Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Seeking methods to solve: $\int_0^\infty \frac{\ln(t)}{t^n + 1}\:dt$ Seeking Methods to solve the following two definite integrals:
\begin{equation}
I_(n) = \int_0^\infty \frac{\ln(t)}{t^n + 1}\:dt \qquad J(n) = \int_0^\infty \frac{\ln^2(t)}{t^n + 1}\:dt
\end{equation}
For $n \in \mathbb{R},\:n \gt 1$
The method I took... | Here is an alternative strategy that can be used to find $I^{(1)}_n$ (your $I_n$), $I^{(2)}_n$ (your $J_n$), and $I^{(p)}_n$ (the general case where $p \in \mathbb{N}$).
Writing
$$I^{(1)}_n = \int_0^\infty \frac{\ln t}{1 + t^n} \, dt = \int_0^1 \frac{\ln t}{1 + t^n} \, dt + \int_1^\infty \frac{\ln t}{1 + t^n} \, dt.$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3073077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Determining the area of a right triangle, perimeter given, hypotenuse value given in terms of one of the legs. The problem states:
Right Triangle- perimeter of $84$, and the hypotenuse is $2$ greater than the other leg. Find the area of this triangle.
I have tried different methods of solving this problem using Pyt... | Let the sides of the right triangle be $x,y,x+2$.
Given,
$2x+y=82 \tag{1}$
$x^2 + y^2 = (x+2)^2 \tag{2}$
$$\implies x^2 + y^2 = x^2 +4x+4 $$
$$\implies y^2 = 4x+4 $$
Now, substitute the value of $x$ from equation (1) in terms of $y,$ you will get a quadratic equation in $y$ whose roots can be easily found and henc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3076504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Candy machines and optimal strategy in terms of expected value Problem
We have three candy machines: call them G (good), B (bad) and M (mixed) . G always gives you a candy when you put 1\$. B never gives you a candy when you put 1\$. M gives you a candy with probability $1/2$ when you put 1\$. You want a candy and you ... | Clearly you always want $k = 1$. If you have tried two machines and gotten candy from neither, the last one must be the Good one, so there's no reason not to try that next. This leaves our choice of $n$. Intuitively it's clear that this one should be 1 as well: after failing to get a candy from the machine once, it's e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3076954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to compute $u_x(\sqrt {(x^2+y^2)})=$? This is in reference to this question
$g(z) = f(|z|)$ is not holomorphic for a non constant function $f$
If $u$ is a function of single variable $x$ and if I want to differentiate $u(\sqrt {(x^2+y^2)}$ with respect to $x$
then what will be $u_x(\sqrt {(x^2+y^2)})=$?
It is give... | What happens when I do this with $u(x)=x^2+x$. Write
$$f(x,y)=u\left(\sqrt{x^2+y^2}\right)=x^2+y^2+\sqrt{x^2+y^2}.$$
Then
$$\frac\partial{\partial x}f(x,y)=2x+\frac{x}{\sqrt{x^2+y^2}}.$$
But $u'(x)=2x+1$, and
$$u'\left(\sqrt{x^2+y^2}\right)\frac\partial{\partial x}\sqrt{x^2+y^2}
=\left(2\sqrt{x^2+y^2}+1\right)\frac{x}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show using the formal, limited based definition of integral I am currently trying to figure out the following: for $f(x) = 2x - 5$, I want to show that, using the formal, limited based definition of integral, $\int_{3}^{7}f(x) = 20$ (both the domain and codomain of $f(x)$ are $\mathbb{R}$). However, I already seem to g... | Divide the interval from x= 3 to x= 7 into n equal length subintervals. Each has length (7- 3)/n= 4/n. The x values at the endpoints of the subintervals are $x_0= 3$, $x_1= 3+ 4/n= (3n+ 4)/n$, $x_2= 3+ 2(4/n)= (3n+ 8)/n$, …, $x_i= 3+ i(4/n)= (3n+ 4i)/n$, …, $x_n= 3+ n(4/n)= 7$. At each i, $f(x_i)= 2x_i- 5= 2(3n+ 4i)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving $\binom{n + m}{r} = \sum_{i = 0}^{r} \binom{n}{i}\binom{m}{r - i}$ To prove $$\binom{n + m}{r} = \sum_{i = 0}^{r} \binom{n}{i}\binom{m}{r - i},$$
I demonstrated that the equality is true for any $n,$ for $m = 0, 1,$ and for any $r < n + m$ simply by fixing $n$ and $r$ and inserting $0,1$ for $m.$ Then, I procee... | I think the formula is wrong. If you have $n$ white balls and $m$ red balls then, to choose $r$ balls from then you have to choose $i$ white balls and $r-i$ red balls for some $i$ between $0$ and $\min \{r,n\}$. Hence the sum should be over $0\leq r \leq \min \{r,n\}$. However, if you define $\binom {n} {i}$ to be $0$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3081297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Compute $ \lim\limits_{n \to \infty}\frac{\sqrt{3n^2+n-1}}{n+\sqrt{n^2-1}}$
Compute $$ \lim\limits_{n \to \infty}\frac{\sqrt{3n^2+n-1}}{n+\sqrt{n^2-1}}$$
I did the following:
$$ \lim\limits_{n \to \infty}\frac{\sqrt{\frac{3n^2}{n^2}+\frac{n}{n^2}-\frac{1}{n^2}}}{\frac{n}{n^2}+\sqrt{\frac{n^2}{n^2}-\frac{1}{n^2}}} = \... | $$
\lim_{n\to\infty}\frac{\sqrt{3n^2 + n-1}}{n+\sqrt{n^2-1}}
$$
Factor out $n$, thus:
$$
\frac{\sqrt{3n^2 + n-1}}{n+\sqrt{n^2-1}} = \frac{n\sqrt{3 + {1\over n} - {1\over n^2}}}{n\left(1 + \sqrt{1- {1\over n^2}}\right)} = \frac{\sqrt{3 + {1\over n} - {1\over n^2}}}{1 + \sqrt{1- {1\over n^2}}}
$$
So your limit becomes:
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3081675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Find range of $x$ if $\log_5\left(6+\frac{2}{x}\right)+\log_{1/5}\left(1+\frac{x}{10}\right)\leq1$
If $\log_5\left(6+\dfrac{2}{x}\right)+\log_{1/5}\left(1+\dfrac{x}{10}\right)\leq1$, then $x$ lies in _______
My Attempt
$$
\log_5\bigg(6+\dfrac{2}{x}\bigg)+\log_{1/5}\bigg(1+\dfrac{x}{10}\bigg)=\log_5\bigg(6+\dfrac{2}{x... | You forgot to check when $$6+\dfrac{2}{x}>0$$ and $$1+\dfrac{x}{10}>0$$
is true!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3084524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Proving $3n^2 - 4n \in \Omega(n^2)$ Proving $3n^2 - 4n \in \Omega(n^2)$
Attempt:
$3n^2 - 4n \geq cn^2$
$n(3n-4) \geq cn^2$
$(3n-4) \geq cn$
$3n - 4 - cn \geq 0$
$n(3-c) \geq 4$
$n \geq \frac{4}{3-c}$
would do I choose for $n_0$ and $c$ to satisfy this proof?
| $3n^2 - 4n \in \Omega(n^2)$ means:
there are $c>0$ and $n_0 \in \mathbb N$ such that $3n^2-4n \ge cn^2$ for $n \ge n_0$.
Since $\frac{3n^2-4n}{n^2}= 3-\frac{4}{n} \to 3$ as $n \to \infty$, there is $n_0$ such that $\frac{3n^2-4n}{n^2} \ge 2$ for $n \ge n_0$.
Hence $3n^2-4n \ge 2n^2$ for $n \ge n_0$.
This shows that $3n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3085545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to prove binomial coefficient $ {2^n \choose k} $ is even number? Prove:
${2^n \choose k}$ (for integers $k$ & $n$ : $0<k<2^n$) is even number.
I have tried induction but was unable to get any useful results.
| Provided that you already know $\binom{2^n}{k}$ is an integer, it suffices to show the numerator has more factors of $2$ than the denominator. We have:
$$\binom{2^n}{k}=\frac{(2^n)(2^n-1)\cdots(2^n-k+1)}{k!}.$$
There are at least $n$ factors of $2$ in the numerator because $2^n$ is a factor. So now we need to count the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3087700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 1
} |
$\sqrt{2x^2 + 3x +5} + \sqrt{2x^2-3x+5}=3x$ Problem:
solve equation
$$\sqrt{2x^2 + 3x +5} + \sqrt{2x^2-3x+5}=3x$$
I don't look for easy solution (square booth side and things like that...) I look for some tricks for "easy" solution because:
I would like to use substitution, but we have $3x$ and $-3x$, but I can't see... | Render
$(\sqrt{2x^2 + 3x +5} + \sqrt{2x^2-3x+5})(\sqrt{2x^2 + 3x +5} - \sqrt{2x^2-3x+5})=6x$
from the difference of squares factorization. So then
$\color{blue}{\sqrt{2x^2 + 3x +5} + \sqrt{2x^2-3x+5}=3x}$
$\color{blue}{\sqrt{2x^2 + 3x +5} - \sqrt{2x^2-3x+5}=2}$
Add them up:
$2\sqrt{2x^2 + 3x +5} = 3x+2$
Then square an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3090819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Show that if $a,b,c\in \mathbb{N}$ and ${a^2+b^2+c^2}\over{abc+1}$ is an integer it is the sum of two nonzero squares I was reading about the fascinating problem in the IMO ($1988 $ #$6$) that asks:
Let $a$ and $b$ be positive integers such that $(ab+1) | (a^2+b^2)$. Show that ${a^2+b^2}\over{ab+1}$ is a perfect square... | I PROVED THE CONJECTURE
$$ x^2 + y^2 + z^2 = k + kxyz $$
Notice that we cannot have a solution with $x<0$
Got the other part. Let $x \geq y \geq z \geq 1$ be a solution giving $k$ that minimizes $x+y+z$ among positive solutions giving $k.$ Vieta jumping is just the new solution
$$ (x,y,z) \mapsto (kyz-x,y,z) $$
Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3094426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What is the value of $1+\frac{1}{2^2}+ \frac{1}{4^2}+ \frac{1}{8^2}+... $? I need to find the value of the following series:
$$ 1+\frac{1}{2^2}+ \frac{1}{4^2}+ \frac{1}{8^2}+\cdots $$
That is
$$
1+ \frac{1}{2^2}+ \frac{1}{(2^2)^2} +\frac {1}{(2^3)^2}+ \cdots$$
It is the summation of a geometric progression where all ... | $$
\begin{align}
1+\frac{1}{2^2}+ \frac{1}{4^2}+ \frac{1}{8^2}+\cdots
&=\frac{1}{(2^0)^2}+\frac{1}{(2^1)^2}+ \frac{1}{(2^2)^2}+ \frac{1}{(2^3)^2}+\cdots\\
&=\frac{1}{(2^2)^0}+\frac{1}{(2^2)^1}+ \frac{1}{(2^2)^2}+ \frac{1}{(2^2)^3}+\cdots\\
&=\sum\limits_{n=0}^{\infty}\frac{1}{\left(2^2\right)^n}\\
&=\sum\limits_{n=0}^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3096423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding $\int^{\infty}_{0}\frac{\ln^2(x)}{(1-x^2)^2} dx$
Calculate $$\int^{\infty}_{0}\frac{\ln^2(x)}{(1-x^2)^2}dx$$
I have tried to put $\displaystyle x=\frac{1}{t}$ and $\displaystyle dx=-\frac{1}{t^2}dt$
$$ \int^{\infty}_{0}\frac{t^2\ln^2(t)}{(t^2-1)^2}dt$$
$$\frac{1}{2}\int^{\infty}_{0}t\ln^2(t)\frac{2t}{(t^2-1)... | Here is yet another slight variation on a theme.
Let
$$I = \int_0^\infty \frac{\ln^2 x}{(1 - x^2)^2} \, dx$$
then
\begin{align}
I &= \int_0^1 \frac{\ln^2 x}{(1 - x^2)^2} \, dx + \int_1^\infty \frac{\ln^2 x}{(1 - x^2)^2} \, dx = \int_0^1 \frac{(1 + x^2) \ln^2 x}{(1 - x^2)^2} \, dx \tag1,
\end{align}
after a substitution... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3097187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 3
} |
Exponentiation in Non-Commutative Rings I'm fairly new to abstract algebra and in an exercise I was asked under which conditions it is true that $(ab)^n = a^nb^n$, for $a,b \in R$ and $n$ a positive integer, where $R$ is a ring. It can be easily shown that the statement holds if $R$ is commutative but I am stuck on the... | Hagen von Eitzen gave a noncommutative nonunital ring with this property in the comments. However if your ring has a unit, this is impossible.
The proof below is a mess of algebra, so let me briefly explain the idea.
The problem is that the equalities we are given are homogeneous equations of degree at least $4$, like ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3098857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Maths exam question on consecutive Pythagoras triplets Our 13 year old daughter brought home this maths question that she was asked in a "maths challenge" last week:
Q. The values of the adjacent and opposite sides in a right angle triangle of lengths 3, 4, 5 are consecutive. Obtain another triangle with consecutive ad... | $x^2+(x+1)^2=y^2 \implies (x+1)^2 = (y+x)(y-x)$
Note that $a=y+x$ and $b=y-x$ gives $x={a-b\over 2}$ and the equation becomes $(a-b+2)^2=4ab$
$2x^2 > 500^2$, $2(x+1)^2 < 1000^2\implies x \geq 354$ and $x \leq 706$ so $1706\geq a\geq 854$ and $646 \geq b\geq 0$
Let $gcd(a,b)=k$ then $(a_1k-b_1k+2)^2=4a_1b_1k^2$ meaning ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3100922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
How do I rearrange this? Hi please could some one explain the steps taken to rearange the below. I've had a dabble but don't seem to be getting it.
$$\frac{C}{\sqrt{C^2 - V^2}} $$
To
$$\frac{1}{\sqrt{1 - \frac{V^2}{C^2}}}$$
Any help would be greatly appreciated.
Thanks
| $$
\begin{align}
\frac{C}{\sqrt{C^2 - V^2}}
&=\frac{C}{\sqrt{C^2 - V^2}}\cdot\frac{1/C}{1/C}\\
&=\frac{C/C}{\frac{1}{C}\sqrt{C^2\bigg(1 - \frac{V^2}{C^2}\bigg)}}\\
&=\frac{1}{\frac{1}{C}\sqrt{C^2}\sqrt{1 - \frac{V^2}{C^2}}}\\
&=\frac{1}{\frac{C}{C}\sqrt{1 - \frac{V^2}{C^2}}}\\
&=\frac{1}{\sqrt{1 - \frac{V^2}{C^2}}}.
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3102159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
how to find the radius of convergence ? $\sum_{n=1}^{\infty} (\frac{x^n}{n} - \frac{x^{n+1}}{n+1})$ How can i find the radius of convergence ? i dont know where to start i cant use $\frac{a_n}{a_{n+1}}$ test here.
$\sum_{n=1}^{\infty} (\frac{x^n}{n} - \frac{x^{n+1}}{n+1})$
the question looks simple but i dont know how ... | This is harder, but you can use the ratio test by combining the two fractions under the same denominator.
The $n$th term is equal to: $$\frac{x^n(n+1) - x^{n+1}(n)}{n(n+1)}.$$
Now, for:
$$\lim_{n \to \infty} \left(\frac{x^{n+1}(n+2) - x^{n+2}(n+1)}{(n+1)(n+2)} \cdot \frac{n(n+1)}{x^n(n+1) - x^{n+1}(n)} \right) < 1$$
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3105040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\frac{2a}{a^2-4} - \frac{1}{a-2}-\frac{1}{a^2+2a}$
Evaluate
$$\dfrac{2a}{a^2-4} - \dfrac{1}{a-2}-\dfrac{1}{a^2+2a}$$
We have to see what their common term is. Therefrom, we can evaluate the simplified expression by canceling out.
$$a^2 - 4 = (a)^2 - (2)^2 = (a-2)(a+2) \tag {1}$$
$$a^2 +2a = a(a+2)\tag{2}$$... | Let’s simplify the first fraction.
$$\frac{2a}{a^2-4}=\frac{2a}{(a+2)(a-2)}=\frac{a+2+a-2}{(a+2)(a-2)}$$ which becomes $$\frac{a+2}{(a+2)(a-2)}+ \frac{a-2}{(a+2)(a-2)}=\frac{1}{a+2}+\frac{1}{a-2}$$ so that the entire expression is reduced to $$\frac{1}{a+2}-\frac{1}{a(a+2)}=\frac{1}{a+2}(1-\frac1a)=\frac{a-1}{a(a+2)}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3105387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
$\sum_{k=1}^{n-1}\left(1-e^{\frac{2\pi ki}{n}}\right)^{-1}$ How can I go about computing
$$
\sum_{k\ =\ 1}^{n - 1}
\left(1 - \mathrm{e}^{\large 2\pi k\mathrm{i}/n}\right)^{-1}\
{\Large ?}
$$
I originally thought that it was supposed to be the reciprocal of the sum, and I ended up with $1/n$, but now I realized that it ... | The answer is $\frac{n-1}{2}$. There are multiple ways to approach it, here is one way;
Let $\displaystyle \alpha_k = e^{i \frac{2\pi k}{n}}$ for $k = 1, \dots, n-1$. We seek to evaluate,
$$\sum_{k=1}^{n-1} \frac{1}{1-\alpha_k}, \ \ (*) $$
We know that $\alpha_1, \dots , \alpha_{n-1}$ are roots of the polynomial,
$$P(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3106914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Better proof for $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$ Prove this $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$
My attempt
Proof - by using [axiomdistributive] and [axiommulcommutative]:
$$\begin{split}
&(x+y)(x^2 - xy + y^2)\\
&= (x+y)x^2 - (x+y)xy + (x+y)y^2\\
&= (x^3+x^2y) - (x^2y+xy^2) + (xy^2+y^3)\\
&= x^3 + x... | In this case of $n=3$ your result follows from this other theorem by considering $x^3+y^3=x^3-(-y)^3$. The pattern with odd $n$ is that $-y^n=(-y)^n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3110478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
prove that : $\dfrac{a^2}{2}+\dfrac{b^3}{3}+\dfrac{c^6}{6} \geq abc$ for $a ,b ,c \in \mathbb{R}^{>0}$ prove that : $\dfrac{a^2}{2}+\dfrac{b^3}{3}+\dfrac{c^6}{6} \geq abc$ for $a ,b ,c \in \mathbb{R}^{>0}$ .
I think that must I use from $\dfrac{a^2}{2}+\dfrac{b^2}{2} \geq ab$ but no result please help me .!
| I naturally want to generalize this.
Here's a first cut.
$\begin{array}\\
\dfrac{a^u}{u}+\dfrac{b^v}{v}
+\dfrac{c^{uv}}{uv}
&=\dfrac{va^u+ub^v+c^{uv}}{uv}\\
&=\dfrac{va^u+ub^v+c^{uv}}{u+v+1}\dfrac{u+v+1}{uv}\\
&\ge (a^{uv}b^{uv}c^{uv})^{1/(u+v+1)}\dfrac{u+v+1}{uv}\\
&= (abc)^{uv/(u+v+1)}\dfrac{u+v+1}{uv}\\
&=\dfrac{(ab... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3110691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
how to solve this two limit tasks Hello i stumbled across this two limits task and i cant find an answer to them:
*
*Find the limit depending on the parameter $A$
$$\lim \limits_{x\to\infty}\left(\left(\sqrt{x+1} - \sqrt[4]{x^2 + x + 1} \right) \cdot x^A\right)$$
I tried by multipliying with
$$\frac{\sqrt{x+1} ... | You should be aware of Taylor expansion. Substitute $x=1/t$, so $t\to0^+$ and the limit can be rewritten as
$$
\lim_{t\to0^+}\frac{\sqrt{1+t}-\sqrt[4]{1+t+t^2}}{t^{A+1/2}}
$$
The numerator can be rewritten as
$$
1+\frac{1}{2}t-1-\frac{1}{4}{t}+o(t)
$$
So the limit is $1/2$ when $A+1/2=1$. What if $A>-1/2$ or $A<-1/2$?
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3119427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $a_n=n\left(1-\frac{1}{n}\right)^{n[\log n]}$, prove $1\leqslant \liminf a_n$ and $\limsup a_n\leqslant e.$
Prove that:
$$\limsup n\left(1-\frac{1}{n}\right)^{n[\log n]} \leqslant e$$
and
$$\liminf n\left(1-\frac{1}{n}\right)^{n[\log n]} \geqslant 1.$$
Attempt. Since for $n>2$:
$$1<[\log n]<\log n+1\leqsla... | First, note that $\log n-1 < [\log n] \leq \log n$. Now, $0<1- \frac{1}{n}<1$, so we have
$$ n \left( 1- \frac{1}{n} \right)^{n \log n} \leq n \left( 1- \frac{1}{n} \right)^{n [\log n]} < n \left( 1- \frac{1}{n} \right)^{n (\log n - 1)}. $$
We now want to use the fact that $\left(1- \frac{1}{n}\right)^{n} \to e^{-1}$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3121393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Strategy for solving $\sqrt{3}\cos(2x) - \sin(x)\cos(x) = 1$ I believe the way to solve the following equation is to use the "R-formula":
$$ \sqrt{3}\cos(2x) - \sin(x)\cos(x) = 1 $$
If so, it should be rewritten as:
$ R\cos(x-\mathcal{L})$ or $R(\cos x\cos\mathcal{L}+\sin x\sin\mathcal{L})$.
With coefficients equated a... | One more answer won't do any harm, hopefully. Here's another way to look at this "thing".
Once you replaced $\sin x \cos x$ with $\frac{1}{2}\sin 2x$, your equation becomes
$$\sqrt 3 \cos 2x - \frac{1}{2}\sin 2x = 1.$$
Recalling that cosine and sine are just abscissa and ordinate of points on the circumference of radiu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3123017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Am I properly simplifying this geometric progression? I'm studying recurrence relations and am given:
$T(n) = 2 \cdot T(n-1) - 1$
with an initial condition that $T(1) = 3$.
I worked through the first few recurrences:
$T(n-1) = 2^2 \cdot T(n-2) - 2 - 1$
$T(n-2) = 2^3 \cdot T(n-3) - 2^2 - 2 - 1$
and so forth, and came th... | You've gotten an answer to your specific question, but I want to explain how to correctly solve the question in a mathematically rigorous manner, and not appealing to handwaving. (Every occurrence of "···" is an instance of lack of rigour.)
Working 1
$T(n) - 2·T(n-1) = -1$.
$2 · ( T(n-1) - 2·T(n-2) ) = 2·-1$. // We ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3124039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 3,
"answer_id": 0
} |
Find maximum value by using AM-GM inequality I have a problem: Find the maximum value of $P=\frac{x^3y^4z^3}{(x^4+y^4)(xy+z^2)^3}+\frac{y^3z^4x^3}{(y^4+z^4)(yz+x^2)^3}+\frac{z^3x^4y^3}{(z^4+x^4)(zx+y^2)^3}$ with $x,y,z>0$.
Is there anyway to solve this problem by using the AM-GM inequality ? Thank for your answer.
The ... | For $x=y=z=1$ we get $P=\frac{3}{16}.$
We'll prove that it's a maximal value.
Indeed, by AM-GM $$\sum_{cyc}\frac{x^3y^4z^3}{(x^4+y^4)(xy+z^2)^3}\leq\sum_{cyc}\frac{x^3y^4z^3}{(x^4+y^4)(2\sqrt{xyz^2})^3}=\sum_{cyc}\frac{\sqrt{x^3y^5}}{8(x^4+y^4)}.$$
Let $\sqrt{\frac{x}{y}}=a$, $\sqrt{\frac{y}{z}}=b$ and $\sqrt{\frac{z}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3127089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find $m>1/2$ that minimizes the area of the triangle formed by lines $y=10-2x$, $y=mx$, $y=-x/m$ I am tackling this problem below:
A triangle is formed by the three lines
$$\begin{align}
y &=10-2x \\
y &= mx \\
y &=-\frac{x}{m}
\end{align}$$ where $m>\frac{1}{2}$. Find the value of $m$ for which the area of the ... | Let $\ell_1$ be the line with equation $y = 10-2x$, let $\ell_2$ be the line with equation $y = mx$, and let $\ell_3$ be the line with equation $y = \frac{-x}{m}$. Then the lines $\ell_1$ and $\ell_2$ intersect at
$$P = \left( \frac{10}{m+2}, \frac{10m}{m+2} \right),$$
and the lines the lines $\ell_1$ and $\ell_3$ in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3130918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
In an equilateral $\triangle ABC$ : $ DB^2 + DC^2 + BC^2 = 100 $ I have a question that:
There is a point $D$ inside the equilateral triangle ABC. If
$$ DB^2 + DC^2 + BC^2 = 100 $$
and the area of $DBC$ is $ 5 \sqrt{3} $, find $AD^2$.
This is what I tried: $DB = x ,\: DC = y,\: BC = a$. Then $$ x^2 + y^2 + a^2 = 100 ... | Denote $x= AD$, $y= BD$, $z=CD$, $a=BC(=AB=AC)$ and $\alpha=\angle BDC$. So we have: $$(*)\ y^2+z^2+a^2=100,$$
by the area of $\triangle DBC$ condition we have:
$$(**)\ yz\sin\alpha=10\sqrt 3,$$
and by cosine rule on $\triangle DBC$ we have:
$$(***)\ a^2=y^2+z^2-2yz\cos\alpha.$$
Consider the rotation $\mathcal R_{C,60^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3134514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Compute $\lim\limits_{x\to \infty} \left(\sqrt{\left(x+\frac{1}{\sqrt x}\right) ^3} - \sqrt{x^3} \right) $ Compute $\lim\limits_{x\to \infty} \left(\sqrt{\left(x+\frac{1}{\sqrt x}\right) ^3} - \sqrt{x^3} \right) $.
I tried multiplying by the conjugate and I get to an ugly expression. What should I do?
| So, you arrived at
$$\begin{align}
\sqrt{(x+1/\sqrt x)^3}-\sqrt{x^3}&=\frac{3x^{3/2}+3+x^{-3/2}}{\sqrt{(x+1/\sqrt x)^3}+\sqrt{x^3}}\\\\
&=\frac{3x^{3/2}+3+x^{-3/2}}{x^{3/2}\left(1+\sqrt{(1+1/x^{3/2})^3}\right)}
\end{align}$$
Now let $x\to \infty$. Can you finish now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3136527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Variance of a Brownian motion Let $\{X(t), t \geq 0\}$ be a Brownian motion with drift parameter $\mu = 3$ and variance parameter $\sigma^2 = 9$. If $X(0) = 10$, find $P(X(0.5) > 10)$.
First, I calculated the expectation and variance of $X(0.5)).$ Since $X(0.5) - X(0)$ is normal with mean $1.5$ and variance $4.5$, it f... | It should be $4.5$. There may be an error in the answer key. The variance of the deterministic part is $0$ and doesn't add to the variance of the increment like you wrote.
At $X(0.5)$, we have $X(0.5) = 10 + 3 * 0.5 + 3\sqrt{0.5}Z$, where $Z$ is a standard normal random variable ($m = 0$, $s^2 = 1$). And so to calcula... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3137411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove $\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\ge2$ where a, b, c and d are positive real numbers
I have to prove the following inequality using the Cauchy-Schwarz inequality:
$$\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\ge2$$
where a, b, c and d are positive real numbers.
But I am ... | Use $$xy\leq {(x+y)^2\over 4}$$
if $x,y\geq 0$, so
$$\frac{a}{b+c}+\frac{c}{d+a} = {a(a+d)+c(b+c)\over (a+d)(b+c)} \geq 4{a^2+c^2+ad+bc\over (a+b+c+d)^2}$$
and similary $$\frac{b}{c+d}+\frac{d}{a+b}\ge 4{b^2+d^2+ab+dc\over (a+b+c+d)^2}$$
So $$...\geq 4{a^2+c^2+ad+bc+b^2+d^2+ab+dc\over (a+b+c+d)^2}\geq 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3137621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find the functions $f:\mathbb{R}\rightarrow\mathbb{R}$, that have primitives, such that $F(x-1)+f(1-x)=x^3$ Find the functions $f:\mathbb{R}\rightarrow\mathbb{R}$, that have primitives $F:\mathbb{R}\to\mathbb{R}$, such that $F(x-1)+f(1-x)=x^3$ for any $x\in\mathbb{R}$. For $1-x$ instead of $x$, $F(-x)+f(x)=(1-x)^3$, $F... | So, we're solving for differentiable $g$ such that $g(-x) + g'(x) = (1 - x)^3$. Note that $g'(x) = (1 - x)^3 - g(-x)$, and hence $g$ is twice differentiable with
$$g''(x) = -3(1 - x)^2 + g'(-x) = -3(1 - x)^2 + (1 + x)^3 - g(x)$$
That is, $g$ must satisfy the differential equation
$$y'' + y = -3(1 - x)^2 + (1 + x)^3.$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3138756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to solve these trigonometry equations? I have to work with the following 5 equations:
*
*$(1-\tan^2x)(1-\tan^22x)(1-\tan^24x)=8$
*$(2\cos 2x+1)(2\cos 6x+1)(2\cos 18x+1)=1$
*$\dfrac{\cos 2x}{\sin 3x}+\dfrac{\cos 6x}{\sin 9x}+\dfrac{\cos 18x}{\sin 27x}=0$
*$\dfrac{\cos x}{\sin 3x}+\dfrac{\cos 3x}{\sin 9x}+\dfra... | In the first you can get it by the following way:
$$LS=\frac{2\tan{x}}{\tan{2x}}\cdot\frac{2\tan{2x}}{\tan{4x}}\frac{2\tan{4x}}{\tan{8x}}=\frac{8\tan{x}}{\tan{8x}}$$ and the rest is smooth.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3140794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find sum $ \sum\limits_{k=2}^{2^{2^n}} \frac{1}{2^{\lfloor \log_2k \rfloor} \cdot 4^{\lfloor \log_2(\log_2k )\rfloor}} $ Calculate sum
$$ \sum_{k=2}^{2^{2^n}} \frac{1}{2^{\lfloor \log_2k \rfloor} \cdot 4^{\lfloor \log_2(\log_2k )\rfloor}} $$
I hope to solve this in use of Iverson notation:
my try
$$ \sum_{k=2}^{2^{2^n... | Let's write
$$\sum_{k=2}^{2^{2^n}} \frac{1}{2^{\lfloor \log_2(k) \rfloor}4^{\lfloor \log_2(\log_2(k))\rfloor}} = \sum_{i=0}^{n-1} \sum_{k=2^{2^i}}^{2^{2^{i+1}}-1} \frac{1}{2^{\lfloor \log_2(k) \rfloor}4^{\lfloor \log_2(\log_2(k))\rfloor}} + \frac{1}{2^{2^n}4^{n}}$$
$$= \sum_{i=0}^{n-1} \sum_{k=2^{2^i}}^{2^{2^{i+1}}-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3141510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
$a,b,c\in\Bbb R^+, x,y,z\in \Bbb R, $ show that $\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c} \ge \frac{(x+y+z)^2}{a+b+c}$ $a,b,c\in\Bbb R^+, x,y,z\in \Bbb R, $ show that $\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c} \ge \frac{(x+y+z)^2}{a+b+c}$ (use Cauchy–Schwarz inequality)
I have trouble finding the two vectors. Is it $(... | Hint: Use Cauchy Schwarz in Engelform:
$$\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\geq \frac{(x+y+z)^2}{a+b+c}$$
It is equivalent to $${z}^{2}{a}^{2}b+{y}^{2}{a}^{2}c+{z}^{2}a{b}^{2}-2\,abcxy-2\,abcxz-2\,a
bcyz+{y}^{2}a{c}^{2}+{x}^{2}{b}^{2}c+{x}^{2}b{c}^{2}
\geq 0$$
and this is $$a(bz-cy)^2+b(az-xc)^2+c(bx-ay)^2\geq 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3142310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Confusion in this limits problem
Evaluate $$\lim\limits_{n\to \infty} \frac{1^2 + 2^2 +...+n^2}{n^3}$$
This is the original method to solve this is:
Taking summation of the square numbers $1^2 + 2^2 + 3^2 +...+n^2 = \frac{1}{6}n(n+1)(2n+1)$
$$\lim\limits_{n\to \infty} \frac{1^2 + 2^2 +...+n^2}{n^3} = \frac{\frac{1}{6... | Converting the limit into an integral is one right way to evaluate it.
$$\lim_{n\to \infty}\dfrac{1^2+2^2+\cdots+n^2}{n^3}=\lim_{n\to \infty}\dfrac{1}{n}\sum_{k=1}^{n}\left(\dfrac{k}{n}\right)^2=\int_{0}^{1}x^2\mathrm dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3145200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
How to show that $\cos(x+y)=\cos(x)\cos(y)−\sin(x)\sin(y)$ by using power series? Original question:Proofs of $\cos(x+y) = \cos x\cos y - \sin x \sin y$
I would like to know an answer to the question linked above by using power series.
I tried to expand the $(x+y)^{2n}$ by using the binomial theorem but it didn't lead ... | Let
$$C_k = \begin{cases}
1 & k \equiv 0 \pmod 4 \\
0 & k \equiv 1 \pmod 4 \\
-1 & k \equiv 2 \pmod 4 \\
0 & k \equiv 3 \pmod 4 \\
\end{cases}$$
$$S_k = \begin{cases}
0 & k \equiv 0 \pmod 4 \\
1 & k \equiv 1 \pmod 4 \\
0 & k \equiv 2 \pmod 4 \\
-1 & k \equiv 3 \pmod 4 \\
\end{cases}$$
Then the series is
$$\cos(t) = \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3146117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Define the sequences $x_n$& $y_n$ for $n \geq 1$ using the relations $x_n = x_{n+1} +2y_{n+1}$ & $y_n = y_{n-1} +x_{n-1}$ $?$ Define the sequences $x_n$& $y_n$ for $n \geq 1$ using the relations $x_n = x_{n+1} +2y_{n+1}$ & $y_n = y_{n-1} +x_{n-1}$ $?$
$:$
$n \geq 1$ , $x_0,y_0 \epsilon \mathbb{Z}_{+} $
$x_n = x_{n+... | Hint.
We can arrange it as
$$
\left[
\begin{array}{c}
x_n\\
y_n
\end{array}
\right] =
\left[
\begin{array}{cc}
-1 & -2\\
1 & 1
\end{array}
\right]\left[
\begin{array}{c}
x_{n-1}\\
y_{n-1}
\end{array}
\right]
$$
NOTE
For $A = \left[
\begin{array}{cc}
-1 & -2\\
1 & 1
\end{array}
\right]$
we have
$$
\left[
\begin{array}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3149276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Value of $(p,q)$ in indefinite integration Finding value of $(p,q)$ in $\displaystyle \int\frac{2x^7+3x^2}{x^{10}-2x^5+1}dx=\frac{px^3+qx^8}{x^{10}-2x^5+1}+c$
what i try
$\displaystyle \int\frac{2x^7+3x^2}{x^{10}-2x^5+1}dx$
put $x=1/t$ and $dx=-1/t^2dt$
$\displaystyle -\frac{2t^5+3x^{10}}{t^{10}-2t^5+1}dt$
How do i so... | The best way I can seem to think of is differentiating both sides of the equality that gives you the following: $$\dfrac{\mathrm d}{\mathrm dx}\int\dfrac{2x^7+3x^2}{x^{10}-2x^5+1}\mathrm dx=\dfrac{\mathrm d}{\mathrm dx}\left(\dfrac{px^3+qx^8}{x^{10}-2x^5+1}\right)$$
$$ \begin{equation}\begin{aligned} \dfrac{\mathrm d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3150009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Generating function with tough restrictions
In how many ways can a coin be flipped $25$ times in a row so that exactly $5$ heads occur and no more than $7$ tails occur consecutively?
For the heads, I think that it is $\binom{25}{5}$, but I do not know what to do with the tails restriction.
Not sure how to approach t... | More generally, we might ask how many sequences of flips of length $n$ have $m$ heads and no more than seven tails in a row. We will find a bivariate generating function for this problem.
First suppose we drop the constraint that the sequence must contain exactly $m$ heads, retaining the requirement that there be no m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Prove that $(a - 1)^3 + (b - 1)^3 + (c - 1)^3 \ge -\dfrac{3}{4}$ where $a+b+c=3$
If $a$, $b$, $c$ are non-negative numbers such that $a + b + c = 3$ then prove
$$(a - 1)^3 + (b - 1)^3 + (c - 1)^3 \ge -\frac{3}{4}$$
Here's what I did.
Let $c \ge a \ge b$.
We have that
\begin{align*}
(c - 1)^3 + (a - 1)^3 &= (c + a ... | My sketchy proof:
First you need to prove that if $x+y+z=0$, then $x^3 + y^3 + z^3 = 3xyz$
Since $a+b+c=3$, we have $(a - 1)+(b-1)+(c-1)=0$, therefore, $S = (a - 1)^3+(b-1)^3+(c-1)^3=3(a - 1)(b-1)(c-1)$
Replace $c = 3 - a -b$, we have $S = 3(a-1)(b-1)(2-a-b)$
Expand this, we have $S/3 = a^2 + b^2 + 4ab - 3(a+b) - ab(a+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3152853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Prove that $ \ln\left( \frac{1+x}{x}\right) >\frac{1}{1+x}$? How can one prove that
$$\ln\left( \frac{1+x}{x}\right) >\frac{1}{1+x} \, ?$$
| Considering $f(t) = \log t$ and applying Lagrange's theorem in the interval $[x,x+1], x > 0$ you get
$$
f(x+1) - f(x) = f'(\xi_x) (x+1-x) = \frac{1}{\xi_x}, \quad \xi_x \in ]x,x+1[
$$
But, since $f(x+1)-f(x)= \log (x+1)-\log x = \log \left(\frac{x+1}{x}\right)$ and $\frac{1}{\xi_x} > \frac{1}{x+1}$ you can conclude tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3156979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Proving the identity $(\tan^2(x)+1)(\cos^2(-x)-1)=-\tan^2(x)$ Proving the trigonometric identity $(\tan{^2x}+1)(\cos{^2(-x)}-1)=-\tan{^2x}$ has been quite the challenge. I have so far attempted using simply the basic trigonometric identities based on the Pythagorean Theorem. I am unsure if these basic identities are un... | Recall:
$$\begin{align}
\tan^2(x) - \sec^2(x) &= -1 \\
\sin^2(x) + \cos^2(x) &= 1
\end{align}$$
Thus,
$$\begin{align}
\tan^2(x) + 1 &= \sec^2(x)\\
\cos^2(x) - 1 &= -\sin^2(x)
\end{align}$$
We also note that $\cos(x)$ is an even function, and thus $\cos(-x) = \cos(x)$. Thus, the formula becomes:
$$(\tan^2(x) + 1)(\cos^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3157886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
series sum $\frac{1}{2}\cdot \frac{1}{2}k^2+\frac{1}{2}\cdot \frac{3}{4}\cdot k^4+\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot k^6+\cdots $ Finding value of $\displaystyle \int^{\pi}_{0}\ln(1+k\cos x)dx$ for $0<k<1$
what I try
Let $\displaystyle I =\int^{\pi}_{0}\ln(1+k\cos x)dx$
put $\displaystyle x\rightarrow ... | For any $R\geq 1$,
$$(R+e^{i\theta})(R+e^{-i\theta}) = (R^2+1)+2R\cos\theta \tag{1}$$
$$ 1+\frac{2R}{R^2+1}\cos\theta = \frac{R^2}{R^2+1}\left(1+\frac{e^{i\theta}}{R}\right)\left(1+\frac{e^{-i\theta}}{R}\right)\tag{2}$$
$$ \log\left(1+\tfrac{2R}{R^2+1}\cos\theta\right) = \log\left(\tfrac{R^2}{R^2+1}\right)+2\sum_{n\geq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3161755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Proving by induction of $n$ that $\sum_{k=1}^n \frac {k+2}{k(k+1)2^{k+1}} = \frac{1}{2} \ - \frac1{(n+1)2^{n+1}} $
$$
\sum_{k=1}^n \frac {k+2}{k(k+1)2^{k+1}} = \frac{1}{2} \ - \frac1{(n+1)2^{n+1}}
$$
Base Case:
I did $n = 1$, so..
LHS-
$$\sum_{k=1}^n \frac {k+2}{k(k+1)2^{k+1}} = \frac3{8}$$
RHS-
$$\frac{1}{2} \ - \fr... | Your error is just after the sixth step from the bottom:
$$\frac{1}2 -\frac {1}{(n+1)2^{n+1}} +\frac{1}{(n+1)2^{n+1}} - \frac{1}{(n+2)2^{n+2}}=\frac{1}2 -\frac{1}{(n+2)2^{n+2}}$$
Then you are done.
You accidentally added the two middle terms instead of subtracting.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3162553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How to perform polynomial long division on 1/(1 - x)? How do I perform polynomial long division on $\frac{1}{1 - x}$ to obtain the sequence $1 + x + x^2 + x^3 + \cdots$?
In this video, the teacher went about it in the following way...
$$
\require{enclose}
\begin{array}{r}
1 + x + x^2 + x^3 + \cdots \\
1 - x \enclose... | You do it when $x$ is small compared to $1$. That is not as silly a remark as it sounds like it is. If you stop the video's approach part way through you get a remainder term just like you might with the top down approach. You could write $$\frac 1{1-x}=1+x+x^2+\frac {x^3}{1-x}$$
This is an algebraic fact, valid for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3164061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
can not find root of the equation $(0,0)$ with semi-major axis a and semi-minor axis $b$">
I want find out for each point $(x,y)$ in the white region, the nearest point $(a \cos \theta, b \sin \theta )$ on the ellipse curve.I tried with the following approach.
$$
\text {distance} = \sqrt{(x-a \cos \theta)^2+(y-b \sin \... | You have $axsec(\theta)- bycsc(\theta)= a^2- b^2$. Let $z= sec(\theta)$. Since $csc(\theta)= \frac{1}{sec(\theta)}$, the equation becomes $axz- \frac{by}{z}= a^2- b^2$. Muliply on both sides to get $axz^2- by= (a^2- b^2)z$.
We can write that as $(ax)z^3+ (b^2- a^2)z- by= 0$ and solve it using the quadratic formula: ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3165762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Sum of Infinite series $\frac{1.3}{2}+\frac{3.5}{2^2}+\frac{5.7}{2^3}+\frac{7.9}{2^4}+......$ Prove that the sum of the infinite series $\frac{1.3}{2}+\frac{3.5}{2^2}+\frac{5.7}{2^3}+\frac{7.9}{2^4}+......$ is 23.
My approach
I got the following term
$S_n=\sum_1^\infty\frac{4n^2}{2^n}-\sum_1^\infty\frac{1}{2^n}$.
For ... | Hint. One may used this
$$\sum_{n\ge1}n^{2}x^n=\frac{x}{(1-x)^{2}}+\frac{2x^{2}}{(1-x)^{3}},\qquad |x|<1,$$ proved for example here.
Hope you can take from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3165878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int_a^b\frac{1}{ x^2}dx$ using limit of a sum definition
From the definition of a definite integral as the limit of a sum,
evaluate $$\int_a^b\frac{1}{ x^2}dx$$
Step 1
To simplify working with $x^2$, divide the interval $[a,b]$ using variable length intervals: $$[a,b]=\bigcup_{j=1}^n\bigg[a+\frac{\sqrt ... |
We obtain for $0<a\leq b$:
\begin{align*}
\color{blue}{\int_a^b\frac{1}{x^2}\,dx}&=\lim_{n\to\infty}\sum_{j=1}^nf\left(a+j\frac{b-a}{n}\right)\frac{b-a}{n}\\
&=\lim_{n\to\infty}\sum_{j=1}^n\frac{1}{\left(a+j\frac{b-a}{n}\right)^2}\cdot\frac{b-a}{n}\\
&\,\,\color{blue}{=\lim_{n\to\infty}\frac{n}{b-a}\sum_{j=1}^n\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3168351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why do these laurent series approaches conflict? I was working on a problem of finding the Laurent series of $\frac{1}{z-3}$ that converges where $|z-4| > 1$
So I had one approach, let $u=z-4$ then:
$$\frac{1}{z-3} = \frac{1}{1+u} $$
$$ = \frac{1}{u} - \frac{1}{u^2} + \frac{1}{u^3}...$$
$$ = \frac{1}{z-4} - \frac{1}... | $$
\begin{align}
\frac1{z-4}\frac1{1+\frac1{z-4}}
&=\frac1{z-4}\left(1-\frac1{z-4}+\frac1{(z-4)^2}-\dots\right)\\
&=\frac1{z-4}-\frac1{(z-4)^2}+\frac1{(z-4)^3}-\dots
\end{align}
$$
The series starts with $\frac1{z-4}$, not $\frac1{(z-4)^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3175400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Numbers of roots in Quadratic Equations If $a,b,c,d$ are real numbers, then show that the equation $$(x^2 +ax -3b)(x^2-cx+b)(x^2 -dx +2b)=0$$ has at least two real roots.
| Just line them up and shoot them.
The roots of $x^2 +ax -3b=0$ will also be roots of $(x^2 +ax -3b)(x^2-cx+b)(x^2 -dx +2b)=0$. So if $x^2 +a -3b = 0$ has two real roots then $(x^2 +ax -3b)(x^2-cx+b)(x^2 -dx +2b)=0$ will have at least two real root.
The roots of $x^2 +ax -3b=0$ will, by the quadratic formula be, $\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3177482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Basic combinations logic doubt in probability
"If $3$ students are chosen at random from a class with $6$ girls and $4$ boys, what is the probability that all $3$ students chosen will be girls?"
$\left(\dfrac{6}{10}\right)\left(\dfrac{5}{9}\right)\left(\dfrac{4}{8}\right)$
So why can't we use that logic to answer thi... | What you calculated is the probability of selecting two red marbles, two blue marbles, and one green marble in that order. However, if we select blue, green, red, red, blue, we still get two red marbles, two blue marbles, and one green marble.
To correct your attempt, we must multiply by the number of orders in whic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3177997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Show that the wave equation takes the form $\frac{\partial^2 u}{\partial r \partial s}$ Show that the wave equation $\frac{\partial^2 u}{\partial t^2} - a^2 \frac{\partial^2 u}{\partial x^2} = 0$ takes the form $\frac{\partial^2 u}{\partial r \partial s} = 0$ under the change of variable $r = x + at$, $s = x - at$.
I ... | First for $r =x + at$ and $s=x-at$, we have $x=\frac{1}{2}(r+s)$ and $t=\frac{1}{2a}(r-s)$. Using chain rule for partial derivatives.
For $u:=u(r,s)$, we have
\begin{align*}
u_s &= u_x \frac{dx}{ds} + u_t \frac{dt}{ds} = \frac{1}{2}u_x -\frac{1}{2a} u_t \\
&\Rightarrow u_{rs} = \frac{1}{2}\bigg(u_{xx} \frac{dx}{dr} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3183240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $y^2-x^2\bigm|2^ky-1$ and $2^k-1\bigm|y-1$ then $y=2^k$ and $x=1$ Suppose that $k\geq2$ and $0<x<y$ and $y^2-x^2\bigm|2^ky-1$ and $2^k-1\bigm|y-1$. Is it necessarily the case that $x=1$ and $y=2^k$?
Equivalently (I prove equivalence at the end): Suppose that $k\geq2$ and $m\geq1$ and suppose that there are two posit... | Too long to comment:
It is necessary that
$$y=\left\lfloor 1+\frac{x-1}{2^k-1}\right\rfloor(2^k-1)+1$$
Proof :
We can write
$$y-1=m(2^k-1)\tag1$$
where $m$ is a positive integer.
Also, $$y^2-x^2\mid 2^ky-1$$
implies $$2^ky-1-(y^2-x^2)\ge 0\tag2$$
From $(1)(2)$, we get
$$2^k(m2^k-m+1)-1-(m2^k-m+1)^2+x^2\ge 0,$$
i.e.
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3184704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 0
} |
Prove with induction $\sum_{k=1}^n\frac{1}{(2k-1)(2k+1)(2k+3)}=\frac{n(n+2)}{3(2n+1)(2n+3)}$ Prove with induction the identity
$\sum_{k=1}^n\frac{1}{(2k-1)(2k+1)(2k+3)}=\frac{n(n+2)}{3(2n+1)(2n+3)}$
How can I solve this problem?
Should i set k= (p+1) and n = (p+1), then try to get the left side equal to the right side?... | When $n=1$,
$$\begin {align}
\sum_{k=1}^n\frac 1{(2k-1)(2k+1)(2k+3)}&=\frac 1{(2.1-1)(2.1+1)(2.1+3)}\\
&=\frac 1{15}\\
&=\frac{1(1+2)}{3(2.1+1)(2.1+3)}\\
&=\frac{n(n+2)}{3(2n+1)(2n+3)}
\end {align}$$
Assume the result to be true for $n=m$
We show it is also true for $n=m+1$,
$$\begin {align}
\sum_{k=1}^{m+1}\frac 1{(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3186411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
If $\tan 9\theta = 3/4$, then find the value of $3\csc 3\theta - 4\sec 3\theta$.
If $\tan9\theta=\dfrac{3}{4}$, where $0<\theta<\dfrac{\pi}{18}$, then find the value of $3\csc 3\theta - 4\sec 3\theta$.
My approach:-
$$\begin{align*} \tan9\theta &=\frac{3}{4} \\[6pt] \implies \theta & = \frac{37^{\circ}}{3} \end{align... | Consider the right triangle with sides $3,4,5$, whose angle opposite the side $3$ is $9\theta$. Then:
$$\sin 9\theta =\frac35 \Rightarrow 5\sin 9\theta=3; \\
\cos 9\theta =\frac45 \Rightarrow 5\cos 9\theta =4;\\
3\csc 3\theta - 4\sec 3\theta=\frac{3}{\sin 3\theta}-\frac4{\cos 3\theta}=\frac{3\cot 3\theta-4\sin 3\theta}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3187142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Solve $x^{5} \equiv 2$ mod $221\ $ [Taking modular $k$'th roots if unique] We know that $221 = 17*13$. So we can check if the system has roots to both of those equations separately, which it does:
$x^{5} \equiv 2$ mod $13$ has the solution $6 + 13n$ and $x^{5} \equiv 2$ mod $17$ has the solution $15 + 17n$.
I got thes... | We work all the time modulo $221$ from now on tacitly. Since $2$ is relatively prime to $221$, a/the solution $x$ of $x^5=2$ (modulo $221$ - i write equivalences modulo $221$ as equalities from now on) is also relatively prime to $221$. The Euler indicator function of $221$ is $\varphi(221)=\varphi(17\cdot13)=\varphi(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3187793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find the sum: $\sum_{n=1}^{\infty} \ln\left(\frac{b+n+1}{a+n+1}\right)$ Let $0<a<b$, I would like to compute the sum
$$\sum_{n=1}^{\infty} \ln\left(\frac{b+n+1}{a+n+1}\right).$$
But first I am worrying that a test convergence might lead to the divergence of this series
What do I miss here?
$$\begin{split}\sum_{n=1}^{... | Btw., the series does not only seem divergent but it is divergent indeed as can be seen as follows:
$$\ln\left(\frac{b+n+1}{a+n+1}\right) = \ln (b+n+1) - \ln(a+n+1) \stackrel{a < \xi_n < b}{=}(b-a)\frac{1}{\xi_n+n+1} \geq (b-a)\frac{1}{\lceil b\rceil+n+1}$$
Hence, the given sum has a divergent tail of the harmonic seri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3188733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the binomial coefificient of $x^8$ in $(1+x^2-x^3)^9$ I was trying to solve it using the multinomial theorem.
I was trying to find which combinations could give me such $x^8$
and I came to the conclusion that it only occours when i take
$(x^2)^4$ or $(-x^3)^2*(x^2)^1$
therefore: $\binom{9!}{0!*4!*0!} + \binom{9!}... | A comparison. It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$.
Multinomial theorem:
\begin{align*}
\color{blue}{[x^8]}&\color{blue}{\left(1+x^2-x^3\right)^9}\\
&=[x^8]\sum_{{k_1+k_2+k_3}\atop{k_1,k_2,k_3\geq 0}}\binom{9}{k_1,k_2,k_3}1^{k_1}\left(x^2\right)^{k_2}\left(-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3189991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Generating function of binomial coefficients We want to evaluate the sum $$\sum_{L=0}^{\infty}\frac{1}{2}L(L+1)x^L$$
From this set of notes (page 2, equation 8) we find the formula $$\sum_{n=0}^{\infty}\binom{n}{k}y^n = \frac{y^n}{(1-y)^{n+1}}$$ which suggests that I can do $$\frac{1}{x}\sum_{L=0}^{\infty}\binom{L+1}{2... | In my humble opinion, a simpler way would to consider
$$S=\sum_{L=0}^{\infty}\frac{1}{2}L(L+1)x^L\implies T=2S=\sum_{L=0}^{\infty}L(L+1)x^L$$
$$T=\sum_{L=0}^{\infty}L(L-1+2)x^L=\sum_{L=0}^{\infty}L(L-1)x^L+2\sum_{L=0}^{\infty}Lx^L$$
$$T=x^2\sum_{L=0}^{\infty}L(L-1)x^{L-2}+2x\sum_{L=0}^{\infty}Lx^{L-1}$$
$$T=x^2 \left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3195701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to find the correct orthogonal eigenvectors for a basis of an eigenvalue? I have an exercise where I found the two correct eigenvalues $\lambda_{1} = 0$ and
$\lambda_{2} = 6$.
The algebraic multiplicity of $\lambda_{2}$ is 2. Now I try to find $E_{\lambda_{2}}$ with the following matrix. I've already done the elem... | If you just want a vector that satisfies $x=-y+2z$ and is orthogonal to $\begin{pmatrix}-1\\1\\0\end{pmatrix}$, then solve $x=-y+2z$ and (from the dot product being $0$) $-x+y=0$. You'll find that $x=y=z$ and then might as well pick $\begin{pmatrix}1\\1\\1\end{pmatrix}$. As you mentioned, if you wanted an orthonormal b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3196257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
$11x + 13 \equiv 4$ (mod 37) $11x + 13 \equiv 4$ (mod 37)
My "solution" to the problem
$11x + 13 \equiv 4$ (mod 37) $\rightarrow$
$11x + 13 = 4 + 37y $
$11x - 37y = - 9$
Euclid's algorithm.
$37 = 11*3 + 4$
$11 = 4*2 + 3$
$4 = 3*1 + 1 \rightarrow GCD(37,11) = 1$
$3 = 1*3 + 0$
Write as linear equation
$1 = 4 - 1*3$
$3 ... | We are looking for a solution to $11x\equiv-9\pmod{37}$.
First, solve $11x+37y=1$. As shown in this answer, we use the Extended Euclidean Algorithm:
$$
\begin{array}{r}
&&3&2&1&3\\\hline
1&0&1&-2&3&-11\\
0&1&-3&7&-10&37\\
37&11&4&3&1&0
\end{array}
$$
Thus,
$$
\begin{array}{c}
3\cdot37-10\cdot11=1\\
\Downarrow&\text{mul... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3198371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
what is the Prime factorisation of 28! Can someone check my solution for this question?
$28! = 2^a \times 3^b \times 5^c \times 7^d \times 11^e \times 13^f \times 17^g \times 19^h \times 23^i$
To find a:
$28/2 = 14$
$28/2^2 = 7$
$28/2^3 = 3$
$28/2^4 = 1$
Where $a= 14+7+3+1= 25$ giving $2^{25}$
I repeated this method t... | Your answer is correct and your method is fine, but not your choice of notation. It is not true that $\dfrac{28}{2^3}=3$. You are interested in integer division here, and therefore what you should write is $\left\lfloor\dfrac{28}{2^3}\right\rfloor=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3202148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int _0^{2\pi }\frac{\cos (n\theta) }{a+\cos\theta}\,d\theta$ with $a>1$, $n\in \mathbb{N}-\left\{0\right\}$
Evaluate $$\int _0^{2\pi }\frac{\cos (n\theta) }{a+\cos\theta}\,d\theta,\quad\,a>1$$
I wrote $$f\left(z\right)=\frac{\frac{1}{2}\left(z^n+z^{-n}\right)}{\frac{iz^2}{2}+aiz+\frac{i}{2}}$$
The discrimi... | As cosine is an even function, integral from o to 2$\pi$ can be written as 2 times the the integral from 0 to $\pi$, So you would be left with the upper semicircle and hence take $\frac{-ai + i \sqrt{a^2 -1}}{2}$. This is because in the upper plane imaginary part $Im(z) >= 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3204616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Help with proof that if q is a prime divisor of $\frac{n^p-1}{n-1}$, then either q=p or $q\equiv1 \pmod p$ Here is the proof:
Why does $(\frac{n^p-1}{n-1},n-1)=(p,n-1)$? Unless n=1 I think it should be the geometric sum $\sum_{i=0}^{p-1} n^i$.
| Assuming $p$ is and odd prime, this is a result of the more general fact
$$\gcd\left(\frac{a^p + b^p}{a + b}, a + b\right) = \gcd(p, a + b)$$ when $\gcd(a,b) = 1$, which can be shown using the binomial theorem, for instance:
$$\begin{align*}\frac{a^p + b^p}{a + b} & = \frac{((a + b) - b)^p + b^p}{a + b} \\ & = \frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3210539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What's the remainder when $x^{7} + x^{27} + x^{47} +x^{67} + x^{87}$ is divided by $x ^ 3 - x$ What's the remainder when $x^{7} + x^{27} + x^{47} +x^{67} + x^{87}$ is divided by $x ^ 3 - x$ in terms of $x$?I tried factoring $x$ from both polynomials but I don't know what to do next since there'd be a $1$ in the second ... | Doing long division:
$$P(x)=\frac{x^{7} + x^{27} + x^{47} +x^{67} + x^{87}}{x^3-x}=\frac{x^{86}+x^{66}+x^{46}+x^{26}+x^6}{x^2-1}=\\
\frac{\sum_{i=0}^{9}(x^{86-2i}-x^{84-2i})+2\sum_{i=0}^{9}(x^{66-2i}-x^{64-2i})+3\sum_{i=0}^{9}(x^{46-2i}-x^{44-2i})+4\sum_{i=0}^{9}(x^{26-2i}-x^{24-2i})+5\sum_{i=0}^{2}(x^{6-2i}-x^{4-2i})+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3212443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Smallest square to cover a right triangle I believe smallest possible triangle to cover a square is well studied... But how about square covering a triangle? I read the post here, but I think that I don't want such a general question (and complicated solution), just right triangle.
I first encounter the problem, which ... | This answer shows that $\frac{144}{\sqrt{193}}$ is the smallest possible side length.
(This is an answer to the question "why is it the smallest possible square covering the triangle?", but I have to say that this is not an answer to the question "Why do we need only to consider these three cases?". Anyway, I hope this... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3212942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 5,
"answer_id": 2
} |
Proving $\tan^{-1}\frac{1}{2\cdot1^{2}}+\tan^{-1}\frac{1}{2\cdot2^{2}}+\cdots+\tan^{-1}\frac{1}{2\cdot n^{2}}=\frac{\pi}{4}-\tan^{-1}\frac{1}{2n+1}$
By using Mathematical Induction, prove the following equation for all positive integers $n$:
$$\tan^{-1}\frac{1}{2 \cdot 1^{2}} + \tan^{-1}\frac{1}{2 \cdot 2^{2}} +\cdo... | I don't know what you did wrong but you should get the required result using the given identity:
$$\begin{align}
\arctan{\left(\frac{1}{2k+1}\right)}-\arctan{\left(\frac{1}{2(k+1)^2}\right)}
&=\arctan{\left(\frac{\frac{1}{2k+1}-\frac{1}{2(k+1)^2}}{1+\left(\frac{1}{2k+1}\right)\left(\frac{1}{2(k+1)^2}\right)}\right)}\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3214532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What is $\cos\left(\frac{\arctan(x)}{3}\right)$ and $\sin\left(\frac{\arctan(x)}{3}\right)$? I know that $$\cos(\dfrac{\pi}{3} - \arctan(x))= \dfrac{1}{2\sqrt{(1+x^2)}} + \dfrac{\sqrt{3}x}{2\sqrt{(1+x^2)}}$$
$\cos\left(\dfrac{\pi}{3} - \dfrac{\arctan(x)}{3}\right)$ = ?
$\cos\left(\dfrac{\pi}{3} - \dfrac{\arctan(x)}{3}\... | Let us care about
$$t:=\tan\left(\frac{\arctan(x)}3\right), $$
using the fact that
$$\tan\left(3\frac{\arctan(x)}3\right)=x.$$
By the triple angle formula, this equation writes
$$\frac{3t-t^3}{1-3t^2}=x$$
or
$$t^3-3xt^2-3t+x=0.$$
We depress it with $u:=t-x$, giving
$$u^3-3(x^2+1)u-2x(x^2+1)=0.$$
Now the discriminant is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3215525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Describe the region $\{|z^2 - 1|<1\}$ of the complex plane Since it's complex analysis, I assume there is an easy solution without much algebra. But even in my algebra heavy solution there should be a more streamlined approach... Either approach is appreciated as an answer, or verification that my approach is the sta... | I can't say if this is "easier" or not, but here is my approach.
Let $w = z^2$. Then $\{w: |w-1|<1\}$ is a circle of radius $1$ centered at $(1,0)$. The mapping $w: z \mapsto z^2$ transforms the original set into this circle.
In polar form, $z = r_ze^{i\theta_z}$, and $w = r_we^{i\theta_w} = r_z^2e^{i2\theta_z}$.
You c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3217213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving $\cos^2x+\sin^2y=1$ is reflexive, symmetric, transitive. I want to make sure that I got the hang of the following relations.
For reflexivity, if $x=y,\cos^2x+\sin^2y=\cos^2x+\sin^2x=1 \implies xRx$, then it is reflexive.
For symmetry, $xRy\implies\cos^2x+\sin^2y=1$ and $yRx\implies\cos^2 y+\sin^2x=1 \implies$ s... | Reflexive
$cos^2x+sin^2x=1$
symmetric
Suppose that $cos^2x+sin^2y=1$, $cos^2y+sin^2x=1-sin^2y+1-cos^2x=2-(cos^2x+sin^2y)=2-1=1$
Transitive
$cos^2x+sin^2y=1, cos^2y+sin^2z=1$
$cos^2x+sin^2z=cos^2x+1-cos^2y=cos^2x+sin^2y=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3217317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $a \equiv 9 \pmod {12}$, find all possible values for $\gcd(a^2+21a+72,252)$ We know that
$$a^2+21a+72 \equiv 9^2 + 21 \cdot 9 + 0 \equiv 6 \pmod {12}$$
So we know that that expression, let's say $\alpha$, is such that $12 \mid \alpha - 6$. But then $12 \mid 2(\alpha - 6)=2a+12$, which means that $\alpha$'s prime fa... | Note $\ (f,\,\overbrace{ 9\cdot 7\cdot 4}^{\large 252}) = (f,9)(f,7)(f,4) =\, 9\cdot \color{#c00}1\cdot\color{#0a0}2\ $ for $\ f = a^2+21a+72\ $ by
$3\mid a\ \Rightarrow\ 9\mid a^2+21a+72\ \Rightarrow\ (f,9) = \color{}9$
$\!(f,7)\, =\, (f\bmod 7,\,7)\, =\, (a^2\!+2,\,7) =\,\color{#c00} 1,\ $ by $\ x^2\not\equiv -2\pmo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3226356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to factorized this 4th degree polynomial? I need your help to this polynomial's factorization.
Factorize this polynomials which doesn't have roots in Q.
$ \ f(x) = x^4 +2x^3-8x^2-6x-1 $
P.S.) Are there any generalized method finding 4th degree polynomials factor?
| Just working through the hint in the accepted answer, and verified with the Wolfram result in the second answer. Just for me, and maybe a check/guide for someone else:
!$x^4+2x^3+x^2−9x^2−6x−1 \\
(x^2+x)^2 - (3x+1)^2 \\
\\
((x^2+x) + (3x+1)) ((x^2+x) - (3x+1)) \\
(x^2+4x+1) (x^2-2x-1) \\
\\
\text{quadratic formula f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3226755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
"Dividing" matrix by a vector If I have $\mathbf{Ax}=\mathbf{c}$ where $\mathbf{x} =
\left[\begin{array}{r}
x_1\\
x_2\\
x_3\\
x_4\\
x_5
\end{array}\right]$ and $\mathbf{c} = \left[\begin{array}{r}
2x_1-4x_2-x_3-3x_4+2x_5\\
-x_1+2x_2+x_3+x_5\\
x_1-2x_2-x_3-3x_4-x_5\\
-x_1+4x_2-x_3+5x_5
\end{array}\right]$.
To solve the... | Notice that $$Ax = A \cdot x =\left[\begin{array}{ccccc}
a_{11} & a_{12} & a_{13} & a_{14} & a_{15}\\
a_{21} & a_{22} & a_{23} & a_{24} & a_{25}\\
\cdots & \cdots & \cdots & \cdots & \cdots\\
a_{41} & \cdots & \cdots & \cdots & a_{45}
\end{array}\right]\cdot\left[\begin{array}{c}
x_{1}\\
x_{2}\\
x_{3}\\
x_{4}\\
x_{5}
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3226876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Splitting / rearranging sigma sums I am struggling to understand the concept of sigma sum rearrangement. In fact, I don't even know what to call it. That being said, if anybody can recommend sources for me to study this from or let me know what you call this type of problem so that I can study it more, that would be gr... | If you look closely, the equation change from 5 to 6 is the reverse of what's happening in 4 to 5. Let's take a look at what's happening to the indices with an example.
Let $n=3$, and we'll figure out what pairs of $(j,k)$ are valid for the index set in each equation.
In equation $4$, going through in order, we have $(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3227235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
using trigonometric identities For proving $$\frac {16}{\cos (4x)+7} =\frac{1}{\sin^4x +\cos^2x} +\frac{1}{\sin^2x +\cos^4x} $$
I tried to use that:
\begin{align}
\sin^4 x +\cos^4 x&=\sin^4 x +2\sin^2x\cos^2 x+\cos^4 x - 2\sin^2x\cos^2 x\\
&=(\sin^2x+\cos^2 x)^2-2\sin^2x\cos^2 x\\
&=1^2-\frac{1}{2}(2\sin x\cos x)^2\\
&... | $\displaystyle \sin^4x+\cos^2x=\frac{(1-\cos2x)^2}{4}+\frac{1+\cos 2x}{2}=\frac{3+\cos^22x}{4}=\frac{3+\frac{1+\cos 4x}{2}}{4}=\frac{7+\cos4x}{8}$
$\displaystyle \sin^2x+\cos^4x=\sin^4\left(\frac \pi2-x\right)+\cos^2\left(\frac \pi2-x\right)=\frac{7+\cos4\left(\frac \pi2-x\right)}{8}=\frac{7+\cos4x}{8}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3229789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Can we show a sum of symmetrical cosine values is zero by using roots of unity? Can we show that
$$\cos\frac{\pi}{7}+\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{5\pi}{7}+\cos\frac{6\pi}{7}=0$$
by considering the seventh roots of unity? If so how could we do it?
Also I have observed that
$$\cos\f... | Note that $$\cos(\pi - \alpha)= - \cos(\alpha)$$ Therefore $$\cos(\frac{\pi}{7})+\cos(\frac{2\pi}{7})+\cos(\frac{3\pi}{7})+\cos(\frac{4\pi}{7})+\cos(\frac{5\pi}{7})+\cos(\frac{6\pi}{7})=$$
$$\cos(\frac{\pi}{7})+\cos(\frac{2\pi}{7})+\cos(\frac{3\pi}{7})-\cos(\frac{3\pi}{7})-\cos(\frac{2\pi}{7})-\cos(\frac{\pi}{7})=0$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3234931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Finding $\lim_{x \to 0+}\frac{x^{(\sin x)^x}-(\sin x)^{x^{\sin x}}}{x^3}$ Problem
$$\lim_{x \to 0+}\frac{x^{(\sin x)^x}-(\sin x)^{x^{\sin x}}}{x^3}.$$
Attempt
\begin{align*}
&\lim_{x \to 0}\frac{x^{(\sin x)^x}-(\sin x)^{x^{\sin x}}}{x^3}\\
=&\lim_{x \to 0}\frac{\exp [(\sin x)^x\ln x]-\exp [{x^{\sin x}\ln(\sin x)]}}{x^3... | My Solution
Recall the following formulas
$$\sin x=x-\frac{1}{3!}x^3+\frac{1}{5!}x-\cdots\tag 1$$
$$e^x=1+x+\frac{1}{2}x^2+\cdots\tag 2$$
$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots \tag 3$$
which are the starting point we need to rely on.
From$(1)$,
\begin{align}
\frac{\sin x-x}{x}=\dfrac{x-\dfrac{1}{3!}x^3+o(x^4)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3237802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Solve $\sqrt{1 + \sqrt{1-x^{2}}}\left(\sqrt{(1+x)^{3}} + \sqrt{(1-x)^{3}} \right) = 2 + \sqrt{1-x^{2}} $ Solve $$\sqrt{1 + \sqrt{1-x^{2}}}\left(\sqrt{(1+x)^{3}} + \sqrt{(1-x)^{3}} \right) = 2 + \sqrt{1-x^{2}} $$
My attempt:
Let $A = \sqrt{1+x}, B = \sqrt{1-x}$ and then by squaring the problematic equation we get:
$$(... | As $-1\le x\le1$
WLOG $x=\cos2t,0\le2t\le\pi,\sin2t=\sqrt{1-x^2}$
$$\implies\sqrt{1+\sin2t}[(2\cos^2t)^{3/2}+(2\sin^2t)^{3/2}]=2+\sin2t$$
As $\sin t,\cos t\ge0$ and $(\sin t+\cos t)^2=1+\sin2t$
$$2\sqrt2(\cos t+\sin t)(\cos^3t+\sin^3t)=2+\sin2t$$
$$\sqrt2(1+\sin2t)(2-\sin2t)=2+\sin2t$$ which is on rearrangement, a Qua... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 3
} |
Proof that $ \frac{3\pi}{8}< \int_{0}^{\pi/2} \cos{\sin{x}} dx < \frac{49\pi}{128}$ Proof that
$$
\frac{3\pi}{8} <
\int_{0}^{\pi/2} \cos\left(\sin\left(x\right)\right)\,\mathrm{d}x <
\frac{49\pi}{128}
$$
Can somebody give me some instruction how to deal with inequality like that? My current idea is:
I see $\frac{3\pi}{... | Follow-up to @rtybase:
$$
\cos x\le 1-\frac{x^2}2+\frac{x^4}{24}\implies
\int_0^{\pi/2}\cos\sin xdx<\int_0^{\pi/2}\left(1-\frac{\sin^2 x}2+\frac{\sin^4 x}{24}\right)dx=\frac{49\pi}{128}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3240437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Help with $-\int_0^1 \ln(1+x)\ln(1-x)dx$ I have been attempting to evaluate this integral and by using wolfram alpha I know that the value is$$I=-\int_0^1 \ln(1+x)\ln(1-x)dx=\frac{\pi^2}{6}+2\ln(2)-\ln^2(2)-2$$
My Attempt:
I start off by parametizing the integral as $$I(a)=\int_0^1 -\ln(1+x)\ln(1-ax)dx$$
where $I=I(1)$... | Proceed as follows
\begin{align}
&\int_0^1 \ln(1-x)\ln(1+x)\,dx \\
=&\int_0^1\ln2 \ln(1-x)dx +\int_0^1 {\ln(1-x)\ln\frac{1+x}2dx}, \>\>\>{IBP} \\
=& -\ln2 +\int_0^1 x\left( \frac{\ln\frac{1+x}2}{1-x} -\frac{\ln(1-x)}{1+x}\right) dx\\=&-\ln2 +1 -\underbrace{\int_0^1 \ln\frac{1+x}2dx}_{=\ln2-1}+\int_0^1\frac{\ln(1-x)}{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3246020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 2
} |
Find $a,b,c,d$ such that $2^a + 2^b + 2^c = 4^d$ Let $a,b,c,d$ be whole numbers that satisfy
$$2^a + 2^b + 2^c = 4^d$$
What values of $(a,b,c,d)$ would make this equation true?
Here is my work so far.
Without loss of generality, assume $a\ge b\ge c$. Then one trivial solution by inspection is $(1,0,0,1)$. Playing aro... | Obviously, $d>0$. WLOG, assume that $a\ge b\ge c$
$2^c(2^{a-c}+2^{b-c}+1)=2^{2d}$
$2^{a-c}+2^{b-c}+1\ge 3$. Therefore, $b-c$ must be $0$ as otherwise $2^{a-c}+2^{b-c}+1$ will be an odd number larger than $1$.
$2^{a-c}+2^{b-c}+1=2^{a-c}+2=2(2^{a-c-1}+1)$ and hence $2^{a-c-1}+1$ must be even. $a-c=1$.
We have $2^c\times ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3247351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
} |
Find $\lim_{x\to\infty}1+2x^2+2x\sqrt{1+x^2}$ Consider the function
$$f(x)=1+2x^2+2x\sqrt{1+x^2}$$
I want to find the limit $f(x\rightarrow-\infty)$
We can start by saying that $\sqrt{1+x^2}$ tends to $|x|$ when $x\rightarrow-\infty$, and so we have that
$$\lim_{x\rightarrow-\infty}{(1+2x^2+2x|x|)}=\lim_{x\rightarrow-\... | $$\lim_{x\to-\infty}(1+2x^2+2x\sqrt{1+x^2})$$$$=\lim_{x\to-\infty}(\sqrt{1+x^2}+x)^2$$$$=\lim_{x\to-\infty}\frac{1}{(\sqrt{1+x^2}-x)^2}$$$$=\lim_{x\to\infty}\frac{1}{(\sqrt{1+x^2}+x)^2}$$$$=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3249693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Rationalizing $\frac{1}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}$ Problem
Rationalize $\frac{1}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}$
So, I'm training for the Mexican Math Olimpiad. This is one of the algebra problems from my weekly training.
Before this problem, there was other very similar, after proving it, there's an ... | You can proceed by using $x^3-y^3=(x-y)(x^2+xy+y^2).$
If $a+b+c=0$ and $a^2+b^2+c^2-ab-ac-bc=0$ so the last gives
$$(a-b)^2+(a-c)^2+(b-c)^2=0$$ or $a=b=c,$ which gives $a=b=c=0,$ which is impossible.
Thus, your trick works for all reals $a$, $b$ and $c$ such that $$a+b+c\neq0$$ and
$$(a-b)^2+(a-c)^2+(b-c)^2\neq0,$$ w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3250105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
When differentiating $\sum\limits_{n=0}^{\infty} \left(\frac{x}{2}\right)^n$ When differentiating $\sum\limits_{n=0}^{\infty} \left(\frac{x}{2}\right)^n$, gives $\sum\limits_{n=1}^{\infty} \left(\frac{nx^{n-1}}{2^n}\right)$. so, $n =0$ , becomes $n = 1$. Then, if we were to differentiate $\sum\limits_{n=1}^{\infty} ... | When in doubt, write it out. You don't need to memorize formulas for how the limits change in differentiation of series. There are too many cases to consider anyway. Just look at the first few terms to match the lowest index.
For instance:
$$
\sum_{n=0}^{\infty} \left(\frac{x}{2}\right)^n
= \sum_{n=0}^{\infty} \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3252115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
For which $n$ is $\frac{n!}{4}$ equal to $\left\lfloor \sqrt{\frac{n!}{4}}\right\rfloor\left(\left\lfloor\sqrt{\frac{n!}{4}}\right\rfloor + 1\right)$? For how many values of $n$, is $\frac{n!}{4}$ equal to $\left\lfloor \sqrt{\frac{n!}{4}}\right\rfloor\left(\left\lfloor\sqrt{\frac{n!}{4}}\right\rfloor + 1\right)$?
Fur... | It's not a full solution, but maybe it will help:
Let $x=\sqrt{\frac{n!}{4}}$.
$$ x^2 = \lfloor x\rfloor \cdot (\lfloor x\rfloor + 1)$$
$$ x^2 + \frac14 = (\lfloor x\rfloor + \frac12)^2$$
$$ -\frac12 + \sqrt{x^2+ \frac14} = \lfloor x\rfloor$$
$$ -1 + \sqrt{n!+ 1} = 2\lfloor x\rfloor \in 2\mathbb N$$
So we get a necess... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3254631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How can i get the least $n $ such that $17^n \equiv 1 \mod(100$)? When I solve the problem:
$17^{2018}\equiv r \pmod{100} $
used Euler theorem since $\gcd (17,100)=1$ and so
$\phi(100)=40$
and so
$17^{40}\equiv 1 \pmod{100}$
But i also found that: $17^{20}\equiv 1 \pmod{100}$
How can i get the least n such that $17^{... | $$17^n\equiv (10+7)^n\equiv 7^n+n\cdot 7^{n-1}\cdot 10 \mod 100 $$
Coincidentally,
$$7^4 \equiv 1 \mod 100$$
$$\rightarrow 7^{-1}\equiv 7^3 \equiv 43 \mod 100$$
So we need to find the smallest $n$ such that,
$$7^n+n\cdot 7^{n-1}\cdot 10 = 7^n+n\cdot 7^{n}\cdot 43\cdot10\equiv 7^n(1+30\cdot n)\equiv 1\mod 100$$
$$7^{-n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3254898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
How to solve this congruence of triangles geometrically?
Find x.
Solution with trigonometry:
First there are the missing angles ...
Superior = $180 - 13 - 32 = 135$
Bottom = $180 - 13 - 24 = 143$
By the breast theorem we have to
$$\frac{9 \sqrt{2}}{sen(13)}=\frac{Diagonal}{sen(135)} \to Diagonal=\frac{9 \sqrt{2}... | In the figure $ x = 3 $. However, the lenght that you seek is $ 5x = 15 $
In your case $$x = \frac{9 \sqrt{2} \cdot sen(135)}{sen(143)} = \frac{9 \sqrt{2} \cdot \frac{1}{\sqrt 2}}{\frac{3}{5}} = 15 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3255695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Trig identities analogous to $\tan(\pi/5)+4\sin(\pi/5)=\sqrt{5+2\sqrt{5}}$ The following trig identities have shown up in various questions on MSE:
$$-\tan\frac{\pi}{5}+4\sin\frac{2\pi}{5}=\tan\frac{\pi}{5}+4\sin\frac{\pi}{5}=\sqrt{5+2\sqrt{5}}$$
$$-\tan\frac{2\pi}{7}+4\sin\frac{3\pi}{7}=-\tan\frac{\pi}{7}+4\sin\frac{2... | The case of multiples of $\pi/11$ actually involves five "symmetrically equivalent" forms:
$4\sin(5\pi/11)-\tan(2\pi/11)=\sqrt{11}$
$4\sin(\pi/11)+\tan(4\pi/11)=\sqrt{11}$
$4\sin(4\pi/11)-\tan(5\pi/11)=-\sqrt{11}$
$4\sin(2\pi/11)+\tan(3\pi/11)=\sqrt{11}$
$4\sin(3\pi/11)+\tan(\pi/11)=\sqrt{11}$
All are derivable from th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3257939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 1
} |
How is Wolfram Alpha and the reduction formula arriving at a different result for the integral of $\int \sec^4 x\,dx$ than naive $u$-substitution? I calculated the following on paper for the value of $\int \sec^4 x\,dx$.
$$\int \sec^4 x\,dx=\int \sec^2 x \sec^2 x\,dx=\int (\tan^2 x + 1)(\sec^2 x)\,dx.$$
Let $u = \tan x... | This is because you made a slight error in $$\frac13\tan^3x+\tan x+C=\frac13\tan x(\tan^2x+3)+C\ne\frac13\tan x(\tan^2x+1)+C.$$ Then you get \begin{align}\frac13\tan x(\tan^2x+3)+C&=\frac13\tan x(\tan^2 x+1)+\frac23\tan x+C\\&=\frac13\tan x\sec^2x+\frac23\tan x\cos^2x\sec^2x+C\\&=\frac13\tan x\sec^2x+\frac23\tan x\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3258309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Find the positive integers pair $(x, y)$ such that $xy^2 + y + 1 \mid x^2y + x + y$.
Find the positive integers pair $(x, y)$ such that $xy^2 + y + 1 \mid x^2y + x + y$.
We have that $xy^2 + y + 1 \mid x^2y + x + y \implies xy^2 + y + 1 \mid y(x^2y + x + y) - x(xy^2 + y + 1)$
$\iff xy^2 + y + 1 \mid y^2 - x \implies ... | You have obtained that $xy^2+y+1 \mid x^2+y+1$, so $x \geq y^2$.
If $x=y^2$, then obviously all pairs of the form $(k^2,k) $ are solutions.
If $x >y^2$, then we let $x=My^2+N$ with $N<y^2$. If $N=0$ then $x=My^2$ so $My^4+y+1 \mid M^2y^4+y+1$ and so $My^4+y+1 \mid My+M-y-1$, which is impossible due to bounding reasons... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3258799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Wish to evaluate $\int_{0}^{1}\frac{x^4-3x^2+2x}{\sin(\pi x)} dx$ We want to evaluate this integral,$\newcommand{\cosec}{\operatorname{cosec}}$
$$\int_{0}^{1}\frac{x^4-3x^2+2x}{\sin(\pi x)}\mathrm dx\tag1$$
$$\int_{0}^{1}[x^4\cosec(\pi x)-3x^2\cosec(\pi x)+2x\cosec(\pi x)]\mathrm dx$$
Apply integration by parts, first ... | Too long for comments.
@user10354138 provided a very nice solution for the problem.
Since similar questions would probably happen, I tabulated the expressions of
$$I_n=\int_0^1 x^{n-2}\,\frac{x^2-x}{\sin(\pi x)}\,dx$$
$$\left(
\begin{array}{cc}
n & I_n \\
2 & -\frac{7 \zeta (3)}{\pi ^3} \\
3 & -\frac{7 \zeta (3)}{2 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3260759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Compute $\int\limits_{-1/2}^{1/2}\frac{dx}{\sqrt{1-x^2}\arccos^2x}$ Compute $$\int\limits_{-1/2}^{1/2}\frac{dx}{\sqrt{1-x^2}\arccos^2x}.$$
I started by making the substitution $\arccos x = t$. Hence, $-\frac{dx}{\sqrt{1-x^2}}=dt$.
Now I get that $$\int\limits_{-1/2}^{1/2}\frac{dx}{\sqrt{1-x^2}\arccos^2x}=-\int\limits_... | Setting $\arccos x = t$ gives us $\frac{-dx}{\sqrt{1-x^2}} = dt$
So our integral is then:
$$\int_{2\pi/3}^{\pi/3} \frac{-dt}{t^2}$$
$$=\int_{\pi/3}^{2\pi/3} \frac{dt}{t^2}$$
$$= \frac{3}{\pi}-\frac{3}{2\pi}$$
$$=\frac{3}{2\pi}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3263310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
The integral $\int_{-\infty}^{\infty}\frac{ a^2x^2 dx}{(x^2-b^2)^2+a^2x^2}=a\pi,~ a, b \in \Re ?$ This integral
$$\int_{-\infty}^{\infty}\frac{ a^2x^2 dx}{(x^2-b^2)^2+a^2x^2}=a\pi, ~ a, b \in \Re$$
looks suspiciously interesting as it is independent of the parameter $b$. The question is: What is the best way of provin... | How about this, please check it critically
Denote the integral as $$I=\int_{-\infty}^{\infty}\frac{ a^2x^2 dx}{(x^2-b^2)^2+a^2x^2}.$$
$$(x^2-b^2)^2+a^2x^2=(x^2-r^2)(x^2-s^2) \Rightarrow r=(d+ia)/2,s=(d-ia)/2, d=\sqrt{4b^2-a^2}.$$
$$I=\frac{a^2}{r^2-s^2} \int_{-\infty}^{\infty} \left( \frac{r^2}{x^2-r^2}-\frac{s^2}{x^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3267110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
If $a_n=\left[\frac{n^2+8n+10}{n+9}\right]$, find $ \sum_{n=1}^{30} a_n$ I am working on a problem, assuming high school math knowledge.
Let ${a_n}$ be the sequence defined by $$a_n=\left[\frac{n^2+8n+10}{n+9}\right]\,,$$ where $[x]$ denotes the largest integer which does not exceed $x$. Find the value of $ \sum\limit... | Hint
\begin{align*}
\frac{n^2+8n+10}{n+9}&=\frac{n^2+9n-n-9+19}{n+9}\\
&=n-1+\frac{19}{n+9}\\
\therefore \left\lfloor\frac{n^2+8n+10}{n+9}\right\rfloor&=n-1+\left\lfloor \frac{19}{n+9}\right\rfloor
\end{align*}
As $n$ varies from $1$ to $30$, we will have
\begin{align*}
\left\lfloor\frac{19}{\color{red}{1}+9}\right\rfl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3267245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
unknown polynomial divided by $x^2(x-1)$, find the remainder. I took an exam today and there's a problem stuck in my head; I still can't figure out yet.
Here's the question (just the concept as I can't remember precisely).
An unknown polynomial divided by $(x-1)^2$ leaves the remainder of $x + 3$ (not sure about the n... | I am going to follow your approach. We have
$$f(x)=(x-1)^2p(x)+x+3$$
So, $f(1)=4$.
We also have that
$$f(x)=x^2q(x)+2x+4$$
And we want to find $a,b,c$ in
$$f(x)=(x-1)x^2g(x)+ax^2+bx+c$$
Plugin $x=1$ you get $4=a+b+c$.
Plugin $x=0$ doesn't give us enough information since this only gives us the remainder after dividin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3271609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Show that $\int_{0}^{\pi/2} \frac{\sin x~dx}{1+\sin 2x}=\frac{\coth^{-1} \sqrt{2}}{\sqrt{2}}$ The integral $$\int_{0}^{\pi/2} \frac{\sin x~dx}{1+\sin 2x}=\frac{\coth^{-1} \sqrt{2}}{\sqrt{2}}?$$ can be checked at Mathematica. The question is how to do it by hand?
| Using the substitution $x \mapsto \frac{\pi}2 - x$ we get
$$I = \int_0^{\frac{\pi}2} \frac{\sin x\,dx}{1+\sin2x} = \int_0^{\frac{\pi}2} \frac{\cos x\,dx}{1+\sin2x} = \frac12\int_0^{\frac{\pi}2} \frac{\sin x+\cos x}{1+\sin2x}\,dx = \frac12\int_0^{\frac{\pi}2} \frac{\sin x+\cos x}{(\sin x+\cos x)^2}\,dx$$
so we have
$$I ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3275274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Given two positive numbers $x,\,y$ so that $32\,x^{6}+ 4\,y^{3}= 1$. Prove that $\frac{(2\,x^{2}+ y+ 3)^{5}}{3(x^{2}+ y^{2})- 3(x+ y)+ 2}\leqq 2048$ . Given two positive numbers $x,\,y$ so that $32\,x^{6}+ 4\,y^{3}= 1$. Prove that
$$p(x)\equiv p= \frac{(2\,x^{2}+ y+ 3)^{5}}{3(x^{2}+ y^{2})- 3(x+ y)+ 2}\leqq 2048$$
My s... | By AM-GM
$$1=32x^6+4y^3=32x^6+2\cdot\frac{1}{2}+4y^3+2\cdot\frac{1}{2}-2\geq$$
$$\geq3\sqrt[3]{32x^6\cdot\left(\frac{1}{2}\right)^2}+3\sqrt[3]{4y^3\cdot\left(\frac{1}{2}\right)^2}-2=6x^2+3y-2,$$
which gives $$2x^2+y\leq1.$$
Id est, $$\frac{(2x^2+y+3)^5}{3(x^2+y^2)-3(x+y)+2}\leq\frac{4^5}{3(x^2+y^2)-3(x+y)+\frac{3}{2}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3275539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
What is the least common factor in equation $\frac{5}{x+4}=4+\frac{3}{x-2}$ I am attempting to solve for x $\frac{5}{x+4}=4+\frac{3}{x-2}$
I know that I need to find the least common denominator. In this case, since I cannot see a clear relationship among them all I think it's just the product of all 3 denominators:
$\... | Yes, you are on the right path. Yes, you do get a quadratic. The quadratic that you should get is $10-6x-4x^2$ and its roots are $1$ and $-\frac52$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3277024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.