Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Proving $\frac{1}{\cos^2\frac{\pi}{7}}+ \frac {1}{\cos^2\frac {2\pi}{7}}+\frac {1}{\cos^2\frac {3\pi}{7}} = 24$ Someone gave me the following problem, and using a calculator I managed to find the answer to be $24$.
Calculate $$\frac {1}{\cos^2\frac{\pi}{7}}+ \frac{1}{\cos^2\frac{2\pi}{7}}+\frac {1}{\cos^2\frac{3\pi}{7... | First notice that since $\cos(x)=\cos(\pi-x)$, we have
$$1+2\left(\frac{1}{\cos(\frac{\pi}{7})^2}+\frac{1}{\cos(\frac{2\pi}{7})^2}+\frac{1}{\cos(\frac{3\pi}{7})^2}\right)=\sum_{k=0}^6 \frac{1}{\cos(\frac{k\pi}{7})^2}$$
Now, $x\mapsto 2x$ is a bijection of the integers mod $7$, so we may make the summands $\cos(\frac{2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
The sum of more than two consecutive natural numbers cannot be prime.
The sum of more than two consecutive natural numbers cannot be prime.
Is the statement true and is there any way to prove it?
I was able to prove that the sum of an odd amount of consecutive numbers cannot be prime:
So, since the sum of consecuti... | For a sum of three or more consecutive positive integers
$S = x + (x+1) + (x + 2) + ..... + (x + n -1)$ $x > 0; n > 2$
$S = (x + n-1) + (x+n - 1) + (x + n - 2) + ..... + (x + 1)+x$
Add 'em together.
$2S = (2x + n -1) + (2x + n-1) + .... (2x + n-1) = n(2x + n-1)$
Case one: $n$ is even. Then $S=\frac n2(2x + n -1)$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Fully simplify $\sqrt {14 + 8\sqrt {3}}$. Please help me with this problem: Fully simplify $\sqrt{14 + 8\sqrt {3}}$.
I've tried to assume that the radical is in the form of $a+b\sqrt{3}$, so $a+b\sqrt{3}=\sqrt {14 + 8\sqrt {3}}$, then I squared both sides and got $$a^2+3b^2+2ab\sqrt{3}=14+8\sqrt{3}.$$
So I assumed tha... | Note
$$\sqrt {14 + 8\sqrt {3}}=\sqrt{2(7+2\sqrt{12})}
=\sqrt{2(\sqrt4+\sqrt3)^2}=\sqrt2(2+\sqrt3)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
How to simplify this expression I'm trying to find a way to simplify $\sqrt{5+\sqrt{24}}$. I know that this expression is equivalent to $\sqrt{2}+\sqrt{3}$ because they are both roots of the equation: $x^4-10x^2+1$ (and the decimal equivalents are the same using a calculator). My question is , how does one get from $\s... | If you don't want guess-and-check or hope-you-see-it approaches, suppose that $a+\sqrt{b} = \sqrt{5+\sqrt{24}}$ for some $a,b$. (Note that $a$ could be a square root.. it often works out where one of the two terms is an integer and the other is a root - which is why I wrote it in this way.) Squaring both expressions gi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1864474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Indefinite integral of a rational function with linear denominator: $ \int \frac{ x^7}{(x+1)}{dx} $ $$ \int \frac{ x^7}{(x+1)}{dx} $$
$$ \int \frac{ \left(x^7 + x^6 - x^6 - x^5 + x^5 + x^4 -x^4 - x^3 + x^3 + x^2 - x^2 -x^1 + x^1 +1 -1\right ) }{\left(x+1\right)}{dx}$$
$$ \int { \left(x^6 - x^5 + x^4 - x^3 + x^2 -x +... | Let
$$I=\int\frac{x^7}{x+1}dx$$
Apply $x+1\to y$ to get
$$I=\int\frac{(y-1)^7}{y}dy$$
Using the binomial theorem
$$I=\int\frac{1}{y}\sum_{k=0}^{7}\binom{7}{k}y^k(-1)^{7-k}dy$$
$$=\int\sum_{k=0}^{7}\binom{7}{k}y^{k-1}(-1)^{7-k}dy$$
$$=\sum_{k=0}^{7}\int\binom{7}{k}y^{k-1}(-1)^{7-k}dy$$
$$=C-\ln|y|+\sum_{k=1}^{7}\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Decompose $5^{1985}-1$ into factors
Decompose the number $5^{1985}-1$ into a product of three integers, each of which is larger than $5^{100}$.
We first notice the factorization $x^5-1 = (x-1)(x^4+x^3+x^2+x+1)$. Now to factorize $x^4+x^3+x^2+x+1$ we get $$(x^2+ax+1)(x^2+bx+1) = x^4+(a+b)x^3+(ab+2)x^2+(a+b)x+1 = x^4+x... | I would try Aurifeuillian factorization here.
Let $f(x)=x^4+x^3+x^2+x+1$ be the fifth cyclotomic polynomial. The Aurifeuillian stuff says that $f(5x^2)$ factors into a product as follows
$$
f(5x^2)=\left(25 x^4-25 x^3+15 x^2-5 x+1\right) \left(25 x^4+25 x^3+15 x^2+5
x+1\right).
$$
Call those two factors on the RHS $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 2
} |
If $x$ is positive, then why does $\frac{1}{\sqrt{x+1} + \sqrt{x}} = \sqrt{x+1} - \sqrt{x}$? Given that $x$ is positive, $\frac{1}{\sqrt{x+1} + \sqrt{x}} = \sqrt{x+1} - \sqrt{x}$
I've been trying to convert the left side of the equation to the right side:
$$ \frac{1}{\sqrt{x+1} + \sqrt{x}}$$
But then how can I flip th... | You need to use a little trick with fractions. Recall that for any number $n$ that is not zero, $\frac{n}n=1$, and 1 times any number is itself.
So start with the left side: $\sqrt{1+x}-\sqrt{x}$. If we choose our $n$ to be $\sqrt{1+x}+\sqrt{x}$, then:
$$\sqrt{1+x}-\sqrt{x}=\frac{\sqrt{1+x}-\sqrt{x}}1\cdot\frac{\sqrt{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Solution of Inequality $ \frac{1}{x-6}\le 3$
Solve the inequality: $\displaystyle \frac{1}{x-6}\le 3$
solution:
\begin{align*}\frac{1}{x-6}& \le 3 \\ x-6& \le \frac{1}{3}
\\x& \le 6+\frac{1}{3}\\
x&\le19/3\end{align*}
but, for values of $x\le 6$ also, inequality holds true as left hand side provides some negative val... | If $x < 6$, then $x - 6$ is negative, so multiplying both sides by $x-6$ reverses the inequality. Thus $1 \ge 3(x - 6)$, or $19 \ge 3x$, which is true for all $x < 6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1868762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 2
} |
How to factorise $(x-1)^2 - (x-5)^2$? My attempt:
$a = (x-1)$
$c = (x-5)$
$a^2 - c^2$
which is equal to:
$$((x-1) - (x-5))((x-1)+(x-5))$$
But the correct answer is :
$8(x-3)$
Can you explain, please?
| Your doing is correct, you just need to expand the expressions inside the parenthesis:
\begin{align}((x-1) - (x-5))((x-1)+(x-5))&=(x-1-x+5)(x-1+x-5)\\ & =4(2x-6)\\ & =8(x-3)\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
How to solve $\int \frac{1}{x^2+4x+7} dx$? How to solve $\int \frac{1}{x^2+4x+7} dx$?
I think the first step is to write it in the following form: $$\int \frac{1}{(x+2)^2+3} dx$$
| With the substitution $u=\frac{x+2}{\sqrt{3}}$:
\begin{align*}\int \frac{1}{(x+2)^2+3} \mathrm{d}x &= \int \frac{1}{3(\frac{x+2}{\sqrt{3}})^2+3} \mathrm{d}x \\ &= \frac13\int \frac{1}{(\frac{x+2}{\sqrt{3}})^2+1} \mathrm{d}x \\ &= \frac{1}{3}\sqrt{3}\int \frac{1}{u^2+1} \mathrm{d}u \\ &= \frac{1}{3}\sqrt{3} \arctan(u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1870787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Proving that $\lim_{(x,y) \to (0,0)} (x^2 +y^2 -x^3 y^3)/(x^2 +y^2) =1$ How can I go about proving that $$\lim_{(x,y) \to (0,0)} \frac{x^2 +y^2 -x^3 y^3}{x^2 +y^2} = 1 ?$$
I checked some lines along $x, y$ and $x=y$ and it all gave $1$
| An alternative solution without polar coordinate:
$$\lim_{(x,y) \to (0,0)} \frac{x^2 +y^2 -x^3 y^3}{x^2 +y^2} = \lim_{(x,y) \to (0,0)} 1- \frac{x^3 y^3}{x^2 +y^2}$$
To show $\lim_{(x,y) \to (0,0)} \frac{x^3 y^3}{x^2 +y^2} = 0$ is equivalent to show $\lim_{(x,y) \to (0,0)} |\frac{x^3 y^3}{x^2 +y^2}| = 0 $:
Since $$0 \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1874517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Proof Verification - Spivak's Calculus - Chapter 1 Question 1.3 In this post, I asked for help getting started on a proof in Spivak's Calculus Fourth Edition, Chapter 1 Question 1.3
Prove that if $x^2 = y^2$, then $x = y$ or $x = −y$ using only the following properties of numbers:
A lot of awesome people responded ... | In logic, the statement $p \to q$ does not necessarily mean $q \to p$. ( That is, if $p$ implies $q$, $q$ does not necessarily imply $p$. In other words, if $p$ is true, $q$ is also true--but there are some cases in the universe of discourse when $q$ is true but $p$ is not. )
In the case of this particular question fro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving that $\frac{\sqrt{1-x^2_{1}}+\sqrt{1-x^2_{2}}+\ldots+\sqrt{1-x^2_{17}}}{x_{1}+x_{2}+\ldots+x_{17}}\geq 4$
If $x_{1},x_{2},x_{3},x_{4},\ldots,x_{17}\in \left[0,1\right]$ and $x^2_{1}+x^2_{2}+x^2_{3}+\ldots+x^2_{17} = 1$ then prove that $$\frac{\sqrt{1-x^2_{1}}+\sqrt{1-x^2_{2}}+\ldots+\sqrt{1-x^2_{17}}}{x_{1}+x_... | Let $x_i=\frac{a_i}{\sqrt{17}}$.
Hence, we need to prove that $\sum\limits_{i=1}^{17}\sqrt{17-a_i^2}\geq4\sum\limits_{i=1}^{17}a_i$, where $\sum\limits_{i=1}^{17}a_i^2=17$.
Indeed, we need to prove that $\sum\limits_{i=1}^{17}\left(\sqrt{17-a_i^2}-4a_i\right)\geq0$ or
$\sum\limits_{i=1}^{17}\frac{1-a_i^2}{4a_i+\sqrt{17... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the remainder of $2^n+n^2$ modulus 6
Find the remainder of $2^n+n^2$ modulus 6 given that $2^n+n^2$ is a prime and $n\geq2$($n$ positive integer)
I tried to solve this but failed!I just know that $n$ must be odd. No progress at all!!
| $n$ can't be even because then $n^2+2^n$ is even and larger than $2$.
Therefore $n$ is odd and $2^n\equiv 2\bmod 3$. Notice that $n^2$ can only be $1$ or $0\bmod 3$. In the first case $3$ divides $2^n+n^2$ and $n>3$, implying $2^n+n^2$ is not prime.
We conclude $n^2\equiv 0\bmod 3$. Therefore $2^n+n^2$ is odd and $2\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1879053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Which number is more near to $S$ Which number is more near to $S$?
$S=\frac{(2^3-1)(3^3-1)\dots (100^3-1)}{(2^3+1)(3^3+1)\dots (100^3+1)}$
1.$0.6$
2.$0.67$
3.$0.667$
4.$0.6667$
I personaly can not do anything special I have tried to calculate some easy case but no resaults any hints?
| \begin{align*}
\prod_{k=2}^{n} \frac{k^3-1}{k^3+1} &=
\prod_{k=2}^{n} \frac{(k-1)(k^2+k+1)}{(k+1)(k^2-k+1)} \\
&= \prod_{k=2}^{n} \frac{k-1}{k+1} \times \frac{k^2+k+1}{(k-1)^2+(k-1)+1} \\
&= \left(
\frac{1}{3} \cdot \frac{2}{4} \cdot \frac{3}{5} \cdot \frac{4}{6} \cdot
\ldots \cdot \frac{n-3}{n-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluation of $I=\int \frac{1}{\sin (x-a) \sin(x-b) \sin(x-c)}.dx$ Evaluate the given integral:
$$I=\int \frac{1}{\sin (x-a) \sin(x-b) \sin(x-c)}.dx$$
When it was $I=\int \frac{1}{\sin (x-a) \sin(x-b)}.dx$, I solved it by multiplying and dividing by $\sin (b-a)$. But I am not getting how to proceed here..
Could someone... | I looked at the partial feaction decomposition method. It is not particularly elegant, but it reduces the calculus problem to a linear algebra one. It is likely a more efficient method for this integral exists, but I am not sure what it would be.
$$\int\frac{dx}{\sin(x-a)\sin(x-b)\sin(x-c)}$$
$$=\int\left(\frac{A}{\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
What is the square root of $3 + 2\sqrt{10}i$? I need to compute the square root of $3 + 2\sqrt{10}i$.
I know how to solve it, but for some reason I'm not getting the correct answer. I attempted to solve it like this:
$$
\sqrt{3 + 2\sqrt{10}i} = x + iy
\quad \longrightarrow \quad
3 + 2\sqrt{10}i = x^2 - y^2 +2x... | You want $z=a+bi$ where $a,b\in\mathbb{R}$ such that
$$z^2=a^2-b^2+2abi=3+2\sqrt{10}i.$$
Comparing coefficients, you need $a^2-b^2=3$ and $ab=\sqrt{10}$. So $a=\sqrt{5}$ and $b=\sqrt{2}$, or $a=-\sqrt{5}$ and $b=-\sqrt{2}$. Thus the square root of $3+2\sqrt{10}$:
$$\pm(\sqrt{5}+\sqrt{2}i).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Evaluating the indefinite integral $\int\frac{x\arctan x}{(1+x^2)^2}dx$ The integral again is $$\int\frac{x\arctan x}{(1+x^2)^2}dx$$
I'm actually pretty much just stumped on this one. It looks like it should be amenable to a substitution kinda like $\arctan x$
but that obviously does not work.
I also thought maybe by ... | Here is a way by direct integration by parts:
Let $u = \arctan x, dv = \frac{x}{(1+x^2)^2}dx$, then $du = \frac{1}{x^2+1}dx, v = -\frac{1}{2(x^2+1)}$.
Then $$\int\frac{x\arctan x}{(1+x^2)^2}dx =-\frac{1}{2(x^2+1)}\arctan x+\int \frac{1}{2(x^2+1)}\frac{1}{x^2+1}dx=\int \frac{1}{2(x^2+1)^2}dx$$
Try some trigonometri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1886601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How many $5$-digit numbers can be made from digits in the number $75226522$? I'm having troubles with this one:
How many $5$-digit numbers can be made from digits in the number $75226522$ ?
So there is: one seven, one six, two fives, and four twos in that number.
So I guess I need to use combination with repetition but... | Add up the following:
*
*The number of ways to rearrange $\color\red{2222}\color\green{ 5}\color\orange{ }\color\lightblue{ }$, which is $\frac{5!}{\color\red{4!}\color\green{1!}\color\orange{ }\color\lightblue{ }}= 5$
*The number of ways to rearrange $\color\red{2222}\color\green{ }\color\orange{6}\color\lightb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1886901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Functional Equation $f(f(x))=\frac{x}{2}+3$ Find all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ of class $C^1$ where $f\circ f(x)=\dfrac{x}{2}+3$.
I need help with this problem.
| Let $f^{\circ n}$ be the shorthand for composing $f$ with itself for $n$ times.
Let $g(x) = f(x+6)-6$. Since $f \in C^1$, so does $g$. Notice
$$g\left(\frac{x}{2}\right) = f\left(\frac{x+6}{2}+3\right)-6 = f(f^{\circ 2}(x+6)))-6
= f^{\circ 2}(f(x+6)) - 6\\ = \frac{f(x+6)}{2} + 3 - 6 = \frac{f(x+6)-6}{2} = \frac{g(x)}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1887028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Prove that $\lnot (p \implies q)$ is equivalent to $p \land \lnot q$? By equivalent I mean the biconditional, as in
$$\lnot (p \implies q) \iff p \land \lnot q$$
Given the definition of implication, I understand why this is true, but I need a bit of help showing this with a formal proof using rules like $\lor-\text{El... | $$\newcommand{\DeductionBox} [1]{\begin{array} {l|} #1 \\ \hline \end{array}}$$
The reverse direction is straightforward:
$$\begin{array} {l}
\begin{array} {rlr}
(1) & p \land \lnot q & \text{Given} \\
(2) & p & \land\text{-Elimination of }(1) \\
(3) & \lnot q &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1887802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Which of the following sets are orthonormal? I have three sets to determine whether they're orthonormal or not. These are;
(a) $\{1,\frac{x}{\sqrt2}+\frac{x^3}{\sqrt2},\frac{x^2}{\sqrt2}-\frac{x^4}{\sqrt2}$}
(b) $\{\frac{x}{2}-\frac{x^2}{2}-\frac{x^3}{2}, -\frac{x}{2}-\frac{x^2}{2}+\frac{x^4}{2}, \frac{x^2}{2}-\frac{x^... | In polynomial space, the standard orthonormal basis (with respect to the inner-product you gave above) is $\{1, x, x^2, x^3, ...\}$ and so, for example, a polynomial such as $2 + 3x^2 - 4x^4$ can be represented by the vector $(2, 0, 3, 0, -4, 0, ...)$ since
$$2\cdot 1 + 0 \cdot x + 3 \cdot x^2 + 0 \cdot x^3 + (-4) \cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $x$ and $y$ where $20!=\overline{24329020081766xy\dots}$ Find $x$ and $y$ where $20!=\overline{24329020081766xy\dots}$(without using calculator.)
My attempt:I first find how many zeroes does it have:
$$\left\lfloor {\frac{20}{5}} \right\rfloor=4.$$
It can be solved easily if we know that after $y$ there are only t... | Some judicious pairing of the numbers from $1$ to $20$ leads to the rough estimate
$$\begin{align}
20!&=(20\cdot1)(17\cdot3)(19\cdot2)(12\cdot10)(15\cdot4)(18\cdot5)(16\cdot6)(14\cdot7)(13\cdot8)(11\cdot9)\\
&\approx20\cdot50\cdot40\cdot120\cdot60\cdot100\cdot100\cdot100\cdot100\cdot100\\
&\approx1000\cdot5000\cdot60\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Counting multiple of $3$ in a triangular arrangement of the integers. I write
:$$\mathbb{N}_\triangle = \begin{matrix}
&&&&&21&\ldots \\
&&&&15&20&\ldots \\
&&&10&14&19&\ldots \\
&&6&9&13&18&\ldots \\
&3&5&8&12&17&\ldots \\
1&2&4&7&11&16&\ldots
\end{matrix}$$
Can we det... | The number of multiples of $3$ in the interval $[1,N]$ is $\lfloor N/3\rfloor$.
Moreover the $n^{th}$ column of $\mathbb{N}_\triangle$ is the set of consecutive integers from $T_{n-1}+1$ to $T_n$ with $T_n=n(n+1)/2$.
Hence, the number of multiples of $3$ in the $n^{th}$ column of
$\mathbb{N}_\triangle$ is
$$f_3(n)=\lf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
coefficient of $x^9$ in binominal expansion Find the co efficient of $x^9$ in the expansion of $$\left[\frac{(1+x)^3}{(1-x)^3}\right]^\frac{1}{2}$$
I get answer as $\frac{3}{128}$
But in my book the answer is given as $\frac{1}{128}$
I was very much confused .someone help me please.
| You could notice that $$\left(\frac{(1+x)^3}{(1-x)^3}\right)^\frac{1}{2}=\left(\frac{1+x}{1-x}\right)^\frac{3}{2}=\left(1+2\sum_{i=1}^\infty x^i\right)^\frac{3}{2}$$ Now, being patient, you could use the generalized binomial theorem.
If you are not, Taylor series (tedious too) would lead to
$$\left(\frac{(1+x)^3}{(1-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1889664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
A simpler way to solve this determinant equation-$2$
Question Statement:-
Solve the following determinant equation
$$\begin{vmatrix}
x & 2 & 3 \\
6 & x+4 & 4 \\
7 & 8 & x+8 \\
\end{vmatrix}=0$$
My Solution:-
$$\begin{vmatrix}
x & 2 & 3 \\
6 & x+4 & 4 \\
7 & 8 & x+8 \\
\end{vmatrix}=
\begin{vmatrix}
x+13 & x+14 ... | Your solution is fine but here is a more simple way to solve this problem:
$$\begin{vmatrix}
x & 2 & 3 \\
6 & x+4 & 4 \\
7 & 8 & x+8 \\
\end{vmatrix} $$
= $$-\begin{vmatrix}-\begin{pmatrix}
0 & 2 & 3 \\
6 & 4 & 4 \\
7 & 8 & 8 \\
\end{pmatrix} - x I)\end{vmatrix} = 0$$ where $I$ is the identity matrix.
It lookes similar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1890082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $b^2-4ac \equiv 0 \pmod{p}$
Let $a,b,c,r$ be integers and $p \geq 5$ be a prime such that $ax^2+bx+c \equiv a(x-r)^2 \pmod{p}$. Prove that $b^2-4ac \equiv 0 \pmod{p}$. Does the same hold if $p = 3$?
I wasn't sure how to go about proving this. We can rearrange the congruence to get $$x(b+2ar)+c-ar^2 \equiv... | Since $ax^2+bx+c \equiv a(x-r)^2 \pmod{p}$ is a "forall" statement for all the x in least residue system with respect to p,
$x(b+2ar)+c-ar^2 \equiv 0 \pmod{p}$ also satisfies this condition.
Since $0$ and $1$ are in the least residue system with respect to p, Therefore
$$0 \times (b+2ar)+c-ar^2 \equiv c-ar^2 \equi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1892486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
find $f^{(15)}(0)$ if $f = \frac{10x^2+12x+4}{(x+2) (x^2 +1)}$ $$f = \frac{10x^2+12x+4}{(x+2)(x^2 +1)}=\frac{4}{x+2}+\frac{6x}{x^2+1}=\frac{2}{1-\left(-\frac x 2 \right)}+\frac{6x}{1-(-x^2)}$$
So we end up with $$2 \sum_{n=0}^{15} \left(-\frac x 2 \right)^n + 6x \sum_{n=0}^{15} (-x^2)^n$$ but because we are looking for... | We have
\begin{align*}
f(x) = \frac{4}{x+2} + \frac{3}{x+i} + \frac{3}{x-i}
\end{align*}
and hence
\begin{align*}
f^{15}(x) = -(15!)(4(x+2)^{-16} + 3(x+i)^{-16} + 3(x-i)^{-16}
\end{align*}
and
\begin{align*}
f^{15}(0) = -(2^{-14}+6)\cdot 15!
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1893250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Show that, for all $n > 1: \log \frac{2n + 1}{n} < \frac1n + \frac{1}{n + 1} + \cdots + \frac{1}{2n} < \log \frac{2n}{n - 1}$ I'm learning calculus, specifically limit of sequences and derivatives, and need help with the following exercise:
Show that for every $n > 1$,
$$\log \frac{2n + 1}{n} < \frac1n + \frac{1}{n + ... | For the LHS it is $$\frac { 1 }{ n } +\frac { 1 }{ n+1 } +\cdots +\frac { 1 }{ 2n } =\int _{ n }^{ n+1 }{ \frac { 1 }{ n } dx } +\int _{ n+1 }^{ n+2 }{ \frac { 1 }{ n+2 } dx } +...+\int _{ 2n }^{ 2n+1 }{ \frac { 1 }{ 2n } dx } >\\ >\int _{ n }^{ n+1 }{ \frac { 1 }{ x } dx } +\int _{ n+1 }^{ n+2 }{ \frac { 1 }{ x } dx }... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1895591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
If $A+B+C=\frac {\pi}{2}$ then prove that.... If $A+B+C=\frac {\pi}{2}$ thrn prove that:$$\tan (A+B) [\tan A-\tan B]=\sqrt {1+\cot^2 C} (\sec A\cos B-\cos A\sec B)$$
My attempt
Here, $$A+B+C=\frac {\pi}{2}$$
$$A+B=\frac {\pi}{2} - C$$
Now,
$$L.H.S.=\tan(A+B) [\tan A-\tan B]
= \tan\left(\frac{\pi}{2} - C\right) \left(\f... | \begin{align*}
\cot C \frac{\sin (A-B)}{\cos A \cos B} &= \frac{\cos C}{\sin C}\frac{\sin (A-B)}{\cos A \cos B} \\
&= \frac{\sin(A+B)}{\cos(A+B)}\frac{\sin (A-B)}{\cos A \cos B} \\
&=\frac{\frac{\cos 2B - \cos 2A}{2}}{\sin C(\cos A \cos B)}\\
&= \frac{\cos^2 B - \cos^2 A}{\sin C \cos A \cos B}\\
&= \frac{1}{\sin C}(\se... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1895699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I find the tangent of an equation, when the $x$-coordinate is $1$? Find the equation of the tangent to the graph $y = x^2 + 4x + 3$, whose $x$ coordinate is $1$.
Write your answer in form $y = mx + c$
I got the equation $y = 2x^2 + 2x + 2$.
Is this correct?
| $$y=x^2+4x+3$$
$$\frac{dy}{dx}=2x+4$$
At $x=1$, y is 8 and the derivative is 6, so write
$y=mx+c\rightarrow8=6(1)+c$; we get $c=2$. Hence the tangent at $x=1$ is $y=6x+2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1896046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What is the probability of selecting three different integers, without replacement, from the numbers $1$ to $15$ so their sum is divided by $3$?
What is the probability of selecting three different integers, without replacement, from the numbers $1$ to $15$, inclusive, so that their sum can be exactly divided by $3$?
... | Correct. Here it is a high-tech alternative. Let we consider:
$$q(x,y)=(1+yx)(1+yx^2)(1+yx^3)\cdot\ldots(1+yx^{15}).$$
If we take the coefficient of $y^3$ in the expansion of $q(x,y)$ we get a polynomial of the $x$-variable, that evaluated at $x=1$ gives the number of subsets of $\{1,2,\ldots,15\}$ with three elements:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1896828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the co-efficient of $x^3y^3zw^2$ in the expression of $(x-y+2z-2w)^9$.
Find the co-efficient of $x^3y^3zw^2$ in the expression of $(x-y+2z-2w)^9$.
I have clearly understood how to use the bin theorem for $(a+b)^n$, but am unable to extend the theorem to $4$ terms .
| It is also possible to apply the binomial theorem iteratively. We focus on one term and keep the other terms together. This way we reduce the number of terms step by step by one, till we have only two left at the end.
To keep the calculations manageable it is convenient to use the coefficient of operator $[x^j]$ to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1897814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Sum from $n=1$ to $N$ of $\sin(nx)/2^n$ The question states:
I'm not sure what to do here. I used the geometric sum to do $2^{-n}$, but I can't think of a way to sum $\sin(n\theta)$ to get the required RHS.
Further working
Apologies for maybe being slow or missing something obvious here... I don't know how to get i... | The solution below is just to give a basic proof without using Euler's identity.
\begin{align*}
&\quad \sum_{n=1}^N \frac{\sin(n\theta)}{2^n} \cdot 2^N(5-4\cos\theta) \\
&= \left( \sum_{n=1}^{N} 2^{n-1}\sin(N-n+1)\theta \right)(1+4-4\cos \theta) \\
&= \left( \sum_{n=1}^{N} 2^{n-1}\sin(N-n+1)\theta \right) + \left( \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Finding the determinant of a skew-symmetric matrix $K$
Find the determinant of the skew-symmetric matrix $K$
$$K = \begin{bmatrix}
0 & 1 & 3\\
-1 & 0 & 4 \\
-3 & -4 & 0 \\
\end{bmatrix}$$
My Attempted Solution:
I performed the following row operations to reduce $K$ into upper-triangular form $U$
$R_2 \leftrightarrow... | $$\begin{array}{rl} \det \left[\begin{array}{cc|c} 0 & 1 & 3\\ -1 & 0 & 4 \\ \hline -3 & -4 & 0\end{array}\right] &= \det \begin{bmatrix} 0 & 1\\ -1 & 0\end{bmatrix} - \det \left(\begin{bmatrix} 0 & 1\\ -1 & 0\end{bmatrix} - \begin{bmatrix} 3\\ 4\end{bmatrix} \begin{bmatrix} 3\\ 4\end{bmatrix}^T\right)\\ &= 1 - \det \b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1904093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Prove that $\cos^6 \beta - \sin^6 \beta=\frac {1}{16} (15 \cos 2\beta + \cos 6\beta)$.
Prove the given trigonometric identity:
$$\cos^6 \beta - \sin^6 \beta=\frac {1}{16} (15 \cos 2\beta + \cos 6\beta)$$
My Approach:
\begin{align*}
\text{L.H.S.} &=\cos^6 \beta - \sin^6 \beta\\
&=(\cos^2 \beta)^3 - (\sin^2 \beta)^3\... | Use $$\cos2A=2\cos^2A-1=1-2\sin^2A$$ to find
$$8(\cos^6 \beta - \sin^6 \beta)=(2\cos^2\beta)^3-(2\sin^2\beta)^3=(1+\cos2\beta)^3-(1-\cos2\beta)^3=?$$
Finally use $$\cos3B=4\cos^3B-3\cos B\iff4\cos^3B=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1905142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
If the sum of the digits of $a+b, a+c, a+b$ are bounded by $k$, what is the maximum sum of digits of $a+b+c$?
Question: For $n \in \mathbb{N}$, let $S(n)$ denote the sum of the base 10 digits of $n$. Assuming that $a,b,c \ge 0$ and
$$
S(a+b), S(b+c),S(a+c) \le k,
$$
What is the maximum possible value of $S(a+b+c)$... | We will show that the maximum is indeed $15k - 9$.
First, let's investigate the inequalities which you used, and in particular, what the effect is if either is not an equality.
In your previous answer, it was shown that $S(x) \leq 5S(2x)$ essentially by noting that this is true for each digit, and then noting that $S(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1905610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\frac{x^{3333}+x^{333}+x^{33}+x^{3}+1996}{100(x^2+x)}$ if $x^2+x+1=0$
Find $$\frac{x^{3333}+x^{333}+x^{33}+x^{3}+1996}{100(x^2+x)}$$ if $x^2+x+1=0$
My work so far:
1)$x^2+x=-1$, then $100(x^2+x)=-100$
2)$x^2+x+1=0$
$x=\frac{-1\pm\sqrt{3}i}{2}$
| $x^2+x+1=0\implies$
$x=-(-1)^{1/3},(-1)^{2/3}\implies$
$x^3=1\implies$
$\cfrac{x^{3333}+x^{333}+x^{33}+x^{3}+1996}{-100}=\cfrac{1^{1111}+1^{111}+1^{11}+1^{1}+1996}{-100}=-20$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1906012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to understand the following result? I am beginning to study probability and I found the following problem:
What is the probability of to get a number twice after two throwings?
The answer was that since we have six distinct results in a fair die every possible result has $1/6$ of probability of happen and since tha... | The probability of drawing a $1$ is $1/6$, because there is a single favorable outcome and six possible ones.
The probability of drawing two $1$s in a row is $1/36=1/6^2=(1/6)^2$ because there are now thirty-six possible outcomes. The product appears for combinatorial reasons. Specifically, the set of possible cases is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1907412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can we find incentre of a triangle by using equation of lines? My question is, in the manner which we find the orthocentre of a triangle merely by using it's side's line equations, can we also find the incentre?
For example, consider,
$$3x+4y-7=0$$
$$4x-3y+19=0$$
$$18x-6y+7=0$$
We can find the orthocentre of the tria... | The intersection point (Incenter) of the internal bisectors can be obtained through a formula with the cofactors, coefficients and constants of the equations.
$A·x+B·y+C=0$
$D·x+E·y+F=0$
$G·x+H·y+I=0$
$M=\left( \begin{array}{} A & B & C \\ D & E & F \\ G & H & I \\ \end{array} \right) $
$p=\sqrt{A^2+B^2}$
$q=\sqrt{D^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1908347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Prove the limit of a function with two variables is zero I have to prove as homework that
$$\lim_{(x,y)\to (0,0)} \frac{\sin\left(x^2y\right)}{x^2+y^2}=0$$
and that
$$\lim_{(x,y)\to (0,0)} \frac{\tan\left(x^3\right)}{x^2+y^2}=0.$$
For the second limit here's what I did:
let $\epsilon >0$ we want $\delta >0$ such that i... | Hint: Use the fact that $\sin(x)$ and $\tan(x)$ are odd to reduce to the case that $x,y\ge0$. Then
$$
x^2\le x^2+y^2\quad\text{and}\quad y\le\left(x^2+y^2\right)^{1/2}
$$
therefore,
$$
\frac{x^2y}{x^2+y^2}\le\left(x^2+y^2\right)^{1/2}
$$
Then use the fact that for $x\ge0$, $\sin(x)\le x$.
For the second inequality, yo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1910570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How to come up with ways of simplifying equations? Consider the cubic equation
$$ x^3 + bx^2 + cx + d = 0$$
If one replaces $x$ with $x - {b\over 3}$ then this reduces to
$$ x^3 + px + q = 0$$
If one knows what to substitute with then checking the claim is easy. But my question is:
How does one see which substitution ... | Recall that $-b$ is the sum of the roots of $ x^3 + bx^2 + cx + d = 0.$ That is, if the roots are $r_{1},\,r_{2},\,r_{3},$ then $(r_1 + r_2 + r_3) = -b.$ See Vieta's formulas if this is new to you.
Now note that if the graph is shifted right by $\frac{b}{3},$ which can be done by replacing $x$ with $x - \frac{b}{3},$ t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1910915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Determine all real $x$ that satisfies $\sqrt{3-x} - \sqrt{x+1} > \frac{1}{2}$ I have this equation that I have to solve:
Determine all real $x$ that satisfies $\sqrt{3-x} - \sqrt{x+1} > \frac{1}{2}$
Maybe it involves means? The square roots are leaning towards it... Could someone help me?
| Note that
*
*$\sqrt{3-x}$ is defined only for $x\le3$
*$\sqrt{x+1}$ is defined only for $x\ge-1$
*The first square root is strictly decreasing and the second strictly increasing, so the whole LHS must be strictly decreasing (to see why, graph the functions)
Thus it remains to solve for $x$ in the case of equalit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1911239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Positive integers $a$ and $b$ are such that $a + b =\frac{ a}{b} +\frac{ b}{a}$. What is the value of $a^2 + b^2$? I tried to take $a = 1$ and $b = 2$ and arrive at the conclusion that $a^2 + b^2 = 2$ which indeed is the answer.
However, I wanted to prove this equation algebraically without using the replacement techni... | If $a>1$ or $b>1$, then it follows that $a>\frac{a}{b}$ or $b>\frac{b}{a}$ so $$a+b=\frac{a}{b}+\frac{b}{a}<a+b$$
Which is a contradiction.
Thus we conclude that $a,b \le 1$, and thus $a=b=1$. Therefore, the answer is $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1912782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
homework - probability problem A dice is thrown 3 times and the sum of the 3 numbers thrown is 15. Then, what is the probability that the first throw was a four?
My Attempt:
The sample space for the above problem, $S = 6^3$.
Combinations of thrown numbers such that their sum is $15$ and the first digit is $4 = [\{4, 5,... | Since the sum of the 3 numbers thrown is 15 it follows that the sample space is given by 10 elements: 15=6+6+3 ($3!/2!=3$ ways), 15=6+5+4 (3!=6 ways), 15=5+5+5 ($3!/3!=1$ way).
There are two cases when $4$ is the first value 4,5,6 and 4,6,5.
Therefore the probability is $p=2/10=1/5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1913428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Symmetric and antisymmetric matrices as subspaces of $M_{3 \times 3}(\Bbb R)$ Show that the set $S$ compound of $3 \times 3$ symmetric matrices and the set $A$ of the antisymmetric matrices, are subspaces of $M_{3\times 3}(\Bbb R)$. Determine basis for $S$ and $A$. Show, using a $\dim$ relation, that $S \oplus A = M_{3... | Any $n \times n$ matrix $B$ can be written uniquely as a sum of a symmetric matrix and a skew symmetric matrix:
\begin{align*}
B = \frac{1}{2}(B+B^t) + \frac{1}{2}(B-B^t)
\end{align*}
where $B^t$ denotes the transpose of $B$. Thus $M_{n\times n}(\mathbb{R}) = S + A$. If $B \in S \cap A$, then $B^t = B$ and $B^t = -B$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1913820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
RMO inequality problem If $a,b,c,d,e>1$, then prove that:
$$\frac{a^2}{b-1} + \frac{b^2}{c-1} + \frac{d^2}{e-1} + \frac{c^2}{a-1} + \frac{e^2}{d-1} \ge 20. $$
I don't know how to begin. What should be the approach?
| Note that $x^2-4(x-1) = (x-2)^2\ge 0$ for all $x\in\mathbb{R}$, and hence $x^2\ge 4(x-1)$ for all $x\in\mathbb{R}$. As such, we have
$$ \frac{a^2}{b-1} + \frac{b^2}{c-1} + \frac{d^2}{e-1} + \frac{c^2}{a-1} + \frac{e^2}{d-1} \ge 4\left(\frac{a-1}{b-1} + \frac{b-1}{c-1} + \frac{c-1}{a-1} + \frac{d-1}{e-1} + \frac{e-1}{d-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1913873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
} |
Find the ratio in which the perpendicular Find the ratio in which the perpendicular from $(4,1)$ to the line segment joining the points $(2,-1)$ and $(6,1)$ divides the segment.
My Approach:
Equation of the line joining the points $(2,-1)$ and $(6,1)$ is given by:
$$y-y_1=\frac {y_2-y_1}{x_2-x_1} (x-x_1)$$
$$y+1=\frac ... | You have a line $x-2y-4=0$, and you can fix it to the form $y=\frac{1}{2}x-2$.
The perpendicular line will have a slope of $-\frac{1}{a}$ or in this case $-2$ and will pass via $(4,1)$.
Hence:
$$1=4\times (-2)+b \implies b=9$$
So the perpendicular line:
$$y=-2x+9$$
Now let us find the point these two line intersect:
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1914268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve $3^x \equiv 43 \pmod {97}$ using Pohlig-Hellman Problem: Solve $3^x \equiv 43 \pmod {97}$ using the Pohlig-Hellman algorithm.
Here is my solution, but the problem is that I don't get it right at the end.
First of all, I calculate $\phi(97) = 96 = 2^5 \cdot 3$.
Now, let $x=a_0 + 2^5a_1=a_0+32a_1$. Then
$$ {3^{(a_0... | $3^{70} \mod 97$ IS 43, and your solution is fine. The only problem is that it's not unique - because the order of $3 \mod 97$ is 48 instead of 96.
So morally speaking, you should look at $x \mod 48$ instead. Using Chinese remainder theorem, you would split that up into $x \mod 16$ and $x \mod 3$. Your calculations we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1915508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $A+B+C+D=2\pi$, prove that: If $A+B+C+D=2\pi$, prove that: $$\cos A+\cos B+\cos C+\cos D=4\cos\frac {A+B}{2}\cdot\cos\frac {A+C}{2}\cdot\cos\frac {B+C}{2}$$.
My Approach:
Here,
$$A+B+C+D=2\pi$$
$$A+B=2\pi - (C+D)$$
$$ \sin(A+B)=\sin(2\pi-(C+D))$$
$$\sin(A+B)=-\sin(C+D)$$
Again,
$$\cos(A+B)=\cos(2\pi-(C+D))$$
$$\cos(... | $$cos\frac{C+D}{2}=cos\frac{2\pi-(A+B)}{2}=-cos\frac{A+B}{2}$$
$$cos\frac{C-D}{2}=cos\frac{C+A+B+C-2\pi}{2}=-cos\frac{A+C+B+C}{2}$$
Now go to your last line
$$2cos\frac{A+B}{2}(cos\frac{A-B}{2}+cos\frac{A+C+B+C}{2})$$
Use the rule to turn an addition of cosinuses to a multiplication and you will get your proof
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1916560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Solve algebraically and graphically: $\arcsin(x) + \arccos\left(\frac{x}{2}\right) = \frac{5\pi}{6}$ Solve:
$$\arcsin(x) + \arccos\left(\frac{x}{2}\right) = \frac{5\pi}{6}$$
I think the algebraic solution should start like:
$$\arcsin(x) = \frac{5\pi}{6} - \arccos\left(\frac{x}{2}\right)$$
$$x = \sin\left(\frac{5\pi}{6}... | $$\arcsin x + \arccos\frac{x}{2} = \frac{5\pi}{6}$$
$$\arcsin x = \frac{5\pi}{6} - \arccos\frac{x}{2}$$
$$x = \sin{\left(\frac{5\pi}{6} - \arccos\frac{x}{2}\right)}$$
$$x=\sin\frac{5\pi}6\cos \left(\arccos\frac{x}{2}\right)-\cos\frac{5\pi}6 \sin\left(\arccos\frac{x}{2}\right)$$
$$x=\frac12\cdot\frac x2+\frac{\sqrt3}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1916903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove $\int_0^{\pi/2}x^2\sqrt{\cot x} \ dx=\frac{\sqrt2}{8}\left( \frac{5\pi^{3}}{12}-\pi^2\ln2-\pi \ln^22 \right)$ I came across the follwing integral:
$$\int_0^{\pi/2}x^2\sqrt{\cot x} \ dx=\frac{\sqrt2}{8}\left( \frac{5\pi^{3}}{12}-\pi^2\ln2-\pi \ln^22 \right)$$
I try to do it like the following:
Consider
$$I(a,b)=\i... | A slightly easier form for $I(a, \frac{1}{2})$ is
$$I(a, \frac{1}{2}) = 2^{a-\frac{3}{2}} \beta\left(\frac{1+2a}{4}\right) (1 + \sin \pi a + \cos \pi a), $$
where $\beta(p) = \Gamma(p)^2/\Gamma(2p)$ is the central beta function. (This can be easily obtained by applying the Euler reflection formula and the Legendre dupl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1921384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 2,
"answer_id": 0
} |
Why is it intuitively "obvious" that: $2^0+2^1+2^2+...+2^n=2^{n+1}-1$ Why is it "obvious" that: $$2^0+2^1+2^2+...+2^n=2^{n+1}-1$$
And how about: $$2^{-1}+2^{-2}+2^{-3}+...+2^{-n}=1-2^{-n}$$
I am looking for some intuition about these facts.
Is there a similar pattern when the base is $3$?
| Adding another answer to the very good proofs below : you are already familiar with this concept each time you use numbers.
$999_{10} + 1 = 1000_{10}$ is the same than $(9 \times 10^2)+(9 \times 10^1)+(9 \times 10^0) +1 = 10^3$.
If you set $b=9_{10}$ you wrote $(b-1) \times b^2+ (b-1) \times b^1+(b-1) \times b^0 +1 = b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1921892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Existence of $A$ and $B$ such that $A^2-4A+4I=0$, $A+B=\begin{pmatrix} 4 & 1 \\ -3 & 4\end{pmatrix}$ and $AB=$...
Question: Prove or disprove that $A$ and $B$ exist such that
\begin{align}
&A^2-4A+4I=0\\
&A+B=\begin{pmatrix} 4 & 1 \\ -3 & 4\end{pmatrix}\\
&AB=\begin{pmatrix} 1 & 1 \\ -9 & 3\end{pmatrix}\\
\end{align... | The second two equations are of the form $A+B=C$ and $AB=D$.
Then $A^2+AB=AC$ and so $A^2=AC-AB=AC-D$.
Substitute for $A^2$ in the original equation to get
$$AC-D-4A+4I=0$$
or
$$A(C-4I)=D-4I.$$
So if the inverse of $C-4I$ exists, this gives you a unique $A$. From this you can compute $B=C-A$ (and verify that indeed $AB... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1923301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
limit of $\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4}$ I have question. I want to solve this limit. it's $\frac{0}{0}$ so we have to change it. there is two way with two different value.
$\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4}$
First way:
before that we know that $\lim_{x\to 0} \frac{\sin x}{... | Note that we have
$$\begin{align}
\frac{\sin^2(x)-x^2\cos^2(x)}{x^4}&=\frac{\left(\frac{\sin(x)}{x}\right)^2-1}{x^2}+\frac{\sin^2(x)}{x^2}\\\\
&=\color{blue}{\underbrace{\left(\frac{\sin(x)}{x}+1\right)}_{\to 2\,\,\text{as}\,\,x\to 0}}\left(\frac{\sin(x)-x}{x^3}\right)+\color{red}{\underbrace{\frac{\sin^2(x)}{x^2}}_{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1924213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 3
} |
find value of determinant $f(n)=β^n+α^n$
then $$
\begin{vmatrix}
3 & 1+f(1) & 1+f(2) \\
1+f(1) & 1+f(2) & 1+f(3) \\
1+f(2) & 1+f(3) & 1+f(4) \\
\end{vmatrix}
$$
I don't quite understand which property of determinant can we use here? i tried using linear property but i did not get... | There is a trick, indeed. The values defined by $f(n)=\alpha^n+\beta^n$ obey the recurrence relation
$$ f(n+2) - (\alpha+\beta)\,f(n+1) + (\alpha\beta)\,f(n) = 0$$
hence the original determinant equals
$$ (1-\alpha)(1-\beta)\det\begin{pmatrix}1+f(0) & 1+f(1) & 1+f(2) \\ 1+f(1) & 1 + f(2) & 1+f(3) \\ 1 & 1 & 1\end{pmatr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1924279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solution of simultaneous equations using properties of symmetric polynomials
Solve the following system of equations.
$$ x + y + z = 5 $$
$$ \frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx} = 5 $$
$$ x^3 + y^3 + z^3 = 53 $$
How would I use properties of symmetric polynomials (e) to solve this?
| .From $\frac{1}{yz} + \frac{1}{xz} + \frac{1}{xy} = 5$, we get:
$$
\frac{x+y+z}{zyx} = 5 \implies zyx = 1
$$
Now, $$
x^3+y^3+z^3 - 3xyz= (x+y+z)(x^2+y^2+z^2 - xy-yz-zx) \\= (x+y+z)((x+y+z)^2-3(xy+yz+zx))
$$
Solving using values, this gives $xy+yz+zx = 5$.
Now, note that:
$$
(t-x)(t-y)(t-z) = t^3 - (xy+yz+zx)t^2 + (x+y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1924707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
expanding and factoring in algebra Expanding very simple process, just open round brackets and multiply each with each for example
$(x-6)(x+3) = x^2 + 3x - 6x -18 = x^2 -3x -18$
Factoring is confused me a little for example I have $x^2 -3x -18$ I try to brute force all combinations in my mind and get the result like th... | Here is another alternative
$$
\begin{align}
x^2-3x-18 &= \frac{1}{4}\bigg(4x^2-4\cdot 3x-4\cdot 18\bigg )\\
&= \frac{1}{4}\bigg((2x)^2-2\cdot 3(2x)-72\bigg )\\
\end{align}$$
Now suppose we define $t=2x$
$$\begin{align}
\frac{1}{4}\bigg(t^2-2\cdot 3t-72\bigg )&= \frac{1}{4}\bigg(t^2-2\cdot 3t+9-9-72\bigg )\\
&= \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1925037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove $2^{n+3}|(p^2-1)$ Olympiad problem:
let odd prime nunmber $p$ such $$p|(2+\sqrt{3})^{2^n}+(2-\sqrt{3})^{2^n}$$
show that:
$$2^{n+3}|(p^2-1)$$
I want to take advantage of this conclusion：
if prime factor $ p $ of a Fermat number $ 2^{2^n} + 1 $ ,then $ p \equiv 1 \pmod{2^{n + 2}} $
But I failed
| If $p\equiv 1\pmod{2^{n+2}}$, then either $p\equiv 1\pmod{2^{n+3}}$ or $p\equiv 2^{n+2}+1\pmod{2^{n+3}}$. In either case, squaring yields $p^2\equiv 1\pmod{2^{n+3}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1925109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluation of $\int^{\frac{\pi}{2}}_{0}\frac{\cos^6 x}{\sin x+\cos x}dx$
Evaluation of $\displaystyle \int^{\frac{\pi}{2}}_{0}\frac{\cos^6 x}{\sin x+\cos x}dx$
$\bf{My\; Try::}$ Let $\displaystyle I = \int^{\frac{\pi}{2}}_{0}\frac{\cos^6 x}{\sin x+\cos x}dx = \int^{\frac{\pi}{2}}_{0}\frac{\sin^6 x}{\sin x+\cos x}dx$ ... | Let
$$ A=\int^{\frac{\pi}{2}}_{0}\frac{\cos^6 x}{\sin x+\cos x}dx, B=\int^{\frac{\pi}{2}}_{0}\frac{\sin^6 x}{\sin x+\cos x}dx. $$
Clearly $A=B$ by changing variable from $x$ to $\frac{\pi}{2}-x$. Now
\begin{eqnarray}
A+B&=&\int^{\frac{\pi}{2}}_{0}\frac{\cos^6 x+\sin^6x}{\sin x+\cos x}dx\\
&=&\int^{\frac{\pi}{2}}_{0}\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1929308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Polar to cartesian form of r=sin(3θ) $$\sin(2θ) = \sinθ\cosθ + \cosθ\sinθ = 2\sinθ\cosθ $$
$$\cos(2θ) = \cosθ\cosθ - \sinθ\sinθ = \cos²θ - \sin²θ $$
$$\sin(3θ) = \sin(2θ+θ) $$
$$= \sin(2θ)\cosθ + \cos(2θ)\sinθ $$
$$= 2\sinθ\cos²θ + \cos²θ\sinθ - \sin³θ
$$ $$= 2\sinθ(1-\sin²θ) + (1-\sin²θ)\sinθ - \sin³θ
$$ $$= 2\sinθ ... | $$\sin(3\theta) = 3\sin\theta - 4\sin^3\theta$$
$$\sin^3\theta = \sin\theta\cdot \sin^2\theta$$
$$r = 3\sin\theta - 4\sin\theta\sin^2\theta$$
Hence
$$r^2 = 3r\sin\theta\left(1 - \frac{4}{3}\sin^2\theta\right)$$
$$r^2 = 3y\left(1 - \frac{4}{3}\frac{r^2}{r^2}\sin^2\theta\right) = 3y\left(1 - \frac{4}{3r^2}y^2\right)$$
So... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1930388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Show that $arg(z_1) \equiv \arg(z_2) \pmod{2\pi}$ assuming $|a|^2 - 4|b| \leq 0$ and $\arg b \equiv 2\arg a \pmod{2\pi}$ Let $a, b \in \mathbb{C}^{*}$ and $(E)$ the following equation : $z^2 - az + b = 0$.
Assuming :
$\left\{
\begin{equation}
\begin{aligned}
& \arg b \equiv 2\arg a \pmod{2\pi} \\
& |a|^2 - 4|b| \geq 0
... | I think the condition you want is actually $|a|^2-4|b|\ge 0$ (otherwise, take $a = b = 2$, and the roots of $z^2-2z+2$ are actually $1\pm i$, which have different arguments)
If the roots have the same argument, then they should have the same argument as their sum, i.e. $a$. As such, let $y = z/a$. We then have
$$ z^2-a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1930601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the relation between $\lfloor x+y \rfloor$ and $\lfloor x \rfloor + \lfloor y\rfloor$ for any reals $x,y$ What is the relation between $\lfloor x+y \rfloor$ and $\lfloor x \rfloor + \lfloor y\rfloor$ for any reals $x,y$?
My effort:
We have $x+y-1\lt \lfloor x+y\rfloor \leq x+y$. What can I do after this? P... | For some real number $0 \le \alpha < 1,\quad x = \lfloor x \rfloor + \alpha$
That is really all you need to show that $$x-1 < \lfloor x \rfloor \le x < \lfloor x \rfloor + 1$$
and
$$\lfloor \lfloor x \rfloor + y \rfloor = \lfloor x \rfloor + \lfloor y \rfloor$$
One thing you can use it for is, for some $0 \le \alpha, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1931463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
A similar Somos sequence problem prove $A_{2n}B_{n+3}$ is integer sequence I have read some interesting with Somos sequence,Recently I met a similar question：
For a fixed positive integer $k$, there are two sequences $A_n$ and $B_n$.
They are defined inductively, by the following recurrences.
$$A_1 = k,A_2 = k,A_{... | Considering the sequence (for $n \ge 2$):
$$
A_{2(n-1)} B_{(n-1)+3}, \space A_{2n} B_{n+3}, \space A_{2(n+1)} B_{(n+1)+3} \space \rightarrow \space A_{2n-2} B_{n+2}, \space A_{2n} B_{n+3}, \space A_{2n+2} B_{n+4}
$$
According to definitions, we have:
$$
{ A_{2n+3} = A_{2n+2} A_{2n+1} = (A_{2n+1} A_{2n}) A_{2n+1} = A^2_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1932492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\lim_{x\to 0} \frac{1-\cos x\sqrt{\cos 2x}\sqrt[3]{\cos 3x}}{x^2}$ I want to evaluate:
$$\lim_{x\to 0} \frac{1-\cos x\sqrt{\cos 2x}\sqrt[3]{\cos 3x}}{x^2}.$$
Here's what I did. We know that as soon as ${x\to 0}$
$$1 - \cos x = \frac{x^2}{2} + O(x^4)$$
Therefore
$$\cos x = 1 - \frac{x^2}{2} + O(x^4)$$
Now... | Since in a neighbourhood of the origin $\cos(x)= 1-\frac{x^2}{2}+o(x^2)$, it follows that $\cos(2x)= 1-2x^2+o(x^2)$ and $\cos(3x)= 1-\frac{9x^2}{2}+o(x^2)$. Since $\sqrt{1+z}=1+\frac{z}{2}+o(z)$ and $\sqrt[3]{1+z}=1+\frac{z}{3}+o(z)$ we get that:
$$ \cos(x)= 1-\frac{x^2}{2}+o(x^2),\quad \sqrt{\cos(2x)}= 1-x^2+o(x^2),\q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1933292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Question about Diophantine equations? Solve the Diophantine equation.
$6x+ 9y = 15$
We can see that $x = 1$ and $y=1$ solves the equation, so all solutions are given by $x = 1 + 3n$ and $y = 1 - 2n, n \in \mathbb{Z}$. By using:
$$x=x_0 + n \times \frac{b}{gcd(a,b)}$$ and $$y=y_0 - n \times \frac{a}{gcd(a,b)}$$
Howe... | For any $x_0,y_0$ that are solutions, you can use the formula:
$$x=x_0+n\cdot\frac{b}{\gcd(a,b)}\ ,\ y=y_0-n\cdot\frac{a}{\gcd(a,b)}$$
But you can also change the sign, and use the formula like this:
$$x=x_0-n\cdot\frac{b}{\gcd(a,b)}\ ,\ y=y_0+n\cdot\frac{a}{\gcd(a,b)}$$
It doesn't matter.
Now, the textbook used $(1,1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1934048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Are there different integers so that $ab=c^2$ and $a^2+b^2=2d^2$? I wish to find 2 integers $a,b\ \ (a\neq b)$ so that $GM(a,b)$ and $RMS(a,b)$ are integers.
So I write down:
$$\sqrt{ab}=c\ \ \rightarrow ab=c^2\\\sqrt{\frac{a^2+b^2}{2}}=d\ \ \rightarrow a^2+b^2=2d^2$$
By lots of trial and error I think there aren't any... | If $a$ and $b$ are equal, then of course the above are satisfied.
$a^2+b^2 + 2ab = (a+b)^2 = 2(d^2+c^2)$ and $(a-b)^2 = 2(d^2-c^2)$. Hence, both $2(d^2-c^2)$ and $2(d^2+c^2)$ are perfect squares.
Let $e^2 = 2(d^2-c^2)$ and $f^2=2(d^2+c^2)$. Then, $ e^4 - f^4 = (e^2-f^2)(e^2+f^2) = 16d^2c^2$. In other words,
$$
e^4 = f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1935328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\sin^4\theta+\cos^4\theta=\frac{3+\cos4\theta}{4}$ I have a question that goes exactly like this:
By considering $(\sin^2\theta+\cos^2\theta)^2$ and $(\sin^2\theta+\cos^2\theta)^3$ prove that
$$
\text{a) }\sin^4\theta+\cos^4\theta=\frac{3+\cos4\theta}{4},\qquad\text{b) }\sin^6\theta+\cos^6\theta=\frac{5+3\c... | There is a general systematic method to prove this kind of linearisation identities. It's not as swift as the previous answers, bit it's purely automatic.
You use the Euler formulae
$$\cos \theta = \frac{e^{i\theta} + e^{-i\theta}}2 \qquad \text{et}\qquad \sin \theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}$$
and you ex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1936121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 4
} |
Proving nth difference of squares I'm a little stuck on proving $$(b^n-a^n)=(b-a)(a^0b^{n-1}+a^1b^{n-1}+ \cdots + a^{n-1}b^{0}).$$A solution I came across gave this as an answer:
$$(b-a)(b^n + b^{n-1} a+...+ba^{n-1} +a^{n} )\\
=(b-a)b^n + (b-a)b^{n-1} a+...+(b-a)ba^{n-1} +(b-a)a^n\\
=b^n+1 -b^n a+b^n a-b^{n-1} a^2 +...... | To prove that
$$\begin{align}
b^n-a^n &= (b-a)(a^0b^{n-1}+a^1b^{n-1} + \cdots + a^{n-1}b^{0})\\
&=(b-a)\sum_{k=0}^{n-1}b^ka^{n-1-k}
\end{align}$$
It suffices to show that $$b^n-1 = (b-1)\sum_{k=0}^{n-1}b^k,$$
as follows:
$$\begin{align}
(b-1)\sum_{k=0}^{n-1}b^k &=b\sum_{k=0}^{n-1}b^k-\sum_{k=0}^{n-1}b^k\\
&=\sum_{k=0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1938027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to show this conjecture：$\frac{1}{\sqrt{a_{1}}}+\frac{1}{\sqrt{a_{2}}}+\cdots+\frac{1}{\sqrt{a_{n}}}\ge\frac{2n}{\sqrt{a_{1}}+\sqrt{a_{n}}}$ I have prove this inequality
$$\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}>2\sqrt{n+1}-2$$
because
$$\sum_{k=1}^{n}\dfrac{1}{\sqrt{k}}>\sum_{k=1}^{n}\d... | Using the same method, I can show the following slightly weaker result.
Let $a_i = a_0 + id$, for $1 \le i \le n+1$. Since $d > 0$, we have $a_{i+1} > a_i$. Hence
$$ \frac{1}{\sqrt{a_i}} > \frac{2}{\sqrt{a_{i+1}} + \sqrt{a_i}} = \frac{2(\sqrt{a_{i+1}} - \sqrt{a_i})}{a_{i+1} - a_i} = \frac{2(\sqrt{a_{i+1}} - \sqrt{a_i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How can I evaluate: $\lim_{x \to \infty} \sqrt{x^2 + 4x} - \sqrt{x^2 - 5x}$ I need to find this limit:
$$\lim_{x \to \infty} \sqrt{x^2 + 4x} - \sqrt{x^2 - 5x}$$
The answer I got from using the limit laws is $\sqrt{\infty} - \sqrt{\infty}$.
How do I proceed now?
Added I took the conjugate of the function and I got a new... | Multiply and divide by $$\sqrt{x^2+4x} +\sqrt{x^2-5x}$$
$$\lim_{x\to \infty} \frac{9x}{\sqrt{x^2+4x} +\sqrt{x^2-5x}}$$
$$\lim_{x\to \infty} \frac{9}{\frac{\sqrt{x^2+4x}}{x} +\frac{\sqrt{x^2-5x}}{x}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1941003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Estimate $ \int_\sqrt{n}^{\sqrt{n+1}} \sin (2\pi x^2) \; dx$ using the Trapezoid rule As a follow-up to my previous question. What is the integral of $\sin (2\pi x^2)$ as $x \in [\sqrt{n}, \sqrt{n+1}]$ we get this:
$$ \int_\sqrt{n}^{\sqrt{n+1}} \sin (2\pi x^2) \; dx \tag{$\ast$}$$
Since $\sin 2\pi n = 0$ for all $n \i... | One approach, following @Leg's answer:
$$I = \int_0^{1} \sin(2\pi t) \dfrac{dt}{2\sqrt{n+t}}=\int_0^{1/2} \sin(2\pi t) \dfrac{dt}{2\sqrt{n+t}}+\int_{1/2}^{1} \sin(2\pi t) \dfrac{dt}{2\sqrt{n+t}}$$
$$=\int_0^{1/2} \sin(2\pi t) \dfrac{dt}{2\sqrt{n+t}}-\int_{0}^{1/2} \sin(2\pi t) \dfrac{dt}{2\sqrt{n+t+1/2}}$$
$$=\int_0^{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1943837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show that $1+5^n + 5^{2n} + 5^{3n} + 5^{4n}$ is never prime for n > 0 integer Show that $1+5^n + 5^{2n} + 5^{3n} + 5^{4n}$ is never prime for n > 0 integer
Trying cases on a computer I can see that the equation is divisible by 11 when $n \neq 5^k$. That's easy to prove if you write the sum as a fraction using the fact ... | Case $1:n=2k$
Then our sum is $\frac{5^{10k}-1}{5^{2k}-1}=\frac{(5^{5k}-1)(5^{5k}+1)}{5^{2k}-1}$. Since the denominator is smaller than each of the factors we conclude the number is not prime.
Case $2: n=2k+1$
Let $a=5^{2k+1}$ and notice $1+a+a^2+a^3+a^4=(a^2+3a+1)^2-5a(a+1)^2$.
Since $a=5^{2k+1}$ this is equal to: $(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1944434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Nth power of the following matrix The matrix is
$ \begin{pmatrix}0&1\\-1&0\end{pmatrix}^n $
for $n=2 \implies \left(\begin{matrix}-1 & 0\\ 0 &-1\end{matrix}\right)$ for $n=3 \implies \begin{pmatrix}1&0\\0&1\end{pmatrix}$ for $n=4 \implies \begin{pmatrix}-1&0\\0&-1\end{pmatrix}$
so I assume that for every $n=2k $, wh... | $${{\left( \begin{matrix}
\cos \,\phi & -\sin \,\phi \\
\sin \,\phi & \cos \,\phi \\
\end{matrix} \right)}^{n}}=\left( \begin{matrix}
\cos \,n\phi & -\sin \ n\phi \\
\sin \,n\phi & \cos \,n\phi \\
\end{matrix} \right)$$
We know $A=\begin{pmatrix}0&\!-1\\1&0\end{pmatrix}$
then set $\phi=\frac{\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1944689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Arrangements of a,a,a,b,b,b,c,c,c in which no three consecutive letters are the same Q: How many arrangements of a,a,a,b,b,b,c,c,c are there such that
$\hspace{5mm}$ (i). no three consecutive letters are the same?
$\hspace{5mm}$ (ii). no two consecutive letters are the same?
A:(i). 1314. ${\hspace{5mm}}$ (ii). 174.
I t... | (i)
For the first question, your idea of using the Inclusion-Exclusion Principle is the correct approach.
The number of distinguishable arrangements of the letters $a, a, a, b, b, b, c, c, c$ is
$$\frac{9!}{3!3!3!}$$
where the factors of $3!$ in the denominator are attributable to the fact that interchanging identical ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1944807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 0
} |
Submonoid of $SL(2,\mathbb{Z})$ generated by two matrices. I'm trying to figure out which kind of matrices can be written as $A^{k_{1}}B^{k_{2}}...A^{k_{n}}B^{k_{n}}$ where
A=$\begin{pmatrix} 1&1\\0&1\end{pmatrix}$ and B=$\begin{pmatrix} 2&1\\1&1\end{pmatrix}$.
I did it for $C=\begin{pmatrix} 1&2\\0&1\end{pmatrix}$ an... | Regarding $B^k$:
$B$ is Hermitian and thus diagonalizable. See this for details. So, we have that:
$$B = \overbrace{\begin{pmatrix}
\frac{1+\sqrt{5}}{2} & \frac{1-\sqrt{5}}{2} \\ 1 & 1
\end{pmatrix}}^{P}
\overbrace{\begin{pmatrix}
\frac{3+\sqrt{5}}{2} & 0 \\ 0 & \frac{3-\sqrt{5}}{2}
\end{pmatrix}}^{D}
\overbrace{\beg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1945144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Show that $\sum_{k=0}^{n}\binom{n}{k}^2=\frac{n+1}{n}\sum_{k=1}^{n}\binom{n}{k}\binom{n}{k-1}$ Show that $\sum_{k=0}^{n}\binom{n}{k}^2=\frac{n+1}{n}\sum_{k=1}^{n}\binom{n}{k}\binom{n}{k-1}$
I came across this result
while trying to solve this:
inductive proof for $\binom{2n}{n}$
My proof is cumbersome,
so I hope that
s... | Apply four equation:
$$\sum_{k=0}^\infty\binom{r}{k}\binom{s}{n-k}=\binom{r+s}{n} [1]$$
and
$$\binom{n}{k}=\frac{n}{n-k}\binom{n-1}{k} [2]$$
and
$$\binom{n}{k}=\binom{n}{n-k} [3]$$
and (for k!= 0)
$$\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1} [4]$$
Apply [1] and [2] we have:
$$\sum_{k=0}^{n}\binom{n}{k}^2 = \b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1947123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 5
} |
Find the equation of the circle passing ... Find the equation of the circle passing through the points $P(5,7)$, $Q(6,6)$ and $R(2,-2)$.
My Attempt:
Let the equation of the circle be:
$$x^2+y^2+2gx+2fy+c=0$$
The point $P(5,7)$ lies on the circle then,
$$5^2+7^2+10g+14f+c=0$$
$$10g+14f+c=-74$$-----(1)
The point $Q(6,6)$... | Subtract $2^{nd}$ equation from $3$ times the $3^{rd}$ to get
$$
c-12f=36
$$
Then subtract $2$ times the first from $5$ times the third to get:
$$
3c-48f=148
$$
From the last two, subtract $3$ times the first from the second to get $f=-10/3$. Plugging back in one them gives $c=-4$. Now pluging in one of the fir... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1948723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
How many natural solutions to $x_1+x_2+x_3+x_4=100$ with $x_1 \neq x_2 \neq x_3 \neq x_4$? I have invented a little exercise of combinatorics that I can't solve (without brute forcing): in how many ways can I choose 4 non-negative integers, the sum of which is 100 and they are all different? So:
$x_1+x_2+x_3+x_4=100$
$... | Here is a solution with generating functions. Let $a_n$ be the number of objects you wish to count, namely the ordered quadruples of distinct non-negative integers with a sum of $n$, and let $f(x)=\sum_{n\ge0}a_nx^n$ be the ordinary generating function of this sequence. Since the integers in the sum are distinct, this ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1950308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Bulgarian Mathematical Olympiad Round 4 problem 1. Problem: Prove that there exists a unique triple of positive integers greater than 1 such that product of any two increased by one is divisible by the third number.
I found $(2,3,7)$ to satisfy this but cant prove this. Please help.
| Clearly all numbers must be pairwise coprime, let them be $a<b<c$
Notice that $abc| ab+ac+bc+1$ and so $abc\leq ab+ac+bc+1$.
Notice that if $a\geq 3$ then $abc\geq 3bc> ab+ac+bc+1$ and so $a=2$.
It follows that $b|2c+1$ and $c|2b+1$ and so $bc|2b+2c+1$ In particular $bc\leq 2b+2c+1$.
It $b\geq 4$ then $bc\geq 4c>2b+2c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1952881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What will be the value of the following determinant without expanding it? $$\begin{vmatrix}a^2 & (a+1)^2 & (a+2)^2 & (a+3)^2 \\ b^2 & (b+1)^2 & (b+2)^2 & (b+3)^2 \\ c^2 & (c+1)^2 & (c+2)^2 & (c+3)^2 \\ d^2 & (d+1)^2 & (d+2)^2 & (d+3)^2\end{vmatrix} $$
I tried many column operations, mainly subtractions without any succ... | I wanted to present a different view of the problem that takes advantage of the sameness of the rows in a less computational way.
We can view the entries of each row as the evaluations of 4 polynomials of degree 2: $x^2$, $(x+1)^2$, $(x+2)^2$ and $(x+3)^2$ at a point.
Since the set of polynomials of degree at most 2, $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1953843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 7,
"answer_id": 1
} |
Proof $\sum_{n=0}^\infty \frac{n^2+3n+2}{4^n} = \frac{128}{27}$
$\sum_{n=0}^\infty \frac{n^2+3n+2}{4^n} = \frac{128}{27}$ Given hint: $(n^2+3n+2) = (n+2)(n+1)$
I've tried converting the series to a geometric one but failed with that approach and don't know other methods for normal series that help determine the actu... | One may start with the standard finite evaluation:
$$
1+x+x^2+...+x^n=\frac{1-x^{n+1}}{1-x}, \quad |x|<1. \tag1
$$ Then by differentiating $(1)$ we have
$$
1+2x+3x^2+...+nx^{n-1}=\frac{1-x^{n+1}}{(1-x)^2}+\frac{-(n+1)x^{n}}{1-x}, \quad |x|<1, \tag2
$$ by differentiating once more one gets
$$
2\times 1+3\times 2 x^2+..... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1955505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 5
} |
Conditional probability selecting a number and flipping a fair coin. One of the numbers 1, 2 or 3 is selected at random. Then a fair coin is flipped that number of times. What is the probability that the number 3 was selected given:
*
*no heads on the coin flip(s)
*1 head
*2 heads
*3 heads
I know for three head... | I think I solved my own problem here:
For the first case of no heads:
*
*There is $\frac{1}{2}$ chance to get no heads if the number picked is 1.
*There is $\frac{1}{4}$ chance to get no heads if the number picked is 2.
*There is $\frac{1}{8}$ chance to get no heads if the number picked is 3.
So we can set this ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1957101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find all $0^\circ\leq A\leq360^\circ$ with $\tan A + \tan 2A + \tan 3A = 0$ solve:
$\tan A+ \tan 2A+ \tan 3A=0$
My Attempt:
$$\tan A+\tan 2A+\tan 3A=0$$
$$\tan A+\frac {2\tan A}{1-\tan^2A}+\frac {3\tan A-\tan^3A}{1-3\tan^2A}=0$$
$$\frac {\tan A-\tan^3A+2\tan A}{1-\tan^2A}+\frac {3\tan A-\tan^3A}{1-3\tan^2A}=0$$
What sh... | $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1958135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Equation for Distance of the Straight line from the Origin. By reduction of the equation $ax + by + c = 0 $ of a straight line to the normal form , we get
$$\left(\frac{-a}{\sqrt{a^2 + b^2}}\right)x + \left(\frac{-b}{\sqrt{a^2 + b^2}}\right)y = \frac{c}{\sqrt{a^2 + b^2}}$$
And,
$$p= \frac{∣c∣}{\sqrt{a^2 + b^2}}$$
And ... | By simple identification, the two equations are identical when
$$-\frac a{\sqrt{a^2+b^2}}=\cos\theta,\\
-\frac b{\sqrt{a^2+b^2}}=\sin\theta,\\
\frac c{\sqrt{a^2+b^2}}=p.$$
This is coherent as you verify $\cos^2\theta+\sin^2\theta=1$. If $c$ is negative, change all signs.
As one can verify by substitution,
$$x=p\cos\th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1958856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
A function with domain that is asking for sum of a+b+c+d Let $f$ be a function such that $$ \sqrt {x - \sqrt { x + f(x) } } = f(x) , $$for $x > 1$. In that domain, $f(x)$ has the form $\frac{a+\sqrt{cx+d}}{b},$ where $a,b,c,d$ are integers and $a,b$ are relatively prime. Find $a+b+c+d.$
| Let $y = f(x)$. Our first goal is to find all pairs of real numbers $(x,y)$ that satisfy
[\sqrt{x - \sqrt{x + y}} = y.]We observe that $y$ must be nonnegative. Squaring both sides, we get
[x - \sqrt{x + y} = y^2,]so
[\sqrt{x + y} = x - y^2.]Squaring both sides again, we get
[x + y = x^2 - 2xy^2 + y^4,]so
[y^4 - 2xy^2 -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1960328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integrating $\frac{x}{1+x^2}$ I need to integrate $\frac{x}{1+x^2}$ and I tried it like this:
$$f(x)=\frac{x}{1+x^2}=\frac{x}{(x+1)(x-1)}=\frac{x+1-1}{(x+1)(x-1)}= \frac{x+1}{(x+1)(x-1)}-\frac{1}{(x+1)(x-1)}=\frac{1}{x-1}-\frac{1}{1+x^2}$$
After I simplified it I integrate it!
$$\int{f(x)}=\int{\frac{1}{x-1}}-\int{\fra... | Easiest way is to use $$\int \frac{f'(x)}{f(x)} dx = \log(f(x)) + c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1960445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Prove $(3+\sqrt{11})^{1/3}$ is irrational. I can't say I've gotten very far. You can show $3 + \sqrt{11}$ is irrational, call it $a$. Then I tried supposing it's rational, i.e.:
$a^{1/3}$ = $\frac{m}{n}$ for $m$ and $n$ integers.
You can write $m$ and $n$ in their canonical factorizations, then cube both sides of the ... | Rational Root Theorem
If $P(x) = a_nx^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + \cdots + a_1x + a_0$, where all of the $a_i$ are integers, and $P\left( \dfrac uv \right)=0$ where $u$ and $v$ are integers, then $v \mid a_n$ and $u \mid a_0$.
Let $x = (3 + \sqrt{11})^{\frac 13}$. Then
\begin{align}
x^3 &= 3 + \sqrt{11} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1961614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Existence of integer solutions to $a^2 + b^2 + c^2 = d^2$ for specified $d$ Previous questions contained cubes instead of squares, that was my mistake, sorry.
How to tell if $a^2 + b^2 + c^2 = d^2$ has any integer, non-zero solutions for specified positive integer $d$? Is there any criterium or algorithm for this?
EDIT... | 1) As regards the original post: existence of integer solutions to
$a^3 + b^3 + c^3 = d^3$ for a given $d$.
Note that the following identity holds:
$$(9t^4)^3+(1-9t^3)^3 +(3t-9t^4)^3=1.$$
Therefore, given a positive integer $d$ there is always an integer solution,
$$(9dt^4)^3+(d(1-9t^3))^3+(d(3t-9t^4))^3 = d^3.$$
For m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1964884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Need help in indefinite integral elimination I'm trying to get a easy and short answer for that indefinite integral:
$$\int\frac{\sqrt{x^2+x+1}}{(x+1)^2}\,dx$$
But everything I find is very big and difficult to understand (too much so it seems like a wrong result). Is there any way to solve it in a more beautiful way?
| Let's calculate by parts first
$\int{\frac{\sqrt{x^2+x+1}}{(1+x)^2}dx} = -\frac{\sqrt{x^2+x+1}}{1+x} + \int{\frac{1}{x+1}\cdot\frac{2x+1}{2\sqrt{x^2+x+1}}dx}\\
\int{\frac{\sqrt{x^2+x+1}}{(1+x)^2}dx} = -\frac{\sqrt{x^2+x+1}}{1+x} +\int{\frac{(2x+2)-1}{2(x+1)\sqrt{x^2+x+1}}dx}\\
\int{\frac{\sqrt{x^2+x+1}}{(1+x)^2}dx} = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1967094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the range of $a$ such $|f(x)|\le\frac{1}{4},$ $|f(x+2)|\le\frac{1}{4}$ for some $x$.
Let $$f(x)=x^2-ax+1.$$
Find the range of all possible $a$ so that there exist $x$ with
$$|f(x)|\le\dfrac{1}{4},\quad |f(x+2)|\le\dfrac{1}{4}.$$
A sketch of my thoughts: I write
$$f(x)=\left(x-\dfrac{a}{2}\right)^2+1-\dfrac... | To start off a lot of time can be saved by only handling $a\geq0$ since both $|x^2-ax+1|\leq \frac{1}{4}$ and $|(x+2)^2-a(x+2)+1|\leq\frac{1}{4}$ are odd inequalities all of solutions that we find for $a\geq0$ will be the same except with opposite signs.
As a first step we should try to isolate $a$ from both of the ine... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1967726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 10,
"answer_id": 2
} |
Internal angle bisectors and right-angled triangle? Imagine a right angled triangle. Two of its internal bisectors are of length $7cm$ and $4cm$, respectively. The first internal angle bisector belongs to the angle of the side $a$ and the second one belongs to the angle of the side $b$ Calculate the length of the hypot... | Case 1: Suppose that $a,b$ are the legs of the right triangle. Let then $c$ be the length of the hypotenuse, and $C = \frac{\pi}{2}, A, B$ the respective angles.
It follows by inspection that:
$$ 7 = \frac{c \; \cos A}{\cos A/2} \quad \text{and} \quad 4 = \frac{c \; \cos B\; }{\cos B/2}$$
Since $\cos \frac{x}{2}=\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1969553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find a line which is tangent to the curve $y=x^4-4x^3$ at 2 points How can I solve this? Should I set the two points as $a^4-4a^3$ and $b^4-4b^3$?
| Suppose a line $y=mx+c$ is tangent to the curve $y=x^4-4x^3$ at two points. Then the difference between these two equations must be a quartic polynomial with two double roots:
$$x^4-4x^3-mx-c=(x-a)^2(x-b)^2$$
$$=x^4-2(a+b)x^3+(a^2+4ab+b^2)x^2-2ab(a+b)x+(ab)^2$$
Comparing coefficients, we have
$$2(a+b)=4$$
$$a^2+4ab+b^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1970630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Fourier series of $\cos\mu x$ Find the Fourier series for the $2\pi$-periodic function $F(x)=\cos\mu x$, $x\in[-\pi,\pi]$ and where $\mu \notin \mathbb Z$. Simplify the result for $\mu=\frac{1}{4}$. Hence show that
$$\pi\sqrt{2}=A-\sum^\infty_{n=1}\frac{(-1)^n}{Bn^2-\frac{1}{8}},$$
where $A$ and $B$are integers to be... | Hint: Once you get the correct Fourier cosine series
$$\cos(\mu x)=\frac{\sin(\mu \pi)}{\mu \pi}+\frac{2\mu\sin(\mu\pi)}{\pi}\sum_{n\geq 1}\frac{(-1)^n}{\mu^2-n^2}\cos(nx)\tag{1}$$
by setting $\mu=\frac{1}{4}$ you get
$$\cos\frac{x}{4}=\frac{2\sqrt{2}}{\pi}+\frac{\sqrt{2}}{4\pi}\sum_{n\geq 1}\frac{(-1)^n}{\frac{1}{16}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1971226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Discrete mathematics, divisibility How can I prove that for all $n\in\mathbf{N}$ that $6 | n^5 + 5n$?
I tested for $n = 2$ and got $6 | 32 + 10 = 42$.
| Use modular arithmetic and solve for the finite number of cases $n \equiv 0, 1, 2, 3, 4, 5 \pmod{6}$.
For example, $4 \cdot 4 = 16 \equiv 4 \pmod{6}$, so that $4^5 \equiv 4 \pmod{6}$, and $n^5 + 5n \equiv 4 + 20 \pmod{6} \equiv 24 \pmod{6} \equiv 0 \pmod{6}$.
The five other cases are similar.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1974310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Compute $ \sqrt{n^2 +3} - \sqrt{n^2 +2} $. I have a question concerning determining a basic limit of $ \sqrt{n^2 +3} - \sqrt{n^2 +2} $ as $n \rightarrow \infty $. Upon using the difference of squares, this yields: $ \frac{1}{\sqrt{n^2 +3} + \sqrt{n^2 +2}} $Now the limit seems intuitive, but I cant find a formal way of ... | Let $f(x) = \sqrt{1+3x} -\sqrt{1+2x}$.
Then $f(0) = 0$ and $f'(0) = {1 \over 2}$, hence for $x$ sufficiently
small, we have $|f(x)| \le |x|$.
Hence for $n$ sufficiently large, we have
$|\sqrt{n^2+3} -\sqrt{n^2+2} | = n |f({1 \over n^2})| \le {1 \over n}$,
from which the limit follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1976042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
General form of a matrix that is both centrosymmetric and orthogonal A centrosymmetric matrix is of the form $JAJ=A$ where $J$ is the counter identity matrix, i.e. $$J=
\begin{bmatrix}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0 \\
\end{bmatrix}
$$And an orthogonal matrix is $A$ such tha... | Note that $J^2 = I$ and so the condition $JAJ = A$ is equivalent to the condition $JA = AJ$. That is, $A$ and $J$ should commute. The matrix $J$ is orthogonally diagonalizable with eigenvalues $1,-1$ and eigenspaces
$$ V_{1} = \operatorname{span} \{ e_2, \frac{1}{\sqrt{2}}(e_1 + e_3) \}, V_{-1} = \operatorname{span} \{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1976654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that $ 2/1!+4/3!+6/5!+8/7!+\dots = e$
Prove that $ 2/1!+4/3!+6/5!+8/7!+\dots = e$
My attempt:
I googled the problem, and I found that $\sum_{(n=1)}^∞ \cfrac{2 n}{(2 n-1)!} = e$
I also found that $\sum _{n=0}^{\infty \:}\cfrac{2n+2}{\left(2n+1\right)!}$ is equal to $e$.
How can I prove this?
| Let $f(x)= x \sin(x)$. Then
$$
f(x) = \frac{x^2}{1!} - \frac{x^4}{3!} + \frac{x^6}{5!} -\frac{x^8}{7!} + \cdots
$$
and so
$$
f'(x) = \frac{2x}{1!} - \frac{4x^3}{3!} + \frac{6x^5}{5!} -\frac{8x^7}{7!} + \cdots
$$
Therefore
$$
f'(i) = \frac{2i}{1!} + \frac{4i}{3!} + \frac{6i}{5!} +\frac{8i}{7!} + \cdots = iS
$$
On the ot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1979848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Find matrix $A$. How many solutions are there? The matrices below is matrix multiplcation of $AB=C$ but someone has changed the order of the coloumns in $C$ and erased what was in $A$. What is matrix $A$? Is it the only one? I have no idea how to start. I appreciate all the help I can get.
$$\begin{bmatrix} & & & \\ ... | There is a unique solution to this problem. The problem states
that $AB=C$ holds, but that someone has changed the order of the
columns in $C$ and erased what was in $A$. We are told what the matrix
$B$ is, but not told what $C$ is. We are only told what $C$ looks like
after its columns have been permuted. That is, we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1980645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Quadratic equation with absolute values Solve $|x^{2}+x-2|+|x^{2}-x-2|=2$
My attempt:
$|x^{2}+x-2|=
x^{2}+x-2, x\in (-\infty,-2]\cup[1,\infty)$
$|x^2+x-2|=-x^{2}-x+2, x\in(-2,1)\\$
$|x^{2}-x-2|=x^{2}-x-2, x\in (-\infty,-1]\cup[2,\infty)$
$|x^{2}-x-2|=-x^{2}+x+2, x\in (-1,2)\\$
$1)$ $x\in(-\infty,-2]\cup[1,\infty)$ a... | The derivative of the function is
$$(2x+1)\text{sign}|x^2+x-2|+(2x-1)\text{sign}|x^2-x-2|.$$
Depending on the signs, this reduces to $\pm4x$ or $\pm2$ so that the only possible root occurs at $x=0$. The changes of sign can occur at $-2,-1,1,2$ so that the function is certainly monotonous in the intervals
$$]-\infty,-2,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1981701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"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.