Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Line through two given skew lines and origin Question
Find the direction cosines of the line through the origin which intersects each of the lines
$\displaystyle \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $\displaystyle \frac{x+2}{4} = \frac{y-3}{3} = \frac{z-4}{2}$.
My Attempt
Projection of a vector connectin... | You are correct that there is no such line other than a mistake you made in calculation. Here is another approach I used to validate.
An intersection point on line $1$ is $P (2s+1, 3s+2, 4s+3)$ and on line $2$, $Q (4t-2, 3t + 3, 2t + 4)$.
Now as the the intersecting line passes through origin $O$, we must have, $\vec {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4207398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Calculate the limit $\lim_{n\rightarrow\infty}\left(\frac{\prod_1^n(1 +r/n^2)}{e^{1/2}}\right)^n$ How to calculate the limit
$$\lim_{n\rightarrow\infty}\left(\frac{\prod\limits_{r=1}^n\left(1 +\frac{r}{n^2}\right)}{e^{1/2}}\right)^n?$$
I tried using the sandwich theorem by taking logarithm of the limit and using the ex... | Using the fact that $\log(1+x)=x-\frac{x^2}{2}+\mathcal{O}(x^3)$ for small $x$, we find
\begin{align*}
\left( {\prod\limits_{r = 1}^n {\left( {1 + \frac{r}{{n^2 }}} \right)} } \right)^n & = \exp \left( {n\sum\limits_{r = 1}^n {\log \left( {1 + \frac{r}{{n^2 }}} \right)} } \right) \\ &= \exp \left( {n\left( {\sum\limit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4209348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Showing $ n=\sum_{k=1}^{(n+1)/2}\sin{\frac{2\pi k}{n+2}}\sin{\frac{\pi(n-2(k-1))}{n+2}}\sec^2{\frac{\pi(n-2(k-1))}{2n+4}}$ for natural $n$ Good afternoon,
I am a little confused and intrigued by this finite summation formula I came up with.
If $n$ is a natural number then
$$
n=\sum_{k=1}^{(n+1)/2}\sin{\frac{2\pi k}{n+2... | We have, $$\theta=\frac {\pi(n+2-2k)}{2n+4}$$
Then $\frac {2k\pi}{n+2}=\pi-2\theta $
Thus, the series is equivalent to:
$$S=\sum_{k=1}^{\frac {n+1}{2}} \sin (\pi-2\theta)\sin 2\theta \sec^2 \theta $$
Using $\sin 2\theta=2\cos \theta \sin \theta$ and $\sin (\pi-2\theta)=\sin 2\theta$, we get:
$$S=\sum 4\sin^2 \theta =\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4213142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
If $| f(p + q) – f(q)| \le \dfrac pq$ for all $p$ and $q\in \mathbb Q$ & $q\ne 0$, show that $\sum_{i=1}^k| f(2^k ) – f(2^i ) |\le \dfrac{k(k – 1)}2$ The following question is taken from the practice set of JEE exam.
If $| f(p + q) – f(q)| \le \dfrac pq$ for all $p$ and $q \in \mathbb Q$ & $q \ne 0$, show that $\sum_... | After the hints given by @Amit_Rajaraman and @leoli1, I am attempting the solution:
$|f(2^k)-f(2^i)|\le|f(2^k)-f(2^{k-1})|+|f(2^{k-1})-f(2^{k-2})|+...+|f(2^{i+1})-f(2^i)|$
$|f(2^k)-f(2^i)|\le\dfrac{2^k-2^{k-1}}{2^{k-1}}+\dfrac{2^{k-1}-2^{k-2}}{2^{k-2}}+...+\dfrac{2^{i+1}-2^i}{2^i}$
$|f(2^k)-f(2^i)|\le(2-1)+(2-1)+...(k-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4213585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $n(n+1)(n+2)$ where $n\in N$ and $n+1=k^2$ ($k\in N$) is divisible by $60$. My Attempt
Let $N=n(n+1)(n+2)$
Now $n=k^2-1 \implies N=(k^2-1)k^2(k^2+1)$
Now if $k=2m \implies N=(4m^2-1)4m^2(4m^2+1)$ which is divisible by $3$ and $4 \implies$ It is divisible by $12$.
Now if $k=2m+1 \implies N=(4m^2+4m)(4m^2+4m+1... | To conclude you have to show that it is divisible by $5$ so you have to see what the possible remainders of $n+1$ when dividing by $5$ are, and conclude that at least one of $n,n+1$ and $n+2$ is a multiple of $5$.
If $n+1$ is a square then $n+1\equiv -1,0$ or $1\bmod 5$. It follows at least one among $n,n+1,n+2$ is $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4215470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is there an analytical solution to $\int_{0^{-}}^{t} d{s} \ T\left(\frac{x}{\sqrt{2t}}, \sqrt{\frac{s}{2t-s}}\right)$? I am trying to solve the integral
\begin{align}
\int_{0^{-}}^{t} d{s} \ \ T\left(\frac{x}{\sqrt{2t}}, \sqrt{\frac{s}{2t-s}}\right)
\end{align}
where $T$ is Owen's T function. I have been trying to subs... | Let
$$ f \colon \mathbb{R} \times (0,\infty) \to (0,\infty) \, , \, f(x,t) = \int \limits_0^t \operatorname{T} \left(\frac{x}{\sqrt{2 t}},\sqrt{\frac{s}{2t-s}}\right) \, \mathrm{d} s \, .$$
The substitution $s = t u$ yields $f(x,t) = t g\left(\frac{x}{2\sqrt{t}}\right)$ with $g \colon \mathbb{R} \to (0,\infty)$ given b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4215629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Minimum of $\sqrt{a^2+b^2}+\sqrt{(a-1)^2+b^2}+\sqrt{a^2+(b-1)^2}+\sqrt{(a-1)^2+(b-1)^2}.$
Let $a$ and $b$ be real numbers. Find the minimum of $$\sqrt{a^2+b^2}+\sqrt{(a-1)^2+b^2}+\sqrt{a^2+(b-1)^2}+\sqrt{(a-1)^2+(b-1)^2}.$$
The problem is from an inequality book. While I was doing the problem myself, I got the follow... | Assuming you got the minimum $0$ and you wanna ask what is your mistake.
Then each term reduces to $\sqrt{\dfrac{1}{2}}$
So $\sqrt{\dfrac{1}{2}}+\sqrt{\dfrac{1}{2}}+\sqrt{\dfrac{1}{2}}+\sqrt{\dfrac{1}{2}} = \sqrt{2}$
Just to help you help here is another helpful solution-
In the cartesian system of coordinates conside... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4215799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$\underbrace{\sum_{x = 1}^{r} \ldots \sum_{x = 1}^{r}}_{\text{n times}} x = ?$ Let $f(n,r) = \sum_{x = 1}^{r} f(n - 1,x)$ with $f(0,x) = x$.
We get
$$f(n,r) = \underbrace{\sum_{x_{n-1} = 1}^{x_n = r} \ldots \sum_{x = 1}^{x_1}}_{\text{n times}}x\text{ .}$$
This sum popped up when I was trying to find a formula for the n... | There is a somewhat combinatorial approach to this problem that uses hockey-stick identity.
Note that $f(0,x)=\binom{x}{1}$. We have that
$$f(1,r)=\sum_{x=1}^r \binom{x}{1}$$
Using hockey-stick identity, this is
$$f(1,r)=\binom{r+1}{2}$$
We are motivated to proceed by induction. We hypothesize that $f(n,r)=\binom{r+n}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4218063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to simplify this equation? I'm trying to simplify equation
$$n = \Big\lceil\sqrt{2x +\frac14} - \frac12 \Big\rceil$$
into
$$n = \Big\lfloor\sqrt{2x} + \frac12 \Big\rfloor$$
where $x$ is an integer. Indeed, both equations seem to output the same result when testing for some values. Thus far, I've done:
\begin{align*... | Is it possible that that there is an integer $k$ so that $2x < k^2 < 2x +\frac 14$?
As $x$ and $k^2$ are integers the answer to that is obviously no.
Let's get a little tighter. Is it possible there is "half-integer" $\frac k2$ where $k\in \mathbb Z$ and $2x < (\frac k2)^2 < 2x+\frac 14$.
That is $8x < k^2 < 8x + 1$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4218328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
show this indentity $\sum_{k=1}^{2n-1}\frac{\sin{\frac{k^2\pi}{2n}}}{\sin{\frac{k\pi}{2n}}}=n$ let $n$ be postive integers.show that
$$\sum_{k=1}^{2n-1}\dfrac{\sin{\frac{k^2\pi}{2n}}}{\sin{\frac{k\pi}{2n}}}=n$$
Try:I can show $n$ is smaller number.
let $LHS=f(n)$.when $n=1$ it is clear
$$f(1)=\dfrac{\sin{\pi/2}}{\sin{\... | Let $\omega = e^{i\pi/2n}$. If we write $S$ for the sum, then from
$$ \frac{\sin(\pi k^2/2n)}{\sin(\pi k/2n)} = \frac{\omega^{k^2} - \omega^{-k^2}}{\omega^k - \omega^{-k}} = \omega^{k(-k+1)} \frac{(\omega^{2k})^k - 1}{\omega^{2k} - 1} = \sum_{j=0}^{k-1} \omega^{k(2j-k+1)}, $$
we may recast $S$ as the double sum
$$ S =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4218841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\sum_{r =1}^{n -1}{\frac{n(n+1)-r(2r+3)-1}{(r+1)^2}\cdot{}^{n+r}{C_r}\cdot {}^n{C_r}}$
Evaluate $\sum_{r =1}^{n -1}{\frac{n(n+1)-r(2r+3)-1}{(r+1)^2}\cdot{}^{n+r}{C_r}\cdot {}^n{C_r}}$
I noted that
(A) ${}^{20}{C_{10}} - 110$, if $n=10$
(B) ${}^{22}{C_{11}} - 110$, if $n=11$
(C) ${}^{18}{C_{9}} - 100$, if $n... | Note that
$$\begin{align}\frac{n(n+1)-r(2r+3)-1}{(r+1)^2}&=\frac{n^2+n-2r^2-3r-1}{(r+1)^2}
\\&=\frac{n^2-r^2+n-r-(r^2+2r+1)}{(r+1)^2}\\
&=\frac{(n+r+1)(n-r)}{(r+1)^2}-1\end{align}$$
which implies
$$\begin{align}\frac{n(n+1)-r(2r+3)-1}{(r+1)^2}&\binom{n+r}{r}\binom{n}{r}\\
&=\frac{n+r+1}{r+1}\binom{n+r}{r}\frac{n-r}{r+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4219810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Roots of polynomials with repeated roots Let $a$ and $b$ be real numbers. Consider the cubic equation $$x^3+2bx^2-ax^2-b^2=0$$
(i) Show that if $x=1$ is a solution of the cubic then $$ -1+\sqrt{2}\leq b\leq1+\sqrt{2} $$
(ii) Show that there is no value of $b$ for which $x=1$ is a repeated root of the cubic
I need some ... | By polynomial long division, you get that$$x^3+2bx^2-ax^2-b^2=(x-1)^2(x+2b+2)+(-a^2+4b+3)x-(b^2+2b+2).$$So, if $1$ was a repeated root of $x^3+2bx^2-ax^2-b^2$, that is, if$$(x-1)^2\mid x^3+2bx^2-ax^2-b^2,$$the polynomial $(-a^2+4b+3)x-(b^2+2b+2)$ would be the null polynomial. But, for every $b\in\Bbb R$, $b^2+2b+2=(b+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4220912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve in $\mathbf{N}$ the equation $9x^2+p=y^2$ In these days, I have been trying to solve this problem:
Let $p \in \mathbf{N}$ a positive large integer ($> 10^9$). Find all $x, y \in\mathbf{N}$ such that:
$$9x^2+p=y^2$$
The first approach that I have tried is the following. We know that:
$$(t+n)^2-t^2=t^2+2\cdot t \... | An experimental approach:
Here I find one solution for this equation which a family of solutions can be based on:
$9x^2=y^2-p=(y-\sqrt p)(y+\sqrt p)$
we can can construct following system of equations:
$\begin{cases}y-\sqrt p=9\\y+\sqrt p=x^2\end{cases}$
Subtracting first equation from second one we get:
$2\sqrt p=x^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4222108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
A simple general expression for a definite integral Let
$$I_k = \int_0^{\frac{\pi}{2}} \sin x \cos^{2k} x \sqrt{1 + \sin x} \, dx,$$
where $k = 0,1,2,\ldots$. I wish to find a simple general expression for $I_k$ in terms of $k$. Simple here is the operative word.
Making a tangent half-angle substitution of $t = \tan \f... | Too long for a comment.
I did some back-of-the-envelope computation and obtained the following conjecture:
$$ I_k = 2 \sum_{n=0}^{k+1} \sum_{l=n}^{\left\lfloor\frac{n+k+1}{2}\right\rfloor} \frac{(-1)^{n}}{2n+2k+1} \binom{n+k+1}{2l}\binom{l}{n}. $$
The idea is that the following identity is likely to hold for $|z| < 1$:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4223426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Integrate $ \int \frac{2\tan x+1}{\sqrt{\tan^2 x+2\tan x+2}} \, \mathrm dx $ I tried writing $\tan^2 x+2\tan x+2$ as $(\tan x+1)^2+1$ and letting $\tan\theta = \tan x+1$ yields
$$ \int \frac{2\tan\theta-1}{\tan^2\theta-2\tan\theta+2}\,\sec\theta\,\mathrm d\theta $$
Then letting $ t = \tan(\frac{\theta}{2})$ yields
$$ \... | Let $t=\tan x$ to rewrite the integral as
$$ I=\int \frac{2\tan x+1}{\sqrt{\tan^2 x+2\tan x+2}} dx
=\int \frac{2t+1}{(1+t^2)\sqrt{t^2+2t+2}} dt\tag1
$$
Note that
$$
\bigg( \tan^{-1} \frac{at-a^{-1}}{\sqrt{t^2+2t+2}}\bigg)’
= \frac{a^{-1} t +ab }{(t^2+b)\sqrt{t^2+2t+2}},\>\>\>\>\>
b= \frac{1+2a^2}{a^2+a^4}$$
Set $b=1$ t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4223889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
$\lim\limits_{n\rightarrow\infty} \frac{n}{\log n}(n^{\frac{1}{n}}-1)$ How to find $\lim\limits_{n\rightarrow\infty} \frac{n}{\log n}(n^{\frac{1}{n}}-1)$? Here is my attempt.
Put $f(x)=\frac{x}{\log x}(x^{\frac{1}{x}}-1)$ for $x>1$. Then
\begin{aligned}
\lim\limits_{n\rightarrow\infty} \frac{n}{\log n}(n^{\frac{1}{n}}... | Let $n=e^x.$ Then $n^{1/n}-1=$ $e^{x/e^x}-1=[x/e^x][1+(x/e^x)/2!+(x/e^x)^2/3!+(x/e^x)^3/4!+...]$
which lies between $x/e^x=(\log n)/n$ and $(x/e^x)(1+(x/e^x))=(\log n)/n+((\log n)/n)^2.$
Because if $x\ge 0$ then $0\le x/e^x<1$ so $$0\le (x/e^x)/2!+(x/e^x)^2/3!+(x/e^x)^3/4!+...\le$$ $$\le [(x/e^x)/2!][1+1/2^1+1/2^2+..... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4225441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Prove that $a^a+b^b\ge a^b+b^a>1$ if $a,b> 0$. I succeeded in proving the second part of the inequality by showing that for every real number$x,y ∈ (0, 1)$, we have
$x^y≥ \frac x{x + y − xy}.$
By Bernoulli’s inequality we have
$x^{1−y}= (1 + x − 1)
^{1−y}
≤ 1 + (x − 1)(1 − y) = x + y − xy,$
$\implies$ $x^y ≥ \frac ... | First, suppose $a>b>1$. Consider the function $f(x):=x^a-x^b$.
It suffices to show that $f$ is monotonically increasing on $[1,\infty[$.
But we can write $f(x)=x^b(x^{a-b}-1)$ which is the product of the two strictly positive strictly increasing functions on $]1,\infty[$. So $f$ is monotonically increasing in this cas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4226202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Bounds of an Expression
if $(a,b,c)>0$
and $abc(a+b+c)=3$
Then what can you say about the bounds of
$(a+b)(b+c)(c+a)$ ?
Hint:
think Geometrically!
My Approach
1.assumed $a\geq b \geq c$
2.Tried to think geometrically
3.used the Hadwiger-Finsler inequality
and modified it to get $(a+b+c) \geq \sqrt{4A\sqrt{3}+2(... | The following inequality is true.
$$\sqrt[3]{\frac{(a+b)(a+c)(b+c)}{8}}\geq\sqrt[4]{\frac{abc(a+b+c)}{3}}.$$
For the proof use $$(a+b)(a+c)(b+c)\geq\frac{8}{9}(a+b+c)(ab+ac+bc).$$
It gives a lower bound.
The upper bound is $+\infty$.
Can you end it now?
We need to prove that
$$\left(\frac{(a+b)(a+c)(b+c)}{8}\right)^4\g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4226354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find the quadratic equation from given relatioship the quadratic equation whose roots are a and b where $a^2 +b^2=5$ and
$3(a^5+b^5)=11(a^3+b^3)$
What I Tried
$a^2 +b^2=5$ $(a+b)^2-2ab=5$ $(\text{sum of roots})^2 -2(\text{products of roots})=5$
$3(a^5+b^5)=11(a^3+b^3)$ $a^3(3a^2-11)=b^3(11-3b^2)$
| Hint: If $a+b = 0$, then you have $a = \pm \sqrt{\frac{5}{2}}, b = \mp \sqrt{\frac{5}{2}}$. From this you can find the equation. If $a + b \neq 0\implies 3(a^4-a^3b+a^2b^2-ab^3+b^4)=11(a^2-ab+b^2)= 11(5-ab)$. Using $a^4+b^4 = (a^2+b^2)^2 - 2a^2b^2 = 25 - 2(ab)^2, a^3b + ab^3 = ab(a^2+b^2) = 5ab$. Thus you do have a qua... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4228648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to find the maximum value of $\frac{2\sin\theta\cos\theta}{(1+\cos\theta)(1+\sin\theta)}$? Let $\theta\in\left(0,\frac{\pi}{2}\right)$, then find the maximum value of $\dfrac{2\sin\theta\cos\theta}{(1+\cos\theta)(1+\sin\theta)}$
I found the derivative, which is equal to $\dfrac{(\cos\theta−\sin\theta)(\sin\theta+\c... |
Step 1: Find the critical points of $f(\theta)$:
For this, it is necessary to find the derivative of the function and do $f'(\theta)=0$, such that:
\begin{align*}
f'(\theta)=&-\dfrac{2\cos^2\theta\sin\theta}{(cos\theta+1)(\sin\theta+1)^2} +\\
&+\dfrac{2\cos^2\theta}{(\cos\theta+1)(sin\theta+1)}+ \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4230143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Exponent of Matrix A when A is not diagonalizable I want to compute the Matrix $A^{19}$ where
$$A = \begin{pmatrix}
6&1&0\\0&6&1\\0&0&6\end{pmatrix}$$
Since the only eigenvalue of the matrix is 6, it's not diagonalizable. How do you proceed to calculate the exponent matrix of A then?
| $ A = \begin{bmatrix} 6 && 1 && 0 \\ 0 && 6 && 1 \\ 0 && 0 && 6 \end{bmatrix} $
We can express $A^n$ as follows:
$A^n = \alpha_0 I + \alpha_1 A + \alpha_2 A^2 $
The above equation is satisfied by $\lambda = 6$, i.e.
$\lambda^n = \alpha_0 + \alpha_1 \lambda + \alpha_2 \lambda^2 $
And since it is a repeated eigenvalue, w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4233454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$\sum_{cyc}^{}\frac{a(a+b+c)}{9b^2+(a+b+c)^2} \geq \frac{1}{2}$ I was attempting this question: https://www.youtube.com/watch?v=600X-ZGNBbk
Which is the following:
$$\sum_{cyc (a,b,c)}{}\frac{a}{b^2+1} \geq \frac{3}{2}$$
Where $a+b+c = 3$ and $a,b,c > 0 $
The video itself used a solution where they added $3$ and then s... | By AM-GM $$\sum_{cyc}\frac{a}{b^2+1}=3+\sum_{cyc}\left(\frac{a}{b^2+1}-a\right)=3-\sum_{cyc}\frac{ab^2}{1+b^2}\geq$$
$$\geq3-\sum_{cyc}\frac{ab^2}{2b}=3-\frac{1}{2}\sum_{cyc}ab\geq3-\frac{3}{2}=\frac{3}{2}.$$
Another way.
After full expanding we need to prove that:
$$\sum_{cyc}(10a^6+42a^5b+24a^5c-21a^4b^2+105a^4c^2+38... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4235853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
USAMO 1973 (Simultaneous Equations)
Determine all the roots, real or complex, of the system of simultaneous equations (USAMO 1973/4)
$$x+y+z=3$$
$$x^2+y^2+z^2=3$$
$$x^3+y^3+z^3=3$$
Multiply equation I by 2 and subtract it from Equation II:
$$x^2-2x+y^2-2y+z^2-2z=-3$$
Complete the square for all three variables:
$$(x-... | $$(x+y+z)^2=9=x^2+y^2+z^2+2(xy+yz+zx)\implies xy+yz+zx=3$$
$$(x+y+z)^3=27=6xyz+3(x+y+z)(x^2+y^2+z^2)-2(x^3+y^3+z^3)=6xyz+27-6\implies xyz=1$$
Hence by Viète's formulas $x,y,z$ are the roots of $t^3-3t^2+3t-1=(t-1)^3$, so $x=y=z=1$ is the only solution.
Note that Viète's formulas are valid even in the complex plane; the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4239747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Proving a function is increasing in $n$. I am trying to prove that the function $\frac{n}{2n+1}$, defined for $n \in \mathbb{N}$, decreases in $\mathbb{N}$. I attempted it by induction, but I'm not convinced that I fully need induction. Why can I not prove that for an arbitrary $n$, $f(n) \leq f(n+1)$ and deduce that, ... | As Kavi Rama Murthy and "user" point out, you don't need induction to show that $f(n + 1) \geq f(n)$. But (according to the usual definition of "increasing"), you need to show more than that; that for any $m$ and $n$ with $n \geq m$, $f(n) \geq f(m)$. And to derive that rigorously from $f(n + 1) \geq f(n)$, you do ne... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4243093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Number of roots of the equation $ax^2+ bx + c = 0$ in $(1,2)$
Let $a, b, c \in R, a \ne 0$ such that $a$ and $4a + 3b + 2c$ have the same sign. Then the
number of roots of the equation $ax^2+ bx + c = 0$ that lie(s) in $(1,2)$ is(are)?
I began by writing that $4a^2+3ab+2ac>0$. I tried finding the sign of $f(1)\cdot f... | Let's start from $4a^2+3ab+2ac=a(4a+3b+2c)>0$.
Then we have
$8a^2+6ab+4ac>0$
Case 1: $-\frac{b}{2a}<2$
$\implies 8a^2+6ab+b^2 > b^2-4ac$
$\implies \frac{8a^2+6ab+b^2}{4a^2} > \frac{b^2-4ac}{4a^2}$ , since $a^2 > 0, a \in \mathcal{R} \wedge a \neq 0$
$\implies 2+\frac{3b}{2a}+\frac{b^2}{4a^2} > \frac{b^2-4ac}{4a^2}$
$\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4244775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\int_0^1\frac{x^{n+1}}{x+1}dx<\frac{1}{2(n+1)}$ Prove that $$\int\limits_0^1\frac{x^{n+1}}{x+1}dx<\frac{1}{2(n+1)}$$
On simplifying by parts we get:
$$\int\limits_0^1\frac{x^{n+1}}{x+1}dx=\frac{1}{2(n+2)}+\int\limits_0^1\frac{x^{n+2}}{(x+1)^2(n+2)}dx$$
Thus if we prove that$$\max\left(\displaystyle\int\limi... | Simply observe that for $x \in [0,1]$ and $n \in \mathbb Z^+$,
$$0 \le (1-x)x^n = (x+1)x^n - 2x^{n+1},$$ the first inequality arising from the fact that $x \ge 0$ and $1-x \ge 0$. Consequently,
$$0 \le \frac{x^{n+1}}{x+1} \le \frac{x^n}{2},$$ and integrating gives the desired inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4245686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Are $a=3, b=7$ the only solutions to $\sqrt{3}+\sqrt[3]{7}=\sqrt{a}+\sqrt[3]{b}$ for $a,b \in \mathbb{Q}$? So I had a lesson about calculating surds e.g. $\sqrt{5+2\sqrt{6}}$, then the teacher wrote the steps like that:
For some $a,b\in\mathbb{Q}_{\ge 0}$
$$\sqrt{5+2\sqrt{6}}=\sqrt{a}+\sqrt{b} \\
5+2\sqrt{6}=a+b+2\sqr... | By doing some calculations you can see that $\sqrt{3}+\sqrt[3]{7}$ is a root of the polynomial
$$
p(x) = x^6-9 x^4-14 x^3+27 x^2-126 x+22.
$$
Furthermore, $\sqrt{a}+\sqrt[3]{b}$ is a root of
$$
q_{a,b}(x) = x^6 -3 a x^4 -2 b x^3 +3 a^2 x^2 -6 a b x -a^3+b^2.
$$
The polynomial $p$ is irreducible. This means that it is t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4246704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate$J(k) = \int_{0}^{1} \frac{\ln^2x\ln\left ( \frac{1-x}{1+x} \right ) }{(x-1)^2-k^2(x+1)^2}\text{d}x$ I am trying evaluating this
$$J(k) = \int_{0}^{1} \frac{\ln^2x\ln\left ( \frac{1-x}{1+x} \right ) }{(x-1)^2-k^2(x+1)^2}\ \text{d}x.$$
For $k=1$, there has
$$J(1)=\frac{\pi^4}{96}.$$
Maybe $J(k)$ doesn't ... | A solution idea of calculating the integral by Cornel Ioan Valean
The post is extremely short since I have no time, but once you know what to do, all is trivial. So, what do to? Observe that any $J(2n,k)$ is half the real part of the integral over the positive real line.
$$\int_{0}^{1} \frac{\log(x)^{2n}\log\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4247363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
} |
$\sum_{i = 0}^{K - 1} {[\frac{1}{i + 1}\prod_{j=i+2}^{K}{(1 - \frac{1}{jq})}]} = q$ I've come across an interesting sum of products series. I know the sum, I can more or less see why it is so, but I'm stuck at a certain point of the proof.
Let $0 <q \leq 1$ be some fraction, $K$ some large positive integer.
$\sum_{i = ... | If you are familiar with Pochhammer symbols
$$\prod_{j=i+2}^{K}\left(1 - \frac{1}{jq}\right)=\frac{\left(i+2-\frac{1}{q}\right)_{K-i-1}}{(i+2)_{K-i-1}}$$
$$\sum_{i = 0}^{K - 1}\frac{\left(i+2-\frac{1}{q}\right)_{K-i-1}}{(i+2)_{K-i-1}}=q+\frac{q^2 \,\Gamma \left(K-\frac{1}{q}+1\right)}{\Gamma (K+1)\, \Gamma
\left(-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4250987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the minimum of $a^4+b^4+c^4+d^4+a^2+b^2+c^2+d^2$.
Let $a,b,c,d$ be real numbers such that $a+b+c+d=0$ and $abcd=1$. Find
the minimum value of $a^4+b^4+c^4+d^4+a^2+b^2+c^2+d^2$.
By $\text{Vieta}$'s theorem, $a,b,c,d$ are the roots of the equation $x^4+sx^2+tx+1=0$. Thus
\begin{align*}
&~~~~~~a^4+b^4+c^4+d^4+a^2+b... | Let $f(a,b,c,d,\lambda,\mu)=\sum\limits_{cyc}(a^4+a^2)+\lambda(a+b+c+d)+\mu(abcd-1).$
Thus, in the minimal point we need $$\frac{\partial f}{\partial a}=\frac{\partial f}{\partial b}=\frac{\partial f}{\partial c}=\frac{\partial f}{\partial d}=\frac{\partial f}{\partial \lambda}=\frac{\partial f}{\partial a}=\frac{\part... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4251279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Given the triangle ABC. Let BC = a, AC = b, AB = c. Find the minimum value of the following expression Given the triangle ABC. Let BC = a, AC = b, AB = c. Find the minimum value of the following expression:
a) $$P=\frac{4a}{b+c-a} + \frac{9b}{c+a-b} + \frac{16c}{a+b-c}$$
b) $$P=\frac{a^3}{2a+bc} + \frac{b^3}{2b+ac} + \... | The first problem.
Let $a=y+z$, $b=x+z$ and $c=x+y$.
Thus, $x$, $y$ and $z$ are positives and by AM-GM we obtain:
$$\frac{4a}{b+c-a} + \frac{9b}{c+a-b} + \frac{16c}{a+b-c}=\frac{4(y+z)}{2x}+\frac{9(x+z)}{2y}+\frac{8(x+y)}{z}=$$
$$=\frac{2y}{x}+\frac{9x}{2y}+\frac{2z}{x}+\frac{8x}{z}+\frac{9z}{2y}+\frac{8y}{z}\geq6+8+12... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4252731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
What is the measure of the PQ segment in the figure below? Given a semi-circumference of diameter AB and center «O» and
let P and Q be points on this curve, we draw $QH\perp OB (H \in OB)$ .
Calculate "PQ", if AP = 2; QH = 6, $ m\measuredangle PAQ=3m\measuredangle QAB$
My progress:(hard question)
$\triangle BAQ: 6^2 ... | Here is a solution using trigonometry. Say radius of the circle is $r$ and $O$ is the center of the circle.
$\angle BQH = x, \angle QOB = 2x$
So, $a = \frac{6}{\cos x}$
$\frac{a}{2} = OB \sin x \implies r = \frac{6}{\sin 2x}$
But in $\triangle ABP, AB = 2 r = \frac{2}{\cos 4x}$
Equating, $6 \cos 4x = \sin 2x$
$6 - 12 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integrate $\int_{0}^{\infty}\frac{\tan^{-1}x}{x\left(1+x^{2}\right)}dx$ Q:
Integrate $I=\int_{0}^{\infty}\frac{\tan^{-1}x}{x\left(1+x^{2}\right)}dx$
My Approach:
Put $$\tan^{-1}x=t\to x=\tan t$$
Also we have, $$\frac{dx}{1+x^{2}}=dt$$
We get, $$I=\int_{0}^{\frac{\pi}{2}}\frac{t}{\tan t}dt$$
I'm stuck here, how do I p... | Another way using Feynmann trick
$$\frac d {da}\int_0^\infty \frac{\tan ^{-1}(a x)}{x \left(x^2+1\right)}\,dx=\int_0^\infty \frac{dx}{\left(x^2+1\right) \left(a^2 x^2+1\right)}$$
$$\frac{1}{\left(x^2+1\right) \left(a^2 x^2+1\right)}=\frac{a^2}{\left(a^2-1\right) \left(a^2x^2+1\right)}-\frac{1}{\left(a^2-1\right) \left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
How to solve $ \sin(x) - \cos(x) -3\sin(2x) +3\cos(2x) +\sin(3x) -\cos(3x) =0$ Hello friends please help me to solve this equation.
$ \sin(x) - \cos(x) -3\sin(2x) +3\cos(2x) +\sin(3x) -\cos(3x) =0.$
Here is what I have done so far
$ \Rightarrow \sqrt2(\frac{1}{\sqrt2}\sin x - \frac{1}{\sqrt2}\cos x ) - 3\sqrt2(\frac... | You changed a couple of $3x$ terms into $x$, so your attempt is faulty.
Consider instead that
$$
\sin 3x+\sin x=2\sin2x\cos x,\qquad
\cos 3x+\cos x=2\cos2x\cos x
$$
so your equation becomes
$$
3(\cos2x-\sin2x)-2\cos x(\cos2x-\sin2x)=0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4254449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that there are don't exist integers $a,b,c$ such that for every integer $x$ the number $A=(x+a)(x+b)(x+c)-x^3-1$ is divisible by $9$. Problem: Prove that there are don't exist integers $a,b,c$ such that for every integer $x$ the number $$A=(x+a)(x+b)(x+c)-x^3-1$$ is divisible by $9$.
We can consider $0\le a,b,c<9... | Perhaps a slight hint should do good, it seems (not entirely sure if this should work properly, but let's see how far insights take us).
As you saw in my comment, $x$ can either be $9k-1$ or an integer root of $x^2 - x + 1 - 9k = 0$. Let's take the first case.
For $x = 9k -1$, we see $9 \mid x^3 + 1$, so turning our fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4255970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Challenging series of a fraction with a cosine and a variable Is it possible to express the following series explicitly (e.g. as a polynomial in $\alpha$):
$$f(\alpha )=\sum_{k=1}^{\infty} \frac{\cos(\alpha2\pi k)}{(2\pi k)^2((2\alpha k)^2-1)^2} , $$
where $0\leq\alpha\leq1$ ?
Do you have any idea on how to approach th... | $$S_0(a)=\sum_{k=1}^{\infty} \frac{\cos(2\pi ka)}{(2\pi k)^2((2ka)^2-1)^2}=\frac{1}{4\pi^2}\frac{1}{16a^4}\Re \frac{1}{2}S(a)$$
where
$$S(a)=\sum_{\binom{k=-\infty}{k\neq0}}^{\infty}\frac{e^{2\pi ika}}{k^2(k-1/2a)^2(k+1/2a)^2} ,\,\,a\in(0;1), \,\,a\neq\frac{1}{2n} \text(n - integer)$$
To evaluates $S(a)$ we go in the c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4257183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Analytic expressions for two definite integrals containing trigonometric functions and a real parameter I encountered these two integrals in my research about the physics of Josephson junctions:
$$\int^{\pi}_{-\pi}\frac{dx}{\sqrt{1+\sqrt{1-D\sin^2\frac{x}{2}}}} $$
$$\int^{\pi}_{-\pi}\sqrt{1+\sqrt{1-D\sin^2\frac{x}{2}}}... |
Define the two functions $\mathcal{I}:(0,1)\rightarrow\mathbb{R}$ and $\mathcal{J}:(0,1)\rightarrow\mathbb{R}$ via the respective integrals,
$$\mathcal{I}{\left(D\right)}:=\int_{-\pi}^{\pi}\mathrm{d}\omega\,\frac{1}{\sqrt{1+\sqrt{1-D\sin^{2}{\left(\frac{\omega}{2}\right)}}}}$$
and
$$\mathcal{J}{\left(D\right)}:=\int_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4257421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is the relationship between the length of the circumradius and the inradius in $ \triangle ABC $? For reference: In a right angle triangle ABC, an interior bisector BD is traced, where I is or incenter, $\measuredangle B = 90 ^ o$ and $3BI = 4ID$. Find the relationship between the circumraio and inraio lenght of $... | Using standard notations for side lengths,
Please note that in $\triangle ADB$, as $AI$ is the angle bisector,
$AD:c = 3:4 \implies AD = \frac{3c}{4}$
Similarly in $\triangle CBD, CD = \frac{3a}{4}$
So, $b = 2R = AD + CD = \frac{3}{4} (a + c)$
$a + c = \frac{8R}{3} \tag1$
Now use the fact that $\triangle ABC$ is a righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4263318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Show $\frac{n^2}{2^{\sqrt{\log(n)}}} \geq \frac{n}{2}. $ So we know that $\frac{n^2}{2} \geq \frac{n}{2}$, but I'm stuck proving that $\frac{n^2}{2} \geq \frac{n^2}{2^{\sqrt{\log n}}}\geq \frac{n}{2}$. Am I missing something?
| Let $\sqrt {\log n}=u,\thinspace u≥0$, then we have
$$\begin{align}&\frac{n^2}{2^{\sqrt{\log(n)}}} ≥ \frac{n}{2}\\
\iff &\frac{n^2}{2^{\sqrt{\log(n)}}} ≥ \frac{n^2}{2n}\\
\iff &n≥2^{\sqrt{\log(n)}-1}\\
\iff &e^{u^2}≥2^{u-1}\\
\iff &e^{u^2}>e^{u-1}>2^{u-1}.\end{align}$$
This means, the equality is impossible.
Therefore,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4264808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Evaluating $\int \frac{1}{a+b\sec(x)}dx$ I tried to find a closed-form expression for the integral
$$\int\frac{1}{a+b\sec(x)}\>dx$$ and, afterwards, set $a=0$, $b=1$ to recover the result for the simplified integral
$$\int\cos(x)\>dx=\sin(x)+C$$
The above integral was inspired by Américo Tavares’s rather illuminating s... | As @Hussain-Alqatari commented, using immediately the tangent half angle substitution, you have
$$I=\int \frac{dx}{a+b\sec(x)}=\int \frac{2(1-t^2)}{ (b-a)t^4+2 b t^2+(a+b)}\,dt$$ Assuming $b\neq a$
$$I=\frac 2{b-a}\int \frac {1-t^2}{(t^2+1)(t^2-\alpha)}\,dt \qquad \text{with} \qquad \alpha=\frac{a+b}{a-b}$$ Using parti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4266408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Find the minimum natural number $n$, such that the equation $\lfloor \frac{10^n}{x}\rfloor=1989$ has integer solution $x$
Find the minimum natural number $n$, such that the equation $\lfloor \frac{10^n}{x}\rfloor=1989$ has integer solution $x$.
My work-
$\frac{10^n}{x}-1<\lfloor \frac{10^n}{x}\rfloor≤\frac{10^n}{x}\... | $\frac{10^4}5=2000>1989$
$\Rightarrow x>5 $ for n=4
let $x=5.1$ for $10^4$ or 51 for $10^5$ we have:
$\frac{10^5}{51}=1960$
$\Rightarrow 5.1>x>5$ for $n=5$
let $x=50.1$ for $10^5$ or 501 for $10^6$ we have:
$\frac{10^6}{501}=1996$
$\Rightarrow 502>x>501$ for n=6
let $x=502$ we have:
$\frac {10^6}{502}=1992$
$\Rightarr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4275620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $a$ for which $y\ge x^2+4a^2,x\ge y^2+4a^2$ has only one solution For what $a$ does this system of inequalities have only one solution?
$$
\begin{cases}y\ge x^2+4a^2\\x\ge y^2+4a^2\end{cases}
$$
So, I have tried getting $y$ from the first equation then putting it into second, but then I have two quadratic and $x$,... | We have,
$$\begin{align}&\begin{cases}y^2≥\left(x^2+4a^2\right)^2≥0\\y^2≤x-4a^2,\thinspace x≥4a^2\end{cases}\\\\
\implies &\left(x^2+4a^2\right)^2≤y^2≤x-4a^2\\
\implies &\left(x^2+4a^2\right)^2≤x-4a^2 \\
\implies &16a^4+a^2(8x^2+4)+(x^4-x)≤0\\
\implies &\left(4a^2+\left(x^2+\frac 12\right)\right)^2≤\left(x+\frac 12\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4275782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Integrating a two complex functions I need to evaluate the norm $||G(s)||_2$ where $G(s) = \frac{\alpha_1 s+\alpha_o}{s^2+\beta_1s+\beta_o }$ is a transfer function.
And
$||G(s)||^2_2$ = $\int_{-\infty}^{\infty} trace {G(j\omega) G^*(j\omega)} \,d\omega$
I am able to get to this point
$||G(s)||^2_2$ = $\int_{-\infty}^{... | I am assuming that all the constants are $\gt 0$.
We start by noticing that the integrand is even, thus
$$I =2 \int\limits_0^\infty\frac{\alpha_1^2x^2+\alpha_0^2}{x^4+(\beta_1^2-2\beta_0)x^2+\beta_0^2 }\, \mathrm dx $$
Now, using change of variables $x^4=\beta_0^2 t^4$, we get
$$\begin{align} I &= \frac2{\beta_0^{3/2}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4276967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to prove:$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c\ge\sum_{cyc}{\sqrt{2(a^2+b^2)}}$ Problem: Let $a,b,c>0. $ Prove that: $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c\ge\sqrt{2(a^2+b^2)}+\sqrt{2(b^2+c^2)}+\sqrt{2(c^2+a^2)}$$
I have seen problem before, and I tried to prove: $$2(a+b+c)\ge\sqrt{2(a^2+b^2)}... | Also, we can use $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq\frac{37(a^2+b^2+c^2)-19(ab+ac+bc)}{6(a+b+c)}$$ and
$$\sqrt{2(a^2+b^2)}\leq\frac{3a^2+2ab+3b^2}{2(a+b)}.$$
By this way we can get something stronger:
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c\g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4277130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$\int_a^{a + n \cdot h} (x-a)(x-(a+h))(x-(a+2h))\cdots(x-(a+n\cdot h))\, \text{d}x = ?$ I would like to dissolve this expression:
\begin{align}
f(h) = \int_a^{a + n \cdot h} (x-a)(x-(a+h))(x-(a+2h))\cdots(x-(a+nh)) \ \text{d} x
\end{align}
I have a guess by working out the first two expressions that the following appea... | A closed form can be given in terms of the Gregory co-efficients. Note that we can write $$b_n=\int_{0}^{n}x\left(x-1\right)\left(x-2\right)...\left(x-n\right)dx$$
We have, $$\int_{ }^{ }x\left(x-1\right)\left(x-2\right)...\left(x-n\right)dx=\psi_{n+2}\left(x\right)(n+1)!\tag{1}$$
Where, $\psi_n(x)$ is the $n$-th Bern... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4277463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find $\lim_{n\to \infty}\frac{-7n^5-6n^4+122n^3-15n^2+7n-2}{21n^5+12n^3-501}$ I am stuck with this question, I do not know what I'm doing wrong. I am dividing everything by $n^5$ and getting end result as $\frac{-7}{21}$ but the result in book answer says -$\frac{1}{3}$. Can someone advice? My calculation:
$${\displays... | You are correct, infact we have
$$\frac{-7}{21} = \frac{-7}{7 \times 3}=\frac{-1}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4281341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What's the measure of the segment $PQ$ in the triangle below? For reference:
NIn figure CQ is bisector, and
PE = EC, AC = b and BC = a. Calculate PQ.(Answer: $\frac{ab}{a+b}$)
My progress:
$(Bissector ~Th.):\triangle ABC:\\ \frac{BQ}{AQ}=\frac{a}{b} \\
By~ proportionality: \frac{BQ}{BQ+AQ}=\frac{a}{a+b}=\frac{BQ}{AB}... |
Draw segment $QR = QP$ then,
$\angle PRQ = \angle PCA \implies AC \parallel QR$. Also, $CQ$ is angle bisector of $\angle ACB$.
So, $ \displaystyle \frac{QR}{AC} = \frac{BQ}{BA} = \frac{BC}{BC+AC} = \frac{a}{a+b}$
$ \displaystyle \implies PQ = QR = \frac{ab}{a+b}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4282081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Maximum eigenvalue of an adjacency matrix Let $G = (V,E)$ be an undirected graph without multiple edges.
Additionally, let $|V| = n + 1$, where $n \geq 3$.
$G$ is created in such a way, that $n$-first vertices create a cycle and $n+1$ vertex is connected only to the previous $n$-vertices.
Prove that the maximum eigenva... | Hint: using the symmetries in the graph, try out vectors that equal 1 on the first $n$ components and equal a variable $x$ on the last component. The eigenvalue equation gives you two equations and two unknowns; without giving too much away, you’ll end up with a quadratic that you can solve to get the desired eigenvalu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4285818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Apply the Cauchy–Goursat theorem to compute the integral $\oint_C\frac{\sinh(z)}{z^{4}(1-z^{2})} dz$ where $C$ is the circle $|z|=2$. Apply the Cauchy–Goursat theorem to compute the integral
$$
\oint_C\frac{\sinh(z)}{z^{4}(1-z^{2})}dz,
$$
where $C$ is the circle $|z|=2$.
Solution:
$z^{4}(1-z^{2})=0 \Rightarrow$
$z=0,z=... | You need to find the residue at $0$ ,$1$ and $-1$.
For the residue at $0$ you need to calculate the coefficient of $\frac{1}{z}$ in the Laurent Series expansion around $0$.
To do that use :-
$\sinh(z)=\frac{e^{z}-e^{-z}}{2}=z+\frac{z^{3}}{3!}+\frac{z^{5}}{5!}+....$.
And $\frac{1}{1-z^{2}}=1+z^{2}+z^{4}+z^{6}+....$
Comb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4288497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\frac1{12}(\sin^2A+\sin^2B+\sin^2C)\leq \sin^2\frac12B\sin^2\frac12C+\sin^2\frac12C\sin^2\frac12A+\sin^2\frac12A\sin^2\frac12B$ for $A+B+C=180^\circ$
If $A+B+C=180^{\circ}$, prove that
$$\begin{align}
\tfrac{1}{12}(\sin^2 A+\sin^2 B+\sin ^2C)
&\leq
\sin^2\tfrac{1}{2}B\sin^2\tfrac{1}{2}C \\[0.5em]
&+\sin^2\tfrac{1}{2... | Is known that
$$\sin^2u+\sin^2v=1-\dfrac{\cos2u+\cos2v}2=1-\cos(u-v)\cos(u+v).$$
Let
$$\alpha=\dfrac12\angle A,\quad \beta=\dfrac12\angle B.$$
Then
$$\delta=12(\sin^2\alpha\sin^2\beta+(\sin^2\alpha+\sin^2\beta) \cos^2(\alpha+\beta))-(\sin^22\alpha+\sin^22\beta+\sin^22(\alpha+\beta))$$
$$=4\big(3\sin^2\alpha\sin^2\beta-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4289933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
} |
Eliminate $\theta$ from $4x=5\cos\theta -\cos 5\theta$ and $4y=5\sin\theta -\sin 5\theta$ Eliminate $\theta$ from the equations.
$$4x=5\cos\theta -\cos 5\theta$$
$$4y=5\sin\theta -\sin 5\theta$$
Alternative forms are
$$x=5\cos^3\theta-4\cos^5\theta$$
$$y=5\sin^3\theta-4\sin^5\theta$$
and $$x=\cos^3\theta(3-2\cos2\theta... | These are the parametric equations of an epicycloid, where the large circle has radius $R=1$, and the smaller exterior rolling circle has radius $r=\frac{1}{4}$.
A possible simplification is writing
$$4a=4(x+i y) = 5 z - z^ 5\\
4b=4(x-i y) = \frac{5}{z} - \frac{1}{z^5}$$
where $z = \cos \theta + i \sin \theta$
and get ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4290398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Existence of integer coefficient polynomial such that $f(\sqrt{2}+\sqrt{3})=\sqrt{2}$ The original problem asked to "Show that there exists a polynomial $f(x)$ with rational coefficients such that $f(\sqrt{2}+\sqrt{3})=\sqrt{2}$."
I was wondering if there exists an integer coefficient polynomial. Intuitively, I think t... | We know that $\sqrt2 + \sqrt3$ is a root of the polynomial $$(x-\sqrt2-\sqrt3)(x-\sqrt2+\sqrt3)(x+\sqrt2-\sqrt3)(x+\sqrt2-\sqrt3) = x^4 - 10x^2+1.$$
So in any polynomial $f$ of degree at least $4$, we can replace the leading term $a_n x^n$ by $a_n(10x^{n-2} - x^{n-4})$, reducing the degree while not changing the value ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4292677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Number of solutions of $\cos^5x+\cos^5\left( x+\frac{2\pi}{3}\right) + \cos^5\left( x+\frac{4\pi}{3}\right) =0$
Solve in the interval $[0,2\pi]$ : $$\cos^5x+\cos^5\left( x+\frac{2\pi}{3}\right) + \cos^5\left( x+\frac{4\pi}{3}\right)=0 $$
I tried expanding the L.H.S by applying the formula of $\cos(A+B)$ but it result... | Use $\cos (A+B)$ formula to show for a natural number $n$,
$$\cos nx + \cos \left(nx+\frac{2n\pi}{3}\right) + \cos \left(nx+\frac{4n\pi}{3}\right) = \begin{cases} 0 , \quad \text{when n is coprime to 3} \\ 3\cos nx , \quad \text{when n is a multiple of 3} \end{cases}$$
Combining this with the identity from other answer... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4293179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Find $a\in \mathbb{R}$ such that a root of $ax^3-13x^2+(15a)x-25$ is $2+i$ Find $a\in \mathbb{R}$ such that a root of the polynomial $$p(x)=ax^3-13x^2+(15a)x-25$$ is $2+i$
Solution:
$q(x)=\frac{p(x)}{a}$
$=x^3-\frac{13}{a}x^2+15x-\frac{25}{a}$
$p(x)$ and $q(x)$ have the same roots, call them $r_1, r_2, r_3$
$r_1=2+i$. ... | Plug in $2+i$ into $p(x)$. Note that $(2+i)^2=3+4i$, $(2+i)^3=2+11i$. So $$p(2+i)= a*(2+11i) -13*(3+4i)+15a*(2+i)-25=
(32a-64)+(26a-52)i =(a-2)(32+26i)=0.$$ So $a=2$.This is much easier than what is in the OP and gives the right result. You do not need the assumption that $a$ is real.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4295130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Exponential diophantine equation $2^x+7^y=9^z$. The challenge is to solve this equation $2^{x}+7^{y}=9^{z}$ in positive integers. The obvious solution is $x=y=z=1$. Using brute force, I found $3$ possible solutions:
\begin{eqnarray*}
(x_1,y_1,z_1)&=&(3,0,1),\\
(x_2,y_2,z_2)&=&(1,1,1)\\
(x_3,y_3,z_3)&=&(5,2,2).\\
\end{e... | here is a method I learned from Exponential Diophantine equation $7^y + 2 = 3^x$ and polished. I have answered many questions of the form $A^u - B^v = C$ for given positive integers $A,B,C$
This answer finishes $x=1.$
The following shows that, with $x = 1,$ the largest solution to $9^z - 7^y = 2 $ is $z=y=1.$ Fir... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4295884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Given $\cos(5\theta)=0$, prove that $\cos(\frac{\pi}{10})\cos(\frac{3\pi}{10}) = \frac{\sqrt{5}}{4}$ Q: (a) By comparing the expressions for $(\cos(\theta) + \sin(\theta)^5$ given by De Moivre's theorem and by the binomial theorem prove that $\cos(5\theta) = 16\cos^5(\theta)-20\cos^3(\theta) + 5\cos(\theta)$
(b) By con... | I don't know if this is what is intended, but here is where I went with this problem.
let $u = \cos \theta$
$u(16 u^4 - 20 u^2 + 5) = 0$
From the binomial theorem:
$u^2 = \frac {20 \pm \sqrt {80}}{32}\\
u^2 = \frac {5 \pm \sqrt {5}}{8}$
Yes, $u = 0$ is a solution associated with $\theta = \frac {\pi}{2}$ but that isn't... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4299942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the maximum value of $P= \frac2{1+a^2}-\frac2{b^2+1}+\frac3{c^2+1}$ The problem is, given that $a,b,c>0$ and $abc+a+c=b$, then find the maximum value of
\begin{align}
P= \frac2{1+a^2}-\frac2{b^2+1}+\frac3{c^2+1}
\end{align}
One thing I know for sure is that $a,c$ are symmetric, or so I thought.
In the condition, c... | You have two ways to solve the problem without changing variables.
First way
From the constraint, we have, for example, $b=-\frac{a+c}{a c-1}$. Replacing in $P$ and simplifying
$$P=\frac{a^2 \left(3-2 c^2\right)+4 a c+2 c^2+3}{\left(a^2+1\right)
\left(c^2+1\right)}$$ Computing the partial derivatives
$$\frac{\partia... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4300568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
how to express the $n$th power of the cosine as a series of cosines? Which is the correct way for expressing the $n$th power of a cosine as a series of cosines without any exponent?
By using the Euler's formula
$\cos^n{(\theta)}=\left( \frac{e^{j\theta}+e^{-j\theta}}{2} \right)^n=
\frac{1}{2^n}\left... | We start from
$$ s_n = \frac{1}{2^n} \sum_{k=0}^n \binom{n}{k} z^{2k-n} $$
Case 1: Let $n = 2m$
$$ s_{2m} = \frac{1}{2^{2m}} \sum_{k=0}^{2m} \binom{2m}{k} z^{2(k-m)} $$
$$ s_{2m} = \frac{1}{2^{2m}} \sum_{k=0}^{m} \binom{2m}{k} z^{2(k-m)} + \frac{1}{2^{2m}} \sum_{k=m+1}^{2m} \binom{2m}{k} z^{2(k-m)} $$
$$ s_{2m} = \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4300731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
About an inequality wich have a link with $\sqrt{\dfrac{a^b}{b}}+\sqrt{\dfrac{b^a}{a}}\ge 2$ Hi it's a follow up of show this inequality $\sqrt{\frac{a^b}{b}}+\sqrt{\frac{b^a}{a}}\ge 2$:
Problem :
Let $a,x>0$ then (dis)prove :
$$\left(\frac{1}{2}a^{x}+\frac{1}{2}x^{-1}\right)^{\frac{1}{5}}\cdot\left(\frac{2}{a^{-x}+x}\... |
Your function looks like this on the reals:
So it is reasonable to expand to small $a$ and $x$ and to examine the curve where the value $2$ is taken. For example $a(x,2)$.
The problem is fundamental because $a^x$ and $x^a$ appear in the composite function. The function is rational and factored in both summands.
Speci... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4304209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to show that the following sequence is monotinacly increasing? given $$a_{1}=1, \quad a_{n+1}=\frac{2\left(2 a_{n}+1\right)}{a_{n}+3}$$ I am trying to show $a_{n}\leq a_{n+1}$
I am trying to show in induction, so my assumption is that $a_{n}\leq a_{n+1}$ and I want to show $a_{n+1}\leq a_{n+2}$ the only thing I cou... | Notice that $a_n$ had better never be $-3$ or the recursion fails to be defined. We wish to show \begin{align*}
a_n &\leq a_{n+1} \\
&= \frac{2(2a_n+1)}{a_n+3} \\
&= \frac{4a_n+2}{a_n+3}
\end{align*}
so
\begin{align*}
\begin{cases}
4a_n + 2 \geq a_n(a_n+3) ,& a_n+3 > 0 \\
4a_n + 2 \leq a_n(a_n+3) ,& a_n+3 < ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4305223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
If $ab>0$, show that $\dfrac{a}{b}+\dfrac{b}{a}\ge2$. If $ab>0$, show that $\dfrac{a}{b}+\dfrac{b}{a}\ge2$.
I am really sorry for the basic question, but I want to make things clear for me. By the AM-GM inequality we have $$\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}\cdot\dfrac{b}{a}}=2$$ Is this enough? I mean we ... | I have followed another reasoning to prove the statement.
Case 1: $a=b$
$$\frac{a}{a}+\frac{a}{a}=2\quad \blacksquare$$
Case 2: $a\neq b$
Notice that $ab>0$ is the same as saying $a,b\neq 0$, and both have the same sign (which later becomes irrelevant).
$$\frac{a}{b}+\frac{b}{a}=\frac{a^2+b^2}{ab}=\frac{a^2+2ab+b^2-2ab... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4309396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
What's the measure of the radius of the circle below? For reference: In the figure, calculate $R$. If : $\overset{\LARGE{\frown}}{AB}= 120°$, $CM = 1$ and $MB = 2$
(Answer: $\frac{3}{4}(\sqrt{3}+1$))
My progress:
Draw OP
Th.Chords:
$CM.MP = BM.AM \implies 1.[(R-1)+(R)] = 2.AM\\
\boxed{2AM = 2R-1}: \boxed{R=AM+ \frac{1... |
Extend $\overline{CMO}$ to meet the circle at E and Extend $\overline{AO}$ to meet the circle at $D$. As $\triangle ABD$ is a 30-60-90 triangle $AB=R\sqrt{3}$. Since $BM=2$ and $AM=R\sqrt{3}-2$.
Now, look at $\triangle ACM$ and $\triangle BEM$ they are similar.
$\frac{1}{2}=\frac{R\sqrt{3}-2}{2R-1}$
After solving for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4310815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Given the PDF of $X$ which is right hand continuous, compute PDF of $Y$ = CDF$(x)$ Let x have a piecewise right hand continuous pdf, $f_X(x)$, which is defined as follows.
$$f_X(x) = \begin{cases}
\frac{1}{2}(x+2)^2 & -2 \leq x < -1 \\
-\frac{1}{2}x +\frac{1}{12}\delta (x+1) & -1 \leq x < 0 \\
\frac{1}{2}x & 0 \leq x <... | To explain the solution step by step, a couple of drawings are needed but with this sketch I am sure you will be able to conclude by yourself.
First observe, that IF your X rv was continuous, $Y\sim U(0;1)$ by integral transform, being
$$F_Y(y)=F_X\left[F_X^{-1}(y) \right]=y$$
Your case is very similar but you have 2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4310981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Playing tennis until one of the players wins 3 times (Binomial distribution) Question
Nadal and Federer playing tennis against each other.
Nadal's probability to win one match is $\frac{2}{3}$, independently from the previous results.
The two are playing until one of them wins 3 matches.
What is the probability for Nad... | The final score could be 3-2 for Nadal in which case we must have seen 5 experiments with 3 Nadal victories and 2 Federer ones. So $\binom{4}{2}=6$ ways to "place" two of those victories among the first $4$ games and then the final game must be Nadal's. So the probability of that happening is
$$6 \times (\frac{2}{3})^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4312216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
inhomogen linear equation system problem I need to find all the answers of this linear equation system.
$$\begin{vmatrix}
1 & 2 & 1 & 3 & 0 & =\beta \\
2&3&2&5&1&=0\\
2&1&1&4&1&=0\\
3&3&2&7&1&=1\\
\end{vmatrix} $$
I tried to solve it like that:
Row 4- row 3->
$$\begin{vmatrix}
1 & 2 & 1 & 3 & 0 & =\beta \\
2&3&2&5&1&=0... | As I wrote in the comment. From your work we conclude that the linear system might only possess a solution for $\beta = 1$. As it turns out the solutions for this case are of the form
$$\left(\begin{matrix}
1-2\cdot x_4 \\
2-x_4+x_5 \\
-4+x_4-2\cdot x_5 \\
x_4 \\
x_5
\end{matrix}\right)$$
for $x_4, x_5 \in \mathbb{R}$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4313227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Let $f: (\Bbb Z_{28}, +)\to(\Bbb Z_{16}, +)$ be a group homomorphism such that $f(1)=12$. Find $\ker f$. Let $f: (\Bbb Z_{28}, +)\to(\Bbb Z_{16}, +)$ be a group homomorphism such that $f(1)=12$. Find $\ker f $.
1- $\langle 2\rangle $
2-$\langle 4\rangle $
3-$\langle 7\rangle $
4-$\langle 1\rangle $
I think first of all... | We have
$$\begin{align}
\color{red}{12+12}\pmod{16}&=f(1)+f(1)\\
&=f(1+1)\\
&=f(2)\\
&=\color{red}{8}\pmod{16},
\end{align}$$
$$\begin{align}
\color{green}{8+12}\pmod{16}&=f(2)+f(1)\\
&=f(2+1)\\
&=f(3)\\
&=\color{green}{4}\pmod{16},
\end{align}$$
and
$$\begin{align}
\color{blue}{4+12}\pmod{16}&=f(3)+f(1)\\
&=f(3+1)\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4313370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Seeking for help to find a formula for $\int_{0}^{\pi} \frac{d x}{(a-\cos x)^{n}}$, where $a>1.$ When tackling the question, I found that for any $a>1$,
$$
I_1(a)=\int_{0}^{\pi} \frac{d x}{a-\cos x}=\frac{\pi}{\sqrt{a^{2}-1}}.
$$
Then I started to think whether there is a formula for the integral
$$
I_n(a)=\int_{0}^{\p... | One way using complex analysis:
Since the integrand is even:
$$
I=\int_{0}^{\pi} \frac{d x}{(a-\cos x)^n} = \frac{1}{2} \int_{-\pi}^{\pi} \frac{d x}{(a-\cos x)^n}
$$
Since $a>1$, we can expand the integrand with the generalized binomial theorem:
$$ I = \frac{1}{2} \int_{-\pi}^{\pi} \frac{d x}{(a-\cos x)^n} = \frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4315858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Counting routes from home to office if you only return home if you realize you have forgotten something before you reach the office The following is how you can go to the office from home:
Home $\to$ Four Roads $\to$ Schools $\to$ Three Roads $\to$ University $\to$ Five Roads $\to$ Parks $\to$ Two Roads $\to$ Offices
Y... | Keep in mind that you add when two events are mutually exclusive, meaning that they cannot occur at the same time. However, when choices can be made independently (such as choosing how to travel from home to school, school to home, then from home to school to university to park to office), you should be multiplying.
C... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4316563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I know if the determinant is positive? I have the following problem.
Let $ A = \begin{pmatrix}a&b&c&d\\-b&a&d&-c\\-c&-d&a&b\\-d&c&-b&a\end{pmatrix}$
Find $\det(A)$.
I calculated $$AA^T = \begin{pmatrix}a^2+b^2+c^2+d^2&0&0&0\\0&a^2+b^2+c^2+d^2&0&0\\0&0&a^2+b^2+c^2+d^2&0\\0&0&0&a^2+b^2+c^2+d^2\end{pmatrix}$$ and g... | the determinant is the sum of 4! terms, one of which is $a^4$, so we choose the + sign, not the - sign.
alternatively, if we don't know the Leibniz formula, sub a=1 and b=c=d=0, then A is the identity matrix, so $\det(A)=1$, so we choose the + sign.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4323006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the largest $t$ such that for all positive $x, y, z$ the following inequality is satisfied Find the largest $t$ such that for all positive $x, y, z$ the following inequality is satisfied:
$(xy+xz+yz) \left(\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}\right)^2 \geq t$.
If there were such an inequality:
$ t_{1} \leq... | Another way.
By my previous post we need to prove that:
$$\sum_{cyc}xy\left(\sum_{cyc}\frac{1}{x+y}\right)^2\geq\frac{25}{4},$$ which is true by Muirhead:
$$\sum_{cyc}xy\left(\sum_{cyc}\frac{1}{x+y}\right)^2-\frac{25}{4}=$$
$$=\frac{\sum\limits_{sym}\left(4x^5y-x^4y^2-3x^3y^3+x^4yz+14x^3y^2z+\frac{19}{3}x^2y^2z^2\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4324192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding $\lim\limits_{n→∞}\frac{a^n+b^n}{a^n-b^n}$ with $a ≠ b$
Find $\lim\limits_{n→∞}\dfrac{a^n+b^n}{a^n-b^n}$ with $a ≠ b$.
Notice that I don't have $a>b$ or $b>a$, and also no $b≠0$ or $a≠0$, so I don't know if i even can use my solution.
My solution :
\begin{gather*}
\frac {a^n+b^n}{a^n-b^n} = \frac {a^n}{a^n-b^... | Alternative approach:
First of all, I am interpreting the problem to represent that $n$ is restricted to the positive integers. This is because, apparently, it is permissible to have either $a$ or $b$ negative.
If (for example), $a = -1.5$, you run into problems evaluating $a^n$ for $n \in \Bbb{R^+}$. For example, is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4326321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Determining $A$ such that $\lim \limits_{x \to\infty }(\sqrt{Ax^2+2x}−5x)$ exists and is finite
Determine the value of the real number $A$ so that a finite limit exists, and then compute the limit:
$$\lim \limits_{x \to \infty}(\sqrt{Ax^2+2x}−5x)$$
Where would I start with this? I thought I could factor the highest d... | $\begin{aligned} \sqrt{Ax^2+2x}-5x &= (\sqrt{Ax^2+2x}-5x)\frac{\sqrt{Ax^2+2x}+5x}{\sqrt{Ax^2+2x}+5x}\\ &= \frac{Ax^2+2x-25x^2}{\sqrt{Ax^2+2x}+5x}\\ &= \frac{(A-25)x+2}{\sqrt{A+\frac{2}{x}}+5} \end{aligned}$
Notice that, as $x$ goes to $\infty$, the denominator goes to $\sqrt{A}+5$ and the numerator goes to
*
*$+\inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4327130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
projection of $x^2+y^2+z^2-yz=1$ to xoy plane if the surface is projected to yoz plane, I known than the equation of the projection is $y^2+z^2-yz-1=0$. But what what about xoy plane, is the equation of project projection $x^2+y^2=1$?
| $$
x^2+y^2+z^2-yz=1
\\
z^2-yz+\left( x^2+y^2-1 \right) =0
\\
\Delta =-4x^2-3y^2+4\ge 0
\\
x^2+\frac{3}{4}y^2\le 1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4328213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Vanier College Practice test - Prove that $\log_{\frac {1} {\sqrt b} }\sqrt x= -\log_b x$ Source : Colorado.edu, Vanier College, Worksheet Logarithm Function, Question 8 (https://math.colorado.edu/math1300/resources/Exercises_LogarithmicFunction.pdf)
To be proved :
$\log_{\frac {1} {\sqrt b} }\sqrt x= -\log_b x$.
What ... | \begin{align}
\log_{\frac{1}{\sqrt{b}}}{\sqrt{x}} &= \frac{\log{\sqrt{x}}}{\log{\frac{1}{\sqrt{b}}}}\\
&= \frac{\log{\sqrt{x}}}{-\log{{\sqrt{b}}}} \\
&= \frac{\frac{1}{2}\log{{x}}}{-\frac{1}{2}\log{{{b}}}} \\
&= -\frac{\log{x}}{\log{b}} \\
&= -\log_b{x}
\end{align}
In total we use the following two rules:
\begin{align}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4330069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Need some help expressing this as a series, if possible Okay, so I'm working on a question, and after a bunch of simplification and manipulations, I get the following expression. Just wanted to ask if there's any way I can get this in the form of a series, with $n$ going from 0 to $\infty$. a is a random positive integ... | This is just for your curiosity
$U(x)$ is already expressed as a linear combination of two Taylor series (for which you give only the first terms but in which the patterns of the coefficients is quite clear). I suppose that this is the power series solution of a second oreder differential equation which could be intres... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4333787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate Indefinite Integral $\int \frac{x^5(1-x^6)}{(1+x)^{18}}dx$ The following integration is given by Wolfram Alpha
$$\int \frac{x^5(1-x^6)}{(1+x)^{18}}dx=\frac{x^6(28x^4+16x^3+39x^2+16x+28)}{168(1+x)^{16}}.$$
My question is: what is the best (meaning least work), method to achieve this result by hand ? There are ... | Explore the embedded symmetry by denoting
$$f_{n,m}(x)= \frac{x^{n-m}+x^{n+m}}{(1+x)^{2n}}$$
which differentiates as
$$f’_{n,m}(x)
=\frac{(n-m)(x^{-m-1}-x^{m+1}) -2m( x^{-m}-x^{m})-(n+m)(x^{-m+1}-x^{m-1}) }{x^{-n}(1+x)^{2n+2}}
$$
In particular, with $n=8$ and $m=0,1,2$
\begin{align}
f’_{8,0}(x)&= \frac{16(x^7-x^9)}{(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4339068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
The minimal value of a fraction based on a focal chord of an ellipse I came across a very interesting olympiad problem. It goes as follows: suppose you have an ellipse given by $\frac{x^2}{16}+\frac{y^2}{9}=1$ and a line that goes through the point $A(\sqrt{7},0)$. This line also goes through the ellipse at two points ... | This is a focal chord, so the product of the distances $AN$ and $AM$ is a constant
$AN \cdot AM = (a - c) \cdot (a + c) = a^2 - c^2 = b^2$
Thus
$AN \cdot AM = 9 $
The ratio requested is thus $\dfrac{ 9 }{ NM } $
and the distance ${NM}$ is maximum when $N$ and $M$ are on the major axis, in which case ${NM} = 2a = 8$
Thu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4339188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\int_{0}^{\infty} \frac{4\pi}{16\pi^2 + x^2} \left(\frac{1}{x}+\frac{1}{e^{-x}-1}\right) \, dx$ I’m trying to evaluate the following integral:
$$\int_{0}^{\infty} \frac{4\pi}{16\pi^2 + x^2} \left(\frac{1}{x}+\frac{1}{e^{-x}-1}\right) \, dx$$
I’ve tried using contour integration by using a quarter-circle con... | I'm afraid there is no a closed form for this integral.
$$I=\int_0^\infty \frac{4\pi}{16\pi^2 + x^2} (\frac{1}{x}+\frac{1}{e^{-x}-1}) dx=4\pi\int_0^\infty \frac{dx}{16\pi^2 + x^2} (\frac{1}{x}+\frac{1}{2}\coth\frac{x}{2}-\frac{1}{2}) dx$$
$$=-\frac{\pi}{4}+\pi\int_0^\infty \frac{dt}{4\pi^2 + t^2}\Big(\frac{1}{t}-\cot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4341440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Integration of Bounded Region I'm trying to solve this problem:
The region of integration is the triangle $D$ with vertexes $A(0,0),B(1,1),C(10,1)$.
Find the solution of $\iint_D \sqrt{x^2-y^2}\,dx\,dy$.
MY SOLUTION (as @ryang suggested):
We can write the triangle as $D=\{(x,y)\in\mathbb R^2|0\le y\le 1,y\le x \le 10y\... | Riemann integration for the area integrals is nasty as you need to find a way to enumerate the whole area.
I would suggest slightly different way, i.e. randomly picking a point from the triangle:
$$ f(x,y) = \sqrt{x^2-y^2} $$
$$ \int_D {f(x,y)} = A \cdot \frac{\sum_{n=1}^{k}{f(x_n,y_n)}}{k}, (x_n,y_n)\in D, k>1200$$, w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4343599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find the area of region triangular BPM For reference: In triangle $ABC$, the external angle bisector $BE$ and the median $AM$ intersect at $P$. Determine the area of triangular region $BPM$; if $AB =3BC$ and $S_{ABC}=40\ \mathrm{m^2}$.
My progress:
$S_{ABM}=S_{AMC}=\frac{40}{2} = 20\\
\frac{S_{ABP}}{S_{APE}}=\frac{... | The only data of this problem is $AB=3BM$ and area $S=40$ in the triangle $\triangle {ABC}$ in which $M$ is the midpoint of $BC$. There are a lot of such triangles whose shape can be quite different so either the problem is undetermined or there is a family of distinct triangles having area as a very nice invariant. We... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4344346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Make $2$ cubes out of $1729$ unit cubes, expected number of times you have to paint I'm trying to solve question 6 from the PuMac 2007 Combinatorics A competition:
Joe has $1729$ randomly oriented randomly arranged unit cubes, which are initially unpainted. He makes two cubes of sidelengths $9$ and $10$ or of sideleng... | Thankfully we are being asked about the expected value of this random variable, the random variable is a sum of simpler random variables, namely whether each unit face remains unpainted or not, so we can exploit linearity of expectation.
If we are able to find a good expression for the expected value after $k$ steps th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4345812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
} |
$(\frac{a}{b}-1)^2+(\frac{b}{a}+1)^2\ge3$ I tried expanding the parenthesis and then grouping them back somehow but i couldn't get anywhere.
I also tried to apply known inequalities with no success. Can someone help me out?
| Substitute $x=\frac{a}{b}$.
\begin{align*}
\left(\frac{a}{b}-1\right)^2+\left(\frac{b}{a}+1\right)^2 - 3 &= (x-1)^2+\left(1+\frac{1}{x}\right)^2-3 \\
&= \frac{x^2(x^2-2x+1)+(x^2+2x+1)-3x^2}{x^2} \\
&= \frac{x^4-2x^3-x^2+2x+1}{x^2} \\
&= \frac{(x^2-x-1)^2}{x^2} \\
&\geq 0.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4347891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find function $ f $ such that $f(\frac{x-3}{x+1})+f(\frac{x+3}{1-x})=x$ I am looking for functions $ f:\Bbb R \to \Bbb R $ satisfying
$$f\Big(\frac{x-3}{x+1}\Big)+f\Big(\frac{x+3}{1-x}\Big)=x$$
I used the substitution $ x=\cos(2t) $ for $ x\in (0,2\pi) $, to use the identities
$$1+x=2\cos^2(t) \text{ and } 1-x=2\sin^2(... | Substitute in $x\mapsto\frac{x-3}{x+1}$. Then we have
$$f\left(\frac{x+3}{1-x}\right)+f\left(x\right)=\frac{x-3}{x+1}$$
Now substitute again $x\mapsto \frac{x-3}{x+1}$,
$$f(x)+f\left(\frac{x-3}{x+1}\right)=\frac{x+3}{1-x}$$
Now add these two to get
$$2f(x)+f\left(\frac{x+3}{1-x}\right)+f\left(\frac{x-3}{x+1}\right)=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4350435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
A Recurrence Relations problem I need to bound
$$T(n) = 2T\left(\frac{n}{2}\right) + n\cdot \frac{\log \log n}{\log n}$$ as tight as possible.
Here is what I got:
Firstly, we introduce a change of variable. Let $n = 2^k$ while ignoring the rounding, then we have
$$
T(n) = T(2^k) =: g(k).
$$
Expand it further, we get
$... | Here I will present a bound for the sum
$$\sum_{j=1}^k \frac{\log j}{j}.$$
The basic idea is that the partial sum can be asymptotically approximated by the integral of the function $\frac{\log x}{x}$, which is easy to compute. More generally, if a function $f(x)$ is decreasing on $[a,\infty)$, then the following bounds... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4350749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is $\binom{n}{2} \binom{n}{1} + \binom{n}{3} \binom{n}{2} + \ldots + \binom{n}{n-1} \binom{n}{n-2} + \binom{n}{n} \binom{n}{n-1}$?
A bag contains $n$ white and $n$ black balls, all of equal size. Balls are drawn at random. Find the probability that there are both white and black balls in the draw and that the num... | A combinatoric proof : Let :
$$A = \{1, 2, \ldots, 2 k\}, B = \{1, \ldots, k\} \text{ and } C = \{k + 1, \ldots, 2 k\}$$
We need to choose $k - 1$ elements from $A$. There are two ways :
*
* First way : Choose $k - 1$ element directly for $A$ : In this case, we have then :
$$\binom{2 k}{k - 1}$$
ways.
* Second way :... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4352574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Harmonic series with sign alternates every $n$ terms. Let $A(1)=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots$
Let $A(2)=\frac{1}{1}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}+\dots$
Let $A(3)=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\frac{1}{6}+\frac{... | $A_n$ can be expressed as
$$
A_n=\sum_{k=1}^n \sum_{u=0}^\infty \frac{(-1)^u}{nu+k}.
$$
Now, from the Taylor series of $\frac{1}{1+x}$ for $|x|<1$ we have
$$
{
\frac{1}{1+x}=\sum_{u=0}^\infty (-1)^ux^u\implies
\\
\frac{1}{1+x^n}=\sum_{u=0}^\infty (-1)^ux^{nu}\implies
\\
\frac{x^{k-1}}{1+x^n}=\sum_{u=0}^\infty (-1)^ux^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4355000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Calculate the determinant of $a_{ij} = \frac{(1+x)^{i+j-1}-1}{i+j-1}$ There is a question asked by my classmate. Looking forward to some ideas, thanks.
Set $A=\{a_{ij}\}_{n\times n}$, where
$$a_{ij}=\frac{(1+x)^{i+j-1}-1}{i+j-1}.$$
Prove that $\det A=cx^{n^2}$ for some $c$.
I have tried to calculate it, but failed. ... | Firstly, we know there is a nondegenerate matrix $J(n)$ such that
$$(1,x,\dots,x^{n-1})=(1,(1+x),\dots,(1+x)^{n-1})J(n).$$
Then we have
$$
\begin{aligned}
&\begin{pmatrix}
1&x&\cdots&x^{n-1}\\
x&x^2&\cdots&x^{n}\\
\vdots&\vdots&\ddots&\vdots\\
x^{n-1}&x^n&\cdots&x^{2n-2}
\end{pmatrix}
=(1,x,\cdots,x^{n-1})^T(1,x,\cdots... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4357189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
} |
I have a small problem with long division when dividing polynomials of the same degree I was doing long division with $x^2 + 1$, and $3x^2+5$. (the second polynomial is the quotient).
the problem I've found is related to dividing two polynomials of the same degree. Even if I know that the quotient is always a constant ... | You are confusing what you dividing into with what you are dividing by and you are taking the quotient, $3$, and multiply it by what you are dividing into; not what you are dividing by. You must multiply the quotient by what you are dividing by.
Question 1: $\frac {3x^2+ 5}{x^2 + 1}$ then we divide $x^2$ into $3x^2$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4360202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate $\int_0^\infty \frac{\cos(ax)}{1+x^N}dx$ for $a,N\in\mathbb{R}$ and $N> 1$ Interesting challenge: Does anyone know how to evaluate the integral $$\int_0^\infty\frac{\cos(ax)}{1+x^N}\text{d}x$$ for $a,N\in\mathbb{R}$ and $N>1$?
The $N=2$ case is quite easily done using the Residue Theorem as well as for ... | For even $N$ you can use the residue theorem as well. It suffices to consider $a>0$ and calculate the integral
$$\oint_{-\infty}^{\infty} \frac{e^{iax}}{1+x^N} \, {\rm d}x = \int_{0}^{\infty} \frac{2\cos(ax)}{1+x^N} \, {\rm d}x = 2\pi i \sum_{0 \leq k \leq \frac{N-1}{2}} {\rm Res}\left( \frac{e^{iax}}{1+x^N} \right)\Bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4363414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Show that the bissectors of $\angle BHC$ and $\angle BFC$ meet on $BC$ Given an acute triangle $\triangle ABC$ with orthocenter $H$. Let $D = BH \cap AC, E = CH \cap AB$ and $F = (AEDH) \cap (ABC) \neq A$. Show that the inner angle bisectors of $\angle BFC$ and $\angle BHC$ meet on the side $BC$.
It is easy to prove t... |
With the ideas of sirous, we use barycentric calculus. The coordinates of $D$ and $E$ are
\begin{equation}
D\bigg(\frac{a^2+b^2-c^2}{2b^2},0,\frac{-a^2+b^2+c^2}{2b^2}\bigg),\quad E\bigg(\frac{a^2-b^2+c^2}{2c^2},\frac{-a^2+b^2+c^2}{2c^2},0\bigg)
\end{equation}
The circumcircle $(ABC)$ has equation $-a^{2}yz-b^{2}xz-c^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4363558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Composition of piecewise functions - Strange result I'm trying to get $f(g(x))$, where:
$$
f(x)=
\begin{cases}
\sqrt{1-x} &\text{if } x \leq 1 \\
x &\text{if } x > 1
\end{cases}
$$
$$
g(x)=
\begin{cases}
x + 1 &\text{if } x \geq 0 \\
x^2 &\text{if } x < 0
\end{cases}
$$
I
followed these steps:
*
*$g(x \geq 0) = x + 1... | Since$$g(x)=\begin{cases}x+1&\text{ if }x\geqslant0\\x^2&\text{ if }x<0,\end{cases}$$you have$$g(x)\leqslant1\text{ if }x\in[-1,0]\quad\text{and}\quad g(x)>1\text{ if }x<-1\text{ or }x>0.$$So,\begin{align}f\bigl(g(x)\bigr)&=\begin{cases}\sqrt{1-g(x)}&\text{ if }x\in[-1,0]\\g(x)&\text{ if }x<-1\text{ or }x>0\end{cases}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4365986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
About integration of $\frac{1}{(1+a^2 x^2)(1+x^2)}$ in differentiation under integral sign Consider the function $F(a)=\int_0^{+\infty} \frac{\arctan(ax)}{x(1+x^2)} dx$. I assumed $a \ge 0$ because $F$ is odd, I applied the theorem of differentiation under the integral sign and arrive to $F'(a)=\int_0^{+\infty} \frac{1... | The short idea of the story below is to use a further deformation parameter, $b$, and avoid the denominator $(a^2-1)$ which makes problems in $a=\pm1$ by having instead $(a^2-b^2)$. The deformation used is continuous in $b$. Everything transposes on the computational side.
In order to exchange the order of differenti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4367582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Let $f(x) = \sin^{-1}(\frac{2x}{1+x^2})$ Show that $f(x) = 2\tan^{-1}(x)$
Let $$f(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) ~~ -\infty<x<\infty.$$
Show that,
(a) $f(x) = 2\tan^{-1}(x)$ for $-1\leq x \leq 1$ and
(b) $f(x) = \pi-2\tan^{-1}(x)$ for $x \geq 1.$
Proof: I started off by equating $$\sin^{-1}\left(\frac{... | Show that $f'(x)=\frac{2}{1+x^2}$, hence $f(x)$ and $2\arctan(x)$ differ only by a constant. Then substitute $x=0$ to show that this constant is $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4369535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Suppose X$\sim$ Cauchy(0,1). Then what will be the distribution of $\frac{1-X}{1+X}$? In order to find distribution of $\frac{1-X}{1+X}$ below approach I followed,
Let,
\begin{align}
Y = \frac{1-X}{1+X}
\end{align}
Then, cdf of Y is
\begin{align}
F_{Y}(y) = P(Y \leq y)
\end{align}
\begin{align}
= P\left(\fr... | $$P((1-X)/(X+1)\leq t)=\begin{cases}
P(\{X\leq -1\}\cup\{X\geq (1-t)/(1+t)\})&t>-1\\
P(\{(1-t)/(1+t)\leq X\leq -1\}& t<-1
\end{cases}$$
So for $t>-1$
$$P((1-X)/(X+1)\leq t)=\frac{1}{2}+\frac{1}{\pi}\arctan(-1)+\frac{1}{2}-\frac{1}{\pi}\arctan{\frac{1-t}{1+t}}=\\\frac{3}{4}-\frac{1}{\pi}\arctan{\frac{1-t}{1+t}}$$
while ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4375593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Representing the cube of any natural number as a sum of odd numbers I'm expanding my notes on exercises from Donald Knuth's The Art of Computer Programming, and found something rarely mentioned in the Internet, but still useful to prove Nicomachus' Theorem about the sum of cubes.
Knuth phrases this in the following way... | $\sum_{k=0}^{n-1} (n^2 -n + 1 + 2k)$ = $\sum_{k=0}^{n-1} n^2 + \sum_{k=0}^{n-1} (-n + 1 + 2k)$, but the second sum is 0 because of the symmetry relative to 0:
$\sum_{k=0}^{n-1} (-n + 1 + 2k) = \sum_{2k\leq n-1} \left((-n+1+2k) + (n-1-2k)\right) =\sum 0$. What remains is $\sum_{k=0}^{n-1} n^2 = n\cdot n^2 = n^3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4375782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Inverse of a 2x2 matrix with an example Need to find inverse of this matrix:
$
\begin {bmatrix}
1 & 3/5\\
0 & 1\\
\end {bmatrix}
$
This is how it has been solved:
$
\begin {bmatrix}
1 & 3/5\\
0 & 1\\
\end {bmatrix}
$ $
\begin {bmatrix}
x_1 & x_2\\
x_3 & x_4\\
\end {bmatrix}
$ = $
\begin {bmatrix}
1 & 0\\
0 & 1\\
\end {... | One may find the inverse matrix by row operations on the augmented matrix:
$$
\left[\begin{array}{rr|rr}
1 & 3/5 & 1 &0 \\
0 & 1 & 0 &1 \\
\end{array}\right]\overset{R_1'=-3/5R2+R1}{\longrightarrow}
\left[\begin{array}{rr|rr}
1 & 0 & 1 &-3/5 \\
0 & 1 & 0 &1 \\
\end{array}\right],
$$
so $$
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4375939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Newton's Method of Fluxions pg.8 paragraph $25$ question y'all. Newton has this problem. I'm using the free copy downloaded from the Library of Congress website, the original English. In paragraph 25 line 7 he states $y \times b - x = v$ . . . by which I guess he means $y(b-x)$. Then he says to write,$ v/(b-x) = y$... | No, there is no mistake. He has the equation $y^3+\frac{b^2}{b-x}\cdot y^2-x^3=0$. At the next step we substitute $y$ by $\frac{v}{b-x}$
$$\left(\frac{v}{b-x}\right)^3+\frac{b^2}{b-x}\cdot \left(\frac{v}{b-x}\right)^2-x^3=0$$
$$\frac{v^3}{(b-x)^3}+\frac{b^2v^2}{(b-x)^3}-x^3=0$$
Expanding the last summand by $(b-x)^3$.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4380833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
A better way to find the radius of the complex roots of $(z + 1)^5 = 32z^5$? I was gnawing on this problem today:
All the complex roots of $(z + 1)^5 = 32z^5,$ when plotted in the complex plane, lie on a circle. Find the radius of this circle.
I solved this by first dividing
$$
\left(\frac{z+1}{2z}\right)^5 = 1
$$
th... | Alternative approach.
The $5$th roots of $32$ are given by $2 \times k$, where
$k$ is one of the $5$ roots of $z^5 = 1.$
These roots are $~\displaystyle e^{i(2k\pi/5)} ~: ~k \in \{0,1,2,3,4\}.$
From the original problem, you have that
$$\left[\frac{z + 1}{z}\right]^5 = 32.$$
Therefore, $(z + 1) = z \times 2 \times e^{i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4382385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Number of ways to pick $3$ balls from a box with $8$ balls if $3$ of the balls are identical and the other $5$ are all different
A box contains 8 balls, of which 3 are identical and the remaining 5 are different from each other. 3 balls are to be picked out of the box; the order in which they are picked out does not m... | $\mathbf{\text{Generating Function Approach:}}$
The generating function of identical objects : $(1+x+x^2+x^3)$
The generating function of distinct objects : $(\binom{5}{0}x^0+ \binom{5}{1}x^1+\binom{5}{2}x^2+\binom{5}{3}x^3+\binom{5}{4}x^4 +\binom{5}{5}x^5)$
Then , find the coefficent of $x^3$ in the expansion of $$(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4382843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integrate : $\int \frac{1}{x^{2} \sqrt{2x-x^{2}}} dx$ Integrate :
$$I = \int \frac{1}{x^{2} \sqrt{2x-x^{2}}} dx$$
My attempt : substitute $\sin t = x-1$, $u = \tan \frac{t}{2}$
$$I = \int \frac{1}{x^{2} \sqrt{1-(x-1)^{2}}} dx = \int \frac{1}{(\sin t + 1)^{2} \cos t} \cos t dt$$
$$= \int \frac{1}{(\sin t + 1)^{2}} dt = ... | If $x<0$ then $\sqrt{2x-x^{2}}$ does not make sense on $\mathbb{R}$ so I suppose that $x>0$, we have
\begin{align*}
\int\frac{1}{x^{2}\sqrt{2x-x^{2}}}{\rm d}x&=\int \frac{1}{x^{2}\sqrt{1-(x-1)^{2}}}{\rm d}x\\&\overset{x\mapsto x-1}{=}\int \frac{1}{(x+1)^{2}\sqrt{1-x^{2}}}{\rm d}x\\&\overset{x\mapsto \sin x}{=}\int \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4386603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
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.