Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Inverse function of $f(x) = \frac{x+5}{x-2}$ Find the inverse of the function
$f(x) = \frac{x+5}{x-2}$
Here's what I have so far:
$y = \frac{x+5}{x-2}$
$x = \frac{y+5}{y-2}$
$(x)(y-2) = (y+5)$
but this seems to be a dead end.
How should I approach this?
Thank you!
| Almost there:
$$x = \frac{y+5}{y-2}$$
$$x(y-2) = y+5$$
$$xy - 2x = y+5$$
$$xy - y = 5 + 2x$$
$$y(x-1) = 5+2x$$
I'll let you finish.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1511766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How to compute this gross limit.
How do I compute this limit?
$$
\lim_{n \to \infty}
\frac{\left(1 + \frac{1}{n} + \frac{1}{n^2}\right)^n -
\left(1 + \frac{1}{n} - \frac{1}{n^2}\right)^n
}{
2 \left(1 + \frac{1}{n} + \frac{1}{n^2}\right)^n -
\left(1 + \frac{1}{n} - \frac{1}{n^2 + 1}\ri... | Use the fact that $a^n-b^n = (a-b)(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1})$. It is going to be quite lengthy so bear with me: The numerator is
$$
\left(1+\frac{1}{n}+\frac{1}{n^2}\right)^n-
\left(1+\frac{1}{n}-\frac{1}{n^2}\right)^n = \frac{2}{n^2}\sum_{m=0}^{n-1}
\left(1+\frac{1}{n}+\frac{1}{n^2}\right)^{n-1-m}
\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1512063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
Greatest Common Divisor Proof: $\gcd(m^2-n^2, m^2+n^2) = 1$ Prove that if $\gcd(m,n) = 1$ and $m+n \equiv 1 (\text{mod} ~2)$ then $\gcd(m^2-n^2, m^2+n^2) = 1$
Workings:
Suppose that $\gcd(m,n) = 1$ and $m+n \equiv 1 (\text{mod} ~2)$
$\gcd(m^2-n^2, m^2+n^2)$
$= gcd((m-n)(m+n), (m-n)(m+n)+2n^2)$
Now I know that $m+n=1 (\... | Euclid's algorithm gives:
\begin{align*}
\gcd(m^2 - n^2, m^2 + n^2)
&= \gcd(m^2 - n^2, (m^2 + n^2) - (m^2 - n^2)) \\
&= \gcd(m^2 - n^2, 2n^2)
\end{align*}
Since $m + n \equiv 1 \pmod{2}$, $m^2 - n^2 \equiv 1 \pmod{2}$, so we can forget about the $2$.
We get
\begin{align*}
&= \gcd(m^2 - n^2, n^2) \\
&= \gcd(m^2, n^2) \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1513549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find the limit as x approaches infinity $$\lim_{x\to \infty} {\cosh^{-1}(x^{3}) + \coth^{-1}(\sqrt{x^{2}+1}) - 3\sinh^{-1}(x)}$$
Honestly, I don't really know how to approach this. I know the logarithmic formulae for the inverse hyperbolic functions, but that gives me a very complicated expression. Is there an identity... | We can directly use the expression of inverse hyperbolic functions namely
\begin{align}
\sinh^{-1}x &= \log(x + \sqrt{x^{2} + 1})\notag\\
\cosh^{-1}x &= \log(x + \sqrt{x^{2} - 1})\notag\\
\coth^{-1}x &= \frac{1}{2}\log\left(\frac{x + 1}{x - 1}\right)\notag
\end{align}
From the above we see that $\coth^{-1}x \to 0$ as $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1513636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding maximum $b$ in $x^5-20x^4+bx^3+cx^2+dx+e=0$ Let $b, c, d, e$ be real numbers such that the following equation
$$x^5-20x^4+bx^3+cx^2+dx+e=0$$
has real roots only. Find the largest possibe value of $b$.
What I have done is:
Let $x_1, x_2, x_3, x_4, x_5$ be the 5 real roots of the equation. Then we have
$$x_1+x_2+... | Hint: Let $P(x) = x^5-20x^4+bx^3+cx^2+dx+e$. If $P$ has all real roots, then $P'''(x)$ must have two real roots..
$\implies 60x^2-480x+6b$ has real roots $\implies b \le 160$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1514076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Equation of locus Point P$(x, y)$ moves in such a way that its distance from the point $(3, 5)$ is proportional to its distance from the point $(-2, 4)$. Find the locus of P if the origin is a point on the locus.
Answer:
$$(x-3)^2 + (y-5)^2 = (x+2)^2 + (y-4)^2$$
or, $$10x+2y-14=0$$
or, $$5x+y-7=0$$
but answer given is... | $$(x-3)^2 + (y-5)^2 = \lambda((x+2)^2 + (y-4)^2).$$
We express that the curve passes through the orgin:
$$(-3)^2 + (-5)^2 = \lambda((+2)^2 + (-4)^2),$$
hence
$$\lambda=\frac{17}{10}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1517502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Use integration by parts to find the integral $\int\frac{\sqrt {4x^2-9}}{x^2}dx$ $$\int\frac{\sqrt {4x^2-9}}{x^2}dx$$
I tried to solve this using integration by parts, but I come up with something that is much more difficult to solve. How can this be solved?
| $$\int \frac{\sqrt{4x^2-9}}{x^2}dx$$
$$=-\int \sqrt{4x^2-9} \,\ d(\frac{1}{x})$$
Using by parts, $$=-\left[\frac{\sqrt{4x^2-9}}{x}-\int \frac{4}{\sqrt{4x^2-9}}dx\right]$$
$$=-\left[\frac{\sqrt{4x^2-9}}{x}-2\int \frac{d(2x)}{\sqrt{(2x)^2-3^2}}\right]$$
$$=-\left[\frac{\sqrt{4x^2-9}}{x}-2 \ln \left|2x+\sqrt{4x^2-9}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1518793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Does the sequence $a_n = \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)}$ converge?
Does the sequence $a_n = \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)}$ converge?
This is a homework question for an analysis class.
Firstly, I know the sequence is comprised of odd number... | $$a^{ 2 }_{ n }={ \left( \frac { 1\cdot 3\cdot 5\cdot ...\cdot (2n-1) }{ 2\cdot 4\cdot 6\cdot ...\cdot (2n) } \right) }^{ 2 }=\frac { { 1 }^{ 2 }\cdot 3^{ 2 }\cdot 5^{ 2 }\cdot ...\cdot (2n-1)^{ 2 } }{ 2^{ 2 }\cdot 4^{ 2 }\cdot 6^{ 2 }\cdot ...\cdot (2n)^{ 2 } } =\\ =\frac { 1\cdot 3 }{ { 2 }^{ 2 } } \cdot \frac { 3\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1519540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Express $\frac{1-x}{(x-1)^2+y^2}-\bigg(\frac{y}{(x-1)^2+y^2}\bigg)i$ in the form of $f(z)$ I need to express $f(z)$ from the form $\color{blue}{u+vi}$ to the form $\color{blue}z$ for example if: $g(z)=\frac{1}{x+yi}$ so $ =g(z)=\frac{1}{z}$
$$f(z)=\underbrace {\frac{1-x}{(x-1)^2+y^2}}_{=u(x,y)}+\underbrace {\bigg(-\fr... | Hint ( to prove the answer of @RonGordon):
$$
f(z)=\frac{1-x}{(x-1)^2+y^2}-\bigg(\frac{y}{(x-1)^2+y^2}\bigg)i=
$$
$$
=\frac{1}{(1-x)^2+y^2}\left(1-x-i y \right) = \frac{\bar u}{|u|^2}=\frac{1}{u}
$$
for $u=1-x +iy=1-(x-iy)=1-\bar z$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1522760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Fermat's infinite descent for finding the squares that sum to a prime Fermat's theorem on sum of two squares states that an odd prime $p = x^2 + y^2 \iff p \equiv 1 \pmod 4$
Applying the descent procedure I can get to $a^2 + b^2 = pc$ where $c \in \mathbb{Z} \gt 1$
I want $c = 1$, so how do I proceed from here? How do ... | This is how we proceed.
We found out that $22^2+(−1)^2=97∗5$
Notice it's a sum of squares - so let's use the equation from before $x^2 + 1 = 97m$. In general, the $1$ might be something else. In this case, since $(-1)^2 = 1$, nothing really changes and we can use the exact same equation.
In the second iteration:
$m = 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1525113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
If $x$ is a positive integer such that $x(x+1)(x+2)(x+3)+1=379^2$, find $x$
If $x$ is a positive integer such that $x(x+1)(x+2)(x+3)+1=379^2$, find $x$
This is a 1989 ARML problem. One, ugly way to solve this is:
Approximate this as $x^4=379^2$, so $x\approx \sqrt{379}\approx 19$ and guess and check around there to ... | $$x(x+1)(x+2)(x+3) +1 =(x^2 +x)(x^2 +5x +6) +1 = x^4+5x^3 +6x^2 +x^3 +5x^2 +6x +1 =x^4 +6x^3 +11x^2 +6x +1 =x^2 (x^2 +6x +9) +2x(x+3)+1 =(x(x+3) +1)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1528187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
What is the best way to solve an equation of the form $(f(x))^2-a(f(x))+b=x$? On a math contest I was told to solve the equation $$(x^2-3x+1)^2-3(x^2-3x+1)+1=x$$
For this particular problem I simplified by letting $$a\equiv x^2-3x+1$$
Then I continued with $$a^2-3a+1-x=0$$
$$a=\frac{3\pm\sqrt{9-4\left(1-x\right)}}{... | Add $x^2-3x+1$ to both sides, giving
$$(x^2-3x+1)^2 - 2(x^2-3x+1) + 1 = x + x^2-3x+1 = x^2-2x+1,$$
or
$$((x^2-3x+1)-1)^2 = (x-1)^2.$$
Thus $(x^2-3x+1)-1 = \pm (x-1)$. The plus sign gives
$$x^2-4x+1=0\quad\Rightarrow\quad x = 2\pm\sqrt{3}.$$
The minus sign gives
$$x^2-2x-1=0\quad\Rightarrow\quad x = 1\pm\sqrt{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1528652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Bound on difference between irrational square root and rational number I'm trying to show that if $\alpha$ is the square root of a non-perfect square $d$, then for any rational $p/q$ there's a positive real $x$ such that $|\alpha - p/q| > x/q^2$. I'm having trouble starting on the problem, could someone give a hint? Th... | Let $ d $ be a positive integer which isn't a square, and consider a rational $ \frac{p}{q} $ ($ p, q $ integers, $ q > 0 $).
Looking at $ \left| \sqrt{d} - \frac{p}{q} \right| $, it is $ \dfrac{1}{q} |q\sqrt{d} - p | $ $ = \dfrac{1}{q} \dfrac{|q^2 d - p^2|}{|q\sqrt{d}+p|} $. As $ \sqrt{d} $ is irrational, numerator $ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1530454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
When is EF longer than AC? (a generalization) ABC is an isosceles right triangle,
M is on AC,
and EMF is a straight line.
When is EF longer than AC?
]1
Note:
This is a generalization
of the following problem,
which has M in the
center of AC:
Prove that EF is longer than AC
In that case,
I showed that
EF is always l... | We may suppose that $$A(0,1),B(0,0),C(1,0),M(m,1-m),E(0,1-m-am),F\left(m+\frac{m-1}{a},0\right)$$
where $0\lt m\lt 1$ is the $x$ coordinate of $M$ and $a\lt -1$ is the slope of the line $EF$.
Then, we have
$$\begin{align}&|EF|\gt |AC|\\\\&\iff |EF|^2\gt |AC|^2\\\\&\iff \left(m+\frac{m-1}{a}\right)^2+(1-m-am)^2\gt 2\\\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1530698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
An Algebraic Proof that $|y^3 - x^3| \ge |(y - x)|^3/4 $ I can prove this using calculus, but not by simple algebra: can anyone help ?
Calculus Proof:
Fix the separation of $x$ and $y$ so that $y = x + d$ with $d>0$ ($ \implies y > x \implies y^3 > x^3$) and now consider $ f(x) = y^3 - x^3 = (x+d)^3 - x^3$.
$ f(x) = (... | Wlog. $y\ge x$. Then
$$\begin{align}y^3-x^3&\ge \frac{(y-x)^3}{4}&\iff\\
4(y^3-x^3)-(y-x)^3&\ge 0&\iff\\
3y^3+3y^2x-3x^3-3x^2y&\ge 0&\iff\\
3(y+x)^2(y-x)&\ge0\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1533407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
$x^2+(a^2+b^2)x+ab=0$ where $a,b,x\in \mathbb Z$ Trivial solutions include $a=0, x=-b^2$ and $b=0,x=-a^2$. Then $a=b=1,x=-1$ or $a=b=-1, x=-1$. Is there any other? How could you prove there aren't? $(a^2+b^2)^2-4ab=(a^2+b^2)+2ab(ab-1)$ needs to be a square. I tried to look at it $\mod 4$ but it doesn't yield anything u... | Suppose $a,b\ne 0$ and $(a^2+b^2)^2-4ab = c^2$ for some integer $c\ge 0$. We can then write $c = a^2+b^2 - k$ for some $0<k\le a^2+b^2$. Then
$$ 4ab = (a^2+b^2) - c^2 = (a^2+b^2) - (a^2+b^2-k) = k(2a^2+2b^2-k) $$
Note that, for $0<k\le a^2+b^2$, the RHS is increasing in $k$. Hence, if $k\ge 2$, we have
$$ 4ab = k(2a^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1534809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Compute the limit of the sequence of functions of $\lim_{n\to \infty} f_n(x) = \frac{x^2}{x^2+(1-nx)^2}$ Compute the limit of the sequence of functions of
$$
\lim_{n\to \infty} f_n(x) = \frac{x^2}{x^2+(1-nx)^2}.
$$
Attempt.
$$\lim_{n\to \infty} f_n(x) = \lim_{n\to \infty} \frac{x^2}{x^2+(1-nx)^2} = \lim_{n\to \infty} ... | Well $$\lim \frac{x^2}{x^2+n^2x^2-2nx + 1} \ne \lim \frac{x^2}{x^2+\frac{x^2}{n^2}-\frac{2x}{n} + \frac{1}{n}} $$
Note that there is no need to do any algebraic manipulation here; you can just do it by inspection as $n \to \infty$ we immediately see that $$\lim_{n\to \infty} \frac{x^2}{x^2+(1-nx)^2}=0$$
as the denomi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1537927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove by induction $ \sin x + \sin 2x + ... + \sin nx = \frac {\sin (\frac {n + 1} {2} x)} {\sin \frac{x}{2}} \sin \frac{nx}{2} $
Prove by induction
$$ \sin x + \sin 2x + ... + \sin nx = \frac {\sin (\frac {n + 1} {2} x)} {\sin \frac{x}{2}} \sin \frac{nx}{2} $$
What I have for now:
$$ \frac {\sin (\frac {n + 1} {... | You are almost there, but use $\sin 2y = 2\sin y\cos y$, considering your left hand side we get:
$$\frac {\sin y} {\sin \frac{x}{2}} \sin (y - \frac{x}{2}) + \sin2y =
\frac {\sin y} {\sin \frac{x}{2}} \sin (y - \frac{x}{2}) + 2\sin y\cos y = $$
$$\left(\frac {\sin (y-\frac{x}{2})} {\sin \frac{x}{2}} +2\cos y\right)\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1538163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
How to prove that $\frac{1+4n^2}{2+2n^2}$ is a cauchy sequence? I know that the following sequence is a
$$\frac{1 + 4n^2}{2+2n^2}$$
How can I show that it is a cauchy sequence - well it has to cause its convergent but I want to understand it with the cauchy definition.
I know that i have to get the $|a_m - a_n | < \var... | Let $a_n$ the $n$-th term of the sequence, since $$a_n=\frac{1+4n^2}{2+2n^2}=\frac{(4+4n^2)-3}{2+2n^2}=2-\frac{3}{2}\frac{1}{n^2+1}$$
it follows
$$|a_n-a_m|=\frac{3}{2}\left|\frac{1}{n^2+1}-\frac{1}{m^2+1}\right|\le\frac{3}{2}\left(\left|\frac{1}{n^2+1}\right|+\left|\frac{1}{m^2+1}\right|\right)\tag{1}$$
Given $\varep... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1539786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
$\int\frac{(2x+3)dx}{(x^2+2x+3)\sqrt{x^2+2x+4}}$ $\int\frac{(2x+3)dx}{(x^2+2x+3)\sqrt{x^2+2x+4}}$
$I=\int\frac{(2x+3)dx}{(x^2+2x+3)\sqrt{x^2+2x+4}}=\int\frac{(2x+2)dx}{(x^2+2x+3)\sqrt{x^2+2x+4}}+\int\frac{1 dx}{(x^2+2x+3)\sqrt{x^2+2x+4}}$
Let first integral be called $I_1$ and the second integral be called $I_2$.
$I_1... | Hint: the following substitutions reduce the original integral into a pair of very familiar ones.
$$\begin{align}
I
&=\int\frac{2x+3}{\left(x^2+2x+3\right)\sqrt{x^2+2x+4}}\,\mathrm{d}x\\
&=\int\frac{2y+1}{\left(y^2+2\right)\sqrt{y^2+3}}\,\mathrm{d}y;~~~\small{\left[x+1=y\right]}\\
&=\int\frac{2}{w^2-1}\,\mathrm{d}w;~~~... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1544196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\lim_{x\to 0}\frac{\sin(x)\arcsin(x)-x^2}{x^6}$ Step by Step Using L' Hopital Rule The limit to be found is
$$ \lim_{x\to 0}\frac{\sin(x)\arcsin(x)-x^2}{x^6}$$
I've tried l'hopital rule but it gets really messy. I've also tried splitting it into 2 limits but that doesn't work. I can't think of any meaningfu... | Some basic limits can be obtained by repeated use of
l'Hospital's Rule easily,
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{\sin x-x}{x^{3}} &=&-\frac{1}{6},\ \ \ \ \ \ \ \
and\ \ \ \ \ \ \ \ \ \lim_{x\rightarrow 0}\frac{\arcsin x-x}{x^{3}}=\frac{1}{%
6} \\
\lim_{x\rightarrow 0}\frac{\sin x-x+\frac{1}{6}x^{3}}{x^{5}} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1549725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
How does one get from $p+3=3k+2$ then $2^{p+3} \equiv 4 \pmod 7$, to $5 \cdot 2^{p+3} -31 \equiv 3 \pmod7$ How does one get from $p+3=3k+2$ then $2^{p+3}\equiv4 \pmod 7$, to $5 \cdot 2^{p+3} -31 \equiv 3 \pmod7$.
I am just starting with modular arithmetic so any help would be greatly appreaciated.
Credit to user236182 ... | $2^3=8$, so the remainder when you divide $2^3$ by 7 is 1. That means $2^3=1\pmod7$.
$2^{3k}=(2^3)^k$ because of power laws. This equals $(7+1)^k$. Expand this using the Binomial Theorem, and all the terms are multiples of 7 except for the final term which is 1. So $2^{3k}=1\pmod7$.
Lastly, $2^{3k+2}=2^{3k}2^2=4.2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1550027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Locus of the point of intersection of the pair of perpendicular tangents to the circles $x^2+y^2=1$ and $x^2+y^2=7$ Locus of the point of intersection of the pair of perpendicular tangents to the circles $x^2+y^2=1$ and $x^2+y^2=7$ is the director circle of the circle with radius
$(A)\sqrt2\hspace{1cm}(B)2\hspace{1cm}(... | You know it's a circle with center $O$, so you just need a point to find the radius.
You can take a tangent the the first circle: $y=1$ and a perpendicular tangent to the other circle: $x=\sqrt{7}$ , the intersection $(1,\sqrt{7})$ is on the circle: the radius is
$$\sqrt{1^2+\sqrt{7}^2}=\sqrt{8}=2\sqrt{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1551153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $\tan A + \tan B + \tan C = 3 \sqrt{3}$, then $\triangle ABC$ is equilateral.
Given $\triangle ABC$ such that
$$\tan A + \tan B + \tan C = 3 \sqrt{3}$$
prove that $\triangle ABC$ is equilateral.
The full process is needed.
| First we prove $\tan A + \tan B + \tan C = \tan A \tan B \tan C$.
We know that
$A + B + C = 180$
from the angle addition formula, so
$\tan((A + B) + C) = \frac{\tan(A + B) + \tan C}{1−\tan(A + B) \tan C}$
$= \frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1 - \tan A \tan B - \tan A \tan C - \tan B \tan C}$
W... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1551375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the eigenvalues and associated eigenvectors for this matrix $$A = \begin{pmatrix} 1 & -1 & 4 \\ 3 & 2 & -1 \\ 2 & 1 & -1\end{pmatrix}$$
I solved for the determinant of matrix $A$ and got the following:
$\det\begin{bmatrix}
1-\lambda & -1 & 4\\
3 & 2-\lambda & -1 \\
2 & 1 & -1-\lambda
\end{bmatrix} = -\lambda^{3... | To find eigenvalues, note that you want to find the solutions to the characteristic equation given by
$\det (A - \lambda I) = 0$
As you noted, we have the equation
$-\lambda^3 + 2\lambda^2 + 5 \lambda - 6 = 0$
Note that this is just
$-(\lambda - 3)(\lambda + 2)(\lambda -1) = 0$
so your eigenvalues are $\lambda_1 = 3, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1552375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Express $\frac{\sin 7\theta}{\sin \theta}$ in powers of $\sin \theta$ only By using DeMoivre's theorm express
$$\frac{\sin 7\theta}{\sin \theta}$$
in the powers of Sine only
answer given in the book is
$$7-56\sin ^2\theta+112\sin ^4 \theta-64\sin^6 \theta$$
can any one help to solve the question
| $(\cos x+i\sin x)^7=\cos 7x+i\sin7x$.
We have
$$(r+t)^7=r^7+7r^6t+21r^5t^2+35r^4t^3+35r^3t^4+21r^2t^5+7rt^6+t^7$$
hence, making $\sin x=t$ and $\cos x=r$ and having in account the powers of $i$, we get
$$\sin 7x=7r^6t-35r^4t^3+21r^2t^5-t^7$$ so, because of $\cos^2x=1-\sin^2x$,
$$7t-21t^3+21t^5-7t^7-35t^3+70t^5-35t^7+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1553047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How to simplify an expression? I have tried to simplify this expression for quite a long time now but I can't find how to do it.
$$\left(\frac{1}{2+4m}-\frac{1-m}{8m^3+1}:\frac{1-2m}{2m^2-2m+1}\right)\cdot\frac{4m+2}{2m-1}-\frac{1}{1-4m+4m^2}$$
Can someone help me with it?
| $\require{cancel}$
Starting with the original expression:
$$\left(\frac{1}{2+4m}-\frac{1-m}{8m^3+1}:\frac{1-2m}{2m^2-2m+1}\right)\cdot\frac{4m+2}{2m-1}-\frac{1}{1-4m+4m^2}$$
Factoring where possible and changing the order of some terms:
$$\left(\frac{1}{2(2m+1)}+\frac{m-1}{(2m+1)(4m^2-2m+1)}:\frac{-(2m-1)}{2m^2-2m+1}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1554806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Zero divided by zero must be equal to zero What is wrong with the following argument (if you don't involve ring theory)?
Proposition 1: $\frac{0}{0} = 0$
Proof: Suppose that $\frac{0}{0}$ is not equal to $0$
$\frac{0}{0}$ is not equal to $0 \Rightarrow \frac{0}{0} = x$ , some $x$ not equal to $0$ $\Rightarrow$ $2(\frac... | The first fault is in the very statement. You assume that there is a $0/0$ in the first place, you have to have that before you even assume that it's not equal to zero.
Then even if there were a $0/0$ you're assuming that there is a $2{0\over0}$ and then that it is equal to ${2\cdot 0\over0}={0\over0}$.
Finally you ass... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1554929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "52",
"answer_count": 16,
"answer_id": 3
} |
Evaluate $\int e^{2\theta} \sin (3\theta)\ d\theta$
Evaluate $$\int e^{2\theta} \sin (3\theta)\ d\theta .$$
I am little stuck as to what I can do after this point. Please tell me if my method overall is flawed:
| You have some mistake:
$$
\int e^{2\theta}\sin(3\theta) d\theta=\frac{1}{2}e^{2\theta}\sin(3\theta)-\frac{3}{2}\int e^{2\theta}\cos(3\theta) d\theta=
$$
$$
=\frac{1}{2}e^{2\theta}\sin(3\theta)-\frac{3}{4} e^{2\theta}\cos(3\theta) -\frac{9}{4}\int e^{2\theta}\sin(3\theta) d\theta
$$
so:
$$
\left(1+\frac{9}{4} \right)\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1556613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
is $\frac{a-b}{a+b} - \frac{c-d}{c+d}$ equal to $ \frac{a-c}{a+c}$? I am using values out of a cross-correlation analysis and my intuition tells me that the equation in the title is true, however I would like to prove it.
$$ \frac{a-b}{a+b} - \frac{c-d}{c+d} = \frac{a-c}{a+c}$$
Unfortunately my math skills are horrible... | $$\frac{a-b}{a+b}-\frac{c-d}{c+d}=$$
$$\frac{(a-b)(c+d)}{(a+b)(c+d)}-\frac{(c-d)(a+b)}{(c+d)(a+b)}=$$
$$\frac{(a-b)(c+d)-(c-d)(a+b)}{(a+b)(c+d)}=$$
$$\frac{2(ad-bc)}{(a+b)(c+d)}$$
If you would like to look if your identity is correct, set your variables:
Choose $a=2,b=3,c=4,d=5$:
$$ \frac{2-3}{2+3} - \frac{4-5}{4+5} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1556923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Convergence rate of the sequence $a_{n+1} = a_n-a_n^2, a_0=1/2$. The sequence converges to zero at a rate that seems to be slightly faster than $1/n$.
What are the best known results on the convergence rate of this sequence?
| The sequence is a monotonic decreasing one. Also, it is easy to see that it is always positive. Hence, $\{a_n\}$ converges.
Let $N$ be sufficiently large integer so that
$$
a_N - a_{N + 1} < \varepsilon_N,
$$
where $\varepsilon_n \to 0$. But we have
$$
a_N - a_{N + 1} = a_N (1 - 1 + a_N) = a_N^2 < \varepsilon_N.
$$
Hen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1558592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Define $a$ and $b$ rational numbers so they satisfy equation Define $a, b \in \mathbb{Q}$ so that
$$\frac{a}{\sqrt{7 + 4\sqrt{3}}} + \frac{b}{\sqrt{7 - 4\sqrt{3}}} = \sqrt{4 + 2\sqrt{3}}$$
Using $\sqrt{a \pm \sqrt{b}} = \sqrt{\frac{a + \sqrt{a^2 - b}}{2}} \pm \sqrt{\frac{a - \sqrt{a^2 - b}}{2}}$ I got $\frac{a}{2 + \sq... | Hint:
$$\frac{1}{\sqrt{7+4\sqrt{3}}}=2-\sqrt{3}$$
$$\frac{1}{\sqrt{7-4\sqrt{3}}}=2+\sqrt{3}$$
$$\sqrt{4+2\sqrt{3}}=1+\sqrt{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1558762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
What are $\int\sqrt{a^2-x^2}\,\textrm{d}x, \int\sqrt{x^2+a^2}\,\textrm{d}x,\int\sqrt{x^2-a^2}\,\textrm{d}x$? Can someone confirm these equations below? I got it from my college textbook, unfortunately there are no proofs and more importantly I cannot seem to find any other sources that say have these equations.
$\displ... | It's OK to me.
If you just want to confirm it, you may just differentiate each function on the right side to see if you get the integrand on the left hand side.
One may obtain these results by performing respectively the change of variable
$x:=a\sin t$, $x:=a\sinh t$ and $x:=a\cosh t$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1560912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Values of quadratic form on unit circle We have the quadratic form $q(\begin{pmatrix}x\\y\end{pmatrix})=11x^2-16xy-y^2$.
Which values does $q$ take on the unit circle $x^2+y^2=1$?
I know that $q(x,y)$ is given by $q(x,y)=(x,y)\begin{pmatrix}11&-8\\-8&-1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$ and that $\begin... | From Joseph Curwen's hint:
$$\begin{align}q &= 11\cos^2\theta-16\cos \theta\sin \theta-\sin ^2\theta\\
&=12\cos^2\theta-16\cos \theta\sin \theta-1\end{align}$$
Now $$\sin 2\theta = 2\cos\theta\sin\theta\\\cos 2\theta = 2\cos^2 \theta - 1$$
So,
$$\begin{align}q &= 6\cos 2\theta - 8\sin 2\theta + 5\\&=6\cos t - 8\sin t ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1560990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the nth derivative of $y=\frac4{(6x+8)^3}$ Find the $n$-th derivative of $y=\frac{4}{(6x+8)^3}$
\begin{align}
y' ={} & 4(-3)(6) \frac{1}{(6x+8)^4}
\\
y''={} & 4(-3)(-4)(6)^2 \frac{1}{(6x+8)^5}
\\
y'''={} & 4(-3)(-4)(-5)(6)^3\frac{1}{(6x+8)^6}
\end{align}
I recognise the pattern but can't interpret that into a f... | $y=\frac{4}{(6x+8)^3} y'$ so $\frac{\rm d}{\rm dx}(\ln y)=\frac{(6x+8)^3}{4}$.
Now you can find $\ln y$ and hence $y$, and then take the derivatives.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Unitary transformation I have a matrix in following form
$$A=\begin{bmatrix}
1&0&0&0&0&0&0&0\\
0&-1&0&0&0&0&0&0\\
0&0&1&0&0&0&0&0\\
0&0&0&-1&0&0&0&0\\
0&0&0&0&1&0&0&0\\
0&0&0&0&0&-1&0&0\\
0&0&0&0&0&0&1&0\\
0&0&0&0&0&0&0&-1\\
\end{bmatrix}$$
I need some help to reduce the matrix in following form using unitary transform... | Denote by $e_i$ the standard basis for $\mathbb{C}^8$. You are given a matrix $A$ such that
$$ Ae_i = \begin{cases} e_i & i \text{ is odd}, \\
-e_i & i \text{ is even}. \end{cases} $$
and you look for a unitary matrix $U \in M_{8}(\mathbb{C})$ such that
$$ UAU^{*}(e_i) = \begin{cases} e_i & 1 \leq i \leq 4, \\
-e_i & 5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1565652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Convergence of $\frac{1}{5} + \frac{1}{9} + \frac{1}{13} + ... = \sum_{i=1}^{\infty} \frac{1}{1+4i}.$ I've been working on an approximation for a problem in my numerical methods course, and I seemed to have run into the series
$$\frac{1}{5} + \frac{1}{9} + \frac{1}{13} + ... = \sum_{i=1}^{\infty} \frac{1}{1+4i}.$$
I'm ... | Notice that:
$$(\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5})+(\frac{1}{9}+\frac{1}{9}+\frac{1}{9}+\frac{1}{9})+...\ge(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8})+(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12})+..$$
So $4\sum a_n$ diverges, hence also $\sum a_n$ diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1566947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
How is this summation simplified? I'm taking EdX's probability class and one question asked to find the expectation of a uniform random variable (problem described here: https://youtu.be/vB6EKsX12hc).
There's one part of the calculation that I don't understand:
how does the numerator in the summation get reduced to a ... | $\sum_{i=1}^a 1 = a$. Sot there's that.
So we just need to show $\sum_{k=a + 1}^b \frac {b-k +1}{b-a - 1} = \frac{1}{b -a - 1}\sum_{k=1}^{b-a} k$.
Which is $\sum_{k=a + 1}^b \frac {b-k +1}{b-a - 1} =\frac{1}{b -a - 1}\sum_{k=a + 1}^{b} (b - k +1) = \frac{1}{b -a - 1}\sum_{k=1}^{b-a} (b - a - k +1) =\frac{1}{b -a - 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1567032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Polynomial $P(x)$ contains only terms of odd degree.When $P(x)$ is divided by $(x-3)$ ,the remainder is $6$. Polynomial $P(x)$ contains only terms of odd degree.When $P(x)$ is divided by $(x-3)$ ,the remainder is $6$.If $P(x)$ is divided by $(x^2-9)$ then remainder is $g(x).$Find the value of $g(2).$
As $P(x)$ is a po... | Keep in mind, $P(x)$ is an odd polynomial, so $P(-3)=-6=-3a+b$. So we adding that to $6=3a+b$, we get $0=2b$, so $b=0$. That leaves $a=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1567477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to solve the equation of $\sqrt{x}+\sqrt{y}=\sqrt{2205}$ in integers? How to solve the equation of $\sqrt{x}+\sqrt{y}=\sqrt{2205}$ in integers?
How in general to solve the similar equations?
| Simple factorization yields $\sqrt{2205} = 21\sqrt{5} $
$$\sqrt{x}=\sqrt{2205}-\sqrt{y}$$
$$x=2205+y-2.21\sqrt{5}\sqrt{y}$$
$$42\sqrt{5y}=2205+y-x$$
$$y\text{ is in the form of } 5k^2$$
$$y=k\sqrt{5}\Rightarrow\sqrt{x}=l{5}$$
Then we have,
$\sqrt{x}=0,\sqrt{y}=21\sqrt{5}$
$\sqrt{x}=\sqrt{5},\sqrt{y}=20\sqrt{5}$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1575392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Divisibility of $2^n-n^2$ by 7 How many positive integers $n<10^4$ are there such that $2^n - n^2$ is divisible by 7?
| Just write out periods:
$2^n:2,4,1,2,4,1,2,4,1,2,4,1,2,4,1,2,4,1,2,4,1,|2...$ with period $3$
$n^2:1,4,2,2,4,1,0,1,4,2,2,4,1,0,1,4,2,2,4,1,0,|1...$ with period $7$
So in every $21$ numbers there are $6$ numbers.
Now just calculate $10^4\over 21$$=476$ and since $10000$ itself is not included there are $3$ remainder lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1576496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Calculating $\lim_{n \to \infty}\frac{n^{3}}{(3+\frac{1}{n})^{n}}$ I need help to calculate this limit:
$$\lim_{n \to \infty}\frac{n^{3}}{(3+\frac{1}{n})^{n}}$$
| And this is why $\lim_{n\to \infty}\frac{n^3}{3^n} = 0$.
${3^n}=(1+2)^n=1+n2+\frac{n(n-1)}{2}2^2+\frac{n(n-1)(n-2)}{3!}2^3+\frac{n(n-1)(n-2)(n-3)}{4!}2^4+...+2^n>\frac{n(n-1)(n-2)(n-3)}{4!}$
And so,
$0 \le \lim_{n\to \infty}\frac{n^3}{3^n} \le \lim_{n\to \infty}\frac{n^3}{\frac{n(n-1)(n-2)(n-3)}{4!}}= 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1580904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
distinct roots of the equation $A\sin^3 x+B\cos^3 x+C =0$
The number of distinct roots of the equation $A\sin^3 x+B\cos^3 x+C =0$
no two of which differ by $2\pi$ is, Where $A,B,C\in \mathbb{R}$
$\bf{(a)}\;\;\;\;\;\; 3\;\;\;\;\;\; (b)\;\;\;\;\;\; 4\;\;\;\;\;\; (c)\;\;\;\;\;\; 6\;\;\;\;\;\; (d)\;\;\;\;\;\; infinite$
$... | See after simplifying use double angle formula so $(2\sin x+\sin x-\sin 3x)=2\sin x-2\cos 2x \sin x=2 \sin x(1-\cos 2x)$ same goes for other so it becomes $2\cos x(1+\cos 2x)$ thus simplifying further you get $a \sin x(1-\cos 2x)+b\cos x(1+\cos 2x)+2c=0$ now you just need values of $cos,sin$ where $c$ has a considerabl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1580975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find product of solutions of $x^6=-64$
If the six solutions of $x^6=-64$ are written in the form $a+bi$,where
$a$ and $b$ are real, then find the product of those solutions with $a>0$.
The answer in my book is given as $4$ but I don't see why.
Can you give me some hint on that?
| $$x^6=-64\Longleftrightarrow$$
$$x^6=64e^{\pi i}\Longleftrightarrow$$
$$x=\left(64e^{(2\pi k+\pi)i}\right)^{\frac{1}{6}}\Longleftrightarrow$$
$$x=2e^{\frac{1}{6}(2\pi k+\pi)i}\Longleftrightarrow$$
$$x=\begin{cases}2e^{\frac{1}{6}(2\pi\cdot0+\pi)i}\\
2e^{\frac{1}{6}(2\pi\cdot1+\pi)i}\\
2e^{\frac{1}{6}(2\pi\cdot2+\pi)i}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1583626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Formula for $\sum \limits_{n=0}^{\infty} \frac{1}{(n+a)!}$ Is there a closed form for the infinite sum $$\sum \limits_{n=0}^{\infty} \frac{1}{(n+a)!} \mathrm{?}$$ where a is an integer greater than or equal to $0$.
When $a=0$, the sum is just the series expansion for $e$, $\sum \limits_{n=0}^{\infty} \frac{1}{n!}=e$
Wh... | Not clearly more convenient, but an alternative form can be built by first establishing that:
$$\int_0^1\frac{x^n}{n!}e^{-x}\,dx=1-\frac{1}{e}\sum_{k=0}^n\frac{1}{k!}$$
[This is best done by induction - if we call the integral on the left $I_n$, integrating by parts quickly gives us $I_n=I_{n-1}-\frac{1}{e\cdot n!}$.]
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1585822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve the equation: $\cos^2(x)+\cos^2(2x)+\cos^2(3x)=1$ How to solve the equation
$$\cos^2(x)+\cos^2(2x)+\cos^2(3x)=1$$
Can anyone give me some hints?
| $\cos^2(x)+[2 \cos^2x-1]^2 +[4 \cos^3 x-3 \cos x]^2 = 1$, Simplify the equation to get: $ \cos^2 x [8 \cos^4 x - 10 \cos^2 x + 3] = 0$. Hope this will help.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Let $k = 2008^2 + 2^{2008}$. What is the last digit of $k^2 + 2^k$? Let $k = 2008^2 + 2^{2008}$. What is the last digit of $k^2 + 2^k.$
I thought of this
$$2008^2+2^{2008}\pmod{10} ≡ {-2}^2+{2^4}^{502}\pmod{10} ≡ 4+{-4}^{502}\pmod{10} ≡ 4+6^{251} \pmod{10}$$
but I still cannot prove it. Maybe there is a clever solution... | $k$ is a huge even number, hence $k^2+2^k$ is for sure an even number. To compute the last digit of $k^2+2^k$, we just need to understand what $k^2+2^k\pmod{5}$ is. We have $k\equiv 0\pmod{4}$ and
$$ k \equiv 3^2+4^{1004} \equiv -1+1\equiv 0\pmod{5}, $$
hence $k^2+2^k\equiv 0+1\equiv 1\pmod{5}$ by Fermat's little theo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How to Simplify $-\frac14\sin\frac x4 + \frac{\sqrt{3}}4\cos\frac x4 = 0$? A problem requires finding the value of x, such that
$$-\frac14\sin\frac x4 + \frac{\sqrt{3}}4\cos\frac x4 = 0$$
The site reduces the problem to $\tan(\frac x4) = \sqrt{3}$.
How one should do this?
| $$-\frac{1}{4}\sin\left(\frac{x}{4}\right)+\frac{\sqrt{3}}{4}\cos\left(\frac{x}{4}\right)=0\Longleftrightarrow$$
$$\frac{1}{4}\left(\sqrt{3}\cos\left(\frac{x}{4}\right)-\sin\left(\frac{x}{4}\right)\right)=0\Longleftrightarrow$$
$$\sqrt{3}\cos\left(\frac{x}{4}\right)-\sin\left(\frac{x}{4}\right)=0\Longleftrightarrow$$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Solve the equation $27 \sin(x) \cdot \cos^2(x) \cdot \tan^3(x) \cdot \cot^4(x) \cdot \sec^5(x) \cdot \csc^6(x) = 256$.
Solve the equation $27 \sin(x) \cdot \cos^2(x) \cdot \tan^3(x) \cdot \cot^4(x) \cdot \sec^5(x) \cdot \csc^6(x) = 256$.
I was hoping some things would cancel out when I expanded this but nothing. I th... | since
\begin{align*}27\sin{x}\cos^2{x}\tan^3{x}\cot^4{x}\sec^5{x}\csc^6{x}&=27\sin{x}\cos^2{x}\cdot\dfrac{\sin^3{x}}{\cos^3{x}}\cdot\dfrac{\cos^4{x}}{\sin^4{x}}\cdot\dfrac{1}{\cos^5{x}}\cdot\dfrac{1}{\sin^6{x}}\\
&=\dfrac{27}{\cos^2{x}\sin^6{x}}
\end{align*}
so
$$\sin^6{x}\cos^2{x}=\dfrac{27}{256}$$
but on the other ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
} |
$\frac{1}{\sin 8^\circ}+\frac{1}{\sin 16^\circ}+....+\frac{1}{\sin 4096^\circ}+\frac{1}{\sin 8192^\circ}=\frac{1}{\sin \alpha}$,find $\alpha$ Let $\frac{1}{\sin 8^\circ}+\frac{1}{\sin 16^\circ}+\frac{1}{\sin 32^\circ}+....+\frac{1}{\sin 4096^\circ}+\frac{1}{\sin 8192^\circ}=\frac{1}{\sin \alpha}$ where $\alpha\in(0,90^... | This is just the same idea as in lab bhattacharjee's answer, but using the identity from the Weierstrass Substitution
$$
\tan(x/2)=\frac{\sin(x)}{1+\cos(x)}
$$
we get
$$
\begin{align}
\frac1{\tan(x/2)}-\frac1{\tan(x)}
&=\frac{1+\cos(x)}{\sin(x)}-\frac{\cos(x)}{\sin(x)}\\
&=\frac1{\sin(x)}
\end{align}
$$
The rest is the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
How to compute $\lim\limits_{x \to 0_+} \left(\frac{x^2+1}{x+1}\right)^{\frac{1}{\sin^2 x}}$? I have a problem with this limit, I dont know what method to use. After several attempts the eventually find was obtained ($\infty$). But in reality the result sought is $0$.
Can you explain the method and the steps used? Than... | Note that $\frac{x^2+1}{x+1}=\frac{x+1 -(x-x^2)}{x+1}$. If $x$ is small positive, then $x-x^2\gt x/2$ and $x+1\lt 2$. It follows that if $x$ is small positive, then
$$0\lt \frac{x^2+1}{x+1}\lt 1-\frac{x}{4}.$$
Note also that if $x$ is small positive, then $\frac{1}{\sin^2 x}\gt \frac{1}{x^2}$. Thus if $x$ is small pos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1592141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Prove $(1+\sec x)(1+\csc x) > 5$ if $x \in {\left]0,\frac{\pi}{2}\right[}$
Prove $(1+\sec x)(1+\csc x) > 5$ if $x \in {\left]0,\frac{\pi}{2}\right[}$.
I have done the following:
Let $t=\tan \frac{x}{2}\in ]0,1[$, then using the $t$-formulae the LHS becomes
$$(1+\sec x)(1+\csc x) = \frac{2t+1+t^2}{t(1-t^2)}$$
I could... | Let $$y=\left(1+\dfrac{1+t^2}{1-t^2}\right)\left(1+\dfrac{1+t^2}{2t}\right)=\dfrac{2(1+t)^2}{2t(1-t^2)}=\dfrac{1+t}{t(1-t)}$$ as $t+1\ne0$
$$\implies yt^2-t(y-1)+1=0$$
As $t$ is real, the discriminant $$(y-1)^2-4y\ge0$$
As the roots of $y^2-6y+1=0$ are $y=\dfrac{6\pm\sqrt{32}}2=3\pm2\sqrt2$
$y^2-6y+1\ge0\implies$ eith... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1592203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find $\lim_{n\to \infty}r_n$ and $\lim_{n\to \infty}s_n$
A sequence of polynomials is given by $$p_n(x) =a_{n+2}x^2 +a_{n+1}x-a_n$$ for
$n\geq 0$, where $a_0=a_1=1$ and for $n\geq 0$ $$a_{n+2}=a_{n+1} +a_n$$ Denote by $r_n$ and $s_n$ the roots of $p_n(x)=0$, with $r_n\leq s_n$. Find $\lim_{n\to \infty}r_n$ and $\lim... | We can use sum of roots and product of roots.
For a quadratic equation $ax^2+bx+c=0$ with roots $s, t$, we have $s+t=-\frac{b}{a}$ and $st=\frac{c}{a}$.
Since we already have $1$ root, we can substitute it into the product of roots equation: $s_nr_n=\frac{-a_n}{a_{n+2}}$
Hence $s_n=\frac{a_n}{a_{n+2}}$.
Since Fibonacci... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find $\lfloor z \rfloor $ given that $z=(\{\sqrt{3}\}^2-2\{\sqrt{2}\}^2)/(\{\sqrt{3}\}-2\{\sqrt{2}\})$ Let $\lfloor x \rfloor$ denote the greatest integer function, and $\{x\}=x-[x]$ the fractional part of $x$. If $$z=\cfrac{\{\sqrt{3}\}^2-2\{\sqrt{2}\}^2}{\{\sqrt{3}\}-2\{\sqrt{2}\}}$$ find $\lfloor z \rfloor $
My att... | We have
\begin{align*}
z = \cfrac{(\sqrt{3}-1)^2-2(\sqrt{2}-1)^2}{\sqrt{3}-1-2\cdot (\sqrt{2}-1)} &= \frac{3+1-2\sqrt{3}-2(2+1-2\sqrt2)}{\sqrt{3}-2\sqrt2+1}\\
&= \frac{-2\sqrt3 + 4\sqrt2-2}{\sqrt{3}-2\sqrt2+1}\\
&= -2\frac{\sqrt{3}-2\sqrt2+1}{\sqrt{3}-2\sqrt2+1}\\
&= -2.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Trigonometry with Quadratic Equations If $\tan A$ and $\tan B$ are the roots of $x^2+px+q=0$, then prove that
$$\sin^2(A+B)+p \sin(A+B) \cos(A+B) + q \cos^2(A+B) = q$$
I tried the question but with $q$ other terms came associated.
| Hint : Let $E=\sin^2(A+B)+p \sin(A+B) \cos(A+B) + q \cos^2(A+B) $
Divide $E$ by $\cos^2(A+B)$.
Then you will get $$E\sec^2(A+B)= \tan^2(A+B)+p \tan(A+B) + q $$
Since $$\tan(A+B)=\frac{−p}{1−q} \Rightarrow \sec^2(A+B)=1+\frac{p^2}{(1−q)^2}$$, provided $q≠1$.
$$E\sec^2(A+B)= \frac{p^2}{(1−q)^2}-\frac{p^2}{1−q} + q $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Simultaneous Equations (Stuck on the algebra)
Question: Solve the following simultaneous equations for real values of x and y
$$
\left\{
\begin{array}{l}
9^{2x+y} - 9^x \times 3^y = 6 \\
\log_{x+1}(y+3) + \log_{x+1}(y+x+4) = 3
\end{array}
\right.
$$
What I have attempted; for the first equation
$$
9^{2... | we have $$(3^{2x+y})^2-3^{2x+y}=6$$ and $$log_{x+1}(y+3)(y+x+4)=3$$
the last equation can be written as $$(y+3)(y+x+4)=(x+1)^3$$
from the quadratic equation we get $$2x+y=1$$ can you proceed?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
The value of the polynomial at given point. Given that:
$f(x)=x^{10}+2x^9-2x^8-2x^7+x^6+3x^2+6x+1$.
Find the value of $f(x)$ at $x=\sqrt{2}-1$
Answer is an integer. I tried factorization but couldn't proceed towards anything promising.
| Just compute the value for each power $$x=\sqrt{2}-1$$ $$x^2=x \times x=3-2 \sqrt{2}$$ $$x^3=x\times x^2=5 \sqrt{2}-7$$ $$x^6=x^3 \times x^3=99-70 \sqrt{2}$$ $$x^7=x\times x^6=169 \sqrt{2}-239$$ $$x^9=x^7\times x^2=985 \sqrt{2}-1393$$ $$x^{10}=x^7\times x^3=3363-2378 \sqrt{2}$$ With this kind of problem, just work in a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Probability of choosing positive 2-digit integer with 4 in either place GRE study exam guide has following
If an integer is randomly selected from all positive 2-digit integers,
what is the probability that the integer chosen has at least one 4 in
the tens place or the units place?
I understand probability of bei... | A = $4$ in a ten place,
B = $4$ in a unit place,
C = at least one $4$ in a ten or unit place.
$P(A) = \frac{1}{9}$, $P(B) = \frac{1}{10}$, $P(C) = \frac{1}{9} + (1-\frac{1}{9})*\frac{1}{10} = \frac{10 + 8}{90} = \frac{1}{5}$.
Here you go:
You have $10$ cases for event $A$: $40, 41, 42, 43,44,45,46,47,48,49$, and $9$ ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Does $\sum_{n=1}^\infty \frac{2\cdot 4\cdot 6\cdot ...\cdot (2n)}{n^n}$ converge? Does $\sum_{n=1}^\infty \frac{2\cdot 4\cdot 6\cdot ...\cdot (2n)}{n^n}$ converges?
Well, I'm trying the approach using Cauchy's test for convergence. If $a_n >= 0$ and $\lim \sqrt[n]a_n = L $ exists. then, I need to check whether it is bi... | Note that $\frac{2*4*6*...*2n}{n^n}=\frac{n!2^n}{n^n}$.
By the ratio test, $\dfrac{\left(\frac{2}{n+1}\right)^{n+1}(n+1)!}{\left(\frac{2}{n}\right)^{n}(n)!}=\dfrac{2n^n}{(n+1)^n}=2\left(\dfrac{n}{n+1}\right)^n=2\left(1-\frac{1}{n+1}\right)^n\to\frac{2}{e}<1$.
Thus the serie is convergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
solving $\int \cos^2(2x)\sin(2x)\,dx$
$$\int \cos^2(2x)\sin(2x)\,dx$$
I have tried $\int \cos^2(2x)\sin(2x)\,dx=\int (1-\sin^2(2x))\sin(2x)\,dx$
And $\int \cos^2(2x)\sin(2x)\,dx=\int (\cos^2(2x))2\cos x\sin x\, dx$
But no substitution seems to work.
| $$\int \cos ^{ 2 } (2x)sin(2x)dx=-\frac { 1 }{ 2 } \int { \cos ^{ 2 }{ \left( 2x \right) d\left( \cos { \left( 2x \right) } \right) } } =-\frac { \cos ^{ 3 }{ \left( 2x \right) } }{ 6 } +C$$
substitution here is $u=\cos { \left( 2x \right) } $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1598800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Help finding the determinant of a 4x4 matrix? Sorry for the lack of notation but the work should be easy to follow if you know what you are doing. Okay my problem is that the book says it can be done by expanding across any column or row but the only way to get what the book does in their practice example is to choose ... | For M3 it is $-90+100$ rather than $-90-100$ which gives $10 \times 2=20$ as the final answer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 2
} |
Integral $\int\left(\frac{1}{x^4+x^2+1}\right)dx$ Someone can halp me to solve this integral:
$$\int\left(\frac{1}{x^4+x^2+1}\right)$$
solution$$\frac{1}{4}\ln\left(\frac{x^2+x+1}{x^2-x+1}\right)+\frac{1}{2\sqrt3}\arctan\frac {x^2-1}{x\sqrt3}$$
I don't manage using partial fraction because $${x^4+x^2+1}$$ has $\Delta\... | HINT: $$x^4+x^2+1=(x^2+1)^2-x^2=(x^2-x+1)(x^2+x+1)$$ then $$\dfrac
{1}{x^4+x^2+1}=\dfrac{1}{(x^2-x+1)(x^2+x+1)}=-\dfrac12\left(\dfrac{x-1}{x^2-x+1}\right)+\dfrac12\left(\dfrac{x+1}{x^2-x+1}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
The greatest integer less than or equal to the number $R=(8+3\sqrt{7})^{20}$
Given $$R=(8+3\sqrt{7})^{20}, $$ if $\lfloor R \rfloor$ is Greatest integer less than or equal to $R$, then which of the following option(s) is/are true?
*
*$\lfloor R \rfloor$ is an even number
*$\lfloor R \rfloor$ is an odd number
*$R-\... | Hints For (1), (2): Using the binomial expansion twice gives that \begin{align}A := R + (8 - 3 \sqrt{7})^{20} &= (8 + 3 \sqrt{7})^{20} + (8 - 3 \sqrt{7})^{20} \\ &= \sum_{k = 0}^{20} {20 \choose k} 8^{20 - k} (3 \sqrt{7})^k + \sum_{k = 0}^{20} {20 \choose k} 8^{20 - k} (-3 \sqrt{7})^k \\ &= \sum_{k = 0}^{20} {20 \choos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
calculating a limit of sequence with tan Calculating a limit
$$\lim_{n\longrightarrow +\infty} \dfrac{\tan(\frac{\pi n}{2n+1})}{\sqrt[3]{n^3+2n-1}}$$
thanks.
| Another way is based on the asymptotics of the numerator and denominator $$\tan \left(\frac{\pi n}{2 n+1}\right)=\frac{4 n}{\pi }+\frac{2}{\pi }-\frac{\pi }{12 n}+\frac{\pi }{24
n^2}+O\left(\left(\frac{1}{n}\right)^3\right)$$ $$\sqrt[3]{n^3+2 n-1}=n+\frac{2}{3 n}-\frac{1}{3 n^2}+O\left(\left(\frac{1}{n}\right)^3\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Let $x,y \ge 0 $.Find the maximum value of $\cfrac{(3x+4y)^2}{x^2+y^2}$ Let $x,y \ge 0 $.Find the maximum value of $\cfrac{(3x+4y)^2}{x^2+y^2}$
This exercise is from the chapter of my book regarding the Cauchy-Schwarz inequality but I don't know if I've applied it correctly .
The problem asks for the maximum value of t... | You are right till
$$R (x^2+y^2) \ge (3x+4y)^2$$
Now comparing with the CS inequality
$$(3^2+4^2)(x^2+y^2) \ge (3x+4y)^2$$
$\implies R = 3^2+4^2 = 25$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to factor polynomials? I am wondering if there is a methodical, algorithmic, brain-dead way to factor polynomials. For example:
$x^6 - 14x^{5} + 73x^{4} - 188x^{3} + 256x^{2} - 176x^{1} + 48$
can be written as
$(x-1)^2 (x-2)^3 (x-6)$
But how do you actually get there?
| If you will accept a "brain-dead" method, why not just appeal to computer algebra, such as Mathematica?
Factor[x^6 - 14 x^5 + 73 x^4 - 188 x^3 + 256 x^2 - 176 x^1 + 48]
$(x-6) (x-2)^3 (x-1)^2$
or
Factor[-20284203008 + 12410729472 x + 2524112000 x^2 -
1341562640 x^3 - 334176180 x^4 + 12601652 x^5 + 12198347 x^6 +
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1602768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Limit $(n - a_n)$ of sequence $a_{n+1} = \sqrt{n^2 - a_n}$
Consider the sequence $\{a_n\}_{n=1}^{\infty}$ defined recursively by
$$a_{n+1} = \sqrt{n^2 - a_n}$$ with $a_1 = 1$. Compute $$\lim_{n\to\infty} (n-a_n)$$
I am having trouble with this. I am not even sure how to show the limit exists. I think if we know the... | First, we'll show that $0 \le n-a_n \le 2$ for all $n$. Then, we will show that $n-a_n \to \dfrac{3}{2}$ as $n \to \infty$.
Since, $a_1 = 1$, we have $1-a_1 = 0$, so $0 \le 1-a_1 \le 2$, as desired.
Now, suppose $0 \le n-a_n \le 2$ for some positive integer $n$. Then, we have:
$$\sqrt{n^2-n} \le \sqrt{n^2-a_n} \le \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1602850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Evaluating - $\int_{|z| =R,R > 0} z^2 \big( \sin\big(\frac{1}{z}\big) + \sin\big(\frac{2}{z}\big) + ...+ \sin\big(\frac{n}{z}\big) \big) dz $ Let:
$$ I_n =\int_{|z| =R , \text{ }R > 0} z^2 \left( \sin\left(\frac{1}{z}\right) + \sin\left(\frac{2}{z}\right) + ...+ \sin\left(\frac{n}{z}\right) \right)\, dz $$
Knowing tha... | \begin{eqnarray}
I_n&=&\sum_{k=1}^n\int_{|z|=R}z^2\sin\left(\frac{k}{z}\right)\,dz=\sum_{k=1}^n\sum_{m=0}^\infty\int_{|z|=R}z^2\frac{(-1)^m}{(2m+1)!}\frac{k^{2m+1}}{z^{2m+1}}\,dz\\
&=&\sum_{k=1}^n\sum_{m=0}^\infty\int_{|z|=R}\frac{(-1)^m}{(2m+1)!}\frac{k^{2m+1}}{z^{2m-1}}\,dz=\sum_{k=1}^n\int_{|z|=R}\frac{-k^3}{3!}\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1603404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show that $\ln\left(1+\frac1n\right)=\frac1n-\frac{\theta_n}{2n^2}$ I've been asked to show that for any natural number $n$, there exists $\theta_n\in (0,1)$ such that
$$\ln\left(1+\frac1n\right)=\frac1n-\frac{\theta_n}{2n^2}$$
I've tried to use integral identities in order to convert the logarithm into something easie... | $$\ln\left(1+\frac1n\right) = \frac1n + \frac{\theta_n}{2n^2}$$
Write $f(x) = \ln(1+x)$. Then by definition of the derivative $$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2 + o(x^2).$$ Where $o(x^2)/x^2 \to 0$ as $x \to 0$.
$f(0)= \ln(1+0) = 0$.
$f'(x) = \frac{1}{1+x}$ so $f'(0) = 1/(1+0) = 1$.
$f''(x) = \frac{-1}{(1+x)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1603479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Some equations from Russian maths book. Could you please help me with solving these equations. I would like to solve them in the most sneaky way. All of the exercises in this book can be solved in some clever way which I can't often find.
$$
\frac{(x-1)(x-2)(x-3)(x-4)}{(x+1)(x+2)(x+3)(x+4)} = 1
$$
$$
\frac{6}{(x+1)(x+2... | The first equation implies that the product of the distances from $x$ to $-1, -2, -3, -4$ is the same as the product of the distances from $x$ to $1, 2, 3, 4$. (This condition is equivalent to the quotient appearing in the equation being $\pm 1$.) If $x > 0$, the latter distances are smaller than the corresponding form... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
Evaluate Integral Using Change of Variables Let $A = \{(x, y)\in \mathbb{R}^2 | x^2-xy+2y^2<1 \}$; define $f: \mathbb{R}^2 \to \mathbb{R}$ by $f(x, y) = xy$.
Express the integral
$$\int_A f$$
as an integral over the unit ball $B = \{(x, y)\in \mathbb{R}^2 | x^2+y^2<1 \}$.
Someone gave a solution here but it seems to be... | Looks reasonable to me. As a check on your computations, here’s a solution using an approach similar to Ron Gordon’s from the original question.
Examine the matrix $Q$ associated with the quadratic form $x^2-xy+2y^2$: $$
Q = \pmatrix{1&-\frac12\\-\frac12&2}.
$$ $\det Q>0$ and $\operatorname{tr}Q>0$, so the eigenvalue... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to solve $\lim _{x\to \sqrt{2}}\left(\frac{e^{x^2}+e^2(1-x^2)}{[\ln(x^2-3\sqrt{2}x+5)]^2}\right)$? I have a problem with this limit, i have no idea how to compute it. Can you explain the method and the steps used(without Hopital if is possible)? Thanks
$$\lim_{x\to \sqrt{2}}\left(\frac{e^{x^2}+e^2(1-x^2)}{[\ln(x^2-... | Notice, one can easily apply L'Hospital's rule for $\frac 00$ form as follows $$\lim_{x\to \sqrt 2}\left(\frac{e^{x^2}+e^2(1-x^2)}{[\ln(x^2-3\sqrt 2 x+5)]^2}\right)$$
$$\lim_{x\to \sqrt 2}\left(\frac{2xe^{x^2}-2xe^2}{\frac{2\ln(x^2-3\sqrt 2 x+5)}{x^2-3\sqrt 2 x+5}\cdot (2x-3\sqrt 2)}\right)$$
$$\lim_{x\to \sqrt 2}\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1605491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find $\lfloor 1000S \rfloor$ for $S =\sum_{n=1}^{\infty} \frac{1}{2^{n^2}} = \frac{1}{2^1}+\frac{1}{2^4}+\frac{1}{2^9}+\cdots.$
Let $$S = \displaystyle \sum_{n=1}^{\infty} \dfrac{1}{2^{n^2}} = \dfrac{1}{2^1}+\dfrac{1}{2^4}+\dfrac{1}{2^9}+\cdots.$$
Find $\lfloor 1000S \rfloor$
My attempt was to recognize that consec... | The sum of the first three terms is
$$1000\left[\frac{1}{2^{1}}+\frac{1}{2^{4}}+\frac{1}{2^{9}}\right] \simeq 564.453$$
The sum of the rest of the terms is bounded by (using a geometrical series)
$$1000\left[\frac{1}{2^{16}}+\frac{1}{2^{25}}+\frac{1}{2^{36}}+\ldots\right] < 1000\left(\frac{1}{2^{16}} + \frac{1}{2^{16+9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1606189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Laurent expansion of $\frac{1}{z^2}$ I need to find a Laurent expansion of $\frac{1}{z^2}$ with centre in $z_0 = 1$ and $P(1, 2014, 2015)$.
If it was $\frac{1}{z}$, I'd rewrite the fraction like this:
$$
\frac{1}{(z-1) +1 }
$$
But with $z^2$ I'm not sure how to proceed...
My attempt:
According to the given recommendat... | Another way is:
$\dfrac {1}{z^2}=\dfrac {\left(1-\dfrac{1}{z}\right)^2}{(z-1)^2}=
\dfrac {(1-z)^2+2(1-z)^3+3(1-z)^4+4(1-z)^5+ \ldots}{(1-z)^2}=1+2(1-z)+3(1-z)^2+4(1-z)^3+ \ldots$
using the product of convergent power series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1609849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
The sequence $\phi_n= 1+\frac{1}{2^4}+\frac{1}{3^4}+\ldots+\frac{1}{n^4}$ is bounded above . A sequence $(\phi_n)$ is defined as follows : $$\phi_n= 1+\frac{1}{2^4}+\frac{1}{3^4}+\ldots+\frac{1}{n^4}$$
Show that the sequence is convergent.
Because this sequence is monotonic, proving it is bounded above will be sufficie... | Since $t \mapsto \dfrac{1}{t^4}$ is decreasing, we have
$$\frac{1}{(k+1)^4}=\int_{k}^{k+1}\frac{dt}{(k+1)^4}\leq \int_{k}^{k+1}\frac{dt}{t^4},\,\,\, k\geq1,$$ which gives$$\phi_n= 1+\frac{1}{2^4}+\frac{1}{3^4}+\ldots+\frac{1}{n^4}=1+\sum_{k=1}^{n-1}\frac{1}{(k+1)^4}\leq 1+\int_{1}^{n}\frac{dt}{t^4}=\frac{4}{3}-\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1612011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Solving this limit $\lim_{x\to 3}\frac{x^2-9-3+\sqrt{x+6}}{x^2-9}$. The question is $\lim_\limits{x\to 3}\frac{x^2-9-3+\sqrt{x+6}}{x^2-9}$.
I hope you guys understand why I have written the numerator like that. So my progress is nothing but $1+\frac{\sqrt{x+6}-3}{x^2-9}$.
Now how do I rationalize the numerator?
It is g... | The solution goes as follows:
$$\lim_\limits{x\to 3}\frac{x^2-9-3+\sqrt{x+6}}{x^2-9}$$
$$=\lim_\limits{x\to 3}\left(1+\frac{\sqrt{x+6}-3}{x^2-9}\right)$$
$$=\lim_\limits{x\to 3}\left[1+\frac{(\sqrt{x+6}-3)(\sqrt{x+6}+3)}{(x^2-9)(\sqrt{x+6}+3)}\right]$$
$$=\lim_\limits{x\to 3}\left[1+\frac{x+6-9}{(x^2-9)(\sqrt{x+6}+3)}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1615527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Let $f:[1,\infty)\to R$ be a monotonic and differentiable function and $f(1)=1.$If $N$ is the number of solutions of $f(f(x))=\frac{1}{x^2-2x+2}$. Let $f:[1,\infty)\to R$ be a monotonic and differentiable function and $f(1)=1.$If $N$ is the number of solutions of $f(f(x))=\frac{1}{x^2-2x+2}$.Find $N.$
$\frac{1}{x^2-2x... | Let $g(x)=f(f(x))$, then $g'(x)=f'(x)f'(f(x))\geq0$ since $f(x)$ is monotonic function.
Note that $x=1$ is the solution of $g(x)=f(f(x))=\frac{1}{x^2-2x+2}=\frac{1}{(x-1)^2+1}$
Now, since $\frac{1}{(x-1)^2+1}$ is strictly decresing function on interval $[1,\infty)$, and $g(x)$ is nondecreasing on that interval, the so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is there a way to prove vector triple product from quaternion multiplication? For pure imaginary quaternions $u, v, w$, is there a way to prove the vector triple product $u\times(v\times w) = v(u\cdot w) - w(u\cdot v)$ from the relation:
$$uv = -u\cdot v + u\times v \text{ for $u, v \in \mathbb{R}^3$ }$$
I tried many t... | The vector triple product identity follows from quaternion associativity.
A convenient tool for expressing quaternion statements in terms of dot and cross products is to write the quaternion as a real number and a 3-vector, as $a = (\alpha, \mathbf{a})$. Then quaternion multiplication of quaternion $b = (\beta, \mathb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1618508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Complex number - wrong result at the end I need to solve this:
$$ \frac{i^4+3}{i-1}$$
On my book the result should be: $-2-2i$ but I get: $-1-2i$ and I do not understand where the error is.
My steps:
$$ \frac{i^4+3}{i-1} = \frac{i^4+3}{i-1} \cdot \frac{-1-i}{-1-i}$$
$$ \frac{(i^4+3)(-1-i) + (-1-i)(i-1)}{(i-1)(-1-i)}$$
... | By the way, since $i^4=1$ the fraction simplifies as
$$
\frac{i^4+3}{i-1}=\frac{4}{i-1}=\frac{4(i+1)}{-2}=-2(i+1).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1619360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Prove that $\sqrt{6}-\sqrt{2}$ $> 1$. I'm trying to prove that $\sqrt{6}-\sqrt{2}$ $> 1$. I need to admit that I'm completely new to proof writing and I have completely no experience in answering that kind of questions. However, I came up with this:
Since $\sqrt{4}=2$ and $\sqrt{9}=3$ we have a following inequality: $2... | $$\sqrt{6}-\sqrt{2} = \frac{6-2}{\sqrt{6}+\sqrt{2}}= \frac{2}{\frac{\sqrt{6}+\sqrt{2}}{2}}> \frac{2}{\sqrt{\frac{6+2}{2}}}=1$$
$\bf{Added:}$
$$\sqrt{6} - \sqrt{2} > \sqrt{6.25} - \sqrt{2.25} = 2.5 - 1.5 = 1$$
since the function $x\mapsto \sqrt{x}$ is concave.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1620218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Prove that $\sqrt{(1+x)}<1+\frac{1}{2}x$ $\forall x>0$ I was just practicing some real analysis for an upcoming quiz and I just wanted some feedback on if my proof was written sufficiently. The question's above, my answer's below. Thanks!
Pf: Let $x>0,$ $ x \in (0,x)$, and let $f(x) = \sqrt{(1+x)}-(1+\frac{1}{2}x)$. Ob... | Let $x>0$, then $\frac{1}{4}x^2>0$
By adding $1+x$ to both sides, we obtain:
$$1+x+\frac{1}{4}x^2>1+x$$It follows that:$$(1+\dfrac{1}{2}x)^2>1+x$$
Since both sides are positive for all $x>0$, then $$\sqrt{(1+x)}<1+\dfrac{1}{2}x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1620274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
On $4n+1 = x^2, 5n+1 = y^2$ and the Fibonacci numbers While answering this question, I decided to look at the particular case,
$$4n+1 = x^2\\5n+1 = y^2$$
to be solved simultaneously. The solutions for $n,x,y$ are A157459, A007805, and A049629, respectively,
$$\begin{aligned}
n_k &= 72, 23256, 7488432, 2411251920,\dots... | The Fibonacci numbers are given in terms of $\phi$ by
$$
F(n) = \frac{\phi^n - (-\phi)^{-n}}{\sqrt{5}}.
$$
Also note that for even $n$, $(-\phi)^{-n} = \phi^{-n}$. This implies that
\begin{align*}
F(6k) F(6k+6) &= \frac{1}{5} (\phi^{6k} - \phi^{-6k})(\phi^{6k+6} - \phi^{-6k - 6}) \\
&= \frac{1}{5} \left[ \phi^{12k + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1621423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How can I integrate $ \int \frac{x^2}{1 + x^3} $?
$$\int \frac{x^2}{1 + x^3}$$
I found this problem in one of my past papers and didn't understand how the solution to this is $ \frac{1}{3} \ln (1+x^3) $. I would really appreciate it if someone guided me through this. I'm sure this is the answer of the integration bec... | The most efficient solution has already been covered by Yagna, however it can never hurt to have another method. The method I will take here is without question an extremely overly complicated method for this question, but is good practice for integrals of rational functions as a general method.
Here you have a rationa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1623243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is the range of $y$ if $x+y+z=4$ and $xy+yz+xz=5$ for $x, y, z \in\mathbb{R}_+$
What is the range of $y$ if $x+y+z=4$ and $xy+yz+xz=5$ for $x, y, z \in\mathbb{R}_+$
How to explain the following method?
Let $x=z$ then:
$$2x+y=4\quad;\quad 2xy+x^{2}=5$$
$$\implies \left( 4-y\right) y+\left( \dfrac {4-y} {2}\right... | $\bf{Another \; Way::}$
Given $x+y+z = 4\Rightarrow x+z=4-y$ and $xy+yz+zx = 5$
Using $$(x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx)\Rightarrow x^2+y^2+z^2 = 6$$
So we get $x^2+z^2 = 6-y^2$
Now Using $\bf{Weighted\; A.M\geq G.M\;,}$ We get
$$\frac{x^2+z^2}{2}\geq \left(\frac{x+z}{2}\right)^2$$
So $$\frac{6-y^2}{2}\geq \left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1623373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to solve this limit involving sine and log? I've tried L'Hopital's Rule but the differentiated numerator involves cos(1/x) which does not exist when x approaches 0.
$$ \lim_{x\to 0^+} \frac{x^2sin\frac{1}{x}}{\ln(1+2x)}$$
| We know
$$\lim_{x\rightarrow 0} \ x\sin\left(\frac{1}{x}\right) = 0$$
$$\ln(1+2x) = 2x - \dfrac{4x^2}{2} + \dfrac{8x^3}{3} ... $$
So the limit reduces to :
$$\lim_{x\rightarrow 0} \ \frac{x\sin\left(\frac{1}{x}\right)x}{2x - \dfrac{4x^2}{2} + \dfrac{8x^3}{3} ..}$$
$$ = \lim_{x\rightarrow 0} \ \frac{x\sin\left(\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1626179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Evaluating the following limit: $\lim _{x\to \frac{\pi }{4}}\left(\tan\left(2x\right)\tan\left(\frac{\pi }{4}-x\right)\right)$ I don't find the right identities for this
$$\lim _{x\to \frac{\pi }{4}}\left(\tan\left(2x\right)\tan\left(\frac{\pi }{4}-x\right)\right)$$
Someone can help me ?
Thanks.
| Since
\begin{align}
\tan(2x)&=\frac{2\tan x}{1-\tan^2 x}&&&\text{and}&&\tan\left(\frac{\pi}{4}-x\right)&=\frac{\tan\frac{\pi}{4}-\tan x}{1+\tan\frac{\pi}{4}\tan x}=\frac{1-\tan x}{1+\tan x}
\end{align}
we have
\begin{align}
\lim_{x\to\frac{\pi}{4}}\left(\tan\left(2x\right)\tan\left(\frac{\pi }{4}-x\right)\right)&=\lim_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1626834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Given $0This seems true, but I have so far failed to prove it. Any clever estimates I could use?
| Let $x=\sin^2 \alpha$ and $z=\cos^2\beta$, where $0<\alpha,\beta<\frac{\pi}{2}$. Then the condition $0\le z\le1-x$ is equivalent to $\alpha\le\beta$ and the inequality becomes
$$ \sqrt{\sin^2\alpha+\cos^2\beta}\le(1-\sin\alpha)\frac{\cos\beta}{\cos\alpha}+\sin\alpha $$
which is equivalent to
$$ \cos^2\beta\le \frac{(1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1627841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$X^4 - 4Y^4 = -Z^2$ has no solutions in non zero integers I am trying to prove that $X^4 - 4Y^4 = -Z^2$ has no solutions in non zero integers. I know there are similar questions on MS, but that minus signs before the $Z$ gives me a hard time.
For the moment, I converted the equation to the equivalent $X^4 + Z^2 = 4Y^2$... | Your argument is correct, and no quicker method is springing to mind to solve this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1628270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the last $2$ digits of $7^{7^{7^{10217}}}$
Find the last $2$ digits of $$\large7^{7^{7^{10217}}}$$
So far I have: $17\cdot 601=10217$ and $7=7 \pmod{10}$. Any help greatly appreciated.
| Note that $100=4\cdot 25$. We can find $7^{7^{7^{10217}}}$ mod $4$, $25$, then combine this to get the value mod $100$ using the Chinese Remainder Theorem.
$7^4\equiv \left(7^2\right)^2\equiv (-1)^2\equiv 1\pmod{25}$
Therefore $$7^{7^{7^{10217}}}\equiv 7^{7^{7^{10217}}\pmod 4}\equiv 7^{(-1)^{7^{10217}}\pmod 4}\pmod{25}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1631133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $\int_{-\infty}^{\infty}\frac{dx}{\sqrt{x^4+1}}$ converges. Show that
$$
\int_{-\infty}^{\infty}\frac{dx}{\sqrt{x^4+1}}
$$
converges.
I recognized that that since the integrand is even then
$$
\int_{-\infty}^{\infty}\frac{dx}{\sqrt{x^4+1}} = 2\int_{0}^{\infty}\frac{dx}{\sqrt{x^4+1}}
$$
Then, I started with t... | In the regime $|x| \ge 1$, you can compare $(x^4+1)^{-1/2}$ against $x^{-2}$, but in the neighborhood of $0$, the latter is unbounded so you are best off noting that $$\int_{x=-\infty}^\infty (x^4+1)^{-1/2} \, dx = 2 \int_{x=0}^\infty (x^4+1)^{-1/2} \, dx < 2\left( \int_{x=0}^1 (x^4+1)^{-1/2} \, dx + \int_{x=1}^\infty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1631309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
How could we define the factorial of a matrix? Suppose I have a square matrix $\mathsf{A}$ with $\det \mathsf{A}\neq 0$.
How could we define the following operation? $$\mathsf{A}!$$
Maybe we could make some simple example, admitted it makes any sense, with
$$\mathsf{A} =
\left(\begin{matrix}
1 & 3 \\
2 & 1
\end{matr... | I use the well known (and simple) definitions
$$n!=\Gamma (n+1)$$
and
$$\Gamma (A+1)=\int_0^{\infty } \exp (-t) \exp (A \log (t)) \, dt$$
Now, if A is a (square) matrix, all we need is to define the exponential function for a matrix.
This can always be done (in principle) via the power series which only requires to ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1634488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "204",
"answer_count": 7,
"answer_id": 0
} |
If $ A=\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+.........+\frac{1}{100\sqrt{99}}\;,$ Then $\lfloor A \rfloor =$
If $\displaystyle A=\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+.........+\frac{1}{100\sqrt{99}}\;,$ Then $\lfloor A \rfloor =$
Where $\lfloor x \rfloor$ represent floor fu... | Let
$$A = \sum_{k=1}^{99} \frac1{(k+1) \sqrt{k}} $$
You can show that the sum is greater than $1$. First we can lower bound the sum by
$$\sum_{k=1}^{99} \frac1{k (k+1)} = \frac{99}{100} $$
which, while less than one, may be corrected by its second term:
$$\frac{99}{100} + \frac13 \left (\frac1{\sqrt{2}} - \frac12 \rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1634645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 2
} |
Prove inequality $1 < \frac{1}{n} + \frac{1}{n+1} + \ldots + \frac{1}{3n-1} < 2$ Prove the inequality $1 < \frac{1}{n} + \frac{1}{n+1} + \ldots + \frac{1}{3n-1} < 2$ For all $n \in \mathbb{N}$
I've done the right hand side, but can't do the left side of the inequality. For the right:
$\frac{1}{n} + \ldots + \frac{1}{3n... | Hint:
Show $\dfrac{1}{n} \lt \dfrac{1}{n+k}+\dfrac{1}{3n-1-k} \lt \dfrac{2}{n} $ for $0 \le k \lt n$
Then sum over $k$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1635728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
How do I solve the following equation: $z^4+z^3+z+1=0$ Is there an existing method to solve the following equation:
$z^4+z^3+z+1=0$?
| Hint:
$$z^4 + z^3 + z + 1 = z^4 + z + z^3 + 1 = z(z^3+1) + z^3 + 1 = (z+1)(z^3 + 1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1637779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Find all incongruent solutions to $21x \equiv 14 \pmod{91}$ Find all incongruent solutions to $21x \equiv 14 \pmod{91}$.
I am able to work out the solution using Euclidean algorithm techniques, but the signs on the expression do not match up with the initial expression when I check my work. So by the linear congruenc... | By definition, the congruence
$$21x \equiv 14 \pmod{91} \tag{1}$$
is equivalent to the equation
$$21x = 14 + 91t, t \in \mathbb{Z} \tag{2}$$
If we divide each term of equation 2 by $7$, we obtain the equivalent equation
$$3x = 2 + 13t, t \in \mathbb{Z}$$
which is equivalent to the congruence
$$3x \equiv 2 \pmod{13} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1638266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Reasoning Aptitude CSIR 15 A single-celled spherical organism contains $70$% water by volume. If it loses $10$% of its water content, how much would its surface area change by approximately?
*
*$3\text{%}$
*$5\text{%}$
*$6\text{%}$
*$7\text{%}$
| Let $r$ be the original radius the original volume of water is $$V_1=\frac{70}{100}\times \frac{4}{3}\pi r^3=\frac{14}{15}\pi r^3$$ & volume of organic material
$$V_2=\frac{30}{100}\times \frac{4}{3}\pi r^3=\frac{2}{5}\pi r^3$$
& surface area $S_0=4\pi r^2$
when it loses $10$% of water then remaining volume of water ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Basic algebra problem: $ \frac{\frac{1}{x}+\frac{1}{y}}{\frac{1}{x^2}-\frac{1}{y^2}} $ Basic algebra problem I can't seem to figure out: $$ \frac{\frac{1}{x}+\frac{1}{y}}{\frac{1}{x^2}-\frac{1}{y^2}} $$ $x,y \in \mathbb{R}, x^2 \neq y^2, xy\neq0$.
Now I know the result is: $\frac{xy}{y-x}$, but I am not sure how to get... | $$a=\frac1x+\frac1y=\frac{x+y}{xy}$$
$$b=\frac1{x^2}-\frac1{y^2}=\frac{y^2-x^2}{(xy)^2}$$
$$\frac{a}b=\frac{(x+y)xy}{x^2-y^2}=\frac{(x+y)xy}{(y-x)(y+x)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
$p = \sqrt{1+\sqrt{1+\sqrt{1 + \cdots}}}$; $\sum_{k=2}^{\infty}{\frac{\lfloor p^k \rceil}{2^k}} = ? $
Let $p = \sqrt{1+\sqrt{1+\sqrt{1 + \cdots}}}$
The sum $$\sum_{k=2}^{\infty}{\dfrac{\lfloor p^k \rceil}{2^k}}$$
Can be expressed as $\frac{a}{b}$. Where $\lfloor \cdot \rceil$ denotes the nearest integer function. ... | For $n \ge 2$, the numbers $\lfloor \phi^k \rceil$ are equal to the Lucas numbers, which satisfy the Fibonacci recurrence but in which $L_1=1$ and $L_2=3$. The Lucas numbers satisfy
$$L_k = \phi^k + (-1)^k \phi^{-k}$$
and thus have the generating function
$$L(x) = \frac{2-x}{1-x-x^2} $$
so the sum we want is
$$L \left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1644523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
proving an inequality related to $AM\ge GM$ $$a^2+ab+b^2\ge 3(a+b-1)$$
$a,b$ are real numbers
using $AM\ge GM$
I proved that
$$a^2+b^2+ab\ge 3ab$$
$$(a^2+b^2+ab)/3\ge 3ab$$
how do I prove that $$3ab\ge 3(a+b-1)$$
if I'm able to prove the above inequality then i'll be done
| Follow @BarryCipra's hint of letting $a=x+1$ and $b=y+1$:
$$(x+1)^2+(x+1)(y+1)+(y+1)^2 \geq 3((x+1)+(y+1)-1)$$
Expand the left side and simplify the right side:
$$x^2+2x+1+xy+x+y+1+y^2+2y+1 \geq 3(x+y+1)$$
Simplify the left side and distribute the $3$ on the right side:
$$x^2+xy+y^2+3x+3y+3 \geq 3x+3y+3$$
Subtract both... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1644694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that one of the lines represented by $ax^2+2hxy+by^2=0$ will bisect the angle between the coordinate axes if $(a+b)^2=4h^2$. Prove that one of the lines represented by $ax^2+2hxy+by^2=0$ will bisect the angle between the coordinate axes if $(a+b)^2=4h^2$.
Solution
I calculated the two lines represented by $ax^2+... | Let $y=mx$ be the required line. Then we can rewrite the given pair of lines as $$bm^2+2hm+a=0$$.
$$\implies m=\frac{-2h\pm\sqrt{4h^2-4ab}}{2b}$$
$$(a+b)^2=4h^2 \implies 4h^2-4ab=(a-b)^2$$
Substituting in the value of $m$, $$m=\frac{-\vert a+b\vert\pm\vert a-b\vert}{2b}$$
For all values of $a,b$, the above will a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
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.