Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Evaluate $f(x)=x\log{\left|\frac{x+2}{3-x}\right|}$ for $x\rightarrow\infty$ I have the following function:
$$f(x)=x\log{\left|\frac{x+2}{3-x}\right|}$$
I want to find the limit for $x\rightarrow+\infty$.
This is what I do. Since $x>=0$, I can remove the absolute value:
$$f(x)=x\log\left({\frac{x+2}{3-x}}\right)\sim x\... | $\frac{x+2}{3-x} = \frac{5}{3-x} - 1 < 0$ when $x\to \infty$.
Hence $\underset{x\to\infty}{\text{lim}}f(x) = \underset{x\to\infty}{\text{lim}} x\log(1 + \frac{5}{x - 3}) = \underset{x\to\infty}{\text{lim}}x \frac{5}{x - 3} = 5$
where the second to last equality follows from $\log(x + 1) = x + o(x) \text{ for } x\to 0 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2888587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Solve: $\lim_{x\to -\infty} (\sqrt {4x^2+7x}+2x)$ Solve: $$\lim_{x\to -\infty} (\sqrt {4x^2+7x}+2x)$$
My attempt:
Rationalizing:
$$\lim_{x\to -\infty} (\sqrt {4x^2+7x}+2x) *\frac{\sqrt {4x^2+7x}-2x}{\sqrt {4x^2+7x}-2x}$$
$$=\lim_{x\to -\infty} \frac{4x^2+7x-4x^2}{\sqrt {4x^2+7x}-2x}$$
$$=\lim_{x\to -\infty}\frac{7x}{\s... | It's a very common error: $\sqrt{x^2}=-x$, when $x<0$.
I usually suggest the substitution $t=-1/x$, so the limit becomes
$$
\lim_{t\to0^+}\left(\sqrt{\frac{4}{t^2}-\frac{7}{t}}-\frac{2}{t}\right)
=
\lim_{t\to0^+}\frac{\sqrt{4-7t}-2}{t}
$$
which is an easy derivative:
$$
f(t)=\sqrt{4-7t}
\qquad
f'(t)=\frac{-7}{2\sqrt{4-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2891096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Interpolation and Approximation A quadratic polynomial $p(x)$ is constructed by interpolating the data points $(0,1)$, $(1,e)$, $(2,e^2)$. If $\sqrt{e}$ is approximated by using $p(x)$ then its approximated value is.
| Assume $p(x)=ax^2+bx+c$ this quadratic polynomial satisfies the given points
$$p(0)=1=c$$
$$p(1)=e=a+b+c$$
$$p(2)=e^2=4a+2b+c$$
you got $a=\frac{(e-1)^2}{2}$, $b=\frac{(e-1)(3-e)}{2}$, $c=1$
thus your polynomial is $$p(x)=\frac{(e-1)^2}{2}x^2+\frac{(e-1)(3-e)}{2}x+1$$
$$\frac{(e-1)^2}{2}x^2+\frac{(e-1)(3-e)}{2}x+1=0$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2895194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
An Euler type sum: $\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{n\cdot 4^n}{2n \choose n}$, where $H_n^{(2)}=\sum\limits_{k=1}^{n}\frac{1}{k^2}$ I've been trying to compute the following series for quite a while :
$$\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{n\cdot 4^n}{2n \choose n}$$
where $H_n^{(2)}=\sum\limits_{k=1}^{n}\frac{1}{k... | Using the well-known identity
$$\sum_{n=1}^\infty \frac{\binom{2n}n}{4^n}x^n=\frac{1}{\sqrt{1-x}}-1$$
Divide both sides by $x$ then integrate , we get
$$\quad\displaystyle\sum_{n=1}^\infty \frac{\binom{2n}n}{n4^n}x^n=-2\ln(1+\sqrt{1-x})+C $$
set $x=0,\ $ we get $C=2\ln2$
Then
$$\sum_{n=1}^\infty \frac{\binom{2n}n}{n4^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2895620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
Find the general solution (i) $\cot \theta =-\dfrac {1} {\sqrt {3}}$ (ii)$4\cos ^{2}\theta =1$ my attempt for
(i)
$\left. \begin{array} { l } { \cot ( \theta ) = - \frac { 1 \cdot \sqrt { 3 } } { \sqrt { 3 } \sqrt { 3 } } } \\ { 1 \cdot \sqrt { 3 } = \sqrt { 3 } } \end{array} \right.$
$\cot ( \theta ) = - \frac { \sqr... | $$1=4\cos^2t=2(1+\cos2t)$$
$$\iff\cos2t=?$$
$$2t=2n\pi\pm\dfrac{2\pi}3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2897340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Making an expression stable for small values The following expression shows significant numerical differences in a program when I compile in x86 (32 bit) versus x64 (64 bit), when $a$ is small:
$$ \left( \dfrac{1}{a} - b \right) \left( 1- \exp(-a)\right)$$
Is there a way that I can refactor this expression so that it i... | You can approximate
$$
e^x \approx 1+x + \frac{x^2}{2!} + \frac{x^3}{3!}
\qquad
\Rightarrow
e^{-x} \approx 1-x + \frac{x^2}{2!} - \frac{x^3}{3!}
$$
Plugging this into the equation, we get
$$
\left(\frac{1}{x}-b\right) \left(1- \left[ 1-x + \frac{x^2}{2!} - \frac{x^3}{3!}\right]\right) = \left(\frac{1}{x}-b\right) \left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2899268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integrate $\frac{x^2-4x+10}{x^2\sqrt x}$
Find the indefinate integral with respect to $x$ of $$\frac{x^2-4x+10}{x^2\sqrt x}$$
For this problem I first made each individual number in the numerator separate from each other for easier integration and then simplified
$$=\int \left(\frac{x^2}{x^2\sqrt x}-\frac{4x}{x^2\sqr... | It is $x^{-5/2}$ and not $x^{-3/2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2900440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Probability of real roots for $x^2 + Bx + C = 0$ Question: The numbers $B$ and $C$ are chosen at random between $-1$ and $1$, independently of each other. What is the probability that the quadratic equation $$x^2 + Bx + C = 0$$ has real roots? Also, derive a general expression for this probability when B and C are chos... | I assume you mean that $B,\,C\sim U(-1,\,1)$. We'll get the answer as a function of a fixed value for $C$, then average it out. For $C<0$ (which has probability $1/2$), the result is $0$; for $C> 1/4$ (which has probability $3/8$), the result is $1$; for $0\le C\le\frac{1}{4}$ (which has probability $\frac{1}{8}$), the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2900828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Sum the series $\frac{3}{1⋅2⋅4}+\frac{4}{2⋅3⋅5}+ \frac{5}{3⋅4⋅6}+...\text{(upto n terms)}$ $\frac{3}{1⋅2⋅4}+\frac{4}{2⋅3⋅5}+ \frac{5}{3⋅4⋅6}+...\text{(upto n terms)}$
The general term seems to be
$$T_r= \frac{r+2}{r(r+1)(r+3)}.$$
I see no way to telescope this because the factors of the denominator of the general term ... | There still is the brutal way:
$$ \frac{r+2}{r(r+1)(r+3)}\stackrel{\text{PFD}}{=}\frac{2}{3}\cdot\frac{1}{r}-\frac{1}{2}\cdot\frac{1}{r+1}-\frac{1}{6}\cdot\frac{1}{r+3} \tag{1}$$
leads to:
$$ \frac{r+2}{r(r+1)(r+3)} = \frac{1}{6}\int_{0}^{1} x^{r-1}\left(4-3x-x^3\right)\,dx \tag{2}$$
then by summing both sides over $r\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2904182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Evaluate $(\sqrt{3})^{11}$ (Evaluate Square Root $3$ to the power of $11$) I know the answer is $243 \sqrt 3$ but in my maths book they got $(\sqrt 3) (\sqrt 3) (\sqrt 3) (\sqrt 3) (\sqrt 3) (\sqrt 3) (\sqrt 3) (\sqrt 3) (\sqrt 3) (\sqrt 3) (\sqrt 3) (\sqrt 3)$ but then they only took the first $5$ out of the $11$ $(\s... | $$ \sqrt{3}^{11} = \sqrt{3} \cdot \sqrt{3}^{10} = \sqrt{3} \cdot \left(\sqrt{3}^2\right)^5 = \sqrt{3} \cdot 3^5 = 243\sqrt{3}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2905304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Least Common Multiple and the product of a sequence of consecutive integers Let $x>0, n>0$ be integers.
Let LCM$(x+1, x+2, \dots, x+n)$ be the least common multiple of $x+1, x+2, \dots, x+n$.
Let $v_p(u)$ be the highest power of $p$ that divides $u$.
It seems to me that:
$$\frac{(x+1)(x+2)\times\dots\times(x+n)}{\text{... | Your claim is true according to a wiki page.
Changing $n,k$ to $x+n,n-1$ respectively in$$\binom nk\le\frac{\text{LCM}(n-k,n-k+1,\cdots, n)}{n-k}$$
we get
$$\binom{x+n}{n-1}\le\frac{\text{LCM}(x+1,x+2,\cdots, x+n)}{x+1}$$
which is your claim.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2907120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Contour Integral of irrational polynomial from -1 to 1 I've been stuck at htis contour integral problem for a few hours now, and seem to be hitting brick walls.
$$
\int_{-1}^1 \frac{\sqrt{1-x^2}}{1+x^4}dx\,,
$$
I tried a trig substitution $x=\cos{\theta}$ but noticed all the poles were the unit circle and I didn't know... | As pointed out by @paul garrett, dog-bone contour (dumbbell contour) works perfectly. Indeed, consider the contour $\mathcal{C}$ given as follows:
$\hspace{12em}$
With the principal branch cut, as the radius/band-width of $\mathcal{C}$ goes to zero,
\begin{align*}
\oint_{\mathcal{C}} \frac{i\sqrt{z-1}\sqrt{z+1}}{i(z^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Help: $ |\frac{a+1}{a}- (\frac{xz}{y^2})^k|\leq \frac{1}{b}$ In the paper linear forms in the logarithms of real algebraic numbers close to 1, it is written on page $9$ that
On other hand, a short calculation yields
$$ \left|\frac{a+1}{a}- \left(\frac{xz}{y^2}\right)^k\right|\leq \frac{1}{b}$$
Image of the page... | I have got $$\left|\frac{a+1}{a}-\frac{(a+1)(ab^2+1)}{(ab+1)^2}\right|=\left|{\frac { \left( a+1 \right) \left( 2\,ab-a+1 \right) }{a \left( ab+1
\right) ^{2}}}
\right|$$
I have compute $$\frac{1}{4}-f(a,b)^2=\frac{\left(a^3 b^2-2 a^2 b+2 a^2-4 a
b+a-2\right) \left(a^3 b^2+6 a^2 b-2 a^2+4 a
b+a+2\right)}{4 a^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Express $x$ in terms of $a$ and $b$: $\sin^{-1} {\frac{2a}{1+a^2}} + \sin^{-1}{\frac{2b}{1+b^2}} = 2\tan^{-1}x$
Find the value of $x$ from the following equation in terms of $a$ and $b$
$$\sin^{-1} {2a\over{1+a^2}} + \sin^{-1}{2b\over{1+b^2}} = 2\tan^{-1}x$$
I tried to expand the LHS using the formula $$\sin^{-1}c... | Divide both sides by $2$, then using the $\arctan(x)=y \implies x=\tan(y)$ you will get that $$x=\tan\left(\frac{\arcsin\left(\frac{2a}{1+a^2}\right)+\arcsin\left(\frac{2b}{1+b^2}\right)}{2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
continued fraction of $\sqrt{10k+3}$ Could you please help me to find the continued fraction of
$$\sqrt{10k+3}.$$
Where $k$ is a positive integer.
All the best,
| \begin{equation*}
x+1=\sqrt{10k+3}
\end{equation*}
\begin{equation*}
x^{2}+2x+1=10k+3
\end{equation*}
\begin{equation*}
x^{2}+2x=10k+2
\end{equation*}
\begin{equation*}
x\left(x+2\right)=10k+2
\end{equation*}
\begin{equation*}
x=\dfrac{10k+2}{2+x}
\end{equation*}
\begin{equation*}
x=\dfrac{10k+2}{2+\dfrac{10k+2}{2+\dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2916930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Linear transformations defined by $T(v) = Av$. Find all of possible $v$ I'm stuck on a problem. The problem is this:
The linear transformation $T : \Bbb{R}^4 \to \Bbb{R}^2$ is defined by $T(v) = Av$, where
$$A = \begin{bmatrix} 2 & -1 & 0 & 1 \\ 1 & 2 & 1 & -3 \end{bmatrix}$$
Find all vectors $v$ such that: $$T(v) = \b... | Hint: Augment the matrix $A$ with the column $\begin{pmatrix} 1\\2\end{pmatrix}$ and row-reduce the resulting augmented matrix.
So, row reduce $\left (\begin{array}{rrrr|r}2&-1&0&1&1\\1&2&1&-3 &2\end{array}\right ) $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2918172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
maximum value of expression $(\sqrt{-3+4x-x^2}+4)^2+(x-5)^2$ maximum value of $\bigg(\sqrt{-3+4x-x^2}+4\bigg)^2+\bigg(x-5\bigg)^2\;\forall\;x\in[1\;,3]$
what i try
$\displaystyle -3+4x-x^2+16+8\sqrt{-3+4x-x^2}+x^2+25-10x$
$\displaystyle -6x+38+8\sqrt{-3+4x-x^2}$
using derivative it is very lengthy
help me how to solve,... | With a bit of arrangements I did on paper, you could show that
\begin{equation}
f'(x)
=
-\dfrac{6\sqrt{-x^2+4x-3}+8x-16}{\sqrt{-x^2+4x-3}}
\end{equation}
which is zero when
\begin{equation}
6\sqrt{-x^2+4x-3}= -8x+16
\end{equation}
square both sides
\begin{equation}
-36x^2+ 144x - 108 = 64x^2 -256x + 256
\end{equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2918617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Show that $\sum_{\text{cyc}} \frac{1}{b^2+c^2+5bc-a^2} \leq \frac{\sqrt3}{8S}$ for a triangle with sides $a$, $b$, $c$ and area $S$ Let be $a$, $b$, $c$ sides of a triangle and $S$ his area. Prove that $$\sum_{\text{cyc}} \frac{1}{b^2+c^2+5bc-a^2} \leq \frac{\sqrt3}{8S}$$
My idea: $b^2+c^2-a^2 = 2bc \cos A$, so the in... | Now, you can use Jensen because
$$\left(\frac{\sin\alpha}{5+2\cos\alpha}\right)''=\frac{\sin\alpha(10\cos\alpha-17)}{(5+2\cos\alpha)^3}<0$$ for all $0<\alpha<\pi.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2918736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
First derivative of $f(x)= \frac{2}{x+1} +3$ I'm struggling to find the first derivative of $f(x)= \frac{2}{x+1} +3$ using the limit definition of derivative. I keep coming up with $f'(x) = \frac{-5}
{(x+1)^2}$ but I should be getting $f'(x) = \frac{-2}{(x+1)^2}$
\begin{align}
\lim_{x\to 0} &= \frac{(\frac{2}{x+h+1}+3)... | The $3$ in the numerator should cancel out.
Even if you want to keep it for a while, it shouldn't be $3(x+1)$ and $3(x+h+1)$, it should be $3(x+1)(x+h+1)$ for both of them.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2923569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How many even/odd numbers can be formed within a specified range?
Consider
$$A = \{ 1,2,3,4,5,6\}$$
a) How many $3$ digit even numbers between $200$ and $500$ can be formed?
b) How many $3$ digit odd numbers between $240$ and $600$ can be formed?
For the part a, we have that
$$P = (5)(6)(3) = 90$$
For the part b, we ... | For $(a)$, they have first digit $2,3$ or $4$, and last digit $2,4$ or $6$. So, $3\cdot 6\cdot 3=54$.
For $(b)$, first digit can be $2,3,4$ or $5$. Second digit can be $4,5$ or $6$ when the first digit is $2$, and the last digit is $1,3$ or $5$. So $3\cdot3=9$ for when the first digit is two. Then when the first di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2925698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find constant $a$ in way that $\lim_{x\rightarrow -2} \frac{3x^2+ax+a+3}{x^2+x-2}$ has limit Problem
If there exists $a \in \mathbb{R}$ such that:
$$ \lim_{x\rightarrow -2} \frac{3x^2+ax+a+3}{x^2+x-2} $$
has limit in $-2$. If such $a$ exists what is limit in $-2$ ?
Attempt to solve
My idea was first to try factorize ... | You have $3x^2+ax+a+3=0$ and you set $x = -2$ and got $3(-2)^2+a(-2)+(-2)+3=0$
But you should have gotten $3(-2)^2+a(-2)+a+3=0$ .
That is an "$a$" between the $ax$ and the $3$. It is not an "$x$".
So you should have gotten $12 -2a + a + 3 = 0$ or $15 -a = 0$ or $a = 15$
.......
so $3x^2 + 15x+18 = 0$ or in other w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2927062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Proof verification of $\lim_{x\to-2^+}\frac{x^2+x-2}{\sqrt{x+2}}=0$ using $\epsilon$-$\delta$ I am trying to prove that
$$\lim_{x\to-2^+}\frac{x^2+x-2}{\sqrt{x+2}}=0$$
and this is my approach.
Let $\epsilon>0,\delta>0\,$ so that
$$\left|\frac{x^2+x-2}{\sqrt{x+2}}\right|<\epsilon\quad\text{whenever}\quad-2<x<-2+\delta$... | Seems fine, just write it in the forward direction after working backward.
Given $\epsilon >0$, we let $\delta = \min\{ 1, \frac{\epsilon^2}{9}\}$,
then if $-2 < x < -2 + \delta$, we have $-3<x-1< -2+\delta \implies 2<|x-1|<3$.
hence,
$$\left|\frac{x^2+x-2}{\sqrt{x+2}}\right|=\left|\frac{(x-1)(x+2)}{\sqrt{x+2}}\right|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2928373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Is it true that $\frac{1}{\pi^{2n+1}} \int_0^{\theta} \ln^{2n}\left(\frac{\sin x}{\sin\left(\theta-x\right)}\right)\,dx$ is a rational... I was trying to evaluate $\displaystyle \int_0^{\frac{\pi}{6}}\ln^2\left(2\sin x\right)\,dx$ in an elementary way (no complex variable) so i have considered:
$\displaystyle \int_0^{\... | The integral can be modified as
\begin{align}
I&= \frac{1}{\pi^{2n+1}} \int_0^{\theta} \ln^{2n}\left(\frac{\sin x}{\sin\left(\theta-x\right)}\right)\,dx\\
&=\frac{1}{\pi^{2n+1}} \int_{-\theta/2}^{\theta/2} \ln^{2n}\left(\frac{\sin \left( \theta/2+y \right)}{\sin\left(\theta/2-y\right)}\right)\,dy\\
&=\frac{2}{\pi^{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2930728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Closed form of $\int_0^\infty \left(\frac{\arctan x}{x}\right)^ndx$ I know that for $n=1$ the integral is divergent and that for $n=2$ the integral has a closed form. However, I wonder if the general expression has a closed form.
My attempt: $$\int_0^\infty \left(\frac{\arctan(x)}{x}\right)^ndx=\frac{n}{1-n}\int_0^\inf... | As user90369 doesn't have the time apparently I thought I repesent my own solution explicitly for those interested.
Starting with
\begin{align}
\int_{-\pi/2}^{\pi/2} \frac{x^n \left(\cos x\right)^{n-2}}{\left(\sin x\right)^n} \, {\rm d}x &\stackrel{y=2x}{=} 2^{-n+1} \, i^n \int_{-\pi}^{\pi} y^n \, e^{-iy} \, \frac{\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2933377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 3
} |
If $a^2+b^2 \gt a+b$ and $a,b \gt 0$ Prove that $a^3+b^3 \gt a^2+b^2$ I'm not too sure about this, I have been working on for some time and I reached a solution (not really too sure about)
Question: If $a^2+b^2 \gt a+b$ and $a,b \gt 0$ Prove that $a^3+b^3 \gt a^2+b^2$
My solution: Let $a \geq b$
From $a^2+b^2 \gt a+b$ ... | It is correct, but I would explain the "since $a\ge b$ then $a^3 - a^2 > b^3 - b^2$" step a bit more.
\begin{align*}a^2 - a > b^2-b &\iff a(a^2-a) > a(b^2-b) \quad \text{(since $a>0$)} \\ &\iff a(a^2-a)>a(b^2-b)\ge b(b^2-b) \quad \text{(since $b\le a$)} \\ &\iff a^3-a^2 > b^3 -b^2 \\ &\iff a^3 + b^3 > a^2 + b^2\end{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2934061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
How can I show a decreasing derivative? I have a derivative
$f'(x) = -1 + 8 \cos \frac{1}{x} + 4 \sin \frac{1}{x}$
And we have to show that it's decreasing in all intervals on the form $[\frac {6}{(12 n + 11) \pi } , \frac {6}{(12n + 7)\pi}$ for n>= 1
Our hint is to use the derivative such as $\frac {6}{(12n + 7)\pi} ... | Hint: $$x \in \left[\frac {6}{(12 n + 11) \pi } , \frac {6}{(12n + 7)\pi}\right] \iff \frac1x \in \left[\left(2n+\frac76\right)\pi,\left(2n+\frac{11}6\right)\pi\right]$$
We have $x,y \in \left[\frac {6}{(12 n + 11) \pi },\frac {6}{(12n + 7)\pi}\right]$ and $x<y$ iff $\frac1x>\frac1y$ in $\left[\left(2n+\frac76\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2934243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to factor $(1+\frac{1}{x})(-\frac{6}{x^2})+(\frac{6}{x^3})(1+\frac{1}{x})^2$ to get $\frac{6}{x^3}(1+\frac{1}{x})(1+\frac{2}{x})$?
How can I factor this: $$\left(1+\frac{1}{x}\right)\left(-\frac{6}{x^2}\right)+\left(\frac{6}{x^3}\right)\left(1+\frac{1}{x}\right)^2$$
in order to get this result: $$\frac{6}{x^3}\lef... | Hint
$$-\frac{6}{x^2}\left(1+\frac{1}{x}\right)+\frac{6}{x^3}\left(1+\frac{1}{x}\right)^2=\frac{6}{x^2}\left(1+\frac 1x\right)\left(\frac 1x+\frac 1{x^2}-1\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2935936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
If $a_n=\frac{(-1)^n}{\sqrt{1+n}}$ and $c_n=\sum_{k=0}^n a_{n-k}a_k$, does $\sum_{n=0}^\infty c_n$ converge?
Suppose for $n \in \mathbb{N} \cup \{0\}$, $a_n=\frac{(-1)^n}{\sqrt{1+n}}$. Define $c_n=\sum_{k=0}^na_{n-k}a_k$. Does $\sum_{n=1}^\infty c_n$ converge?
What I attempted:-
\begin{equation}
\begin{aligned}
c_n&=... | Hint. Note that
$$d_n=\sum_{k=0}^n\frac{1}{\sqrt{(1+k)(n+1-k)}}\geq \sum_{k=0}^n\frac{2}{(1+k)+(n+1-k)}=\frac{2(n+1)}{n+2},$$
and therefore $c_n=(-1)^n d_n$ does not tend to zero as $n$ goes to infinity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2937987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Proving $a\sqrt{7}>\frac{1}{c}$, where $a$ is an integer and $c = \lceil\frac{a}{\sqrt{7}}\rceil-\frac{a}{\sqrt{7}}$
Let $a \in \Bbb{N}$, and $c=\lceil \frac{a}{\sqrt{7}}\rceil-\frac{a}{\sqrt{7}}$. Prove that $$a\sqrt{7}>\frac{1}{c}$$
So the very original problem sounds like this:
It is given that $a,b \in \Bbb{N... | For the original question the critical thing to realize is that if $7b^2-a^2\gt 0$, it is at least $3$. The squares $\bmod 7$ are $1,2,4$ which gives this, as $a^2$ cannot be $5$ or $6\ \bmod 7$
First we take care of a side case. If $b\sqrt 7 -a \gt 1, b\sqrt 7-a \gt \frac 1a$.
Given $1 \gt b\sqrt 7-a\gt 0$ we ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2939663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Integral $\int\frac{\sqrt{4x^2-1}}{x^3}dx$ using trig identity substitution!
$$\int \frac{\sqrt{4x^2-1}}{x^3}\ dx$$
So, make the substitution
$ x = \sqrt{a \sec \theta}$, which simplifies to $a \tan \theta$.
$2x = \sqrt{1} \sec \theta$,
$ d\theta = \dfrac{\sqrt{1}\sec\theta\tan\theta}{2}$
$\int \dfrac{\sqrt{1}\tan\... | Setting
$x=\frac{\sqrt{\theta^{2}+1}}{2}$
$dx=\frac{\theta}{2\sqrt{\theta^{2}+1}}$,
and the funcion to integrate is:
$\frac{4\theta^{2}}{(\theta^{2}+1)^{2}}$,
whose integral is:
$2. atan(\theta)-\frac{2\theta}{\theta^{2}+1}$.
Substituting the value of
$\theta=\sqrt{4x^{2}-1}$,
we get:
$I=2. atan(\sqrt{4x^{2}-1})-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2940220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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$s_n=s_{n-1}+(n-1)s_{n-2}$ prove $s_n>\sqrt{n!n}$ for $n\ge4$ Define sequece as follows: $s_1=1,~s_2=2, s_n=s_{n-1}+(n-1)s_{n-2}$. I want to prove that $s_n>\sqrt{n!n}$ for $n\ge4$. I'd tried to use traditional induction on $n$, but it involves both two terms that are in front of the current term. Should I use strong i... | Let's try an induction :
You have $s_1 = 1$ and $s_2 = 2$, so $s_3 = 4$ and $s_4 = 10$ and $s_5 = 26$.
Therefore, you have $s_4 = 10 = \sqrt{100} > \sqrt{96} = \sqrt{4! \times 4}$. And $s_5 = 26 = \sqrt{676} > \sqrt{600} = \sqrt{5! \times 5}$.
Now, let's suppose your result is true for two successive ranks $n$ and $n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2942338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Deriving polar graphs There is just some confusion I have with polar graphs, for instance there are some well known general forms of the polar graphs such as the circle, Limaçon, rose, Lemniscate.
The general equation for a circle with radius $\frac{a}{2}$ is given by either:
$$r = a \cos\theta \space \space \text{or}... | Degree 1 polynomials describe lines
Degree 2 polynomials describe lines the conic sections, parabola, circle, ellipse, hyperbola.
Degree 3 polynomials describe "elliptic curves"
You will need a $4^{th}$ degree$^+$ polynomial.
A cartiod
$r = a(1+\cos \theta)\\
r^2 = a(r+r\cos \theta)\\
x = r\cos \theta, r = \sqrt {x^2 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2942427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integrate $\int \frac {dx}{\sqrt {(x-a)(x-b)}}$ where $b>a$ Integrate: $\displaystyle\int \dfrac {dx}{\sqrt { (x-a)(x-b)}}$ where $b>a$
My Attempt:
$$\int \dfrac {dx}{\sqrt {(x-a)(x-b)}}$$
Put $x-a=t^2$
$$dx=2t\,dt$$
Now,
\begin{align}
&=\int \dfrac {2t\,dt}{\sqrt {t^2(a+t^2-b)}}\\
&=\int \dfrac {2\,dt}{\sqrt {a-b+t^2}... | Set
$$J=\int{\frac{dx}{\sqrt{{{x}^{2}}-\left( a+b \right)x+ab}}}$$
Use the following substitution:
$$u=\frac{2x-\left( a+b \right)}{\sqrt{{{x}^{2}}-\left( a+b \right)x+ab}}$$
once get:
$$du=\left( -\frac{{{\left( 2x-\left( a+b \right) \right)}^{2}}}{2{{\left( \sqrt{{{x}^{2}}-\left( a+b \right)x+ab} \right)}^{3}}}+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2944173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
convergence of the series $\sum u_n, u_n = \frac{n^n x^n}{n!}$ for $x>0$ While checking the convergence of the series $\sum u_n, u_n = \frac{n^n x^n}{n!}$ for $x>0$, we used ratio test to say that for $0< x < \frac1e$ $\sum u_n$ is convergent and for $\frac1e < x<\infty$ $\sum u_n$ is divergent.
We use Logarithimic Tes... | $$
n+n^2\log\frac{n}{n+1} = n-n^2\log\frac{n+1}{n} = n-n^2\log\left(1+\frac{1}{n}\right)
$$
So do some asymptotics. As $n \to +\infty$, we have
\begin{align}
\log\left(1+\frac{1}{n}\right) &= \frac{1}{n}-\frac{1}{2n^2}+o\left(\frac{1}{n^2}\right)
\\
n^2\log\left(1+\frac{1}{n}\right) &= n-\frac{1}{2}+o(1)
\\
n-n^2\log\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2944305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that the quantity is an integer I want to prove that $\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6} \in \mathbb{Z}, \forall n \geq 1$.
I have thought to use induction.
Base Case: For $n=1$, $\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6}=\frac{1}{3}-\frac{1}{2}+\frac{1}{6}=0 \in \mathbb{Z}$.
Induction hypothesis: We suppose t... | Yes, we can. If $f(n)=\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6}$, then $f(n+1)-f(n)=n^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2946269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 5
} |
Solving the system $a^2-c^2=x^2-z^2$, $ab=xy$, $ac=xz$, $bc=yz$ I've stumbled upon these equations, and am struggling to find a manual way to solve this in $\Re$:
$$\begin{align}
a^2-c^2&=x^2-z^2 \\
ab&=xy \\
ac&=xz \\
bc&=yz\end{align}$$
I've used Wolfram Alpha to compute this and I found that this is only possible ... | This is an homogeneous system so making $y = \lambda x$ and $z = \mu x$ we have
$$
a^2-c^2=x^2(1-\mu^2)\\
ab = x^2\lambda\\
ac = x^2\mu\\
bc = x^2\lambda\mu
$$
and making now $A = \frac ax,B = \frac bx, C = \frac cx$
$$
A^2-C^2=1-\mu^2\\
AB = \lambda\\
AC = \mu\\
BC = \lambda\mu
$$
so we conclude easily
$$
A^2B^2C^2=\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2946743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Convergence for series failed using "Ratio Test" $$\sum_{n=1}^{\infty}\frac{1\cdot 3\cdot 5\cdot ...\cdot (2n-1)}{1\cdot 4\cdot 7\cdot ...(3n-2)}$$
Using Ratio test: $$\lim_{n\rightarrow \infty}\frac{\frac{2(n+1)-1}{3(n+1)-2}}{\frac{2n-1}{3n-2}}$$
which equals to : $$\lim_{n \to \infty}\frac{6n^{2}-n-2}{6n^{2}-n-1}$$
t... | $$a_k=\prod_{k=1}^{n}\frac{2k-1}{3k-2}=\left(\frac{2}{3}\right)^n\cdot\frac{\Gamma\left(\frac{2}{3}\right)^2}{\pi 2^{2/3}}\cdot B\left(n+\tfrac{1}{2},\tfrac{1}{6}\right)\tag{1}$$
implies
$$\begin{eqnarray*} \sum_{k\geq 1}a_k &=& \frac{\Gamma\left(\frac{2}{3}\right)^2}{\pi 2^{2/3}}\int_{0}^{1}\sum_{n\geq 1}\left(\frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2948533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find the order of the matrices $A$ and Matrice $B$?
Find the order of the matrices $A$ and $B$ in the Group $GL_2(\mathbb{F}_7)$: $$A=\begin{bmatrix} 1 &1 \\ 0& 1 \end{bmatrix}\;\text{and}\; B=\begin{bmatrix} 2 &0 \\ 0 & 1 \end{bmatrix}$$
For $A$ I take $
\pmatrix{1&x\\0&1}\pmatrix{1&y\\0&1}=\pmatrix{1&x+y\\0... | For your matrix $A$, we have $$\begin{pmatrix}1 &1 \\0 &1 \end{pmatrix}^n=\begin{pmatrix}1 &n \\0 &1 \end{pmatrix}$$
so, we have to find a least $n$ for which $$\begin{pmatrix}1 &n \\0 &1 \end{pmatrix}=\begin{pmatrix}1 &0 \\0 &1 \end{pmatrix}$$ Now, clearly for $n=1,2,\cdots,6$ it does't happen but in the seventh power... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2949070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Rational fraction expression for triangular powers of 2 Why does the following pattern hold?
$$1=1=2^0$$
$$\frac{2(2^2-1)}{3}=2=2^1$$
$$\frac{3^2(3^2-1)(3^2-2^2)}{3^2 \times 5}=8=2^3$$
$$\frac{4^2(4^2-1)^2(4^2-2^2)(4^2-3^2)}{3^3 \times 5^2\times 7}=64=2^6$$
$$\frac{5^3(5^2-1)^2(5^2-2^2)^2(5^2-3^2)(5^2-4^2)}{3^4 \times ... | a partial answer:
$$
\frac{5^3(5^2-1)^2(5^2-2^2)^2(5^2-3^2)(5^2-4^2)}{3^4 \times 5^3 \times 7^2 \times 9}
$$
may be re-written as
$$
\frac{\frac{9!}{0!}\frac{7!}{2!}\frac{5!}{4!}}{\frac{9!}{4!2^4}\frac{7!}{3!2^3}\frac{5!}{2!2^2}\frac{3!}{1!2^1}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2957325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Find $A+2B$ if $3\cos^2A+2\cos^2B=4$ and $\frac{3\sin A}{\sin B}=\frac{2\cos B}{\cos A}$
$A$ and $B$ are positive acute angles satisfying $3\cos^2A+2\cos^2B=4$ and $\dfrac{3\sin A}{\sin B}=\dfrac{2\cos B}{\cos A}$, then find the value of $A+2B$ ?
My Attempt
$\cos2B=2\cos^2B-1=3-3\cos^2A$ and $\sin2B=2\sin B\cos B=3\s... | Hint:
Using double angle formula,
$$2\sin2B=6\sin A\cos A=3\sin2A$$
$$2\cos2B=6(1-\cos^2A)=6(2\sin^2A)=3(1-\cos2A)$$
Squaring and adding we get $\dfrac49=2-2\cos2A=2(2\sin^2A)$
$\implies\sin A=\dfrac13,\cos A=+\dfrac{2\sqrt2}3$ as $0<A<\dfrac\pi2$
$$\sin2B=3\sin A\cos A=\dfrac{2\sqrt2}3,$$
$$\cos2B=3\sin^2A=\dfrac13$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2962706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Is $\lim_{n\to\infty} \left( (\sum^n_{m=1} \frac{1}{\sqrt{m}}) - \sqrt{n} \right)$ convergent?
$\lim_{n\to\infty} \left( (\sum^n_{m=1} \frac{1}{\sqrt{m}}) - \sqrt{n} \right)$ : convergent?
My attempt
$$
(\sum^n_{m=1} \frac{1}{\sqrt{m}}) - \sqrt{n} = (\frac{1}{\sqrt{1}} - \frac{1}{\sqrt{n}}) + (\frac{1}{\sqrt{2}} - \f... | The limit is not convergent. You can use the integral test to see this.
\begin{align}
\sum_{m=1} ^n \frac{1}{\sqrt{m}} & \geq \int_{1}^{n} \frac{1}{\sqrt{x}} \ dx \\ & = 2\sqrt{n} - 2 .
\end{align}
So we can conclude that
$$\sum_{m=1} ^n \frac{1}{\sqrt{m}} - \sqrt{n} \ \geq \sqrt{n}- 2.$$
Letting $n \to \infty$ givs... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2963418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Minimum value of the given expression If I have $a>0$ and $ b >0 $ and $a+b=1$ then how can I find the minimum value of $(a+1/a)^2 + (b+1/b)^2$. I just tried to do it by expanding the expression just because I knew the graph of $x^2+1/x^2 $ and for that I had the minimum value as 2 but since there are two variables $a... | This problem is as old as a volkswagen beetle classic ( that I have ,haha ). My answer is problably one of the $20$ different "proofs" that you can find here or google. I try to be a little different if possible...
Using Cauchy-Schwarz inequality twice: $(1^2+1^2)\left(\left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2964932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $\frac{2x + 6} {(x + 2)^2}- \frac{2} {x + 2} = \frac{a} {(x + 2)^2}$, then what is $a$? $$\frac{2x + 6} {(x + 2)^2}- \frac{2} {x + 2}$$
The expression above is equivalent to $$\frac{a} {(x + 2)^2}$$
where $a$ is a positive constant and $x \neq -2$.
What's the value of $a$?
| There is no need to do any algebra.
Consider
$$
\frac{2x + 6} {(x + 2)^2}- \frac{2} {x + 2}
=
\frac{a} {(x + 2)^2}
$$
Since this equality holds for all values of $x\ne -2$,
set $x=0$ and get
$$
\frac{6} {2^2}- \frac{2} {2}
=
\frac{a} {2^2}
$$
which gives $a=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2966822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How to obtain the lower bound of $\dfrac{n}{\sqrt[n]{n!}}$ by Taylor's series? We want to prove $$\lim_{n \to \infty}a_n=\lim_{n \to \infty}\frac{n}{\sqrt[n]{n!}}=e.$$
I have some solutions for this, but I want to find another method applying the squeeze theorem. Thus, a natrual thought is to find the upper bound and t... | \begin{align*}
e^n&=1+n+\frac{n^2}{2!}+\cdots+\frac{n^n}{n!}+\cdots\\
&<(n+1)\cdot\frac{n^n}{n!}+\frac{n^n}{n!}\cdot\left[\frac{n}{n+1}+\frac{n^2}{(n+1)(n+2)}+\cdots\right]\\
&< (n+1)\cdot\frac{n^n}{n!}+\frac{n^n}{n!}\cdot\left[\frac{n}{n+1}+\frac{n^2}{(n+1)^2}+\cdots\right]\\
&=(n+1)\cdot\frac{n^n}{n!}+\frac{n^n}{n!}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2967367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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How can I find $\gcd(n^a-1,m^a-1)$? From Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$ , we have $$\gcd(a^n-1,a^m-1)=a^{\gcd(n,m)}-1$$
for every positive integers $a,n,m$.
I reversed $a$ with $n,m$, and I had this question:
Find $\gcd(n^a-1,m^a-1)$ for every positive integers $a,n,m$
My attempt is to find... | Welcome. If n and m are as following forms:
$n=k b+1$ ⇒ $n≡1\mod b$⇒ $n^a≡1\mod b$
$m=t b+1$ ⇒ $m≡1 \mod b$⇒$m^a≡1\mod b$
Then the common divisor will be:
$gcd(n^a-1, m^a-1)=b$
In other words if $n-1$ and $m-1$ have a common divisor like b then $n^a-1$ and $m^a-1$ have a common divisor like b. This can also seen in fol... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2968044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Integration-by-part Here is the given question
$$\int{\frac{x^3}{\sqrt{1+x^2}}dx}$$
I solved using integration by part as follow:
$$
\int\frac{x}{\sqrt{1+x^2}}x^2\,dx = x^2\int\frac{x}{\sqrt{1+x^2}}\,dx - \int\biggl(\frac{dx^2}{dx}\int\frac{x}{\sqrt{1+x^2}}dx\biggr)dx\tag{i}\label{i}
$$
Solving for $\int \frac{x}{\sqr... | It should be
$$\int{\frac{x}{\sqrt{1+x^2}}*x^2dx} = x^2\sqrt{1+x^2} - 2\cdot \frac{(1+x^2)^{3/2}}{3} + C$$
Use
$$x^2\sqrt{1+x^2}=(x^2+1-1)\sqrt{1+x^2}=(1+x^2)^{3/2}-(1+x^2)^{1/2}$$ to match with the given answer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2970470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Die is rolled until 1 appears. What is the probability of rolling it odd number of times? Problem: Die is rolled until 1 appears. What is the probability of rolling it odd number of times?
So, so far I have this:
$\frac{1}{6}$ - this is a probability of rolling "1" on first try
$\frac{5}{6} \cdot \frac{5}{6} \cdot \f... | Yet another wording for the same argument. See if this thinking/wording is convincing.
$P_m = $ prob of the first $1$ occurring on the $m$-th roll $= (\frac 56)^{m-1}\cdot \frac 16$.
Rolling the first one on the $m$th roll and rolling the first one on the $k$th roll are mutually exclusive events. So
$P_{m\lor k} =$ t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2972532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Bezout's Identity: Finding A Pair of Integers
There is only one integer $x$, between 100 and 200 such that the integer pair $(x, y)$ satisfies the equation $42x + 55y = 1$. What's the value of $x$ in this integer pair?
We know that
$$\begin{align} x &= x_0 + 5t \\ y &= y_0 - 4t \end{align}$$
But we need to know what ... | First, you find a solution to $42x + 55y = 1$
\begin{array}{c|ccc|l|l}
& 55 & 1 & 0 & &55 = 55(1)+42(0) \\
-1 & 42 & 0 & 1 & (55,1,0)+(-1)(42, 0, 1)=(13,1,-1) & 42 = 55(0)+42(1) \\
\hline
-3 & 13 & 1 & -1 & (42, 0, 1) + (-3)(13,1,-1)=(3,-3,4)& 13 = 55(1)+42(-1) \\
-4 & 3 & -3 & 4 & (13,1,-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2973025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Proving that a sequence $a_n: n\in\mathbb{N}$ is (not) monotonic, bounded and converging $$a_n = \left(\dfrac{n^2+3}{(n+1)^2}\right)\text{ with } \forall n\in \mathbb{N}$$
$(0\in\mathbb{N})$
Monotonicity:
To prove, that a sequence is monotonic, I can use the following inequalities:
\begin{align}
a_n \leq a_{n+1}; a_n <... | In order to analyse the sequence, $$ a_n = \left(\dfrac{n^2+3}{(n+1)^2}\right)\text{ with } \forall n\in \mathbb{N}$$ We can look at the function, $$f(x) =\frac {x^2+3}{(x+1)^2}$$
$$f'(x) = \frac { x^2-2x-3}{(x+1)^4} >0 \text { for x>3 }$$ Therefore the sequence is increasing for $n>3$
Note that $$\lim _{x\to \infty ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
If $2 \mid F_n$, then $4 \mid F_{n+1}^2-F_{n-1}^2$, where $F_n$ is $n$-th Fibonacci number
I want to show that
*
*If $2 \mid F_n$, then $4 \mid F_{n+1}^2-F_{n-1}^2$
*If $3 \mid F_n$, then $9 \mid F_{n+1}^3-F_{n-1}^3$
where $F_n$ is the $n$-th Fibonacci number.
I have tried the following so far:
Since $F_... | We have more general answer:
If $k\mid a-b$ then $k^2\mid a^k-b^k$
Proof: Write $a-b=km$ for some integer $m$. Then we have
$$a^k-b^k=(a-b)(\underbrace{a^{k-1}+a^{k-2}b+a^{k-3}b^2+...+b^{k-1}}_E)$$
\begin{eqnarray}E&\equiv_k& b^{k-1}+b^{k-2}b+b^{k-3}b^2+...+b^{k-1}\\
&\equiv_k&k\cdot b^{k-1} \\
&\equiv_k&0 \\
\end{e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2974491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Diagonalizing a matrix with complex numbers hope everyone is doing well.
I'm a bit stuck on the following problem:
$T: C^2 \rightarrow C^2$, where $T$ is defined by $T(x,y)=(2x + (3-3i)y, (3+3i)x + 5y$
So I perform the usual calculations to diagonalize the matrix, right up until I find that the roots of the characteris... | We may write
$T(x,y)=(2x + (3-3i)y, (3+3i)x + 5y \tag 1$
as
$T \begin{pmatrix} x \\ y \end{pmatrix} = \begin{bmatrix} 2 & 3 - 3i \\ 3 + 3i & 5 \end{bmatrix} \begin{pmatrix} x \\ y \end{pmatrix}, \tag 2$
and then he characterisic polynomial of $T$ is
$\det (T - \lambda I) = \det \left ( \begin{bmatrix} 2 & 3 - 3i \\ 3 +... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Prove that $\sqrt[3]{5} + \sqrt{2}$ is irrational I tried with both squaring and cubing the statement, it got messy, here's my latest attempt:
Assume for the sake of contradiction: $\sqrt[3]{5} + \sqrt{2}$ is rational
$\sqrt[3]{5} + \sqrt{2}$ = $\frac{a}{b}$ $a,b$ are odd integers $> 0$ and $ b\neq 0$
${(\sqrt[3]{5}... | Assume that $$ \sqrt[3]{5} + \sqrt{2}=r$$ where r is a rational number.
We have $$ \sqrt[3]{5} =r-\sqrt{2}$$
Raise to the third power to get $$5=r^3-3r^2 \sqrt 2 +6r - 2\sqrt 2 $$
Solving for $\sqrt 2$ we get $$ \sqrt 2 = \frac {5-r^3-6r}{-3r^2-2}$$
The $RHS$ is a rational number which is impossible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2975313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
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Proof that $\sin^{2n + 1}(x) = \sum_{i = 0}^{n} a_{i} \sin\left(\left(2i + 1\right)x\right)$ In playing around on wolframalpha, I noticed that $\sin^{2n + 1}(x)$ where $n \in \mathbb{N} \cup \{0\}$ takes the form:
$$\sin^{2n + 1}(x) = \sum_{i = 0}^{n} a_{i} \sin\left(\left(2i + 1\right)x\right)$$
Where $a_{i} \in \math... | Just a thought that came to mine on how this could be proven
Employ proof my induction
Step (1) : Prove true for $n = 0 $
$$\sin^{1}(x) = 1\cdot\sin(1\cdot x) = \sum_{i = 0}^{0} a_{i}^{(0)}\sin\left(\left(2i + 1\right)x\right)$$
Hence true.
Step (2) : Inductive Step - Assume true for $n = k$:
$$\sin^{2k + 1}(x) = \su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2981137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Asymptotic behavior of solution to nonlinear ODE I am trying to determine the asymptotic behavior as $t \to \infty$ of the solution to the IVP:
$$y''(t) - y(t) + \frac{1}{[y(t)]^3} = 0\\y(0) = 1;\, y'(0) = 1$$
Without the nonlinearity, we have solutions of the form $e^{\pm t}$. I think the nonlinear solution has simil... | Let's see if we can solve this exactly.
Inverting the function gives us:
$$-\frac{t''}{t'^3} - y + \frac{1}{y^3} = 0 \\ t''+\left(y-\frac{1}{y^3} \right) t'^3=0$$
$$t'(y)=f(y)$$
$$f'=\left(\frac{1}{y^3}-y \right) f^3$$
$$-\frac{1}{2 f^2}=-\frac{1}{2 y^2}-\frac{y^2}{2}+C$$
$$\frac{1}{f^2}=\frac{1}{y^2}+y^2+C$$
$$f^2=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2982356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Double summation $\sum_{m,n=1\, m\neq n}^\infty{\frac{m^2+n^2}{mn(m^2-n^2)^2}}$ As a follow to this answer I came across the double sum $$\sum_{m,n=1\, m\neq n}^\infty{\frac{m^2+n^2}{mn(m^2-n^2)^2}}.$$
But unfortunately I do not have skills in techniques to handle double summation .
Help appreciated.
I've made some re... | Partial Fractions gives
$$
\frac{m^2+n^2}{\left(m^2-n^2\right)^2}=\frac{1/2}{(m-n)^2}+\frac{1/2}{(m+n)^2}\tag1
$$
First, we will compute
$$
\begin{align}
\sum_{\substack{m,n=1\\m\ne n}}^\infty\frac1{mn(m-n)^2}
&=2\sum_{n=1}^\infty\sum_{m=n+1}^\infty\frac1{mn(m-n)^2}\tag{2a}\\
&=2\sum_{n=1}^\infty\sum_{m=1}^\infty\frac1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2986764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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Inverse of tridiagonal Toeplitz matrix Consider the following tridiagonal Toeplitz matrix. Let $n$ be even.
$${A_{n \times n}} = \left[ {\begin{array}{*{20}{c}}
{0}&{1}&{}&{}&{}\\
{1}&{0}&{1}&{}&{}\\
{}&{1}&{\ddots}&{\ddots}&{}\\
{}&{}&{\ddots}&{\ddots}&{1}\\
{}&{}&{}&{1}&{0}
\end{array}} \right]$$
What is the inverse... | Firstly Matrix is Toeplitz. This means it represents multiplication by power series expansion. This means matrix inversions corresponds to multiplicative inversion
Therefore, consider $$x+x^{-1}=\frac{x^2+1}x$$ Now it's multiplicative inverse:
$$(x+x^{-1})^{-1}=\frac x{x^2+1}$$
Now you can expand with geometric series ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2991218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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How to show $\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+...+\frac{n}{2^n}<2$ using induction I´d like to show that $$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+...+\frac{n}{2^n}+\frac{n+1}{2^{n+1}}<2$$ using the fact that $$\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+...+\frac{n}{2^n}<2$$
I guess the answer use transitivity of natural numbe... | Hint:
$$\begin{aligned}
&\frac{1}{2}+\frac{2}{4}+\frac{3}{8}+...+\frac{n}{2^n}+\frac{n+1}{2^{n+1}}\\
=&\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{2^n}+\frac{1}{2^{n+1}}\right) \\
&+\left(\frac{1}{4}+\frac{2}{8}+...+\frac{n-1}{2^n}+\frac{n}{2^{n+1}}\right)\\
\end{aligned}$$
Now try to show each bracket is le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2992278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluating $\frac{\sin A + \sin B + \sin C}{\cos A + \cos B + \cos C}$ for a triangle with sides $2$, $3$, $4$
What is
$$\frac{\sin A + \sin B + \sin C}{\cos A + \cos B + \cos C}$$
for a triangle with sides $2$, $3$, and $4$?
One can use Heron's formula to get $\sin A$, etc, and use $\cos A = (b^2+c^2-a^2)/(2bc... | There are well-known identities for $\triangle ABC$
with the angles $A,B,C$,
sides $a,b,c$,
semiperimeter $\rho=\tfrac12(a+b+c)$,
area $S$,
radius $r$ of inscribed and
radius $R$ of circumscribed circles,
\begin{align}
\sin A+\sin B+\sin C
&=\frac\rho{R}
\tag{1}\label{1}
,\\
\cos A+\cos B+\cos C
&=\frac{r+R}{R}
\tag{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2992867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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Minimize the expression This was a problem given in a no calculator math contest (HMMT):
What is the minimum value of $(xy)^2+(x+7)^2+(2y+7)^2$ where $x$ and $y$ are reals. My friends and I discussed about using calculus, but we reasoned that there must be a faster trick (calc would take too long). Any ideas?
| Since another user has given the essential hint of this problem. Let's us look at a generalization of the problem and spell out some details.
For any $a,b,c > 0$, consider the function
$$G(x,y) = x^2y^2 + (ax+c)^2 + (by+c)^2$$
Expanding RHS out, we get
$$\begin{align}G(x,y)
&= x^2y^2 + a^2x^2 + b^2y^2 + 2c(ax+by) + 2c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2993602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find the sum of the series $\sum_{n=1}^{\infty} \frac{2n+3}{n(n+1)(n+2)}$
Find the sum of the series $\sum_{n=1}^{\infty} \frac{2n+3}{n(n+1)(n+2)}$
My attempt:
I tried partial fractions decomposition and I get :
$$\sum_{n=1}^{\infty} \frac{2n+3}{n(n+1)(n+2)}=\sum_{n=1}^{\infty}\frac {3}{2n}-\frac 1{n+1}-\frac 1{2(n+... | Good answers have already been given; here's some intuition:
$$\sum_{n=1}^{\infty}\frac{3}{n}-\frac{2}{n+1} - \frac{1}{n+2} = 3\left(1+\frac{1}2 +\frac{1}3 + \ldots\right) - 2\left(\frac{1}2 +\frac{1}3 + \ldots\right)-1\left(\frac{1}3 + \ldots\right)= 3+\frac{3}{2}+3\left(\frac{1}3 + \ldots\right) -1 -2\left(\frac{1}3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2998674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solving $8x^3 - 6x + 1$ using Cardano's method Solve for the first root of $8x^3 - 6x + 1 = 0$
After solving I get $\sqrt[3]{\frac{-1 + \sqrt{3}i}{16}} - \sqrt[3]{\frac{1 + \sqrt{3}i}{16}}$, which is not a solution to the cubic equation:
Here's how I come up with:
Using Cardano's method:
let $x = y - \frac{b}{3a}$
Then... | You may have a sign error. I render $s^3$ as $-(1-i\sqrt{3})/16$ but you seem to have $+(1-i\sqrt{3})/16$. But even with the sign correction (which probably does get you a good answer), your approach is not best. See below.
When you have an expression with two different complex cube roots there are nine choices for ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Sandwich Theorem not working?
This is the limit I need to solve:
$$\lim_{n \to \infty} \frac{(4 \cos(n) - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n \sin(n))(n + 2)^4}$$
I simplified it to this:
$$\lim_{n \to \infty} \frac{2(4 \cos(n) - 3n^2)}{(6n^3 + 5n \sin(n))}.$$
At this point I want to use the Sandwich Theorem on the N... | We have that
$$\frac{(-4 - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n \sin(n))(n + 2)^4)}\le \frac{(4 \cos(n) - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n \sin(n))(n + 2)^4)}\le \frac{(4 - 3n^2)(2n^5 - n^3 + 1)}{(6n^3 + 5n \sin(n))(n + 2)^4)}$$
and we can conclude by squeeze theorem since for both bounds
$$\frac{(\pm4 - 3n^2)(2n^5 - n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3004776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Complex root of equation 1 let $a,b$ & $c$ are the roots of cubic $$x^3-3x^2+1=0$$
Find a cubic whose roots are $\frac a{a-2},\frac b{b-2}$ and $\frac c{c-2} $
hence or otherwise find value of $(a-2)(b-2)(c-2)$
| You know that: $$a+b+c=3$$ $$ab+bc+ac=0$$ $$abc=-1$$
Also, $$(a-2)(b-2)(c-2)=abc-2(ab+bc+ac)+4(a+b+c)-8$$
So $$(a-2)(b-2)(c-2)=-1-2(0)+4(3)-8=3$$
For cubic with roots as $$\frac a{a-2},\frac b{b-2},\frac c{c-2}$$
Find out $$\frac a {a-2}+\frac b{b-2}+\frac c{c-2}$$
and $$\frac a{a-2}\cdot \frac b{b-2} + \frac b{b-2}\... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Probability that 2 heads do not come consecutively.
A fair coin is tossed $10$ times. Then the probability that two heads do not appear consecutively is?
Attempt:
Cases are:
Given condition cannot be met with $10, 9, 8, 7$ or $6$ heads.
1) 5 heads + 5 tails.
First fulfilling the essential condition we get:
$\mathrm{H... | Another way: Let the number of solutions for $n$ tosses be $a_n$. For $n \ge 2$, the solutions are either T plus a solution for $n - 1$ tosses, or HT plus a solution for $n - 2$ tosses. So $a_n = a_{n-1} + a_{n-2}$. Since $a_1 = 2$ and $a_2 = 3$ we see that $a_n$ is the Fibonacci number $F_{n+2}$, which makes the proba... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3009989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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How to calculate $\int_{-\infty}^{+\infty} \frac{x \sin x}{x^2+4x+20}\, dx$? I get the answer that it equals $\frac{\pi(2\cos2+\sin2)}{2e^4}$ by mathematica.
But I don't know how it can equal this form.
My idea is to construct equations so that problems can be transformed.Like these:
$$\int_{-\infty}^{+\infty} \frac{x ... | As you've already noted, the substitution $y=x-2$ gives$$\int_{\mathbb{R}}\frac{x\sin xdx}{x^{2}+4x+20} =2C_{0}\sin 2-2S_{0}\cos 2+S_{1}\cos 2-C_{1}\sin 2$$with $$C_{n}:=\int_{\mathbb{R}}\frac{y^{n}\cos ydy}{y^{2}+16},\,S_{n}:=\int_{\mathbb{R}}\frac{y^{n}\sin ydy}{y^{2}+16},$$i.e. $$C_{n}+iS_{n}=\int_{\mathbb{R}}\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3012688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Show that these non-linear recursions produce integers only The recurrence is of third order:
Start with
\begin{align*}a_0(x)&=1\\ a_1(x)&=1\\ a_2(x)&=x \end{align*} and then
\begin{align*}a_{n+3}(x)&=\frac{a_{n+2}^2(x)-a_{n+1}^2(x)}{a_{n}(x)}.\end{align*}
The calculation and factorization of $a_n(x)$ for $3\le n... | Here is another proof that $a_n$ is integer, which is somewhat more elementary as it does not require the knowledge of Chebyshev Polynomials. It follows the same lines of reasoning as in Malouf, J.L., An integer sequence from a rational recursion, Discrete Mathematics 110 (1992)
257-261.
We suppose (induction hypothes... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3013282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How do I show that the system is hyperbolic if $u^2 + v^2 > c^2$ I know that for a system to be hyperbolic it must have 2 real distinct eigenvalues $\lambda$ where $\det(B-\lambda A)=0$. My system of equations are:
\begin{aligned}
(pu)_x + (pv)_y &= 0 \\
p(uu_x + vu_y) + c(p)^2p_x &= 0 \\
p(uv_x +vv_y)+c(p)^2p_y &= 0
\... | In order for the system to be hyperbolic, it needs to have $n$ distinct real eigenvalues if you are working with $n$ dependent variables. In this case, $n=3$.
The first thing to do is to write it in the form
$$\mathbf A\frac{\partial \mathbf u}{\partial x} + \mathbf B\frac{\partial \mathbf u}{\partial y} = \mathbf c$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3015087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Real solution of $(\cos x -\sin x)\cdot \bigg(2\tan x+\frac{1}{\cos x}\bigg)+2=0.$
Real solution of equation $$(\cos x -\sin x)\cdot \bigg(2\tan x+\frac{1}{\cos x}\bigg)+2=0.$$
Try: Using Half angle formula
$\displaystyle \cos x=\frac{1-\tan^2x/2}{1+\tan^2 x/2}$ and $\displaystyle \sin x=\frac{2\tan^2 x/2}{1+\tan^2 ... | HINT:
Since $\sin x=\frac{2t}{1+t^2}$, the equation becomes $$\left(\frac{1-t^2-2t}{1+t^2}\right)\left(\frac{2t+1+t^2}{1-t^2}\right)+2=0\implies -t^4-4t^3-4t^2+1=2t^4-2$$ or $$3t^4+4t^3+4t^2-3=0$$ from which a root is $t=-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3015467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to show that matrix $A,B$ are not similar? Let $A, B$ be $4\times 4$ square matrix given by,
$$A=\begin{bmatrix}0 &0&1&0\\0&0&0&1\\0&0&0&0\\0&0&0&0\\\end{bmatrix}, B=\begin{bmatrix}0 &0&1&0\\0&0&0&0\\0&0&0&1\\0&0&0&0\\\end{bmatrix}$$
\begin{array}{|c|c|c|}
\hline
\text{Property}& A & B\\\hline
\text{Determinant}&... | According to the comments, $A^2,B^2$ are not similar because
$\begin{pmatrix}1&0\\0&0\end{pmatrix}\begin{pmatrix}0&1\\0&0\end{pmatrix}\not= 0$.Yet, even if the previous product had been zero, $A,B$ would not have been similar. More generally
let $A=\begin{pmatrix}0&M\\0&0\end{pmatrix},B=\begin{pmatrix}0&U\\0&V\end{pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3016055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Minimum value of the given function
Minimum value of $$\sqrt{2x^2+2x+1} +\sqrt{2x^2-10x+13}$$ is $\sqrt{\alpha}$ then $\alpha$ is________ .
Attempt
Wrote the equation as sum of distances from point A(-1,2) and point B(2,5) as
$$\sqrt{(x+1)^2 +(x+2-2)^2} +\sqrt{(x-2)^2 + (x+2-5)^2}$$
Hence the point lies on the line ... | Hint.
Calling $f(x) = \sqrt{x^2+(x+1)^2}$ we seek for
$$
\min_x f(x) + f(x-3)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3022822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Proving $(\sec^2x+\tan^2x)(\csc^2x+\cot^2x)=1+2\sec^2x\csc^2x$ and $\frac{\cos x}{1-\tan x}+\frac{\sin x}{1-\cot x} = \sin x + \cos x $
Prove the following identities:
$$(\sec^2 x + \tan^2x)(\csc^2 x + \cot^2x) = 1+ 2 \sec^2x \csc^2 x
\tag i$$
$$\frac{\cos x}{1-\tan x} + \frac{\sin x}{1-\cot x} = \sin x + \cos x
\... | (i)
\begin{align*}
(\sec^2x + \tan^2x)(\csc^2x + \cot^2x) & = (1 + \tan^2x + \tan^2x)(1 + \cot^2x + \cot^2x)\\
& = (1 + 2\tan^2x)(1 + 2\cot^2x)\\
& = 1 + 2\cot^2x + 2\tan^2x + 4\\
& = 5 + 2(\csc^2x - 1) + 2(\sec^2x - 1)\\
& = 5 + 2\csc^2x - 2 + 2\sec^2x - 2\\
& = 1 + 2\csc^2x + 2\sec^2x\\
& = 1 + 2\left(\frac{1}{\sin^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3027602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Number of one-one function from sets $A$ to $B$. Let $A=\{1,2,3,4,5\}$ and $B=\{0,1,2,3,4,5\}$. Number of one-one function from $A$ to $B$ such that $f(1) \neq 0$ and $f(i)\neq i$ for $i={1,2,3,4,5}$ is _______ .
So I know one one function means for each $x$ there will be only one $y$.
But here if I take for $1$ from ... | Definition. A number $c$ is said to be a fixed point of a function if $f(c) = c$.
Let $A = \{1, 2, 3, 4, 5\}$; let $B = \{0, 1, 2, 3, 4, 5\}$.
If there were no restrictions, we would have $6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 = 6!$ ways to map the elements in the domain to distinct elements in the codomain. Hence, there... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3028709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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What are the Legendre symbols $\left(\frac{10}{31}\right)$ and $\left(\frac{-15}{43}\right)$? I have the following two Legendre symbols that need calculated:
$\left(\frac{10}{31}\right)$ $=$ $-\left(\frac{31}{10}\right)$ $=$ $-\left(\frac{1}{10}\right)$ $=$ $-(-1)$ $=$ $-1$
$\left(\frac{-15}{43}\right)$ $=$ $\left(\fra... | Here is an approach I would take. Note that
$$\left(\frac{10}{31}\right)=\left(\frac{-21}{31}\right)=\left(\frac{-1}{31}\right)\left(\frac{3}{31}\right)\left(\frac{7}{31}\right)=(-1)\Biggl(-\left(\frac{31}{3}\right)\Biggr)\Biggl(-\left(\frac{31}{7}\right)\Biggr)\,.$$
That is, $$\left(\frac{10}{31}\right)=-\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3029147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the minimum value of $\sqrt {2x^2+2y^2} +\sqrt {y^2+x^2-4y+4} +\sqrt {x^2+y^2-4x-4y+8}$ Given that $0\lt x\lt 2$ and $0\lt y\lt 2$ then find the minimum value of $$\sqrt {2x^2+2y^2} +\sqrt {y^2+x^2-4y+4} +\sqrt {x^2+y^2-4x-4y+8}$$
My try:
On factorisation we need minimum value of $$\sqrt {2x^2+2y^2} +\sqrt {(y-2)... | You can dress up your geometric argument as an inequality in $\mathbb{C}$.
Let $z=x+iy$, $a=2i$, $b=2-2i$. Then
\begin{align}
\sqrt{2x^2+2y^2} &= \lvert (1-i)z\rvert\\
\sqrt{y^2+x^2-4y+4} &= \lvert z-a\rvert\\
\sqrt{x^2+y^2-4x-4y+8} &= \lvert -iz-b\rvert
\end{align}
So the triangle inequality gives
$$
\lvert (1-i)z\rv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3030812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Simplify third degree polynomial equations. Given an equation:
$6x^3 - 13x^2 + 8x + 3 = 0$
Broken down to get one form
$(x + {3\over2})$
How can you divide the prior equation to know it will simplify to
$(x + {3\over2})(6x^2 + 4x + 2) = 0$
| It looks like you have typo. If you take the polynomial $$p(x)=6 x^3 + 13 x^2 + 8 x + 3,$$ then $-3/2$ is a root.
This guarantees that $x+3/2$ is a factor of $p$. This follows easily from the Division Algorithm: if you write $$p=(x+3/2) q(x)+r(x),$$ with $\deg r<\deg (x+3/2)=1$, then $r$ is constant; from $p(3/2)=0$ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3030976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How many non-negative solutions for $x_{1} + x_{2} + x_{3} + x_{4} = 40$ where $2 \leq x_{1} \leq 8, x_{2} \leq 4, x_{3} \geq 4, x_{4} \leq 5$? My solution:
We have:
$x_{1} + x_{2} + x_{3} + x_{4} = 40$ where $2 \leq x_{1} \leq 8, x_{2} \leq 4, x_{3} \geq 4, x_{4} \leq 5$
$\Leftrightarrow x_{2} + x_{3} + x_{4} = 40 - ... | It's the coefficient of $x^{40}$ of the product polynomial
$$(x^2+x^3+x^4 +x^5 + x^6 + x^7 + x^8)(1+x^1+x^2+x^3 +x^4)(x^4 + x^5 + \ldots)(1+x^1+x^2+x^3 +x^4 + x^5)$$
Or equivalently the coefficient of $x^{34}$ of
$$(1+x+x^2+x^3+x^4 +x^5 + x^6 )(1+x^1+x^2+x^3 +x^4)(1 + x + x^2 + \ldots)(1+x^1+x^2+x^3 +x^4 + x^5)$$
which... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3032553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Extreme values of ${x^3 + y^3 + z^3 - 3xyz}$ subject to ${ax + by + cz =1}$ using Lagrange Multipliers
If ${ax + by + cz =1}$, then show that in general ${x^3 + y^3 + z^3 - 3xyz}$ has two stationary values ${0}$ and $\frac{1}{(a^3+b^3+c^3-3abc)}$, of which first is max or min according as ${a+b+c>0}$ or ${< 0}$ but se... | Without Lagrange Multipliers.
Making the change of variables $y = \lambda x, z = \mu x$ and substituting we get
$$
\min\max x^3(1+\lambda^3+\mu^3-3\lambda\mu)\ \ \mbox{s. t. }\ \ x(a+\lambda b+\mu c) = 1
$$
or equivalently
$$
\min\max f(\lambda,\mu) = \frac{1+\lambda^3+\mu^3-3\lambda\mu}{(a+\lambda b+\mu c)^3}
$$
whose... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3039184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Lagrange Error Value for $f(x)=\frac{1}{x}$ I'm working through an example of the Taylor Polynomial and I'm stuck at nearly the last part. I have a function $f(x)= \frac{1}{x}, x>1$ and I am asked to find the Taylor Polynomial to the $3.$ degree at $a=5$ and then find the Lagrange Remainder between the interval $[4,6]$... | Consider the function
$$f(x) = \dfrac{1}{x}$$
Find the third degree Taylor polynomial for $f(x)$ centered at $x = 3$
$$\begin{array}{|c|c|}
\hline \text{Derivatives} & \text{Values at x} =5 \\\hline
f(x) = \dfrac{1}{x} & \dfrac{1}{5} \\\hline
f'(x) = -\dfrac{1}{x^2} & -\dfrac{1}{25} \\\hline
f''(x)=\dfrac{2}{x^3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3043240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Finding the Laurent series (complex numbers) I have
$$
f(z)={\frac{1}{z(1-z)}}
$$
Need to find the Laurent series around $z=0, z=1, z=\infty$.
I did
$$
{\frac{1}{z(1-z)}} = {\frac{A}{z}}+{\frac{B}{1-z}}
$$
and found $A=1, B=1$. Therefore we get
$$
{\frac{1}{z}}+{\frac{1}{1-z}} = {\frac{1}{z}} + \sum z^n
$$
But in the b... | We have
$$
\eqalign{
& {1 \over {z\left( {1 - z} \right)}} = \cr
& = \left\{ \matrix{
- \left( {{1 \over z} + {1 \over {\left( {1 - z} \right)}}} \right)\quad
\Rightarrow \quad - {1 \over z} - \sum\limits_{0\, \le \,n} {z^{\,n} } \quad \left| {\,z \to 0} \right. \hfill \cr
{1 \over {\left( {z - 1} \rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3045468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Integral$\int_{\pi/6}^{\pi/2} \frac{\sin^{1/3} (x)}{\sin^{1/3} (x)+\cos^{1/3} (x)}$ $$\int_{\pi/6}^{\pi/2} \frac{\sin^{1/3} (x)}{\sin^{1/3} (x)+\cos^{1/3} (x)}$$
I tried using the substitution $t^3=\tan x$. Which gives me
$$\int\frac{3t^3}{(t+1)(t^6+1)}$$
How should I proceed?
| As Lord Shark the Unknown commented, the first thing to do is partial fraction decomposition
Since $t^6+1=(t^2+1)(t^4-t^2+1)$, then
$$\frac{3t^3}{(t+1)(t^6+1)}=-\frac{3}{2 (t+1)}-\frac{t+1}{2 \left(t^2+1\right)}+\frac{2 t^3-t^2-t+2}{t^4-t^2+1}$$I suppose that the last term will impose to work with the roots of unity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3046039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to show $7^{th}$ degree polynomial is non-positive in $[0,1]$
Let $0\le x\le 1$, show that inequality
$$99x^7-381x^6+225x^5-415x^4+157x^3-3x^2-x-1\le 0$$
This problem comes from the fact that I solved a different inequality.I tried to solve it by factorizing it to see if I could get symbols.but I failed.
this ... | Because by AM-GM $$1+x+3x^2-157x^3+415x^4-225x^5+381x^6-99x^7=$$
$$=1+x+3x^2-157x^3+370x^4+x^4(45-225x+282x^2)+99x^6(1-x)\geq$$
$$\geq370x^4+3x^2+x+1-157x^3=6\cdot\frac{185}{3}x^4+3x^2+x+1-157x^3\ge$$
$$\geq9\sqrt[9]{\left(\frac{185}{3}x^4\right)^6\cdot3x^2\cdot x\cdot1}-157x^3=\left(9\sqrt[9]{\left(\frac{185}{3}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3046317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Divisor of $x^2+x+1$ can be square number? $$1^2+1+1=3$$
$$2^2+2+1=7$$
$$8^2+8+1=73$$
$$10^2+10+1=111=3\cdot37$$
There is no divisor which is square number.
Is it just coincidence? Or can be proved?
*I'm not english user, so my grammer might be wrong
| The square of any prime $p\equiv1\pmod3$ appears as a factor of $x^2+x+1$ for some choice of $x$.
This is seen as follows.
The multiplicative group $\Bbb{Z}_{p^2}^*$ of coprime residue classes modulo $p^2$ is known to be cyclic of order $p(p-1)$. It follows that there is an element of order three in that group. Let th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3047388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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How to solve equations of this type My math teacher gave us some questions to practice for the midterm exam tomorrow, and I noticed some of them have the same wired pattern that I don't know if it has a name:
Q1. If: $x - \frac{1}{x} = 3$ then what is $x^2 + \frac{1}{x^2}$ equal to?
The answer for this question is 11, ... | Hint:
1: $x - \dfrac{1}{x} = 3 \implies x^2 + \dfrac{1}{x^2} - 2 = 9$
Now use, $\big(x + \dfrac{1}{x}\big)^2 = x^2 + \dfrac{1}{x^2} + 2$
2: $\dfrac{x}{x+y} = \dfrac{1}{1 + \dfrac{y}{x}}$
3: $x^4 + y^4 = 6 * x^2 * y^2 \implies \dfrac{x^2}{y^2} + \dfrac{y^2}{x^2} = 6$, solve for y/x or x/y use that to get the value of fi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Solve the system $x^2(y+z)=1$ ,$y^2(z+x)=8$ and $z^2(x+y)=13$ Solve the system of equations in real numbers
\begin{cases} x^2(y+z)=1 \\ y^2(z+x)=8 \\z^2(x+y)=13 \end{cases}
My try:
Equations can be written as:
\begin{cases}\frac{1}{x}=xyz\left(\frac{1}{y}+\frac{1}{z}\right)\\
\frac{8}{y}=xyz\left(\frac{1}{x}+\frac{1}... | Alternatively, denote $y=ax, z=abx$. Then:
$$\begin{cases} x^2(y+z)=1 \\ y^2(z+x)=8 \\z^2(x+y)=13 \end{cases} \Rightarrow \begin{cases} ax^3(1+b)=1 \\ a^2x^3(1+ab)=8 \\a^2b^2x^3(1+a)=13 \end{cases}.$$
Divide $(2)$ by $(1)$ and $(3)$ by $(2)$:
$$\begin{cases} a(ab+1)=8(1+b) \Rightarrow b=\frac{8-a}{a^2-8} \\ 8b^2(1+a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3050498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\int_0^1\frac{\ln{(x^2+1)}}{x+1}dx$ Any idea on how to solve the following definite integral?
$$\int_0^1\frac{\ln{(x^2+1)}}{x+1}dx$$
I have tried to parameterize the integral like $\ln{(a^2x^2+1)}$ or $\ln{(x^2+a^2)}$, which don't seem to work.
| \begin{equation}
I = \int_0^1 \frac{\ln\left(x^2 + 1\right)}{x + 1}\:dx
\end{equation}
Here let:
\begin{equation}
I(t) = \int_0^1 \frac{\ln\left(tx^2 + 1\right)}{x + 1}\:dx
\end{equation}
We observe that $I = I(1)$ and $I(0) = 0$. Thus,
\begin{align}
I'(t) &= \int_0^1 \frac{x^2}{\left(tx^2 + 1\right)\left(x + 1\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3054362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
minimum value of $I(l)$ in definite integration If $\displaystyle I(l)=\int^{\infty}_{0}\frac{x^l}{2x^6+4x^5+3x^4+5x^3+3x^2+4x+2}dx.$ for $0<l<5.$
Then value of $l$ for which $I(l)$ is minimum
What i tried
Put $\displaystyle x=\frac{1}{t}$ and $\displaystyle dx=-\frac{1}{t^2}dt$
$\displaystyle I(l)=\int^{\infty}_{0}\fr... | Denote $P(x)$ the polynomial of denominator, $$I'(l)=\int_0^\infty\frac{x^l\ln x}{P(x)}dx\\
=\int_0^1\frac{x^l\ln x}{P(x)}dx+\int_1^\infty\frac{x^l\ln x}{P(x)}dx\\
=\int_0^1\frac{x^l\ln x}{P(x)}dx-\int_0^1\frac{x^{4-l}\ln x}{P(x)}dx\text{ (Sub $x\mapsto 1/x$)}\\
=\int_0^1\frac{(x^l-x^{4-l})\ln x}{P(x)}dx$$
$x^l-x^{4-l}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3054634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find $|2a + 3b + 2 \sqrt{3}(a \times b)|$ where $a,b$ are perpendicular unit vectors. Find $|2a + 3b + 2 \sqrt{3}(a \times b)|$ where &a,b$ are perpendicular unit vectors.
My attempt$:$
$(|2a+3b+2\sqrt(3)(a \times b)|)^{2}$ = $ 4a^{2} + 9b^{2} + 12 + 8\sqrt{3} a.\hat{n} +12\sqrt{3}b.\hat{n}$
$( a.b = 0, |a| = |b| = 1),... | $\{a,b,a\times b\}$ is an orthogonal set in $\mathbb{R}^3$ so the Pythagorean theorem gives:
$$|2a + 3b + 2\sqrt{3}(a\times b)|^2 = |2a|^2 + |3b|^2 + |2\sqrt{3}(a\times b)|^2 = 4 + 9 + 12 = 25$$
Hence $|2a + 3b + 2\sqrt{3}(a\times b)| = 5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3054735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prove that $a_n \in [0,2)$
Let $(a_n)_{n \in \mathbb{N}}$ be a sequence, with $a_0=0$, $a_{n+1}=\frac{6+a_n}{6-a_n}$.
Prove that $a_n \in [0,2)$ $\forall n \in \mathbb{N}$
Here's what I did:
I tried to prove this by induction:
Base case:
$0 \leq a_0 (=0) < 2$.
Inductive step:
Suppose that $0 \leq a_n < 2$
So $$\beg... | The function $f(x) = \frac{6+x}{6-x}$ is strictly increasing, which you can check by taking the derivative:
$$f'(x) = \frac{(6-x) + (6+x) }{(6-x)^2} = \frac{12}{(6-x)^2} > 0$$
Hence if you assume $0 \le a_n < 2$, you get
$$a_{n+1} = \frac{6+a_n}{6-a_n} \ge \frac{6+0}{6-0} = 1 \ge 0$$
$$a_{n+1} = \frac{6+a_n}{6-a_n} < \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Evaluate $ \lim\limits_{n \to \infty}\sum\limits_{k=2}^{n} \frac{1}{\sqrt[k]{n^k+n+1}+1} $ $$ \lim_{n \to \infty}\sum_{k=2}^{n} \frac{1}{\sqrt[k]{n^k+n+1}+1} $$
I expect the squeeze theorem helps us solving this but I can't find the inequality.
The result should be $1$.
| Consider
$s(n)
=\sum_{k=2}^{n} \dfrac{1}{\sqrt[k]{n^k+f(n)}+1}
$
where
$f(n) \ge 0$ and
$f(n)/n^{c} \to 0$
for some $c > 0$.
$\begin{array}\\
s(n)
&=\sum_{k=2}^{n} \dfrac{1}{\sqrt[k]{n^k+f(n)}+1}\\
&\lt\sum_{k=2}^{n} \dfrac{1}{\sqrt[k]{n^k}}\\
&=\sum_{k=2}^{n} \dfrac{1}{n}\\
&\to \ln(n)-1+\gamma\\
\end{array}
$
Similar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3057044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding $B^*$, the dual basis Find a basis $B$ for
$$V = \left\{ \left[
\begin{array}{cc}
x\\
y\\
z
\end{array}
\right] \in \mathbb{R}^3 \vert x+y+z = 0\right\}$$ and then find $B^*$, the dual basis for $B$.
The way we learned it was that given a basis, we build a matrix A whose columns is the vectors, find $A^{... | Be
\begin{equation}
V = \left\lbrace \
\left[\begin{array}{c}
x \\
y \\
z \\
\end{array}\right] \in \mathbb{R}^3 : x+y+z = 0 \right\rbrace
\end{equation}
Then $z=-x-y$, thus basis $B$ is:
\begin{equation}
B = \left\lbrace\
\left[\begin{array}{c}
1 \\
0 \\
-1 \\
\end{array}\right]
,
\left[\begin{array}{c}
0 \\
1 \\
-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3059814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is $x^4 + 2x^2 - x + 1$ irreducible in $\mathbb Z_7[x]$? How would one do this?
I know since it doesn't have roots it can only be divisible by irreducible polynomials of degree 2.
How would I prove that it is or it isn't?
There are a lot of polynomials with this conditions.
| The product of all the monic, irreducible polynomials over $\mathbb{F}_7$ wih degree $\leq 2$ is given by $x^{49}-x$, so if we prove that $f(x)=(x^2+1)^2-x$ and $x^{48}-1$ are coprime over $\mathbb{F}_7[x]$ we are done. Let us compute $x^{48}-1\pmod{f(x)}$, for instance through the Brauer chain $x^4\to x^8\to x^{16}\to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3060034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\ X_i $ is discrete random variable, Compute $\ \sum X_i = 97 $ Let $\ X_1, X_2, , \dots , X_{10} $ be a discrete random variable with uniform distribution between $\ 0 $ to $\ 10 $. Compute $\ P\{ \sum_{i=1}^{10} \ X_i = 97 \} $, the variables are independent.
My attempt:
So I can either get 97 by having nine 10's an... | Your solution is almost correct.
The correct solution is:$$11^{-10}\left(\frac{10!}{9!1!}+\frac{10!}{8!1!1!}+\frac{10!}{7!3!}\right)$$
Observe that $8$ and $9$ are distinct numbers.
So e.g. $(10,10,9,10,10,10,10,10,8,10)$ and $(10,10,8,10,10,10,10,10,9,10)$ are distinct possibilities.
That is why $\binom{10}{2}=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3060409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Sum of series $\sqrt{1+\frac1{n^2}+\frac1{(n+1){}^2}}$ How to solve this sum?$$\sqrt{1+\frac1{1^2}+\frac1{2^2}}+\sqrt{1+\frac1{2^2}+\frac1{3^2}}+\cdots+\sqrt{1+\frac1{19^2}+\frac1{20^2}}$$
I assumed it to be a sum of $\sqrt{1+\dfrac{1}{n^2} + \dfrac{1}{(n+1)^2}}$ but It complicates the powers and I am unable to factori... | $$1+\dfrac1{n^2}+\dfrac1{(n+1)^2}=\dfrac{n^2+(n+1)^2+(n^2+n)^2}{n^2(n+1)^2}=\dfrac{n^4+2n^3+3n^2+2n+1}{(n^2+n)^2}$$
$n^4+2n^3+3n^2+2n+1=(n^2)^2+2n^2\cdot1+2n\cdot1+(n)^2+1^2+2n^2\cdot n=(n^2+n+1)^2$
Now $$\dfrac{n^2+n+1}{n(n+1)}=1+\dfrac1{n(n+1)}=1+\dfrac{n+1-n}{n(n+1)}=?$$
See also: Telescoping series
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find all $p \in \mathbb{Z}$ such that $ p^2+ 4p + 16 $ is a perfect square I've tried some things involving modular residues but they don't seem to work. Anyone know how to do this?
| So $p^2 + 4p + 16 = m^2$
$p^2 + 4p + 4 =m^2 - 12$
$(p+2)^2 = m^2 - 12$
$m^2 - (p+2)^2 = 12$
$(m + p + 2)(m - p - 2) = 12$
So $m+p+2 = k$ and $m-p-2 = j$ for two integers so that $kj = 12$.
Also note, $k + j = 2m$.
So $k + j$ is even. So $\{k,j\} = \{\pm 2, \pm 6\}$ and .... thats it. All other factors involve an odd ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $ \int\limits_0^\infty\frac{x^2}{e^{2 x} (1-x)^2+e^{-2 x} (1+x)^2}\,dx=\dfrac{3\pi}{8} $ Is this conjecture true?
Conjecture:
$$
\int_0^\infty\frac{x^2}{e^{2 x} (1-x)^2+e^{-2 x} (1+x)^2}\,dx=\dfrac{3\pi}{8}
$$
I found it myself based on numerical evidence. Need help in analytical proof. Thanks.
| It is easier than I exspected in the first place. However, first of all lets denote your integral as
$$\mathfrak I~=~\int_0^\infty \frac{x^2}{e^{2x}(1-x)^2+e^{-2x}(1+x)^2}\mathrm d x$$
We may rewrite the integral a little bit. According to Trebor's answer we have to exspect an inverse tangent with an argument of the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3062665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Repeatedly dividing $360$ by $2$ preserves that the sum of the digits (including decimals) is $9$ Can someone give me a clue on how am I going to prove that this pattern is true or not as I deal with repeated division by 2? I tried dividing 360 by 2 repeatedly, eventually the digits of the result, when added repeatedly... | Define the digit sum of the positive integer $n$ as the sum $d(n)$ of its digits. The reduced digit sum $d^*(n)$ is obtained by iterating the computation of the digit sum. For instance
$$
n=17254,\quad
d(n)=1+7+2+5+4=19,
\quad
d(d(n))=1+9=10,
\quad
d(d(d(n)))=1+0=1=d^*(n)
$$
Note that $d(n)\le n$, equality holding if a... | {
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"url": "https://math.stackexchange.com/questions/3064917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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"answer_id": 2
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Simplify $\frac{\sqrt{mn^3}}{a^2\sqrt{c^{-3}}} * \frac{a^{-7}n^{-2}}{\sqrt{m^2c^4}}$ to $\frac{\sqrt{mnc}}{a^9cmn}$ I need to simplify $$\frac{\sqrt{mn^3}}{a^2\sqrt{c^{-3}}} \cdot \frac{a^{-7}n^{-2}}{\sqrt{m^2c^4}}$$
The solution provided is: $\dfrac{\sqrt{mnc}}{a^9cmn}$.
I'm finding this challenging. I was able to m... | Recall that $a^{-x}=\frac{1}{a^x}$, $a^{1/x}=\sqrt[x]{a}$, and $a^na^m=a^{m+n}$
Always convert everything to exponents first, then use arithmetic
$$\frac{\sqrt{mn^3}}{a^2\sqrt{c^{-3}}}\frac{a^{-7}n^{-2}}{\sqrt{m^2c^4}}=m^{1/2}\ n^{3/2}\ a^{-2}\ c^{3/2}\ a^{-7}\ n^{-2}\ m^{-2/2}\ c^{-4/2}$$
Rearrange
$$m^{1/2}\ n^{3/2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3066413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $3^{2n} +7$ is divisible by 8
Prove by induction that $3^{2n} +7$ is divisible by 8 for $n \in \Bbb N$
So I think I have completed this proof but it doesn't seem very thorough to me - is my proof valid?
If $n=1$ then $3^{2n} +7 = 16 = 2(8)$ so true when $n=1$
Assume true for $n=k$ so $$ 8\vert 3^{2k} +7$$
... | An option:
Step $n+1$:
$3^{2n+2}+7 =9 \cdot 3^{2n} +7=$
$(8+1)\cdot 3^{2n} +7=$
$(3^{2n}+7) +8;$
The first term is divisible by $8$ by hypothesis, so is the second term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3067826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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What is the probability that no cup is on a saucer of the same colour? A tea set comprises four cups and saucers in four distinct colours.
If the cups are placed at random on the saucers, what is the probability that no cup is on a saucer of the same colour?
MY ATTEMPT
As far as I understand, the given event's probabi... | There are $4!$ ways to arrange the cups on the saucers. From these, we wish to exclude those arrangements in which a cup is placed on a saucer of its own color.
If cup $C_i$ sits on saucer $S_i$, there are $3!$ ways to arrange the remaining three cups on the remaining three saucers, so $|C_i| = 3!$, $1 \leq i \leq 4$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3071423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Generalization of $(a+b)^2\leq 2(a^2+b^2)$ We know that, $(a+b)^2\leq 2(a^2+b^2)$. Do we have anything similar for $$\left(\sum_{i=1}^N a_i\right)^2.$$
where $a_i\in \mathbb{R}\ \ \ \ \forall\ i\in \{1,\cdots,N\}$.
For $n=3$, we get
\begin{equation}
\begin{aligned}
(a_1+a_2+a_3)^2&\leq 2\left((a_1+a_2)^2+a_3^2\right)
\... | It's C-S:
$$n(a_1^2+a_2^2+...+a_n^2)=$$
$$=(1^2+1^2+...+1^2)(a_1^2+a_2^2+...+a_n^2)\geq(a_1+a_2+...+a_n)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3072599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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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.