Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How to evaluate this definite integral? Here is the integral we have to evaluate:
$$\int_0^4\sqrt{x^2+4}\,dx+\int_2^{2\sqrt{5}}\sqrt{x^2-4}\,dx$$
After observing, i realise that i can't evaluate these integrals from area of circle, I say that $u=\sqrt{x^2+4}$. Then i can say $dx=\frac{udu}{\sqrt{u^2-4}}$. The first t... | Let $f(x)=\sqrt{x^2+4}$ and $f^{-1}(x)=\sqrt{x^2-4}$. Note that
$$f(4)=2\sqrt{5} \\
f(0)=2
$$
We know that the integral of an inverse function can be calculated with
$$\int_{f(k)}^{f(l)}f^{-1}(x)\,dx=lf(l)-kf(k)-\int_k^lf(x)\,dx$$
So the integral
$$\int_0^4\sqrt{x^2+4}\,dx+\int_2^{2\sqrt{5}}\sqrt{x^2-4}\,dx$$
can be ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2215942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
System of 5x5 non-linear equations from Lagrange multiplier problem I have a Lagrange multiplier problem, and I've got down to the system of equations, but solving these has proved to be a nightmare for me.
I have five equations:
$$x+y+z=0$$
$$x^2+y^2+z^2=24$$
$$y=\lambda_1+2x\lambda_2$$
$$x=\lambda_1+2y\lambda_2$$
$$2... | One method is to eliminate the constraints. We have $x+y+z=0$ so $z=-x-y$. Also $x^2+y^2+z^2=1$ so
\begin{eqnarray*}
xy+yz+zx&=&-12. \\
f =xy+2z &=& -12+z(2-x-y)=12+z(2+z)=(z+1)^2-13
\end{eqnarray*}
So the extreme point occurs at $z=-1$. Now lets calulate $x$ & $y$
\begin{eqnarray*}
x+y&=&1 \\
xy &=& -11 \\
(x-y)^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2216573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Abstract Algebra; Galois field extension and automorphisms I am really curious...
Can $\sqrt{4+\sqrt{7}}$ be written in the form $\sqrt a + \sqrt b $ with rational numbers $a$ and $b$ ?
I was thinking that we could try to brute force it by equating $\sqrt{4+\sqrt{7}} = x$ and then try to work out the $a$ and $b$ ? Any ... | Wikipedia describes a simple method to write
$$
\sqrt{a+b \sqrt{c}\ } = \sqrt{d}+\sqrt{e}
$$
with
$$
d=\frac{a + \sqrt {a^2-b^2c}}{2},
\qquad
e=\frac{a - \sqrt {a^2-b^2c}}{2}
$$
This works iff $a^2 - b^2c$ is a square.
For $\sqrt{4+\sqrt{7}}$ we have $a^2 - b^2c=9$ and so
$$
\sqrt{4+\sqrt{7}} = \sqrt{\frac72}+\sqrt{\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2216795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Does the sum of digits squared of a number when we keep on doing it lead to a single digit number? When we add the digits of number for eg $2478125$
$$2+4+7+8+1+2+5=29;\\
2+9=11;\\
1+1=2$$
and when we square the digits and add them following this rule
$$2^2+4^2+7^2+8^2+1^2+2^2+5^2=163;\\
1^2+6^2+3^2=46; \\
4^2+6^2=52;\... | The answer is yes, it's the subject of this year's french olympiads : for one or two digits numbers, you prove by hand (or computer) verification that the resulting sequence is either stationnary at $1$, or loops indefinitely from $4$ to $16$, $37$, $58$, $89$, $145$, $42$, $20$, $4$ and so on.
Then you prove it for 3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2217832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Expand binomially to prove trigonometric identity Prompt: By expanding $\left(z+\frac{1}{z}\right)^4$ show that $\cos^4\theta = \frac{1}{8}(\cos4\theta + 4\cos2\theta + 3).$
I did the expansion using binomial equation as follows
$$\begin{align*}
\left(z+\frac{1}{z}\right)^4 &= z^4 + \binom{4}{1}z^3.\frac{1}{z} + \bino... | HINT:
Set $z=e^{i\theta}$ and use Euler's formula.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2217925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Compute Flux (Divergence Theorem) Given that $\textbf{F} = \langle x\cos^2z, y\sin^2z, \sqrt{x^2+y^2}\:z \rangle $ and let E be the solid cone above the $xy$-plane and inside z = $1 -\sqrt{x^2+y^2}.$
I'm trying to use the divergence theorem to compute the flux.
$$\iint_{D} \textbf{F} \cdot \textbf{N} \: dS = \iiint_... | Since $\cos^2z + \sin^2z = 1$, you must have $\nabla \cdot \mathbf{F} = 1+r$. In this case, your integral becomes
$$
\iiint_E\nabla \cdot \mathbf{F} \, \mathrm{d}V = \int_0^{2\pi} \int_0^1 \int_0^{1-r} (1+r)r\, \mathrm{d}z\, \mathrm{d}r\, \mathrm{d}\theta = \frac{\pi}{2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2219432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve $a_n=7a_{n-1}-12a_{n-2}+3^n$ $a_0=1\:a_1=2$
Using generating functions I get:
$f\left(x\right)-1-2x=7xf\left(x\right)-7x-12x^2f\left(x\right)+\frac{1}{1+3x}-1-3x$
$f\left(x\right)=\frac{3}{14\left(1+3x\right)}-\frac{13}{2\left(3x-1\right)}\:+\:\frac{40}{7\left(4x-1\right)}$
Now how can I express $a_n$ from this?
| Hint. One may just recall that
$$
\frac{1}{1-u}=\sum_{n=0}^\infty u^n,\qquad |u|<1,
$$ then applying it to
$$
f\left(x\right)=\frac{3}{14\left(1+3x\right)}+\frac{13}{2\left(1-3x\right)}\:-\:\frac{40}{7\left(1-4x\right)}
$$ respectively with $u$ equals to $-3x$, $3x$ and $4x$.
Can you finish it?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2222800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $a$ , $b$, $c$ are positive integers satisfying the following $(a^2 +2)(b^2+3)(c^2+4)=2014$ What is the value of $ a+b+c $ If $a$ , $b$, $c$ are positive integers satisfying the following
$(a^2 +2)(b^2+3)(c^2+4)=2014$
What is the value of $ a+b+c $
I need details because i don't know how to solve simlar probl... | $2014 = 2*19*53$
So each of the $a^2 + 2, b^2+3, c^2 + 4$ are some combination of $2, 19, 53$.
As none of them can equal 1, the must be that one of them equals $2$, another $19$ and the third $53$. What possible numbers work? (Note: just by looking only one can be small enough to equal $2$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2225235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Show the value of determinant is 0 I dont have any idea how to show the value of the following determinant is 0 without expanding the determinants
$$
\begin{vmatrix}
1 & a & a^2-bc \\
1 & b & b^2-ca \\
1 & c & c^2-ab \\
\end{vmatrix}
$$
Any ideas?
| $\Delta=\begin{vmatrix}
1 & a & a^2-bc \\
1 & b & b^2-ca \\
1 & c & c^2-ab \\
\end{vmatrix}=\begin{vmatrix}
1 & a & a^2 \\
1 & b & b^2 \\
1 & c & c^2\\
\end{vmatrix}-\begin{vmatrix}
1 & a & bc \\
1 & b & ca \\
1 & c & ab \\
\end{vmatrix}=\Delta_1-\Delta_2$, say.
Now, $\Delta_2=\begin{vmatrix}
1 & a & bc \\
1 & b & ca \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2225907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Compute $\prod_{k=5}^{\infty}\frac{k^4-17k^2+16}{k^4-8k^2+16}$ Compute the product: $$\prod_{k=5}^{\infty}\frac{k^4-17k^2+16}{k^4-8k^2+16}$$
I was able to factor in the following manner:
$$ \frac{k^4-17k^2+16}{k^4-8k^2+16}=\frac{(k-1)(k+1)(k-4)(k+4)}{(k-2)^2(k+2)^2}$$
but what should I do now?
| Take the partial products up to $N$ and use the fact that $$\prod_{k=5}^N \frac{k-1}{k-2} = \frac{4}{3} \cdot \frac{5}{4} \cdot ... \cdot \frac{N-1}{N-2} = \frac{N-1}{3},$$ and similarly $$\prod_{k=5}^N \frac{k+1}{k+2} = \frac{6}{N+2}, \quad \prod_{k=5}^N \frac{k-4}{k-2} = \frac{2}{(N-2)(N-3)}, \quad \prod_{k=5}^N \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2226363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
use polar coordiantes and the Poincare-Bendixson Theorem. I am searching for some help with this question in review for a upcoming exam, I have been having some trouble understanding the workings that have been provided and am just looking for some clarification.
The question says to show that,
$$x'=x+y-x(x^2+2y^2)$$
$... | We calculate
$$r'=\dfrac{xx'+yy'}{r} = \dfrac{1}{2} r^3 \cos 2 \theta-\dfrac{3}{2}r^3 + r = -r\left(\left(\dfrac{3}{2} - \dfrac{1}{2} \cos 2 \theta \right)r^2 - 1 \right) = -r(F(\theta)r^2-1)$$
where
$$F(\theta) = \left(\dfrac{3}{2} - \dfrac{1}{2} \cos 2 \theta\right) = 1 + \sin^2 \theta$$
Update
From $\cos 2 \theta = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2229907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving Recurrence Relation $a_n=6a_{n-1} - 9a_{n-2}$ for $n\geqslant2$ So the problem is to solve this recurrence relation with the initial conditions $a_0 = 2, a_1 = 21$.
$a_n=6a_{n-1} - 9a_{n-2}$ for $n\geqslant2$
And also find the value of $a_{2016}.$
Here is my solution but I'm not entirely sure if it's correct. W... | Generatingfunctionologically, define $A(z) = \sum_{n \ge 0} a_n z^n$, shift the recurrence by 2, multiply by $z^n$, sum over $n \ge 0$ and recognize resulting sums:
$\begin{align*}
\sum_{n \ge 0} a_{n + 2} z^n
&= 6 \sum_{n \ge 0} a_{n + 1} z^n - 9 \sum_{n \ge 0} a_n z^n \\
\frac{A(z) - a_0 - a_1 z}{z^2}
&= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2231362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the minimal polynomial of $\zeta_9+\zeta^{-1}_9$ over $\Bbb Q$ Knowing the degree of $\Bbb Q(\zeta_9+\zeta^{-1}_9)$ over $\Bbb Q$ is 3, now I want to find the minimal polynomial of $\zeta_9+\zeta^{-1}_9$ over $\Bbb Q$. I tried to use the relation $\zeta_9$ is the root of $x^6＋x^3＋1$ and $x^8＋x^7+x^6+x^5+x^4+x^3+x^... | $\zeta_9$ is a root of $p(x)=\frac{x^9-1}{x^3-1}=x^6+x^3+1$ and
$$ \frac{p(x)}{x^3} = 1+x^{3}+x^{-3} = 1+\left(x+\frac{1}{x}\right)^3- 3\left(x+\frac{1}{x}\right) $$
hence $\zeta_9+\zeta_9^{-1}=2\cos\frac{2\pi}{9}$ is a root of $\color{red}{z^3-3z+1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2232285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the value of $(\log_{a}b + 1)(\log_{b}c + 1)(\log_{c}a+1)$ if $\log_{b}a+\log_{c}b+\log_{a}c=13$ and $\log_{a}b+\log_{b}c+\log_{c}a=8$ Let $a,b$, and $c$ be positive real numbers such that
$$\log_{a}b + \log_{b}c + \log_{c}a = 8$$
and
$$\log_{b}a + \log_{c}b + \log_{a}c = 13.$$
What is the value of
$$(\log_{a}b... | As lab bhattacharjee commented, convert to natural logarithms and get
$$\log_{a}b + \log_{b}c + \log_{c}a = 8\implies \frac{\log (b)}{\log (a)}+\frac{\log (a)}{\log (c)}+\frac{\log (c)}{\log (b)}=8$$ $$\log_{b}a + \log_{c}b + \log_{a}c = 13\implies \frac{\log (a)}{\log (b)}+\frac{\log (c)}{\log (a)}+\frac{\log (b)}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2236306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 2
} |
Finding the exact value of a radical How do I show that $\sqrt{97 +56\sqrt3}$ reduces to $7 +4\sqrt3?$. Without knowing intitially that it reduces to that value.
| If you suspect that $\sqrt{97+56\sqrt3}$ (or similar) equals something
like $a+\sqrt b$ with $a$, $b$ rational you must have
$$97+56\sqrt 3=a^2+b+2a\sqrt b.$$
To match up the surds you'll have to have $a^2+b=97$
and $2a\sqrt b=56\sqrt3$. Thus $a^2b=28^2\times 3=2352$. Therefore
$a^2$ and $b$ are the solutions of the qu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2238274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Triangles of the form $a^n+b^n=c^n$
Let $\Delta$ be a triangle with (real) sides length $a,b,c>0$. From this question, we know that $$a+b=c \qquad \iff \qquad \Delta \text{ is a straight line}$$ Now, we know from the Pythagorean theorem that
$$ a^2+b^2=c^2 \qquad \iff \qquad \Delta \text{ is a right triangle}$$
I... | If $x = \frac{a}{c}$, $y = \frac{b}{c}$ (note: both are in $(0, 1)$)
then for $n \gt 2$,
$x + y \gt x^2 + y^2 > x^n + y^n = 1$
Thus $a + b \gt c$ and $a^2 + b^2 \gt c^2$ and so $a,b,c$ form the sides of a triangle and the triangle must be acute.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2239150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 0
} |
How to prove that $|\int_{x}^{x+1} \sin(t^2)dt| \leq \frac{1}{x}$? I have $F(x) = \int_{x}^{x+1} \sin(t^2)dt$, how can i proof that $|F(x)| \leq \frac{1}{x}$ for all $x > 0 $? I have no idea how to do it.
| Observe that
$$
\left|\int_x^{x+1}\sin(t^2)dt\right|=\left|\int_x^{x+1}\frac{t}{t}\sin(t^2)dt\right|.
$$
Using integration by parts, we have
\begin{align*}
u&=\frac{1}{t}&dv&=t\sin(t^2)dt\\
du=&-\frac{1}{t^2}&v&=-\frac{1}{2}\cos(t^2).
\end{align*}
Therefore,
\begin{align*}
\left|\int_x^{x+1}\frac{t}{t}\sin(t^2)dt\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2243337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Probability that a passenger will catch the bus A bus and a passenger arrive at a bus stop at a uniformly distributed time over the interval $0$ to $1$ hour. Assume the arrival times of the bus and passenger are independent of one another and that the passenger will wait up to $15$ minutes for the bus to arrive. What i... | There is a $\frac{3}{4}$ chance the bus will be later than 10:15. Therefore there is a $\frac{1}{4}$ of an hour window of the person $Y2$, arriving on time. If $Y1 < \frac{1}{4}$ and $Y2 > \frac{1}{4}$ the probability of the person missing the bus is certainly $1$.
If $Y1 (\frac{1}{4})$ AND $Y2 (\frac{1}{4})$ arrives ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2246057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find $ \lim\limits_{{n \to \infty}} \frac1{2^n} \sum\limits_{k=1}^n \frac1{\sqrt{k}} \binom nk$ Find
$$\lim_{n \to \infty} \frac {1}{2^n} \sum_{k=1}^n \frac{1}{\sqrt{k}} \binom{n}{k}.$$
First time I thought about Stirling's approximation but didn't get anything by applying it. I would also think about a Riemann Sum, bu... | Let $a_k = k^{-1/2}$. Notice that $(a_k)$ decreases to $0$. Then for each fixed $m \geq 1$ and for all $n \geq m$,
$$ \frac{1}{2^n} \sum_{k=1}^{n} \binom{n}{k} a_k \leq \frac{1}{2^n} \underbrace{\sum_{k=1}^{m} \binom{n}{k} (a_k - a_m)}_{= \mathcal{O}(n^m)} + \frac{1}{2^n} \underbrace{\sum_{k=1}^{n} \binom{n}{k} a_m}_{=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 6,
"answer_id": 5
} |
Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.
I spent a lot of time trying to solve this and, having consulted some ... | We can minimize $x^2 + y^2 + z^2$ subject to the constraint $x+y+z = 1$ using Lagrange multipliers, we then find that $x = y = z = \frac{1}{3}$, therefore $x^2 + y^2 + z^2\geq\frac{1}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 15,
"answer_id": 5
} |
Using eigenvectors to find the general solution from a system of equations \begin{bmatrix}-13&40&-48\\-8&23&-24\\0&0&3\end{bmatrix}
Consider the matrix above. This corresponds to
\begin{align*}
x_{1}'&=-13x_{1}+40x_{2}-48x_{3}\\
x_{2}'&=-8x_{1}+23x_{2}-24x_{3}\\
x_{3}'&=3x_{3}
\end{align*}
What I want to do is use eig... | Solving the System of linear Equations
So it seems you want to solve the system of equations
$$
Ax =
\begin{pmatrix}-13&40&-48\\-8&23&-24\\0&0&3\end{pmatrix}
\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}
=
\begin{pmatrix} x'_1 \\ x'_2 \\ x'_3 \end{pmatrix}
= x'
$$
Where the $x'$ is given and the $x_j$ are looked for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
What are the solutions to: $x(x-3)=x^2 -4$? What are the solutions to: $x(x-3)=x^2 -4$ ?
Subtracting $x^2$ from both side results in the loss of the solution $(x=\infty)$.
My book says the solutions are $(x=\frac{4}{3},\infty)$
But is $(x=-\infty)$ another solution?
Is the answer $(x=\frac{4}{3},\pm\infty)$ or $(x=\fra... | $$\begin{align}
x(x-3)&=x^2 -4 \\ \\
\iff x^2-3x &= x^2 - 4\\ \\\iff 3x-4&=0\\ \\ \iff 3x&=4 \\ \\ \iff x &= \frac 43
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2249482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
How many zeros of $z^{17}+z-1$ lie in the region $|z|<1$? General problem:
Conjecture (B. Galo and Jack D'Aurizio) Let $p$ be a prime number then the number of zeros of $z^{p}+z-1$ that lie in the region $|z|<1$ is:
1) $⌊\frac{p}{3}⌋-2$ (where $⌊*⌋$ is the lower integer function). If $ p≡5 (mod6)$ but $p\neq 5$.
2) $⌊... | Visualiztion
Start with
$$
f(z) = z^{17}+z-1
$$
The Re and Im components are shown next.
The $0$ contours, the line where
$$
\color{blue}{\text{Re } f= 0}, \qquad \color{red}{\text{Im } f= 0}
$$
are shown below against the unit circle. The roots where $f(z)=0$ are where the blue and red contours intersect. We see six ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2254126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Solve the differential equation IVP
Solve $\displaystyle x \frac{dy}{dx} = 2y + x^2, y(1) = 2$
We can use the homogeneous form, $y = vx$, so $dy/dx = xdv/dx + v$
$x^2dv/dx + vx = 2x(v + x) \implies x^2 dv/dx = vx + x \implies x dv/dx = v + 1 \implies (v+1) dv = xdx \implies v^2/2 + v = x^2 + C$
Then we would just su... | \begin{eqnarray}
xy'&=&2y+x^2\\
x^2y'-2xy&=&x^3\\
\dfrac{x^2y'-2xy}{x^4}&=&\dfrac{1}{x}\\
\left(\dfrac{y}{x^2}\right)'&=&\dfrac{1}{x}\\
\int\left(\dfrac{y}{x^2}\right)'&=&\int\dfrac{1}{x}\\
\dfrac{y}{x^2}&=&\ln x+C\\
y&=&x^2\ln x+2x^2
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2255634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Probability of drawing a spade on the first draw, a heart on the second draw, and an ace on the third draw My answer is
$(\frac{1}{52})(\frac{1}{51})(\frac{2}{50})$ + $2(\frac{1}{52})(\frac{12}{51})(\frac{3}{50})$ + $(\frac{12}{52})(\frac{12}{51})(\frac{4}{50})$
= $\frac{1}{204}$
But the answer in the textbook is
$(\fr... | $$P(A) = \frac{n(A)}{n(\Omega)}$$
Here, $$n(A) = n(\text{ace of spade}) \times n(\text{ace of heart}) \times n(\text{other aces}) + \\ n(\text{spade other than ace}) \times n(\text{ace of heart}) \times n(\text{other aces}) + \\ n(\text{ace of spade}) \times n(\text{ heart other than ace}) \times n(\text{other aces}) +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2258643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find $y'$ and $y''$ $\log(\sqrt{x^2 + y^2})=\arctan{\frac{y}{x}}$
Let $F(x,y)= \log(\sqrt{x^2 + y^2})-\arctan{\frac{y}{x}}$
I can't do it right, because I don't understand the formula $\frac{\partial F}{\partial x}+\frac{\partial F}{\partial z}\frac{\partial z}{\partial x}=0$
I think $z=y(x)$. So should I get something... | Frankly, I would just differentiate directly, using the chain rule, on both sides. And, before differentiating, reduce the left side: $\log(\sqrt{x^2+ y^2})= \frac{1}{2}\log(x^2+ y^2)$
The derivative of $\frac{1}{2}\log(x^2+ y^2)$ is
$$
\frac{1}{x^2+ y^2}(x+ yy')= \frac{x+ yy'}{x^2+ y^2}.
$$
The derivative of $\arcta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2258961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Which is greater $x_1$ or $x_2$? $$x_1=\arccos\left(\frac{3}{5}\right)+\arccos\left(\frac{2\sqrt{2}}{3}\right)$$
$$x_2=\arcsin\left(\frac{3}{5}\right)+\arcsin\left(\frac{2\sqrt{2}}{3}\right)$$ We have to find which is greater among $x_1$ and $x_2$
If we add both we get $$x_1+x_2=\pi$$ If we use formulas we get
$$x_1=\a... | (I'm going to do this in degrees rather than radians as this is very intuitive about right triangles).
Obviously if $\theta = \arccos \frac ah$ then $\theta$ represents an angle of a right triangle with adjacent side $a$ and hypotenuse $h$ (and opposite side $o = \sqrt {h^2 - a^2}$). So $\arcsin \frac ah = 90 - \theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2259157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Finding a sum of $1+\frac{1}{4\cdot2^{4}}+\frac{1}{7\cdot2^{7}}+\frac{1}{10\cdot2^{10}}+\cdots$ I need someone to find a mistake in my soliution or maybe to solf it much more easily... I have got a sum $$1+\frac{1}{4\cdot2^{4}}+\frac{1}{7\cdot2^{7}}+\frac{1}{10\cdot2^{10}}+\cdots$$ and need to evaluate it. So here's my... | It's worth cleaning things up a bit:
$$1+{1\over4\cdot2^4}+{1\over7\cdot2^7}+{1\over10\cdot2^{10}}+\cdots={1\over2}+{1\over2}\left(1+{1\over4\cdot2^3}+{1\over7\cdot2^6}+{1\over10\cdot2^9}+\cdots \right)\\={1\over2}+{1\over2}f(1)$$
where
$$f(x)=x+{x^4\over4\cdot2^3}+{x^7\over7\cdot2^6}+{x^{10}\over10\cdot2^9}+\cdots$$
N... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2260413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Finding a function from a Maclaurin expansion Given the following Maclaurin expansion:
$$1+\frac{x}{2!}+\frac{x^2}{4!} + \frac{x^3}{6!} + \ldots + \frac{x^r}{(2r)!}+ \ldots$$
How would you go about finding the function? All I can see to do is use forms of trial and error.
| So you're looking for a closed form for the series:
$$\sum_{r=0}^{\infty} \frac{x^r}{(2r)!} = 1+\frac{x}{2!}+\frac{x^2}{4!} + \frac{x^3}{6!} + \ldots $$
You probably know that:
$$e^x = \sum_{r=0}^{\infty} \frac{x^r}{r!} = 1+x+\frac{x^2}{2!} + \frac{x^3}{3!} + \ldots $$
And evaluating this at $\sqrt{x}$ makes the powers... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2261878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the number of common, odd positive divisors of 27, 900 & 20, 700. Question as in title,
Find the number of common, odd positive divisors of $27, 900$ & $20, 700$.
explaining the steps in your method would be much appreciated.
I prime factorized and got that the common factor is $2^2\cdot3^2\cdot5^2$, then my n... | Factorize $27{,}900$ and $20{,}700$ :
$$27{,}900=2^2 \times 3^2 \times 5^2 \times 31 ~;~ 20{,}700=2^2 \times 3^2 \times 5^2 \times 23$$
For their common odd divisors ; you need total number of divisors of $3^2 \times 5^2$.
For counting number of divisors of $3^2 \times 5^2$, you can have $0 , 1$ or $2$ power of $3$ fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2263043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding $\lim_{n \to \infty} x_n$ Let $x$ be a positive real number. A sequence $x_n$ of real numbers is defined as follows:
$$x_1=\frac 12\left(x+\frac5x\right)$$ and $$x_{n+1}=\frac 12\left(x_n+\frac 5{x_n}\right) \quad\forall n\geq 1$$
The question is to show that $$\frac{x_n-\sqrt 5}{x_n+\sqrt 5}=\left(\frac{x-\sqr... | The limit, if it exists, should be a fixed point of $f(x) = \frac{1}{2} \left( x + \frac{5}{x} \right)$. Rearranging $x = \frac{1}{2} (x + \frac{5}{x})$ gives $x^2 - 5 = 0$ or $x = \pm \sqrt{5}$, and a negative limit is impossible as the series is manifestly positive. (And it's easy to show that the limit exists: note ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2265615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Infinite Sum Calculation: $\sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} = \frac{3}{4} \sum_{n=1}^{\infty} \frac{1}{n^2}$ Problem
Show the following equivalence: $$\sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} = \frac{3}{4} \sum_{n=1}^{\infty} \frac{1}{n^2}$$
Good afternoon, dear StackExchange community. I'm studying real analysis ... | You are almost done,:
$$\sum_{n=1}^{\infty}\frac{1}{n^2} = 1 + \sum_{k=1}^{\infty} \frac{1}{(2k)^2} + \sum_{k=1}^{\infty}\frac{1}{(2k+1)^2}= \sum_{k=1}^{\infty} \frac{1}{(2k)^2} + \sum_{k=0}^{\infty}\frac{1}{(2k+1)^2}$$
Notice that :
$$\sum_{k=1}^{\infty} \frac{1}{(2k)^2}=\sum_{k=1}^{\infty} \frac{1}{4k^2}=\frac14\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2268605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Differentiation - Power rule Find derivative of this function
$$g(t)=-3t(6t^4-1)^{4/3}$$
I have tried it till the answer:
$ -3t.\frac{4}{3} (6t^4 - 1)^{\frac{4}{3}-1} .\frac{d}{dt}(6t^4 -1) $
$-3t x \frac{4}{3} (6t^4-1)^\frac{1}{3} ( 6X4 t^3 -0 ) $
$ -4t (6t^4 -1)^\frac{4}{3} (24t^3) $
$-96t(6t^4-1)^\frac{1}{3} $
Ho... | after the product and the chain rule we get
$$-3(6t^4-1)^{4/3}-3t\frac{4}{3}(6t^4-1)^{1/3}\cdot 24t^3$$
after simplifying this derivative we get
$$-3 \sqrt[3]{6 t^4-1} \left(38 t^4-1\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2268700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Higher order interpolation Let we have some points $(x_i,y_i),~i=1,2,...n,$ that are scattered on a $2D$ domain like square. And $u_h$ is the solution of a numerical method at these points. Now, if I want to construct a higher order interpolation of $u_h$ at these points, how can I do that? Should I find $u_h^{new}$ w... | The griddata function is finding a surface. Here is a way to see how the fit works.
Problem statement
The data is a sequence of $m$ points of the form $$\left\{ x_{k}, y_{k}, z_{k}
\right\}_{k=1}^{m}$$
The model surface is an approximation of order $d$. Consider $d=2$:
$$
z(x,y) = a_{00} + a_{10}x + a_{01}y + a_{20}x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2269391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$\sqrt{40-9x}-2\sqrt{7-x}=\sqrt{-x}$ In MAO 1991,
Find $2x+5$ if $x$ satisfies $\sqrt{40-9x}-2\sqrt{7-x}=\sqrt{-x}$
My attempt,
I squared the the equation then I got $144x^2+1648x+4480=144x^2-1632x+4624$, which results $x=-9$, and $2x+5=-13$.
I want to ask is there another way to solve this question as my method is v... | Substitution $3.5-x=t$ seems to help because $-x$ and $7-x$ are symmetric around the point $3.5$ you get
$$\sqrt{9t+8.5}=\sqrt{t-3.5}+2\sqrt{t+3.5}\\9t+8.5=t-3.5+4t+14+4\sqrt{t^2-3.5^2}\\4t-2=4\sqrt{t^2-3.5^2}\\(2t-1)^2=4(t^2-3.5^2)\\4t^2-4t+1=4t^2-4\cdot 3.5^2\\4t=4\cdot 3.5^2+1\\4t=(2\cdot 3.5)^2+1\\4t=50\\t=12.5\\x=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2276148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
For integer m greather than 2, $\frac{1}{m} + \frac{1}{m+2}$, the numerators and denomitors are primitive pythagorean triples $a$ and $b$ For $m = 2$, the fraction is $\frac{3}{4}$. for $m=3$, the fraction is$\frac{8}{15}$. I was wondering why numerators and demoninators of $\frac{1}{m} + \frac{1}{m+2}$ show primitive... | Notice that $$\frac{1}{m}+\frac{1}{m+2}=\frac{2m+2}{m^2+2m}.$$ We can check that $$(m^2+2m)^2+(2m+2)^2=m^4+4m^3+4m^2+4m^2+8m+4=m^4+4m^3+8m^2+8m+4=(m^2+2m+2)^2.$$ Note that this is always a primitive triple if you put the fraction in lowest form, but $2m+2$ and $m^2+2m$ are not coprime in general (and thus the triple $2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2277117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find all solutions to $|x^2-2|x||=1$ Firstly, we have that $$\left\{
\begin{array}{rcr}
|x| & = & x, \ \text{if} \ x\geq 0 \\
|x| & = & -x, \ \text{if} \ x<0 \\
\end{array}
\right.$$
So, this means that
$$\left\{
\begin{array}{rcr}
|x^2-2x| & = & 1, \ \text{if} \ x\geq 0 \\
|x^2+2x| & = & 1, \ \tex... | As G.H.lee already pointed out, you did casework $x\geq 0$ and $x<0$ to simplify the expression but you completely disregarded it later. A correct way would be something like this:
$$|x^2-2|x||=1\implies
\begin{cases}|x^2-2x| = 1,& x\geq 0\\
|x^2+2x| = 1,& x < 0 \end{cases} \implies
\begin{cases} x^2-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2280212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Triangular series perfect square formula 8n+1 derivation In triangular series
$$1 $$
$$1+2 = 3$$
$$1+2+3 = 6$$
$$1+2+3+4 =10$$
$$\ldots$$
Triangular number in 8n+1 always form perfect square .
Like $8\cdot 1+1 = 9 , 8\cdot 3+1 = 25$ .
How this formula is derived ?
| Let$$\begin{align}
A\cdot \overbrace{\frac {n(n+1)}2}^{T_n}+1&=(Bn+C)^2\\
\frac A2 n^2+\frac A2n+1&=B^2n^2+2BCn+C^2\\
\end{align}$$
Equating coefficients gives $C=1, A=4B, A=2B^2$, solving for which gives $B=2, A=8$.
Hence
$$8T_n+1=(2n+1)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2281906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find the cubic whose roots are $\alpha^3, \beta^3, \gamma^3$
Let $x^3+ax^2+bx+c=0$ are $\alpha, \beta, \gamma$. Find the cubic whose roots are $\alpha^3, \beta^3, \gamma^3$
My attempt,
As I know from the original equation,
$\alpha+\beta+\gamma=-a$
$\alpha\beta+\beta\gamma+\alpha\gamma=b$
$\alpha\beta\gamma=-c$
I'... | If we replace $x$ by $x^{1/3}$, we obtain $x + ax^{2/3} + bx^{1/3} + c = 0$. This equation does have the required roots, but it's not a polynomial.
To get the good ol' cubic form, write it as $x + c = -(ax^{2/3} + bx^{1/3})$ and cube both sides so that$$(x + c)^3 = -(ax^{2/3} + bx^{1/3})^3 = -(a^3 x^2 + b^3 x + 3abx(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2285064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Laurent series expansion of a given function
Let $\sum_{-\infty}^{\infty} a_n z^n$ be the Laurent series expansion of $f(z)=\frac{1}{2z^2-13z+15}$ in an annulus $\frac{3}{2}< \vert z \vert < 5$.
What is the value of $\frac{a_1}{a_2}$?
My attempt:
First note that $f$ is analytic in the given annulus.
$f(z)=\frac{1}{2z... |
The function
\begin{align*}
f(z)&=\frac{1}{2z^2-13z+15}\\
&=-\frac{1}{7}\cdot\frac{1}{z-\frac{3}{2}}+\frac{1}{7}\cdot\frac{1}{z-5}\\
\end{align*}
has two simple poles at $\frac{3}{2}$ and $5$.
Since we want to find a Laurent expansion with center $0$, we look at the poles $\frac{3}{2}$ and $5$ and see they determ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2285139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
3-cocycle in $\mathbb{Z}/n\mathbb{Z}$ It is well known that $3$-cocycles in $H^3(\mathbb{Z}/n\mathbb{Z};U(1))$ are of the form
$$w(a,b,c)=\exp(\frac{2\pi ik}{n^2}a(b+c-[b+c])),$$
for $k\in\mathbb{Z}_n$, $a,b,c\in\mathbb{Z}_n$ and $[b+c]=b+c\mod{n}$.
(See https://mathoverflow.net/questions/121065/explicit-3-cocycle-of-a... | The mistake in the original question is the following: $w'(a+b, c, d) = [a + b]\cdot (c + d - [c + d])$ instead of $(a + b)\cdot (c + d - [c + d])$. So, you should make each argument modulo $n$ before using the formula of the 3-cocycle. With this changes the formula is correct. Indeed:
\begin{multline}
w'(b, c, d) - w'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2286199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
How many solutions does the equation $x^2-y^2 = 3^8\cdot 5^6\cdot 13^4$ have? Given that $x$ and $y$ are positive integers. Problem: How many solutions does the equation $x^2-y^2 = 3^8\cdot 5^6\cdot 13^4$ have? Given that $x$ and $y$ are positive integers.
I tried a similar approach to the ones described here, but to ... | Factor
$$ (x+y)(x-y) = 3^8 \cdot 5^6 \cdot 13^4 $$
The number of ways to split up $3^8$ in a product is 9. The number of ways to split up $5^6$ in a product is 7. The number of ways to split up $13^4$ in a product is 5. Collectively this means $3^8 \cdot 5^6 \cdot 13^4$ can be split into a product in $9\cdot7\cdot5=315... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2286419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to find sum and convergence of series $\sum_{n = 1}^{\infty}\frac{1}{(2n-1)^{2}\cdot(2n+1)^{2}}$ I laid this shot that:
$$\frac{1}{(2n-1)^2(2n+1)^2}=1/4\, \left( 2\,n+1 \right) ^{-2}-1/4\, \left( 2\,n-1 \right) ^{-1}+1/
4\, \left( 2\,n-1 \right) ^{-2}+1/4\, \left( 2\,n+1 \right) ^{-1}
$$
But I don't know what to d... | By partial fraction decomposition
$$ \frac{1}{(2x-1)^2 (2x+1)^2} = \frac{\frac{1}{16}}{\left(x-\frac{1}{2}\right)^2}+\frac{\frac{1}{8}}{x-\frac{1}{2}}-\frac{\frac{1}{8}}{x+\frac{1}{2}}+\frac{\frac{1}{16}}{\left(x+\frac{1}{2}\right)^2} \tag{1}$$
hence it follows that:
$$ \sum_{n\geq 1}\frac{1}{(2n-1)^2 (2n+1)^2}=\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2289063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding $\lim_{(x,y)\to(0,0)} \frac{e^{-\frac{1}{x^2+y^2}}}{x^6+y^6}$
$$\lim_{(x,y)\to(0,0)} \frac{e^{-\frac{1}{x^2+y^2}}}{x^6+y^6}$$
How should I approach this?
| I would suggest using polar coordinates:
\begin{align}
x&=r\cos(\theta)\\
y&=r\sin(\theta)\\
r^2&=x^2+y^2.
\end{align}
Since $x^6+y^6=(x^2+y^2)(x^4-x^2y^2+y^4)$, we could also compute:
$$
\lim_{r\rightarrow 0^+}\frac{e^{-\frac{1}{r^2}}}{r^6(\cos^4(\theta)-\cos^2(\theta)\sin^2(\theta)+\sin^4(\theta))}.
$$
Since $\cos^6(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2290396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$ \lim_{x \to \infty } ( \sqrt[3]{4 x^{a} + x^{2} } - \sqrt[3]{ x^{a} + x^{2} } )^{x-[x]} $ fine limit :
$$ \lim_{x \to \infty } ( \sqrt[3]{4 x^{a} + x^{2} } - \sqrt[3]{ x^{a} + x^{2} } )^{x-[x]} $$
such that : $$ a \in (0,2)$$
and :
$[x]: \ \ $ floor function
My Try :
$$f(x):=( \sqrt[3]{4 x^{a} + x^{2} } - \sqrt[3]... | The expression under limit operation say $F(x) $ is of the form $\{f(x) \} ^{g(x)} $ and clearly $f(x) >0$ for all $x>0$ and $0\leq g(x) <1$ for all $x>0$. It is now clear that $F(x) $ lies between $1$ and $f(x) $ and it takes the value $1$ when $x$ is a positive integer. Moreover if $x$ is near a positive integer and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2290519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
how many $3\times 3$ matrices with entries from $\{0,1,2\}$.
How many $3 × 3$ matrices $M$ with entries from $\left\{0, 1, 2\right\}$ are there for which taken from
the sum of the main diagonal of $M^TM$ is $5$.
Attempt: Let $M = \begin{pmatrix}
a & b & c\\
d & e & f\\
g & h & i
\end{pmatrix}$. where $a,b,c,d,e,f... | Either $5=1^2+1^2+1^2+1^2+1^2+0^2+0^2+0^2+0^2$ or $5=1^2+2^2+0^2+0^2+0^2+0^2+0^2+0^2+0^2$
thus there are two cases: 5 of them are 1 and the other 4 are 0, or 1 is 1, 1 is 2, and the other 7 are 0.
Case 1: there are $\binom{9}{5} = 126$ possibilities.
Case 2: there are $\binom{9}{1}\binom{8}{1}\binom{7}{7} = 72$ possibi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2291491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Factorise $x^5+x+1$
Factorise $$x^5+x+1$$
I'm being taught that one method to factorise this expression which is $=x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1$
$=x^3(x^2+x+1)-x^2(x^2+x+1)+x^2+x+1$
=$(x^3-x^2+1)(x^2+x+1)$
My question:
Is there another method to factorise this as this solution it seems impossible to invent it?
| Let $f (x)=x^5+x+1$
$$f'(x)=5x^4+1>0$$
$f $ is stricly increasing at $\mathbb R $, thus it has only one real root $\alpha \in (-1,0) $.
the factorisation will be of the form
$$(x-\alpha)(x^2+ax+b)(x^2+cx+d) $$
with $a^2-4b <0$ and $c^2-4d <0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2293493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 6
} |
Calculate $\sum_{n=1}^{\infty}{c_n\over n}$ when $c_1=2, c_2=1, c_3=-3, c_{n+3}=c_n, n\in \mathbb{N}$ Let $(c_n)$ be defined as $c_1=2, c_2=1, c_3=-3, c_{n+3}=c_n, n\in \mathbb{N}$. I want to get $\sum_{n=1}^{\infty}{c_n\over n}$. I would appreciate any suggestions how to start on this. It is my suspicion that I would ... | I'll expand on the (very clever!) method by Jack, giving some more details that may not have been obvious from his answer.
$$\begin{align}\sum_{n=1}^\infty\frac{c_n}{n}&=\sum_{k=0}^\infty\left(\frac{2}{3k+1} + \frac{1}{3k+2} - \frac{3}{3k+3}\right) \\
&=\sum_{k=0}^\infty\left(2 \int_0^1 x^{3k}dx + \int_0^1 x^{3k+1}dx -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2296836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Calculation related to second partial derivative Problem: let $u(x,t) = \frac{1}{2\sqrt{\pi t}} e^{-\frac{x^2}{4t}}$, prove when $t \gt 0$, we have $|\frac{\partial^2u}{\partial x^2}| \le Ct^{-3/2}e^{-\lambda\frac{x^2}{t}}$, where $C \gt 0$ and $\lambda \le \frac{1}{4}$.
My Attempt:
\begin{align}
\frac{\partial u}{\par... | We start from the question simplification by equivalent transformations, for all real $x$ and positive $t$ and $C$.
$$\left|\frac{\partial^2u}{\partial x^2}\right| \le Ct^{-3/2}e^{-\lambda\frac{x^2}{t}}$$
$$\left|\left(\frac{x^2}{2t}-1\right)\frac{1}{4\sqrt{\pi}t^{3/2}}e^{-\frac{x^2}{4t}}\right|\le Ct^{-3/2}e^{-\lambda... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2300845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find all prime $a, b, c$ such that $ab+bc+ac > abc$ Find all primes $a, b, c$ such that $ab + bc + ac > abc$
| Hint:
The above inequation is equivalent to
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} > 1$$
where $a,b,c$ are primes
Complete Solution:
WLOG we can assume $a \geq b \geq c$
This means that $c<3$ ie. $c=2$
Now, we have $$\frac{1}{a}+\frac{1}{b} > \frac{1}{2}$$
Since $a \geq b$, we have $b<4$ ie $b \in \{2,3\}$
I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2304534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What is the correct way of expressing the general solution of $\sin x=0,1,-1$ and $\cos x=0,-1,1$? The general solution of $\sin x = \sin y$ is $x = n\pi + (-1)^ny$. Then if $\sin x = 1$ or $\sin x = \sin(\frac{\pi}{2})$ or $x = n\pi+(-1)^n\frac{\pi}{2}$. Is this correct?
Also, the general solution of $\cos x = \cos y$... | The clearest way of writing the solutions are
$$
\sin{x} = \sin{y} \Leftrightarrow x = y + n2\pi \text{ or } x = \pi - y + n2\pi$
$$
$$
\cos{x} = \cos{y} \Leftrightarrow x = \pm y + n2\pi
$$
where $n$ is a whole number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2305235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integer values of $(\frac{19}{7}-\sum_{k=0}^N \frac{1}{k!})^{-1}$ The approximation for $e$ involved in An integral for $2\pi+e-9$
exactly matches a truncation of factorial reciprocal series
$$\sum_{k=0}^5 \frac{1}{k!} =\frac{163}{60}$$
In contrast, the difference
$$\frac{19}{7}-\sum_{k=0}^N \frac{1}{k!}$$
never cance... | For ease of notation I'll denoting $\sum_{k=0}^N \frac{1}{k!}$ as $e_N$ for now.
Note that $(19/7 - e)^{-1} \approx -250.24...$, or in particular it's somewhere in-between $-250$ and $-251$. So if we can show that for $N \ge 8$ the sum $(\frac{19}{7}-e_N)^{-1}$ is in between $-250$ and $-251$ then from that point it ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2306631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving $(a-1)^{p+1}+a^p=(a+1)^{p-1}$
Suppose that $a,p$ are nonnegative integers such that $p$ is prime and $p\nmid (a-1)$. If $(a-1)^{p+1}+a^p=(a+1)^{p-1}$, find the sum of all possible values of $a$.
We can't have $p = 2$ since the equation $$(a-1)^3+a^2 = (a+1)$$ has no integer solutions. We now take two cases:
C... | Hint
$$1 \geq \frac{(a-1)^{p+1}}{(a+1)^{p-1}}=(a-1)^2 (1-\frac{2}{a+1})^{p-1}\geq (a-1)^2 (1-\frac{2(p-1)}{a+1})$$
with the last inequality following by Bernoulli.
Thus
$$a+1 \geq (a-1)^2(a+3-2p) \\
2 =a+1-(a-1)\geq a+1-(a-1)^2\geq (a-1)^2(a+2-2p)$$
This implies that either $(a+2-2p) \leq 0$ or $a-1=1$ and $(a+2-2p) \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2307251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integer solutions to $y=(2^x-1)/3$ where $x$ is odd For the equation $y=(2^x-1)/3$ there will be integer solutions for every even $x$.
Proof: When $x$ is even the equation can be written as $y=(4^z-1)/3$ where $z=x/2$.
$$4^z =1 + (4-1)\sum_{k=0}^{z-1} 4^k$$
If you expand that out you get:
$$4^z=1+(4-1)4^0+(4-1)4^1+(4-1... | If $x$ is odd, then $x=2k+1$ and then
$$2^{2k+1}-1=4^k\cdot 2-1\equiv1^k\cdot 2-1\equiv1\pmod 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2311019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find integral solutions - rational points of $4p=x^2+27y^2$ for $p>3$ prime Let $p\geqslant 5$ be a prime. I want to solve $$4p=x^{2}+27y^{2} \tag{1}\quad x,y\in \Bbb N$$
This comes when considering the discriminant of a cubic polynomial $T^3 -pT-yp$ with cyclic group.
Are there only elementary ways to find solutions?... | All primes $1 \pmod 3$ are integrally represented by $x^2 + xy + 7 y^2,$ shorthand $\langle 1,1,7 \rangle.$ Discriminant $-27.$
The primitive quadratic forms of discriminant $-108$ are
$$ \langle 1,0,27 \rangle, \; \; \langle 4,2,7 \rangle, \; \; \langle 4,-2,7 \rangle $$
Among primes $p \equiv 1 \pmod 3,$ we have $p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2314640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Finding the orthogonal projection of $(2,1,3)$. $V=\mathbb R^3,u=(2,1,3) $ and $W$={$(x,y,z):x+3y-2z=0$}.
Vectors ortogonal to $W$=Span{$(1,3,-2)$},Vectors on $W$=Span{$(-1,1,1)$}.
So,orthogonal basis is {$(1,3,-2),(-1,1,1)$}.Then,orthonormal basis is
{$\frac{1}{\sqrt {14}}(1,3,-2),\frac{1}{\sqrt {3}}(-1,1,1)$}.
So or... | Let $(2,1,3)+k(1,3,-2)$ be the orthogonal projection.
Thus, $$(2+k,1+3k,3-2k)(1,3,-2)=0$$ or
$$2+k+3+9k-6+4k=0,$$
which gives $k=\frac{1}{14}$ and since $\left(2+\frac{1}{14},1+\frac{3}{14},3-\frac{1}{7}\right)=\frac{1}{14}(29,17,40)$,
we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2315222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Maximization of multivariable function $ M = \frac{a+1}{a^{2}+ 2a+2} + \frac{b+1}{b^{2}+ 2b+2} + \frac{c+1}{c^{2}+ 2c+2}$ subject to constraint This is purely out of curiosity. I have this set of a high school entrance Math exam in Vietnam (a special high school for gifted kids, the exam was held 3 days ago, maybe 4, t... | Here Laplace multipliers, derivatives and gradients do not exist. Only multiplying and checking if something simplifies.
In this case, symmetry for simplifying the problem into:
$$
\mathcal{M}: 3{x+1 \over x^2+2x+2}\\
st. x^3+3x^2=2
$$
This holds because every term on the function is a sum of independent terms. In here... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2315548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
$P[X\geq m/2]\geq\frac{1}{m+1}$ if $\mathbb{E}[X]=\frac{m}{2}$ Let $m\gt 0$ and let $X$ be a random variable on the set $\{0,\dots,m\}$.
Assuming that $\mathbb{E}[X]=\frac{m}{2}$, prove that:
$$P[X\geq m/2]\geq\frac{1}{m+1}$$
I tried induction but it didn't work, know I am trying to prove it by contradiction, someone g... | If $P(X \ge m/2) < \frac{1}{m+1}$, then
\begin{align}
E[X] &= \sum_{k < m/2} k P(X=k) + \sum_{k \ge m/2} kP(X=k)
\\
&\le \left(\left\lceil\frac{m}{2}\right\rceil - 1\right) P(X < m/2) + m P(X \ge m/2)
\\
&= \left(\left\lceil\frac{m}{2}\right\rceil - 1\right)(1-P(X \ge m/2)) + mP(X \ge m/2)
\\
&= \left(\left\lceil\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2317618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Infinite zeros in infinite series The problem:
Given that
$$\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \ldots $$
Prove
$$\frac{\pi}{3} = 1 + \frac{1}{5} - \frac{1}{7} - \frac{1}{11} + \frac{1}{13} + \frac{1}{17} + \ldots$$
My solution:
We know
$$
\begin{align}
\frac{\pi}{4} & = 1 - \f... | *
*Yes, that is ok.
*Yes it is. The partial sum of $\sum_k (a_k+b_k)$ is the partial sum of $\sum_ka_k$ plus the partial sum of $\sum_k b_k.$ The result follows from the sum property for limits.
*Yes. Adding zeros will only delay the inevitable convergence of the sequence of partial sums. Where you insert zeros, the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2320456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
"answer_id": 2
} |
Solve $\frac{a+b-x}c + \frac{a+c-x}b + \frac{b+c-x}a -\frac{4abc}{a+b+c} = -7$ for $x$ I have been scratching my head for solving this equation but I am unable to do this. Even I am unable to get how to use the hint. A way to solve this would be of great help
Solve for $x$ :-
$$\frac{a+b-x}c + \frac{a+c-x}b + \frac{b+... | Let $a^{-1}+b^{-1}+c^{-1}=k$. Your equation becomes
$$
\frac{a}{c}+\frac{b}{c}+
\frac{a}{b}+\frac{c}{b}+
\frac{b}{a}+\frac{c}{a}-
\frac{4abc}{a+b+c}+7
=kx
$$
The first six terms on the left-hand side can be rewritten as
$$
\frac{a^2b+ab^2+a^2c+ac^2+b^2c+cb^2}{abc}
$$
Now, set $S=a+b+c$, $Q=ab+bc+ca$ and $P=abc$. The nu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Proof that $\log_a b \cdot \log_b a = 1$
Prove that $\log_a b \cdot \log_b a = 1$
I could be totally off here but feel that I have at least a clue.
My proof is:
Suppose that $a = b$, then $a^{1} = b$ and $b^{1} = a$ and we are done.
Suppose now that $a \neq b$. We wish to show that: $\log_b a = \frac{1}{\log_a b}$.
... | Your argument does not prove the identity. I can substitute any two values for $\log_a b$ and $\log_b a$ and get the "proof"
\begin{align}
1 &= \frac1x \cdot \frac1y \cdot \left( x \cdot y \right) \\
&= \left( x \cdot \frac1x \right)\left( y \cdot \frac1y \right) \\
&= \left( x \cdot \frac1x \right) \cdot 1 \\
&= 1 \cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find extremas of $f(x,y) = xy \ln(x^2+y^2), x>0, y>0$ As the title says I need to find extreme values(maximum and minimum) of
$$f(x,y) = xy \ln(x^2+y^2), x>0, y>0$$
I don't understand how to find critical points of this problem.
I start with finding partial derivative and set derivatives equal to zero. And that is wher... | We look for points for wich $\nabla f=0$
The method you describe is this:
$\dfrac{\partial f}{\partial x}=y\ln(x^2+y^2)+xy\dfrac{2x}{x^2+y^2}$
$\dfrac{\partial f}{\partial y}=x\ln(x^2+y^2)+xy\dfrac{2y}{x^2+y^2}$
(You need to practice the derivatives :)
Setting the partials to zero we can get the coordinates for those p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solving equation of type $a\cos x+b\cos y-c=0$ and $a\sin x+b\sin y-d=0$ Here's the questions
There are two equations:
$a\cos x+b\cos y-c=0 $ and $ a\sin x+b\sin y-d=0$ .
For instance
What is the value of $x$ and $y$ in following question?
$$2\cos x+3\cos y-2=0$$
$$2\sin x+3\sin y-8=0$$
| I'll assume $a > 0$ and $b > 0$.
Let $R = \sqrt{c^2 + d^2}$, so that $c = R \cos(\theta)$ and $d = R \sin(\theta)$ for some $\theta \in [0,2\pi)$. The points $[0,0]$, $[a \cos(x), a \sin(x)]$ and $[c, d]$ form a triangle with sides of length $a$, $b$ and $R$.
We need $R \le a + b$, $a \le R + b$ and $b \le R + a$ for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Show that $(2,0,4) , (4,1,-1) , (6,7,7)$ form a right triangle What I tried:
Let $A(2,0,4)$, $B(4,1,-1)$, $C(6,7,7)$ then
$$\vec{AB}=(2,1,-5), \vec{AC}=(4,7,3), \vec{BC}=(2,6,8)$$
Then I calculated the angle between vectors:
$$\begin{aligned}
\alpha_1 &= \cos^{-1}\left(\frac{(2,1,-5)(4,7,3)}{\sqrt{2^2+1^2+(-5)^2}\sqrt{... | The distances satisfy the Pythagorean theorem.
$d(A, B) = \sqrt{2^2 + 1^2 + 5^2} = \sqrt{30}$
$d(A, C) = \sqrt{4^2 + 7^2 + 3^2} = \sqrt{74}$
$d(B, C) = \sqrt{2^2 + 6^2 + 8^2} = \sqrt{104}$
And indeed:
$\sqrt{30}^2 + \sqrt{74}^2 = \sqrt{104}^2$
Therefore it is a right triangle (link).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 4,
"answer_id": 0
} |
Prove $\cos6x=32\cos^{6}x-48\cos^{4}x+18\cos^{2}x-1 $ So far I've done this:
LHS $ =\cos^{2}3x-\sin^{2}3x$
$={(4\cos^{3}x-3\cos{x})}^2 -{(3\sin{x}-4\sin^{3}x)}^2$
$=16\cos^{6}x+9\cos^{2}x-24\cos^{4}x-9\sin^{2}x-16\sin^{6}x+24\sin^{4}x$
I can tell I'm going in the right direction but how should I proceed further?
EDIT ... | Uain Prosthaphaeresis Formulas, $$\cos6x+\cos 2x=2\cos4x\cos2x\iff\cos6x=\cos2x(2\cos4x-1)$$
$\cos4x=2\cos^22x-1$ and $\cos2x=2\cos^2x-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Proof/Derivation of Closed form of Binomial Expression $\sum\limits_{k=0}^{2n}(-1)^k\binom{2n}{k}^2$ The binomial expression given as follows:
$$\sum_{k=0}^{2n}\left(-1\right)^{k}\binom{2n}{k}^{2}$$
results nicely into the following closed form:
$$(-1)^{n}\binom{2n}{n}$$
I wish to know how exactly is it being done? I h... | Think generating functions. The sum is the $x^n$ coefficient of
$$\sum_{k=0}^n\binom nkx^k\sum_{l=0}^n(-1)^l\binom nlx^l.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
$x^4 -ax^3 +2x^2 -bx +1$ has real root $\implies$ $a^2+b^2 \ge 8$ it is requested to show that if the quartic polynomial $f(x) \in \mathbb{R}[x]$, defined by:
$$
f(x) = x^4 -ax^3 +2x^2 -bx +1,
$$
has a real root, then
$$
a^2 +b^2 \ge 8
$$
this question was asked by @medo, then deleted a few minutes ago. however having ... | We have from Cauchy-Schwarz inequality: $((1+x^2)^2)^2=(x^4 + 2x^2 + 1)^2 = (ax^3+bx)^2 \le (a^2+b^2)(x^6+x^2) = (a^2+b^2)x^2(x^4+1)\implies a^2+b^2 \ge \dfrac{(1+x^2)^4}{x^2(x^4+1)}\ge 8 \iff (1+x^2)^4 \ge 8x^2(1+x^4) \iff (1+y)^4 \ge 8y(1+y^2), y = x^2 \ge 0$. Lastly, consider $f(y) = (1+y)^4 - 8y(1+y^2), y \ge 0\imp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2327987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Help to understand polynomial factoring I'm following some proof, but got stuck at how the factoring works. I can follow this part:
$$\begin{align*}
1^3 + 2^3 + 3^3 + \cdots + k^3 + (k+1)^3 &= \frac{k^2(k+1)^2}{4} + (k+1)^3\\
&= \frac{k^2(k+1)^2 + 4(k+1)^3}{4}\\
\end{align*}$$
The next two steps are not so clear to me ... | The first step that is troubling you is just factorization by $(k+1)^2$.
Consider the expression of the numerator:
$$k^2(k+1)^2+4(k+1)^3$$
which is equal to
$$(k+1)^2\times k^2+(k+1)^2\times4(k+1)=(k+1)^2(k^2+4(k+1))=(k+1)^2(k^2+4k+4).$$
The next step needs you to remark that
$$(k+2)^2=k^2+2\times k\times 2+2^2=k^2+4k+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2332018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
If $A$ and $B$ are matrices such that $AB^2=BA$ and $A^4=I$ then find $B^{16}$.
If $A$ and $B$ are matrices such that $AB^2=BA$ and $A^4=I$, then find $B^{16}$.
My Method:
Given $$AB^2=BA \tag{1}$$ Post multiplying with $B^2$ we get
$$AB^4=BAB^2=B^2A$$ Hence
$$AB^4=B^2A$$ Pre Multiplying with $A$ and using $(1)$ we g... | $$B^2=A^4B^2=A^3BA.$$
Thus,
$$B^4=A^3BAA^3BA=A^3B^2A=A^2BA^2.$$
Hence, $$B^8=A^2BA^2A^2BA^2=A^2B^2A^2=ABA^3$$
and from here
$$B^{16}=ABA^3ABA^3=AB^2A^3=BA^4=B$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2332199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 5,
"answer_id": 2
} |
Inequality involving the sum of square roots of distances from a fixed point in triangle to triangle sides Let $P$ be a point of a triangle $ABC$, and $d_1=dist(P,AB)$, $d_2=dist(P,BC)$, $d_3=dist(P,CA)$.
Prove that $\sqrt{d_1}+\sqrt{d_2}+\sqrt{d_3}\le \sqrt{\frac{9}{2}R}$
| I re-define $d_1$ as $d(P,BC)=d(P,a)$ and so on for the sake of symmetry.
By the Cauchy-Schwarz inequality
$$\begin{eqnarray*}\sqrt{d_1}+\sqrt{d_2}+\sqrt{d_3} &=& \frac{1}{\sqrt{a}}\sqrt{ad_1}+\frac{1}{\sqrt{b}}\sqrt{bd_2}+\frac{1}{\sqrt{c}}\sqrt{cd_3}\\ &\leq& \sqrt{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)(ad_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2332884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding minima and maxima to $f(x,y) = x^2 + x(y^2 - 1)$ in the area $x^2 + y^2 \leq 1$ I'm asked to find minima and maxima on the function
$$f(x,y) = x^2 + x(y^2 - 1)$$
in the area
$$x^2 + y^2 \leq 1.$$
My solution:
$$\nabla (f) = (2x + y^2 - 1, 2xy)$$
Finding stationary points
$$2xy = 0$$
$$2x + y^2 - 1 = 0$$
gets... | You have a small problem. When you consider $f(x,1-x^2)$ it should be considered over the closed interval $x \in [-1,1]$, which means its endpoints must also be included into the possible list of points.
$f(-1)=2$...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2333967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Proving $\frac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} = \left(\tan\frac{x}{2}\right)^2$
How do I prove this equality?
$$\frac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} = \left(\tan\frac{x}{2}\right)^2$$
I have come this far by myself:
$$\begin{array}{llll}
\dfrac{2\sin x- \sin 2x}{2\sin x+ \sin 2x} &= \dfrac{2\sin x- 2\sin x... | Hint
If we cannot use the half angle, let us use the double angle formulae letting $x=2y$ $$\frac{1-\cos(x)}{1+\cos(x)}=\frac{1-\cos(2y)}{1+\cos(2y)}$$ Simplify and when finished reset $y=\frac x2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
If $z^4 + \frac1{z^4}=47$ then find the value of $z^3+\frac1{z^3}$ If $z^4 + \dfrac {1}{z^4}=47$ then find the value of $z^3+\dfrac {1}{z^3}$
My Attempt:
$$z^4 + \dfrac {1}{z^4}=47$$
$$(z^2+\dfrac {1}{z^2})^2 - 2=47$$
$$(z^2 + \dfrac {1}{z^2})^2=49$$
$$z^2 + \dfrac {1}{z^2}=7$$
How do I proceed further??
| Use the same trick again to get
\begin{eqnarray*}
(z+\frac{1}{z})^2-2=7 \\
z+\frac{1}{z}=3
\end{eqnarray*}
\begin{eqnarray*}
(z+\frac{1}{z})(z^2+\frac{1}{z^2})=(z^3+\frac{1}{z^3})+(z+\frac{1}{z})=21 \\
z^3+\frac{1}{z^3}=\color{red}{18}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 3
} |
calculating a limit using taylor series $$\lim_{x\rightarrow0}\frac{\cos(x) \sin(x)-x}{\sin^3 x}$$
I tried to use the identity $\cos(x) \sin(x)=\frac{1}{2}\sin(2x)$
and then use the taylor $\lim_{x\rightarrow0}\frac{x+\alpha(x)}{x^3+\beta(x)}$ where $\alpha(x), \beta(x)$ are the remainders which leads me to infinity bu... | $$\lim _{x\rightarrow 0} \frac {\cos x \sin x -x }{\sin^3 x} =\lim _{x\rightarrow 0} \frac { \sin 2x -2x }{2\sin^3 x} =\lim_{x\rightarrow 0} \frac { 2x-\frac {8x^3}{3!} +o(x^3) -2x }{ 2\sin^3 x} =\\
=\lim_{x\rightarrow 0} \frac {-\frac {8x^3}{3!} +o(x^3)}{2\sin^3 x} =\lim_{x\rightarrow 0} -\frac {8x^3}{12\sin^3 x} +\li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2336091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Given triangle $ABC$ with $CE=6 cm $ , $BD=9 cm $ , $\angle A = 60$ , CE , BD are medians. Find the area of $ ABC $
Given triangle $ABC$ with $CE=6 cm $ , $BD=9 cm $ , $\angle A = 60$ , CE , BD are medians.
Find the area of $ ABC $
This problem is too difficult for me , the teacher said it is a challenge pr... | I tried to find an elegant geometric solution but couldn't. Here is the gung-ho algebraic version.
First by cosine rule about $A$ in $\Delta ADB$, and $\Delta AEC$, we get $$b^2 + c^2/4 -bc/2 = 36$$ and $$b^2/4 + c^2 -bc/2 = 81$$ From these we get by subtracting $$c^2=b^2 + 60 $$ Substituting this in the first equati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2336445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Proof $13 \mid (k\cdot 2^n+1)$ if $n\equiv2 \pmod{12}$ and $k\equiv3 \pmod{13}$?
Proof $13 \mid (k \cdot 2^n + 1)$ if $n\equiv2 \pmod{12}$ and $k\equiv3 \pmod{13}$
Hint: for $k$ odd: $2^n \equiv-k' \pmod p$ and $kk' \equiv1 \pmod p$
My thoughts:
$13\mid(k-3) \Rightarrow k=13a+3$
and
$12|(n-2) \Rightarrow n=12b+2$
s... | ${\rm mod}\ 13\!:\!\! \overbrace{\color{#c00}{ 2^{\large n}}\!\equiv 2^{\large 2+12j}}^{\large\quad\ \ \color{#c00}n\ \equiv\ 2\pmod{\!12}}\!\!\!\!\!\!\!\equiv 2^{\large 2}\color{#0a0}{(2^{\large 12})}^{\large j}\!\equiv 4\color{#0a0}{(1)}^{\large j}\!\equiv \color{#c00}4\,\ $ by $\ \rm\color{#0a0}{ Fermat}$
therefore ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2337268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Expand $\frac{1}{z-2}$ on $|z|>2$
Expand $\frac{1}{z-2}$ around $|z|>2$
So there is no singularity points
$$\frac{1}{(z-2)}=-\frac{1}{2-z}=-\frac{1}{2}\frac{1}{1-\frac{z}{2}}=-\frac{1}{2}\sum_{n=0}^{\infty} \frac{z^n}{2^n}=\sum_{n=0}^{\infty}-\frac{z^n}{2^{n+1}}$$
but we can also write:
$$\frac{1}{(z-2)}=\frac{1}{z}... | Yes, you are on the right track. The Laurent expansion of $\frac{1}{z-a}$ in $|z|>R\geq |a|$ is,
$$\frac{1}{z-a}=\frac{1}{z}\cdot \frac{1}{1-\frac{a}{z}}=\frac{1}{z}\sum_{n=0}^{\infty} \frac{a^n}{z^n}=\sum_{n=0}^{\infty} \frac{a^n}{z^{n+1}}$$
Hence, for $|z|>R=2$,
$$\frac{1}{(z-1)(z-2)}=\frac{1}{z-2}-\frac{1}{z-1}=\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
If the roots of an equation are $a,b,c$ then find the equation having roots $\frac{1-a}{a},\frac{1-b}{b},\frac{1-c}{c}$. Actually I have come to know a technique of solving this kind of problem but it's not exactly producing when used in a certain problem.
Say we have an equation
$$2x^3+3x^2-x+1=0$$
the roots of this ... | First consider the polynomial with roots $a,b,c$:
$$g(x)=(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+ac+bc)x-abc$$
and let
$$r=a+b+c \quad\quad s=ab+ac+bc \quad\quad t=abc$$
And we accordingly write
$$g(x)=x^3-rx^2+sx-t$$
Now we multiply the linear factors of the expression having the desired roots:
$$f(x)=\Big(x-\frac{1-a}{a}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Solve an indefinite integral Find the value of
$$ \int\frac{\sqrt{1+x^2}}{1-x^2}dx $$
My Attempt: I tried to arrange the numerator as follows,
$$ \sqrt{1+x^2} = \sqrt{1-x^2 + 2x^2} $$
but that didn't help me.
Any help will be appreciated.
| Hint 1: Using $x=\tan(u)$, we get
$$
\begin{align}
\int\frac{\sqrt{1+x^2}}{1-x^2}\,\mathrm{d}x
&=\int\frac{\sec^3(u)}{1-\tan^2(u)}\,\mathrm{d}u\\
&=\int\frac{\sec(u)}{\cos^2(u)-\sin^2(u)}\,\mathrm{d}u\\
&=\int\frac{\sec^2(u)}{1-2\sin^2(u)}\,\mathrm{d}\sin(u)\\
&=\int\frac{\mathrm{d}\sin(u)}{\left(1-2\sin^2(u)\right)\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Find $\arctan(\sqrt{2})-\arctan\left(\frac{1}{\sqrt{2}}\right)$ I did it as follows: $$\arctan(\sqrt{2})-\arctan\left(\frac{1}{\sqrt{2}}\right)=\tan\Bigg(\arctan(\sqrt{2})-\arctan\left(\frac{1}{\sqrt{2}}\right)\Bigg)=\frac{\sqrt{2}-\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}\sqrt{2}}=\frac{\sqrt{2}}{4}.$$ But there is no ... | $$\tan[\arctan(\sqrt{2})-\arctan\left(\frac{1}{\sqrt{2}}\right)]=\frac{\sqrt{2}-1/\sqrt{2}}{1+1}=\frac{\sqrt{2}}{4},$$
on simplifying we see that
$$\arctan(\sqrt{2})-\arctan\left(\frac{1}{\sqrt{2}}\right) =\arctan(\frac{\sqrt{2}}{4})= \sin^{-1} \frac{1}{3}.$$
BTW, this identity suggests a Ruler & Compass construction t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2342039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to solve$\int_0^a\frac{(a-x)^{n-1}}{(a+x)^{n+1}}\, dx$? How can I solve the following integral:
$$\int_0^a\frac{(a-x)^{n-1}}{(a+x)^{n+1}}\, dx.$$ I'm not posting my work because I don't even know where to start with this integral.
| Integrating by parts:
$$I_n=\int_0^a\frac{(a-x)^{n-1}}{(a+x)^{n+1}}\, dx = $$
$$(a-x)^{n-1}\cdot \left(-\frac{1}{n(a+x)^n}\right) \bigg |_0^a-\int_0^a \left(-\frac{1}{n(a+x)^n}\right)\cdot (-1)(n-1)(a-x)^{n-2}dx=$$
$$\frac{1}{na}-\frac{n-1}{n} \cdot I_{n-1}, \ \ \ I_1=\frac{1}{2a}.$$
$$I_2=\frac{1}{2a}-\frac{1}{2}I_1=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2346096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Algebraic identities Given that $$a+b+c=2$$ and $$ab+bc+ca=1$$
Then the value of
$$(a+b)^2+(b+c)^2+(c+a)^2$$ is how much?
Attempt:
Tried expanding the expression. Thought the expanded version would contain a term from the expression of $a^3+b^3+c^3-3abc$, but its not the case.
| If you expand you find its $a^2+b^2+c^2+2ab+2bc+2xa+a^2+b^2+c^2$. Now square first equation and we have $a^2+b^2+c^2+2ab+2bc+2ac=4. ..(3)$ thus now we want value of $4+a^2+b^2+c^2$ .rearranging $3$ and using your equation $2$ we have $a^2+b^2+c^2+2 (ab+bc+ac)=4$ thus $a^2+b^2+c^2+2=4$ thus value of expression we want i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2346162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 4
} |
How to recognize a sum as a Riemann Sum Evaluate $$\frac{1}{1}+\frac{1}{2}-\frac{2}{3}+\frac{1}{4}+\frac{1}{5}-\frac{2}{6}+\frac{1}{7}+\frac{1}{8}-\frac{2}{9}+\cdots+\frac{1}{3n+1}+\frac{1}{3n+2}-\frac{2}{3n+3}+\cdots$$
answer choices:
a) $\ln 2$
b) $\ln 3$
c) $e^2$
d) $\dfrac 9 {25}$
Looking at the answer choices, I a... | Since $-\frac{2}{3n+3} = \frac{1}{3n+3} - \frac{1}{n+1}$, the general term can be written
$$a_n = \left(\frac{1}{3n+1} + \frac{1}{3n+2} + \frac{1}{3n+3}\right) - \frac{1}{n+1}$$
so the sum (up to the $n$'th term) is the difference between two harmonic numbers
$$\left(1 + \frac{1}{2}+\ldots + \frac{1}{3n+3}\right) - \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2347522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
How many such three digit numbers are possible? Certain $3$-digit numbers have the following characteristics:
*
*all three digits are different
*the number is divisible by $7$
*the number on reversing the digits is also divisible by $7$
How many such numbers are there? I have tried using a brute force approach and... | If the number is $10^2a+10b+c$ as digits then considering $(10^2a+10b+c) - (10^2c+10b+a)$ leads to realising $99(a-c)$ must be divisible by $7$ and cannot be zero and, since $99$ is coprime to $7$.
*
*$1b8$ but $108 \equiv 3 \pmod 7$ requiring $10b \equiv 4 \pmod 7$ making the only possible value of $b=6$, which is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to obtain the sum of the following series? $\sum_{n=1}^\infty{\frac{n^2}{2^n}}$ It seems that I'm missing something about this.
First of all, the series is convergent: $\lim_{n\rightarrow\infty}\frac{2^{-n-1} (n+1)^2}{2^{-n} n^2}=\frac{1}{2}$ (ratio test)
What I tried to do is to find a limit of a partial sum $\li... | Note that
$$
n^2=\binom{n+1}{2}+\binom{n}{2}\tag{1}.
$$
Also
$$
\frac{1}{(1-x)^k}=\sum_{n=0}^\infty \binom{k+n-1}{k-1}x^n\tag{2}.
$$
for $k\geq 1$ (by repeatedly differentiating the geometric series or by the extended binomial theorem). In particular
$$
\frac{1}{(1-x)^3}=\sum_{n=0}^\infty \binom{n+2}{2}x^n\tag{3}.
$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
Simplifying the Solution to the Cubic I am trying to solve the cubic. I currently have that, for $ax^3+bx^2+cx+d=0$, a substitution to make this monic. Dividing by $a$ gives
$$x^3+Bx^2+Cx+D=0$$
where $B=\frac{b}{a}, C=\frac{c}{a}, D=\frac{d}{a}$. Then, with the substitution $x=y-\frac{B}{3}$, I got
$$y^3+\left(C-\fr... | $m=\sqrt[3]{\frac{-q\pm\sqrt{q^2-\frac{4p^2}{27}}}{2}}$
Lets choose the positive root for m.
$m=\sqrt[3]{\frac{-q + \sqrt{q^2-\frac{4p^2}{27}}}{2}}$
and we know that this solves:
$q=-(m^3+n^3)$
So lets plug it into $m^3$
$q=\frac{q - \sqrt{q^2-\frac{4p^2}{27}}}{2} - n^3$
$n^3 = \frac{-q - \sqrt{q^2-\frac{4p^2}{27}}}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\pi^2/8 = 1 + 1/3^2 + 1/5^2 + 1/7^2 + \cdots$ Attempt: I found the Fourier series for $f(x) = \begin{cases} 0,& -\pi < x < 0 \\ x/2,& 0 < x < \pi \end{cases}$
a) $a_0 = \frac{1}{2\pi}\int_0^{\pi} r\,dr = \pi/4$
$a_n = \frac{1}{2\pi}\int_0^r \frac{r\cos(nr)}{2}dr = \frac{(-1)^n - 1}{2\pi n^2}$
$b_n = \frac... | You can prove $$\sum \frac{1}{n^2}=\frac{\pi^2}{6}$$ using Fourier series.
Hence, $$\sum\frac{1}{(2n)^2}+\sum\frac{1}{(2n+1)^2}=\frac{\pi^2}{6}$$
Therefore,
$$\frac14 \sum \frac{1}{n^2}+\sum\frac{1}{(2n+1)^2}=\frac{\pi^2}{6}$$
This shows $$\sum\frac{1}{(2n+1)^2}=\frac{\pi^2}{6}-\frac{\pi^2}{24}=\frac{\pi^2}{8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
What is the value of the expression $\sin\frac{2\pi}{7}\sin\frac{4\pi}{7}+\sin\frac{4\pi}{7}\sin\frac{8\pi}{7}+\sin\frac{8\pi}{7}\sin\frac{2\pi}{7}$? This is rather a simple problem that I'm posting ; looking forward not for the solution of it but the different ways it could be solved.
What is the value of
$\sin\frac{... | As you observed,
$$\sin\alpha\sin\beta = \frac{1}{2}[\cos(\alpha - \beta) - \cos(\alpha + \beta)]$$
Applying that observation here yields
\begin{align*}
\sin\left(\frac{2\pi}{7}\right)&\sin\left(\frac{4\pi}{7}\right) + \sin\left(\frac{4\pi}{7}\right)\sin\left(\frac{8\pi}{7}\right) + \sin\left(\frac{8\pi}{7}\right)\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2352527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
What is the probability of the sum of four dice being 22? Question
Four fair six-sided dice are rolled. The probability that the sum of the results being $22$ is $$\frac{X}{1296}.$$ What is the value of $X$?
My Approach
I simplified it to the equation of the form:
$x_{1}+x_{2}+x_{3}+x_{4}=22, 1\,\,\leq x_{i} \,\,\le... | The bad combinations criteria is atleast one $x_i \geq 7$.
The number of bad combinations when:
*
*One of $x_i$ is forced to be greater than or equal to $7$ is $\binom{4}{1}\binom{12 + 4 - 1}{12} = 1820$.
*Two of $x_i$'s are forced to be greater than or equal to $7$ is $\binom{4}{2}\binom{6+4-1}{6} = 504$.
*Three ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2352721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 11,
"answer_id": 8
} |
Roots of $abc^2x^2 + 3a^2 c x + b^2cx-6a^2 -ab +2b^2 = 0$ are rational We have to show that roots of $$abc^2x^2 + (3a^2 c + b^2c)x-6a^2 -ab +2b^2 = 0$$ are rational.
This can be possible if the discriminant is a perfect square. SO I tried converting it into perfect square but failed:
$$\text{Discriminant}=(3a^2c+b^2c... | Starting with your discriminant:
$$
c^2(9a^4+b^4+10a^2b^2+24a^3b-8ab^3)
$$
we see that we can easily ignore the $c^2$ since that is already a square. Now, we look at the degrees of all the terms and notice that they are all degree $4$. Therefore, the factorization (if it exists) has to be of the form
$$
(9a^4+b^4+10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2354744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
If $a,b,c$ are distinct positive numbers , show that $\frac{a^8 + b^8 + c^8}{a^3 b^3 c^3} > \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ If $a,b,c$ are distinct positive numbers , show that $$\frac{a^8 + b^8 + c^8}{a^3 b^3 c^3} > \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.$$ I am thinking of Tchebycheff's inequality for this ... | We need to prove that
$$\sum_{cyc}(a^8-a^3b^3c^2)\geq0$$ or
$$\sum_{cyc}(2a^8-2a^3b^3c^2)\geq0$$ or
$$\sum_{cyc}(a^8-a^6b^2-a^2b^6+b^6+a^6c^2-2a^3b^3c^2+b^6c^2)\geq0$$ or
$$\sum_{cyc}((a^2-b^2)^2(a^4+a^2b^2+b^4)+c^2(a^3-b^3)^2)\geq0,$$
which is obvious.
More way.
By AM-GM:
$$\sum_{cyc}a^8=\frac{1}{8}\sum_{cyc}(3a^8+3b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2355748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Evaluate the given limit: $\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$
Evaluate the following limit.
$$\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$$
My Attempt:
$$=\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx})$$
$$=\lim_{x\to \infty} (\sqrt {x-a} - \sqrt {bx}) \times \dfrac {\sqrt {x-a} + \sqrt {bx}}{\sqrt {x-... | Alternatively:
$$\lim_\limits{x\to\infty} \sqrt{x-a}-\sqrt{bx}=\lim_\limits{x\to\infty} \sqrt{x}-\sqrt{bx}=(1-\sqrt{b})\lim_\limits{x\to\infty} \sqrt{x}=\begin{cases} 0, \ if \ b=1 \\ \infty, \ if \ b\ne 1. \end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2362432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Simplifying $\frac{16^{x+1}+20\ (4^{2x})}{2^{x-3}8^{x+2}}$ I'm struggling trying to simplify $$\frac{16^{x+1}+20\ (4^{2x})}{2^{x-3}8^{x+2}}$$
Here's my attempt:
All of these coefficients can be converted to base $2$ exponentials.
$$\frac{2^{4^{(x+1)}}+20^\ (2^{2^{(2x)}})}{2^{x-3}2^{3^{(x+2)}}}$$
Doing some algebra:
$$\... | why not use base 16? You have
$$
\frac{16^{x+1}+20\ (4^{2x})}{2^{x-3}8^{x+2}} = \\
\frac{16 \cdot 16^{x}+20 \cdot 16^{x}}{16^{x} 2^{-3}8^{2}} = \\
\frac{36}{8} = \frac{9}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Double Series equals Gamma function I want to show $$\sum_{n,m=0}^\infty \frac{\Gamma(n+m+3x)}{\Gamma(n+1+x)\Gamma(m+1+x)}\cdot \frac{1}{3^{n+m+3x-1}} = \Gamma(x)$$ for $x>0$ and I don't know how.
| All credit goes to @Nemo. As he pointed out in mathoverflow.net, \begin{align*}
&\sum\limits_{n,m=0}^\infty \frac{\Gamma(n+m+3x)}{\Gamma(n+1+x)\Gamma(m+1+x)} \frac{1}{3^{n+m+3x-1}} \\
&= \int\limits_0^\infty t^{n+m+3x-1}e^{-t} \sum\limits_{n,m=0}^\infty \frac{1}{\Gamma(n+1+x)\Gamma(m+1+x)} \frac{1}{3^{n+m+3x-1}}\\
&\ov... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2364283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Finding solutions for expression to be a perfect square Determine all integers such that $$ n^4- n^2 +64$$ is the square of an integer.
The first two lines of the solution given in the textbook is as below:
Since $n^4-n^2+64>(n^2-1)^2. $ For some non negative integer $k$,
$n^4-n^2+64=(n^2+k)^2$.
I fail to understa... | This is a brute force approach.
The square previous to $\left(n^2\right)^2=n^4$ is $\left(n^2-1\right)^2=n^4-2n^2+1$. Since it is impossible for
$$
n^4-n^2+64\le n^4-2n^2+1\implies n^2\le-63
$$
we are left with
$$
n^4-n^2+64=n^4\implies\color{#C00}{n^2=64}
$$
or
$$
n^4-n^2+64=n^4+2n^2+1\implies n^2=21
$$
or
$$
n^4-n^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2369252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 5
} |
Solving the the Integral: $\int\frac{1}{u^2-2}du$ I am attempting to solve the following integral, in which I am not sure, how to proceed:
$\int\frac{1}{u^2-2}du$
I have tried the following, but I am not sure if it is correct:
$y = u^2-2$
$dy = 2u*du$
$du = \frac{dy}{2u}$
Therefore:
$\int\frac{1}{u^2-1}du = \int\frac{1... | Another approach: Let $u=\sqrt2\sec t$, so $du=\sqrt2\sec t\tan t\,dt$. Then we have:
$$\begin{align}
\int\frac1{u^2-2}du &= \int\frac{\sqrt2\sec t\tan t}{2\sec^2t-2}dt\\
&= \frac1{\sqrt2}\int\frac{\sec t \tan t}{\sec^2 t-1}dt\\
&= \frac1{\sqrt2}\int\frac{\sec t\tan t}{\tan^2 t}dt\\
&= \frac1{\sqrt2}\int\frac{\sec t}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2372489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
generating function formula Suppose $\lambda$ is a young tableau, with one color boxes.
Prove the generating function,
$F(x) = \Sigma_{\lambda} x^{i} = \prod_{n \geq 1} \frac{1}{1-x^n} $, where $i$ is the number of boxes in the young tableau.
attempt:
$\Sigma_{\lambda} x^{i} = \prod_{n \geq 1} \frac{1}{1-x^n} = \... | Let $p(n,k)$ denote the number of partitions of $n$ with largest part at most $k$. Then, of the $p(n,k)$ such partitions, there are some that have a partition with part $k$ and some that don't. Those that don't have part $k$ are all precisely the partitions of $n$ with largest part at most $k-1$, so there are $p(n,k-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2372759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is an intuitive approach to solving $\lim_{n\rightarrow\infty}\biggl(\frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2}+\dots+\frac{n}{n^2}\biggr)$?
$$\lim_{n\rightarrow\infty}\biggl(\frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2}+\dots+\frac{n}{n^2}\biggr)$$
I managed to get the answer as $1$ by standard methods of so... | Intuition should say:
the denominator grows with $n^2$, the numerator grows with $n$. However, the number of fractions also grows by $n$, so the total growth of the numerator is about $n^2$.
And that's where intuition stops. From here on, you go with logic and rigor, not intuition.
And it gets you to
$$\frac1{n^2} + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2373357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 11,
"answer_id": 0
} |
if $x+y+z=1$ then find the maximum of the form
Let $x>0$, $y>0$ and $z>0$ such that $x+y+z=1$. Find the maximal value of
$$x\sqrt{y}+y\sqrt{z}$$
I think $x=y=\dfrac{4}{9},z=\dfrac{1}{9}$ then the maximum $\dfrac{4}{9}$,but how to use AM-GM prove it?
| By AM-GM
$$x\sqrt{y}+y\sqrt{z}=\sqrt{y}\left(x+2\sqrt{\frac{y}{4}\cdot{z}}\right)\leq\sqrt{y}\left(x+\frac{y}{4}+z\right)=$$
$$=\sqrt{y}\left(1-y+\frac{y}{4}\right)=\frac{4}{9}-\frac{1}{36}\left(\left(3\sqrt{y}\right)^3+16-36\sqrt{y}\right)\leq$$
$$\leq\frac{4}{9}-\frac{1}{36}\left(3\sqrt[3]{\left(3\sqrt{y}\right)^3\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2374389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$x^2 + y^2 = 10 , \,\, \frac{1}{x} + \frac{1}{y} = \frac{4}{3}$ I couldn't solve the above equation. Below, I describe my attempt at solving it.
$$x^2 + y^2 = 10 \tag{1}$$
$$\frac{1}{x} + \frac{1}{y} = \frac{4}{3} \tag{2}$$
Make the denominator common in the RHS of $(2)$.
$$\frac{y + x}{xy} = \frac{4}{3} \tag{2.1}$$
Mu... | from the second equation we get
$$x+y=\frac{4}{3}xy$$ and $$x,y\ne 0$$
solving this equation for $y$ we get $$y=\frac{x}{\frac{4}{3}x-1}$$
this can we insert in the first equation
$$x^2+\left(\frac{x}{\frac{4}{3}x-1}\right)^2=10$$
simplifying and factorizing we get
$$2\, \left( x-1 \right) \left( x-3 \right) \left( 8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2374762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"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.