Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Integrate $\int\frac{\cos^2x}{1+\tan x}dx$ Integrate $$I=\int\frac{\cos^2x}{1+\tan x}dx$$
$$I=\int\frac{\cos^3xdx}{\cos x+\sin x}=\int\frac{\cos^3x(\cos x-\sin x)dx}{\cos^2x-\sin^2x}=\int\frac{\cos^4xdx}{1-2\sin^2x}-\int\frac{\cos^3x\sin xdx}{2\cos^2x-1}$$
Let $t=\sin x,u=\cos x,dt=\cos xdx,du=-\sin xdx$
$$I=\underbra... | Decompose the integrand as follows\begin{align}
\frac{\cos^2x}{1+\tan x}&= \frac{\cos^3 x \ (\cos x +\sin x)}{(\cos x+\sin x)^2}\\
&=\frac{(1+\cos2x)\ (\cos 2x + 1+ \sin2x)}{4(1+\sin 2x)}\\
&=\frac14\left( 1+\cos2x + \frac{\cos^2 2x+\cos2x}{1+\sin2x}\right)\\
&=\frac14\left( 2+\cos2x -\sin 2x +\frac{\cos2x}{1+\sin2x}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/930684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Variational problem concerning variances Let $\phi$ be the family consisting of all random variables $X$ such that $P(X\in [0,1])=1$, $EX=\frac{1}{3}$, $P(X<\frac{1}{4})<\frac{1}{2}$, $P(X>\frac{1}{4})\geq\frac{1}{2}$. Calculate $\sup \{Var(X):X \in \phi\} - \inf \{Var(X):X \in \phi\} .
$
| Let us put $I=\inf({\sf Var}(X)|X\in\phi)$ and $S=\sup({\sf Var}(X)|X\in\phi)$. We will
compute $I$ and $S$.
Trivially $I=0$ (take $X$ constant equal to $\frac{1}{3}$).
Let us now compute $S$. To ease notation, we introduce the
characteristic functions $\alpha=1_{X\in[0,\frac{1}{4})}$ and $\beta=1_{X\in[\frac{1}{4},1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/932269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
PEMDAS question: $F(x) = 3x^2 - x+2$. Find $[f(a)]^2$ How should I go about doing this? $(3a^2-a+2)^2$?
Thus, $9a^4-a^2+4$
| Hint For any real numbers $x,y$ we have $$\color{green}{(x-y)^2 = x^2-2xy+y^2}, \quad \color{blue}{(x+y)^2 = x^2+2xy+y^2},$$
and note that with $\color{green}{x = (2-a)}$ and $y = 3a^2$ we get
$$(3a^2-a+2)^2=\color{blue}{(x+y)^2} = \ldots$$
Can you continue from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/934165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Weighted Coin Toss Probablity Suppose two weighted coins are tossed. The first is weighted so that it comes up heads with probability $\frac{1}{3}$. The second is weighted so that it comes up heads with probability $\frac{1}{4}$. What is the probability that when both coins are tossed, one comes up heads and the other ... | *
*$P(HH)=\dfrac{1}{3}\cdot\dfrac{1}{4}=\dfrac{1}{12}$
*$P(HT)=\dfrac{1}{3}\cdot\dfrac{3}{4}=\dfrac{1}{4}$
*$P(TH)=\dfrac{2}{3}\cdot\dfrac{1}{4}=\dfrac{1}{6}$
*$P(TT)=\dfrac{2}{3}\cdot\dfrac{3}{4}=\dfrac{1}{2}$
So the answer to your question is $P(HT)+P(TH)=\dfrac{1}{4}+\dfrac{1}{6}=\dfrac{5}{12}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/935636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Modular calculus and square I want to prove that $4m^2+1$ and $4m^2+5m+4$ are coprimes and also $4m^2+1$ and $4k^2+1$ when $k\neq{m}$ and $4m^2+5m+4$ and $4k^2+5k+4$ when $k\neq{m}$. Firstly : Let $d|4m^2+1$ and $d|4m^2+5m+4$ then $d|4m^2+5m+4-(4m^2+1)=5m+3$ and $d|5m^2+3m$ thus $d|5m^2+3m-(4m^2+5m+4)=m^2-2m-4$ and $d|... | You've already found $d$ divides $5m+3$
Again, $d$ divides $4m(5m+3)-5(4m^2+1)=12m-5$
$\implies d$ divides $12(5m+3)-5(12m-5)=61$
Now, $4m^2+1\equiv0\pmod{61}\iff m\equiv\pm36\pmod{61}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/936312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Without using Taylor expansion or L'Hospital rule evaluate the limit:
$$L=\lim_{x\to0}\frac{e^x-1-x-x^2/2}{x^3}$$
I give the two alternate ways so that no one gives them again in their answer:
Using L'Hospital:
$$L=\lim_{x\to0}\frac{e^x-x-1}{3x^2}=\lim_{x\to0}\frac{e^x-1}{6x}=\lim_{x\to0}\frac{e^x}{6}=\frac16$$
Usin... | I will deal with the case $x \to 0^{+}$ and leave the $x \to 0^{-}$ case for OP and other readers. Let us use the formula $$e^{x} = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n}$$ We then have $$\begin{aligned}L &= \lim_{x \to 0^{+}}\dfrac{e^{x} - 1 - x - \dfrac{x^{2}}{2}}{x^{3}}\\
&= \lim_{x \to 0^{+}}\dfrac{\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/936542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Derivative of $f(x) = x^2 \sin(1/x)$ using the derivative definition derivative of $f(x) = x^2 \sin(1/x)$ using the derivative definition
When not using the derivative definition I get $\cos (1/x) + 2x \sin(1/x)$,
which WolframAlpha agrees to.
However when I try solving it using the derivative definition:
$$\lim_ {h\to... | You're doing wrongly the computation:
\begin{align}
f(x+h)-f(x)&=(x+h)^2\sin\frac{1}{x+h}-x^2\sin\frac{1}{x}\\
&=x^2\left(\sin\frac{1}{x+h}-\sin\frac{1}{x}\right)+2hx\sin\frac{1}{x+h}+h^2\sin\frac{1}{x+h}
\end{align}
When you divide by $h$ you get
$$
\frac{f(x+h)-f(x)}{h}=
\frac{x^2}{h}\left(\sin\frac{1}{x+h}-\sin\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/940264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Calculus I - Simple Difference Quotient Question The problem is to calculate the difference quotient of $f(x) = \sqrt{x^2 +2x+1}$. But $\sqrt{x^2+2x+1}= \sqrt{x+1}^2 = x+1$ so can I just take the difference quotient of $x+1$? If not, how do I simplify $\frac{\sqrt{x^2+2xh+h^2+2x+2h+1} - \sqrt{x^2+2x+1}}{h}$ ?
| Not observing the fact that $\sqrt{x^2+2x+1}=x+1$, that lonely $h$ will factor out of the denominator in your other form if you multiply the entire mess by
$$\frac{\sqrt{x^2+2xh+h^2+2x+2h+1} + \sqrt{x^2+2x+1}}{\sqrt{x^2+2xh+h^2+2x+2h+1} + \sqrt{x^2+2x+1}}.$$
Things will suddenly start to cancel in the numerator signif... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/941054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
manipulations with Taylor expansions for log and sinh How could we derive the equality
$$ \frac14 \sum_{m=1}^\infty \frac1m \frac{1}{\sinh^2 \frac{m\alpha}2} = - \sum_{n=1}^\infty n\log (1-q^n)$$
where $q=e^{-\alpha}$ ?
| $$ \begin{align} \sum_{m=1}^{\infty} \frac{1}{m} \frac{1}{\sinh^{2} \frac{m \alpha}{2}} &= \sum_{m=1}^{\infty} \frac{1}{m} \frac{4}{(e^{m \alpha /2}-e^{- m \alpha /2})^{2}} \\ &= 4 \sum_{m=1}^{\infty} \frac{1}{m} \frac{e^{-m \alpha }}{(1-e^{-m \alpha})^{2}} \\ &=4 \sum_{m=1}^{\infty} \frac{1}{m}\sum_{n=1}^{\infty} n(e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/942008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Looking for the logic of a sequence from convolution of probability distributions I am trying to detect a pattern in the followin sequence from convolution of a probability distribution (removing the scaling constant $\frac{6 \sqrt{3}}{\pi }$: $$\left(\frac{1}{\left(x^2+3\right)^2},\frac{2
\left(x^2+60\right)}{\left... | Are you sure about the $564x^6$ term? All polynomial coeffs are divisible by $k$ with the exception of that one.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/942204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Using a given identity to solve for the value of an expression This problem caught my eye in the book yesterday. Till now I still get stuck. Here it is:
If $$\frac{x}{x^2+1}=\frac{1}{3},$$ what is the value of $$\frac{x^3}{x^6+x^5+x^4+x^3+x^2+x+1}?$$
The denominator is a cyclotomic polynomial which can be expressed as... | It can also solved by this equation $x^2+1=3x$.
$$\frac{x^3}{x^6+x^5+x^4+x^3+x^2+x+1}=\frac{x^3}{3x^5+x^5+x^3+3x+x}=
\frac{x^3}{4x^5+4x^3+4x-3x^3}=\frac{x^3}{12x^4+4-3x^3}= \cdots $$
The steps is natural..., I omit; however the answer given by Jack is better.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/942914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Help with dy/dx of natural log
Could someone explain how I would work this problem with steps?
| It is easier to simplify first:
$$
y = \ln x - \ln(1+x^2).
$$
Then
\begin{align*}
\frac{dy}{dx} & = \frac{d}{dx}\ln x - \frac{d}{dx} \ln(1+x^2).
\end{align*}
The first derivative is $1/x$. For the second derivative, we need to use the chain rule:
\begin{align*}
\frac{d}{dx} \ln(1+x^2) & = \frac{d\ln(1+x^2)}{d(1+x^2)} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/945201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Trigonometric Identity involving $\sin^2$ and $\cos^2$ I was trying to prove this trigonometric identity, it looks like using the elementary relations should be enough, but I still can't find how:
$$\frac{1}{2}\sin^2 a \ \sin^2 b + \cos^2a \ \cos^2 b = \frac{1}{3} + \frac{2}{3} \biggl( \frac{3}{2}\cos^2 a - \frac{1}{... | The left hand side is
$$\begin{align}\frac 12\sin^2 a\sin^2b+\cos^2 a\cos^2 b&=\frac 12(1-\cos^2 a)(1-\cos^2b)+\cos^2a\cos ^2b\\&=\frac 12-\frac 12\cos^2a-\frac12\cos ^2b+\frac 32\cos^2a\cos^2b.\end{align}$$
The right hand side is
$$\frac 13+\frac 23\left(\frac 32\cos^2a-\frac 12\right)\left(\frac 32\cos^2b-\frac 12\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/945899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
A triangle determinant that is always zero How do we prove, without actually expanding, that
$$\begin{vmatrix}
\sin {2A}& \sin {C}& \sin {B}\\
\sin{C}& \sin{2B}& \sin {A}\\
\sin{B}& \sin{A}& \sin{2C}
\end{vmatrix}=0$$
where $A,B,C$ are angles of a triangle?
I tried adding and subtracting from the rows and columns and ... | $$\begin{pmatrix}
\sin(2A) & \sin C & \sin B \\
\sin C & \sin (2B) & \sin A \\
\sin B & \sin A & \sin (2C) \\
\end{pmatrix} =
\begin{pmatrix}
\sin(2A) & \sin (\pi-A-B) & \sin (\pi-A-C) \\
\sin (\pi-A-B) & \sin (2B) & \sin (\pi-B-C) \\
\sin (\pi-A-C) & \sin (\pi-B-C) & \sin (2C) \\
\end{pmatrix} $$
$$=
\begin{pmatrix}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/946524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 4,
"answer_id": 0
} |
How do I verify the solution for this problem? (Advanced Calculus) I'm working on a problem in Rudin (Chapter 1, Exercise 1.19)...
Question:
Suppose $\mathbf a\in R^k, \mathbf b\in R^k.$ Find $\mathbf c \in R^k$ and $r>0$ such that
$|\mathbf{x-a}| = 2|\mathbf{x-b}|$
if and only if $|\mathbf{x-c}| = r$
Solution: $3 \ma... | Working with the first equation, notice that:
\begin{align*}
&~|\mathbf{x-a}| = 2|\mathbf{x-b}| \\
&\iff |\mathbf{x-a}|^2 = 4|\mathbf{x-b}|^2 \qquad\text{since norms are nonnegative} \\
&\iff (\mathbf x - \mathbf a) \cdot (\mathbf x - \mathbf a) = 4[(\mathbf x - \mathbf b) \cdot (\mathbf x - \mathbf b)] \\
&\iff \mathb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/952200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Trigonometric Integral $\int _0^{\pi/4}\:\frac{dx}{\cos^4x-\cos^2x\sin^2x+\sin^4x}$ Q. $$\int _0^{\pi/4}\:\frac{1}{\cos^4x-\cos^2x\sin^2x+\sin^4x}$$
My method:
$$\int_0^{ \pi /4} \frac{dx}{(\cos^2x+\sin^2x)^2-3\cos^2x \sin^2x}=\int_0^{ \pi /4} \frac{dx}{1-3\cos^2x\sin^2x} $$
Dividing numerator and denominator by $\cos^... | When you divide the numerator and the denominator by $\cos^2 x$, you should get $$\displaystyle \int^{\pi/4}_0 \dfrac{\sec^2 x dx}{\sec^2 x - 3\sin^2 x}$$ instead of $$\displaystyle \int^{\pi/4}_0 \dfrac{\sec^2 x dx}{\sec^2 x - 3 \tan^2 x}$$
because
$$\begin{array}{rcll}
\displaystyle \int^{\pi/4}_0 \dfrac{dx}{1 - 3\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/952307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Find the range of $f(x) =(x-1)^{1/2} + 2\cdot(3-x)^{1/2}$ How to take out the range of the following function :
$$f(x) =(x-1)^{1/2} + 2\cdot(3-x)^{1/2}$$
I am new to functions hence couldn't come up with a solution.
| Your function
$$f(x) =\sqrt{x-1} + 2\sqrt{3-x}$$
is defined for all $x \in [1, 3]$, otherwise the square root are not defined for real number.
We can check what happens on the border of the segment $[1, 3]$:
*
*$f(1) = 2\sqrt{2}$;
*$f(3) = \sqrt{2}$.
Also, we can find the local maxima and minima of $x$ in the se... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/953842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Series question,related to telescopic series, 1/2*4+ 1*3/2*4*6+ 1*3*5/2*4*6*8 ...infinity The series is $$\frac{1}{2*4}+ \frac{1*3}{2*4*6}+ \frac{1*3*5}{2*4*6*8}+....$$
It continues to infinity.I tried multiplying with $2$ and dividing each term by$(3-1)$,$(5-3)$ etc,starting from the second term which gives me $$\fr... | We have
$$
\begin{align}
\frac{1\cdot 3\cdot 5 \cdots(2n-1)}{2\cdot 4\cdot 6\cdots(2n+2)}&=\frac{1\cdot 2\cdot 3\cdot 4 \cdot 5\cdot6\cdots(2n-1)\cdot 2n}{(2\cdot 4\cdot 6\cdots (2n))^2(2n+2)}\\\\
&=\frac{(2n)!}{(2^{n} \cdot 1\cdot 2\cdot 3 \cdot 4 \cdots n)^2 \cdot (2n+2)}\\\\
& =\frac{(2n)!}{2^{2n} (n!)^2 \cdot (2n+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/954922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Problem with Smith normal form over a PID that is not an Euclidean domain This is an homework exercise of the Algebra lecture.
I need to evaluate the Smith normal form of the following matrix
$$A:=\begin{pmatrix}1 & -\xi & \xi-1\\2 \xi&8&8\xi+7\\\xi& 4 & 3\xi +2 \end{pmatrix} \in M(3\times 3;\Bbb{Z}[\xi]),$$
wh... | $\xi^2=\xi-5$, and therefore $4+\xi^2=\xi-1$ (which is a prime element). On the other side, $2\xi+3=2(\xi-1)+5=2(\xi-1)+\xi-\xi^2=-(\xi-1)(\xi-2)$, so $b_{33}\mid b_{22}$.
The SNF of your matrix is $$\begin{pmatrix}1 & 0 & 0\\0&\xi-1&0\\0& 0 & (\xi-1)(\xi-2) \end{pmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/956591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Evaluation of $\int \frac{x\sin(\sin x)}{x+5} \ dx$ How do we find $$\int \frac{x\sin(\sin x)}{x+5} \ dx\ ,$$ is there any way to take that $\sin x$ out from parent $\sin(\cdot)$ ?
| Let $u=x+5$ ,
Then $x=u-5$
$dx=du$
$\therefore\int\dfrac{x\sin(\sin x)}{x+5}dx$
$=\int\dfrac{(u-5)\sin(\sin(u-5))}{u}du$
$=\int\sin(\sin(u-5))~du-\int\dfrac{5\sin(\sin(u-5))}{u}du$
For $\int\sin(\sin(u-5))~du$ ,
$\int\sin(\sin(u-5))~du$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\sin^{2n+1}(u-5)}{(2n+1)!}du$
$=-\int\su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/960558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Write it as an element of this ring? Since the degree of the irreducible polynomial $x^3+2x+2$ over $\mathbb{Q}[x]$ is odd, it has a real solution , let $a$. I am asked to express $\displaystyle{\frac{1}{1-a}}$ as an element of $\mathbb{Q}[a]$.
I have done the following:
Since $a$ is a real solution of $x^3+2x+2$, we h... | Without any trick: you know that $(1-a)^{-1}=\alpha+\beta a+\gamma a^2$ for some $\alpha,\beta,\gamma\in\mathbb{Q}$, that is
$$
(1-a)(\alpha+\beta a+\gamma a^2)=1
$$
This becomes
$$
\alpha+\beta a+\gamma a^2-\alpha a-\beta a^2-\gamma a^3=1
$$
Since $a^3=-2a-2$ this becomes
$$
(\alpha+2\gamma)+(\beta-\alpha+2\gamma)a+(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/961296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the limit of $\sqrt{x^2+x-1} +x$ as $x\to-\infty$ Find the limit of $\sqrt{x^2+x-1} +x$ as $x\to-\infty$.
My solution:
multiplying by: $\displaystyle\frac{\sqrt{x^2+x-1}-x}{\sqrt{x^2+x-1}-x}$
Which gives us: $\displaystyle\frac{x-1}{\sqrt{x^2+x-1}-x}$
dividing by $\sqrt{x^2}$ gives:
$\displaystyle \frac{1}{\sqrt{... | Your mistake at this step:
dividing by $\sqrt{x^2}$ gives: $\displaystyle \frac{1}{\sqrt{1}-1}$
Since $x\to -\infty$, then $\sqrt{x^2}=-x$. So $\frac{x-1}{\sqrt{x^2+x-1}-x}=\frac{-1+\frac 1x}{\sqrt{1+\frac 1x-\frac{1}{x^2}}+1}.$
Take careful!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/961741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Inequality: $ \frac{a}{1+9bc+k(b-c)^2}+\frac{b}{1+9ca+k(c-a)^2}+\frac{c}{1+9ab+k(a-b)^2}\geq\frac{1}{2} $ I was trying to solve this inequality, but I wasn't able to do so:
Find the maximum number $k\in\mathbb R$ such that:
$$ \frac{a}{1+9bc+k(b-c)^2}+\frac{b}{1+9ca+k(c-a)^2}+\frac{c}{1+9ab+k(a-b)^2}\geq\frac{1}{2} $$
... | Plugging, $a=1/2,b=1/2,c=0 \implies k \le 4$.
I will show that $k=4$.
$$ \frac{a}{1+9bc+4(b-c)^2}+\frac{b}{1+9ca+4(c-a)^2}+\frac{c}{1+9ab+4(a-b)^2}\ge \frac{1}{2} \\ \Longleftrightarrow \frac{a^2}{a+9abc+4a(b-c)^2}+\frac{b^2}{b+9abc+4b(c-a)^2}+\frac{c^2}{c+9abc+4c(a-b)^2} \ge \frac{1}{2}$$
By cauchy it suffice to show... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/962146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Question on an algebraic expression Why this is true?
$$
\frac{\frac{1}{\sqrt{x}}-1}{x-1} = -\frac{1}{\sqrt{x}+x}
$$
| When you have complicated fractions, it is always a good idea to try making common denominators where applicable. The numerator is a bit complicated in this case, so let's work with that:
$$\frac{1}{\sqrt{x}} - 1 = \frac{1}{\sqrt{x}} - \frac{\sqrt{x}}{\sqrt{x}} = \frac{1-\sqrt{x}}{\sqrt{x}}.$$
Now let's replace $\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/963132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\lfloor x\rfloor + \lfloor y\rfloor \leq \lfloor x+y\rfloor$ for every pair of numbers of $x$ and $y$ Give a convincing argument that $\lfloor x\rfloor + \lfloor y\rfloor \leq \lfloor x+y\rfloor$ for every pair of numbers $x$ and $y$. Could someone please explain how to prove this? I attempted to say that the largest ... | Let $x = a + m$ and $y = b + n$ where $a, b$ are integers and $m,n$ are real numbers such that $0 \le m,n < 1$. Then,
$$\lfloor x \rfloor + \lfloor y \rfloor = a + b$$
while
$$\begin{align}\lfloor x + y \rfloor &= \lfloor a + b + m + n \rfloor\\
&= a + b + \lfloor m + n \rfloor\\
& \ge a + b\\
&= \lfloor x \rfloor + \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/963245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
study of two sequences I need to study wether thos two sequences converge or not.
1) $u_n=n\sum_{k=1}^{2n+1} \frac{1}{n^2+k}$
2) $v_n=\frac{1}{n}\sum_{k=0}^{n-1} \cos(\frac{1}{\sqrt{n+k}})$
For the first i get it converges to $0$ by just expressing the sum as a fraction (not very smart imo).
| How can $u_n$ converge to zero?
$$ u_n = n\sum_{k=1}^{2n+1}\frac{1}{n^2+k}\geq n\cdot\frac{2n+1}{(n+1)^2} \geq 2-\frac{3n+2}{(n+1)^2}.$$
On the other hand,
$$ u_n \leq n\cdot\frac{2n+1}{n^2+1} \leq 2+\frac{n-2}{n^2+1},$$
hence:
$$\lim_{n\to +\infty} u_n = 2.$$
For the second sum, since $\cos x \geq 1-\frac{x^2}{2}$ ove... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/963807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving that $\sum \overline{PV_i}^2=\frac{nl^2}{4}\left(1+2\cot^2 \frac{\pi}{n}\right)$ Given that $P$ is any point on the incircle of a regular $n$-sided polygon with edge length $l$ and vertices $V_1,V_2...V_n$, how do we prove that $$\displaystyle\sum_{i=1}^{n} \overline{PV_i}^2=\dfrac{nl^2}{4} \left(1+2\cot^2 \dfr... | Let $W_i$ be the midpoint of the segment $\overline{V_iV_{i+1}}$ (where $V_{n+1}=V_1$). Apply the Apollonius' theorem on the triangle $\triangle PV_iV_{i+1}$ with the midpoint $W_i$. We have
$$\overline{PV_i}^2+\overline{PV_{i+1}}^2=2(\overline{PW_i}^2+\overline{V_iW_i}^2)=2\overline{PW_i}^2+\frac{l^2}{2}$$
Summing thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/963987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\sin(2\arcsin x + \arccos x) = \sqrt{1-x^2}$ I'm trying to prove that
$$
\sin(2\arcsin x + \arccos x) = \sqrt{1-x^2}.
$$
Let $\alpha = 2 \arcsin x$ and $\beta = \arccos x$; meaning $\sin\alpha = \frac{x}{2}, \cos\beta = x$. We know that:
$$
\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta.
$... | $$\arcsin x+\arccos x=\frac\pi2$$
So, we have $$\sin\left(\frac\pi2+\arcsin x\right)=\cos(\arcsin x)$$
Now $\cos(\arcsin(x)) = \sqrt{1 - x^2}$. How?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/967041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Justify: $R$ is commutative
Let $R$ be a ring. If $(ab)^n=a^nb^n$ holds for all $n$ and all $a, b\in R$, then justify that $R$ is commutative ring.
No idea how to prove it. But also fail to get one counter example. Please help
| By assumption
$$
(a(b+1))^2=a^2(b+1)^2=a^2b^2+2a^2b+a^2
$$
and by distrubutive law
$$
(a(b+1))^2=(ab+a)^2=(ab)^2+aba+a^2b+a^2
$$
From this it follows that
$$
a^2b=aba
$$
and this holds for all $a,b \in R$. Especially for $a+1$ and $b$
$$
(a+1)^2b=(a+1)b(a+1)=(ab+b)(a+1)=aba+ab+ba+b
$$
but compute the left side to get
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/971443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find $\int \frac{x^2}{2x-1}\mathrm{d}x$ Use the substitution $2x-1$ to find $\int \frac{x^2}{2x-1}\mathrm{d}x.$
I started out by using the substitution to find $\mathrm{d}x$ in terms of $\mathrm{d}u$.
$$
\frac {\mathrm{d}u}{\mathrm{d}x} = 2
$$
$$
\mathrm{d}x = \frac {1}{2}\mathrm{d}u
$$
Now the question looks like th... | HINT
Here it is another way to solve it: consider the following algebraic manipulation
\begin{align*}
\frac{x^{2}}{2x-1} & = \frac{1}{4}\times\frac{4x^{2}}{2x-1} = \frac{1}{4}\times\frac{(4x^{2}-1) + 1}{2x-1} = \frac{1}{4}\times\left[\frac{4x^{2}-1}{2x-1} + \frac{1}{2x-1}\right]\\\\
& = \frac{1}{4}\times\left[2x+1 + \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/972215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Modular Arithmetic: using congruence to find remainder How do I use the fact that if $a = b \pmod n$ and $c = d\pmod n$ then $ac = bd\pmod n$ to find the remainder when $3^{11}$ is divided by $7$?
| We know that:
$3\equiv3\pmod7$. Multiplying both sides by 3, we get $3^2\equiv9\pmod7$, which simplifies to:
$3^2\equiv2\pmod7$. Multiplying both sides by 3 again (and simplifying):
$3^3\equiv6\pmod7$
$3^4\equiv4\pmod7$
$3^5\equiv5\pmod7$
$3^6\equiv1\pmod7$
$3^7\equiv3\pmod7$
$3^8\equiv2\pmod7$
$3^9\equiv6\pmod7$
$3^{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/972801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
A double Summation involving 7th roots of unity Is there possibly a closed form for $$\sum_{m=1}^{\infty} \left(\sum_{k=1}^{6} \dfrac{1}{m-\alpha^k}\right)^2$$ where $\alpha=e^{2\pi i/7}$ ?
I'm having problems evaluating the first sum, let alone the second. I tried some algebraic manipulation, but I could only find tha... | The six numbers $m-\alpha^k$ are the roots of $f(X) = \frac{(m-X)^7 - 1}{m-X-1} = X^6 + a_5 X^5 + \cdots + a_1 X + a_0$, so the sum of their reciprocals is $-\frac{a_1}{a_0}$.
Letting $X=0$, we have $a_0 = \frac{m^7-1}{m-1}$. Taking a derivative, we have $a_1 = -\frac{6m^7 -7m^6+1}{(m-1)^2}$, therefore:
$$s_m = \sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/975354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluate and find the principal value of $(-1+i)^ {2-i}$ Can anyone please help me evaluate and find the principal value of
$(-1+i)^{2-i}$
I got up to
$=e^{2-i}(ln(-1+i))$
$=e^{(2-i)(1/2 ln(2)+i(3pi/4))}$
| \begin{align}
(-1+i)^{2-i} &= \left[ e^{\pi i} \, \sqrt{2} \, e^{- \pi i/4} \right]^{2 - i} \\
&= e^{2 \pi i + \pi} \, e^{(2-i) \ln(\sqrt{2})} \, e^{- \pi i/2 - \pi/4} \\
&= 2 \, e^{ 3 \pi/4} \left( e^{i(3 \pi + \ln(2))/2} \right)\\
&= 2 \, e^{ 3 \pi/4} \left[ \cos\left( \frac{3 \pi + \ln(2)}{2} \right) + i \,
\sin\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/975553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Using $\lim\limits_{x \to 0} \frac{\sin x}{x} = 1$ evaluate the limit Use $$\lim_{x\to 0} \frac{\sin x}{x} = 1$$
to evaluate the limit
$$\lim_{x\to 0} \frac{x\tan (x^2)}{\cos(3x)\sin^3(2x)}$$
I'm really not sure how to go about this apart from trig identities and using the limit $=1$ to simplify it. Can anyone even get... | $$\lim_{x\to0}\frac{x\tan(x^2)}{\cos(3x)\sin^3(2x)}=\lim_{x\to0}\frac{1}{\cos(3x)}\cdot\frac{x^3}{x^2}\cdot\frac{\tan(x^2)}{\sin^3(2x)}=$$
$$\lim_{x\to0}\frac{1}{\cos(3x)}\cdot\frac{x^3}{\sin^3(2x)}\cdot\frac{\tan(x^2)}{x^2}=\lim_{x\to0}\frac{1}{\cos(3x)}\cdot[\frac{x}{\sin(2x)}]^3\cdot\frac{\tan(x^2)}{x^2}$$
Now you h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/977789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Finding all natural $n$ such that $2^n+2^{2n} +2^{3n}$ has only $2$ prime factors. Find all natural $n$ such that $2^n+2^{2n} +2^{3n}$ has only $2$ prime factors.
I've tried checking the first 6-7 $n$'s on wolframalpha, but I don't see any patterns for even nor odd $n$'s. At first I thought for all odd $n$'s it was di... | $2^n + 2^{2n}+2^{3n} = 2^n(1+2^n + 2^{2n}) = 2^n p^a$ for some prime $p$ and positive integer $a$.
Then, $p^a = 1+2^n + 2^{2n}$.
Suppose $a>1$.
If $a$ is even, $a = 2b$, meaning $p^a - 1 = (p^b - 1)(p^b+1) = 2^n(1+2^n)$, which is impossible (you can see that easily). Then $a$ must be odd.
Then, $2^n + 2^{2n}= (p-1)(p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/978167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Is the derivative of the characteristic polynomial equal to the sum of characteristic polynomial of principal submatrices? Let $A$ by an $n \times n$ matrix over the complex numbers and let $\phi(A,x) = \det(xI-A)$ be the characteristic polynomial of $A$. Let $B_i$ be the principal submatrix of $A$ formed by deleting ... | It is true.
Proof:
$$
\phi(A,x) =
\begin{pmatrix}
x-a_{1,1} & -a_{1,2} & \cdots & -a_{1,n} \\
-a_{2,1} & x-a_{2,2} & \cdots & -a_{2,n} \\
\vdots & \vdots & \ddots & \vdots \\
-a_{m,1} & -a_{m,2} & \cdots & x-a_{m,n}
\end{pmatrix}
$$Let us write the derivative using multilinearity:
$$
\phi'(A,x) =
\begin{pma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/978815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Proof by induction for $ 2^{k + 1} - 1 > 2k^2 + 2k + 1$ for $k > 4$ I was given this proof for hw.
Prove that $ 2^{k + 1} - 1 > 2k^2 + 2k + 1$
So, far I've gotten this
Basis:
$k = 5$, $2^{5 + 1} - 1 > 2\cdot5^2 + 2\cdot5 + 1$ => $63 > 61$ (So, the basis holds true)
Hypothesis:
for all $k > 4$, $ 2^{k + 1} - 1 > 2k^2 +... | Hypothesis:
$$\begin{align} \forall k > 4,&\quad\; 2^{k + 1} - 1 > 2(k)^2 + 2(k) + 1 \\ &\iff 2^{k+1} \gt 2k^2 + 2k + 2 \end{align}$$
Inductive step:
$$\begin{align} 2^{k + 1 + 1}& = 2\cdot 2^{k+1}\\
& \gt 2\cdot (2k^2 + 2k+2) \\
& = 4k^2 + 4k + 4\\
& = (2k^2 + 4k + 2) + (2k^2 + 2)\\
& = 2(k+1)^2 + 2k^2 + 2 \\
& \g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/979355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Show that inequality holds How would you show that the following inequality holds? Could you please write your reasoning by solving this problem too?
$a^2 + b^2 + c^2 \ge ab + bc + ca$ for all positive integers a, b, c
I have tried:
$a^2 + b^2 + c^2 - ab - ab + ab \ge bc + ca $
$(a^2 -2ab + b^2) + c^2 + ab \ge bc + ca... | Hint
$$0\leq (a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/982891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Find the tangen to $\cos(\pi \cdot x)$ I have the following assignment.
Find the tangent to $y=f(x)=\cos(\pi \cdot x)$ at $x=\displaystyle\frac{1}{6}$.
First step would be to take the derivative of $f(x)$
$f'(x)= -\pi \sin(\pi \cdot x)$
Then I put the $x$-value into $f'(x)$ to find the slope
$f'(\displaystyle\frac{1}{6... | Start with your answer: $y = -\dfrac{\pi x}{2} + \dfrac{\sqrt{3}}{2} + \dfrac{\pi}{12}$
Multiply both sides by $12$: $12y = -6\pi x + 6\sqrt{3} + \pi$
Add $6 \pi x$ to both sides: $6\pi x + 12y = 6\sqrt{3} + \pi$
Therefore your answer is equivalant to the book's answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/984565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Show that $F = \{a + b\sqrt{5} | a, b ∈ \mathbb Q\}$ is a field Question:
Show that $F = \{a + b\sqrt{5} | a, b ∈ \mathbb Q\}$ is a field under the
operations - addition and multiplication where addition is given by:
$(a + b\sqrt{5})+(c + d\sqrt{5}) = (a + c) + (b + d)\sqrt{5}$ and
multiplication is given by $(a... | We just need to simplify that expression so that it has the form:
$$
\alpha + \beta\sqrt{5}
$$
for some $\alpha,\beta \in \mathbb Q$. Indeed, we can do this via conjugates:
\begin{align*}
\frac{1}{a + b\sqrt 5}
&= \frac{1}{a + b\sqrt 5} \cdot \frac{a - b\sqrt 5}{a - b\sqrt 5} \\
&= \frac{a - b\sqrt 5}{a^2 - 5b^2} \\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/985020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Find $D$ in RSA cryptosystem For the following encryption key $(n, E)$ in the RSA cryptosystem, compute $D$.
$(n, E)= (451, 231)$
So I know $n=11*41$, so $m=400$. Now $D=$ inverse of $231 \ (mod \ 400)$. However I am not sure how to find the inverse of $231$ in $mod\ 400$. Help with this would be great.
| Expanding on the comment, the Euclidean algorithm gives:
\begin{align*}
400 &= 231 \times 1 + 169.\\
231 &= 169 \times 1 + 62.\\
169 &= 62 \times 2 + 45.\\
62 &= 45 \times 1 + 17.\\
45 &= 17 \times 2 + 11.\\
17 &= 11 \times 1 + 6.\\
11 &= 6 \times 1 + 5.\\
6 &= 5 \times 1 + 1.
\end{align*}
Then, we do this in reverse..... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/986623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to find $α^2(β^4 +γ^4 +δ^4)+β^2(γ^4 +δ^4 +α^4)+γ^2(δ^4 +α^4 +β^4)+δ^2(α^4 +β^4 +γ^4)$
How to do the part (iv) . Please help.
Here are my answers to the first parts:
(i)
α a root of given equation $\implies \alpha^4-5 \alpha^2 + 2 \alpha -1 = 0$
$\implies \alpha^{n+4} - 5 \alpha^{n+2} + 2\alpha^{n+1} -\alpha^n=... | Note that $\beta^4 + \gamma^4 + \delta^4 = S_4 - \alpha^4$, and similarly for the other quantities in parentheses. Substitute this in your desired expression, and expand it. You can then write the entire expression in terms of various $S$'s.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/987105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluation of $\int_0^{\pi/4} \sqrt{\tan x} \sqrt{1-\tan x}\,\,dx$ How to evaluate the following integral
$$\int_0^{\pi/4} \sqrt{\tan x} \sqrt{1-\tan x}\,\,dx$$
It looks like beta function but Wolfram Alpha cannot evaluate it. So, I computed the numerical value of integral above to 70 digits using Wolfram Alpha and I u... | Here I present a completely elementary proof using only high school level techniques.
With the obvious substitution $\displaystyle \large t = \tan{x}$
We proceed as follows:
Using the Euler Substitution of the third kind, the following substitution is obtained:
$\displaystyle \large \sqrt{t-t^2} = tz \Rightarrow t = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/989021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 8,
"answer_id": 7
} |
Algorithm - iteration method
$$\begin{align} &T(n)=n+T(n-1)= \\
&= n + (n-1)+T(n-2)= \\
&= n + (n-1)+(n-2)+T(n-3)= \\
&= n + (n-1)+(n-2)+...+2+T(1)= \\
&= \frac{n(n+1)}{2}-1+T(1) \\
& \mathrm{Hence,} \ T(n)=\frac{n^2+n}{2}-1+T(1)=\Theta(n^2)
\end{align} $$
What is $n(n+1)$?
| $T(n) = n + (n-1) + (n-2) + ... + 2 + T(1)$
$\quad\quad\; = n + (n-1) + (n-2) + ... + 2 + 1 - 1 + T(1)$
$\quad\quad\; = (\sum_{i=1}^{n} i ) - 1 + T(1)$
$\quad\quad\; = \frac{n(n+1)}{2} - 1 + T(1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/989515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If I roll three dice at the same time, how many ways the sides can sum up to $13$? If I rolled $3$ dice how many combinations are there that result in sum of dots appeared on those dice be $13$?
| It's the coefficient of $x^{13}$ in the product $(x+x^2 + x^3 + x^4+ x^5 + x^6)^3$.
To see this, note that to compute that coefficient we have to identify all ways we can form $x^{13}$ by picking one term from each of the three terms $(x + x^2 + x^3 + x^4 + x^5 + x^6)$ we have; we could have $x$ from the first, $x^6$ f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/992125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 0
} |
How to find $\omega^7$ and $\omega^6$ from $\omega^5+1=0$
I did the first parts :
$$\omega= (\cos \pi + i \sin \pi)^\frac{1}{5} \implies \omega^5 = \cos \pi + i \sin \pi=-1$$
$\omega=-1$ is a root so :
$$\omega^5-1= (\omega+1) (\omega^4-\omega^3+\omega^2-\omega+1)=0$$
$$\omega^4-\omega^3+\omega^2-\omega=-1$$
$$\o... | For the quadratic equation part, note that if $(x-a)(x-b)=0$ the roots of the equation are $a$ and $b$.
Expand the brackets and you get the equation $x^2-(a+b)x+ab=0$ with roots $a$ and $b$.
This means that if you know the sum of the roots $a+b$ and the product of the roots $ab$, you can write down a quadratic equation... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/994000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Partial Fractions Decomposition I am failing to understand partial fraction decomposition in cases like the following:
Provide the partial fraction decomposition of the following:
$$\frac{x+2}{(x-4)^3(x^2 + 4x + 16)}$$
I see this and I think of
$$\frac{A}{x-4} + \frac{Bx+C}{(x-4)^2} + \frac{Dx^2 + Ex + F}{(x-4)^3} + \... | You have the correct understanding that the numerator is one degree lower than the denominator. For example:
$$\frac{A}{x+1}$$
$$\frac{Ax+B}{x^2+3x+2}$$
$$\frac{Ax^2+Bx+C}{x^3+2x^2+5x+1}$$
However, repeated terms are an exception. The numerator is one degree less than the repeated polynomial in the denominator. You hav... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/994252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Test for convergence for $\ln \frac{n^2}{n^2-1}$ I've tried to figure out if this converges using the comparison test, and the ratio test, but with no luck: $\sum^\infty_{n=2} \ln(n^2/(n^2-1))$.
I'd appreciate any help
| Here is a way
to show convergence
without finding the sum.
If $x > 0$,
$\ln(1+x)
=\int_1^{1+x} \frac{dt}{t}
$.
Therefore
(the $1+x$ in the denominator
of the next integral
is intentional)
$\int_1^{1+x} \frac{dt}{1+x}
< \ln(1+x)
< \int_1^{1+x} \frac{dt}{1}
$
or
$\frac{x}{1+x}
< \ln(1+x)
< x
$.
Since
$\frac{n^2}{n^2-1}
=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/996099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Is the Pattern in the Number of Digits in the Bernoulli Numbers Showing Something Significant For the first couple of powers of $10$, the number of digits in these show a certain pattern, is this a coincidence or is their a reasonable explanation. Specifically if we look at
$$ \lfloor \log_{10}\left|B\left(10^n\right)\... | The reason is the asymptotic equality(1)
$$\lvert B(2k)\rvert \sim \frac{2\cdot(2k)!}{(2\pi)^{2k}}.$$
Thus we obtain
$$\log_{10} \lvert B(2k)\rvert = \log_{10} \left((2k)!\right) + \log_{10} 2 - 2k\log_{10}(2\pi) + o(1),$$
and using Stirling's approximation
$$\log (n!) = \left(n+\tfrac{1}{2}\right)\log n - n + \tfrac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/996696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find all values of $\alpha$ so that all solutions approach $0$ as $x \to \infty$ Given the equation
$x^2y′′+\alpha xy′+4y=0$
find all values of α so that all solutions approach zero as $x \to \infty$.
Anyone have advice for this question?
So I get $y = c_1 x^{\frac{1}{2}\sqrt{a^2 -2a -15}-a+1} + c_1 x^{\frac{1}{2}\sqr... | As Dr. Sonnhard Graubner said:
In
$x^2y′′+\alpha xy′+4y=0$,
set
$y = x^r
$.
Then,
writing $a$ for $\alpha$
because I am lazy,
$y'
=rx^{r-1}
$
and
$y''
=r(r-1)x^{r-2}
$
so
$0
=r(r-1)x^{r}+arx^{r}+4x^r
= x^r(r(r-1)+ar+4)
= x^r(r^2+(a-1)r+4)
$
or
$r^2+(a-1)r+4=0$.
Therefore,
$r
=\dfrac{-(a-1)\pm\sqrt{(a-1)^2-16}}{2}
=\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/997253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
If $|z +\frac{1}{z}|=a$ find extreme values of $|z|$ Let $a> 0$. Knowing that $z$ is complex number with $|z +\frac{1}{z}|=a$ find extreme values of $|z|$.
My partial solution:
$$|z +\frac{1}{z}|=a \iff (z +\frac{1}{z})(\overline{z} +\frac{1}{\overline{z}})=a^2\iff |z|^2+\frac{1}{|z|^2} + (\frac{z}{\overline{z}}+\f... | You can simply use the fact that:
$|z|+\frac{1}{|z|} \geq |z+\frac1z| \geq ||z|-\frac{1}{|z|}|$
Substituting $|z+\frac1z|=a$, we get two inequalities:
$|z|+\frac{1}{|z|} \geq a$ and $a\geq ||z|-\frac{1}{|z|}|$
Taking $|z| = t$$\geq1$, we get:
$t^2 - at + 1\geq0$ and $t^2 - at - 1\leq0$
For 1st inequality let $t_1$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/997308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Evaluation of $\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\sum_{l=0}^{\infty}\frac{1}{3^{i+j+k+l}}\;\;,$ Where $i\neq j \neq k\neq l$
Evaluation of following Infinite Geometric series.
$(a)\;\; \displaystyle \sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\frac{1}{3^{i+j}}\;\;,$ Where $i\neq j\;\;\;\;\;\;\;\;\;\; ... | We have:
$$(a)\;\; S_2 = \sum_{i=0}^\infty\sum_{j=0}^\infty\frac{1}{3^{i+j}} (i \ne j) = 2\sum_{i< j}\frac{1}{3^{i+j}} = 2\;I_2$$
$$(b)\;\; S_3 = \sum_{i=0}^\infty\sum_{j=0}^\infty\sum_{k=0}^\infty\frac{1}{3^{i+j+k}} (i \ne j \ne k) = 6\sum_{i<j<k}\frac{1}{3^{i+j+k}} = 6\;I_3$$
$$(c)\;\; S_4 = \sum_{i=0}^\infty\sum_{j=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/999350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Questions about the computation of a limit. By L'Hôpital's rule, it is easy to see that
$$
\lim_{x \to 0} \frac{\tan x - x}{x^2 \sin x} = 1/3.
$$
But if we use the following procedure to compute the limit, we get a wrong answer. The procedure is in the following.
$$
\lim_{x \to 0} \frac{\tan x - x}{x^2 \sin x} =
\li... | Another solution could be to double check the result using Taylor series $$f(x)=\frac{\tan x - x}{x^2 \sin x}$$ and use sucessively $$\tan(x)=x+\frac{x^3}{3}+O\left(x^4\right)$$ $$\sin(x)=x-\frac{x^3}{6}+O\left(x^4\right)$$ So, the numerator is $$\tan(x)-x=\frac{x^3}{3}+O\left(x^4\right)$$ and denominetaor is $$x^2\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How to prove the inequality $ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$ How to prove the inequality $$ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$$
for $a,b,c>0$ and $abc=1$?
I have tried prove $\frac{a}{\sqrt{1+a}}\ge \frac{3a+1}... | By the same way as Mr. Mike$,$ it's enough to prove $$2\left(\sum_{cyc}(x^2+3yz)\right)^3\geqslant 9\sum xy(x+y)(2z+x+y)^3$$
Or
$$\frac18\sum \left( 16\,{x}^{4}+100\,{z}^{4}+104\,{x}^{3}y+243\,{y}^{2}{z}^{2
}+330\,{z}^{3}x+416\,{y}^{2}zx+342\,{z}^{2}xy \right) \left( x-y
\right) ^{2}+$$
$$+\frac18\sum x{y}^{2} \left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1000997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 7,
"answer_id": 2
} |
If $f$ is continuous and $\,f\big(\frac{1}2(x+y)\big) \le \frac{1}{2}\big(\,f(x)+f(y)\big)$, then $f$ is convex Let $\,\,f :\mathbb R \to \mathbb R$ be a continuous function such that $$
f\Big(\dfrac{x+y}2\Big) \le \dfrac{1}{2}\big(\,f(x)+f(y)\big) ,\,\, \text{for all}\,\, x,y \in \mathbb R,
$$
then
how do we prove th... |
I would like to properly show by induction that if $ m\in\{0,1,2,\ldots,2^{n}\} $ then
$$
f\left(\frac{m}{2^n}x+\Big(1-\frac{m}{2^n}\Big)y\right)\le \frac{m}{2^n}f(x)
+\Big(1-\frac{m}{2^n}\Big)f(y).
$$
and then the result will follows by contuinity since $$ m=\lfloor2^nt\rfloor \in\{0,1,2,\ldots,2^{n}\} ~~~and~~\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1002248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 5,
"answer_id": 2
} |
Prove that for any positive integer $n$, $2\sqrt{n+1}-2\le 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\dots+\frac{1}{\sqrt{n}}\le 2\sqrt{n}-1$ Prove that for any positive integer n,$$ 2\sqrt{n+1} - 2 \le 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \dots + \frac{1}{\sqrt{n}} \le 2\sqrt{n} - 1 $$
Could anyone give me a h... | Hint. The double inequality
$$
2\sqrt{n+1} - 2 \le 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + .. + \frac{1}{\sqrt{n}} \le 2\sqrt{n} - 1
$$
can be proved inductively. For $n=1$ is clear. Assume that it holds for $n=k$. Then
$$
2\sqrt{k+1} - 2 +\frac{1}{\sqrt{k+1}}\le 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1002337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove inequalities $2\sqrt{n+1}-2 \leq 1 + 1/\sqrt{2} + 1/\sqrt{3} + .... + 1/\sqrt{n}\leq 2\sqrt{n}-1$ Prove that for any positive integer $n$,
$$
2\sqrt{n+1}-2 \leq 1 + 1/\sqrt{2} + 1/\sqrt{3} + .... + 1/\sqrt{n}\leq 2\sqrt{n}-1
$$
the middle part can be expressed as a sum. of $1/\sqrt{x}$ from 1 to $n$, which leads ... | Hint: $2\sqrt{k+1} - 2\sqrt{k}<\dfrac{1}{\sqrt{k}} = \dfrac{2}{2\sqrt{k}} < \dfrac{2}{\sqrt{k}+\sqrt{k-1}} = 2\sqrt{k} - 2\sqrt{k-1}$. Now let $k$ runs from $1$ to $n$, and add up.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Describe how the number of solutions to an equation system depend on the variable $a$. Describe how the number of solutions to the system:
\begin{eqnarray*}
x+y+z &=&1 \\
2x+y+a^2z &=&a \\
x+3y+3z &=&3
\end{eqnarray*}
is dependant on the value of $a$.... | If $A$ denotes the $3\times 3$-matrix of your equation $Ax=b$, then $\det(A)=-2(a-1)(a+1)$. Hence if the field has not characteristic two, and $(a-1)(a+1)\neq 0$, then we have a unique solution. You have already solved this case correctly. Note that you have divided by $2$ here.
For $char(K)=2$ we always have $\det(A)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1004292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$10\sin(x)\cos(x) = 6\cos(x)$ In order to solve
$$10\sin x\cos x = 6\cos x$$
I can suppose: $\cos x\ne0$ and then:
$$10\sin x = 6\implies \sin x = \frac{6}{10}\implies x = \arcsin \frac{3}{5}$$
And then, for the case $\cos x = 0$, we have:
$$x = \frac{\pi}{2} + 2k\pi$$
Is my solution rigth? Because I'm getting another... | $$10\sin x\cos x=6\cos x\iff 10\sin x\cos x-6\cos x=0\iff 2(\cos x)(5\sin x -3)=0\stackrel{:2}{\iff} (\cos x)(5\sin x -3)=0\iff \cos x=0 \; \; \lor \;\; 5\sin x-3=0$$
Is it any more clear now? Also, that $\cos x=0$ in your solution has to be $x=\dfrac{\pi}{2}+k\pi\, ,k\in \Bbb Z$ and your $\sin x=0$ has to include solu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1004592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Limit of $\left(\frac {2}{3}\right)^n \cdot n^4 \cdot \frac {1- 1/ {n^4}} {4+ n^7/ {3^n}}$ as $n$ tends to infinity I already took some steps to simplifying the original question and im stuck at this point:
$$ \lim_{n \to \infty} \left(\frac {2}{3}\right)^n \cdot n^4 \cdot \frac {1- \frac {1} {n^4}} {4+ \frac {n^7} {3^... | HINT:
We have $$\frac{2^n(n^4-1)}{4\cdot3^n+n^7}<\frac{2^n}{4\cdot3^n}<\frac14\left(\frac23\right)^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1007559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Bijection, and finding the inverse function I am new to discrete mathematics, and this was one of the question that the prof gave out. I am bit lost in this, since I never encountered discrete mathematics before. What do I need to do to prove that it is bijection, and find the inverse? Do I choose any number(integer) a... | A function is bijective if it is injective (one-to-one) and surjective (onto).
You can show $f$ is injective by showing that $f(x_1) = f(x_2) \Rightarrow x_1 = x_2$.
You can show $f$ is surjective by showing that for each $y \in \mathbb{R} - \{2\}$, there exists $x \in \mathbb{R} - \{-1\}$ such that $f(x) = y$.
I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1008840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}\ge \frac32$
$a,b,c$ are positive reals with $abc = 1$. Prove that
$$\frac{1}{a^3(b+c)}+\frac{1}{b^3(a+c)}+\frac{1}{c^3(a+b)}\ge \frac32$$
I try to use AM $\ge$ HM.
$$\frac{\dfrac{1}{a^3(b+c)}+\dfrac{1}{b^3(a+c)}+\dfrac{1}{c^3(a+b)}}3\ge \frac{3}{... | Hint: Let $x=a^{-1},y=b^{-1},z=c^{-1}$. Rewrite $a^3=\dfrac{a^2}{bc}$, the inequality becomes
$$\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\ge\frac32,$$
where $xyz=1$. That should be easy by Cauchy-Schwarz.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1009022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
What is the limit for the radical $\sqrt{x^2+x+1}-x $? I'm trying to find oblique asmyptotes for the function $\sqrt{x^2+x+1}$ and I manage to caclculate that the coefficient for the asymptote when x approaches infinity is 1.
But when i try to find the m-value for my oblique asymptote by taking the limit of:
$$
\lim_... | I think there's a mistake in your last expression. We have
$$ \sqrt{x^2+x+1}-x = \frac{x^2+x+1-x^2}{\sqrt{x^2+x+1}+x} = \frac{1+\frac{1}{x}}{\sqrt{1+\frac{1}{x} + \frac{1}{x^2}} + 1} $$
Therefore we have $$\lim(\sqrt{x^2+x+1}-x)= \lim\frac{1+\frac{1}{x}}{\sqrt{1+\frac{1}{x} + \frac{1}{x^2}} + 1} = \frac{1}{2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1009110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Limit of $\prod\limits_{k=2}^n\frac{k^3-1}{k^3+1}$ Calculate $$\lim_{n \to \infty} \frac{2^3-1}{2^3+1}\times \frac{3^3-1}{3^3+1}\times \cdots \times\frac{n^3-1}{n^3+1}$$
No idea how to even start.
| An overkill. Since we have $$\prod_{k\geq2}\frac{k^{3}-1}{k^{3}+1}=\prod_{k\geq2}\frac{\left(k-1\right)\left(k^{2}+k+1\right)}{\left(k+1\right)\left(k^{2}-k+1\right)}=\prod_{k\geq2}\frac{\left(k-1\right)\left(k-\frac{-1+i\sqrt{3}}{2}\right)\left(k-\frac{-1-i\sqrt{3}}{2}\right)}{\left(k+1\right)\left(k-\frac{1+i\sqrt{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1009566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Number of real roots of the equation $(6-x)^4+(8-x)^4 = 16$
Find number of real roots of the equation $(6-x)^4+(8-x)^4 = 16$
$\bf{My\; Try::}$ Let $f(x) = (6-x)^4+(8-x)^4\;,$ and we have to find real values of $x$ for
which $f(x) = 16$. Now we will form Different cases.
$\bf{\bullet \; }$ If $x<6$ or $x>8\;,$ Then $... | You can exploit the symmetry of the situation to simplify things if you set $y=7-x$ so that the equation becomes $(y-1)^4+(y+1)^4=16$ or $2y^4+12y^2+2=16$ or $y^4+6y^2-7=0$
This factorises as $(y^2-1)(y^2+7)=0$ with real solutions $y=\pm 1$ corresponding to $x=6, x=8$
Note this symmetric trick is bound to work, becaus... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1010622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
What can we say If lim U(n+1) - U(n) = 0? So i proved that $$\lim_{n\to \infty} U(n+1) - U(n) = 0$$ and we also have $a< U(n) < b$
What can we say?
Does this prove that the serie converge?
| Consider the series
$$1-\frac{1}{2}-\frac{1}{2}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}-\frac{1}{4}-\frac{1}{4}-\frac{1}{4}-\frac{1}{4}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}- \frac{1}{6}-\frac{1}{6}-\cdots.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1011202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\lim_{x \to 0} \frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{x^3}$ What I attempted thus far:
Multiplying by conjugate
$$\lim_{x \to 0} \frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{x^3} \cdot \frac{\sqrt{1 + \tan x} + \sqrt{1 + \sin x}}{\sqrt{1 + \tan x} + \sqrt{1 + \sin x}} = \lim_{x \to 0} \frac{\tan x - \s... | $\begin{align}
\lim_{x \to 0} \frac{\sec x - 1}{x^2 \cdot (\sqrt{1 + \tan x} + \sqrt{1 + \sin x})} &= \lim_{x\to 0}\frac{1}{\sqrt{1 + \tan x} + \sqrt{1 + \sin x}}\cdot\lim_{x\to 0}\frac{\sec x-1}{x^2}\\
&=\frac12 \cdot \lim_{x\to 0}\frac{\sec x-1}{x^2}
\end{align}$
This last limit should yield to L'Hopital's rule, or y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1011290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Triangle in Triangle I have the lengths of three sides of an acute triangle ABC as shown below. Assume a point P on the side AB such that, if Q is the projection of P onto BC, R is the projection of Q onto CA, P becomes the projection of R onto AB. How can I Find the length PB.
| Here is another approach, which works directly and systematically. Call the angles of the triangle $A, B, C$ and the sides opposite those angles $a, b, c$. We use the cosine rule $a^2=b^2+c^2-2bc \cos A$ so that $\cos A=\cfrac {a^2+b^2+c^2-2a^2}{2bc}$ and if I define $2d^2=a^2+b^2+c^2$ then $\cos A=\cfrac {d^2-a^2}{bc}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1011699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Base for Vector Space I have problem with one assignment and I am pretty sure that it's not that hard. So, I have 4 vectors :
$$
v_{1}=\begin{pmatrix}
1 \\ 1 \\ 1
\end{pmatrix},
v_{2}=\begin{pmatrix}
0 \\ 3 \\ 1
\end{pmatrix},
v_{3}=\begin{pmatrix}
1 \\ -2 \\ 0
\end{pmatrix},
v_{4} =\be... | Here is one general method that involves matrix reduction (I assume you've covered that?)
Set up the matrix: $$\begin{bmatrix} 1 & 0 & 1 & -2 \\ 1 & 3 & -2 & 1 \\ 1 & 1 & 0 & -1 \end{bmatrix}$$
Then use column reduction:
$$\begin{bmatrix} 1 & 0 & 1 & -2 \\ 1 & 3 & -2 & 1 \\ 1 & 1 & 0 & -1 \end{bmatrix} \sim \begin{bmat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1012119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding the sum of x of two power series. Could someone give me a hint on finding the sum of all $x$ for the following power series:
$$
\sum_{n=1}^{\infty}(-1)^{n+1} \frac{x^{2n+1}}{2n+1}
$$
I am pretty sure we need to compare this with $$arctan (x) = \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{2n+1} $$
I'm just not sur... | This is not the full answer,this is work in progress anyway it might help author to get idea.
$$\sum_{n=2}^
\infty \frac{2^nx^n}{n+1}-\frac{nx^n}{n+1}=\frac{1}{x}\sum_{n=2}^\infty\frac{2^nx^{n+1}}{n+1}-\frac{1}{x}\sum_{n=2}^\infty\frac{nx^{n+1}}{n+1}=\frac{1}{x}(-x-x^2+\sum_{n=0}^\infty\frac{2^nx^{n+1}}{n+1})-\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1012333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
If $\sin A+\sin B =a,\cos A+\cos B=b$, find $\cos(A+B),\cos(A-B),\sin(A+B)$ If $\sin A+\sin B =a,\cos A+\cos B=b$,
*
*find $\cos(A+B),\cos(A-B),\sin(A+B)$
*Prove that $\tan A+\tan B= 8ab/((a^2+b^2)^2-4a^2)$
| From the two given conditions you get $$\tan \left(\frac{A+B}{2}\right)=\frac{a}{b}$$
From that we get $$\cos (A+B)=\frac{b^2-a^2}{b^2+a^2}\\
\sin (A+B)=\frac{2ab}{a^2+b^2}$$
And if you square and add both sides of the two given equations from that you'll get $$\cos (A-B)=\frac{a^2+b^2-2}{2}$$
Now, we get $$\cos^2A+\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1012928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Given a vector field $\mathbf{H}$, find a vector field $\mathbf{F}$ and a scalar field g, such that $\mathbf{H}$ = curl(F) + ∇(g). Let$\;\mathbf{H}(x,y,z) = x^2y\mathbf{i}+y^2z\mathbf{j}+z^2x\mathbf{k}$. Find a vector field $\mathbf{F}$ and a scalar field g, such that $\mathbf{H}$ = curl(F) + ∇(g).
I took divergence on... | Below a slightly different approach which first gives $\mathbf{F}$
and then $G$.
The problem at hand is quite similar to that of finding the vector and
scalar potentials in terms of the electric and magnetic fields in
electrodynamics. In that case there are standard solutions. But here, since $%
\mathbf{H}$ is not deca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1013503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Flipping coins probability of $6$ flips having more heads than $5$ flips. I have $6$ fair coins and you have $5$ fair coins. We both flip our own coins and observe the number of heads we each have. What is the probability that I have more heads than you?
Not sure how to start this, any help please?
| An answer using generating functions:
$$G(x)=\left(\frac{1}{2}+\frac{x}{2}\right)^6 \left(\frac{1}{2}+\frac{1}{2 \
x}\right)^5$$
Expanding that I get:
$$\frac{231}{1024}+\frac{1}{2048 x^5}+\frac{11}{2048 \
x^4}+\frac{55}{2048 x^3}+\frac{165}{2048 x^2}+\frac{165}{1024 \
x}+\frac{231 x}{1024}+\frac{165 x^2}{1024}+\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1015345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
prove that $2^n+2^{n-1}+2^{n-2}+8^n-8^{n-2}$ is a multiple of 7 Prove that a number $2^n+2^{n-1}+2^{n-2}+8^n-8^{n-2}$ is a multiple of 7 for every natural $n\ge2$. I am not sure how to start.
| The base case is true, $T_0=4+2+1+64-8=63$ is a multiple of $7$.
Now let us relate $T_{n+1}$ to $T_n$:
$$T_{n+1}=2.2^n+2.2^{n-1}+2.2^{n-2}+8.8^n-8.8^{n-2}.$$
Subtracting $T_n$, we have
$$T_{n+1}-T_n=2^n+2^{n-1}+2^{n-2}+7.8^n-7.8^{n-2}.$$
We can discard the last two terms, obviously multiples of $7$. Remains to prove th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1015460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Need help in figuring out what I am doing wrong when solving for n.. Here is the expression that I am trying to solve for n:
$$ \frac{4}{16+n} = \frac{10}{16+n} \frac{10}{16+n}$$
I am doing the following:
\begin{align}
\frac{4}{16+n} & = \frac{100}{(16 + n)^2} \\[8pt]
\frac{4}{16+n} & = \frac{100}{16^2 + 32n + n^2} \\[... | You can cancel one of the $16 + n$ terms on each side, greatly simplifying the problem. Multiply both sides by $16 + n$. This gives
$$4 = \frac{10 \times 10}{16 + n}$$
Multiply both sides by $16 + n$ again. This gives
$$4(16 + n) = 100$$
Divide both sides by $4$:
$$16 + n = 25$$
Subtract $16$ from both sides:
$$n = 9$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1016234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Solving $\int \frac{1}{\sqrt{x^2 - c}} dx$ I want to solve $$\int \frac{1}{\sqrt{x^2 - c}} dx\quad\quad\text{c is a constant}$$
How do I do this?
It looks like it is close to being an $\operatorname{arcsin}$?
I would have thought I could just do:
$$\int \left(\sqrt{x^2 - c}\right)^{-\frac12}\, dx=\frac{2\sqrt{c+x^2}}{... | $$ \int \frac{1}{\sqrt{x^2-c}}dx = \int \frac{1}{\sqrt{c\left(\frac{x^2}{c}-1\right)}}dx=\frac{1}{\sqrt{c}} \int \frac{1}{\sqrt{\left(\frac{x}{\sqrt{c}}\right)^2-1}}dx $$
Let $u=\frac{x}{\sqrt{c}}$, then $du=\frac{1}{\sqrt{c}} dx$. So now we have
$$ \int \frac{1}{\sqrt{u^2-1}}du =\mathrm{arcosh}(u)+C = \mathrm{arcosh}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1016585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Proving a Triangle inequality If $ a^2 + b^2 > 5c^2 $ in a triangle ABC then show c is the smallest side.I tried to solve this by cosine rule but i was not able to find the answer
| First, recall that in any triangle, the triangle inequality states that $a+b>c$, $b+c>a$, and $c+a>b$.
Combining this with your condition $a^2+b^2>5c^2$, we have
$$5c^2<a^2+b^2<(b+c)^2+b^2=2b^2+2bc+c^2,$$
hence $$2c^2<b^2+bc.$$
Now, suppose it were the case that $b\leq c$. Then $2c^2<b^2+bc\leq 2c^2$, a contradiction.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1017077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Upper bound for the widest matrix with no two subsets of columns with the same vector sum Over at PPCG there is an ongoing contest going on to find the largest matrix without a certain property, called property $X$. The description is as follows (copied from the question).
A circulant matrix is fully specified by its f... | I think any nxm circulant matrices (where m=n+1) of this form will not have property X, with all entries 0 or 1
$$\begin{pmatrix}1&0&1&1&\cdots&1\\
1&1&0&1&\cdots&1\\
1&1&1&0&\cdots&1\\
\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\
1&1&1&1&\cdots&0\end{pmatrix}$$
So the score of a matrix of this form is $$S=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1017169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
"answer_id": 0
} |
Find $t$ in $i = 50\sin\left(120\pi t -\frac{3\pi}{25}\right)$ where $i = 25$
An alternating current generator produces a current given by the equation
$$i = 50\sin\left(120\pi t - \frac{3\pi}{25}\right)$$
where $t$ is the $\text{time}$ in $\text{seconds}$.
(Q) Find the first two values of $t$ when $i = 25 \text{ ... | Hint: Let $x=120\pi t-3\pi/25$. If $x=\theta $ is one
solution of the equation $y=\sin x$ then another solution is $x=\pi -\theta $.(*)
Complete solution using your work. If the hint isn't enough, hover your mouse over the grey to see
You have correctly found the first angle $\theta >0$ for which $\sin \theta =1/2$. It... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1018518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
When does $x^3+y^3=kz^2$? For which integers $k$ does
$$
x^3+y^3=kz^2
$$
have a solution with $z\ne0$ and $\gcd(x,y)=1$? Is there a technique for counting the number of solutions for a given $k$?
| $\qquad\qquad$ Too long for a comment: Positive coprime integers x and y less than $6,000$ :
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad$ $k=1$
$\qquad\qquad\qquad\qquad$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad$ $k=2$
$\qquad\qquad\qquad\qquad$
$\qquad\qquad\qquad\qquad\qquad\qquad\qqua... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1018893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Evaluate the integrals in $\int_{0}^{1} \frac {x^4(1-x)^4}2\, dx \le \int_{0}^{1} \frac {x^4(1-x)^4}{1+x^2}\, dx \leq \int_{0}^{1} {x^4(1-x)^4}\, dx$ Note that when $0\le x \le 1$ we have $$\frac 12 \le \frac 1 {1+x^2} \le 1.$$
Hence, $$\int_{0}^{1} \frac {x^4(1-x)^4}2\, dx \le \int_{0}^{1} \frac {x^4(1-x)^4}{1+x^2}\, ... | To evaluate the first and third integrals, simply expand:
$$x^4 (1 - x)^4 = x^4 - 4 x^5 + 6 x^6 - 4 x^7 + x^8.$$ Then, integrate term by term.
To evaluate the second integral, apply polynomial long division, which will give that the middle integrand has the form
$$\frac{x^4 (1 - x)^4}{1 + x^2} = p(x) - \frac{4}{1 + x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find maximum/minimum for $\cos(2x) + \cos(y) + \cos(2x+y) $ I have not been able to find the critical points for $\cos(2x) + \cos(y) + \cos(2x+y) $
| This might be a solution.
Set $X=2x$ and $\cos(X)=(1-t^2)/(1+t^2),\cos(y)=(1-s^2)/(1+s^2)$,$\sin(X)=(2t)/(1+t^2),\cos(y)=(2s)/(1+s^2)$, then the original expression becomes:
$$\cos(X) + \cos(y) + \cos(X+y)=f(s,t)=\frac{3 - s^2 - 4 s t - t^2 - s^2 t^2}{(1+s^2)(1+t^2)}$$
where
$$f(s,t)=3 - s^2 - 4 s t - t^2 - s^2 t^2$$
F... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1020318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $ \int_0^3 \frac{x^3}{1-x^4}\, dx. $ Evaluate
$$
\int_0^3 \frac{x^3}{1-x^4}\, dx.
$$
I evaluated the integral and got $\left[\dfrac{-\ln(1-x^4)}{4}\right]_0^3$ which ended up diverging. Any help is appreciated!
| The integral gives up easily to direct attack, but with a little twist.
Use the obvious substitution $u=x^4$:
$$\int_0^3 \frac{x^3}{1-x^4}\, dx=\frac{1}{4} \int_0^{81} \frac{du}{1-u}=$$
Now it makes sense to separate the integral into two parts, so the denominator is positive for each part:
$$=\frac{1}{4} \int_0^1 \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1021086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate $\lim\limits_{x\to 0} \frac{\sin x - x + x^3/6}{x^3}$ I'm unsure as to how to evaluate:
$$\lim\limits_{x\to 0} \frac{\sin x - x + \frac{x^3}{6}}{x^3}$$
The $\lim\limits_{x\to 0}$ of both the numerator and denominator equal $0$. Taking the derivative of both ends of the fraction we get:
$$\lim\limits_{x... | Using the identity $sin \ x =3sin(\frac{x}{3})-4 sin^{3}(\frac{x}{3})$, our limit $l=lim \frac{sin \ x -x +\frac{x^{3}}{6}}{x^{3}}$ becomes,
$l= \frac{1}{9} lim \frac{sin \ \frac{x}{3} -\frac{x}{3} +(\frac{x}{3})^{3}.\frac{1}{6}}{x^{3}}+ \frac{4}{27}- lim \ 4 \frac{sin^{3}(\frac{x}{3})}{x^{3}}$, which means $l=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1022218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
Write an expression for $(\cos θ + i\sin θ)^4$ using De Moivre’s Theorem. Obtain another expression for $(\cos θ + i \sin θ)^4$ by direct multiplication (i.e., expand the bracket). Use the two expressions to show
$$
\cos 4\theta = 8 \cos^4 \theta − 8 \cos^2 \theta + 1,\\
\sin 4\theta = 8\cos^3\theta \sin\theta − 4 \co... | By De Moivre’s theorem,
$$
(\cos\theta+i\sin\theta)^4=\cos 4\theta+i\sin 4\theta.
$$
On the other hand
$$
(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4
$$
Expand the binomial and equate the real and imaginary parts. Where you find $\sin^2\theta$, substitute $1-\cos^2\theta$.
You have, almost correctly,
$$
\cos4\theta+i\sin4\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1026401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If the number $x$ is algebraic, then $x^2$ is also algebraic Prove that if the number $x$ is algebraic, then $x^2$ is also algebraic. I understand that an algebraic number can be written as a polynomial that is equal to $0$. However, I'm baffled when showing how $x^2$ is also algebraic.
| Basically $P(x)\cdot P(-x)= Q(x^2)$, a polynomial in $x^2$. A similar trick using $n$-th roots of $1$ allows us to get $P(x) \cdot P_1(x) = Q(x^n)$ for some $Q$ so an equation for $x^n$. To get an equation satisfied by a general $\phi(x)$ ( $\phi(x) = x^2$ in the above example) we can use the companion matrix as @Juli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1026778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 6,
"answer_id": 4
} |
Find exact value of $\cos (\frac{2\pi}{5})$ using complex numbers.
Factorise $z^5-1$ over the real field.
Show that $\cos \frac{2\pi}{5}$ is a root of the equation $4x^2+2x-1=0$ and hence find its exact value.
I have worked out that $$ z^5-1=(z-1)(z^4+z^3+z^2+z+1)\\=(z-1)(z^2-2z\cos72^o+1) (z^2+2z\cos 72^o+1) $$
| Find the polynomials that cancel $\cos\left(\frac{2\pi}{5}\right)$ :
We know that the solution of $z^5=1$ are given by $\alpha^k=e^{\frac{2ik\pi}{5}}$, $k=0,1,2,3,4$. We can observe that $\alpha^{4}=\alpha^{-1}$ and $\alpha^{3}=\alpha^{-2}$. Moreover,
$$(*):\quad 0=1+\alpha+\alpha^2+\alpha^3+\alpha^4=1+\alpha+\alpha^{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1027418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I express the sum $(1+k)+(1+k)^2+\ldots+(1+k)^N$ for $|k|\ll1$ as a series? Wolfram Alpha provides the following exact solution
$$ \sum_{i=1}^N (1+k)^i = \frac{(1+k)\,((1+k)^N-1)}{k}.$$
I wish to solve for $N$ of the order of several thousand and $|k|$ very small (c. $10^{-12}).$ When I do this on a computer in... | for $k<1$ we can use
$$
(1+k)^N \approx 1+ kN + \frac{N(N-1)}{2}k^2 + \frac{N(N-1)(N-2)}{3!}k^3
$$
thus
$$
\begin{align}
\sum (1+k)^i &\approx&\frac{(k+1)(1+ kN + \frac{N(N-1)}{2}k^2 + \frac{N(N-1)(N-2)}{3!}k^3-1)}{k}\\
&=&(k+1)\left(1+\frac{N-1}{2}k+\frac{(N-1)(N-2)}{3!}k^2\right)
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1029172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Partial fraction of $\frac{2x^2-9x-9}{x^3-9x}$ I'm doing some questions from Anton, 8th edition, page 543, question 13. I've found a answer but it does not match with the answer given at the last pages.
Questions asks to solve $\int{\frac{2x^2-9x-9}{x^3-9x}}$
So:
$$\frac{2x^2-9x-9}{x(x^2-9)}$$
then:
$$\frac{2x^2-9x-9}{... | You're answer is fine and is equivalent to the author's.
To see why, recall:
$$\ln a - \ln b = \ln\left(\frac ab\right)$$
$$\ln a + \ln b = \ln(ab)$$
$$2 \ln a = \ln(a^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1029824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
The integral $\int_{|z|=2}\log\frac{z+1}{z-1}dz$ Let $\log$ be the branch of the logarithm that extends the usual real logarithm, and consider on $D=\Bbb C\smallsetminus [-1,1]$ the function $$f(z)=\log\frac{z+1}{z-1}$$
I have to find the integral of $f$ around the circle $|z|=2$. Now, as an example, consider the integ... | As mike said in the comments above, I think the answer is $4 \pi i$.
Since $1+ \frac{1}{z}$ and $1- \frac{1}{z}$ both lie in the right half-plane for $|z| >1$,
$$- \pi < \text{Arg} \left(1 + \frac{1}{z} \right) - \text{Arg} \left(1- \frac{1}{z} \right) \le \pi \ \ \text{for} \ \ |z| >1.$$
Thus
$$ \begin{align} f(z) =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1030445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Finding a line through 4 other lines! This one's probably easy, but I'm dreadfully stuck and can't seem to figure out a decent method.
I have the following lines:
$$a: \vec{x}(\lambda)= \left( \begin{array}{ccc}
4 \\
-2 \\
-2 \end{array} \right) + \lambda\left( \begin{array}{ccc}
1 \\
-1 \\
-1 \end{array} \right)
... | Note that $a$ and $d$ are parallel, so whatever line it is, it needs to lie in the plane containing $a$ and $d$. Considering it also needs to intersect $b$ and $c$, try figuring out which two points those two lines intersect the $ad$-plane.
Edit: Here's a full answer.
$ad$-plane: The $ad$-plane's normal vector is ortho... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1031622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate $\int_0^1 (\arctan x)^2 \ln(\frac{1+x^2}{2x^2}) dx$
Evaluate
$$
\int_{0}^{1} \arctan^{2}\left(\, x\,\right)
\ln\left(\, 1 + x^{2} \over 2x^{2}\,\right)\,{\rm d}x
$$
I substituted $x \equiv \tan\left(\,\theta\,\right)$ and got
$$
-\int^{\pi/4}_{0}\theta^{2}\,{\ln\left(\, 2\sin^{2}\left(\,\theta\,... | Let $I$ be our integral
Since $\displaystyle \small\int \ln\left(\frac{1+x^2}{2x^2}\right)\ dx=x\ln\left(\frac{1+x^2}{2x^2}\right)+2\arctan(x)$, so by integration by parts we have
$$I=2\left(\frac{\pi}{4}\right)^3-4\underbrace{\int_0^1\frac{\arctan^2(x)}{1+x^2}\ dx}_{\frac13\left(\frac{\pi}{4}\right)^3}-\underbrace{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1032483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 5,
"answer_id": 4
} |
Find the domain of $\frac{x}{\sqrt{6x^2+3x+3/4}+x}$. My attempt:
Let's assume that $\sqrt{6x^2+3x+\frac{3}{4}}+x$ $>$ 0$$
\rightarrow\sqrt{6x^2+3x+\frac{3}{4}} > -x
\rightarrow {6x^2+3x+\frac{3}{4}} > x^2\\
\rightarrow {5x^2+3x+\frac{3}{4}} > 0\\$$
If x $<$ 0 then $x^2$ $>$ 0, and ${5x^2+3x+\frac{3}{4}} > 0$... | All we need is $\sqrt{6x^2+3x+\dfrac34}+x\ne0$ and of course $6x^2+3x+\dfrac34\ge0$
$6x^2+3x+\dfrac34=6\left(x+\dfrac14\right)^2+\dfrac34-\dfrac6{16}>0$ for all real $x$
Now $\sqrt{6x^2+3x+\dfrac34}+x=0\iff x=-\sqrt{6x^2+3x+\dfrac34}\le0$
$\sqrt{6x^2+3x+\dfrac34}=-x$
Squaring we get $5x^2+3x+\dfrac34=0$ but $5x^2+3x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1033629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Which rational primes less than 50 are also Gaussian primes? Which rational primes less than 50 are also Gaussian primes?
My attempt: First we need to list all of the rational prime numbers that are less than $50$
$1,2,3,5,7,11,13,17,19,23,29,31,37,39,41,43,47$
A Gaussian prime is defined as a Gaussian integer that is ... | What might really help you understand here is the concept of conjugate pairs. The conjugate of $a - bi$ is $a + bi$. And $(a - bi)(a + bi) = a^2 + b^2$, which should look very familiar because it's the norm function.
Given $p$ a "rational prime" (for lack of a better term) with nonzero real part but no imaginary part, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1034815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Solve the recurrence relation $a_n=3a_{n-1}+n^2-3$, with $a_0=1$. Solve the recurrence relation $a_n=3a_{n-1}+n^2-3$, with $a_0=1$.
My solutions: the homogeneous portion is $a_n=c3^n$, and the inhomogeneous portion is $a^*_n=-1/2n^2-3/4n+9/8$.
This results in a final recurrence relation of
$$a_n=-\frac{1}{8}3^n-\frac{... | Well you've made a mistake,I'm not sure where is your mistake anyway here's is a different solution
$$a_n=3a_{n-1}+n^2-3\\a_{n-1}=3a_{n-2}+n^2-2n-2\\a_n-a_{n-1}=3a_{n-1}+n^2-3-3a_{n-2}-n^2+2n+2\\a_n=4a_{n-1}-3a_{n-2}+2n-1\\a_{n-1}=4a_{n-2}-3a_{n-3}+2n-3\\a_n-a_{n-1}=4a_{n-1}-3a_{n-2}+2n-1-4a_{n-2}+3a_{n-3}-2n+3\\a_n=5a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1035625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
What is the maximum value of $f(\theta) = \sin\theta \cos\theta$ What is the maximum value of $f(\theta) = \sin\theta \cos\theta$ ?
| Hint $1$:
$\sin \theta \cdot \cos \theta = \dfrac{\sin 2\theta}{2}$
Hint $2$:
$\sin \theta \cdot \cos \theta \leq \dfrac{\sin^2 \theta + \cos^2 \theta}{2} = \dfrac{1}{2}$.
Choose the hint you like most...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1036366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Inequalities with arctan I don't understand how to solve inequalities with arctan, such as:
$$\arctan\left(\frac{1}{x^2-1}\right)\ge \frac{\pi}{4} $$
If someone could solve this and give me a very brief explanation of what they did, I'd be thankful.
| Here are the steps
$$\arctan\left(\frac{1}{x^2-1}\right)\ge \frac{\pi}{4} $$
$$\frac{1}{x^2-1}\ge \tan\left(\frac{\pi}{4}\right)$$
$$\frac{1}{x^2-1}\ge 1$$
$$\frac{1}{x^2-1}-1\ge 0$$
$$\frac{1-x^2+1}{x^2-1}\ge 0$$
$$\frac{2-x^2}{x^2-1}\ge 0$$
$$\frac{\left(\sqrt{2}-x\right)\left(x+\sqrt{2}\right)}{(x+1)(x-1)}\ge 0$$
Ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1038315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve $\lfloor \sqrt x \rfloor = \lfloor x/2 \rfloor$ for real $x$ I'm trying to solve $$\lfloor \sqrt x \rfloor = \left\lfloor \frac{x}{2} \right\rfloor$$
for real $x$.
Obviously this can't be true for any negative reals, since the root isn't defined for such.
My approach is the following: Let $x=:n+r$, $n \in \math... | Clearly $x$ can’t be negative, and $\sqrt{x}$ and $\frac{x}2$ can’t be too far apart. They’re equal for $x=4$, and after that $\frac{x}2$ exceeds $\sqrt{x}$ by more and more as $x$ increases. Thus, we should focus initially on relatively small non-negative values of $x$. Points where one or the other floor increases ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1038396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 4
} |
Evaluate $\int^\pi_{-\pi}(1-a\cos\theta)^{-b-2}\log(1-a\cos\theta)d\theta$
$$ \int^\pi_{-\pi} \left(1-a\cos\theta\right)^{-b-2}
\log\left(1-a\cos\theta\right)d\theta$$
Under the condition $0<a<1$ and $b>0$.
Mathematica found the following form.
$$ 2\pi\left(a+1\right)^{-b-2}
\bigg({}_2F_1\left(\frac{1}{2},b+2;1;... | Consider $\int_{-\pi}^\pi(1-a\cos\theta)^{-b-2}~d\theta$ ,
$\int_{-\pi}^\pi(1-a\cos\theta)^{-b-2}~d\theta$
$=\int_{-\pi}^0(1-a\cos\theta)^{-b-2}~d\theta+\int_0^\pi(1-a\cos\theta)^{-b-2}~d\theta$
$=\int_\pi^0(1-a\cos(-\theta))^{-b-2}~d(-\theta)+\int_0^\pi(1-a\cos\theta)^{-b-2}~d\theta$
$=\int_0^\pi(1-a\cos\theta)^{-b-2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1044120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Using Ratio test/Comparison test I have $\displaystyle\sum_{n=1}^{\infty}{\frac{(2n)!}{(4^n)(n!)^2(n^2)}}$ and need to show whether it diverges or converges.
I attempted to use the ratio test, but derived that the limit of $\dfrac{a_{n+1}}{a_n}=1$ and hence the test is inconclusive.
So I now must attempt to use the c... | See below for my solution. Definition:
$$\lim_{x\rightarrow\infty} \frac{a_{n+1}}{a_n}=\lim_{n\rightarrow\infty} \frac{\frac{(2n+2)!}{4^{n+1}\cdot((n+1)!)^2 \cdot (n+1)^2}}{\frac{(2n)!}{4^n\cdot (n!)^2 \cdot n^2}}$$
Expand the factorials, to prepare for removal due to ratio.
$$=\lim_{n\rightarrow\infty} \frac{\frac{(2n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1045156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
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.