}}{2} \approx 1.618 $. $\textbf{True}$
$(2)$ Next: Suppose for some $\textbf{k}$ $\geq n$ we have $f_k \leq (\frac{1+\sqrt{5}}{2})^{k-1}$
$(3)$ We want to show: $f_{k+1} \leq (\frac{1+\sqrt{5}}{2})^{k}$
$(4)$ Well, $f_{k+1} =... | {
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Number of subsets without consecutive numbers Consider $S=\{1,2,\ldots,15\}$. Let $X$ denote the number of subsets of $S$ of four elements which contain no consecutive numbers.
The claim is that $X$ equals the coefficient of $x^{14}$ in $\dfrac{x^6}{(1-x)^5}$. How would I prove this?
| Call the elements of your subset $a_1$, $a_2$, $a_2$, $a_4$ in order. Your restrictions translate into:
\begin{align}
x_1 &= a_1 \ge 1 \\
x_2 &= a_2 - a_1 \ge 2 \\
x_3 &= a_3 - a_2 \ge 2 \\
x_4 &= a_4 - a_3 \ge 2 \\
x_5 &= 15 - a_4 \ge 0
\end{align}
The definitons amount to $x_1 + x_2 + x_3 + x_4 + x_5 = 15$. Set up g... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Need help to solve taylor series of $e^{\sin x}$ How to derive the taylor series of $e^{\sin x}$, up to $x^5$?
i just don't know how to get the answer
$$f(x) = 1 + x + \frac{x^2}{2} - \frac{x^4}{8} -\frac{x^5}{15}$$
really need some help. Thanks
| If
$$
f(x) = a_1x + \frac{a_2}{2}x^2 + \frac{a_3}{6} x^3 + \frac{a_4}{24} x^4 + \cdots + \frac{a_n}{n!} x^n + \cdots
$$
then
$$
e^{f(x)} = 1 + a_1 x + \frac{a_2+a_1^2}{2!} + \frac{a_3 + 3a_2a_1+a_1^3}{6} x^3 + \frac{a_4+4a_3a_1 + 3a_2^2 + 6a_2a_1 + a_1^4}{24} x^4 + \cdots
$$
The pattern is this: Consider the last case... | {
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"timestamp": "2023-03-29T00:00:00",
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Taylor expansion of $\frac{1}{2-z-z^2}$ The problem is:
Find the Taylor expansion of $f(z):= \dfrac{1}{2-z-z^2}$ on the disc $|z| < 1$
So far I have used partial fractions to obtain $f(z) = \dfrac{1}{3}\left(\dfrac{1}{1-z} + \dfrac{1}{2+z}\right)$ which I then rewrite as $f(z) = \dfrac{1}{3}\left(\dfrac{1}{1-z} + \dfra... | $\frac{1}{2-z-z^2}=\frac{ 1 }{ 2} \frac{1}{1-\frac{1}{2}z-\frac{1}{2}z^2}=\frac {1}{2}\sum _{ n = 0 } ^ \infty(\frac { 1 } { 2 } (z+z^2))^{n}$.
Done.
Note that Since $|z|\lt 1 $, it follows that $|\frac {1}{2}(z + z ^ 2 )|\lt \frac {1}{2}(|z|+|z ^ 2|)\lt1$ so it is justified to use geometric series above.
| {
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Compute summation with a relative error of O(n^-2) $a(n) = \sum_{i \geq 0} a_i n^{-i}$, how can we compute the value of $a(n)^n$ with a relative error of $O(n^{-2})$?
| Since
$$
a(n) = a_0 + a_1 n^{-1} + a_2 n^{-2} + O(n^{-3})
$$
we have
$$
\begin{align}
\log a(n) &= \log a_0 + \log\left(1 + \frac{a_1}{a_0} n^{-1} + \frac{a_2}{a_0} n^{-2} + O(n^{-3})\right) \\
&= \log a_0 + \frac{a_1}{a_0} n^{-1} + \frac{a_2}{a_0} n^{-2} + O(n^{-3}) \\
&\qquad - \frac{1}{2} \left(\frac{a_1}{a_0} n^{-1... | {
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I'm stuck. Can somebody help me solve this problem. The sum of our present ages is $41$. If I live $17$ more years, after doubling my present age, I will be $9$ years less than your present age. What are our present age?
| $y=\text{Your age, and }x=\text{Mine}$.
$$x+y=41\\
\text{And also }17+2x=y-9\\
\implies 2x-y=-26$$
Let me do this a bit differently - probably not meant for precalculus, but here it goes anyway.
$$\begin{pmatrix}1&1\\2&-1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}41\\-26\end{pmatrix}$$
Then,
$$\begin{... | {
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How to evaluate $\int\frac{x}{\sqrt{x^2+x+1}} \, dx$ using trigonometric substitution? I am pretty sure that my answer is correct but given answer for the exercise from textbook Calculus James Steward was slightly different. Any idea to solve this:
$$\int\frac{x}{\sqrt{x^2+x+1}} \, dx$$
The given answers: $\sqrt{x^2+x+... | First split off the bit that does not need a trigonometric substitution:
$$\int\frac{x\,dx}{\sqrt{x^2+x+1}}=\int\frac{x+\frac{1}{2}}{\sqrt{x^2+x+1}}dx
-\frac{1}{2}\int\frac{dx}{\sqrt{x^2+x+1}}\ .$$
You should see that the first integral is now easy. For the second, complete the square:
$$\int\frac{dx}{\sqrt{x^2+x+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/739042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proof polynomial has only one real root. I need to prove that this polynomial equation:
$$x^5-(3-a)x^4+(3-2a)x^3-ax^2+2ax-a=0\quad\text{ for }\quad a\in(0,\frac{1}{2}).$$
has only one root. That it has one real root is obvious because it is of odd degree. But Descartes rules here fails to bound the number of roots to o... | I try to factorize your formula :
$x^5-(3-a)x^4+(3-2a)x^3-ax^2+2ax-a=0\quad\text{ for }\quad a\in(0,\frac{1}{2})$
$x^5-(3-a)x^4+(3-2a)x^3-ax^2+2ax-a=x^5+(3-a)(x^3-x^4)-ax^{3}-ax^{2}+2ax-a$
$=x^5+(1-a)(x^3-x^4)-ax^{3}-ax^{2}+2ax-a+2x^{3}-2x^{4}$
$=x^5-1+1+(1-a)(x^3-x^4)-a(x^{3}-x^{2})+2ax+2ax^{2}-a+2x^{3}-2x^{4}$
$=x^5-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/739341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Invalid subtraction when solving system of equations? I'm trying to solve these two equations:
$$\begin{cases}
1-4x(x^2+y^2)=0 \\
1-4y(x^2+y^2)=0
\end{cases}$$
and I tried to do it by subtracting the first equation from the second, yielding $(4x-4y)(x^2+y^2)=0$. Clearly this is satisfied when $x=y$, which gives $(x,y)=... |
$$\begin{cases}
1-4x(x^2+y^2)=0 \\
1-4y(x^2+y^2)=0
\end{cases}$$
Let $S$ the set of solutions.
You can summarise the logic steps:
*
*If $(x,y)$ is a solution:
$$4x(x^2+y^2)=4y(x^2+y^2)
\implies x^2+y^2=0 \text{ or }x=y
\\
\implies (x,y)=(0,0) \text{ or }
\\
x=y=\frac{1}{4(x^2+y^2)}
=\frac{1}{8x^2}\implies x=y=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/739693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Limit of: $\lim_{x\to0}\frac{1-\sqrt[3]{\cos{x}}}{1-\cos{\sqrt[3]{x}}}$ I calculated this limit with L'Hospital's rule, but at the end it got rather complicated.
$\lim_{x\to0}\frac{1-\sqrt[3]{\cos{x}}}{1-\cos{\sqrt[3]{x}}}$
Is there some other more effective way for this limit?
| Taylor expansions yield the result without much difficulty. We have
$$\sqrt[3]{1+y} = 1 + \frac{y}{3} + O(y^2),$$
and
$$\cos z = 1 - \frac{z^2}{2} + O(z^4),$$
hence
$$\sqrt[3]{\cos x} = \sqrt[3]{1 - x^2/2 + O(x^4)} = 1- \frac{x^2}{6} + O(x^4),$$
so
$$\frac{1-\sqrt[3]{\cos x}}{1-\cos \sqrt[3]{x}} = \frac{\frac{x^2}{6} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/739919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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if $abc=1$, then $a^2+b^2+c^2\ge a+b+c$ This is supposed to be an application of AM-GM inequality.
if $abc=1$, then the following holds true: $a^2+b^2+c^2\ge a+b+c$
First of all,
$a^2+b^2+c^2\ge 3$
by a direct application of AM-GM.Also,we have
$a^2+b^2+c^2\ge ab+bc+ca$
Next,we consider the expression
$(a+1)(b+1)(c+1)... | Alternatively, $a^2+a^2+a^2+a^2+b^2+c^2 \geq 6 \sqrt[6]{a^8b^2c^2} = 6 \sqrt[6]{a^6} = 6|a| \geq 6a$ by AM-GM. Adding the analogous inequalities $a^2+4b^2+c^2 \geq 6b$ and $a^2+b^2+4c^2 \geq 6c$ gives the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/740518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Extra help on inequality Someone very helpfully provided an answer to an inequality. See Hard Olympiad Inequality However I don't get part of their answer. How did they get the last factorization??? Thanks so much for any help.
Would Wolfram Alpha help? I plugged it in it is still loading
| Below is a rather long proof, but every step is straightforward. For any function $\phi$ in three variables $a,b,c$, let us denote
$$
\begin{array}{lcl}
\Sigma_{cyc}\phi&=&\phi(a,b,c)+\phi(b,c,a)+\phi(c,a,b) \\
\Sigma_{all}\phi&=&\phi(a,b,c)+\phi(a,c,b)+\phi(b,a,c)+\phi(b,c,a)+\phi(c,a,b)+\phi(c,b,a)
\end{array}\tag{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/740832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
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