Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Value of $y=\sqrt{4 + \sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+\ldots}}}}$ I was given this problem on series by a friend.
If
$$y=\sqrt{4 + \sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+\ldots}}}}$$
then how to solve such problem.
I don’t want the full answer, rather, insights, mathematical facts, theorems, and relationships that would h... | Too long for a comment, so
$\begin{array}\\
y
&=\sqrt{4 + \sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+...............}}}}\\
&=2\sqrt{1 + \frac14\sqrt[3]{4+\sqrt[4]{4+\sqrt[5]{4+...............}}}}\\
&=2\sqrt{1 + \frac14\sqrt[3]{4}\sqrt[3]{1+\frac1{\sqrt[3]{4}}\sqrt[4]{4+\sqrt[5]{4+...............}}}}\\
&=2\sqrt{1 + \frac1{4^{2/3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1843974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 0
} |
To prove the given inequality Question:-
If $a,b,c$ are positive real numbers which are in H.P. show that $$\dfrac{a+b}{2a-b} + \dfrac{c+b}{2c-b} \ge 4$$
Attempt at a solution:-
I tried it by AM-GM inequality, but got stuck at a step. My attempt was as follows:-
$$\dfrac{ \dfrac{a+b}{2a-b} + \dfrac{c+b}{2c-b}}{2} \ge ... | So, why don't we rewrite the HP condition in a nicer way, as
$$\frac{2}{b} = \frac{1}{a} + \frac{1}{c}$$
and let's multiply the numerator and denominator of the first fraction by $2/b$. We get
$$ \frac{a+b}{2a-b} = \frac{a(1/a+1/c)+2}{2a(1/a+1/c)-2} = \frac{3+a/c}{2a/c}=\frac{3c+a}{2a}$$
Similarly, the other one becom... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1844773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solve the system for the given parameter a \begin{align}
ax+y+z&=1,\\
2x+2ay+2z&=3\\
x+y+az&=1
\end{align}
I tried forming the system matrix and discuss it using its rank, but I'm not sure how to row reduce:
$$\begin{pmatrix}a&1&1&1\\ 2&2a&2&3\\ 1&1&a&1\end{pmatrix}$$
| Consider the matrix $$A=\begin{pmatrix}
a&1&1\\2&2a&2\\1&1&a
\end{pmatrix}.$$ We have that $\det(A)=2(a-1)^2(a+2)$. Hence if $a\neq 1$ and $a\neq -2$, then $A$ is invertible and the solution of the system of equations is $A^{-1}b$, where $b=\begin{pmatrix}1\\3\\1
\end{pmatrix}$.
If $a=1$, then the system of equations i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1845651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Constant such that $\max\left(\frac{5}{5-3c},\frac{5b}{5-3d}\right)\geq k\cdot\frac{2+3b}{5-c-2d}$ What is the greatest constant $k>0$ such that
$$\max\left(\frac{5}{5-3c},\frac{5b}{5-3d}\right)\geq k\cdot\frac{2+3b}{5-c-2d}$$
for all $0\leq b\leq 1$ and $0\leq c\leq d\leq 1$?
The right-hand side looks like a weighted... | The original problem is equivalent to finding the minimum value of
$$
\max (\frac{5}{5-3c}, \frac{5b}{5-3d}) \cdot \frac{5-c-2d}{2+3b}
$$
in the region $\{(b, c, d) \mid 0 \leq b \leq 1 \wedge 0 \leq c \leq d \leq 1\}$. We consider two cases below, namely,
*
*Case 1: $\frac{5}{5-3c} \leq \frac{5b}{5-3d}$;
*Case 2: ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1848959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Find the value of $a$ for which v is in the set H = span
Find the value of $a$ for which $ v = \begin{bmatrix}10\\-6\\6\\a\end{bmatrix}$ is in the set $H = span \{ \begin{bmatrix}-5\\2\\-5\\-3\end{bmatrix}, \begin{bmatrix}0\\-2\\-1\\1\end{bmatrix}, \begin{bmatrix}0\\0\\-3\\-1\end{bmatrix}\}$
$a = ?$
I found:
$10 = -... | Note that the augmented system
$$
\left[
\begin{array}{rrr|r}
-5 & 0 & 0 & 10 \\
2 & -2 & 0 & -6 \\
-5 & -1 & -3 & 6 \\
-3 & 1 & -1 & a
\end{array}
\right]
$$
can be row-reduced with the following steps:
*
*scale row 1 by $-1/5$
*add $-2$ times row 1 to row 2
*add $5$ times row 1 to row 3
*add $3$ times row 1 to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1852746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find $\sin \theta $ in the equation $8\sin\theta = 4 + \cos\theta$ Find $\sin\theta$ in the following trigonometric equation
$8\sin\theta = 4 + \cos\theta$
My try ->
$8\sin\theta = 4 + \cos\theta$
[Squaring Both the Sides]
=> $64\sin^{2}\theta = 16 + 8\cos\theta + \cos^{2}\theta$
=> $64\sin^{2}\theta - \cos^{2}\theta= ... | $$8\sin(\theta)- \cos(\theta) =4$$
Dividing by $\sqrt{1^2 +8^2}$
$$\frac{8\sin(\theta)}{\sqrt{65}}- \frac{\cos(\theta)}{\sqrt{65}} =\frac{4}{\sqrt{65}}$$
Let $\sin(\alpha) = \frac{1}{\sqrt{65}}$, we $\cos(\alpha) = \frac{8}{\sqrt{65}}$.
Thus
$$\sin(\theta)\cos(\alpha)-\cos(\theta)\sin(\alpha) = \frac{4}{\sqrt{65}}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Integrate $\int \frac{x^2-2}{(x^2+2)^3}dx$ $$\int \frac{x^2-2}{(x^2+2)^3}dx$$
What I did :
Method $(1) $ Re writing $x^2-2 = (x^2+2)-4 $ and partial fractions.
Method $(2) $ Substituting $x^2 = 2\tan^2 \theta $
Is there any other easy methods ?
Some substitution ?
| Another approach:
Let:
$$I(c)=\int \frac{x^2-2}{x^2+c} dx$$
$$=\int \frac{x^2+c-c-2}{x^2+c} dx$$
The above step is can be avoided by long division, if it you see it as coming out of the blue. Anyways continuing:
$$=\int \left(1-\frac{c+2}{x^2+c} \right) dx$$
$$=x-\frac{(c+2) \arctan (\frac{x}{\sqrt{c}})}{\sqrt{c}}+C_1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Circle Geometry Question 1) In triangle $ABC$, $AB = 10$, $AC = 8$, and $BC = 6$. Let $P$ be the point on the circumcircle of triangle $ABC$ so that $\angle PCA = 45^\circ$. Find $CP$.
Diagram(1)
2) Let $B$, $C$, and $D$ be points on a circle. Let $\overline{BC}$ and the tangent to the circle at $D$ intersect at $A$. I... | 1) Since the angle $\angle PCA = 45^\circ$ and the $\triangle ACB$ is clearly Pythagorean (double of the minimal of sides $3,4,5$) the diameter of the circle is equal to
$\overline{AB}$ and because the angle $\angle APB$ subtends an arc of $90^\circ$ the quadrilatere $ACBP$ is a rectangle. Hence $$\overline{CP}=\overl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1854369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integrate $\int\sin^2x\cos4x\,dx$ I'm having a difficult time solving this integral.
I tried integrating by parts:
$\int\sin^2x\cos4x\,dx$
$u=\sin^2x$, $dv=\cos4x\,dx$
I used the power reducing formula to evaluate $\sin^2x$
$du = 2\sin x\cos x\,dx$, $v=1/4\sin4x$
$uv - \int\ v\,du$
$\dfrac{1}{4}\sin^2x\sin4x - \dfrac{... | Integrals of this sort can be done using some useful identities that are immediate corollaries of the angle-sum formulas $\cos A \cos B-\sin A \sin B=\cos (A+B)$ and $\sin A \cos B+\cos A \sin B=\sin (A+B),$ in conjunction with $\sin (-A)=-\sin A$ and $\cos (-A)=\cos A.$
We have $\cos (A+B)=\cos A \cos B -\sin A \si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1858641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Solving a system of linear congruences
Find all positive integer solutions to \begin{align*}x &\equiv -1 \pmod{n} \\ x&\equiv 1 \pmod{n-1}. \end{align*}
I rewrote the system as $x = nk_1-1$ and $x = (n-1)k_2+1$. Thus, we have $nk_1-1 = (n-1)k_2+1$ and so $n(k_1-k_2) = 2-k_2 \implies n = \frac{2-k_2}{k_1-k_2}$. How do... | As $n,n-1$ are relatively prime the Chinese Remainder Theorem guarantees a unique solution $\pmod {n(n-1)}$ . Indeed we have, $$x\equiv 2n-1 \pmod {n(n-1)}$$
To see this, solve the first congruence to get $x=-1+mn$. Substituting that into the second gives $$-1+mn\equiv 1 \pmod {n-1}\implies mn\equiv 2 \pmod {n-1}\imp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1858745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Problem calculating the sine of a matrix Given the matrix $A=\begin{pmatrix}-\frac{3\pi}{4} & \frac{\pi}{2}\\\frac{\pi}{2}&0\end{pmatrix}$, I want to calculate the sine $\sin(A)$.
I do so by diagonalizing A and plugging it in the power series of the sine:
\begin{align}
\sin (A) = \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!... | The problem is that Wolfram Alpha interprets "sin(A)" for a matrix A (or array of however many dimensions, or list of list of lists, or what have you) as meaning simply the result of applying sin component-wise.
This is not what you intended, and you did your intended calculation perfectly fine.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Number of ways of getting same number when four people throw a die once Four people are rolling a die once. How many ways
*
*None of them get same number
*Exactly two of them get same number
*Two of them get the same number
*Three of them get same number
*All of them get same number
Solution:
*
*None of th... | Your answers to the first two questions are correct, as is your answer to the last question.
If we interpret the third question to mean at least two of them get the same number, then we can compute the answer by subtracting the number of cases in which none of them get the same number from the total number of outcomes.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1861960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I find the derivative of $(1 +1/x)^x $ I tried one approach but the correction in the book shows me a total different answer.
Here's what I did:
$(1+ 1/x)^x=xln(1+1/x)$
Thus, now we try to find the derivative of a multiplication:
$ u(x)=x$
$(u(x))'=1$
$v(x)=ln(1+1/x)$
$(v(x))'= -1/(x^2) +1/x$
And so:
$(uv)'=u'... | Let $f(x) = (1+1/x)^x$. Then $$\ln(f(x)) = x \ln(1 + 1/x) = x \ln( (x+1)/x) = x \ln(x+1) - x \ln(x).$$
Taking derivatives on both sides gives:
$$\frac{f'(x)}{f(x)} = \ln(x+1) + \frac{x}{x+1} - \ln(x) - 1.$$
We wish to find $f'(x)$, so we solve for it and subsitute the value of $f(x)$.
$$f'(x) = \left(1 + \frac{1}{x} \r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding $S$, $D$, and $S^{-1}$ such that $A = SDS^{-1}$
Let $A = \begin{bmatrix}18&12\\-40&-26\end{bmatrix}$Find $S$, $D$, and $S^{-1}$ such that $A = SDS^{-1}$
So I did $\det(A-\lambda I)$ to get the char. poly. eqn. and got eigenvalues $\lambda_1 = -6, \lambda_2 = -2$ then:
I did $(A - \lambda_1 I)$ to get $\begi... | The characteristic polynomial of $A=\left[\begin{array}{rr}
18 & 12 \\
-40 & -26
\end{array}\right]$ is
$$
\chi_A(t)=\det(tI-A)=\det\left[\begin{array}{rr}
t - 18 & -12 \\
40 & t + 26
\end{array}\right]=t^{2} + 8t + 12=(t + 2) \cdot (t + 6)
$$
The eigenvalues of $A$ are thus $\lambda=-2$ and $\mu=-6$.
To compute a basi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
integrate $\int \frac{\sin^3x \cos^2x}{1+\sin^2x}dx$
$$\int \frac{\sin^3x \cos^2x}{1+\sin^2x}dx$$
$$\int \frac{\sin^3x \cos^2x}{1+\sin^2x}dx=\int \frac{\sin^2x \cos^2x \sin x }{1+\sin^2x}dx=\int \frac{(1-\cos^2x )\cos^2x \sin x }{1+1-\cos^2x}dx=\int \frac{(1-\cos^2x )\cos^2x \sin x }{2-\cos^2x}dx$$
$u=cosx$
$du=-... | Observe that, $$\displaystyle\int\dfrac{\sin^3 x\cos^2 x}{1+\sin^2 x}dx=\displaystyle\int\dfrac{\sin^3 x}{1+\sin^2 x}dx-\displaystyle\int\dfrac{\sin^5 x}{1+\sin^2 x}dx$$
Transformation of $\displaystyle\int\dfrac{\sin^3 x}{1+\sin^2 x}dx$
Observe that $\cos^2 x=z\implies \sin x\ dx=-\dfrac{dz}{2}$. Then the integral b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1864228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How would I calculate points on the Peano curve? The Peano curve is often given as an example of a space filling curve which maps the unit line to the unit square. So, it is a function of the form $[0,1] \rightarrow [0,1]^2$? In which case can I evaluate it for given numbers like 0.2, 0.5 and 0.7? How would I calculate... | The approach I would take is to define the curve construction by a set of affine similarity transformations, each corresponding to a mapping from the unit square to part of the curve. Write the length parameter fraction in base(number of transformations) to index a sequential compositions of transformations, and take ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1864996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
On real part of the complex number $(1+i)z^2$
Find the set of points belonging to the coordinate plane $xy$, for which the real part of the complex number $(1+i)z^2$ is positive.
My solution:-
Lets start with letting $z=r\cdot e^{i\theta}$. Then the expression $(1+i)z^2$ becomes $$\large\sqrt2\cdot|z|^2\cdot e^{{i}\... | I would find the general solutions of the inequation first:
\begin{align*}
\cos\Bigl(2\theta+\frac\pi4\Bigr)>0&\iff-\frac\pi2<2\theta+\frac\pi4<\frac\pi2\iff-\frac{3\pi}4<2\theta<\frac\pi4\color{red}{\mod2\pi}\\
&\iff-\frac{3\pi}8<\theta<\frac\pi8\color{red}{\mod\pi}
\end{align*}
Now that if conventionally, we choose $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1868205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find Area Enclosed by Curve I want to find the area enclosed by the plane curve $x^{2/3}+y^{2/3}=1$. My attempt was to set $x=\cos^3t, \ y=\sin^3t$ so:$$x^{2/3}+y^{2/3}=\cos^2t+\sin^2t=1$$
Then the area is $$2A=\oint_Cxdy-ydx=3\oint_C\cos^3ty'dy+\sin^3tx'dx=3\int_0^{2\pi}\cos^2t\cdot \sin^2tdt=\frac{3\pi}{4}\implies A=... | In simple cartesian coordinates, through the substitution $x=z^{3/2}$ and Euler's Beta function,
$$ A = \int_{0}^{1}(1-x^{2/3})^{3/2}\,dx = \frac{3}{2}\int_{0}^{1}z^{1/2}(1-z)^{3/2}\,dz=\frac{3}{2}\,B\left(\frac{3}{2},\frac{5}{2}\right)\tag{1}$$
hence:
$$ A = \frac{3\,\Gamma\left(\frac{3}{2}\right)\Gamma\left(\frac{5}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1868613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Series question involving a cubic polynomial The question asks: Consider the polynomial
$\displaystyle{\,\mathrm{f}\left(X\right) = X^{3} -6X^{2} + mX - 6}$, where $m$ is a real parameter.
a. Show that:
$\displaystyle{{1 \over x_{1}x_{2}} + {1 \over x_{1}x_{3}} + {1 \over x_{2}x_{3}} = 1}$ where $\displaystyle{x1,x2,x... | Let a,b,c the roots. Note that $\frac{1}{ab}+ \frac{1}{ac}+ \frac{1}{bc}= \frac{a+b+c}{abc}=*$.
Now, $f=(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc$.
And then $*=6/6=1$.
For b, note that $m=ab+bc+ca$
If (suposse) $n, n+1, n+2$ are the roots, their sum is 6 by a), and then $3n+3=6\to n=1$.
$m=1*2+2*3+3*1=11$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that $a_n$ is a perfect square
*
*Let $\,\,\,\left(a_{n}\right)_{\ n\ \in\ \mathbb{N}\,\,\,}$ be the sequence of integers defined recursively by
$$
a_{1} = a_{2} = 1\,,\qquad\quad a_{n + 2} = 7a_{n + 1} -a_{n} - 2\quad
\mbox{for}\quad n \geq 1
$$
*Prove that $a_{n}$ is a perfect square for every $n$.
... | A brute-force approach
$$b_{n} = 3b_{n-1} - b_{n-2}$$
Squaring on both sides,
$$b_n^2 = 9b_{n-1}^2+b_{n-2}^2 -6b_{n-1}b_{n-2} $$
Similarly,
$$b_{n-1}^2 = 9b_{n-2}^2+b_{n-3}^2 -6b_{n-2}b_{n-3} $$
Subtracting,
$$b_n^2 - b_{n-1}^2 = 9b_{n-1}^2 - 8b_{n-2}^2 - b_{n-3}^2 -6a_{n-2}(b_{n-1}-b_{n-1}) $$
Since,
$$3b_{n-2} = b_{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Solving an equation involving an integral: $\int_0^1\frac{ax+b}{(x^2+3x+2)^2}\:dx=\frac52.$ Determine a pair of number $a$ and $b$ for which
$$\int_0^1\frac{ax+b}{(x^2+3x+2)^2}\:dx=\frac52.$$
I tried putting $x$ as $1-x$ as the integral wouldn't change but could not move forward from there so can you please suggest ... | Hint. One may observe that
$$
(x^2+3x+2)=(x+1)(x+2)
$$ leading to the following partial fraction decomposition
$$
\begin{align}
\frac{ax+b}{(x^2+3x+2)^2}&=\frac{ax+b}{(x+1)^2(x+2)^2}
\\\\&=\frac{-a+b}{(x+1)^2}+\frac{3 a-2 b}{x+1}+\frac{-2 a+b}{(x+2)^2}+\frac{-3 a+2 b}{x+2}.
\end{align}
$$ Integrating each term gives
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1871337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Legendre symbol simplification I saw a simplification using the Legendre symbol which said $$\left(\dfrac{19}{29}\right) = \left(\dfrac{10}{19}\right) = \left(\dfrac{2}{19}\right) \cdot \left(\dfrac{5}{19}\right) = -1.$$ My question is how did they get $\left(\dfrac{19}{29}\right) = \left(\dfrac{10}{19}\right)$ and fin... | For the first question
$$
\left(\dfrac{19}{29}\right) = \left(\dfrac{29}{19}\right)
= \left(\dfrac{10}{19}\right)
$$
since $29$ is $1$ mod $4$.
For the second,
$$
\left(\dfrac{5}{19}\right) =
\left(\dfrac{19}{5}\right) =
\left(\dfrac{-1}{5}\right) = 1
$$
since $5$ is $1$ mod $4$
and
$$
\left(\dfrac{2}{19}\right) = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1872092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Show that $a_n$ is decreasing
$a_1 = 2, a_{n+1} = \frac{1}{3 - a_n}$ for $n \ge 2$. Show $a_n$ is decreasing.
First we need to show $a_n > 0$ for all $n$.
$a_2 = 1/2$ and $a_3 = 2/5$ and $a_4 = 5/13$
One way we can do this is by showing $3- a_n > 0$. Thus suppose it holds for $n$ then we need to show $\frac{3(3 - a_... | Approach $\boldsymbol{1}$:
Note that
$$
\begin{align}
a_{n+1}-a_n
&=\frac1{3-a_n}-a_n\\
&=\frac{a_n^2-3a_n+1}{3-a_n}\\
&=\frac{\left(a_n-\frac32\right)^2-\frac54}{3-a_n}\tag{1}
\end{align}
$$
If $\frac{3-\sqrt5}2\le a_n\le\frac{3+\sqrt5}2$, then $(1)$ is negative and
$$
\frac{3-\sqrt5}2\le\frac1{3-a_n}\le\frac{3+\sqrt5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1872768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Show that $3^n+4^n+\dots+(n+2)^n=(n+3)^n$ has no answers for $n\ge 6$. Considering
$$3^n+4^n+\dots+(n+2)^n=(n+3)^n$$
Clearly $n=2$ and $n=3$ are solutions of this equation and this equality does not hold for $n=4$ and $n=5$.
How can I show this equation has no solutions for $n>5$.
Thanks.
| Let $n\ge5$ and suppose $$ \sum_{k=3}^{n+2}k^n<(n+3)^n.$$Then $$ \sum_{k=3}^{n+3} k^n <2(n+3)^n,$$ and we deduce \begin{align} \sum_{k=3}^{n+3} k^{n+1}<2(n+3)^{n+1}&<(n+4)^{n+1} \\ \left(1+\frac1{n+3}\right)^{n+1}\ge\left(\frac98\right)^6&>2, \end{align} where he have used the increasingness of $f(x)=\left(1+\frac1{x+3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
How to factorize the polynomial $a^6+8a^3+27$?
I would like to factorize $a^6+8a^3+27$.
I got different answers but one of the answers is
$$(a^2-a+3)(a^4+a^3-2a^2+3a+9)$$
Can someone tell me how to get this answer? Thanks.
| $$a^6+8a^3+27= (a^2)^3+3^3+(-a)^3-3\cdot a^2\cdot3(-a)$$
Now $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$
See : If $a,b,c \in R$ are distinct, then $-a^3-b^3-c^3+3abc \neq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Integration by means of partial fraction decomposition I'm trying to solve this indefinite integral by means of partial fraction decomposition:
$\int\dfrac{x+1}{\left(x^2+4x+5\right)^2}\ dx$.
The denominator has complex (but not real) roots because $\Delta<0$; so, according with my calculus book, i try to decompose the... | As suggested in the comment directly use Trigonometric substitution
As $x^2+4x+5=(x+2)^2+1^2,$ let $\arctan(x+2)=y\implies x+2=\tan y$
$$2I\int\dfrac{x+1}{(x^2+4x+5)^2}dx=\int(\tan y-1)\cos^2y\ dy$$
$$4I=2\int(\sin2y-1-\cos2y)dy=-\sin2y-2y-\cos2y+K$$
Now $\sin2y=\dfrac{2\tan y}{1+\tan^2y}=\cdots$
and $\cos2y=\dfrac{1-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find the value $\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{5 - 2{\sqrt6}}}{1+ \sqrt{6}}\right)$ The value of $$\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{5 - 2{\sqrt6}}}{1+ \sqrt{6}}\right)$$is equal to
*
*$\frac{\pi}{6}$
*$\frac{\pi}{4}$
*$\frac{\pi}{3}$
*$\frac... | We have for all $x >0$, $\tan^{-1} x + \tan^{-1} \frac1{x} = \pi/2$ (hint: take the derivative of LHS). Hence, the obtained expression is just:
$$\frac{\pi}2 - \tan^{-1} \sqrt 3 = \frac{\pi}2 - \frac{\pi}3 = \frac{\pi}6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
how can I prove this $\sum_{k=1}^{n}\frac{1}{4k^{2}-2k}=\sum_{k=n+1}^{2n}\frac{1}{k}$ How can I prove the following equation?:
$$s=\sum_{k=1}^{n}\frac{1}{4k^{2}-2k}=\sum_{k=n+1}^{2n}\frac{1}{k}$$
Simplifying both terms of the equation:
$$\sum_{k=1}^{n}\frac{1}{4k^{2}-2k} = \sum_{k=1}^{n}\frac{1}{(2k-1)2k} =(\sum_{k=1}^... | The core part of the induction argument spelled out more explicitly:
\begin{align}
\sum_{i=1}^{k+1}\frac{1}{2i(2i-1)}&= \sum_{i=1}^k\frac{1}{2i(2i-1)}+\frac{1}{2(k+1)(2k+1)}\\[1em]
&= \sum_{i=k+1}^{2k}\frac{1}{i}+\frac{1}{2(k+1)(2k+1)}\\[1em]
&= \sum_{i=k+2}^{2k+2}\frac{1}{i}+\frac{1}{k+1}-\frac{1}{2k+1}-\frac{1}{2k+2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1876148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
$\tan^{-1}x$, $\tan^{-1}y$, $\tan^{-1}z$ are in arithmetic progression, as are $x$, $y$, $z$. Show ...
$\tan^{-1}x, \tan^{-1}y, \tan^{-1}z $ are in arithmetic progression, as are $x$, $y$, $z$. (We assume $y \ne 0,1,-1$.) Show:
*
*$x$, $y$, $z$ are in geometric progression
*$x$, $y$, $z$ are in harmonic progression... | HINT:
$$\tan^{-1}z-\tan^{-1}y=\tan^{-1}y-\tan^{-1}x$$
$$\implies\tan(\tan^{-1}z-\tan^{-1}y)=\tan(\tan^{-1}y-\tan^{-1}x)$$
$$\iff\dfrac{z-y}{1+yz}=\dfrac{y-x}{1+xy}$$
As $x,y,z$ are in A.P.,$y-x=z-y$
Case$\#1:$If If $y-x=z-y=0$
Case$\#2:$ If $y-x=z-y\ne0, 1+yz=1+xy\iff y(z-x)=0$
$\implies y=0$ or $z-x=0\iff z=x$
Can you... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1876639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Inequality with triangle sides Let $a,b,c$ be the sides of a triangle. Show that:
$$(\sqrt a + \sqrt b - \sqrt c)(\sqrt a - \sqrt b + \sqrt c)(-\sqrt a + \sqrt b + \sqrt c) \ge \sqrt {(a+b-c)(a-b+c)(-a+b+c)}.$$
| Let $\sqrt{a}=x$, $\sqrt{b}=y$ and $\sqrt{c}=y$.
Hence, we need to prove that $\prod\limits_{cyc}(x+y-z)^2\geq\prod\limits_{cyc}(x^2+y^2-z^2)$.
We'll prove that the last inequality is true for all reals $x$, $y$ and $z$.
Indeed, we can assume that $\prod\limits_{cyc}(x^2+y^2-z^2)>0$ and from here
we obtain $x^2+y^2-z^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1876723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integrate $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx$
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{\sec^2x}{\sqrt[3]{\tan\ x}}dx$$
$$f(x) = (\tan \ x)^{\frac{2}{3}}, \ f'(x) = \frac{2}{3} \cdot (\tan \ x)^{-\frac{1}{3}} \cdot \sec^2x$$
$$\therefore \int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\fra... | Your solution was fine. Just needed to have put $$\frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}f'(x)dx$$ instead of $$\frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\frac{f'(x)}{f(x)}dx$$which makes your solution $$\frac{3}{2}(f(\frac{\pi}{3})-f(\frac{\pi}{4}))$$where f(x) is the function you defined, namely $$f(x) =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find $\int \frac{x^2}{(x \cos x - \sin x)(x \sin x + \cos x)}dx $ Find $$\int \frac{x^2}{(x \cos x - \sin x)(x \sin x + \cos x)}dx $$
Any hints please?
Could'nt think of any approach till now...
| $$ \frac{x^2}{(x \cos x - \sin x)(x \sin x + \cos x)} = \frac{x \cos x}{ x \sin x + \cos x } + \frac{x \sin x}{x \ cos x - \sin x} $$
Why I get this:
$$\frac{x^2}{(x \cos x - \sin x)(x \sin x + \cos x)} = \frac{A(x)}{ x \sin x + \cos x } + \frac{B(x)}{x \ cos x - \sin x} \\\ \implies A(x)( x \cos(x)- \sin(x))+ B(x) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Row and Column rotation matrices Today I started studying Rotation matrices and derived a rotation matrix like this:
\begin{equation*}
\begin{pmatrix}
x & y
\end{pmatrix}
\begin{pmatrix}
\cos(\beta) & \sin(\beta) \\
-\sin(\beta) & \cos(\beta)
\end{pmatrix}
\end{equation*}
When I was goo... | The difference is that you used a row vector and the other expression uses a column vector. And they are the same if you transpose one of them.
\begin{align}
\left(
\begin{pmatrix}
x & y
\end{pmatrix}
\begin{pmatrix}
\cos(\beta) & \sin(\beta) \\
-\sin(\beta) & \cos(\beta)
\end{pmatrix}
\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1880676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
show that $19-5\sqrt[3]{2}-8\sqrt[3]{4}$ is a unit in $\mathbb{Z}[\sqrt[3]{2}]$ I found by numerical experiment the norm of $19-5\sqrt[3]{2}-8\sqrt[3]{4}$ (result of multiplying conjugates) is: $$0.9999999999989706-4.4408920985006262 \times 10^{-16}i$$ but I am betting this is just $1$. How can I show this is a unit i... | Or,
Given integers $a,b,c,$ and cubic form
$$ f(a,b,c) = a^3+2 b^3-6 a b c+4 c^3 = \left|\begin{bmatrix} a & 2c & 2b\\b & a & 2c\\ c & b & a\end{bmatrix}\right|, $$
That is because $f(a,b,c) = \det(aI + b X + c X^2),$ where
$$ X = \begin{bmatrix} 0 & 0 & 2\\1 & 0 & 0\\ 0 & 1 & 0\end{bmatrix}. $$ Then $X^3 = 2 I$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1881822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
An inequality involving hyperbolic sine functions The following problem arised in my research work and has been challenging me for several days: Prove that for all $r\in (0,1)$ and for all $x>0$, we have $$\frac{\sinh(r(2-r)x)}{r(2-r)x} \bigg[ \frac{\sinh((1-r)x)}{(1-r)x} \bigg]^2 > \frac{\sinh(x)}{x} \frac{\sinh((1-r)... | First, let's replace your $r$ with $y=1-r$ $\qquad - \quad$ this makes the arguments a little nicer:
$$\frac{\sinh((1-y^2)x)}{(1-y^2)x} \left[ \frac{\sinh(yx)}{yx} \right]^2 > \frac{\sinh(x)}{x} \frac{\sinh(y^2x)}{y^2x}$$
Now we have the imaginary $\operatorname{sinc}$ functions here, and it makes sense to use the infi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int_{3}^{\infty}\frac{dx}{x^2-x-2}$ $$\int_{3}^{\infty}\frac{dx}{x^2-x-2}=\int_{3}^{\infty}\frac{dx}{(x-2)(x+1)}$$
$$\frac{1}{(x-2)(x+1)}=\frac{A}{(x-2)}+\frac{B}{(x+1)}$$
$$1=Ax+A+Bx-2B$$
$$1=(A+B)x+A-2B$$
$A+B=0\iff A=-B$
$-3B=1$
$B=-\frac{1}{3}$, $A=\frac{1}{3}$
$$\int_{3}^{\infty}\frac{dx}{(x-2)(x+1)}=\f... | $$\frac{1}{3}\int_{3}^{+\infty}\left(\frac{1}{x-2}-\frac{1}{x+1}\right)\,dx =\frac{1}{3}\int_{1}^{4}\frac{dz}{z}=\frac{\log 4}{3}=\color{red}{\frac{2}{3}\log 2}\tag{1}$$
since:
$$ \int_{3}^{M}\frac{dx}{x-2}=\int_{1}^{M-2}\frac{dz}{z},\qquad \int_{3}^{M}\frac{dx}{x+1}=\int_{4}^{M+1}\frac{dz}{z}\tag{2} $$
and:
$$ \left|\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Why does this trick gives precisely the formulas for the sum of the $n$ first naturals and the $n$ first squares? I've learned a cool trick several days ago. Suppose I want to find a polynomial that gives me:
$$f(1)=1, \quad f(2)=5,\quad f(3)=14,\quad f(4)=30\tag{1}$$
I could do the following: Take a polynomial of de... | This is an example of working out Ahmed's answer, which is good and contains a lot of information.
This trick works best if you start from $0$. To find the polynomial $f(x)$ with $$f(0) = 0, \; f(1) = 1, \; f(2) = 5, \; f(3) = 14, \; f(4) = 30,$$ form a triangle of differences
\begin{align*} &0\quad \quad 1 \quad\quad ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1883015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 2
} |
Laurent series of $f(z) = \frac{z^2+1}{\sin(z)}$ How does one find the first four terms of the Laurent series of $f(z) = \frac{z^2+1}{\sin(z)}$? My approach was this:
$(z^2+1) = f(z)\sin(z) = \left(\sum\limits_{n=-\infty}^\infty c_n z^n \right)\sin(z)=\left(\sum\limits_{n=-\infty}^\infty c_n z^n \right)\left(z-\frac{z^... | Your approach is correct, but without determining the order of the pole $z=0$ you are stuck because each coefficient of $f \cdot \sin$ will, in principle, be a sum with infinitely many terms. Finding the order of the pole of $f$ will allow us to kill infinitely many of these, leaving only finitely many of them, as you ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1883785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Does $2764976 + 3734045\,\sqrt[3]{7} -2707603\,\sqrt[3]{7^2} = 0$? In the process of my numerical computations I have found a very special identity:
*
*$\;\;1264483 + 1707789 \,\sqrt[3]{7} - 1238313\,\sqrt[3]{7^2} = 9.313225746154785 \times 10^{-10}$
*$
-1500493 - 2026256\,\sqrt[3]{7} + 1469290\,\sqrt[3]{7^2}
= 9.... | To expand on the comments of Fabio and Jyrki:
You're claiming that $\sqrt[3]7$ is a root of the quadratic equation
$$2764976 + 3734045x -2707603x^2 = 0$$
but that leads to a contradiction.
The solutions of a quadratic with integer coefficients can be written as
$$x = \frac{p \pm \sqrt q}{r} $$
for some integers $p, q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 0
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Evaluation of this series $\sum\limits_{n=1}^{\infty }{\frac{{{\left( -1 \right)}^{n+1}}}{4{{n}^{2}}-1}}=??$ I start first to use $u=n+1$ then
$$\sum_{u=2}^{\infty}
{\frac{{{\left( -1 \right)}^{u}}}{4{{\left( u-1 \right)}^{2}}-1}}.$$
| Note $\sum_{u=2}^{\infty}
{\frac{{{\left( -1 \right)}^{u}}}{4{{\left( u-1 \right)}^{2}}-1}}=\sum_{n=1}^{\infty}
{\frac{{{\left( -1 \right)}^{n+1}}}{4{n^{2}}-1}}$. Let
$$ f(x)=\sum_{n=1}^{\infty}
{\frac{{{\left( -1 \right)}^{n+1}}}{4n^2-1}}x^{2n+1}. $$
Then $f(1)=\sum_{n=1}^{\infty}
{\frac{{{\left( -1 \right)}^{n+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Prove a series that equals to $\frac{e}{e-1}$ Prove that
$$
\lim_{N\to\infty}\sum_{k=0}^\infty \left(1+\frac{k}{N}\right)^{-N}=\frac{e}{e-1}
$$
I think $\sum_{k=0}^\infty \left(1+\frac{k}{N}\right)^{-N}$ should be a Riemann sum of a function but could find it. What is the trick in this question?
In addition, the equat... | Note that
$$\left(1+\frac{k}{N}\right)^N = \sum\limits_{j=0}^{N}{{N\choose j}\left(\frac{k}{N}\right)^j} \ge 1 + {N\choose 1}\frac{k}{N} + {N\choose 2}\frac{k^2}{N^2} = 1 + k + \frac{N-1}{2N}k^2 \ge 1 + k + \frac{k^2}{4}$$
for $N\ge 2$. It follows that for any $M\ge 1$ we have
$$\sum\limits_{k=0}^{\infty}{\left(1+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1885755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Definite integral and limit $\lim_{n \rightarrow \infty} (n((n+1)I_{n}-\frac{\pi}{4})$ Given $I_{n} = \int_{0}^{1} x^{n} \arctan(x)dx $
Calculate:
$\lim_{n \rightarrow \infty} (n((n+1)I_{n}-\frac{\pi}{4})$
| $$
\begin{align}
I_n
&=\int_0^1x^n\arctan(x)\,\mathrm{d}x\\
&=\int_0^1\sum_{k=0}^\infty(-1)^k\frac{x^{n+2k+1}}{2k+1}\,\mathrm{d}x\\
&=\sum_{k=0}^\infty(-1)^k\frac1{(n+2k+2)(2k+1)}\\
&=\frac1{n+1}\sum_{k=0}^\infty(-1)^k\left(\frac1{2k+1}-\frac1{n+2k+2}\right)\\
&=\frac\pi{4(n+1)}-\frac1{n+1}\sum_{k=0}^\infty\frac{(-1)^k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1888859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Gauss circle problem : a simple asymptotic estimation.
Find the number of integer points lying in or inside a circle of radius $n\in \mathbb N$ centered at the origin.
The problem asks for all $(a,b)\in \mathbb Z^2$ such that $a^2+b^2\leq n^2$. Looking at such points lying strictly in the North-East quadrant, pick an... | The estimate $\displaystyle \frac 1n \sum_{k=0}^n \sqrt{1-\frac{k^2}{n^2}} = \int_0^1 \sqrt{1-t^2}dt + o(1)$ can easily be refined .
Indeed, $$\begin{align}0\leq \frac 1n \sum_{k=0}^n \sqrt{1-\frac{k^2}{n^2}} - \int_0^1 \sqrt{1-t^2}dt &= \sum_{k=0}^{n-1} \int_{k/n}^{(k+1)/n} \left(\sqrt{1-\frac{k^2}{n^2}} - \sqrt{1-t^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to solve this determinant
Question Statment:-
Show that
\begin{align*}
\begin{vmatrix}
(a+b)^2 & ca & bc \\
ca & (b+c)^2 & ab \\
bc & ab & (c+a)^2 \\
\end{vmatrix}
=2abc(a+b+c)^3
\end{align*}
My Attempt:-
$$\begin{aligned}
&\begin{vmatrix}
\\(a+b)^2 & ca & bc \\
\\ca & (b+c)^2 & ab \\
\\bc & ab & (c+a)^2 ... | Let me try. You have $$LHS = [(a+b)(b+c)(c+a)]^2 + 2(abc)^2 - \sum b^2c^2(b+c)^2.$$
Note that $(a+b)(b+c)(c+a) = \sum bc(b+c) + 2abc$.
Then, $$LHS = \left(\sum bc(b+c)\right)^2 + 6(abc)^2 + 4abc\left(\sum bc(b+c)\right) - \sum b^2c^2(b+c)^2$$
$$=2\sum a^2bc(a+b)(a+c) + 6(abc)^2 + 4abc\left(\sum bc(b+c)\right)$$
$$ =2ab... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
finding square root I want to know how to find the square root/cube root of a number ending with zero.
For instance, $\sqrt{1024}$ : though it's a perfect square, there is a shortcut. Like wise how can I find the $\sqrt{2240}$ or any number ending with zero?
Thanks in advance!
|
By $$(10a+b)^2 = a\times 100a+(10\times 2a+b)\times b$$
We can use long division,
\begin{array}{rrrr}
& & 3 & 2 \\
& & -- & -- \\
3 & \sqrt{} & 10 & 24 \\
& & 9 & \\
& & -- & -- \\
& & 1 & 24 \\
62 & & 1 & 24 \\
& ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Expectation value of trials needed to get $k$ consecutive outcomes Suppose that independent trials, each of which is equally likely to have any of $m$ possible outcome, are performed until the same outcome occurs $k$ consecutive times. If $N$ denotes the number of trials, show that
$$E[N] = \frac{m^k-1}{m-1}$$
This is ... | We use the following generating function for $k\ge 2:$
$$G(z, u) = z^k\times \sum_{q\ge 0} u^q (z+z^2+\cdots+z^{k-1})^q
\\ = z^k \sum_{q\ge 0} u^q z^q (1+z+\cdots z^{k-2})^q
\\ = z^k
\sum_{q\ge 0} u^q z^q \frac{(1-z^{k-1})^q}{(1-z)^q}
\\ = z^k \frac{1}{1-uz(1-z^{k-1})/(1-z)}
= z^k \frac{1-z}{1-z-uz(1-z^{k-1})}.$$
As a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1890121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the limit of the sequence involving integral
Let $f:[0,1] \rightarrow [0,1]$ increasing function and
$a_n=\int_{0}^{1} \frac {1+(f(x)^n}{1+(f(x))^{n+1}} dx \tag 1$
Prove $a_n$ is convergent and find the limit.
It's easy to prove $a_n \ge 1$ and $a_n$ is decreasing, therefore is convergent. By taking $f$ identic... | A different approach, not using dominated convergence:
First, I claim that $r^n - r^{n+1} \le \frac{(n-1)^{n-1}}{n^n} $ for all $r\in [0,1]$. To see this, we note that $r^n - r^{n+1} = (n-1)^{n-1}\left(\frac{r}{n-1}\right)^{n-1}(r-1)$, and by AM-GM
$$ \left(\frac{r}{n-1}\right)^{n-1}(1-r) = \left(\sqrt[n]{\underbrace{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1891377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How many possible ways of organizing the first round ($5$ matches) are there?
$10$ players play table tennis. How many possible ways of organizing the first round ($5$ matches) are there?
My attempt:
we can choose 2 players form 10 in the first match and 2 players from 8 (expect the first 2 players), and 2 players fr... | Double Factorial
Number the players from $1$ to $2n$. Player $1$ can play $2n-1$ people. For each of those choices, the next lowest unpaired player can play $2n-3$ people. For each of the choices so far, the next lowest unpaired player can play $2n-5$ people. Etc. Thus, there are
$$
\begin{align}
(2n-1)(2n-3)(2n-5)\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1895081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Solve this $6$-th degree polynomial equation: $x^6-x^5-8x^4+2x^3+21x^2-9x-54=0$ The question is as follows:
Solve the equation: $x^6-x^5-8x^4+2x^3+21x^2-9x-54=0$, one root being $\sqrt{2}+i$
It's trivial that another root is $\sqrt{2}-i$. But, I can go no further. Can anyone please help me how to solve this?
| In addition to $\sqrt{2} -i$, we have that $-\sqrt{2}+i$ and $-\sqrt{2}-i$ are also guaranteed to be roots. (These are the remaining Galois conjugates, which always take the specified forms in the case of a sum of square roots like $\sqrt{2}+i$.)
To determine the other roots, divide your given polynomial $f(x)$ by $$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1896768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
The rank of a matrix, dependent on the value of $t$ I'm trying to analyse the rank of the following matrix, for $t\in \mathbb{R}$.
$$\begin{bmatrix}
t+3 & 5 & 6 \\
-1 & t-3 & -6 \\
1 & 1 & t+4
\end{bmatrix}$$
With $R_1\leftrightarrow R_3$, $-(t+3)R_1+R_3 \rightarrow R_3$, $R_1+R_2 \rightarrow R_2$, and fina... | \begin{align}
\begin{bmatrix}
t+3 & 5 & 6 \\
-1 & t-3 & -6 \\
1 & 1 & t+4
\end{bmatrix}
&\to
\begin{bmatrix}
1 & 1 & t+4 \\
-1 & t-3 & -6 \\
t+3 & 5 & 6 \\
\end{bmatrix}
&&R_1\leftrightarrow R_3
\\
&\to
\begin{bmatrix}
1 & 1 & t+4 \\
0 & t-2 & t-2 \\
t+3 & 5 & 6 \\
\end{bmatrix}
&&R_2\ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1897312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding the limit $\lim_{n\rightarrow\infty}(1-\frac{1}{3})^2(1-\frac{1}{6})^2(1-\frac{1}{10})^2...(1-\frac{1}{n(n+1)/2})^2$ Any hints on how to find the following limit ?.
I haven't been able to figure it out still.
$$
\lim_{n \to \infty}\left(\, 1 - \frac{1}{3}\, \right)^{2}
\left(\, 1 - \frac{1}{6}\, \right)^{2}\lef... | It may be interesting to see that, using the the Gamma function in the form $$\Gamma\left(z\right)=\lim_{n\rightarrow\infty}\frac{n^{z-1}n!}{z\left(z+1\right)\cdots\left(z+n-1\right)}$$ we can prove that $$\prod_{n\geq0}\frac{\left(n+a\right)\left(n+b\right)}{\left(n+c\right)\left(n+d\right)}=\frac{\Gamma\left(c\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1897569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to compute the Pythagorean triple by one of the numbers that belonged to it? I have a positive number $n>2$. How to compute the Pythagorean triple containing $n$? $n$ may be the hypotenuse and leg.
| If $a$ is the smallest leg then you can find others in very simple way-
If $a$ is odd then -
$$b = \dfrac{n^2-1}{2};\; c = \dfrac{n^2+1}{2}\\
[\dfrac{n^2-1}{2}]^2 + n^2 = [\dfrac{n^2+1}{2}]^2$$
If $a$ is even then -
$$b = \dfrac{n^2}{4}-1;\; c = \dfrac{n^2}{4}+1\\
[\dfrac{n^2}{4}-1]^2 + n^2 = [\dfrac{n^2}{4}+1]^2$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Mathematical induction problem: $\frac12\cdot \frac34\cdots\frac{2n-1}{2n}<\frac1{\sqrt{2n}}$ This is a problem that I tried to solve and didn't come up with any ideas
.?$$\frac{1}{2}\cdot \frac{3}{4}\cdots\frac{2n-1}{2n}<\frac{1}{\sqrt{2n}}.$$
All I get is $\frac{1}{\sqrt{2n}}\cdot\frac{2n+1}{2n+2}<\frac{1}{\sqrt{2n+2... | An easier proof without induction:
Let $r$ be a positive integer, whence we have by the AM-GM inequality,
$\dfrac{(2r-1)+(2r+1)}{2}>\sqrt {(2r-1)(2r+1)}$
$\Rightarrow \dfrac{2r}{2r-1}>\sqrt {\dfrac{2r+1}{2r-1}}$
$\Rightarrow \displaystyle\prod_{r=1}^n\left( \frac{2r}{2r-1}\right)>\prod_{r=1}^n\left(\sqrt {\frac{2r+1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1899857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
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Find the value of $\tan A + \tan B$, given values of $\frac{\sin (A)}{\sin (B)}$ and $\frac{\cos (A)}{\cos (B)}$ Given
$$\frac{\sin (A)}{\sin (B)} = \frac{\sqrt{3}}{2}$$
$$\frac{\cos (A)}{\cos (B)} = \frac{\sqrt{5}}{3}$$
Find $\tan A + \tan B$.
Approach
Dividing the equations, we get the relation between $\tan A$ and $... | Something is wrong. I calculated
\begin{align}
\sin A &= \dfrac{\sqrt 3}{2}\sin B \\
\cos A &= \dfrac{\sqrt 5}{3}\cos B \\
\sin^2 A + \cos^2 A &= \dfrac 34 \sin^2 B + \dfrac59 \cos^2 B\\
1 &= \dfrac 34 - \dfrac{7}{36} \cos^2 B\\
\cos^2 B = -\dfrac{9}{7}
\end{align}
Which is impossible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1901177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Square root of a matrix. Determine all $A,B \in \mathbf{M}_{2}(\mathbf{R})$ such that $A^2+B^2=\begin{pmatrix} 22 & 44\\
14 & 28 \end{pmatrix}$ and $AB+BA=\begin{pmatrix} 10 & 20\\
2 & 4 \end{pmatrix}$.
I have tried to assume that $A=\begin{pmatrix} a & b\\
c & d \end{pmatrix}$ and $B=\begin{pmatrix} e & f\\
g ... | Here is a general approach to find all square root of a $2\times2$ matrix $A$ which is not a multiple of the identity matrix.
Assume that $X^2 = A$. If we write $t = \operatorname{tr}(A)$ and $d = \det(A)$, then by the Cayley-Hamilton theorem we have
$$ X^2 - tX + dI = 0. $$
Plugging this to $X^2 = A$, we have $A = tX ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1902453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solving the geometric series for q Is there a general way to find the $q > 0$ solving the equation from the geometric series $$1+q+q^2+q^3+\ldots + q^n = a$$ or $$\frac{1-q^{n+1}}{1-q} = a\quad\text{with } q \neq 1$$ for $a > 1$ and $n\in\mathbb N$?
My thoughts: Since polynomials aren't solvable in general for degree ... | Re-arrange
\begin{align*}
q &= \frac{a-1}{a-q^n} \\
&= \frac{a-1}{a-\left(\frac{a-1}{a-q^n}\right)^n} \\
&= \frac{a-1}
{a-\left(
\frac{a-1}
{a-\left(
\frac{a-1}{a-\ddots}
\right)^n}
\right)^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1903215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
What is the expansion $(x+a)^b$? I don't have strong math background. Whats is the expansion of the following equation: $(x+a)^b$.
| This seems a difficult problem for someone without a strong mathematical background. One way to approach this is to arrange $b$ copies of $(x+a)$, as follows:
$$
\underbrace{(x+a)(x+a) \cdots (x+a)}_{\text{$b$ copies}}
$$
The product of all of these sums $(x+a)$ will be a bunch of terms of the form $x^ka^{b-k}$, where... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1904633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Compute the limit of $1 - n \ln \left(\dfrac{2n + 1}{2n - 1}\right)$
Compute the limit: $$\lim_{n \to \infty} \left(1 - n \ln \left(\dfrac{2n + 1}{2n - 1}\right)\right) $$
Can someone help me to solve this limit? I forgot how to manupulate fractions in limit calculus.
| If you want to go beyond the limit itself, write $$1 - n \log \left(\dfrac{2n + 1}{2n - 1}\right)=1-n \log\left(1+ \frac{2}{2n - 1}\right)$$ Now, remembering that, for small values of $x$ $$\log(1+x)=x-\frac{x^2}{2}+O\left(x^3\right)$$ relace $x$ by $\frac{2}{2n - 1}$ which makes $$\log \left(\dfrac{2n + 1}{2n - 1}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1905719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Let a, b, c, x and y be integers. Why are there no solutions for $\sqrt {a^2 + b^2 + c^2} = x$, where $x=2^y$ or $x=5 \times 2^y$ Edit - An addition condition: $a$, $b$, and $c$ do not equal $0$.
I'm really digging into 3D vectors and their properties. I've decided to look and see which combinations of three integers w... | WLOG, $a,b,c,y$ are positive. If there exists a solution to $a^2+b^2+c^2\in \{4^y, 25\cdot 4^y\}$ then for some $y$ then there is a solution with a least value of $\min (a,b,c).$ Such a solution cannot have $a,b,c$ all even, else $(a/2)^2+(b/2)^2+(c/2)^2\in \{4^{y-1},4^{y-1}\cdot 25\},$ which contradicts the minim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1907677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that if $0Prove that if $0<a<b$, then
$a<\sqrt{ab}<\dfrac{a+b}{2}<b.$
I have seen that the inequality $\sqrt{ab}\leq (a+b)/2$ is verified for all $a,b\geq 0$.
| This are $3$ inequalities, and $1$ is already solved. $2$ to go, lets solve them by equivalence transformations:
The first one ($a < \sqrt{ab}$):
\begin{align}
a &< \sqrt{ab} \\
a^2 &< ab \\
a &< b
\end{align}
The last one ($\dfrac{a+b}{2} < b$):
\begin{align}
\dfrac{a+b}{2} &< b \\
a + b &< 2 \cdot b \\
a &< b
\end{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1910149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
simplify $\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}$ simplify $$\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}$$
1.$\frac{3}{2}$
2.$\frac{\sqrt[3]{65}}{4}$
3.$\sqrt[3]{2}$
4.$1$
I equal it to $\sqrt[3]{a}+\sqrt[3]{b}$ but I cant find $a$ and $b$
| Note your number $\alpha$
It turns out that (after some calculation) that $\alpha^3=10-9\alpha$
1 seems to do the trick...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1910728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Secondary school level mathematical induction
*
*It is given that
$$1^3+2^3+3^3+\cdots+n^3=\frac{n^2(n+1)^2}{4}$$
Then, how to find the value of
$2^3+4^3+\cdots+30^3$?
Which direction should I aim at?
*Prove by mathematical induction, that $5^n-4^n$ is divisible by 9 for all positive even numbers $n$.
$$5^n-... | I'll give you a hint for the first one. Since you have asked three different questions, I wait to see if they tell you to split your question in there questions...
1
You already have a formula:
$$1^3 + 2^3 + 3^3 + \ldots + n^3 = \frac{n^2((n+1)^2)}{4}$$
Hence in your case $n = 30$ but your sum starts from $2^3$ and no... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1911131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
How can we show that $\ln{2}=\sum_{n=1}^{\infty}{(-1)^{n-1}\over n}\left({12\over e^{n\pi}-1}+{4\over e^{n\pi}+1}\right)$ $$\ln{2}=\sum_{n=1}^{\infty}{(-1)^{n-1}\over n}\left({12\over e^{n\pi}-1}+{4\over e^{n\pi}+1}\right)\tag1$$
Any hints?
| This is a result of theory of theta functions. Let $0 < q < 1$ and $$a(q) = \sum_{n = 1}^{\infty}\frac{q^{n}}{n(1 - q^{n})}$$ Next we can see that \begin{align}
b(q) &= \sum_{n = 1}^{\infty}(-1)^{n - 1}\cdot\frac{q^{n}}{n(1 - q^{n})}\notag\\
&= \sum_{n \text{ odd}}\frac{q^{n}}{n(1 - q^{n})} - \sum_{n \text{ even}}\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1912428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
when does the following system has infinite solutions I was trying to solve the following exercise but I couldn't get anywhere:
Find the values of a and b for which the following system has infinite solutions:
$2x+3y-z=a$
$3x-by+z=1$
$ax-y-z=2$
I tried to solve the problem using the determinant, I mean we know that $Ax... | Idea
Write $$A=\begin{pmatrix} 2 & 3 & -1 \\ 3 & -b & 1 \\ a & -1 & -1\end{pmatrix}, B=\begin{pmatrix} 2 & 3 & -1 & a \\ 3 & -b & 1 & 1 \\ a & -1 & -1 & 2\end{pmatrix}.$$
The system has infinitely many solutions if and only if $ran(A)=ran(B)<3.$ Since $A$ has a 2-order minor that non vanishes, $$A=\begin{pmatrix} \colo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1913662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Intuition for $\lim_{x\to\infty}\sqrt{x^6 - 9x^3}-x^3$ Trying to get some intuition behind why: $$ \lim_{x\to\infty}\sqrt{x^6-9x^3}-x^3=-\frac{9}{2}. $$
First off, how would one calculate this? I tried maybe factoring out an $x^3$ from the inside of the square root, but the remainder is not factorable to make anything ... | $$\sqrt{x^6-9x^3}-x^3=\frac{x^6-9x^3-x^6}{\sqrt{x^6-9x^3}+x^3}=\frac{-9x^3}{\sqrt{x^6-9x^3}+x^3}$$
As $x$ goes to $\infty$, $\sqrt{x^6-9x^3} \approx x^3.$
More, precisely, divide the numerator and denominator by $x^3$:
\begin{align}
\sqrt{x^6-9x^3}-x^3&=\frac{x^6-9x^3-x^6}{\sqrt{x^6-9x^3}+x^3}\\&=\frac{-9x^3}{\sqrt{x^6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1914948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 0
} |
Computing $\lim_{x \to 0} \frac{\cos x - \sqrt{2 - e^{x^2}}}{\ln{(\cos x) + \frac{1}{2} x \sin x}} \cdot \frac{(x+2)^{2017}}{(x-2)^{2015}}$ I'm studying for an exam, but I have trouble with computing the following limit:
$$\lim_{x \to 0} \frac{\cos x - \sqrt{2 - e^{x^2}}}{\ln{(\cos x) + \frac{1}{2} x \sin x}} \cdot \f... | As pointed out, you can pull out the terms on the right to get a factor of $-4$.
Next, $$\sqrt{2-e^{x^2}} = \sqrt{2 - 1 - x^2 - \frac{x^4}{2} + O(x^6)} = 1 - \frac{x^2}{2} - \frac{3x^4}{8} + O(x^6)\\
\cos x - \sqrt{2-e^{x^2}} = 1 - \frac{x^2}{2} + \frac{x^4}{24} - 1 + \frac{x^2}{2} + \frac{3x^4}{8} + O(x^6) = \frac{5x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1920673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove: $\cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)}=\frac {3}{4} \cos{3A}$ Prove that:
$$\cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)}=\frac {3}{4} \cos{3A}$$
My Approach:
$$\mathrm{R.H.S.}=\frac {3}{4} \cos{3A}$$
$$=\frac {3}{4} (4 \cos^3{A}-3\cos{A})$$
$$=\frac {12\cos^3{A} - 9\cos{A}}{4}$$
Now, please help m... | There's also another more algebraic way. You can easily show (by expansion) that if $$a+b+c=0$$ then $$a^3+b^3+c^3=3abc$$
Since, in your problem, for every $A$ $$\cos{A}+\cos{(A+2\pi/3)}+\cos{(A-2\pi/3)}=0$$ Then you can use the above identity
$$\cos^3{A}+\cos^3{(A+2\pi/3)}+\cos^3{(A-2\pi/3)}=3\cos{A}\cos{(A+2\pi/3)}\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1921191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 0
} |
Symmetric polynomials with Vieta's and Newton's theorems
Let $ x_{1}, x_{2}, x_{3}$ be the solutions of the equation $ x^3 -3x^2 + x - 1 = 0.$
Determine the values of $$\frac{1}{{x_{1}x_{2}}} + \frac{1}{{x_{2}x_{3}}} + \frac{1}{{x_{3}x_{1}}}$$
and also
$$ x_{1}^{3}+x_{2}^{3}+x_{3}^{3}$$
| For the first case, using Vieta's Formula $$x_1x_2x_3=1\implies\dfrac1{x_1x_2}=x_3$$
For the second,
Method$\#1$: observe that $$x_r^3-3x_r^2+x_r-1=0\implies x_r^3=3x_r^2-x_r+1\ \ \ \ (r)$$ for $r=1,2,3$
$(1)+(2)+(3)=?$
Now $(x_1+x_2+x_3)^2=x_1^2+x_2^2+x_3^2+2(x_1x_2+x_2x_3+x_3x_1)$
Method$\#2$:
$$a^3+b^3+c^3-3abc=(a+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1923980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Help on a very interesting integral: $\int_0^1\frac{x^5(1-x)^4}{1+x^3}\,dx$ I've been starting to work on problems with difficult integrals, and I came to
one of my problems where I end up having to integrate
$$\int_0^1 \frac{x^5(1-x)^4}{1+x^3} dx.$$
Wolfram seems to time out on integrals like these, so I decided to ta... | In general, any integral of the form $\int_{0}^{1}\frac{p(x)}{1+x^3}\,dx $ with $p(x)\in\mathbb{Q}[x]$ can be written as a linear combination with rational coefficients of $1$ and
$$ I_1=\int_{0}^{1}\frac{x}{1+x^3}\,dx=\frac{1}{9} \left(\pi \sqrt{3} -3\log 2\right),\qquad I_2= \int_{0}^{1}\frac{x^2}{1+x^3}\,dx=\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1924522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove $(a^2+b^2)(c^2+d^2)\ge (ac+bd)^2$ for all $a,b,c,d\in\mathbb{R}$. Prove $(a^2+b^2)(c^2+d^2)\ge (ac+bd)^2$ for all $a,b,c,d\in\mathbb{R}$.
So $(a^2+b^2)(c^2+d^2) = a^2c^2+a^2d^2+b^2c^2+b^2d^2$
and $(ac+bd)^2 = a^2c^2+2acbd+b^2d^2$
So the problem is reduced to proving that $a^2d^2+b^2c^2\ge2acbd$ but I am not sure ... | Diophantus has already shown that
$$(a^{2}+b^{2})(c^{2}+d^{2})=(ac+bd)^{2}+(ad-bc)^{2}.$$ This proves the inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1925766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
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How to integrate $ \int^a_0{\cfrac{dx}{x \ + \ \sqrt{a^2 \ - \ x^2}}} $? I am having a little problem with my maths homework. The problem is as follows:
\begin{equation}
\int^a_0{\cfrac{dx}{x \ + \ \sqrt{a^2 \ - \ x^2}}}
\end{equation}
I tried to do the following but got stuck halfway:
Let $\ \ x \ = asin\theta, \ he... | Try Euler substitutions. The second one might do the job.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1927726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Basic Complex inequalities It is given that $|z+2+3i| \leq 1$. Show that $3 \leq |2z+1+2i| \leq 7$. I really do not have a clue how to start on this but have proceeded by noting that
$$(2z+1+2i) = z+2+3i + z-1+i$$
and so
$$|2z+1+2i| \leq |z+2+3i| + |z-1+i|$$
But this doesn't get me too far. Can anyone help me complete ... | We have that
$$ |2z+1+2i|=2\left|z+\frac{1}{2}+i\right|=2\left|z+\frac{1}{2}+i\pm(\frac{3}{2}+2i)\right|\\=2\left|(z+2+3i)+\left(-\frac{3}{2}-2i\right)\right|.$$
Note that $|-\frac{3}{2}-2i|=\frac{5}{2}$. Now recall that $||u|-|w||\leq |u+w|\leq |u|+|w|$.
Hence if $|z+2+3i|\leq 1$, then
$$3=2\left(\frac{5}{2}-1\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1928161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Show $F_{n+1} \cdot F_{n-1} = F_n^2 + (-1)^n$ for all $n \in \mathbb{N}$ By calculating for $n\in \{1,2,3,4,5,6,7\}$, I've formulated the rule
\begin{equation}
F_{n+1} \cdot F_{n-1} = F_n^2 + (-1)^n,
\end{equation}
where $F_n$ is the $n$th fibonacci number. I want to show that this is true for all $n \in \mathbb{N}$.... | We can prove the rule without induction using Binet's formula
$$F_n = \frac{\varphi^n - \psi^n}{\sqrt{5}}$$
where
$$\varphi = \frac{1 + \sqrt{5}}{2} \qquad \psi = \frac{1 - \sqrt{5}}{2} = -\frac{1}{\varphi}$$
so that
$$\varphi\psi = -1 \qquad \frac{\varphi}{\psi} + \frac{\psi}{\varphi} = -3 = 2 - 5$$
Then
\begin{align}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1928285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Prove the inequality $\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{a^2+c^2}<1+\frac{\sqrt2}{2}$ Let $a,b,c -$ triangle side and $a+b+c=1$. Prove the inequality
$$\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{a^2+c^2}<1+\frac{\sqrt2}{2}$$
My work so far:
1) $a^2+b^2=c^2-2ab\cos \gamma \ge c^2-2ab$
2) $$\sqrt{a^2+b^2}+\sqrt{c^2+b^2}+\sqrt{... | Let $a=y+z$, $b=x+z$ and $c=x+y$.
Hence, $x$, $y$ and $z$ are positives and we need to prove that:
$$\sum\limits_{cyc}\sqrt{2x^2+y^2+z^2+2xy+2xz}\leq(2+\sqrt2)(x+y+z)$$
By C-S $$\left(\sum\limits_{cyc}\sqrt{2x^2+y^2+z^2+2xy+2xz}\right)^2\leq\sum\limits_{cyc}\frac{2x^2+y^2+z^2+2xy+2xz}{\sqrt2x+y+z}\sum\limits_{cyc}(\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1929245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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How many ways are there to distribute $12$ red jelly beans to four children, a pair of identical female twins and a pair identical male twins?
Question: How many ways are there to distribute $12$ red jelly beans to four children, a pair of identical female twins and a pair identical male twins?
I attempted to solve t... | Instead of Polya enumeration theorem, I found an answer using Burnside's lemma. Define sets
$$
X=\{(f_1,f_2,m_1,m_2):f_1,f_2,m_1,m_2\text{ are integers and }f_1+f_2+m_1+m_2=12\}
$$
and
$$
G=\{e,g_1,g_2,g_3\},
$$
where $e,g_1,g_2,g_3$ are functions from $X$ to $X$ such that
\begin{align}
e(f_1,f_2,m_1,m_2)&=(f_1,f_2,m_1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1930947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Not getting the correct answer for an infinite limit problem The problem:
$\lim _{ x\to -\infty } \frac { \sqrt { 9x^{ 6 }-x } }{ x^{ 3 }+8 }$
My answer: $3$
What I did:
$\\ \lim _{ x\to -\infty } \frac { \sqrt { x^{ 6 }(9-\frac { 1 }{ { x }^{ 5 } } ) } }{ x^{ 3 }(1+\frac { 8 }{ { x }^{ 3 } } ) } \\ \\ \lim _{ x\to... | Hint: use that $\sqrt{x^6}=|x^3|$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1931387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find sum of first $n$ terms of the series : $1+\frac{1^3+2^3}{1+2}+\frac{1^3+2^3+3^3}{1+2+3}+\dots$ The main question is:
Find sum of first $n$ terms of the series : $1+\frac{1^3+2^3}{1+2}+\frac{1^3+2^3+3^3}{1+2+3}+\dots$
My approach:
Initially, nothing clicked, so I went forward with simplifying the series.
So, we g... | Hint
$$1^3+2^3+...+n^3=\frac{n^2(n+1)^2}{4} \\
1+2+3+..+n=\frac{n(n+1)}{2}$$
Edit:
You can then use the identity (thanks to Winther for fixing the typo)
$$n(n+1)= \frac{1}{3} \left[(n+1)^3-n^3 \right]-\frac{1}{3}$$
to evaluate the new sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1931771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Initial-value problem $y' = Ax^py$ with initial condition $y(2)=1$
Assuming $p\neq-1$, find the solution $y=y(x)$ to the initial-value problem:
$$\frac{dy}{dx}= Ax^py$$
$$y(2)=1$$
in terms of $A$ and $p$.
So far I have, after rearranging the problem and integrating, came up with this equation:
$$\ln(y) = \frac... | You have almost correctly solved the problem except the last step.
$$\ln(y) = \frac{Ax^{p+1}}{p+1} + C$$
Plugging in the value $y(2) = 1$, we get $$\ln(1) = \frac{A\cdot2^{p+1}}{p+1} + C \implies C=-\frac{A\cdot(2)^{p+1}}{p+1}$$
So the required solution is $$\ln(y) = \frac{Ax^{p+1}}{p+1}-\frac{A\cdot2^{p+1}}{p+1}= \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1933118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For how many positive integers $a$ is $a^4−3a^2+9$ a prime number? I understand that there are many posts on the problems similar to mine. I have tried my best, but still get different answers from the answer sheet. Can anyone help me? Also is there a simple way to find $a$?
For how many positive integers $a$ is $a^4-... | HINTS
Your factorisation is the key: $a^4 - 3a^2 + 9 \equiv (a^2-3a+3)(a^2+3a+3)$.
Since $a$ is an integer, so are both $a^2-3a+3$ and $a^2+3a+3$.
If both $a^2-3a+3$ and $a^2+3a+3$ are bigger than $1$ then $a^4 - 3a^2 + 9$ will have two positive integer factors larger than one, i.e. $a^4 - 3a^2 + 9$ won't be prime. (Co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1933220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Easier way to calculate the derivative of $\ln(\frac{x}{\sqrt{x^2+1}})$? For the function $f$ given by
$$
\large \mathbb{R^+} \to \mathbb{R} \quad x \mapsto \ln \left (\frac{x}{\sqrt{x^2+1}} \right)
$$
I had to find $f'$ and $f''$.
Below, I have calculated them.
But, isn't there a better and more convenient way to do t... | Take advantage of logarithm properties.
$$\ln\left(\frac{x}{\sqrt{x^2 + 1}}\right) = \ln(x) - \frac12\ln(x^2 + 1)$$
Then the derivative is easy:
$$f'(x) = \frac1x - \frac{x}{x^2 + 1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1933460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Prove the inequality $\frac a{a+b^2}+\frac b{b+c^2}+\frac c{c+a^2}\le\frac14\left(\frac1a+\frac1b+\frac1c\right)$ Let $a,b,c>0; a+b+c=1$. Prove the inequality
$$\frac a{a+b^2}+\frac b{b+c^2}+\frac c{c+a^2}\le\frac14\left(\frac1a+\frac1b+\frac1c\right)$$
My work so far:
I tried AM-GM and used fact $a+b+c=1$.
| Also we can us C-S and $uvw$:
We need to prove that $$\sum\limits_{cyc}\left(\frac{a}{a+b^2}-1\right)\leq\frac{ab+ac+bc}{4abc}-\frac{3}{a+b+c}$$ or
$$\sum\limits_{cyc}\frac{b^2}{a+b^2}+\frac{ab+ac+bc}{4abc}-\frac{3}{a+b+c}\geq0$$
Now by C-S:
$$\sum\limits_{cyc}\frac{b^2}{a+b^2}\geq\frac{(a+b+c)^2}{a+b+c+a^2+b^2+c^2}=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1934520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Easier way to discover the area of a right triangle In the following right triangle: $y-x=5$ and the altitude to the hypotenuse is 12. Calculate its area.
I've managed to discover its area using the following method, but it ends up with a 4th degree equation to solve. Is there an easier way to solve the problem?
$ha=x... | you have to ways to arrive at the area.
$A = \frac 12 xy = 6\sqrt{(x^2 + y^2)}$
and you know: $(x-y) = 5$
$(x-y)^2 = 25\\
x^2 + y^2 - 2xy = 25\\
x^2 + y^2 = 25 + 4A$
Plug this into the equation above for areas.
$A = 6\sqrt{25 + 4A}$
square both sides and solve the quadratic
$A^2 = 36(25 + 4A)\\
A^2 - 144A - 900 = 0\\
(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1936070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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why the sequence $\{b_{n}\}$ can't periodic sequence Let $$a_{1}=1,a_{n+1}-a_{n}=a_{\lfloor\frac{n+1}{2}\rfloor}$$
if $a_{n}=\left(\overline{b_{1}b_{2}\cdots b_{n}}\right)_{10}$
show that the sequence $\{b_{n}\}$ can't periodic sequence
A sketch of my thoughts:
$a_{1},$so $ b_{1}=1$
$a_{2}-a_{1}=a_{1},a_{2}=2,$ so $b_... | Assume $a_n\bmod 10$ is (eventually) periodic with period $p$.
Then so is $a_{n+1}-a_n\bmod 10$, i.e., $a_{\lfloor(n+1)/2\rfloor}\bmod 10$.
If $p$ is even, this implies that $a_n\bmod 10$ is eventually periodic with period $\frac p2$.
Therefore, if the sequence is eventually periodic then its minimal periof $p$ is odd.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1938563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How i can integrate this expression? I've read about integration, and i believe i understood concept correctly. But, unfortunately, the simplest exercise already got my stumbled. I need to find an integral of $x{\sqrt {x+x^2}}$. So i proceed as follows,
By the fundamental theorem of calculus:
$f(x)=\int[f'(x)]=\int[x\s... | I hope you do not mind if I prefer to start from scratch. We have
$$ \int (2x+1)\sqrt{x^2+x}\,dx = C+\frac{2}{3}(x^2+x)^{3/2} \tag{1}$$
and the problem boils down to computing $\int\sqrt{x^2+x}\,dx$. Integration by parts gives
$$ \int \sqrt{x^2+x}\,dx = x\sqrt{x^2+x}-\int\frac{x+2x^2}{2\sqrt{x+x^2}}\,dx \tag{2}$$
hence... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\sum_{k=0}^nk{m+k \choose m}=n{m+n+1\choose m+1}-{m+n+1 \choose m+2}$ Can someone please see the work I have so far for the following proof and provide guidance on my inductive step?
Prove that if $m,n\in\mathbb{N}$, then $\sum_{k=0}^nk{m+k \choose m}=n{m+n+1\choose m+1}-{m+n+1 \choose m+2}$
Base Case. Let ... | $$\sum_{k=0}^nk{m+k \choose m}=n{m+n+1\choose m+1}-{m+n+1 \choose m+2}$$
Inductive Step. 1: $n\longrightarrow0$
$\sum_{k=0}^{n=0}k{m+k \choose m}=0$ and $n{m+n+1\choose m+1}-{m+n+1 \choose m+2}=0-0=0$.
the equation is satisfied.
Inductive Step. 2: $n\longrightarrow n$
$$\sum_{k=0}^nk{m+k \choose m}=$$
By Pascal's ide... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
} |
Simplify the following $(1\cdot2)+(2\cdot3)+(3\cdot4)+\dots+(n\cdot(n+1))$ Simplify the following $(1\cdot2)+(2\cdot3)+(3\cdot4)+\dots+(n\cdot(n+1))$.
How do i do this by using these identities.
$C^r_r + C^{r+1}_r + C^{r+2}_r + ....+C^{n}_r = C^{n+1}_{r+1}$
Or
$C^r_0 + C^{r+1}_1 + C^{r+2}_2 + ....+C^{r+k}_k = C^{r+k+1}... | Note that $2\binom{k}{2}=(k-1)k$. Hence
$$(1\cdot2)+(2\cdot3)+(3\cdot4)+\dots+(n\cdot(n+1))=\sum_{k=2}^{n+1}2\binom{k}{2}=2\binom{n+2}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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If $A < B < C < D$ where $A, B, C, D \in \mathbb{N}$, does the following series of inequalities hold? PROBLEM STATEMENT
If $A < B < C < D$ where $A, B, C, D \in \mathbb{N}$, does the following series of inequalities hold?
$$\frac{A}{D}<\frac{A}{C}<\frac{A}{B}<\frac{B}{D}<\frac{B}{C}<\frac{C}{D}<1$$
$$1<\frac{D}{C}... | You are correct; here's a simpler way to analyze the problem.
From $C<D$ we get $1/D<1/C$, so $A/D<A/C$. Similarly, $A/C<A/B$ and $B/D<B/C$. Also $C/D<1$, since $C<D$. Moreover $B/D<C/D$, because $B<C$.
Note also that taking reciprocals of positive numbers reverses inequalities, so the second set is essentially the sam... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1940125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to calculate the number of positive integral solutions for the equations $\frac{1}{x} + \frac{1}{y} = \frac{1}{ab}$? Given $a$, $b$.Calculate the number of positive integral solutions for the eqations $\frac{1}{x} + \frac{1}{y} = \frac{1}{ab}$, where $a$, $b$ can be up to $1000000.$
| $\frac{1}{x}+\frac{1}{y}=\frac{1}{ab}\implies ab(x+y)=xy\implies(x-ab)(y-ab)=a^2b^2$
Thus, if $a^2b^2=pq$, we can take $x=ab+p,y=ab+q$, and the number of solutions is thus equal to the number of factorisations of $a^2b^2$, also known as $d(a^2b^2)$.
One way of seeing that the only solutions with positive $x,y$ are the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1940785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove that $k^n=\frac{d^n}{dx^n} (1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....+\frac{x^n}{n!})^k|_{x=0}$ $$k^n=\frac{d^n}{dx^n} (1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....+\frac{x^n}{n!})^k|_{x=0}$$
It is easy to show $k=1$ and $k=2$
$k=1$
$$\frac{d^n}{dx^n} (1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....+\frac{x^n}{n!})|_{x=0}=1... | Notice that
$$ \frac{d^n}{dx^n} \Bigg( \sum_{j=0}^{n} \frac{x^j}{j!} \Bigg)^k \Bigg|_{x=0} =
\frac{d^n}{dx^n} \Bigg( \sum_{j=0}^{n} \frac{x^j}{j!} + x^{n+1}f(x) \Bigg)^k \Bigg|_{x=0}$$
for any $n$-times differentiable function $f$. So we may choose $f$ at our convenience. Now if we choose
$$f(x) = \sum_{j=n+1}^{\infty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1943301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
How to evaluate $\int\frac{x^5}{x^3-2x^2-5x+6} dx$? None of substitution, partial fraction, and integration by parts seems to work here.
| First note that by long division,
$$
\frac{x^5}{x^3 - 2x^2 - 5x + 6} = x^2 + 2x + \frac{243}{10(x-3)} - \frac{1}{6(x-1)} - \frac{32}{15(x+2)} + 9.
$$
Now, you can integrate the sum term by term, factoring out the constants to get as your answer
$$
\int \frac{x^5}{x^3 - 2x^2 - 5x + 6} dx = \frac{x^3}{3} + x^2 + 9x - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1944460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Show that the numerator of $1+\frac12 +\frac13 +\cdots +\frac1{96}$ is divisible by $97$
Let $\frac{x}{y}=1+\frac12 +\frac13 +\cdots +\frac1{96}$ where $\text{gcd}(x,y)=1$. Show that $97\;|\;x$.
I try adding these together, but seems very long boring and don't think it is the right way to solving. Sorry for bad engl... | What we need to do is as follows:
$$
\begin{equation}\begin{split}
1 + \frac{1}{2} + \frac{1}{3} + \ldots + \frac{1}{96} & = \Big(1+\frac{1}{96}\Big) + \Big(\frac{1}{2}+\frac{1}{95}\Big) + \ldots + \Big(\frac{1}{48} + \frac{1}{49}\Big) \\
& = \frac{97}{96} + \frac{97}{2*95} + \frac{97}{3*94} + \ldots + \frac{97}{48*49}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1945754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 3
} |
How do I see that every left ideal of a square matrix ring over a field is principal? As the question title suggests, how do I see that every left ideal of a square matrix ring over a field is principal?
| Let $I$ be a left ideal. The first step is that if a matrix is in the ideal then all the rows can be obtained separately:
Say $$\begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix} \in I$$
then
$$\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix} \begin{pmatrix}1&2&3\\4&5&6\\7&8&9\end{pmatrix} =\begin{pmatrix}1&2&3\\0&0&0\\0&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1947139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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How to show that $\inf({\frac{mn}{1+m+2n}}) = 1/4$? I am trying to prove the result:
$\inf({\frac{mn}{1+m+2n}}) = \frac{1}{4}$, where $m, n \in\mathbb{N}$.
I proved existence of infimum, but then I do not know how to prove that $1/4$ is a lower bound and then can not prove that $1/4$ is the greatest lower bound.
| To show it achieves this lower bound, let $m = n = 1$. Then
$$\frac{mn}{1+m+2n} = {1 \over 1+1+2} = \frac{1}{4}.$$
Now we need to show for all other $m,n \in \Bbb N$,
$$\frac{mn}{1+m+2n} \ge \frac14.$$
If $n \ge 2$, then
$$\frac{mn}{1+m+2n} \ge \frac{2m}{5+m} \ge \frac{2}{5+1} = \frac{1}{3} > \frac14.$$
If $m \ge 2$, t... | {
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Contour integral over semicircle Evaluate the integral $$\int_{2}^{i} \frac{dz}{1-z^2}$$
where the path is the semicircle $|z-1|=1$ in $\mbox{Im}(z)\leq0$ and the imaginary axis.
My attempted solution:
The primitive function to the definite integral is $$\frac{1}{2}(\log(z+1)-\log(1-z))$$
I need to find a suitable bra... | We have that
$$\frac{d}{dz}\left(\frac{1}{2}\log\left(\frac{z+1}{z-1}\right)\right)=\frac{1}{1-z^2}$$
where the $\log$ has a branch cut along $[0,+\infty)$. Hence
$$\int_{\gamma} \frac{dz}{1-z^2}=\left[\frac{1}{2}\log\left(\frac{z+1}{z-1}\right)\right]_2^i=\left[\frac{1}{2}\log\left(w\right)\right]_3^{-i}=
\frac{1}{2}... | {
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How to prove that $|z^2| = |z|^2$ where $z = a+bi$? I just started my topic on complex numbers and I'm stuck on this question.
What I have managed to get (I go wrong here, I don't know where though):
$z^2 = (a+bi)^2 = a^2 + b^2$, so $|z^2| = \sqrt{(a^2 + b^2)^2 + 0^2} = \sqrt{a^4+2a^2b^2+b^4}$
and
$|z| = a^2+b^2$ so $|... | If you use $z=r\cdot e^{i\theta}$, the proof is easy.
$|z^2|=|r^2\cdot e^{2i\theta}|=r^2$ and $|z|^2=r^2$.
| {
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"source": "stackexchange",
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How to prove "For every positive integer n, $1^{n}$ + $2^{n}$ + ... + $n^{n}$ < $(n+1)^{n}$." by using induction? How can I prove the following theorem:
For every positive integer $n$, $1^n + 2^n + ... + n^n \lt (n+1)^n$
by using induction?
I have proved that "for every real number $x > 0$ and every non-negative int... | This looks like the same proof as GAVD, but without sigma notation. I didn't use your lemma explicitly either.
Let $P(n)$ be the statement that $1^n + 2^n + 3^n + \dots + n^n < (n+1)^n$. Then $P(1)$ is $1 < 2$, which is definitely true.
Suppose $P(k)$ is true for some $k$. That is, suppose
$$
1^k + 2^k + \dots ... | {
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1967 HSC 4 unit Mathematics Question 2 Screenshot from the examination paper
[...asking about partial fraction decomposition of $$\frac{1 - abx^2}{(1-ax)(1-bx)} $$ and related formulas...]
This question is taken from the New South Wales Higher School Certificate 4 unit (highest level possible) paper of 1967. I wasn't s... | Let us do things formally without worrying about convergent issues. As mentioned by Jack Lam, the partial fraction decomposition of the function is given by
\begin{align}
\frac{1-abx^2}{(1-ax)(1-b)} = \frac{1}{1-bx}+\frac{1}{1-ax}-1.
\end{align}
Formally, we will us the geometric series representation for the two frac... | {
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If $a$, $b$ and $c$ are three positive real numbers, prove that $\frac{a^2+1}{b+c}+\frac{b^2+1}{c+a}+\frac{c^2+1}{a+b} ≥ 3$ If $a$, $b$ and $c$ are three positive real numbers, prove that $$\frac{a^2+1}{b+c}+\frac{b^2+1}{c+a}+\frac{c^2+1}{a+b} ≥ 3$$
I think we are supposed to use the AM-GM inequality here; or maybe thi... | Let us define $t=a+b+c$. First we note that
$$\frac{a^2+1}{b+c}+\frac{b^2+1}{c+a}+\frac{c^2+1}{a+b}=
\left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)+
\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)$$
so by Titu's lemma:
$$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\geq\frac{(a+b+c)^2}{2(a+b+c)}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1955141",
"timestamp": "2023-03-29T00:00:00",
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Find the limit of the sequence $x_{n+1}=\sqrt{3x_n}$
Let $x_{n+1}=\sqrt{3x_n}$ and $x_1=1$. Prove $x_n=3^{1-(\frac{1}{2^{n-1}})}$ for all $n$ and find the limit of $\{x_n\}$.
Notes: The first few terms of the sequence are $1,\sqrt{3},\sqrt{3\sqrt{3}},\sqrt{3\sqrt{3\sqrt{3}}} ...$ I do not know from this how to find t... | Hint
Considering the sequence $$x_{n+1}=\sqrt{3x_n}$$ take logarithms of both sides $$\log(x_{n+1})=\frac{\log(3)}2+\frac 12 \log(x_n)$$ Now define $y_n=\log(x_n)$ which makes the new sequence to be $$y_{n+1}=\frac 12 y_n+\frac{\log(3)}2$$ which is much easier to manipulate.
For the general sequence $$y_{n+1}=a y_n+b\q... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\frac{\sin^3\frac{x}{2}-\cos^3\frac{x}{2}}{2+\sin x}=\frac{\cos x}{3}$ Prove that $\frac{\sin^3\frac{x}{2}-\cos^3\frac{x}{2}}{2+\sin x}=\frac{\cos x}{3}$
My Attempt:
$LHS=\frac{\sin^3\frac{x}{2}-\cos^3\frac{x}{2}}{2+\sin x}=\frac{(\sin\frac{x}{2}-\cos\frac{x}{2})(\sin^2\frac{x}{2}+\cos^2\frac{x}{2}+\sin\fra... | $$\frac{\sin^3(0/2)-\cos^3(0/2)}{2+\sin0}=-\frac{1}{2}\neq\frac{1}{3}=\frac{\cos0}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1957883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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.