Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Computing $\zeta(6)=\sum\limits_{k=1}^\infty \frac1{k^6}$ with Fourier series. Let $ f$ be a function such that $ f\in C_{2\pi}^{0}(\mathbb{R},\mathbb{R}) $ (f is $2\pi$-periodic) such that $ \forall x \in [0,\pi]$: $$f(x)=x(\pi-x)$$
Computing the Fourier series of $f$ and using Parseval's identity, I have computed $\z... | I have posted here in Portuguese a recursive method based on the computation of the Fourier trigonometric series expansion for the function defined in $\left[ -\pi ,\pi \right] $ by $f(x)=x^{2p}$ and extended to all of ${\mathbb R}$ periodically with period $2\pi.$ This is a shorter description than the original. In t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/115981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "47",
"answer_count": 6,
"answer_id": 2
} |
Evaluate $\int_0^1 \frac{\ln(1+bx)}{1+x} dx $
What is $ \displaystyle\int_0^1 \frac{\ln(1+bx)}{1+x} dx $?
I call it $f(b)$ and differentiate with respect to be $b,$ a bit of partial fractions and the $x$ integral can be done. Then I integrate with respect to $b$ and get a bit lost.
Can some of the terms be expressed ... | Let $x = \frac{1}{b} \frac{1-u}{u}$. The integration range $0\leqslant x \leqslant 1$ translates to $\frac{1}{1+b} \leqslant u \leqslant 1$, for $b>0$
$$
\frac{\log(1+ b x)}{1+x}\mathrm{d} x \to \frac{-\log(u)}{1 - (1-b) u} \frac{\mathrm{d} u}{u}
= \frac{\log(u)}{u} \left(1 + \frac{(1-b) u}{1-(1-b) u} \right)
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/117924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
} |
Prove: $\int_0^{\frac{\pi}{2}}t(\frac{\sin nt}{\sin t})^4dt<\frac{\pi^2n^2}{4}$ I have a question about integral. Prove:
$$\int_0^{\frac{\pi}{2}}t\left(\dfrac{\sin(nt)}{\sin(t)}\right)^4dt<\dfrac{\pi^2n^2}{4}$$
I have tried several methods including $\sin(t)\geq\frac{2t}{\pi}$, but I can't work it out.
| Assume that $n\in \mathbb{N}$.
Then the ratio of sines is polynomial in cosines:
$$
\frac{\sin(n t)}{\sin(t)} = \cos((n-1) t) + \cos(t) \frac{\sin((n-1) t}{\sin(t)} = \ldots = \sum_{k=1}^n \cos((n-k) t) \cdot \cos^{k-1}(t) \leqslant n
$$
Since $\sin(t)$ is increasing on the interval $\left(0,\frac{\pi}{2}\right)$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/120469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
How I can find the value of $abc$ using the given equations? If I have been given the value of
$$\begin{align*}
a+b+c&= 1\\
a^2+b^2+c^2&=9\\
a^3+b^3+c^3 &= 1
\end{align*}$$
Using this I can get the value of
$$ab+bc+ca$$
How i can find the value of $abc$ using the given equations?
I just need a hint.
I have tried ... | In general, $$a^n + b^n + c^n = \sum_{i+2j+3k=n} (-1)^j \frac{n}{i+j+k}{i+j+k\choose i,j,k}s_1^is_2^js_3^k$$
where the sum is over non-negative $i,j,k$, and where $s_1=a+b+c$, $s_2=ab+ac+bc$ and $s_3=abc$.
In particular, when $n=3$ there are only three triples $(i,j,k)=(3,0,0),(1,1,0),(0,0,1)$, and you get:
$$a^3+b^3+c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/120536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
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The solution set of the equation $|2x - 3| = - (2x - 3)$ The solution set of the equation $\left | 2x-3 \right | = -(2x-3)$ is
$A)$ {$0$ , $\frac{3}{2}$}
$B)$ The empty set
$C)$ (-$\infty$ , $\frac{3}{2}$]
$D)$ [$\frac{3}{2}$, $\infty$ )
$E)$ All real numbers
The correct answer is $C$
my solution:
$\ 2x-3 = -(2x-3)$... | $|2x-3|=\begin{cases}
3-2x, & \text{if } x \leq \frac{3}{2} \\
2x-3, & \text{if } x > \frac{3}{2}
\end{cases}$
a) $|2x-3|=3-2x$ , hence :
$3-2x=-(2x-3)$
$0=0$ , therefore :
$S_a : x \in \left(-\infty, \frac{3}{2}\right]$
b) $|2x-3|=2x-3$ , hence :
$2x-3=-(2x-3)$
$4x=6$
$x=\frac{3}{2}$ , therefore :
$S_b : x \in \e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/121240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Let $n$ be a positive integer such that $\displaystyle{\frac{3+4+\cdots+3n}{5+6+\cdots+5n} = \frac{4}{11}}$ Let $n$ be a positive integer such that
$$\displaystyle{\frac{3+4+\cdots+3n}{5+6+\cdots+5n} = \frac{4}{11}}$$
then
$$\displaystyle{\frac{2+3+\cdots+2n}{4+5+\cdots+4n}} = \frac{m}{p}.$$
The question further "Is... | At first attempt, I was tempted to do $\displaystyle{3+4+\cdots+3n = \frac{3n(3n+1)}{2}-3}$, but there are $4$ expressions of that sort, it is better to find a general form of it like this
$$
\begin{align*}
k+(k+1)+(k+1)+\cdots+kn &= \frac{1}{2} \left[ kn(kn+1)-k(k-1) \right]\\
&= \frac{1}{2}\left[ k(n+1)(kn-k+1)\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/121558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Finding a basis for the columnspace of a matrix
Find a linearly independent set of vectors that spans the same subspace of $R^4$ as that spanned by the vectors -
$$
\begin{bmatrix}
2 \\
-4 \\
-1 \\
-2 \\
\end{bmatrix}
,
\begin{bmatrix}
7 \\
-2 \\
7 \\
2 \\
\end{bmatrix}
,
\begin{bmatrix}
1 \\
2 \\
... | Start with a matrix whose columns are the vectors you have.
Then reduce this matrix to row-echelon form.
A basis for the columnspace of the original matrix is given by the columns in the original matrix that correspond to the pivots in the row-echelon form.
What you are doing does not really make sense because elementa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/121936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
limit of $\lim_{x \to 0}\left ( \frac{1}{x^{2}}-\cot x\right )$ Help me with that problem, please.
$$\lim_{x \to 0}\left ( \frac{1}{x^{2}}-\cot x\right )$$
| $$\lim\limits_{x \to 0} \left(\frac{1}{x^2} - \frac{1}{\tan x}\right) = \lim\limits_{x \to 0} -\left( \frac{x^4 \tan x - x^2 \tan^2 x}{x^4 \tan^2 x}\right) =
-\lim\limits_{x \to 0} \frac{(x^2\tan x)(x^2-\tan x)}{(x^2 \tan x)(x^2 \tan x)}$$
Cancelling out terms:
$$-\lim\limits_{x \to 0} \frac{x^2 - \tan x}{x^2 \tan x}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/123324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Evaluate and prove by induction: $\sum k{n\choose k},\sum \frac{1}{k(k+1)}$
*
*$\displaystyle
0\cdot \binom{n}{0} + 1\cdot \binom{n}{1} + 2\binom{n}{2}+\cdots+(n-1)\cdot \binom{n}{n-1}+n\cdot \binom{n}{n}$
*$\displaystyle\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3}+\frac{1}{3\cdot 4} +\cdots+\frac{1}{(n-1)\cdot n}$
Ho... | One way to do this:
1) You are looking for $\sum_{k=0}^n k\binom{n}{k}$.
First, look at $f(x)=(1+x)^n=\sum_{k=0}^n \binom{n}{k}x^k$, (by Newton's binomial theorem). Hence $2^n=\sum_{k=0}^n \binom{n}{k}$, by calculating $f(1)$. Now, define $g(x)=\sum_{k=0}^n k\binom{n}{k}x^k$. What you need is $g(1)$. Try expressing $g(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/123655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 5
} |
Compute $\sum_{i=1}^{2n} \frac{x^{2i}}{x^i-1}$ where $\{ x \in \mathbb{C}$ | $x^{2n+1} = 1, x \neq 1\}$ $\{ x \in \mathbb{C}$ | $x^{2n+1} = 1$ , $x \neq 1\}$
Compute $\displaystyle{\sum_{i=1}^{2n} \frac{x^{2i}}{x^i-1}}$
| Since
$$(x^{\small{2n+1}}-1) = (x-1)(x^{\small{2n}}+x^{\small{2n-1}}+\cdots+x^{\small{2}}+x+1)=0$$
Therefore
$$(x^{\small{2n}}+x^{\small{2n-1}}+\cdots+x^{\small{2}}+x+1)=0 \hspace{4pt} ({\text{since}} \hspace{4pt} x \neq 1) \tag{1}$$
$$
\begin{align*}
S = \sum_{i=1}^{\small{2n}} \frac{x^{\small{2i}}}{x^i-1} &= \su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/124470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Limit: How to Conclude I have difficulty to conclude this limit ....; place of my attempts and results, can anyone help?
tanks in advance
$$\lim_{x\to +a}\, \left(1+6\left(\frac{\sin x}{x^2}\right)^x\cdot\frac{\log(1+10^x)}{x}\right),\quad a=+\infty,\,\,\,0,\,\,\,\,-\infty$$
1):$\,\,\,{a=+\infty}$
$$\lim_{x\to +\infty}... | As pointed by David Mitra in comments, the function $$f(x) = \left\{1 + 6\left(\frac{\sin x}{x^{2}}\right)^{x}\frac{\log(1 + 10^{x})}{x}\right\}$$ is not well defined whenever $x \to \pm \infty$ because $\sin x$ becomes negative. The same holds when $x \to 0$. However if we restrict ourselves to $x \to 0^{+}$ then the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/125466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Rearranging a formula, transpose for A2 - I'm lost Given the formula:
$$ q = A_1\sqrt\frac{2gh}{(\frac{A_1}{A_2})^2-1} $$
Transpose for $A_2$
I have tried this problem four times and got a different answer every time, none of which are the answer provided in the book. I would very much appreciate if someone could show ... | $$q = A_1\sqrt\frac{2gh}{(\frac{A_1}{A_2})^2-1}$$
$$q^2=(A_1)^2\frac{2gh}{(\frac{A_1}{A_2})^2-1}$$
$$(\frac{A_1}{A_2})^2-1=(A_1)^2\frac{2gh}{q^2}$$
$$(\frac{A_1}{A_2})^2=(A_1)^2\frac{2gh}{q^2}+1$$
$$\frac{A_1}{A_2}=\sqrt{(A_1)^2\frac{2gh}{q^2}+1}$$
$$A_2=\frac{A_1}{\sqrt{(A_1)^2\frac{2gh}{q^2}+1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/125842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ ?. How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ without using a calculator.
Related question: how do we prove that $\cos(\pi/5)\cos(2\pi/5) = 0.25$, also without using a calculator
| Note that: $\cos{2x} = \cos^{2}{x} - \sin^{2}{x} = 2\:\cos^{2}{x} - 1$. Therefore you have $\cos \frac{2\pi}{5} = 2\:\cos^{2}\frac{\pi}{5} - 1$
Now,
\begin{align*}
\cos\frac{\pi}{5} - \cos\frac{2\pi}{5} = \cos\frac{\pi}{5} - 2\: \cos^{2}\frac{\pi}{5}+1
\end{align*}
This is a quadratic equation of the form $2 x^{2} - x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/130817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 18,
"answer_id": 0
} |
Integral with cube, square and zero starting point I don't know is there is a special rule or trick for this but I am trying to find
$$\int_{0}^1(x^3-3x^2) dx$$
I know that $\dfrac{1}{n}$ is the delta $x$ and this is where I do not know what to do next. I think that I want it to look something like
$$\lim \sum \dfrac{1... | So you have $\Delta x = \frac{1}{n}$.
And so$$\begin{align}
\int_0^1 f(x) dx &= \lim_{n\to \infty} \sum_{k=1}^{n} (\Delta x) f\left(0 + k\Delta x \right) \\
&= \lim_{n\to \infty} \sum_{k=1}^{n} \frac{1}{n} f\left(\frac{k}{n}\right) \\
&= \lim_{n\to \infty} \frac{1}{n}\sum_{k=1}^{n} \left(\frac{k}{n}\right)^3 - 3\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/131565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to prove $n^5 - n$ is divisible by $30$ without reduction How can I prove that prove $n^5 - n$ is divisible by $30$?
I took $n^5 - n$ and got $n(n-1)(n+1)(n^2+1)$
Now, $n(n-1)(n+1)$ is divisible by $6$.
Next I need to show that $n(n-1)(n+1)(n^2+1)$ is divisible by $5$.
My guess is using Fermat's little theorem but ... | $n^5-n=(n-1)n(n+1)(n^2+1)$. Rewrite $n^2+1$ as $5(n-1)+(n^2-5n+6)$ to obtain $n^5-n=5(n-1)^2n(n+1)+(n-3)(n-2)(n-1)n(n+1)$ and use the fact that $n!$ divides the product of n consecutive numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/132210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 21,
"answer_id": 5
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How was Euler able to create an infinite product for sinc by using its roots? In the Wikipedia page for the Basel problem, it says that Euler, in his proof, found that
$$\begin{align*}
\frac{\sin(x)}{x} &=
\left(1 - \frac{x}{\pi}\right)\left(1 + \frac{x}{\pi}\right)\left(1 - \frac{x}{2\pi}\right)\left(1 + \frac{x}{2\pi... | Nearly the same question was posted here recently. I hope this will add a little that is not in the other answers to this present question.
We know that $\dfrac{\sin x}{x}=0$ when $\sin x= 0$ and $x\neq0$, and we know that $\dfrac{\sin x}{x}$ "$=$" $1$ when $x=0$ (I think Euler's way of saying this is that $\sin x = x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/134870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 1
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$\int_0^{2\pi}\sin\frac{x}{2}\cos^2\frac{x}{2}\,dx$ $$\int_0^{2\pi}\sin\frac{x}{2}\cos^2\frac{x}{2}\,dx$$
Tried substitution ($u = \cos\frac{x}{2}$), but I get $-\frac{\cos^3\frac{x}{2}}{3}$ ($-\frac{2}{3}$) instead of the correct answer, which is $1\frac{1}{3}$
| It makes it a little easier to do this simple substitution for you to not mess up with constant multiple like you did (You got an extra $\frac{1}{3}$)
Substitute first $\frac{x}{2}=t$. To see the limits, when $x=2\pi, t=\pi$ and when $x=0, u=0$ and $\mathrm{d}x = 2\mathrm{d}t$
$$
\begin{align*}
\int_0^{2\pi}\sin\frac{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/135963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Definition of derivative $f(x) = \sqrt{3-5x}$ I am not sure how to factor this out
$$f(x) = \sqrt{3-5x}$$
I then make it $f(x) = \frac {\sqrt{3-5(x+h)} - \sqrt{3-5x}}{h}$
I tried to multiply by the first time + the second term from the numerator which I called x and y
$$\frac {x - y}{h} \cdot \frac {x + y}{x+y}$$
which... | $\displaystyle f(x) = \sqrt{3-5x}$. Hence, $\displaystyle f(x+h) = \sqrt{3 - 5(x+h)}$. Therefore, we get that $$f(x+h)-f(x) = \sqrt{3 - 5(x+h)} - \sqrt{3 - 5x} = \frac{(3 - 5(x+h)) - (3 - 5x)}{\sqrt{3 - 5(x+h)} + \sqrt{3 - 5x}} = \frac{-5h}{\sqrt{3 - 5(x+h)} + \sqrt{3 - 5x}}$$
Hence, $$\frac{f(x+h)-f(x)}{h} = - \frac{5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/141200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Help with Cardano's Formula I'm trying to understand how to solve cubic equations using Cardano's formula. To test the method, I expand $(x-3)(x+1)(x+2)=x^3-7x-6$. My hope is that the formula will produce the roots $-1,-2,3$. But the formula seems to make a mess of things: I compute that $\frac{q^2}{4}+\frac{p^3}{2... | I know it has been a lot of time since you asked the question, but it is reasonable to assume that no one has answered you. Ok, here we go. When trying to reduce radicals in general, you try to find values for a, b $\in \mathbb{Q}$ such that $$\sqrt[n]{A \pm B\sqrt[m]{C}} = a \pm b\sqrt[m]{C}.$$Notice that $\sqrt[m]{C}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/145590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
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In what case this equation has a solution(s)? I am trying to solve this problem saying, for what values of $a$ and $b$ (in which they are not zero), the equation $a(1-\sin(x))=b(1-\cos(x))$ has any solution(s)? Thank you very much. :)
| $1- \sin(x) = (\cos(x/2) - \sin(x/2))^2$ and $1 - \cos(x) = 2 \sin^2(x/2)$. This gives us that $$\cot(x/2) - 1 = \pm \sqrt{2b/a}$$ i.e. $$\tan(x/2) = \frac1{1 \pm \sqrt{2b/a}}$$ i.e. $$x = 2 \arctan \left(\frac1{1 \pm \sqrt{2b/a}} \right)$$
Another way to approach is as follows. We have $$a \sin(x) - b \cos(x) = b-a$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/146142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Evaluating $\int_{-20}^{20}\sqrt{2+t^2}\,dt$ I have this integral:
$$\int_{-20}^{20}\sqrt{2+t^2}\,dt$$
I tried solving it many times but without success.
The end result is this:
$$2\left( 10\sqrt{402}+\mathop{\mathrm{arcsinh}}(10\sqrt{2})\right).$$
I can't seem to get this end result. I got a few wrong ones but cant fi... | Here are two more ways to calculate $\int \sqrt{a^2+x^2}\ dx$.
*
*Let us make the substitution $\sqrt{a^2+x^2}=t-x$. After squaring, $a^2=t^2-2xt$, from which
$$
x=\frac{t^2-a^2}{2t}\,.
$$
Using the last expression, one finds
$$
\sqrt{a^2+x^2}=t-x=\frac{t^2+a^2}{2t}\,,
$$
and
$$
dx=\frac{t^2+a^2}{2t^2}dt.
$$
Pluggi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/147788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Summation of Infinite Series $\sum_{n=1}^{\infty} \frac{1}{n 2^{2n+1}}$ Show That :
$$\sum_{n=1}^{\infty} \frac{1}{n 2^{2n+1}} = \ln \left(\frac{2}{\sqrt{3}}\right)$$
I could show convergence. (I dont need to show that this converges). However I couldn't figure how to show the value.
| $$
\begin{align*}
\sum_{n=1}^{\infty} \frac{y^n}{n} &= -\ln \left(1-y\right) \hspace{15pt} {\textit{apply }} \hspace{5pt} y=\frac{1}{x^2}\\
\sum_{n=1}^{\infty} \frac{1}{n x^{2n}} &= -\ln \left(1-\frac{1}{x^2}\right) \\
\sum_{n=1}^{\infty} \frac{1}{n 2^{2n+1}} &= \frac{1}{2}\sum_{n=1}^{\infty} \frac{1}{n 2^{2n}} = -\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/148743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Finding square roots of $\sqrt 3 +3i$ I was reading an example, where it is calculating the square roots of $\sqrt 3 +3i$.
$w=\sqrt 3 +3i=2\sqrt 3\left(\frac{1}{2}+\frac{1}{2}\sqrt3i\right)\\=2\sqrt 3(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})$
Let $z^2=w \Rightarrow r^2(\cos(2\theta)+i\sin(2\theta))=2\sqrt 3(\cos\frac{\pi}... | Wikipedia page on polar form of complex numbers is quite good.
Given a complex number $z = a + i b$, its absolute value $|z| = \sqrt{a^2+b^2}$, naturally the quotient $\frac{z}{|z|}$ has unit absolutely value, hence $\frac{z}{|z|} = \mathrm{e}^{i \theta} = \cos(\theta) + i \sin(\theta)$ for some angle $\theta$.
In the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/148871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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How to solve this equation, when the unknown variable just disappears? This problem
$\sqrt{1-x^2} + \sqrt{3+x^2} = 2$
has the solution $x = 1$ and $x = -1$.
However, I always get stuck like this:
*
*$1-x^2 + 3+x^2 = 4$
*$4 = 4$
How do I isolate that darn unknown?
| $(a+b)^2\neq a^2+b^2$ in general. In your case $\sqrt{1-x^2}+\sqrt{3+x^2}=2$ leads to $1-x^2+2\sqrt{1-x^2}\sqrt{3+x^2}+3+x^2=4$, which gives you $2\sqrt{1-x^2}\sqrt{3+x^2}=0$. Since $\sqrt{3+x^2}\neq0$ for all real $x$, so $\sqrt{1-x^2}=0$. Hence $x=\pm1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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limit of power of fraction of sums of sines Find the following limit:
$$\lim_{n\to\infty} \left(\frac{{\sin\frac{2}{2n}+\sin\frac{4}{2n}+\cdot \cdot \cdot+\sin\frac{2n}{2n}}}{{\sin\frac{1}{2n}+\sin\frac{3}{2n}+\cdot \cdot \cdot+\sin\frac{2n-1}{2n}}}\right)^{n}$$
I thought of some $\sin(x)$ approximation formula, but it... | Let $f : [0, 1] \to [0, \infty)$ be of the class $C^1$ and not identically zero. Then by Mean Value Theorem, we have
$$ \sum_{k=1}^{n} f \left( \tfrac{2k}{2n} \right) = \sum_{k=1}^{n} \left( f \left( \tfrac{2k-1}{2n} \right) + f' (x_{n,k}) \frac{1}{2n} \right) $$
for some $x_{n,k} \in \left(\frac{2k-1}{2n}, \frac{2k}{2... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Linear Congruences I understand how to solve an equation like:
$$17x \equiv 3 \pmod 5$$
But is there a known method for solving a congruence of the form:
$$17 \equiv 3\frac{\sqrt{x}}{2} \pmod 5\quad ?$$
Is it as simple as moving the $x$ through operations and applying the congruence to the new equation?
$$(34/3)-\sqrt{... | You have to be careful with square roots in modular equations, since not every number has a square root (much like you would for integers), and when they do, there is no way to select one the way we do with real number (where we always pick the nonnegative one). That is, $\sqrt{x}$ is not well-defined modulo $5$. For e... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Elementary Real Analysis - Let me know if I'm on the right track So I'm guessing this is a pretty simple example for this topic but I just want to check myself as I'm new to this analysis area and not sure that what I'm saying is mathematically sound..
The question is show that $\lim\limits_{x \to -\infty}\frac{x^2+1}{... | Assume $\epsilon>0$, then
$$\begin{align*} \left|\frac{x^2+1}{x^2-1} - 1\right| < \epsilon
&\Rightarrow \left|\frac{2}{x^2-1}\right| < \epsilon \\
&\Rightarrow \left|\frac{x^2-1}{2}\right| > 1/\epsilon \\
&\Rightarrow |x^2-1| > \frac{2}{\epsilon}. \end{align*}
$$
when $x^2-1 > 0$, $x^2>1 $, $|x| > 1$,
so, $(x^2-1)>2/\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How can we produce another geek clock with a different pair of numbers? So I found this geek clock and I think that it's pretty cool.
I'm just wondering if it is possible to achieve the same but with another number.
So here is the problem:
We want to find a number $n \in \mathbb{Z}$ that will be used exactly $k \in \... | For $n=4$ and $k=5$ here is a solution:
$\frac{4}{\left(4+\left(4 \times \left(4-4\right)\right)\right)}=1$
$\left(4-\left(4 \times \frac{4}{\left(4+4\right)}\right)\right)=2$
$\left(4+\frac{4}{\left(4-\left(4+4\right)\right)}\right)=3$
$\left(4+\left(4+\left(4-\left(4+4\right)\right)\right)\right)=4$
$\left(4-\frac{4}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
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"answer_id": 1
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Sequence convergence and parentheses insertion find an example for a series $a_{n}$
that satisfies the following:
*
*$a_{n}\xrightarrow[n\to\infty]{}0$
*${\displaystyle \sum_{n=1}^{\infty}a_{n}}$
does not converges
*There is a way to insert parentheses so ${\displaystyle \sum_{n=1}^{\infty}a_{n}}$
will converges... | The series $$1-1+\frac{1}{2}+\frac{1}{2}-\frac{1}{2}-\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}-\frac{1}{4}-\frac{1}{4}-\frac{1}{4}-\frac{1}{4}+...$$
is indeed divergent, because it has infinitely many partial sums equal to $1$, and infinitely many partial sums equal to $0$. Inserting parentheses as
$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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simplify $ (-2 + 2\sqrt3i)^{\frac{3}{2}} $? How can I simplify $ (-2 + 2\sqrt3i)^{\frac{3}{2}} $ to rectangular form $z = a+bi$?
(Note: Wolfram Alpha says the answer is $z=-8$. My professor says the answer is $z=\pm8$.)
I've tried to figure this out for a couple hours now, but I'm getting nowhere.
Any help is much app... | $$(-2 + 2\sqrt3i) = 4 \exp\left(\frac{2\pi}{3}i\right) = 4 \cos \left(\frac{2\pi}{3}\right) + 4 \sin \left(\frac{2\pi}{3}\right) i$$
and I would say $$\left(4 \exp\left(\frac{2\pi}{3}i\right)\right)^{\frac{3}{2}} = 4^{\frac{3}{2}} \exp\left(\frac{3}{2} \times \frac{2\pi}{3}i\right) =8 \exp(\pi i) = -8. $$
I think usi... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculating the shortest possible distance between points Question:
Given the points $A(3,3)$, $B(0,1)$ and $C(x,0)$ where $0 < x < 3$, $AC$ is the distance between $A$ and $C$ and $BC$ is the distance between $B$ and $C$. What is x for the distance $AC + BC$ to be minimal?
What have I done?
I defined the function $AC ... | Norbert and Ross Millikan have already suggested one slick solution, and Gerry Myerson the even slicker solution, but you can do it purely algebraically: you ran into trouble because your derivative isn’t quite right. It should be
$$f\,'(x)=\frac{x}{\sqrt{1+x^2}}-\frac{3-x}{\sqrt{18-6x+x^2}}\;,$$
with the second term ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 3
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Summation of $ \frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \cdots$ till $n$ terms What is the pattern in the following?
*
*Sum to $n$ terms of the series: $$ \frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \cdots$$
| Here is the pattern:
\begin{align*}
\frac{1}{2} + \frac{3}{4} + \cdots &= \biggl(1-\frac{1}{2}\biggr) + \biggl(1-\frac{1}{2^2}\biggr) + \cdots \\\ &= (1+1+\cdots + 1) - \biggl(\frac{1}{2}+\frac{1}{2^2} + \cdots +\frac{1}{2^n}\biggr)
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Polynomials in Fourier trigonometric series I'm successively integrating $x^{n} \cos{k x}$ for increasing values of positive integer n. I'm finding:
$\frac{\sin{kx}}{k}$,
$\frac{\cos{kx}}{k^2}+\frac{x\sin{kx}}{k}$,
$\frac{2 x \cos{kx}}{k^2}+\frac{\left(-2+k^2 x^2\right)sin{kx}}{k^3}$,
$\frac{3 \left(-2+k^2 x^2\righ... | The polynomials are recursive in nature, and this behavior is most apparent when the given integral
$$
\int x^n \cos(k x)\,dx
$$
is viewed as the real part of the function
$$
F_{n,k}(x) = \int x^n e^{i k x}\,dx = \int x^n \cos(k x)\,dx + i\!\int x^n \sin(k x)\,dx.
$$
Using integration by parts twice we can derive an in... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is wrong with my solution? $\int \cos^2 x \tan^3x dx$ I am trying to do this problem completely on my own but I can not get a proper answer for some reason
$$\begin{align}
\int \cos^2 x \tan^3x dx
&=\int \cos^2 x \frac{ \sin^3 x}{ \cos^3 x}dx\\
&=\int \frac{ \cos^2 x\sin^3 x}{ \cos^3 x}dx\\
&=\int \frac{ \sin^3 ... | $\int\tan x\,dx=\ln|\sec x|=-\ln|\cos x|+ k $since $-\ln |\cos x|+k= \ln|(\cos x)^-1|+k = \ln|\sec x| +k$
$$\frac{\cos2x}{4}= \frac{1}{2}\cos^2(x) - 1/4$$
$k-1/4 = \text{new constant } C $ and together you have the solution.
Your answer seems to be correct it is just manipulation or different approach to trig identi... | {
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Expanding logarithm into series How to expand $f(x)=\ln\left( x+\sqrt{1+x^2} \right)$ into series at $x_0=0$? I've tried using Taylor's formula but counting consecutive derivatives was inconvenient and I couldn't find the general formula.
| This is a somewhat laborious expansion, but it has a nice closed formula.
We begin with
$$f(x) = \frac{1}{\sqrt{1+x^2}}$$
and note that you function $F$ is such that $F'(x) = f(x)$. We can make use of the Generalized Binomial Theorem, namely
$$(1+\mathrm x)^\alpha = \sum_{n=0}^\infty {\alpha \choose n}\mathrm x^n$$
In... | {
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"source": "stackexchange",
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If $ \frac{1}{c} = \frac{1}{2a} + \frac{1}{2b} $, then either $ a \leqslant c \leqslant b$ or $ b \leqslant c \leqslant a $ For $a, b = 1, 2, 3, \cdots$, let $ \frac{1}{c} = \frac{1}{2a} + \frac{1}{2b} $. Then prove that either $ a \leqslant c \leqslant b$ or $ b \leqslant c \leqslant a $ holds.
| Hint $\ $ For $\rm\ A = 1/a,\ B = 1/b,\ C = 1/c\ $ it becomes obvious, viz.
$$\rm C = \frac{A+B}2\ \Rightarrow\ A \ge C \ge B\ \ or\ \ B \ge C \ge A$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Find the possible value from the following. Find the possible value from the following.
I'm not able to end up on a concrete note, as I'm unable to get the essence of question, still not clear to me.
$x$, $y$, $z$ are distinct reals such that $y=x(4-x)$, $z=y(4-y)$, $x=z(4-z)$. The possible values of $x+y+z$ is:
$$\b... | Composing the functions, we get
$$
\begin{align}
0
&=x^8-16x^7+104x^6-352x^5+660x^4-672x^3+336x^2-63x\\
&=x(x-3)(x^3-7x^2+14x-7)(x^3-6x^2+9x-3)\tag{1}
\end{align}
$$
The roots $x=0$ and $x=3$ lead to indistinct $x$, $y$, and $z$.
$x^3-7x^2+14x-7$ has 3 real roots in $[0,4]$ whose sum is $7$ (the negative of the coeffic... | {
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"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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Find the value of $(a^3 + b^3 + c^3)/(abc)$ if $a/b + b/c + c/a = 1$.
Find the value of $$\frac{a^3+b^3+c^3}{abc}\qquad\text{ if }\quad \frac ab + \frac bc + \frac ca = 1.$$
I tried using Cauchy's inequality but it was of no help. Please guide me.
$a, b, c$ are real.
| There is not enough information to solve this problem. Clearing out denominators, your hypothesis is
$$a^2 c + a b^2 + b c^2 = abc \quad (1)$$
and your desired conclusion is
$$a^3+b^3+c^3=kabc \quad (2)$$
for some constant $k$.
Suppose, for the sake of contradiction, there were a $k$ such that $(1)$ implied $(2)$. Sin... | {
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"timestamp": "2023-03-29T00:00:00",
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How to find the product of n terms of an arithmetic progression where common difference is not unity? How can we find the product of $n$ terms of an arithmetic progression where common difference is not unity?
I just want to know the last $3$ digits of $7 \times 23 \times 39 \times \ldots \times 2071$ where common diff... | There are two (unrelated) questions here. An answer to the first question is
$$\prod_{k=1}^n(ak+b)=a^n\frac{\Gamma\left(n+1+\frac{b}a\right)}{\Gamma\left(1+\frac{b}a\right)}$$
To answer the second question, consider $n=\prod\limits_{k=0}^{129}(7+16k)$. Note that $7+16k=0\pmod{5}$ for every $k=3\pmod{5}$. Using this for... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\int_{0}^{\infty}\frac{\sin^{2n+1}(x)}{x} \mathrm {d}x$ Evaluate Integral Here is a fun integral I am trying to evaluate:
$$\int_{0}^{\infty}\frac{\sin^{2n+1}(x)}{x} \ dx=\frac{\pi \binom{2n}{n}}{2^{2n+1}}.$$
I thought about integrating by parts $2n$ times and then using the binomial theorem for $\sin(x)$, that is, us... | I am just adding the proof of the identity for those who have interest:
$$ \sin^{2n+1} x = \frac{1}{4^n}\sum_{k=0}^{n}(-1)^{n-k}\binom{2n+1}{k}\sin\left(\left(2(n-k)+1\right)x\right). $$
Using the complex representation and the Binomial Theorem, we have
$$\begin{aligned}
\sin^{2n+1}x&=\left(\frac{\mathrm{e}^{ix}-\mathr... | {
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"source": "stackexchange",
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Proof: For all integers $x$ and $y$, if $x^3+x = y^3+y$ then $x = y$ I need help proving the following statement:
For all integers $x$ and $y$, if $x^3+x = y^3+y$ then $x = y$
The statement is true, I just need to know the thought process, or a lead in the right direction. I think I might have to use a contradiction, b... | We have
\begin{eqnarray*}
x^3+x=y^3+y&\Longleftrightarrow& (x^3-y^3)+(x-y)=0\\
&\Longleftrightarrow& (x-y)(x^2+y^2+xy+1)=0.
\end{eqnarray*}
Since $x^2+y^2+xy+1=(x+\frac{y}{2})^2+\frac{3}{4}y^2+1>0$, we get $x=y$.
The hypothesis $x,y$ are integer numbers is redundant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/172261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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"answer_id": 3
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Showing that the last digit of $a$ and $a^{13}$ are the same
For $a \in \mathbb N$, show that the last digit of $a$ and $a^{13}$
are the same.
For example: $2^{13} = 8,192$
$7^{13} = 96,889,010,407$
| The last digit of $a^2$ in the set $\{1,2,3,4,5,0\}$ is
$\{1, 4, 9, 6, 5, 0\}$ (due to redundancy, I omit $6\le a\le 9$). Then
for $a^4$ we get $\{1,6,1,6,5,0\}$, the same for $a^8,a^{12}$ and so forth...
Now multiply $a^4$ once more to get
$$
\{1,6,1,6,5,0\}\cdot\{1,2,3,4,5,10\}=
\{1\cdot 1\; ,6\cdot2\; ,3\cdot1\; ,4... | {
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"timestamp": "2023-03-29T00:00:00",
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Functional Equation $f\big(f(x)\big)=3x$ over natural nubers for strictly increasing $f$
If $f$ is a strictly increasing function from the naturals to the naturals, and $f\big(f(x)\big)=3x$, what are all values of $f(2012)$?
I have only proven that $f(3x)=3f(x)$ but that gets nowhere :(
| Proposition. $f(1) = 2$, $f(2) = 3$.
Proof. $f(1)$ cannot be $1$ otherwise we would have
$$
3 = f(f(1)) = f(1) = 1
$$
So $f(1) > 1$ and since $f$ is strictly increasing, $f(x) > x$ for each $x$.
Being $3 = f(f(1)) > f(1)$, the only remaining possibility is $f(1) = 2$.
Finally, $f(2) = f(f(1)) = 3$. $\blacksquare$
Prop... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the limit without L'Hôpital's theorem I'm trying to find $$ \lim_{ n\to\infty} { {n^2+1}\over {n^3+1}} \cdot {\frac {n} {1}}$$
I know the answer is $1$, but I can't remember how my professor found it so simply without using L'Hôpital's theorem.
Could you please show me the shortcut?
| We consider $\lim_{ n\to\infty} { {n^2+1}\over n^3+1} \cdot {\frac n 1}$.
To do this, we note that $\lim_{ n\to\infty} { {n^2+1}\over n^3+1} \cdot {\frac n 1} = \lim \frac{n^3 + n}{n^3 + 1} = \lim \frac{n^3\left(1 + \frac{1}{n^2}\right)}{n^3\left( 1 + \frac{1}{n^3}\right)} = \lim \frac{1 + \frac{1}{n^2}}{1 + \frac{1}{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/176733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Probability density function of a quotient of two normal random variables I have the following expression:
$$R=\frac{\sigma_1^2\nu_1(t)-\sigma_2^2\nu_2(t)}{\sigma_1^2\nu_1(t)+\sigma_2^2\nu_2(t)}$$
where:
$$[\nu_1(t),\nu_2(t)]$$ are two independent normally distributed random variables.
My question is: how can I find an... | Let $\nu_1$ and $\nu_2$ be independent standard normal random variables. Then $U=\frac{\nu_1}{\nu_2}$ is well known to follow Cauchy distribution with pdf:
$$
f_U(u) = \frac{1}{\pi} \frac{1}{1+u^2}
$$
Let $X = \frac{\sigma_1^2 - U \cdot \sigma_2^2}{\sigma_1^2 + U \cdot \sigma_2^2}$. Assuming $\sigma_1>0$ and $\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove trigonometry identity for $\cos A+\cos B+\cos C$ I humbly ask for help in the following problem.
If
\begin{equation}
A+B+C=180
\end{equation}
Then prove
\begin{equation}
\cos A+\cos B+\cos C=1+4\sin(A/2)\sin(B/2)\sin(C/2)
\end{equation}
How would I begin the problem I mean I think $\cos C $ can be $\cos(180-A+B)... | $\cos A+\cos B+\cos C$
$=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}+1-2\sin^2\frac{C}{2}$ as $\cos2x=1-2\sin^2x$
Now $\cos\frac{A+B}{2}=\cos\frac{180^\circ - C}{2}=\cos(90^\circ-\frac{C}{2})=\sin\frac{C}{2}$
So, $\cos A+\cos B+\cos C$ becomes $2\sin\frac{C}{2}\cos\frac{A-B}{2}+1-2\sin^2\frac{C}{2}$
$=1+2\sin\frac{C}{2}(\cos\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/176892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 2
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Finding the sum of series; sum of squares Is there any way to find the sum of the below series ?
$$\underbrace{10^2 + 14^2 + 18^2 +\cdots}_{41\text{ terms}}$$
I got asked this question in a competitive exam.
| Let $$A=10^2+14^2+18^2+\dots$$ to 41 terms. Then $A=4B$, where $$B=5^2+7^2+9^2+\dots$$ to 41 terms. Let $$C=6^2+8^2+10^2+\dots$$ to 41 terms. Then $$B+C=5^2+6^2+7^2+\dots=(1^2+2^2+3^2+\dots)-(1^2+2^2+3^2+4^2)$$ where the sums go to $86^2$. Also, $C=4D$, where $$D=3^2+4^2+5^2+\dots=(1^2+2^2+3^2+\dots)-(1^2+2^2)$$ where ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/179735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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Summation equation for $2^{x-1}$ Since everyone freaked out, I made the variables are the same.
$$
\sum_{x=1}^{n} 2^{x-1}
$$
I've been trying to find this for a while. I tried the usually geometric equation (Here) but I couldn't get it right (if you need me to post my work I will). Here's the outputs I need:
1, 3, 7, 1... | In this instance, without explicitly using the formula for geometric series,
$$\begin{align*}
\sum_{x=1}^n 2^{x-1} &= 1 + 2 + 2^2 + 2^3 + \cdots + 2^{n-1}\\
&= 1 + (1 + 2 + 2^2 + 2^3 + \cdots + 2^{n-1}) - 1 &\text{add and subtract}~ 1\\
&= (1 + 1) + (2 + 2^2 + 2^3 + \cdots + 2^{n-1}) - 1 &\text{regroup}\\
&= 2 + (2 + 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/180169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
How many integer solutions to a diophantine equation Starting with the equation:
$\frac{1}{a}+\frac{1}{b}=\frac{p}{10^n}$,
I reached the equation:
$10^{n-log(p)} = \frac{ab}{a+b}$.
Now given the positive integer $n$, for what integer values of $p$ would the value of:
$10^{n-log(p)}$,
be rational?
Also, given positive i... | As @Qiaochu Yuan noted,
$10^{n-\log p}=\dfrac{10^n}{p}$ is always a rational number.
Suppose that a positive integer $n$ and an integr $p$ are given,
and let $a$ and $b$ to be integers
satisfying the above relation:
$$
\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{p}{10^n}
\Longleftrightarrow
\\
pab = 10^na + 10^nb
\Long... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/183890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
What is the number of real roots of the polynomial The number of distinct real roots of the equation
$x^9+x^7+x^5+x^3+x+1=0.$
| Let $f(x)=x^9+x^7+x^5+x^3+x+1$, then $f^\prime (x)=9x^8+7x^6+5x^4+3x^2+1\geq1>0.$ Hence $f$ is strictly increasing, so has atmost one real root. Also $f(0)=1>0$ and $f(-1)=-4<0$ and so has a root between $-1$ and $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/185734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proof of $(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2)>\frac{(ab+bc+ca)^3}{3}$ For positive real numbers $a$, $b$ and $c$, how do we prove that:
$$(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2)>\frac{(ab+bc+ca)^3}{3}$$
| By Cauchy-Schwarz:
$(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2) \ge (a^{1.5}b^{1.5}+b^{1.5}c^{1.5}+c^{1.5}a^{1.5})^2$
By the power mean inequality:
$(a^{1.5}b^{1.5}+b^{1.5}c^{1.5}+c^{1.5}a^{1.5})^2 \ge \frac{(ab+bc+ca)^3}{3}.$
(Note: The > sign in your post is false; let $a=b=c.$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/186575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding all $x$ for $\frac{2x - 13}{2x + 3} \lt \frac{15}{x}$ I'm trying to find all $x$ for the inequality $\frac{2x - 13}{2x + 3} \lt \frac{15}{x}$. In order to do this, I want to factor one side so that I can find all values where $x$ determines the term to equal $0$.
$$\frac{2x - 13}{2x + 3} \lt \frac{15}{x} \iff \... | There is at least one incorrect "solution" above.
The usual trick is to multiply both sides by something positive. So multiply both sides by $x^2(2x+3)^2$. Next, eliminate common factors in the numerators and denominators, bring everything over to the left hand side and then factorise. The inequality reduces to
$x(2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/186770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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Number of triples $(a,b,c)$ of positive integers such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{4}$?
What is the number of triples $(a,b,c)$ of positive integers such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{4}$ is:
A) $16$
B) $25$
C) $31$
D) $19$
E) $34$
Note: Through trial and error I've conclude... | Assume that $a\le b\le c$. The average of the fractions $\frac1a,\frac1b$, and $\frac1c$ is $\frac14$, so $\frac1a\ge\frac14$, and $a\le 4$. Clearly $a>1$, so $a$ must be $2,3$, or $4$.
*
*If $a=4$, the only possibility is $a=b=c=4$.
*Suppose that $a=3$. If $b=3$, $\frac1c=\frac34-\frac23=\frac1{12}$, so $c=12$. If... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/188560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
All pairs (x,y) that satisfy the equation $xy+(x^3+y^3)/3=2007$ How we can find the all pairs $(x,y)$ from the integers numbers ,that satisfy the equation :
$$xy+\frac{x^3+y^3}{3} =2007$$
| Observe that the equation is symmetric.
As $3|(x^3+y^3)$, either $(x,y)$ will be $(3a+1,3b-1)$, $(3a-1,3b+1)$ or $(3a,3b)$.
If $(x,y)$ is $(3a+1,3b-1)$, $\frac{x^3+y^3}{3}=3(3a^3+3b^3+3a^2-3b^2+a+b)$
So, 3 must divide $xy$ which is impossible as $xy=(3a+1)(3b-1)$
So, $(x,y)$ will be $(3a,3b)$.
So,$9(ab+a^3+b^3)=200... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/188737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
3 nonlinear equations in 3 variables - with the same variable raised to different powers Hoping someone can give me some guidance on solving the following system of 3 nonlinear equations in 3 variables:
$$\left\{\begin{align*}
&x^2+y^2=100\\
&xy+yz=-102\\
&y^2+z^2=117
\end{align*}\right.$$
Thanks!
| Observe that $y ≠0$ for finite $x,z$
$z=0\implies y^2=117>100\implies x^2=-17\implies x^2z^2=117(-17)≠(-102)^2$
$x=0\implies y^2=100\implies z^2=17 \implies y^2z^2=1700≠(-102)^2$
So $xyz≠0$
$(3)-(1)\implies z^2-x^2=17\implies z+x=\frac{17}{z-x}$
$(2)\implies y(z+x)=-102\implies y=-\frac{102}{\frac{17}{z-x}}=6(x-z)$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/191257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Showing that $ \frac{a}{a^2+3}+\frac{b}{b^2+3}\leq\frac{1}{2}$ for $a,b > 0$ and $ab = 1$ using rearrangement inequalities Please help to solve the following inequality using rearrangement inequalities.
Let $a \gt 0$, $b \gt0$ and $ab=1$. Prove that
\begin{equation}\frac{a}{a^2+3}+\frac{b}{b^2+3}\leq\frac{1}{2}.\e... | An opportunity to show the defects of AM-GM, you get a weaker inequailty.
$\dfrac{a^2+3}{a}=a+\dfrac{3}{a} \ge 2 \sqrt{3} \implies (\dfrac{a^2+3}{a})^{-1} \le \dfrac{1}{2 \sqrt{3}}$
$\dfrac{b^2+3}{b}=b+\dfrac{3}{b} \ge 2 \sqrt{3} \implies (\dfrac{b^2+3}{b})^{-1} \le \dfrac{1}{2 \sqrt{3}}$
When you add them, you have we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/191431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Automorphic numbers
Problem. We say that the $n$-digit number $x$ is automorphic iff $x^2\equiv x \mod(10^n)$. Prove that if $x$ is $n$-digit automorphic number then $(3x^2-2x^3)\mod(10^{2n})$ is $2n$-digit automorphic number. Hint: use Chinese reminder theorem to find the necessary and sufficient condition for number... | To do the computation with modular arithmetic instead of divisibility....
If $x \equiv 0 \pmod{2^n}$, then $x \equiv x_1 2^n \pmod{2^{2n}}$ for some integer $x_1$. Then,
$$ 3x^2 - 2x^3 \equiv 3 x_1^2 2^{2n} - 2 x_1^3 2^{3n} \equiv 0 \pmod{2^{2n}}$$
Similarly, in the case of $x \equiv 1 + x_1 2^n \pmod{2^{2n}}$, then,
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/192720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Calculation of a strange series Is it possible to find an expression for:
$$S(N)=\sum_{k=0}^{+\infty}\frac{1}{\sum_{n=0}^{N}k^n}?$$
For $N=1$ we have
$$S(1) = \displaystyle\sum_{k=0}^{+\infty}\frac{1}{1 + k} = \displaystyle\sum_{k=1}^{+\infty}\frac{1}{k}$$
which is the (divergent) harmonic series. Thus, $S (1) = \infty... | Perform a partial fraction decomposition:
$$
\frac{1}{p(k)} = \frac{1}{1+k+\cdots+k^{n-1}} = \frac{1}{ \prod_{m=1}^{n-1}\left(k-\exp\left(i \frac{2 \pi}{n} m \right)\right)} = \sum_{m=1}^{n-1} \frac{1}{k-\exp\left(i \frac{2 \pi}{n} m \right)} \frac{1}{p^\prime\left(\exp\left(i \frac{2 \pi}{n} m \right)\right)}
$$
Now... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/194096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 0
} |
Inequality. $a\sqrt{2b+c^2}+b\sqrt{2c+a^2}+c\sqrt{2a+b^2} \leq 3\sqrt{3}$ Could you help me please with the following inequality
Let $a$, $b$ and $c$ be non-negative numbers such that $a+b+c=3.$ Prove that:
$$a\sqrt{2b+c^2}+b\sqrt{2c+a^2}+c\sqrt{2a+b^2} \leq 3\sqrt{3}.$$
| Since $f(x)=\sqrt{x}$ is a concave function, by Jensen we obtain:
$$\sum_{cyc}a\sqrt{2b+c^2}=3\sum_{cyc}\frac{a}{3}\sqrt{2b+c^2}\leq3\sqrt{\sum_{cyc}\frac{a}{3}(2b+c^2)}=\sqrt{3\sum_{cyc}(2ab+a^2b)}=$$
$$=\sqrt{3(9-\sum_{cyc}(a^2-a^2b)}=\sqrt{27-\left((a+b+c)(a^2+b^2+c^2)-3\sum_{cyc}a^2b\right)}=$$
$$=\sqrt{27-\sum_{cy... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/195846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate $\frac{1}{b^2}\int_0^\infty z^{-2}\exp(-a z)\sin^2(b z)\, \mathrm dz$? How can I integrate the following:
$$\frac{1}{b^2}\int_0^\infty z^{-2}\exp(-a z)\sin^2(b z)\, \mathrm dz$$
for $a,b>0$? Maple gives a compact result:
$$\frac{1}{b} \tan^{-1}(c) - \frac{1}{ac^2} \ln(1 + c^2)$$
where $c=2b/a$. I can so... | Let $I(a)$ denote the integral. Consider
$$\begin{eqnarray}
I^{\prime\prime}(a) &=& \frac{1}{b^2} \int_0^\infty \exp(-a z) \sin^2(b z) \mathrm{d}z = \frac{1}{2b^2} \int_0^\infty \exp(-a z) (1-\cos(2 b z)) \mathrm{d}z \\
&=& \frac{1}{2b^2} \Re \int_0^\infty (\exp(-a z)-\exp(-(a+ 2i b) z)) \mathrm{d}z = \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/197876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How to prove this inequality $\sqrt{\frac{ab+bc+cd+da+ac+bd}{6}}\geq \sqrt[3]{{\frac{abc+bcd+cda+dab}{4}}}$ How to prove this inequality
$$\sqrt{\frac{ab+bc+cd+da+ac+bd}{6}}\geq \sqrt[3]{{\frac{abc+bcd+cda+dab}{4}}} $$
for $a,b,c,d\gt0$?
Thanks
| I remember seeing this problem in some book a couple of years ago. My attention was drawn by the proof of this inequality, which apparently originates from the '70 GDR mathematical olympiad. The solution involves nothing more than AM-GM (I assume $a,b,c,d$ are nonnegative here!), but some algebraic transformations are ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/197955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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$(1+i)$ to the power $n$
Possible Duplicate:
Complex number: calculate $(1 + i)^n$.
I came across a difficult problem which I would like to ask you about:
Compute
$ (1+i)^n $ for $ n \in \mathbb{Z}$
My ideas so far were to write out what this expression gives for $n=1,2,\ldots,8$, but I see no pattern such that I ca... | I think you are making this much too difficult, because:
$$(1+i)^2 = 2i.$$
So $(1+i)^{2m} = (2i)^m = 2^mi^m$.
That takes care of the even powers of $1+i$.
For the odd powers, just multiply by $1+i$ again: $(1+i)^{2m+1} = (1+i)^{2m}(1+i) = 2^mi^m(i+1)$.
If you want to simplify further, you need to know the remainde... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/199004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Show $8\mid n^2-1$ if $n$ is an odd positive integer. Show that $n^2-1$ is divisible by $8$, if $n$ is an odd positive integer.
Please help me to prove whether this statement is true or false.
| Since $n$ is odd $n=4m+1$ or $n=4m+3$.
In the first case $n^2-1=(n-1)(n+1)=4m\cdot(4m+2)=8m(2m+1)$, while in the second case $n^2-1=(n-1)(n+1)=(4m+2)\cdot(4m+4)=8(2m+1)(m+1)$.
So $n^2-1$ is divisible by 8 if $n$ is odd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/199185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 10,
"answer_id": 7
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A inequality proposed at Zhautykov Olympiad 2008 An inequality proposed at Zhautykov Olympiad 2008.
Let be $a,b,c >0$ with $abc=1$. Prove that:
$$\sum_{\mathrm{cyc}}{\frac{1}{(a+b)b}} \geq \frac{3}{2}.$$
Set $a=\frac{x}{y}$, $b=\frac{y}{z}$, $c=\frac{z}{x}$.
Our inequality becomes:
$$\sum_{\mathrm{cyc}}{\frac{z^2... | Here is a very short one.
Suppose w.l.o.g. $a \ge b \ge c$. Then by the rearrangement inequality,
$$
S = \frac{1}{(a+b)b} +\frac{1}{(b+c)c} + \frac{1}{(a+c)a} \ge \frac{1}{(a+b)c} +\frac{1}{(b+c)a} + \frac{1}{(a+c)b} = T
$$
So
$$
2 S \ge S + T = \frac{b+c}{(a+b)bc} +\frac{c+a}{(b+c)ca} + \frac{a+b}{(a+c)ab} \geq 3 \Bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/202053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
A problem about multiples.
For any positive integers $a$, $ b$, if $ab+1$ is a multiple of $16$, then $a+b$ must be a multiple of $p$. Find the largest possible value of $p$.
I have no idea how to solve this. Please help. Thank you.
| Let $ab\equiv-1\pmod {2^k}$
If $a=1,b\equiv-1= c2^k-1,a+b=c2^k$, the minimum value of $c$ is $1$.
So, $p$ must divide $2^k$, so can not have any odd factor.
Now, $b\equiv -\frac 1 a\pmod {2^k}$ as $ab\equiv-1$
So, $b\equiv -\frac 1 a\pmod {2^t}$
Let $2^t||(a+b)$ clearly, $t\le k$
So,$a\equiv -b \pmod{2^t}\equiv \frac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/206509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Correctness of Fermat Factorization Proof I have asked similar questions regarding this proof. But now I would like to know if my reformulation (after perseverance and different thinking) is correct.
Prove: An odd integer $n \in \mathbb{N}$ is composite $\iff$ $n$ can be written as $n = x^2 - y^2 s.t. y+1 < x$
Proof: $... | Since $n$ is composite $a,b>1$ (WLOG $a\geq b$)
$$x-y=\dfrac{a+b}{2}-\dfrac{a-b}{2}=b>1$$
$$x>y+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/206844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Evaluate $\sum_{k=1}^n\lfloor \sqrt{k} \rfloor$ I know that $\sum_{k=1}^m k^2 = \frac{m(m+1)(2m+1)}{6}$.
How can I use this fact to evaluate $\sum_{k=1}^n\lfloor \sqrt{k} \rfloor$ ?
| When $i^2\leq k<(i+1)^2-1$, $\lfloor \sqrt{k} \rfloor=i$.
So $\lfloor \sqrt{k} \rfloor$ takes $2i+1$ times the value $i$.
If $m^2 \leq n < (m+1)^2$, $\sum_{k=1}^n \lfloor \sqrt{k} \rfloor= (\sum_{k=1}^{m^2-1} \lfloor \sqrt{k} \rfloor)+ (\sum_{k=m^2}^n \lfloor \sqrt{k} \rfloor)=
(\sum_{i=1}^{m-1} (2i+1) \times i) + (n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/207320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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How to find the largest rectangle inside an ellipse I have an ellipse that is defined by center, width and height. The axes of the ellipse parallel to the x and y. I want to find the largest rectangle that completely fits inside this ellipse. Is there an easy way to do this?
And sorry if my terminology is a bit off... ... | Top right corner on the ellipse (centered on origin): $x = a \cos \theta$ and $y=b\sin\theta$
where $0\le \theta \le \frac{\pi}{2}, a=\frac{width}{2}$ and $b = \frac{height}{2}$
So area:$$A(\theta) = 4xy=4ab\sin\theta\cos\theta = 2ab\sin2\theta$$
Maximum of sin is at $2\theta=\frac{\pi}{2}$. So, we have $\theta = \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/210695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Elementary Question about limits Prove
$$
\lim_{x \to 0}\frac{x^3 - \sin^3x}{x - \ln{(1+x)} - 1 + \cos x} = 0
$$
Obviously, Using l'Hôpitals rule, we can evaluate this limit. But, taking derivatives of such functions is such a mess. Anyone sees a trick to do this problem faster? any ideas?
| $$\dfrac{x^3 - \sin^3(x)}{x - \ln(1+x) - 1 + \cos(x)} = \dfrac{x - \sin(x)}{x} \times \dfrac{x^2 + x \sin(x) + \sin^2(x)}{1 - \dfrac{\ln(1+x)}{x}- \dfrac{1-\cos(x)}{x}}$$
$$1 - \dfrac{\ln(1+x)}{x}- \dfrac{1-\cos(x)}{x} = 1 - \left(1 - \dfrac{x}2 +\dfrac{x^2}3 - \cdots \right) - \left( \dfrac{\dfrac{x^2}{2!} - \dfrac{x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/214579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Invertible Matrix Let $A$ be the matrix
$$ A=\left(\begin{array}{cccc}1&0&1&2\\2&3&\beta&4\\4&0&-\beta&-8\\ \beta&0&\beta&\beta \end{array}\right). $$
For what values of $\beta$ is the matrix invertible?
| $$\begin{vmatrix}
1&0&1&2\\
2&3&\beta&4\\
4&0&-\beta&-8\\
\beta&0&\beta&\beta
\end{vmatrix}
=
\beta\begin{vmatrix}
1&0&1&2\\
2&3&\beta&4\\
4&0&-\beta&-8\\
1&0&1&1
\end{vmatrix}
=
\beta\begin{vmatrix}
0&0&0&1\\
2&3&\beta&4\\
4&0&-\beta&-8\\
1&0&1&1
\end{vmatrix}
=
-\beta\begin{vmatrix}
2&3&\beta\\
4&0&-\beta\\
1&0&1
\en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/219458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Differentiating $x^2 \sqrt{2x+5}-6$ How do I differentiate this function: f(x)= $x^2 \sqrt{2x+5}-6$
I had: I had $2x\sqrt{2x+5} + x^2 \dfrac{1}{2\sqrt{2x+5}}$ but the correction model said it was I had $2x\sqrt{2x+5} + x^2 \dfrac{2}{2\sqrt{2x+5}}$
| $2x . \sqrt{2x+5} + \dfrac{x^2}{2} . \dfrac{1}{\sqrt{2x+5}}2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/220858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
For primes $p≡3\pmod 4$, prove that $[(p−1)/2]!≡±1\pmod p$. I know to use Wilson's Theorem and that each element in the second half is congruent to the negative of the first half, but I'm not sure how to construct a proof for it.
| $p-r\equiv -r\pmod p\implies r\equiv-(p-r)$
For uniqueness, $r\le p-r$ or $2r\le p\implies r\le\frac p 2$
So, $1\le r\le \frac{p-1}2$ as $p$ is odd
Putting $r=1,2,3,\cdots,\frac{p-3}2,\frac{p-1}2$ we get,
$1\equiv-(p-1)$
$2\equiv-(p-2)$
...
$\frac{p-3}2\equiv-(p-\frac{p-3}2)=\frac{p+3}2$
$\frac{p-1}2\equiv-(p-\frac{p-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/222372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Inequality. $\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} \geq \frac{ab+bc+ca}{2}$ prove the following inequality:
$$\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} \geq \frac{ab+bc+ca}{2},$$ for $a,b,c$ real positive numbers.
Thanks :)
| I started this proof but then realized that there was a mistake. Below is the wrong argument. Enjoy figuring out the mistake. (The argument can be fixed though.)
\begin{align}
S(a,b,c) & = \dfrac{a^3}{a+b} + \dfrac{b^3}{b+c} + \dfrac{c^3}{c+a}\\
& = \dfrac{a^3+b^3}{a+b} + \dfrac{b^3+c^3}{b+c} + \dfrac{c^3+a^3}{c+a} - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/222934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
$(2^m -1)(2^n-1)$ divides $(2^{mn} -1)$ if and only if $\gcd(m,n) = 1$. If $\gcd(m,n) = 1$ then $(2^m-1)(2^n-1)$ divides $2^{mn} - 1$ because each of $2^m-1$, $2^n-1$ divide $2^{mn}-1$ and $\gcd(2^m-1, 2^n-1) = 2^{\gcd(m,n)}-1 = 1$. How about the converse? If $\gcd(m,n) > 1$, can one show that $(2^m-1)(2^n-1)$ does not... | The following generalization can be proved by a straightforward adaptation of the proof I gave above:
Given integer $a \geq 2$, positive integers $m,n$, and $k$ a positive integer that is a common divisor of $m$, $n$, then
$(a^m-1)(a^n-1)/(a^k-1)$ divides $a^{mn}-1$ if and only if $k = \gcd(m,n)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/223818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
sum of a series Can
\begin{equation}
\sum_{k\geq 0}\frac{\left( -1\right) ^{k}\left( 2k+1\right) }{\left(
2k+1\right) ^{2}+a^{2}},
\end{equation}
be summed explicitly, where $a$ is a constant real number? If $a=0,$ this
sum becomes
\begin{equation}
\sum_{k\geq 0}\frac{\left( -1\right) ^{k}}{2k+1}=\frac{\pi }{4}.
\end{e... | This sum can be evaluated by the same trick as presented here:
math.stackexchange.com.
In order to capture the convergence behavior of the series group consecutive terms to obtain absolute convergence, writing
$$ S(a) = \sum_{m\ge 0} \frac{4m+1}{(4m+1)^2+a^2}
- \sum_{m\ge 0} \frac{4m+3}{(4m+3)^2+a^2} =
\sum_{m\ge 0}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/226308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
The product of digits equal to the sum of digits How to find the number(or numbers ) that has $4$ digits, the product of these digits equal to the sum of these digits ?
| First of all, let's observe that all of the digits of such a number cannot be the same. You can just manually check that numbers $1111$, $2222$ and so on don't suit us. It is also clear that all of the digits should be non-zero.
Now suppose that we have such a number. Let $a,\,b,\,c,\,d$ be its digits written in non-as... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/227515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
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Middle line between points How can we calculate LINE that most fit the points T1(1,0) T2(2,0) T3(-1,1) T4(0,1)
$x= (1,2,-1,0) $
$y= (0, 0, 1, 1) $
$1= (1, 1, 1, 1)$
| a)LINE that most fit the points $T1(1,0) \mspace{10mu} T2(2,0) \mspace{10mu} T3(-1,1) \mspace{10mu} T4(0,1) $
$x = (1 , 2 , -1 , 0) $
$y = (0 , 0 , 1 , 1) $
$1 = (1 , 1 , 1 , 1)$
Solving this: you got sistem of 2 equations:
$a < x, x> + b < 1, x> = < y, x> $
$a< x, 1> + b < 1, 1> = < y, 1>$
$< x, x>= x_1*x_1 + x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/231556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding all real zeros of the polynomial $$x^5 - 5x^4 +6x^3 -30x^2 +8x - 40 = 0$$
So far I have...
$$r/s: +- 1, +- 40, +- 2, +- 20, +- 4, +- 10, +-5, +- 8$$
Only $+ 5$ works.
Then I have $$(x + 5)( ) = x^5 - 5x^4 +6x^3 -30x^2 +8x - 40$$
Then you have to use long devision between $x + 5$ and $ x^5 - 5x^4 +6x^... | here is how to do the long division
x^5 - 5x^4 + 6x^3 - 30x^2 + 8x - 40
x^5 - 5x^4 subtract x^4(x-5)
-----------------------------------
6x^3 - 30x^2 + 8x - 40
6x^3 - 30x^2 subtract 6x^2(x-5)
-----------------------------------
8x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/237389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $(a, b, c)$ is a Pythagorean triple, then $(a+1, b+1, c+1)$ is not a Pythagorean triple. Pythagoras stated that there exist positive natural numbers, $a$, $b$ and $c$ such that $a^2+b^2=c^2$. These three numbers, $a$, $b$ and $c$ are collectively known as a
Pythagorean triple. For example, $(8, 15, 17)$ is one of th... | Suppose both statement are true. Then
$(a+1)^2+(b+1)^2=(c+1)^2$
and as well
$a^2+b^2=c^2$
From 1.
$a^2+2a+1+b^2+2b+1=c^2+2c+1$
$\implies 2a+2b=2c-1$
But this does not satisfy for Pythagorean triplet 3,4,5 or for any other like 5,12,13.
Hence our supposition is wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/239312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Asymptotic behavior of $(1/2 + 2/3 + 3/4 + 4/5 + \cdots + (n-1)/n ) \times n$ I am interested in the following questions:
given:
$$G(n) = \left(\frac12 + \frac23 + \frac34 + \frac45 + \cdots + \frac{n-1}n\right)n$$
*
*what is a $F(n)$ which could be an upper bound (clearly as tight as possible) for $G(n)$ for $n$ ... | Let $$F(n) = \left(\dfrac{2-1}{2} + \dfrac{3-1}{3} + \dfrac{4-1}{4} + \cdots + \dfrac{n-1}{n} \right) $$ we then have that
$$F(n) = \left(1 - \dfrac12 + 1 - \dfrac13 + 1 - \dfrac14 + \cdots + 1 - \dfrac1n \right) = (n-1) - \left( \dfrac12 + \dfrac13 + \cdots + \dfrac1n \right)$$
Now note that $$-\left( \dfrac12 + \dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/239524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Integration by parts, Apostol Integration by parts. Need to show that:
If $I_n(x)=\int_{0} ^{x}t^n(t^2+a^2)^{-\frac{1}{2}}dt$
Then: $nI_n(x) = x^{n-1}\sqrt{x^2+a^2}-(n-1)a^2I_{n-2(x)}$ if $x\geq2$
I can get to the point where:
$I_n=\frac{x^{n+1}}{n+1}(x^2+a^2)^{-\frac{1}{2}}+\frac{1}{n+1}\int_0 ^x (t^nt^2+a^2t^n-a^2t^n... | Let $$I_n(x) = \int_0^x \dfrac{t^n}{\sqrt{t^2 + a^2}} dt = \int_0^x \dfrac{t^{n-1} \cdot t}{\sqrt{t^2 + a^2}} dt$$
Let $u(t) = t^{n-1}$ and $dv(t) = \dfrac{t}{\sqrt{t^2 + a^2}} dt \implies v(t) = \sqrt{t^2+a^2}$. Hence,
\begin{align}
I_n(x) & = \int_0^x u(t) dv(t) = \left. u(t) v(t) \right\vert_{0}^x - \int_0^x v(t)du(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/240438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solution for the value of angle $A$ of a triangle In triangle $\triangle \; ABC$ , if $$2\frac{\cos A}{a} + \frac{\cos B}{b} + 2\frac{\cos C}{c} = \frac{a}{bc} + \frac{b}{ca}$$ find angle $A$.
This is a quiz bee problem sent to me by my friend in FB. He asked me if I can do a solution for it. Well I tried several way... | Using $$\cos C=\frac{a^2+b^2-c^2}{2ab}$$ etc.,
we get, $$\frac2a\frac{b^2+c^2-a^2}{2bc}+\frac1b\frac{a^2+c^2-b^2}{2ac}+\frac2c\frac{a^2+b^2-c^2}{2ab}=\frac{a^2+b^2}{abc}$$
or, $$ 2(b^2+c^2-a^2)+(a^2+c^2-b^2)+2(a^2+b^2-c^2)=2(a^2+b^2)$$
or $b^2+c^2=a^2$ as $abc\ne0$ $a,b,c$ being the sides of triangle.
So, $\cos A=0\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/241946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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$2n+1$ and $n^2+1$ are always coprime or their gcd is $5$ Using a spreadsheet, it can be inferred that when $n≡2[5]$, then $\gcd(n^2+1,2n+1)=5$, else $\gcd(n^2+1,2n+1)=1$.
Indeed, when $n≡2[5]$, $n^2+1$ and $2n+1$ can easily be shown to be multiples of $5$, so their gcd is at least $5$. But then, I can't see how to com... | As $(a,b)=(a,a-nb),$
$(2(n^2+1), 2n+1)=(2(n^2+1)-n(2n+1), 2n+1)=(2-n,2n+1)=(2-n,2n+1+2(2n-1))=(2-n,5)$
So, $(2(n^2+1), 2n+1)=5$ if $5\mid(2-n)$ i.e., if $n\equiv2\pmod 5$
$\implies (n^2+1, 2n+1)=5$ as $(2,2n+1)=1$ is as $(2n+1)$ odd.
If $n\not\equiv2\pmod 5,(2(n^2+1), 2n+1)=1\implies (n^2+1, 2n+1)=1$
Alternatively,
W... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/242610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Parametric Equations: Finding Given Tangent Lines At Points The parametric equations are: $x=2\cot\theta$ and $y=2\sin^2\theta$
$$\frac{dy}{dx}=-2\sin^3\theta\cdot \cos\theta$$
And the coordinates are: $(-\frac{2}{\sqrt{3}},3/2)$, $(0,2)$, and $(2\sqrt{3}, 1/2)$
I wasn't quite sure what to do with those coordinates, co... | For example, with the point $\,\displaystyle{\left(-\frac{2}{\sqrt 3}\,,\,\frac{3}{2}\right)}\,$ , and with $\,t=\theta\,$ , for simplicity:
$$2\cot t=x=-\frac{2}{\sqrt 3}\Longrightarrow \tan t=-\sqrt 3\Longrightarrow t=\frac{\pi}{3}+k\pi\,\,,\,k\in\Bbb Z$$
$$\frac{3}{2}=y=2\sin^2t\Longrightarrow \sin t=\pm\frac{\sqrt ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/245953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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What am I doing wrong in calculating this determinant? I have matrix:
$$
A = \begin{bmatrix}
1 & 2 & 3 & 4 \\
2 & 3 & 3 & 3 \\
0 & 1 & 2 & 3 \\
0 & 0 & 1 & 2
\end{bmatrix}
$$
And I want to calculate $\det{A}$, so I have written:
$$
\begin{array}{|cccc|ccc}
1 & 2 & 3 & 4 & 1 & 2 & 3 \\
2 & 3 & 3 & 3 & 2 & 3 & 3 \\
0 & ... | The method that you're using works just fine for $3\times 3$ matrices, but fails to work with $n\times n$ matrices for other $n$. You're going to have to do it another way.
For example, expanding the deteriminant along the first column, we find that $$\begin{align}\det A &=1\cdot\det\left[\begin{array}{ccc}3 & 3 & 3\\1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/246606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Limit $\lim_{n \rightarrow \infty}\frac{1 \cdot 3 \cdot 5 \dots \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot \dots \cdot (2n)} $ $\displaystyle \lim_{n \rightarrow \infty}\frac{1 \cdot 3 \cdot 5 \dots \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot \dots \cdot (2n)} $
| A hint:
Write $1\cdot3\cdot 5\cdot\ldots\cdot(2n-1)$ and $2\cdot 4\cdot 6\cdot\ldots\cdot (2n)$ in terms of factorials and powers of $2$. Then use Stirling's formula
$$m!=\left({m\over e}\right)^m\ \sqrt{2\pi m}\ \bigl(1+o(1)\bigr)\qquad(m\to\infty)$$
to estimate the various factorials appearing in your expression.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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A five-digit number N.... A five-digit number $N$ is equal to $45$ times the product of its $5$ digits. Find $N$.
Please help. I am not sure how to solve this. I have a feeling it is simple
| Let the digits of $n$ from left to right be $a,b,c,d$, and $e$, so that we’re told that
$$10^4a+10^3b+10^2c+10d+e=45abcde\;.\tag{1}$$
Clearly $n$ is a multiple of $5$, so $e$ is either $0$ or $5$. But obviously none of the digits is $0$, so $e=5$, and $(1)$ becomes
$$10^4a+10^3b+10^2c+10d+5=225abcd$$
or, after dividing... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $x$, $y$, $x+y$, and $x-y$ are prime numbers, what is their sum?
Suppose that $x$, $y$, $x−y$, and $x+y$ are all positive prime numbers. What is the sum of the four numbers?
Well, I just guessed some values and I got the answer.
$x=5$, $y=2$, $x-y=3$, $x+y=7$. All the numbers are prime and the answer is $17$.
Supp... | Note that $x>y$, since $x-y$ is positive. Since $x$ and $y$ are both prime, this means that $x$ must be greater than $2$ and therefore odd. If $y$ were odd, $x+y$ would be an even number greater than $2$ and hence not prime. Thus, $y$ must be even, i.e., $y=2$.
Now we want an odd prime $x$ such that $x-2$ and $x+2$ are... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/250584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "100",
"answer_count": 5,
"answer_id": 2
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How can I find the points at which two circles intersect? Given the radius and $x,y$ coordinates of the center point of two circles how can I calculate their points of intersection if they have any?
| Easy solution is to consider another plane such that the centers are along an axis.
Given the points $(x_1,y_1)$ and $(x_2,y_2)$.
We focus on the center point of both circles given by
$$
\left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right).
$$
The distance between the centers of the circles is given by
$$
R = \sqrt{ (x_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/256100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "59",
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Rotation matrix in terms of axis of rotation How to calculate the rotation matrix in 3D in terms of an arbitrary axis of rotation? Given a unit vector $V=V_{x}e_{x}+V_{y}e_{y}+V_{z}e_{z}$ How to calculate the rotation matrix about that axis?
| I think you need the Rodrigue's rotation matrix composition.
If your unit rotation axis is $\vec{v} = (V_x,V_y,V_z)$ and the rotation angle $\theta$ then the rotation matrix is
$$ R = \boldsymbol{1}_{3\times3} + \vec{v}\times\,(\sin\theta) + \vec{v}\times\vec{v}\times\,(1-\cos\theta) $$
where $\vec{v}\times = \begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/256194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to get from $\frac{x}{x+1}\;$ to $\;1 - \frac{1}{x+1}$? Please show me how to manipulate $\dfrac{x}{x+1}\;\;$ to get $\;\;1 - \dfrac{1}{x+1}$
| How about start with $1 - \frac{1}{x+1}$ to get
$$1 - \frac{1}{x+1} = \frac{x+1}{x+1} - \frac{1}{x+1} = \frac{x}{x+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/259498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Maximization of an integer input function Maximize the value of the function
$$
z=\frac{ab+c}{a+b+c},
$$
where $a,b,c$ are natural numbers and are all lesser than 2010 and not necessarily
distinct from each other.
Please provide a proof, and if possible a general technique.
Thank you.
| $$z=\frac{ab+c}{a+b+c}=\frac{ab-a-b}{a+b+c}+1$$
Clearly, this will be maximum if $c$ minimum $=1$
So, $$z\le \frac{ab-a-b}{a+b+1}+1=\frac{ab+1}{a+b+1} $$
$\frac{a_1b_1+1}{a_1+b_1+1}$ will be greater than $\frac{a_2b_2+1}{a_2+b_2+1}$
if $(a_1a_2-1)(b_1-b_2)+(b_1b_2-1)(a_1-a_2)+a_1b_1-a_2b_2>0$
if $(a_1a_2-1)(b_1-b_2)+(b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/259958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Expectation of joint life span The life span of a particular mechanical part is a random variable described by the following PDF:
If three such parts are put into service independently at t=0, determine a simle expression for the expected value of the time until the majority of the parts will have failed.
I can get th... | The value for $E \left[ \max \left( L_1, L_2 \right) \right]$ is computed in
the following way. First, the distribution of the maximum of two identically
independently distribued random variable $L_1$ and $L_2$ is given by $2 f
\left( \ell \right) F \left( \ell \right)$ where $f \left( \ell \right)$ is
the density and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/260031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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Solutions of Diophantine equations in Natural numbers The one of solution of $x^4 - 2y^2 = -1$ is $x = 1$ and $y = 1$. However, the solution $(1, 1)$ of $x^4 - 2y^2 = 1$ is failed. We know $x = 1$ and $y = 1$ is small integers and we can check by trail method. In case more solutions are existing or not how to check? W... | The only integer solutions of $x^4-2y^2=1$ are $x=\pm 1$, $y=0$.
For $x$ must be odd. Rewrite our equation as $(x^2-1)(x^2+1)=2y^2$. The greatest common divisor of $x^2-1$ and $x^2+1$ is $2$. Since $x^2+1$ has shape $8k+2$, it follows that $x^2+1=2s^2$, and $x^2-1=t^2$ for some integers $s$ and $t$.
The only solution i... | {
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"url": "https://math.stackexchange.com/questions/263647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solutions for $ 2^i = 3^ j - 1 $ or $ 2^i = 3^ j + 1 $ Are there any solutions for $ 2^i = 3^ j - 1 $ or $ 2^i = 3^ j + 1 $,
for $i>3$ and $j>2$ ?
Thanks! $:)$
| Observe that $2$ is a primitive root of $3^k$ where $k$ is natural number(Proof below).
(1)So if $2^i=3^{j+1}+1,2^i\equiv1\pmod {3^{j+1}}$
$\implies \phi(3^{j+1})\mid i \implies 2\cdot 3^j\mid i$
So, the minimum possible integral value of $i$ is $2\cdot 3^j$
If $j=0,i_{min}=2,2^i-3^{j+1}=2^2-3^1=1$
If $j=1,i_{min}=2\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/264337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Evaluating the infinite product $\prod\limits_{k=2}^\infty \left ( 1-\frac1{k^2}\right)$
Evaluate the infinite product
$$\lim_{ n\rightarrow\infty }\prod_{k=2}^{n}\left ( 1-\frac{1}{k^2} \right ).$$
I can't see anything in this limit , so help me please.
| Hint: The "typical" term is $\dfrac{k-1}{k}\dfrac{k+1}{k}$. Express the first few terms in this way, and observe the nice cancellations.
For example, here is the product of the first seven terms:
$$\frac{1}{2}\frac{3}{2}\frac{2}{3}\frac{4}{3}\frac{3}{4}\frac{5}{4}\frac{4}{5}\frac{6}{5}\frac{5}{6}\frac{7}{6}\frac{6}{7}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/265483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 6,
"answer_id": 3
} |
find recurrence relation $T(n)=2T(n/2) +\log_2(n)$ $$\begin{align*}
&T(n) = 2T(n/2) + \log_2(n)\\
&T(1) = 0
\end{align*}$$
$n$ is a power of $2$
solve the recurrence relation
my work so far:
unrolling this, we have
$$\begin{align*}
T(n) &= 4T(n/4) + \log_2(n) -1\\
&= 8T(n/8) + 2\log_2(n) -2\\
&=\log_2(n-1) \log_2(n)... | Substituting $n=2^k$ we have:
$$\begin{align*}T(n)=T(2^k)&=2T(2^{k-1})+k=2(2T(2^{k-2})+k-1)+k=4T(2^{k-2})+3k-1\\
&=4(2T(2^{k-3})+k-3)+3k-1=8T(2^{k-3})+7k-13=...=\\
&= 2^mT(2^{k-m})+\sum_{t=1}^m2^{t-1}(k-t+1)=...=2^kT(1)+\sum_{t=0}^{k-1}2^t(k-t)\\
&=2^k\cdot0+k\sum_{t=0}^{k-1}2^t-\sum_{t=0}^{k-1}t2^t=k\frac{2^k-1}{2-1}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/267571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
trace of the matrix $I + M + M^2$ is
Let $ \alpha = e^{\frac{2\pi \iota}{5}}$ and the matrix
$$ M= \begin{pmatrix}1 & \alpha & \alpha^2 & \alpha^3 & \alpha^4\\
0 & \alpha & \alpha^2 & \alpha^3 & \alpha^4\\
0 & 0 & \alpha^2 & \alpha^3 & \alpha^4 \\
0 & 0 & 0 & \alpha^3 & \alpha^4\\
0 & 0 & 0 & 0 & \alpha^4 \end{p... | Note that the trace of $M$ is $0$, since $1+\alpha+\alpha^2+\alpha^3+\alpha^4= 0$.
Also $M$ is upper triangular so that $M^2$ has diagonal elements which are just the square of the diagonal elements of $M$, i.e. $1,\alpha^2, \alpha^4, \alpha^6, \alpha^8$.
Using the fact that $\alpha^5 = 1$ we see that the trace of $M^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/268029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Why do we choose $3$ to be positive after $\sqrt{9 - x^2}$ in the following substitution? The integral $$\int \frac{\sqrt{9 - x^2}}{x^2}dx$$ is solved in my book by letting $x = 3\sin\theta$ where $-\frac {\pi}{2} \le \theta \le \frac {\pi}{2}$.
Then, $dx = 3\cos\theta\,d\theta$
and, $$\sqrt{9-x^2} = 3|\cos\theta| = ... | Your book is being very careful about the fact that $\sqrt{1-\sin^2\theta} = \cos\theta$ is only true when $\cos \theta \geq 0$. Of course, in this problem it doesn't matter, since $-\frac{\pi}{2} < \theta \leq \frac{\pi}{2}$.
Note that you could just as well have chosen the substitution $x = -3\sin \theta$. It is a go... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/268830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Show that $\lim_{n\rightarrow \infty}(\sqrt[3]{n+\sqrt{n}}-\sqrt[3]{n})=0$ I am asked to prove this statement $^{*}$. I am trying now, but it is getting to small and tiny steps that I even loose my way. my steps are as follows:
$$\lim_{n\rightarrow \infty}(\sqrt[3]{n+\sqrt{n}}-\sqrt[3]{n})=0^{*}$$
$\lim_{n\rightarrow... | What Dennis wrote. Or,
$$
\sqrt[3]{n+\sqrt{n}}-\sqrt[3]{n}=C\int_n^{n+\sqrt{n}}\frac{\mathrm dt}{t^{2/3}}\leqslant C\int_n^{n+\sqrt{n}}\frac{\mathrm dt}{C'n^{2/3}}=C''n^{1/2-2/3}\to0,
$$
because $1/2\lt2/3$. Likewise, for every positive $A$ and $B$,
$$
\sqrt[A]{n+\sqrt[B]{n}}-\sqrt[A]{n}\to0\iff\frac1A+\frac1B\lt1.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/271480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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