Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Solve trigonometric equation $\sin14x - \sin12x + 8\sin x - \cos13x= 4$ I am trying to solve the trigonometric equation
$$ \sin14x - \sin12x + 8\sin x - \cos13x= 4 $$
The exact task is to find the number of real solutions for this equation on the range $[0, 2\pi]$. Thanks.
| Define:
$$\sin(14x) - \sin(12x) + 8\sin x - \cos(13x)- 4=:F(\cos x,\sin x)$$
Then set $t=\tan(x/2)$, we have
$$\cos x=\frac{1-t^2}{1+t^2}, \sin x=\frac{2t}{1+t^2}$$
$$G(t)=F\left(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\right)$$
$$=-\frac{t^2-4t+1}{(1+t^2)^{14}}H(t)$$
$$H(t)=5 - 273t^2 + 15262t^4 - 229086t^6 +
1565135t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/837593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Prove that diophantine equation has only two solutions. I am looking at the following exercise:
$$\text{Prove that the diophantine equation } x^4-2y^2=1 \text{ has only two solutions.}$$
That's what I thought:
We could set $x^2=k$,then we would have $k^2-2y^2=1 \Rightarrow 2y^2=k^2-1 \Rightarrow 2 \mid k^2-1$
Can I use... | $$2y^2=x^4-1=(x^2+1)(x^2-1)=(x^2+1)(x+1)(x-1)$$
Suppose that $2y^2$ has an odd prime factor $p$. Then $p^2$ divides $y^2$, and hence, $p^2$ divides $(x^2+1)(x^2-1)$. Since $p$ can not divide both factors, then $p^2$ divides $x^2+1$ or $p^2$ divides $x^2-1=(x+1)(x-1)$. Thas is, $p^2$ divides $x^2+1$, $x+1$ or $x-1$.
Thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/839586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Visualise $3x^2 - 2xy - 10x +3y^2 -2y + 8=0$ after the $x,y$ term has been eliminated (using rotation) This is a continuation from my previous question. I thought it would be better to start a new one since the old one was answered correctly.
The equation in question is: $3x^2 - 2xy - 10x +3y^2 -2y + 8=0$:
We introduc... | These are the steps. Show us your workings for as far as you've gotten:
*
*Plug $x=(X-Y)/\sqrt{2}$ and $y=(X+Y)/\sqrt{2}$ into the equation.
*Rearrange into the form $Ax^2+Bx^2+Cx+Dy+E=1$.
*Complete the square on the LHS to put it in the form of the equation of an ellipse.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/839702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
To prove $a^2+ab+b^2 \Big|(a+b)^{2n}+a^{2n}+b^{2n}$ whenever $3$ does not divide $n$ If $3$ does not divide a positive integer $n$ , then how to prove that $a^2+ab+b^2 \Big|(a+b)^{2n}+a^{2n}+b^{2n}$ ?
| I will stay in the reals. Note that:
$$(a+b)^2\equiv ab \mod{a^2+ab+b^2}\implies (a+b)^{2n}\equiv (ab)^n \mod{a^2+ab+b^2}$$
So $(a+b)^n+a^{2n}+b^{2n}\equiv a^nb^n+a^{2n}+b^{2n}$. Also, we know that $a^3\equiv b^3$. So setting $n=3q+1$, we have:
$$a^nb^n+a^{2n}+b^{2n}\equiv ab^{6q+1}+a^2b^{6q}+b^{6q+2} \equiv b^{6q}(ab+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/840336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Doubt on solution of PDE To Solve: $\displaystyle (x^2-y^2-z^2)\frac{\partial z}{\partial x}+2xy\frac{\partial z}{\partial y}=2xz$
Subsidiary equation: $\displaystyle \frac{dx}{x^2-y^2-z^2}=\frac{dy}{2xy}=\frac{dz}{2xz}$
Using multipliers x,y and z, we have each fraction=$\displaystyle \frac{xdx+ydy+zdz}{x(x^2+y^2+z^2)... | Starting with
$$ \frac{dx}{x^2-y^2-z^2}=\frac{dy}{2xy}=\frac{dz}{2xz}$$
We obtain:
$$2xdx=(x^2-y^2-z^2)\frac{dz}{z}......(1)$$
$$2ydy=2y^2\frac{dz}{z}......(2)$$
$$2zdz=2z^2\frac{dz}{z}......(3)$$
Adding (1),(2), and (3), we obtain:
$$\frac{2xdx+2ydy+2zdz}{x^2+y^2+z^2}=\frac{dz}{z}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/840588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the last two digits of $9^{{9}^{9}}$ I have to find the last two decimal digits of the number $9^{{9}^{9}}$.
That's what I did:
$$m=100 , \phi(m)=40, a=9$$
$$(100,9)=1, \text{ so from Euler's theorem,we have :} 9^{40} \equiv 1 \pmod{100}$$
$$9^{{9}^{9}} \equiv 9^{9^{2+2+2+2+1}} \equiv 9^{9^2 \cdot 9^2 \cdot 9^2 \c... |
Key remark: By the binomial theorem, for every odd $k$, $(10-1)^k=n+10\cdot k-1$ for some integer $n$ which is a multiple of $100$. Since $10\cdot k=10\cdot\ell(k)\bmod{100}$ where $\ell(k)$ denotes the last digit of $k$, this shows that, for every odd $k$, $9^k=10\cdot \ell(k)-1\bmod{100}$.
First application: $\ell(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/842643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Proving $x+\sin x-2\ln{(1+x)}\geqslant0$
Question: Let $x>-1$, show that
$$x+\sin x-2\ln{(1+x)}\geqslant 0.$$
This is true. See http://www.wolframalpha.com/input/?i=x%2Bsinx-2ln%281%2Bx%29
My try: For
$$f(x)=x+\sin x-2\ln{(1+x)},\\
f'(x)=1+\cos{x}-\dfrac{2}{1+x}=\dfrac{x-1}{1+x}+\cos{x}=0\Longrightarrow\cos{x}=\dfr... | Alternative proof:
Let $f(x) = x + \sin x - 2\ln (1+x)$.
We split into three cases:
*
*$x\in (-1, 0]$:
We have $f'(x) = 1 + \cos x - \frac{2}{1+x} \le 1 + 1 - \frac{2}{1+x} = \frac{2x}{1+x} \le 0$.
As $f(0) = 0$, we have $f(x) \ge 0$.
*$x\in [0, \frac{3}{2}]$:
Since $\cos x \ge 1 - \frac{x^2}{2}$ for $x\in \mathb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/843276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 1
} |
Find all integer solution Find all integer solutions such that $$a+1|2a^2+9$$
Solution. I could solve this by writing $$\frac{2a^2+9}{a+1}=2a-2+\frac{11}{a+1}.$$ So, the only integer solution for the last equation are $a=10, a=-12.$
But, i want to get a solution using divisibility properties.
| By elementary means, since $a+1|2a^2+2a$, it divides $2a^2+9-(2a^2+2a)=9-2a$.
Since $a+1|2a+2$, we have $a+1|(9-2a)+(2a+2)$
Hence $a+1|11$
Hence $a+1=±1$ or $a+1=±11$
Hence $a=-12,-2,0,10$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/846091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Transition probabilities in a finite state machine Assume I have a finite state machine and a bunch of tokens. Transitions happen every time a token is inserted. Transitions are based on the token (i.e. at state S, inserting a blue token would give a different result from, say, inserting a red token). There is one stat... | Let $(X_n)_{n\ge 0}$ be our Markov chain, let $T$ be the set of all tokens and let token $a$ have transition matrix $P_a$. Then $\mathbb{P}(X_1=j|X_0=i,\text{next token is t})=(P_t)_{ij}$.
By the law of total probability
$$\mathbb{P}(X_1=j|X_0=i)=\sum_{t\in T} (P_t)_{ij}\mathbb{P}(\text{next token is t}|X_0=i)$$
I hope... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/849288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluation of $ \int \tan x\cdot \sqrt{1+\sin x}dx$ Calculation of $\displaystyle \int \tan x\cdot \sqrt{1+\sin x}dx$
$\bf{My\; Try::}$ Let $\displaystyle (1+\sin x)= t^2\;,$ Then $\displaystyle \cos xdx = 2tdt\Rightarrow dx = \frac{2t}{\sqrt{2-t^2}}dt$
So Integral is $\displaystyle = \displaystyle 2\int \frac{t^2}{\s... | $$ 2\int\frac{t^4-t^2}{2-t^2}dt = 2\int\frac{\require{cancel}\cancel{(t^2 - 2)}(t^2 + 1)+2}{-(\cancel{t^2 - 2})},dt = \underbrace{-2\int (t^2 + 1)\,dt}_{\text{a cinch}} - \underbrace{2\int \frac {2}{t^2 - 2}\,dt}_{\text{partial fractions}}$$
Alternatively, for the second integral, we can express it as $$+ 2\cdot\frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/849580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
} |
Simplifying polynomial fraction Working through an old book I got and am at this problem: Simplify:
$$\frac{3x^2 + 3x -6}{2x^2 + 6x + 4}.$$
The answer is supposed to be $\frac{3(x - 1)}{2(x - 1)}$. I thought I had all this polynomial stuff figured out well enough, but I'm having trouble seeing how they got to that answ... | Simply factorise
$\dfrac{3x^2 + 3x -6}{2x^2 + 6x + 4} = \dfrac{3(x^2 + x -2)}{2(x^2 + 3x + 2)} = \dfrac{3(x-1)(x+2)}{2(x+1)(x+2)}$
Then cancel common terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/850148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
A closed formulae for the coefficient of $x^k$ in $(x-1)^a(x+1)^b$ Let a,b positive integer
Do you know any closed formulae for the coefficient of $x^k$
in $(x-1)^a(x+1)^b=\sum_{k=0}^{a+b}u(k;a,b)x^k$ ?
I look for an a closed expression of $u(k;a,b)$ involving maybe integral , special function
not the symbol $\sum$
t... | The expansion follows as:
\begin{align}
(x-1)^{a} (x+1)^{b} &= (-1)^{a} (1-x)^{a} (1+x)^{b} \\
&= (-1)^{a} \sum_{r=0}^{a} \binom{a}{r} (-1)^{r} x^{r} \cdot \sum_{s=0}^{b} \binom{b}{s} x^{s} \\
&= (-1)^{a} \sum_{r=0} \sum_{s=0} (-1)^{r} \binom{a}{r} \binom{b}{s} x^{r+s} \\
&= (-1)^{a} \sum_{r=0}^{a|b} (-1)^{r} \left( \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/850334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Compute the determinant $4\times 4$ Compute the determinant:
$$
A= \begin{vmatrix}
1 & 1 & a+1 & b+1 \\
1 & 0 & a & b \\
2 & b & a & b \\
2 & a & a & b \\
\end{vmatrix}
$$
I got (in the end): $\det A = (-a-b)(b+a) = -ab -a^2 -b^2 -ba $.
| You can preform some row operations, they preserve the determinant (but no multiplying a row and $-1$ for every transopsition). This gives me
$$-\begin{vmatrix}
1&0&a&b\\
0&1&1&1\\
0&0&-(a+b)&-2b\\
0&0&-2a&-(a+b)\\
\end{vmatrix}$$
So determinant is $-[(a+b)^2-4ab]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/850663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculation of Trigonometric Limit with Summation. If $\displaystyle f(x)=\lim_{n\rightarrow \infty}\sum_{r=0}^{n}\left\{\frac{\tan \left(\frac{x}{2^{r+1}}\right)+\tan^3\left(\frac{x}{2^{r+1}}\right)}{1-\tan^2\left(\frac{x}{2^{r+1}}\right)}\right\}.$ Then value of $f(x)$ is
$\bf{My\; Try::}$ Let $\displaystyle \left(... | You have the right start, but I prefer if you write $y_r = \frac{x}{2^{r+1}}$ so that $y_0 = \frac{x}{2}$ and you are trying to find
$$
\frac{1}{2} \sum_{r=0}^{\infty} \left( \tan(2y_r) - \tan(y_r)\right)
$$
Since $y_r = 2y_{r+1}$, this becomes
$$
\frac{1}{2} \sum_{r=0}^{\infty} \left(\tan(2y_r) - \tan(2y_{r+1})\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/851841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to solve this linear equation? which has an x on each side I have made this equation.
$5x + 8 = 10x + \dfrac{3}{6}$
And I have achieved this result:
$x = 9$
Is my result correct?
I have already posted two other questions related to this topic, I'm a programmer and am learning Math out of my interest, this is not h... | $$5x+8 = 10x + \frac{3}{6} \overset{(1)}{\iff} 8-\frac{3}{6}=10x-5x \overset{(2)}{\iff} \frac{45}{6} = 5x \overset{(3)}\iff x=\frac{9}{6}\overset{(4)}{=} \frac{3}{2} $$
Steps:
(1) subtract $5x+\frac{3}{6}$ on both sides of the equation
(2) Simplify, note that $8-\frac{3}{6} = \frac{48}{6}-\frac{3}{6} = \frac{45}{6}$
(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/853062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Product of factorials divided by factorial to produce perfect square Let $ S = 1! ~2!~\dotsm ~100! $. Prove that there exists a unique positive integer $k$ such that $S/k!$ is a perfect square.
I thought this was a cute, fun problem and I did solve it, but any alternative methods that you guys would use?
| We can rewrite $S$ as
\begin{align*}
S &= 1^{100} \cdot 2^{99} \cdot 3^{98} \cdot 4^{97} \dotsm 99^2 \cdot 100 \\
&= 1^{100} \cdot 2 \cdot 2^{98} \cdot 3^{98} \cdot 4 \cdot 4^{96} \dotsm 99^2 \cdot 100 \\
&= 1^{100} \cdot 2^{98} \cdot 3^{98} \cdot 4^{96} \dotsm 99^2 \times 2 \cdot 4 \dotsm 100.
\end{align*}Clearly, the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/854808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
What is the sum of this series: $\displaystyle\sum_{n=0}^{\infty} \dfrac{a^n}{(3n)!}$ I tried getting it into a closed form but failed. Could someone help me out?
$$\sum_{n=0}^{\infty} \dfrac{a^n}{(3n)!}$$
| The function $f$ defined by $f(x)=\sum_{n=0}^\infty\frac{x^{3n}}{(3n)!}$ satisfies the differential equation $f^{(3)}(x)=f(x)$. So we can just set out to solve this differential equation with its initial conditions that $f(0)=1$, and $f'(0)=f''(0)=0$.
This homogeneous equation has characteristic polynomial $X^3-1$, wit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/857693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Riccati & Lyapunov equations We know that
Lyapunov equation: $A^TP + PA + Q = 0$
Algebraic Riccati equation: $A^TP + PA + Q + PBR^{-1}B^TP= 0$
It seems that the difference between the two lies in $B = 0$ (zero input) in Lyapunov Eq
and both are infinite horizon in the case above.
Is there any other engineering-sense d... | As mentioned in @bersou's answer, here is how to go from the continuous-time Algebraic Riccati equation $$A^T P + P A - P B R^{-1} B^T P + Q = 0$$ to the continuous-time Lyapunov equation $$(A - BK)^T P + P(A - BK) + Q = 0$$ where $A - BK$ is stable (all of its eigenvalues have negative real part):
\begin{align}
A^T P ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/858750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Evaluation of $\int\sqrt[4]{\tan x}dx$ Evaluation of $\displaystyle \int\sqrt[4]{\tan x}dx$
$\bf{My\; Try}::$ Let $\tan x = t^4\;\;,$ Then $\sec^2 xdx = 4t^3dt$. So $\displaystyle dx = \frac{4t^3}{1+t^8}dt$
So Integral Convert into $\displaystyle 4\int\frac{t^4}{1+t^8}dt = 2\int \frac{(t^4+1)+(t^4-1)}{t^8+1}dt$
So Inte... | You are on the right track, but the remaining computations are tedious. First, note that we have the factorizations $$t^4 + 1 = (t^2 - \sqrt{2} t + 1)(t^2 + \sqrt{2} t + 1),$$ and $$t^8 + 1 = (t^2 - \alpha t + 1)(t^2 + \alpha t + 1)(t^2 - \beta t + 1)(t^2 + \beta t + 1),$$ where $\alpha = \sqrt{2+\sqrt{2}}$ and $\beta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/860315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Trigonometric manipulation From
$$\frac{R\sin(\omega t)-\omega L\cos(\omega t)}{\omega^{2}L^{2}+R^{2}}$$
I have to get
$$\frac{\sin(\omega t-\alpha)}{\sqrt{R^{2}+\omega^{2}L^{2}}}$$
where $\alpha$ is a constant. How do I do this?
| Use the well-known identity $\sin(A\pm B) \equiv \sin A \cos B \pm \sin B \cos A$.
In your case: $\sin(\omega t -\alpha) \equiv \sin\omega t\cos\alpha-\sin\alpha\cos\omega t.$
If you multiply by a constant, say $r$, then: $r\sin(\omega t -\alpha) \equiv (r\cos\alpha)\sin\omega t-(r\sin\alpha)\cos\omega t.$
Can you find... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/861691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Does $K_{12,12}$ decompose into $K_{4,4}-I$ subgraphs?
Q: Does the complete bipartite graph $K_{12,12}$ decompose into $K_{4,4}-I$ subgraphs, where $I$ is a $1$-factor (i.e., a perfect matching)?
The obvious necessary conditions work:
*
*$K_{12,12}$ has $12^2$ edges which is divisible by $12$, the number of edges ... | [Edit (by Rebecca): the second version was correct, but let me tidy things up a bit.]
$$
\begin{array}{|cccccccccccc|}
\hline
5&1&1&1&9&2&2&2&5&5&9&9 \\
1&5&1&1&2&9&2&2&5&5&9&9\\
1&1&6&1&2&2&10&2&6&6&10&10\\
1&1&1&6&2&2&2&10&6&6&10&10\\
11&3&3&3&7&4&4&4&11&11&7&7\\
3&11&3&3&4&7&4&4&11&11&7&7\\
3&3&12&3&4&4&8&4&12&12&8&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/861777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate integral $\int_{0}^{\infty} \left(\frac{1-e^{-x}}{x}\right)^n dx$. First, according to
\begin{align*}
\int_{0}^{\infty} x^{-m}(1-e^{-x})^{n} \, dx
=\frac{n}{1-m}\int_{0}^{\infty} x^{1-m}(1-e^{-x})^{n} \, dx
-\frac{n}{1-m}\int_{0}^{\infty} x^{1-m}(1-e^{-x})^{n-1} \, dx,
\end{align*}
which can be denoted... | A basic idea is to use integration by parts, as you did for $I_{2,n}$.
Let $2 \leq m \leq n$ be integers and denote $f(x) = (1 - e^{-x})^{n}$. We have two observations:
*
*$f^{(k)}(x) = O(x^{n-k})$ near $x = 0$ for $0 \leq k \leq n$.
*$f^{(k)}(x) = O(e^{-x})$ near $x = \infty$ for $1 \leq k \leq n$.
Using this, a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/863895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Prove $\sum\limits_{\mathrm{cyc}}\sqrt{a^2+bc}\leq{3\over2}(a+b+c)$ with $a,b,c$ are nonnegative Hope someone can help on this inequality using nonanalytical method (i.e. simple elementary method leveraging basic inequalities are prefered).
Prove $\sum\limits_{\mathrm{cyc}}\sqrt{a^2+bc}\leq{3\over2}(a+b+c)$ with $a,b,c... | first we check $abc=0$,if $a=0 ,\implies 2\sqrt{bc}\le b+c$,it is true.the "=" will hold when $b=c$
in case $abc \not=0$
let $c=Max${$a,b,c$},
we discuss two cases:
*
*$a^2+b^2+c^2 \ge 2(ab+bc+ac) \implies c\ge a+b+2\sqrt{ab} \implies ab \le \dfrac{c^2}{16}$
LHS $\le\sqrt{a^2+bc}+\sqrt{b^2+ac}+\sqrt{c^2+\dfrac{c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/864002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
How find the value of the $x+y$ Question:
let $x,y\in \Bbb R $, and such
$$\begin{cases}
3x^3+4y^3=7\\
4x^4+3y^4=16
\end{cases}$$
Find the $x+y$
This problem is from china some BBS
My idea: since
$$(3x^3+4y^3)(4x^4+3y^4)=12(x^7+y^7)+x^3y^3(9y+16x)=112$$
$$(3x^3+4y^3)^2+(4x^4+3y^4)^2=9(x^6+y^8)+16(y^6+x^8)+24x^3y^3(... | Another idea for you to think about. We have:
$$\begin{cases}
3x^3+4y^3=7 \\
4x^4+3y^4=16
\end{cases}$$
Break it as:
$$\begin{cases}
3x^3+3y^3 + y^3=7 \\
x^4 + 3x^4+3y^4=16
\end{cases}$$
and factor:
$$\begin{cases}
3(x+y)(x^2 - xy + y^2)=7 - y^3\\
3(x+y)(x^3 - x^2y + xy^2 - y^3)=16 - x^4
\end{cases}$$
And so: $$x+y = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/868635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
} |
How to work out miles between Longitude values based on a Latitude value. We know that when Latitude is 0, the distance between Longitude values is roughly 69 miles.
When the Latitude is +/-90, Longitude values are 0 miles.
At 0 Latitude, the earths circumference is 24,902 miles.
From pole-to-pole, the earths circumfer... | This can be accomplished with Thaddeus Vincenty's inverse solution or with the haversine distance formula. If simplicity and speed of calculation is more important than accuracy, then use the haversine distance formula. Otherwise, go with Vincenty's inverse solution.
Vincenty's Inverse Solution
$\alpha$ length of the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/868868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Sum of series with triangular numbers Can you please tell me the sum of the seires
$ \frac {1}{10} + \frac {3}{100} + \frac {6}{1000} + \frac {10}{10000} + \frac {15}{100000} + \cdots $
where the numerator is the series of triangular numbers?
Is there a simple way to find the sum?
Thank you.
| Your expression is equal to $g(1/10)$, where
$$g(x)=\frac{x}{2}\left((2)(1)+(3)(2)x+(4)(3)x^2+(5)(4)x^3+\cdots\right)$$
Take the power series $1+x+x^2+x^3+\cdots$ for $\frac{1}{1-x}$ and differentiate twice. We get $(2)(1)+(3)(2)x+(4)(3)x^2+\cdots$ if we do it term by term, and $\frac{2!}{(1-x)^3}$ if we do it the usu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/868943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\int \frac{\tan^3x+\tan x}{\tan^3x+3 \tan^2x+2 \tan x+6} dx$ $$\int \frac{\tan^3x+\tan x}{\tan^3x+3 \tan^2x+2 \tan x+6} dx$$
My approaches so far has been using substitution with $\tan x = t$ and $\tan \frac x2 = t$ but the calculations has been harder than I think they should.
I've also tried using ordinary ... | Given
$$
\int \frac{\tan^3(x) + \tan(x)}{\tan^3(x) + 3\tan^2(x) + 2\tan(x)+6} dx.
$$
Consider the substitution
$$
\boxed{\color{red}{\tan(x) = \sqrt{2} \tan(y)}},
$$
so
$$
\Big( \tan^2(x) + 1 \Big) dx = \sqrt{2} \Big( \tan^2(y) + 1 \Big) dy.
$$
so we obtain
$$
\begin{eqnarray}
\int \frac{\tan^3(x) + \tan(x)}{\tan^3(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/869142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Homework - Resolve the recurrence relation What's the closed formula of this recurrence relation?
$$a_n = a_{n-1}+2a_{n-2}+2^n \text{ with } a_0=1, a_1=2 $$
| A generating function approach would make this straightforward:
$$G(x) = \sum_{n=0}^\infty a_n x^n = 1+2x+\sum_{n=2}^\infty a_n x^n$$
$$=1+2x+\sum_{n=2}^\infty (a_{n-1} + 2a_{n-2} + 2^n)x^n$$
$$=1+2x+\sum_{n=1}^\infty a_n x^{n+1} + 2 \sum_{n=0}^\infty a_n x^{n+2} + \sum_{n=2}^\infty 2^n x^n$$
$$=1+2x+x(G(x)-1)+2x^2G(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/869341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
an exercise from apostol analytic number theory this is the second exercise of chapter 3:
if $x\ge 2$ prove that
$$\sum_{n\le x} \frac{d(n)}{n}= \frac{1}{2} {\log^2 x} + 2C\log x +O(1)$$ where $C$ is Euler's constant.
here's what i've done to solve:
as we know that $\sum_{n\le x} {d(n)}= \sum_{d\le x} \sum_{q\le\frac{x... | I have not dug into the details but your error term is too small, it is not possibile to have something better than $O(1)$, since, by exploiting the Dirichlet hyperbola method, that leads to:
$$\sum_{n\leq x}d(n) = x\log x + (2\gamma-1)x + O(\sqrt{x}),\tag{1}$$
it follows that:
$$\begin{eqnarray*}\sum_{n\leq x}\frac{d(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/870373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Use an appropriate Half-Angle Formula to find the exact value of the expression $\cos \left(\frac{9\pi}{ 8}\right)$ I've been having problems all semester understanding radian fractions. If I were to double this fraction wouldn't it be $\frac{18\pi}{8}$. Yet, how does that help me find a reference angle on the unit cir... | From Gerry Myerson's comment, we have
\begin{align*}
\frac{18\pi}{8} &= \frac{9\pi}{4} \\
&=2\pi+\frac{\pi}{4} \\
\cos \frac{18\pi}{8} &= \cos \left(2\pi+\frac{\pi}{4}\right) \\
&= \cos \frac{\pi}{4} \\
&= \frac{\sqrt{2}}{2}
\end{align*}
We then have the half-angle formula
$$
\cos \frac{1}{2}\theta = (-1)^{\lfloor (\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/872145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is remainder when $5^6 - 3^6$ is divided by $2^3$ (method) I want to know the method through which I can determine the answers of questions like above mentioned one.
PS : The numbers are just for example. There may be the same question for BIG numbers.
Thnx.
| We know that
$x^2-y^2 = (x-y)(x+y) $ and $x^3 + y^3 = (x + y)(x^2 - xy + y^2) $
So,
$5^6 - 3^6 = (5^3)^2 - (3^3)^2 = (5^3 + 3^3) (5^3 - 3^3) $
so, $5^6 - 3^6 = (5 + 3) (5^2 - 3\centerdot 5 - 3^2) (5^3 - 3^3) $
Now, as reminder of $(x\centerdot y $ mod $ y )$ is $0$, dividing above term (with $(5+3)$ as one of the fact... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/874476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
A problem on continued fractions Find the value of $x$, if:
$$\large 1+\frac{1}{2+\frac{1}{1+\frac{1}{2+...}}}$$
My attempt:
Noting that:
$$\large x=1+\frac{1}{2+\frac{1}{x}}$$
$$x=\frac{1+\sqrt{3}}{2}$$
question: Is my solution correct?
| If others would like to know how he arrived at that:
Like you stated:
$$\large x=1+\frac{1}{2+\frac{1}{x}}$$
$$\large x= 1 + \frac{1}{\frac{2x+1}{x}}$$
$$\large x = 1 + \frac{x}{2x+1}$$
$$\large 2x^2 + x = 2x + 1 + x$$
$$\large 2x^2 - 2x - 1 =0$$
Quadratic Formula:
$$x = \frac{2 \pm \sqrt{4 + 8}}{4}$$
$$x = \frac{2 \pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/874663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluation of $\int\frac{\sqrt{\cos 2x}}{\sin x}\,dx$
Compute the indefinite integral
$$
\int\frac{\sqrt{\cos 2x}}{\sin x}\,dx
$$
My Attempt:
$$
\begin{align}
\int\frac{\sqrt{\cos 2x}}{\sin x}\,dx &= \int\frac{\cos 2x}{\sin^2 x\sqrt{\cos 2x}}\sin xdx\\
&= \int\frac{2\cos^2 x-1}{(1-\cos^2 x)\sqrt{2\cos^2 x-1} }\sin ... | The magic appears when performing the change $u=\dfrac{\cos(x)}{\sqrt{\cos(2x)}}$ due to the fact that there is $\sin(x)$ on denominator.
The integral is changed to $I=\displaystyle\int\dfrac {\mathop{du}}{(u^2-1)(2u^2-1)}\quad$ with not square root anymore.
Now it is just $\dfrac 1{u^2-1}-\dfrac{2}{2u^2-1}$
And $$I=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/876175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 4
} |
If $(a, b,c)$ is a Pythagorean triple with $b, c$ consecutive integers then $c \mid a^b – 1$, proof/disproof? If $(a, b,c)$ is a Pythagorean triple with $b, c$ consecutive integers then $c \mid a^b – 1$, proof/disproof?
Here are some examples:
$(3, 4, 5)$ is a Primitive Pythagorean Triple (PPT), $3^2 + 4^2 = 5^2$, wher... | Here's a remark that will complete @MorganO's answer. Recall that such triples are generated by positive integers $m>n$ as $(a,b,c)=(m^2-n^2,2mn,m^2+n^2)$. Since we want $b$ and $c$ to be consecutive, we require $$c-b=(m-n)^2=(m-n-1)(m-n+1)+1=1$$ which will only work if $m=n+1$. Thus $$ a = 2n+1,\; b=2n(n+1),\; c=2n^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/876344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Simple series divergence problem I've got a problem here:
$$\sum_{n=1}^{\infty} \frac{5^n}{n(3^{n+1})}$$
I've used the ratio test and essentially did this:
$$\sum_{n=1}^{\infty} \left( \frac{5^{n + 1}}{n (3^{n+1+1})} / \frac{5^n}{n(3^{n+1})}\right) = \frac{5^n\,5}{9(n+1)3^n} \cdot \frac{n\,3^n}{5^n}$$
With a bunch of c... | $$\begin{align}
a_n &= \frac{5^n}{n(3^{n+1})} \\[2em]
\frac{a_{n+1}}{a_n}
&= \frac{5^{n+1}}{(n+1)(3^{n+2})} / \frac{5^n}{n(3^{n+1})} \\[0.5em]
&= \frac{5^{n+1} \cdot n \cdot 3^{n+1}}{5^n \cdot (n+1) \cdot 3^{n+2}} \\[0.5em]
&= \frac{5}{3} \left( \frac{n}{n+1} \right) \\[0.5em]
&= \frac{5}{3} \left( 1 - \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/877160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to do integral $\int_0^T \frac{1}{t\sqrt{t(T-t)}}e^{-\frac{b^2}{2t}}dt$ and $\int_0^T \frac{1}{\sqrt{t(T-t)}}e^{-\frac{b^2}{2t}}dt$? I met these two integrals but don't know how to do them:
$$I_1 = \int_0^T \frac{1}{t\sqrt{t(T-t)}}e^{-\frac{b^2}{2t}}\text{d}t$$
$$I_2 = \int_0^T \frac{1}{\sqrt{t(T-t)}}e^{-\frac{b^2}... | Here are closed forms for the first and second integrals respectively
$$ I_1 = {
\frac {\sqrt {2\pi }}{b\sqrt {T}}}{{\rm e}^{-\,{\frac {{b}^{2}}{2T}}}} $$
$$I_2= \, \pi-
{{\rm erf}\left( {\frac {b}{\sqrt {2T}}}\right)}.$$
You can test them numerically.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/880104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Exact value of expression Let
$$f(x)=\frac{4^x}{4^x+2}$$
and
$$S=\sum_{n=1}^{2005}f\left(\frac{n}{2005}\right)$$
What is the exact value of $S$?
I tried to write $a=4^{\large\frac{1}{2005}}$, then
$$S=\sum_{n=1}^{2005}\frac{a^n}{a^n+2}$$
but I still cannot simplify it. Is there any easy method?
| Since $$f(x)+f(1-x)=\frac{4^x}{2+4^x}+\frac{4^{1-x}}{2+4^{1-x}}=\frac{4^x}{2+4^x}+\frac{2}{4^x+2}=1$$
you have:
$$\begin{eqnarray*}S&=&\sum_{n=1}^{2005}f\left(\frac{n}{2005}\right)=f(1)+\sum_{i=1}^{1002}\left(f\left(\frac{n}{2005}\right)+f\left(\frac{2005-n}{2005}\right)\right)\\&=&f(1)+1002=\frac{2}{3}+1002.\end{eqnar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/881394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\sum \limits_{cyc}\frac {a}{(b+c)^2} \geq \frac {9}{4(a+b+c)}$
Given $a$, $b$ and $c$ are positive real numbers. Prove that:$$\sum \limits_{cyc}\frac {a}{(b+c)^2} \geq \frac {9}{4(a+b+c)}$$
Additional info: We can't use induction. We should mostly use Cauchy inequality. Other inequalities can be used rare... | Well I found a different approach for solving this problem. We could rewrite the inequality from the question as:
$$(a+b+c)\left(\sum \limits_{cyc}\frac {a}{(b+c)^2}\right) \ge \frac{9}{4}$$
By the Cauchy-Schwarz inequality:
$$(a+b+c)\left(\sum \limits_{cyc}\frac {a}{(b+c)^2}\right) \ge \left(\sum \limits_{cyc}\frac {a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/882272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
} |
A certain “harmonic” sum Is there a simple, elementary proof of the fact that:
$$\sum_{n=0}^\infty\left(\frac{1}{6n+1}+\frac{-1}{6n+2}+\frac{-2}{6n+3}+\frac{-1}{6n+4}+\frac{1}{6n+5}+\frac{2}{6n+6}\right)=0$$
I have thought of a very simple notation for "harmonic" sums like these: just write down the numerators. So, for... | In the language of Dirichlet series and the Riemann zeta function I believe this could be counted as an elementary proof:
Add the variable $s$ as an exponent to your series so that it becomes:
$$\sum_{n=0}^\infty\left(\frac{1}{(6n+1)^s}+\frac{-1}{(6n+2)^s}+\frac{-2}{(6n+3)^s}+\frac{-1}{(6n+4)^s}+\frac{1}{(6n+5)^s}+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/883233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 0
} |
Integration of $x\cos(x)/(5+2\cos^2 x)$ on the interval from $0$ to $2\pi$
Compute the integral
$$\int_{0}^{2\pi}\frac{x\cos(x)}{5+2\cos^2(x)}dx$$
My Try: I substitute $$\cos(x)=u$$
but it did not help. Please help me to solve this.Thanks
| Using $\displaystyle\int_a^bf(x)\ dx=\int_a^bf(a+b-x)\ dx,$
$$I=\int_0^{2\pi}\frac{x\cos x}{5+2\cos^2x}dx=\int_0^{2\pi}\frac{(2\pi-x)\cos(2\pi-x)}{5+2\cos^2(2\pi-x)}\ dx=\int_0^{2\pi}\frac{(2\pi-x)\cos x}{5+2\cos^2 x}\ dx$$
$$2I=2\pi\int_0^{2\pi}\frac{\cos x}{5+2\cos^2x}dx$$
$$\implies I=\pi\int_0^{2\pi}\frac{\cos x}{7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/884362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How prove $ \sqrt{2}+\sqrt{3}>\pi$? How prove that $ \sqrt{2}+\sqrt{3}>\pi$? Maybe some easy way?
| This is just an improved version of The Great Seo's answer.
Since:
$$\sum_{n=1}^{+\infty}\frac{1}{n^2\binom{2n}{n}}=\frac{\pi^2}{18},\qquad\sum_{n=1}^{+\infty}\frac{1}{n^4\binom{2n}{n}}=\frac{17\,\pi^4}{3240}$$
you only need to check that:
$$\sum_{n=1}^{+\infty}\frac{\frac{3240}{17}-180\,n^2}{n^4\binom{2n}{n}}<-1$$
tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/884424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 3
} |
prove by induction $7 \mid 3^{3^n}+8$ Okay so ive been trying to prove this for about 5 hours...
really need salvation from the geniouses around here.
prove by induction
$7\mid 3^{3^n}+8$
i really need some directions on what to do here...
| We have to show that :
$$3^{3^n}+8 \equiv 0 \pmod 7 \Rightarrow 3^{3^n}+1 \equiv 0 \pmod 7$$
$$n=1: 3^{3}+1=3^{2+1}+1=3^2 \cdot 3+1=2 \cdot 3+1=7 \equiv 0 \pmod 7 \checkmark$$
$$\text{We suppose that the relation stands for n: } \ 3^{3^n}+1 \equiv 0 \pmod 7$$
For $n+1$:
$$ 3^{3^{n+1}}+1=3^{3^n \cdot 3}+1 \equiv (3^{3^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/885849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Integral of $\frac{ \sqrt{\cos 2 x}}{\sin x}$ I am trying to solve the integral of $\frac{ \sqrt{\cos 2 x}}{\sin x}$. I converted this to $(\cot^2 x - 1)^{1/2}$ but after this I am stuck. I am not able to think of a suitable substitution. Any tips?
| $\bf{My\; Solution::}$ Given $$\displaystyle \int\frac{\sqrt{\cos 2x}}{\sin x}dx = \int\frac{\sqrt{\cos^2 x-\sin^2 x}}{(\cos x+\sin x)-(\cos x-\sin x)}dx$$
So Integral $$\displaystyle =\int \frac{\sqrt{(\cos x+\sin x)\cdot (\cos x-\sin x)}}{(\cos x+\sin x)-(\cos x-\sin x)}dx = \int\frac{\sqrt{\frac{\cos x+\sin x}{\cos ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/887082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Trigonometric simplification for limits: $\lim_{x\to 0} \frac{1-\cos^3 x}{x\sin2x}$ Have to evaluate this limit, but trigonometry part is :(
$$\lim_{x\to 0} \dfrac{1-\cos^3 x}{x\sin2x}.$$
Had written the denominator as $2x\sin x\cos x$, no idea what to do next. Please help...
| $$\lim_{x \to 0}\frac{1-\cos^3x}{2x^2} \cdot \lim_{x \to 0}\frac{2x}{\sin(2x)}$$
The right limit is equal to 1.
For the left limit, you may use the rule of de l'Hôpital:
$$\eqalign{
\lim_{x \to 0}\frac{1-\cos^3x}{2x^2}&=\lim_{x \to 0}\frac{3\cos^2 x \sin x}{4x}\\
&=\lim_{x \to 0}\frac{-6\cos x \sin^2 x + 3 \cos^3 x}{4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/888081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Max of $3$-Variable Function I'm trying the find the maximum of the function
$$f(a,b,c)=\frac{a+b+c-\sqrt{a^2+b^2+c^2}}{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}$$
for all nonnegative real numbers $a, b, c$ with $ab + bc + ca > 0$.
I tried in vain to prove that $\max_{a,b,c}f(a,b,c)=1-\frac{\sqrt{3}}{3}$
| You can assume $a +b +c = 3$ because scaling a solution by a positive number doesn't change the objective function. Then you are left with
$$\frac{3 - \sqrt{a^2 + b^2 + c^2}}{\sqrt{ab} + \sqrt{ac} + \sqrt{bc}}$$
subject to $a + b + c = 3$. Show that $a^2 + b^2 + c^2$ is minimized when $a=b=c$ assuming $a+b+c$ is a cons... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/890381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solving $x^2 - 11y^2 = 3$ using congruences I'm looking to find solutions to $x^2 - 11y^2 = 3$ using congruences.
The question specifically asks "Can this equation be solved by congruences (mod 3)? If so, what is the solution? (mod 4) ? (mod 11) ?"
I know using (mod 4) I can show that there aren't any solutions:
$x^2 ... | $$-11 \equiv -11+16 \equiv 5 \equiv 1 \pmod 4$$
So,the congruence becomes:
$$x^2+y^2 \equiv 3 \pmod 4$$
$$x \equiv 0 \pmod 4 \Rightarrow x^2 \equiv 0 \pmod 4$$
$$x \equiv 1 \pmod 4 \Rightarrow x^2 \equiv 1 \pmod 4$$
$$x \equiv 2 \pmod 4 \Rightarrow x^2 \equiv 0\pmod 4$$
$$x \equiv 3 \pmod 4 \Rightarrow x^2 \equiv 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/893394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How to prove that $\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$ without using Fourier Series Can we prove that $\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$ without using Fourier series?
| Yes. The most common way to do this is attributed to Euler. It does still require Maclaurin series, however.
Consider the Maclaurin polynomial for $\frac{\sin x}{x}$:
$$\frac{\sin x}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \cdots$$
However, note that this is a polynomial $p(x)$ with zeroes $\{\pm k\pi\;|\;k \in \Bbb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/893472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Question on Factoring I have very basic Question about factoring, we know that,
$$x^2+2xy+y^2 = (x+y)^2$$
$$x^2-2xy+y^2 = (x-y)^2$$
But what will
$$x^2-2xy-y^2 = ??$$
$$x^2+2xy-y^2 = ??$$
| You can factor a quadratic trinomial
$$ax^2+bxy+cy^2$$
by finding the roots of
$$\frac{ax^2+bxy+cy^2}{y^2}=a\left(\frac xy\right)^2+b\left(\frac xy\right)+c=0.$$
Then
$$ax^2+bx+c=y^2a\left(\frac xy-r_0\right)\left(\frac xy-r_1\right)=a\left(x-r_0y\right)\left(x-r_1y\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/894392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
If a fair die is thrown three times, what is the probability that the sum of the faces is 9?
If a fair die is thrown thrice, what is the probability that the sum of the faces is 9?
I did like this.
The total number of cases is $6^3=216$
Now,the number of solutions of the equation $x + y + z = 9$ with each of $x,y,z$ ... | Alternatively: given that $x, y, z \in \{1,..6\}, x+y+z=9$
If $x=1$ then $y\in \{2,..6\}$, else if $x\in\{2,..6\}$ then $y\in\{1,.. 8-x\}$. For each such pairing there is one value of $z$.
$$\begin{align}
\sum_{y=2}^{6} 1 + \sum_{x=2}^6 \sum_{y=1}^{8-x} 1 & = 5 + \sum_{x=2}^{6}(8-x)
\\ & = 5+8\times 5 - \frac{6\times ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/895803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
proving that: $(\frac{13}{4})^n\leq a(n)\leq (\frac{10}{3})^n$ Given $a(n)$ number of sequences of length $n$ that are formed by the digits: $0,1,2,3$
such that after the digit $0$ the digit $1$ must immediately follow.
Need to prove that $(\frac{13}{4})^n\leq a(n)\leq (\frac{10}{3})^n$
I don't know how to start, i do... | Using the "$1$ immediatelly after $0$" interpretation:
To contruct a sequence with $n$ symbols, we either take a sequence of $n-1$ symbols and place a $1$, $2$ or $3$ afterwards. Or we take a sequence with $n-2$ symbols and place a $01$. Therefore:
$$a(n) = 3a(n-1) + a(n-2)$$
Where $a(0) = 1$, $a(1) = 3$.
The inequalit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/896148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the sum of a series till $n$ terms Series:
5, 11, 19, 29, 41
Find the sum of the series up to $n$ terms.
Well the method that comes to my mind is to find the nth term of the sequence, and then find their summation. I use the basic formulas, such as sum of series $n^2$, $n^3$, etc.
My question is whether their i... | For a slightly different way of getting the pattern, note that the sequence of first differences is $6,8,10,12,\ldots$ and thus the second differences are constant. So the sequence is quadratic i.e. $a_n=An^2+Bn+C$. Then \begin{align}
a_1&=A+B+C=5,\\\\
a_2-a_1&=3A+B=6,\\\\
(a_{n+2}-a_{n+1})-(a_{n+1}-a_n)&=2A=2,
\end{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/897205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Distinct balls into distinct boxes with a minimal number of balls in each box
Find the number of ways to distribute $8$ distinct balls into $3$ distinct boxes if each box must hold at least $2$ balls.
The stars and bars approach would not work because the balls are non-identical. Stirling Numbers of the second kind w... | $8=4+2+2=2+4+2=2+2+4$ gives $3\times\frac{8!}{4!2!2!}$ possibilities.
$8=2+3+3=3+2+3=3+3+2$ gives $3\times\frac{8!}{2!3!3!}$ possibilities.
So not $3!$ but $3$ as extra factor to avoid double counting.
edit
Label the boxes: box1, box2 and box3.
Number of possibilities to end up with $4$ balls in box1: $\binom{8}{4}\bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/898966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Derivatives of trig polynomials do not increase degree? Let $c = \cos x$ and $s = \sin x$,
and consider a trigonometric polynomial $p(x)$
in $c$ and $s$.
The degree of $p(x)$ is the maximum of
$n+m$ in terms $c^n s^m$.
Is it the case that repeated derivatives of $p(x)$,
expressed again in terms of $c$ and $s$,
nev... | Using complex exponential relations,
$$e^{i\theta} = \cos\theta + i \sin\theta \qquad
\sin\theta = \frac{1}{2i}(e^{i\theta}-e^{-i\theta}) \qquad
\cos\theta = \frac{1}{2}(e^{i\theta}+e^{-i\theta})$$
we see that sine-cosine polynomial of degree $p$ can be written as a linear combination of complex exponentials "as larg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/901071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to prove this inequality without using Muirhead's inequality? I ran into a following problem in The Cauchy-Schwarz Master Class:
Let $x, y, z \geq 0$ and $xyz = 1$.
Prove $x^2 + y^2 + z^2 \leq x^3 + y^3 + z^3$.
The problem is contained in the chapter about symmetric polynomials and Muirhead's inequality.
The proo... | In general, using the power mean inequality,
$$\sqrt[n+1]{\frac{x^{n+1} + y^{n+1} + z^{n+1}}{3}} \ge \sqrt[n]{\frac{x^n + y^n + z^n}{3}}$$
Raising both sides to the same power,
$$\frac{x^{n+1} + y^{n+1} + z^{n+1}}{3} \ge \frac{x^n + y^n + z^n}{3} \cdot \sqrt[n]{\frac{x^n + y^n + z^n}{3}}$$
But, again by the power mean ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/902474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Sum of $1+\frac{1}{2}+\frac{1\cdot2}{2\cdot5}+\frac{1\cdot 2\cdot 3}{2\cdot 5\cdot 8}+\cdots$ I am trying to find out the sum of this
$$1+\frac{1}{2}+\frac{1\cdot2}{2\cdot5}+\frac{1\cdot 2\cdot 3}{2\cdot 5\cdot 8}+\frac{1\cdot 2\cdot 3\cdot 4}{2\cdot 5\cdot 8\cdot 11}+\cdots$$.
I tried with binomial theorem with ration... | Here is an another approach that doesn't add anything much to Jack's answer but is at least mildly entertaining. It does not use the $B$ function but instead takes a detour via a "simple" alternating series. As stated in the comments the sum equals $${}_2F_1(1,1,\tfrac{2}{3};\tfrac{1}{3})$$ and using the transformati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/904810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
How prove this inequality $(1+\frac{1}{16})^{16}<\frac{8}{3}$
show that
$$(1+\dfrac{1}{16})^{16}<\dfrac{8}{3}$$
it's well know that
$$(1+\dfrac{1}{n})^n<e$$
so
$$(1+\dfrac{1}{16})^{16}<e$$
But I found this $e=2.718>\dfrac{8}{3}=2.6666\cdots$
so how to prove this inequality by hand?
Thank you everyone solve it,I w... | We can split the series at index $m \in \mathbb{Z}$ where $0 \le m \le 15$:
$S = \Big(1+\dfrac{1}{16}\Big)^{16} = \displaystyle\sum\limits_{k=0}^{m}{{16 \choose k}\Big(\frac{1}{16}\Big)^k} + \displaystyle\sum\limits_{k=m+1}^{16}{{16 \choose k}\Big(\frac{1}{16}\Big)^k} \tag{1}$
Now let $a_k$ be the terms in the summati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/906643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 4
} |
Quadratic formula and factoring are leading to different answers $$x^{ 2 }-2x-15=0$$
By factoring, I get:
$$(x-5)(x+3)$$
Which has the solutions:
$$x=5, x=-3$$
However when I use the quadratic formula (which is what the book saids to use), I get
$$\frac { 2 \pm \sqrt { 4-(4\cdot1\cdot(-15)) } }{ -2 } =$$
$$\frac { 2... | The quadratic formula is
$$
x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}
$$
The denominator should be $2$ because
\[ a=1 \]
\[ b=-2 \]
\[ c=-15 \]
So then
\[
x=\frac{2\pm \sqrt{4-4(-15)}}{2}= \frac{2\pm \sqrt{64}}{2}= \frac{2\pm 8}{2}=1\pm 4
\]
Thus
$$ x=5 $$
$$ x=-3 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/906818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
solve the diophantine equation: $x^3-3xy^2=z^3$ Let $ x,y,z$ be 3 integers greater than 1,if $x$ and $y$ are relatively prime, solve the diophantine equation: $x^3-3xy^2=z^3$.
| $$x^3-3xy^2=z^3 \Rightarrow x^3-z^3=3xy^2 \Rightarrow 3 \mid x^3-z^3 \Rightarrow x^3 \equiv z^3 \pmod 3$$
From Fermat's theorem:
$$x^3 \equiv x \pmod 3 \\ z^3 \equiv z \pmod 3$$
So,we have:
$$x \equiv \ z \pmod 3 \Rightarrow 3 \mid x-z \Rightarrow x=z+3k,k \in \mathbb{Z}$$
Replace at $x^3-3xy^2=z^3$ and take into consi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/906913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Dividing by $\sqrt n$ Why is the following equality true?
I know I should divide by $\sqrt n$ but how is it done exactly to get the RHS?
$$ \frac{\sqrt n}{\sqrt{n + \sqrt{n + \sqrt n}}} = \frac{1}{\sqrt{1 + \sqrt{\frac{1}{n} + \sqrt{\frac{1}{n^3}}}}}$$
| Well this is calculus. As you said yourself, you divide the numerator and the denominator by $\sqrt{n}$
for the numerator it's easy, as $\frac{\sqrt{n}}{\sqrt{n}}=1$.
for the denominator you get \begin{equation}
\frac{1}{\sqrt{n}}\sqrt{n+{\sqrt{n+{\sqrt{n}}}}}=\sqrt{1+\frac{\sqrt{n+\sqrt{n}}}{n}}
\end{equation}
Keep in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/906983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Short form of few series Is there a short form for summation of following series?
$$\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\alpha^{2n}((2y-1)^{2k+1}+1)}{2^{2n+1}(2n)!k!(n-k)!(2k+1)}$$
$$\sum\limits_{n=0}^\infty\dfrac{\alpha^{2n+1}(\cos^{-1}(2y-1)-\pi)}{2^{4n+3}n!(n+1)!}$$
$$\sum\limits_{n=0}^\infty\s... | For the first series, I was able to wittle it down to a sum of a Struve function and an unevaluated series of hypergeometric functions:
$$\begin{align}
S{(\alpha,y)}
&=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\alpha^{2n}((2y-1)^{2k+1}+1)}{2^{2n+1}(2n)!k!(n-k)!(2k+1)}\\
&=\sum\limits_{n=0}^\infty \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/907243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Splitting $\Phi_{15}$ in irreducible factors over $\mathbb{F}_7$ I have to split $\Phi_{15}$ in irreducible factors over the field $\mathbb{F}_7$. It has been a while that I did this kind of stuff, and to be the honest, I've never really understood this matter. I'd really be grateful if someone could show me how this w... | Every such $\zeta$ can be written uniquely as $\zeta_1\zeta_2$ where $\mathrm{ord}(\zeta_1)=3$ and $\mathrm{ord}(\zeta_2)=5$. The cases where $\mathrm{ord}(\zeta_1)=3$ are $2,4$.
The $\zeta_2$ values are roots of $1+x+x^2+x^3+x^4=(x-\zeta_5)(x-\zeta_5^2)(x-\zeta_5^3)(x-\zeta_5^4)$. So
$$\begin{align}(x-2\zeta_5)(x-2\z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/909127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Computing $\sum_{i=0}^{\infty}\frac{i}{2^{i+1}}$ I came across this while trying to solve Google's boys & girls problem, and although I know now it's not the right approach to take, I'm still interested in summing
$\sum_{i=0}^{\infty}\frac{i}{2^{i+1}}$. Apparently it should be 1..but I'm having a tough time seeing this... | $$\begin{align}
\frac{1}{4}&=1-\frac{3}{4}\\
\frac{1}{4} + \frac{2}{8} &= 1-\frac{4}{8}\\
\frac{1}{4}+\frac{2}{8} +\frac{3}{16}&= 1-\frac{5}{16}
\end{align}
$$
Therefore, try to show that $\frac{n}{2^{n+1}} = \frac{n+1}{2^n}-\frac{n+2}{2^{n+1}}$. That's pretty easy, so this is a telescoping sum:
$$\left(\frac{1}{1}-\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/910021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Calculus (Integration) Is there a simple way to integrate $\displaystyle\int\limits_{0}^{1/2}\dfrac{4}{1+4t^2}\,dt$
I have no idea how to go about doing this. The fraction in the denominator is what's confusing me. I tried U-Substitution to no avail.
| $$\int_0^{\frac{1}{2}} \frac{4}{1+4t^2} dt$$
We set $t=\frac{\tan{u}}{2}$ we have the following:
$t=0: u=0$
$t=\frac{1}{2}: u=\frac{\pi}{4}$
$dt=\frac{1}{2 \cos^2{u}}du$
$$\frac{4}{1+4t^2}=\frac{4}{1+4 \frac{\tan^2{u}}{4}}=\frac{4}{1+\tan^2{u}}=\frac{4 \cos^2{u}}{\sin^2{u}+\cos^2{u}}=4 \cos^2{u}$$
Therefore, we have th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/911470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Conditional probability: At least 3 kings given there are at least 2 kings in the hand of 13. My first "conditional probability" problem. Sorry for all the questions. My instructor doesn't make sense to the class.
A hand of 13 cards is to be dealt at random and without any replacement from an ordinary deck of playing c... | There are $\binom{52}{13}$ ways to select 13 cards from 52 cards.
Consider how you would create a hand of 13 cards with the required number of kings.
There are $\binom{4}{n}$ ways of choosing the required number of kings.
There are then $\binom{48}{13-n}$ ways of choosing the remaining (13 - n) cards.
Let A be the ev... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/915362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is my solution to this differential equation correct?
My answer is: $[(1+x^2)^3]y = \dfrac{(1+x^2)^3}3+C$
But this option is not given, so is it correct?
Thanks
| $$\begin{align}
(1+x^2)\frac{dy}{dx}+6xy&=2x\\
(1+x^2)\frac{dy}{dx}&=2x(1-3y)\\
\int \frac{dy}{1-3y}&=\int\frac {2x}{1+x^2}dx\\
-\frac13\ln(1-3y)&=\ln(1+x^2)\\
(1-3y)(1+x^2)^3&=0\\
(1+x^2)^3&=3y(1+x^2)^3\\
1+3x^2+3x^4+x^6&=3y(1+x^2)^3\\
\frac 13+x^2+x^4+\frac{x^6}3&=y(1+x^2)^3\\
y(1+x^2)^3&=x^2+x^4+\frac{x^6}3+C
\end{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/915448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If $\sec \theta+\tan \theta= \sqrt{3}$ then the positive value of $\sin \theta$ If $\sec \theta+\tan \theta=\sqrt{3}$ then the positive value of $\sin \theta$
Note:
$1/\cos\theta+\sin\theta/\cos \theta=\sqrt{3}$
$\sin\theta=\sqrt{3}\cos \theta-1$
squaring on both sides we get
$\sin^2\theta=$
| Beginning with $1+\sin \theta = \sqrt{3} \cos \theta$, we can use the identity $\sin^2 \theta + \cos^2 \theta = 1$ to obtain $$1+\sin \theta = \sqrt{3(1-\sin^2\theta)} \\ \Rightarrow 1 + 2\sin \theta + \sin^2 \theta = 3-3 \sin^2 \theta \\ \Rightarrow 4\sin^2 \theta + 2 \sin \theta - 2 = 0$$ Now you have an quadratic eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/915618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Linear independence of matrices $I, A, A^2$ I want to prove that $I,A,A^2\:$matrices $\in M_{2\times 2}$ are $\textit {linearly independent}$.
I consider the following matrices and their "corresponding" vectors:
$I=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\to (1,0,0,1)$
$A=\begin{pmatrix} a & b \\ c & d \end{pmatri... | The thing you're trying to prove is not true. Consider the case $A = I$.
More generally, if $A$ is any $2 \times 2$ matrix, then you can compute its characteristic polynomial, $c(x)$, the determinant of $A - xI$. That will be some quadratic polynomial in $x$, like
$$
3x^2 - 22 x^1 + 8x^0.
$$
It turns out that if you p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/916312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to simplify $\sqrt{\sqrt{5}+1} \cdot \sqrt{\sqrt{5}-1}$? This is the original problem:
$\sqrt{\sqrt{5}+1} \cdot \sqrt{\sqrt{5}-1} = x$.
I'm really confused about how to solve this problem, I come as far as saying this: $\sqrt[4]{5} + \sqrt{1}\cdot \sqrt[4]{5}-\sqrt{1}$.
| $$\sqrt{\sqrt{5}+1}\sqrt{\sqrt{5}-1}=\sqrt{(\sqrt{5}+1)(\sqrt{5}-1)}$$ $$=\sqrt{5-\sqrt{5}+\sqrt{5}-1}$$ $$=\sqrt{5-1}$$ $$=\sqrt{4}$$ $$=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/916598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Determine the irrational numbers $x$ such that both $x^2+2x$ and $x^3-6x$ are rational numbers I did not make any progress. The problem is from RMC 2008.
The only idea that I have is:
Try to find sets of irrational numbers such that every number in the set multiplied by another number in the set yields a rational numbe... | Let $x$ be an irrational number with
$$x^{2} + 2x = \frac{p_{1}}{q_{1}},\ x^{3} - 6x = \frac{p_{2}}{q_{2}}$$
for some integers $p_{1}, q_{1}, p_{2}, q_{2}$ such that $q_{1}, q_{2} \neq 0$ and $(p_{1}, q_{1}) = (p_{2}, q_{2}) = 1.$
Then we have
$$2x^{2} - (6 + \frac{p_{1}}{q_{1}})x + \frac{p_{2}}{q_{2}} = 0,$$
so that
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/917723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Inequality: $x^2+y^2+xy\ge 0$ I want to prove that $x^2+y^2+xy\ge 0$ for all $x,y\in \mathbb{R}$.
My "proof": Suppose wlog that $x\ge y$, so $x^2\cdot x\ge x^2\cdot y\ge y^2\cdot y=y^3$ (because $x^2\ge 0$ so we can multiply both sides by it without changing the inequality sign) giving $x^3\ge y^3$. Substracting we ha... | A homogeneous bivariate second-degree polynomial with a negative discriminant can be only everywhere non-negative or everywhere non-positive over $\mathbb{R}^2$. Since in $(x,y)=(1,1)$ the inequality holds as $\geq$, for every $(x,y)\in\mathbb{R}^2$ we have $x^2+xy+y^2\geq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/920605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 6
} |
How prove this inequality with $x+y+z=1$
let $x,y,z>0$,and such $$x+y+z=1$$
show that:$$\dfrac{(1+xy+yz+xz)(1+3x^2+3y^2+3z^2)}{9(x+y)(y+z)(x+z)}
\ge \left(\dfrac{x\sqrt{1+x}}{\sqrt[4]{3+9x^2}}+\dfrac{y\sqrt{1+y}}{\sqrt[4]{3+9y^2}}
+\dfrac{z\sqrt{1+z}}{\sqrt[4]{3+9z^2}}\right)^2$$
My idea: maybe we can prove $LHS\ge 1... | We can show LHS $\ge 1 \iff $
$$ (1+xy+yz+xz)[1+3(x^2+y^2+z^2)]\ge 9(x+y)(y+z)(x+z) $$
Let $p = x+y+z = 1, q = xy+yz+zx \le \frac13, r = xyz$. Then we have the inequality as
$$(1+q)[1+3(p^2-2q)] \ge 9(pq-r) \iff 4+9r \ge 11q + 6q^2$$
By Schur $p^3+9r \ge 4pq \implies 1+9r \ge 4q$, so it is enough to show that $3 \ge 7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/921145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I prove that any unit fraction can be represented as the sum of two other distinct unit fractions? A number of the form $\frac{1}{n}$, where $n$ is an integer greater than $1$, is called a unit fraction.
Noting that
$\frac{1}{2} = \frac{1}{3} + \frac{1}{6}$
and
$\frac{1}{3} = \frac{1}{4} + \frac{1}{12}$,
find a ... | In reply to DavidH's question "is the above decomposition of a unit fraction into a pair of distinct unit fractions unique?":
If $n$ is not prime, then we can write $n = n_1n_2$ with $n_1 \neq 1 \neq n_2$, and then we have the decomposition
$$\frac{1}{n} = \frac{1}{n_1}\frac{1}{n_2} = \frac{1}{n_1}\Big(\frac{1}{n_2+1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/921456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Complex analysis: Rewrite $\cos^{-1}{i}$ in algebraic form I'm stuck in this problem (complex analysis), my answer is not the one reported in the book:
Rewrite $\cos^{-1}{i}$ in the algebraic form. A: $k\pi + i \frac{\ln{2}}{2}\ \forall\ k \in \mathbb{Z}$
So I tried this approach in order to solve it:
*
*As $\cos^... | The answer in the book is wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/922124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How does one simplify the expression $\sqrt[3]{2 \sqrt{2}}\left(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\right)$ I don't know how to solve this problem. What I know is $\sqrt[3]{2 \sqrt{2}}=\sqrt{2}$. But I don't know how to continue.
| $\sqrt[3]{2\sqrt 2}=\sqrt 2$ is correct. Since we have $$2\pm \sqrt 3=\frac{4\pm 2\sqrt 3}{2}=\frac{(\sqrt 3\pm 1)^2}{2}\Rightarrow \sqrt{2\pm \sqrt 3}=\frac{\sqrt 3\pm 1}{\sqrt 2},$$
we have
$$\sqrt 2\left(\sqrt{2-\sqrt 3}+\sqrt{2+\sqrt 3}\right)=\sqrt 2\left(\frac{\sqrt 3 -1}{\sqrt 2}+\frac{\sqrt 3+1}{\sqrt 2}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/926347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Ring Theory question in a GRE practice exam I have a question about a GRE practice problem relating to Rings. The question is as follows:
Suppose that two binary operations, denoted by $\oplus$ and $\odot$ , are defined on a nonempty set $S$, and that the
following conditions are satisfied for all $x, y$, and $z$ in $S... | Here's a counterexample for (I). $\mathrm{Mat}_{2 \times 2}(\mathbb{R})$ equipped with matrix addition and multiplication satisfies all of the above criteria, and we consider the matrices $$A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \: \: B = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$$
For $n = 2$, we have th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/928028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Discrete Math sequence question The question is find $a3$:
$a_0 = 2, a_1 = 4$ and $a_{k+2} = 3a_{k+1}-a_k$ for any integer $k \geq 0 $
I know the answer is 26, although how do you get the answer?
| $ a_0 =2 , a_1 = 4, a_{k+2} - 3a_{k+1} +a_k = 0 $
We can suggest an equation that $ x^2 -3x +1 = 0 $,
Let the roots of the equation are $ x_1 , x_2$
then $x_1 +x_2 = 3$, $x_1 x_2 = 1$
then $a_{k+2} - 3a_{k+1} +a_k = 0 $$ \Leftrightarrow $ $a_{k+2} - (x_1 +x_2 )a_{k+1} + x_1 x_2 a_k = 0$
$\Leftrightarrow$ $ a_{k+2} - x_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/929485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Matrix Algebraic Operations, If AA = AB, does A = B? A and B are 2 x 2 matrices and A is not a zero matrix. How is the following proof incorrect?
Since AA = AB, AA - AB = 0
A (A - B) = 0 and since A does not equal zero, then A - B = 0, therefore A = B.
Thank you
| $$\begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}\begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}=\underbrace{\begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}}_{=A}\underbrace{\begin{pmatrix}3 & 5 \\ 0 & 0 \end{pmatrix}}_{=B}$$
EDIT Here's an example where the product is not $0$.
$$\begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/929741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Other ways to evaluate $\lim_{x \to 0} \frac 1x \left [ \sqrt[3]{\frac{1 - \sqrt{1 - x}}{\sqrt{1 + x} - 1}} - 1\right ]$? Using the facts that:
$$\begin{align}
\sqrt{1 + x} &= 1 + x/2 - x^2/8 + \mathcal{o}(x^2)\\
\sqrt{1 - x} &= 1 - x/2 - x^2/8 + \mathcal{o}(x^2)\\
\sqrt[3]{1 + x} &= 1 + x/3 + \mathcal{o}(x)
\end{align... | Let us start by noting that:
*
*$\sqrt[3]{1-\sqrt{1-x}}\cdot \sqrt[3]{1+\sqrt{1-x}}=\sqrt[3]{x}$
*$\sqrt[3]{\sqrt{1+x}-1}\cdot \sqrt[3]{\sqrt{1+x}+1}=\sqrt[3]{x}$
*$$\sqrt[3]{\frac{(1-\sqrt{1-x})(1+\sqrt{1-x})}{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}}=1$$
Using these algebraic facts, we can write that: $$\begin{align} \lim_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/932411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Formula for $\sum\cos(\pi kt)/(1+k^2)$ Is there an explicit formula for the sum
$$F = \sum_{k=0}^\infty \frac{1}{1+k^2} \cos(\pi k t)$$
This is the green function for the operator $1 + \Delta$ on the circle.
| If $0 \le t \le 2 $, $$\sum_{k=0}^{\infty} \frac{\cos (k \pi t) }{1+k^{2}} = \frac{\pi}{2} \frac{\cosh [\pi(t-1)]}{\sinh \pi} + \frac{1}{2}. $$
One way to evaluate series of the form $ \displaystyle \sum f(k) e^{ik \theta}$, where $0 \le \theta \le 2 \pi$ and $f(z)$ is rational function such that $f(z) = \mathcal{O}(z^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/933412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
finding solutions by factoring How would you find the integer solutions to $a^2-b^2=16$?
I know that the factors of $16$ are $8*2,$ $4*4,$ and $16*1.$ How would I use this?
I know that $a^2-b^2=(a+b)(a-b)=16,$ but how would you find to the solutions to it? I know that you can plug in numbers to find the answers, but i... | We wish to find integer solutions for $a$ and $b$ such that $a^2-b^2=16$.
Factoring the expression we obtain $(a+b)(a-b)=16$.
This means that $(a+b)|16$ and $(a-b)|16$.
From this we can find a set $(a+b)\in \{-16,-8,-4,-2,-1,1,2,4,8,16\}$.
So we obtain the equations
$a+b=-16 \implies a-b=-1$
$a+b=-8 \implies a-b=-2$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/935819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Sum of the infinite series $\frac16+\frac{5}{6\cdot 12} + \frac{5\cdot8}{6\cdot12\cdot18} + \dots$ We can find the sum of infinite geometric series but I am stuck on this problem.
Find the sum of the following infinite series:
$$\frac16+\frac{5}{6\cdot 12} + \frac{5\cdot8}{6\cdot12\cdot18} + \frac{5\cdot8\cdot11}{6\cdo... | $$\frac16+\frac{5}{6\cdot12}+\frac{5\cdot8}{6\cdot12\cdot18}+\frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+\dots=$$
$$=\frac12\cdot\bigg[\frac26+\frac{2\cdot5}{6\cdot12}+\frac{2\cdot5\cdot8}{6\cdot12\cdot18}+\frac{2\cdot5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+\dots\bigg]=$$
$$=\frac12\cdot\bigg[\frac{(3-1)}{(6\cdot1)}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/936146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Modular calculus and square I want to prove that $4m^2+1$ and $4m^2+5m+4$ are coprimes and also $4m^2+1$ and $4k^2+1$ when $k\neq{m}$ and $4m^2+5m+4$ and $4k^2+5k+4$ when $k\neq{m}$. Firstly : Let $d|4m^2+1$ and $d|4m^2+5m+4$ then $d|4m^2+5m+4-(4m^2+1)=5m+3$ and $d|5m^2+3m$ thus $d|5m^2+3m-(4m^2+5m+4)=m^2-2m-4$ and $d|... | You have an error in your calculation.
$8m-88$ should be $4(2m-11)=8m-44$. Then, $(8m+17)-(8m-44)=61$.
The two numbers are not always coprime. For example, for $m=36$,
$$4m^2+1=5185=85\cdot 61,\ \ 4m^2+5m+4=5368=88\cdot 61.$$
In general, for $m=61k-25$ where $k$ is a positive integer, the two numbers have $61$ as a co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/936312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Without using Taylor expansion or L'Hospital rule evaluate the limit:
$$L=\lim_{x\to0}\frac{e^x-1-x-x^2/2}{x^3}$$
I give the two alternate ways so that no one gives them again in their answer:
Using L'Hospital:
$$L=\lim_{x\to0}\frac{e^x-x-1}{3x^2}=\lim_{x\to0}\frac{e^x-1}{6x}=\lim_{x\to0}\frac{e^x}{6}=\frac16$$
Usin... |
$$L=\lim_{x\to0}\frac{e^x-1-x-x^2/2}{x^3}$$
If Limit exists:
$$L=\lim_{x\to0}\frac{e^x-1-x-x^2/2}{x^3}\\
L=\lim_{x\to0}\frac{e^{2x}-1-2x-2x^2}{8x^3}\\
8L=\lim_{x\to0}\frac{e^{2x}-1-2x-2x^2}{x^3}\\
7L=\lim_{x\to0}\frac{e^{2x}-e^x-x-3x^2/2}{x^3}\\
7L=\lim_{x\to0}\frac{e^{4x}-e^{2x}-2x-6x^2}{8x^3}\\
28L=\lim_{x\to0}\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/936542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Recurrence problem with a game of probability Fair coin flipping (50% on both sides)
$P_1$ and $P_2$ plays a few games of fair coin flipping. Assume player $A$ starts with $x$ coins and player $B$ with $y$ coins.
Let $P_n$ denote the probability of player $A$ winning all coins. Find $P_0$ and $P_{x+y}$, then write a di... | Solution
$$P_n = \frac{1}{2}P_{n+1} + \frac{1}{2}P_{n-1}
\implies \frac{1}{2}P_{n+1} - P_n + \frac{1}{2}P_{n-1} = 0$$
Which gives the characteristic equation:
$\alpha^2-2\alpha+1=0 \implies(\alpha-1)^2=0 \implies\alpha=1$ (Double root)
General solution:
$P_{hn} = C_1\alpha_1^n + n\cdot C_2\alpha_2^n = C_1+n\cdot C_2$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/936744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the inverse with respect to the binary operation $a ∗ b = a + b + a^2 b^2$ A binary operation on $\mathbb{R}$: $a * b = a + b + a^2 b^2$
The neutral element I found to be $0$.
Then I need to find an invertible element having two distinct inverses. I don't know where to start for this question.
| If $a = 0$, $b = 0$ is the unique inverse of $a$, since $0 = 0 * b = 0 + b + 0^2b^2 = b$ in this case.
If $a \ne 0$, $a * b = a + b + a^2b^2 = 0$ is a quadratic equation for $b$. It may be written as
$b^2 + \dfrac{1}{a^2}b + \dfrac{1}{a} = 0. \tag{1}$
Applying the quadratic formula to (1) yields
$b = \dfrac{1}{2}(-\df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/940611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove $\sum_{i=0}^{i=x} {x \choose i} {y+i \choose x}+\sum_{i=0}^{i=x} {x \choose i} {y+1+i \choose x}$ How to prove that
$$\sum_{i=0}^{i=x} {x \choose i} {y+i \choose x}+\sum_{i=0}^{i=x} {x \choose i} {y+1+i \choose x}=\sum_{i=0}^{i=x+1} {x+1 \choose i} {y+i \choose x}$$ ?
I tried to break the right side of equation ... | This is very easy to prove using the integral representation of
binomial coefficients. Suppose we are trying to prove that
$$\sum_{k=0}^n {n\choose k} {m+k\choose n}
+ \sum_{k=0}^n {n\choose k} {m+1+k\choose n}
= \sum_{k=0}^{n+1} {n+1\choose k} {m+k\choose n}.$$
Use the two integrals
$${m+k\choose n}
= \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/940716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Solving a PDE given a specific curve and condition Consider the following PDE: $$-3U_x+4U_y=0$$ where $U=3x$ on curve $y=x+1$.
First we invoke the method of characteristics: $$\frac{dx}{-3}=\frac{dy}{4}=\frac{dU}{0}.$$ From this we get $c_1=\frac{x}{3}+\frac{y}{4}$ and $U(x,y)=c_2$. So $F(c_1)=c_2$ or $F(\frac{x}{3... | The general solution of $-3U_x+4U_y=0$
obviously is $U(x,y)=F(\frac{x}{3}+\frac{y}{4}) \space$ where $F$ is any derivable function.
I don't understand what you mean "incorporate the curve". If you are talking of a boundary condition on the curve $y=x+1$, please, make clear what is the boundary condition.
With the con... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/941750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Looking for the logic of a sequence from convolution of probability distributions I am trying to detect a pattern in the followin sequence from convolution of a probability distribution (removing the scaling constant $\frac{6 \sqrt{3}}{\pi }$: $$\left(\frac{1}{\left(x^2+3\right)^2},\frac{2
\left(x^2+60\right)}{\left... | Well, at least the denominator seems to be $(x^2 + 3n^2)^{n+1}$...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/942204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
To prove triangle is equilateral> given one equation and one combined equation. show that the lines $ x^2 - 4xy +y^2 $ and $x +y=3 $ form an equilateral trinagle . Also Find Area.
Here is what I have tried:
$ l1 =x+y=3 $
The combined equation is:
$ x^2 - 4xy + y^2 $
compare it with general formula :
$ ax^2 +2hxy+y^2 $... | Put $y = mx$ in the joint equation. That will produce auxiliary equation $$m^2 - 4m + 1 = 0$$
Solving this will give $$m = \dfrac{4 \pm \sqrt{4^2 - 4}}{2} = \dfrac{4 \pm 2\sqrt3}{2} = 2\pm \sqrt{3}$$
Let $m_1 = 2+\sqrt3, m_2 = 2-\sqrt3$
Gradient of $x+y = 1$ is $-1$, let $m_3 = -1$.
Use $\tan\theta = \left|\dfrac{m_a-m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/945808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show that $\log \left| z \right|$ is harmonic and find its the conjugate harmonic function. Is the form correct for the conjugate harmonic?
Attempt:
First, we are given
\begin{align*}
\log \left| z \right| &= u(x,y) + iv(x,y) = \log \sqrt{x^2 + y^2} + i \cdot 0 \\
u(x,y) &= \log \sqrt{x^2 + y^2} = \frac{1}{2} \log (x^2... | Almost correct. $log|z|$ have a harmonic conjugate iff $z$ is in some simply connected subset of $\mathbb C \backslash \{ 0 \}$ (think of convex subsets). I think you $v$ should be like the $\text{Arg}$ function in
http://en.wikipedia.org/wiki/Argument_(complex_analysis)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/946383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Complex exponent problem Find all numbers in complex plane that solves equation
*
*$e^z=4i$
Since $e^u e^{iv}=re^{i\Theta}$ it must be that $e^u=r \to u=\ln4$ and $v=\Theta+n2\pi \to v=\pi/2+n2\pi$. So the equation holds for points $z=\log w=\ln4+i(\pi/2+n2\pi)$ when $n\in \mathbb{N}$
*$e^{1/z}=-1$
Let's set $... | 1) Solution:
$$e^z=4i<=>$$
$$z=\frac{\log(4i)}{\log(e)}<=>$$
$$z=\ln(4i)<=>$$
$$z=\ln\left(4e^{\left(\frac{1}{2}\pi\right)i}\right)<=>$$
$$z=\frac{1}{2}i\left(4\pi n+\pi -4i\ln(2)\right)$$
(n is the element of Z - the set of integers)
2) Solution:
$$e^{\frac{1}{z}}=-1<=>$$
$$\frac{1}{z}=\frac{\log(-1)}{\log(e)}<=>$$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/949964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integrate by Trig Substitution $$\int_{3/4}^{3\sqrt{3}/4}\frac{x^3}{\sqrt{9-4x^2}}dx$$
So far I have the following:
$$ u=2x\Rightarrow$$
$$ u=a\sin \theta\Rightarrow 3\sin\theta$$
$$2x=3\sin\theta \rightarrow x=\frac32\sin\theta \rightarrow dx=\frac32\cos\theta d\theta$$
$$\int_{3/4}^{3\sqrt{3}/4}\frac{\frac{u}{2}^3}{\... | You're complicating things with the double substitution. Consider
$$
\sqrt{9-4x^2}=3\sqrt{1-\frac{4x^2}{9}}
$$
and directly set $\frac{2}{3}x=\sin\theta$, or $x=\frac{3}{2}\sin\theta$, with $dx=\frac{3}{2}\cos\theta\,d\theta$.
If $x=\frac{3}{4}$, then you have $\frac{3}{4}=\frac{3}{2}\sin\theta$ or $\sin\theta=\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/950109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is it true that $\left\lfloor\sum_{s=1}^n\operatorname{Li}_s\left(\frac 1k \right)\right\rfloor\stackrel{?}{=}\left\lfloor\frac nk \right\rfloor$ While studying polylogarithms I observed the following.
Let $n>0$ and $k>1$ be integers. Is the following statement true?
$$\left\lfloor \sum_{s=1}^n \operatorname{Li}_s\left... | For this proof to work, we have to add the restriction that $k \geq 2$, since $\operatorname{Li}_1\left( \frac{1}{1} \right)$ doesn't even converge anyway.
Now, by the definition of the polylogarithm, we have:
\begin{align}
\sum_{s=1}^n \operatorname{Li}_s\left( \frac{1}{k} \right)
&=\sum_{s=1}^n \sum_{j=1}^\infty {1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/950418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Probability of exactly 2 low rolls in 5 throws of a die If a fair die is rolled 5 times what is the probability that the number shown will be less than 3 exactly 2 times. I have tried to work problem using formula from textbook and still get it wrong
| Since there are two possibilities, we use the binomial probability formula
$$\binom{n}{k}p^kq^{n - k}$$
where $p$ is the probability of the event, $q = 1 - p$ is the probability of the complement of the event, $n$ is the number of trials, and $k$ is the number of times the event occurs during those $n$ trials.
Here... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/951236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Trigonometric Integral $\int _0^{\pi/4}\:\frac{dx}{\cos^4x-\cos^2x\sin^2x+\sin^4x}$ Q. $$\int _0^{\pi/4}\:\frac{1}{\cos^4x-\cos^2x\sin^2x+\sin^4x}$$
My method:
$$\int_0^{ \pi /4} \frac{dx}{(\cos^2x+\sin^2x)^2-3\cos^2x \sin^2x}=\int_0^{ \pi /4} \frac{dx}{1-3\cos^2x\sin^2x} $$
Dividing numerator and denominator by $\cos^... | Lets look at the integral in terms of multiple angles.
Now
\begin{array}
$\cos^4x+\sin^4x-\cos^2x\sin^2x&=&(\cos^2x+\sin^2x)^2-3\sin^2x\cos^2x\\
&=&1-\frac{3}{4}\sin^2 2x\\
&=&1-\frac{3}{8}(1-\cos 4x)\\
&=& \frac{1}{8}(5-3\cos 4x)
\end{array}
Hence
$$\int_0^{\pi/4}\frac{dx}{\cos^4x+\sin^4x-\cos^2x\sin^2x}=\int_0^{\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/952307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Proving $\lim\limits_{n \to \infty}[\sqrt{n^2 + 1} - n] = 0$ I'm trying to prove $\lim\limits_{n \to \infty}[\sqrt{n^2 + 1} - n] = 0$. Is the following a correct proof?
For all $n$ we have $0 \leq \left|\sqrt{n^2 + 1} - n\right| \leq \left|\sqrt{n^2+1} - 1 \right|$. For any $\epsilon > 0$ take $ N = \sqrt{(\epsilon+1)^... | No, it isn't. First of all think: does $\lim \limits_{n \to \infty} \sqrt{n^2 + 1} - 1 = 0$?
Your mistake is that assuming that when we take $n > N$, that we will have $\sqrt{n^2 + 1} - 1 < \epsilon$, but actually when we take $n > N$ we will have $\sqrt{n^2 + 1} - 1 > \epsilon$.
Here would be a correct proof:
Multiply... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/953214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find the range of $f(x) =(x-1)^{1/2} + 2\cdot(3-x)^{1/2}$ How to take out the range of the following function :
$$f(x) =(x-1)^{1/2} + 2\cdot(3-x)^{1/2}$$
I am new to functions hence couldn't come up with a solution.
| As $1\le x\le 3\iff-1\le x-2\le1$ which nicely fits with the range of sinusoidal functions,
let $x-2=\cos2\theta$
So, $f(x)=\sqrt{2\cos^2\theta}+2\sqrt{2\sin^2\theta}=\sqrt2(|\cos\theta|+\sqrt2|\sin\theta|)$
Case $\#1:$
If $\displaystyle0\le\theta\le\dfrac\pi2,$
$\displaystyle f(x)=\sqrt2[\cos\theta+ 2\sin\theta]=\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/953842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
Bivariate distribution with normal conditions Define the joint pdf of $(X,Y)$ as:
$$f(x,y)\propto \exp(-1/2[Ax^2y^2+x^2+y^2-2Bxy-2Cx-Dy]),$$
where $A,B,C,D$ are constants.
Show that the distribution of $X\mid Y=y$ is normal with mean $\frac{By+C}{Ay^2+1}$ and variance $\frac{1}{Ay^2+1}$. Derive a corresponding result f... | We use:
$$f_{X\mid Y}(x\mid Y=y) = \frac{f(x,y)}{f_Y(y)} = \frac{f(x,y)}{\int_{-\infty}^{\infty} f(x,y)dx} $$
We first note that:
$$ -0.5(Ax^2y^2+x^2+y^2-2Bxy-2Cx-2Dy) =$$ $$ -0.5(Ay^2+1)\left(x-\frac{By+c}{Ay^2+1} \right)^2 -0.5\left(y^2-2Dy - \frac{(By+c)^2}{Ay^2+1}\right).$$
This gives us:
$$ \int_{-\infty}^{\infty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/954523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Two differentiation results of $\sin^{-1}(2x\sqrt{1-x^2})$ While trying to differentiate $\sin^{-1}(2x\sqrt{1-x^2})$, if we put
$x = \sin\theta$, we get,
\begin{align*}
y &=\sin^{-1}(2x\sqrt{1-x^2})\\
&= \sin^{-1}(2\sin\theta\sqrt{1-\sin^2\theta})\\
&= \sin^{-1}(2\sin\theta\cos\theta)\\
&= \sin^{-1}(\sin2\theta... | This Question bugged me for long time before I found the linked identity
Setting $y=x$ in Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $,
$$
2\arcsin x\\
\begin{align}
&=\arcsin( 2x\sqrt{1-x^2}) \;\;;2x^2 \le 1 \\
&=\pi - \arcsin(2x\sqrt{1-x^2}) \;\;;2x^2 > 1\text{ and } 0< x\\
&=-\pi - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/956432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Calc the sum of $\sum_{k = 0}^{\infty} \frac{(-1)^k}{k} \sin(2k)$ Solving a bigger problem about Fourier series I'm faced with this sum:
$$\sum_{k = 0}^{\infty} \frac{(-1)^k}{k} \sin(2k)$$
and I've no idea of how to approach this.
I've used Leibniz convergence criterium to verify that the sum should have a value, but I... | Consider the series
\begin{align}
S = \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n} \, \sin(2n).
\end{align}
Method 1
Using the known Fourier series
\begin{align}
x = \frac{2 L}{\pi} \, \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \, \sin\left( \frac{n \pi x}{L} \right)
\end{align}
it can quickly be seen that for $L = \pi$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/956968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
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.