Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Trouble in mapping of möbius transformation Question:-
Show that the transformation $$ w = \frac{2z+3}{z-4}$$ maps the circle $x^2+y^2-4x=0$ onto the straight line $4u+3=0$
My attempt:-
The circle $x^2+y^2-4x=0$ is $|z-2|=2$ . . .$(1)$
So the inverse mapping of the given bilinear transformation is:-
$$z= \frac{4w+3}{... | If $z= x+yi$ then equation of circle is $$|z-2|=2$$ Since $z= (4w+3)/(w-2)$ we get
$$\Big|{4w+3 -2w+4\over w-2}\Big| = 2$$ or $$|2w+7|= |2w-4|$$ dividing this by $2$ we get: $$|w-(-7/2)|=|w-2|$$
so $w$ is on perpendicular bisector between $-7/2$ and $2$ so $w=-{3\over 4}$ or $$4w+3=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3231935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Prove $\frac{b-a}{6}≤\sqrt{1+b} - \sqrt{1+a}≤\frac{b-a}{4}$ if $3\leq aI don't really know if I should use brute force or some kind of theorem, it comes on a calculus past exam and it says:
suppose: $3≤a<b≤8$
prove that $$\frac{b-a}{6}≤\sqrt{1+b} - \sqrt{1+a}≤\frac{b-a}{4}$$
| Introduce:
$$1+a=x^2,\ \ 1+b=y^2$$
Obviously:
$$2\le x<y\le3$$
Notice that:
$$4\le y + x \le 6\tag{1}$$
Inequality now becomes:
$$\frac{y^2-x^2}{6}\le y-x\le\frac{y^2-x^2}{4}$$
$$\frac{y+x}{6}\le 1\le\frac{y+x}{4}$$
...which is true because of (1).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3233155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Area of the base of the prism formed by three planes
The three planes $P_1: kx + y+ z=2$,$P_2:x+y-z=3$,$P_3: x+2z=2$ form a triangular prism and area of the normal section (where the normal section of the triangular prism is the plane parallel to the base of the triangular prism) be $k_1$. Then the value of $2\sqrt{14... | The three plane equations are $2x+y+z-2=0, x+y-z-3=0, x+2z-2=0$.
Consider the family of planes passing through the line of intersection of the first two planes.
$2x+y+x-2 + \lambda(x+y-z-3)=0$. The plane part of this family, and parallel to the third plane is:
$x+2z+1=0$.
The perpendicular distance between this plane a... | {
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"url": "https://math.stackexchange.com/questions/3233592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Mathematical induction proof $n^4-1$ is divisible by $16$ for all odd integers $n$ I'm stuck towards the end of proving this, here's my attempt:
$P(3) = 80/16 = 5$, True
$P(k) = k^4-1$
$P(k+1)= (k+1)^4-1$
Expanded $= k^4+4k^3+6k^2+4k+1-1$
This is where I am stuck at.
Sorry for the sloppy formatting im still reading how... | Without loss of generality we proved the statement for positive odd integers.
Note that a positive odd integer has the form of $2k-1$ where $k\ge 1$
We proceed with induction on k.
Let $P(k)$ be $(2k-1)^4-1$ is a multiple of $16$
For $k=1$, $(2k-1)^4-1=0$ which is a multiple of $16$ so $P(1)$ is true.
Assume that $(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3233758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How to calculate the derivative of $\int_0^x \left(\frac{1}{t}-[\frac{1}{t}]\right)dt$ at $x=0$? Let $F(x):=\int_0^x \left(\frac{1}{t}-[\frac{1}{t}]\right)dt$ ,where $[\frac{1}{t}]$ is the largest integer no more than $\frac{1}{t}$.Prove $F'(0)=\frac{1}{2}$.
I have tried in this way:
\begin{equation}
\begin{aligned}
\l... | I have a solution that utilises the digamma function:
\begin{align} F(\frac{1}{n}) &= \sum_{k=n}^\infty \int_{\frac{1}{k+1}}^{\frac{1}{k}} (\frac{1}{t} - k)dt = \\
&= \sum_{k=n}^\infty \int_{k}^{k+1} \frac{s - k}{s^2} ds = \\
&= \sum_{k=n}^\infty \int_{0}^{1} \frac{s}{(s+k)^2} ds = \\
&= \sum_{k=0}^\infty\int_{0}^{1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3234229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 3
} |
Cubic root of $3 \times 3$ identity matrix Is there a real $3 \times 3$ matrix $A$, such that: $A^3 = I_3$ and has at most one zero entry? If so, how can I find it?
| Since $A=(a_{ij})_{(3\times3)}$ satisfies $A^3-I=0$, so the characteristic equation of $A$ is $x^3-1=0$
Now $x^3-1=0\implies (x-1)(x^2+x+1)=0\implies\begin{vmatrix}
1-x & 0 & 0 \\
0 & -x & 1 \\
0 & -1 & -1-x
\end{vmatrix}=0$
Construct a matrix $A=\begin{pmatrix}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & -1 & -1
\end{pmatrix}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3235131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Calculate a Primitive Polynomial LFSR I tried to search on the internet, to read my course multiple times, but the only thing I see are definitions of the primitive polynomials for an LFSR.
I have an exercise:
Find the primitive polynomial of the LFSR of width 4 with longest possible period.
And I just don't know wher... | This is a problem that is small enough that we can do it by brute force. The quartic polynomial must have $x^4$ and $1$ terms in it, and so we need to look only at $8$ quartics. Of these,
$x^4 + 1, x^4+x^2+1$ can be eliminated as obvious squares, while $x^4+x^3+x^2+1$, $x^4+ x^3+x+1$, $x^4+x^2+x+1$ have an even number ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3236915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Calculating sum of converging series $\sum_{n=1}^{\infty}\frac{1-n}{9n^3-n}$ I have a trouble with calculating the sum of this series:
$$2+\sum_{n=1}^{\infty}\frac{1-n}{9n^3-n}$$
I tried to split it into three separate series like this:
$$2+\sum_{n=1}^{\infty}\frac{1-n}{9n^3-n} =2+\sum_{n=1}^{\infty}\frac{2}{3n+1}+\su... | For the direct evaluation of the limit, you have received the good solution from J.G.
You could also consider the partial sum using, as you did, partial fraction decomposition
$$\frac{1-n}{9n^3-n}=\frac{1}{3 n-1}+\frac{2}{3 n+1}-\frac{1}{n}$$ which makes
$$S_p=\sum_{n=1}^{p}\frac{1-n}{9n^3-n}=\frac{1}{3} \left(\psi \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3237476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
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Limit of powers of $3\times3$ matrix Consider the matrix
$$A = \begin{bmatrix} \frac{1}{2} &\frac{1}{2} & 0\\ 0& \frac{3}{4} & \frac{1}{4}\\ 0& \frac{1}{4} & \frac{3}{4} \end{bmatrix}$$
What is $\lim_{n→\infty}$$A^n$ ?
A)$\begin{bmatrix} 0 & 0 & 0\\ 0& 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$
B)$\begin{bmatrix} \frac{1}{4} &\f... | I’m lazy and prefer not to do tedious matrix inversions and multiplications if I can avoid it. Other answers have explained how to quickly eliminate the given possible solutions based on properties of Markov chains and their associated transition matrices, but one can also reason directly from the eigenvalues of the ma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Evaluate $\lim_{x \to 0}[\frac{\ln(x+\sqrt{1+x^2})}{x}]^{1/x^2}$. Let $\ln(x+\sqrt{1+x^2})=:y$，then $x=\dfrac{1}{2}(e^y-e^{-y}).$ Therefore
\begin{align*}
\lim_{x \to 0}\left[\frac{\ln(x+\sqrt{1+x^2})}{x}\right]^{\frac{1}{x^2}}&=\lim_{y \to 0}\left(\frac{2ye^y}{e^{2y}-1}\right)^{\frac{4}{e^{2y}+e^{-2y}-2}}\\
&=\lim_{y ... | Using $\ln(x + \sqrt{x^2 + 1}) = \sinh^{-1}(x) = x - \frac{x^3}{6} + \frac{3 x^5}{40} + \cdots$ then
$$\ln\left(\frac{\sinh^{-1}(x)}{x}\right) = - \frac{x^2}{6} + \frac{11 x^4}{180} - \frac{191 x^6}{5670} + \cdots$$
which leads to
\begin{align}
\lim_{x \to 0} \left(\frac{\sinh^{-1}(x)}{x}\right)^{1/x^2} &= \lim_{x \to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3238312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve for $x, y \in \mathbb R$: $x^2+y^2=2x^2y^2$ and $(x+y)(1+xy)=4x^2y^2$
Solve the following system of equations. $$\large \left\{ \begin{aligned} x^2 + y^2 &= 2x^2y^2\\ (x + y)(1 + xy) &= 4x^2y^2 \end{aligned} \right.$$
From the system of equations, we have that
$$\left\{ \begin{align*} (x + y)^2 \le 2(x^2 + y^2)... | These are symmetric equations, so denoting $s=x+y,p=xy$ we have
$$s^2-2p=2p^2$$
$$s(1+p)=4p^2$$
Now eliminate $s$:
$$s^2=2p^2+2p=\left(\frac{4p^2}{1+p}\right)^2$$
$$2p(1+p)^3=16p^4$$
If $p=0$ then $s=0$ and obviously $x=y=0$. Otherwise, divide by $p$:
$$(1+p)^3=(2p)^3$$
Assuming we are only solving in the reals:
$$1+p=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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IMO 1984: Prove that $0 ≤ yz + zx +xy −2xyz ≤ \frac {7}{27}$, where $x,y$ and $z$ are non-negative real numbers for which $x + y + z = 1.$ I tried to solve this inequality because I find exotic. Actually, I didn't look at the right solution. Because before I look at the right solution, I want to know if my solution i... | I checked your solution. Your solution is right.
I like the following way.
The homogenization helps.
By AM-GM
$$xy+xz+yz-2xyz=(x+y+z)(xy+xz+yz)-2xyz\geq9xyz-2xyz=7xyz\geq0.$$
Also, $$xy+xz+yz-2xyz\leq\frac{7}{27}$$ it's
$$(xy+xz+yz)(x+y+z)-2xyz\leq\frac{7}{27}(x+y+z)^3$$ or
$$\sum_{cyc}(7x^3-6x^2y-6x^2z+5xyz)\geq0,$$ w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Theorem of Pythagoras - Incorrect Derivation I am trying to solve below question from Coursera Intro to Calculus (link)
A right-angled triangle has shorter side lengths exactly $a^2-b^2$ and
$2ab$ units respectively, where $a$ and $b$ are positive real numbers
such that $a$ is greater than $b$. Find an exact expre... | $$a^4+2a^2b^2+b^4 = (a^2+b^2)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Closed form for $\sum_{n=1}^\infty \frac{4^n}{n^p\binom{2n}{n}}$ By Mathematica, we find $$\sum_{n=1}^\infty \frac{4^n}{n^3\binom{2n}{n}}=\pi^2\log(2)-\frac{7}{2}\zeta(3).$$
How to find the closed form for general series:
$$\sum_{n=1}^\infty \frac{4^n}{n^p\binom{2n}{n}}? \ \ (p\ge 3)$$
| Note that
$$\sum_{k=1}^\infty \frac{(4x)^n}{n^2{{2n}\choose n}}=2\arcsin^2(\sqrt{x}).$$
Hence, for $p=3$ we have the integral form
$$\sum\limits\limits_{n=1}^\infty \frac{4^n}{n^3\binom{2n}{n}}=\int_{0}^1\frac{2\arcsin^2(\sqrt{x})}{x}\,dx.$$
and you should be able to recover the result $\pi^2\ln(2)-\frac{7}{2}\zeta(3)... | {
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"timestamp": "2023-03-29T00:00:00",
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Calculate the minimum value of $\sum_\mathrm{cyc}\frac{a^2}{b + c}$ where $a, b, c > 0$ and $\sum_\mathrm{cyc}\sqrt{a^2 + b^2} = 1$.
$a$, $b$ and $c$ are positives such that $\sqrt{a^2 + b^2} + \sqrt{b^2 + c^2} + \sqrt{c^2 + a^2} = 1$. Calculate the minimum value of $$\frac{a^2}{b + c} + \frac{b^2}{c + a} + \frac{c^2}... | By rearrangement inequality,
$$\sum_{cyc} \frac{a^2}{b+c} \ge \sum_{cyc} \frac{b^2}{b+c}.\tag{1}$$
So
$$2\sum_{cyc}\frac{a^2}{b+c} \ge \sum_{cyc}\frac{a^2+b^2}{b+c}.\tag{2}$$
Now, by C-S
$$\sum_{cyc}(b+c) \sum_{cyc}\frac{a^2+b^2}{b+c} \ge \left(\sum\sqrt{a^2+b^2}\right)^2.\tag{3}$$
It's also clear that
$$\sum_{cyc}(b+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3250563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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nth root of $S_n=\frac{n^n}{(n+1)(n+2)...(n+n)}$ Let $S_n=\frac{n^n}{(n+1)(n+2)...(n+n)}, n\geq 1$, then $S_n^{\frac{1}{n}}$ converges to
*
*$e/2$
*$e/4$
*$e/8$
*$0$
Clearly $S_n^{\frac{1}{n}}=\frac{n}{(n+1)^{\frac{1}{n}}(n+2)^{\frac{1}{n}}...(n+n)^{\frac{1}{n}}}$, then how can we proceed
| We have $$(n+1)^n \lt (n+1)(n+2)...(n+n) \lt(2n)^n$$
and therefore
$$\frac12=\frac{n}{2n}\le S_n^{\frac{1}{n}}\le \frac{n}{n+1}\le 1$$
and further
$$\frac12 \le \lim\limits_{n\to 1}S_n^{\frac{1}{n}}\le 1$$
if that limit exists.
Only one of the proposed solution satisfies these inequalities. From $e=2.7\ldots$ follows $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3252679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Do I have a chance to get a closed form for this integral? I conjecture that
$$\int_{-z}^z \frac{\sin \left(\frac \pi {2z} x + \frac \pi 2 \right)} z \int_z^\infty \exp\left({- \frac {(y-x)^2 \pi^2}{16}}\right) \,d y d x \sim \frac 1 {z^2}$$
when $z\to \infty$.
Maybe there should be also equality.
| Changing $x=zt, y=zt+u$, on can express
\begin{align}
I&=\int_{-z}^z \frac{\sin \left(\frac \pi {2z} x + \frac \pi 2 \right)} z \int_z^\infty \exp\left({- \frac {(y-x)^2 \pi^2}{16}}\right) \,d y d x\\
&=\int_{-1}^1\cos\frac{\pi t}{2}\,dt\int_{z(1-t)}^\infty \exp\left( -\frac{\pi^2}{16} u^2 \right)\,du
\end{align}
It ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3253067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the sum of $\sum_{n=1}^\infty \frac{1}{n(n+2)}x^n$ $$S(x)=\sum_{n=1}^\infty \frac{1}{n(n+2)}x^n$$
The question is divided into three parts:
1. Determine its radius of convergence
2. By using the power series of $\frac{1}{1-x}$, show that for all x $\in$ ]-1,1[ , we get $\ln(1-x)=-\sum_{n}^\infty \frac{1}{n}x^n$
3.... | $$-\log(1-x)=\sum_{n=1}^{\infty}\frac{x^n}{n}$$
multiply by $x$
$$-x\log(1-x)=\sum_{n=1}^{\infty}\frac{x^{n+1}}{n}$$
$$\int_{0}^{x}-x\log(1-x)dx=\int_{0}^{x}\sum_{n=1}^{\infty}\frac{x^{n+1}}{n}dx$$
$$\frac{x(x+2)}{4}-\frac{(x^2-1)}{2}\log(1-x)=\sum_{n=1}^{\infty}\frac{x^{n+2}}{n(n+2)}$$
divide by $x^2$
$$\frac{x(x+2)}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3253257",
"timestamp": "2023-03-29T00:00:00",
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Find maximum and minimum of $\sin x + \sin y$ I am working on my scholarship exam practice but I am stuck on finding the minimum. Pre-university maths background is assumed.
When $x + y = \frac{2\pi}{3}, x\geq0, y\geq0$, the maximum of
$\sin x+\sin y$ is ....., and the minimum of that is .....
Let me walk you thro... | Since $f(x)=\sin{x}$ is a concave function on $\left[0,\frac{2\pi}{3}\right]$ and $\left(\frac{2\pi}{3},0\right)\succ(x,y),$ where $x\geq y,$
by Karamata we obtain:
$$\sin{x}+\sin{y}\geq\sin(x+y)+\sin0=\frac{\sqrt3}{2}.$$
The equality occurs for $x=\frac{2\pi}{3}$ and $y=0$, which says that we got a minimal value.
The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3263271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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$\sqrt{2} x^2 - \sqrt{3} x +k=0$ with solutions $\sin\theta$ and $\cos\theta$, find k
If the equation $\sqrt{2} x^2 - \sqrt{3} x +k=0$ with $k$ a constant has two solutions $\sin\theta$ and $\cos\theta$ $(0\leq\theta\leq\frac{\pi}{2})$, then $k=$……
My approach is suggested below but I am not sure how to continue.
Sin... | I'll use $y$ in place of $\theta$ . We know that sum of roots of $ax^2+bx+c$ is $\frac{-b}{a}$
$$siny+cosy=\frac{\sqrt{3}}{\sqrt{2}}$$ also $siny+cosy=\sqrt{2}sin(y+\frac{\pi}{4})$ thus $sin(y+\frac{\pi}{4})=\frac{\sqrt{3}}{2}=sin(\frac{\pi}{3})=sin(\frac{2\pi}{3})$ thus $y=\frac{\pi}{12}$ or $\frac{5\pi}{12}$ thus $si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3264274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Given remainders from other polynomial divisions. Find the remainder in a polynomial division Let $F(x)$ be a polynomial.
If $F(x)$ is divided by $(x-1)^2$ the remainder will be $x+1$
and if $F(x)$ is divided by $x^2$ the remainder will be $2x+3$.
What is the remainder if $F(x)$ is divided by $x^2(x-1)$?
My solution ... | $\!\!\overbrace{\bmod x\!-\!1\!:}^{\large x\ \equiv\ 1}\,\ \ f\equiv \overbrace{1\!+\!x}^{\Large 2} \equiv \overbrace{2x\!+\!3\! +\! a x^2}^{\Large 2\ +\ 3\ +\ a\ \ \ }\!\iff\! a=-3\!\iff\! f\equiv\overbrace{ 2x\!+\!3\!-\!3x^2}^{\large\rm your\ answer\ \checkmark}\!\!\pmod{\!x^2(x\!-\!1)}$
Update You mention $\!\bmod\!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3265943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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How do I solve quadratic double inequalities? I have two questions involving quadratic double inequalities.
Firstly, what are the steps to get the solution for the following?
$0\le(x+2)^2\le4$
My my thought was to separate the inequality into
$0\le(x+2)^2$ and $(x+2)^2\le4$
Which would then allow me to take the squa... | For real $x,$
$$(x+2)^2\ge0$$
So, the problem reduces to $$(x+2)^2\le4\iff x(x+4)\le0$$
$\implies$
either
$x\ge0$ and $x+4\le0\iff 0\le x\le-4$ which is impossible
or $x\le0$ and $x+4\ge0\implies -4\le x\le0 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3266093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Checking my proof of finding a limit I'm trying to find all the $a\in \mathbb R$ values which we get the following limit:
$$L=\lim_{x\to a} \frac{x^2+ax-2a}{(x^2-1)(x+a)}$$
to be finite. I don't see any problem just inserting $a$ into the limit so we get $\frac{1}{a+1}$ so for $a\neq -1$ we get a finite limit. Is it th... | In this case, as long as we don't have $\frac{0}{0}$ or $\frac{\infty}{\infty}$, yes it is that easy and inserting $a$ is enough. And the answer becomes $a \in \mathbb{R}-\{-1\}$.
EDIT: By the warning from fleablood, we should also consider the case where we can have $\frac{0}{0}$. Here, note that in order for nominato... | {
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"url": "https://math.stackexchange.com/questions/3267679",
"timestamp": "2023-03-29T00:00:00",
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} |
How to calculate the sum of a series, whose next value is dependent on the previous value? Consider the below scenario
A -> Amount ; C->Constant; V1 -> Value of first term ; V2 -> Value of 2nd term and so on up to n terms
Now, I need to get the sum of the below series up to n terms. For e.g.
A*C + (A-V1)*C + (A-V2)*C +... | Noting that:
$$
\begin{aligned}
& \frac{V_0}{A\cdot C} = \frac{0}{A \cdot C} = 0 \\
& \frac{V_1}{A\cdot C} = \frac{\left(A - V_0\right)\cdot C}{A \cdot C} = 1 \\
& \frac{V_2}{A\cdot C} = \frac{\left(A - V_1\right)\cdot C}{A \cdot C} = 1 - C \\
& \frac{V_3}{A\cdot C} = \frac{\left(A - V_2\right)\cdot C}{A \cdot C} = 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3268419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding $\lim_{n\rightarrow\infty }\frac{\frac{1}{2}+\frac{\sqrt 2}{3}+\dots+\frac{\sqrt n}{n+1}}{\sqrt n}$
Find $$\lim_{n\rightarrow\infty }\frac{\frac{1}{2}+\frac{\sqrt 2}{3}+\dots+\frac{\sqrt n}{n+1}}{\sqrt n}$$
My work
$$\lim_{n\rightarrow\infty }\frac{\frac{1}{2}+\frac{\sqrt 2}{3}+\dots+\frac{\sqrt n}{n+1}}{\sq... |
The series
$$\sum_{m=1}^\infty \frac{\sqrt m}{m+1}$$
does not converge so can I say $\lim_{n\rightarrow\infty }\frac{\frac{1}{2}+\frac{\sqrt 2}{3}+\dots+\frac{\sqrt n}{n+1}}{\sqrt n}$ does not exist?
Because Zacky deleted his answer, I'll repeat the useful observation that divergence of a numerator does not mea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3269494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Exploring an inequality between $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} $ and $\frac{3}{1+(abc)^{1/3}}$ if $a,b, c>0$ An AM-HM inequality for three positive numbers leads to
$$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} \ge \frac{3}{{1+\frac{a+b+c}{3}}}. ~~~~(1)$$
Next, the well known AM-GM inequality
$$\frac{a+b+c}... | If $f''(x)>0$ in a domain $D$, the Jensen's inequality claims that the mean of functions is greater or equal to the function of mean:
$$\frac{f(x)+f(y)+f(z))}{3} \ge f\left(\frac{x+y+z}{3} \right) , \forall ~x,y,z \in D.~~~~(1)$$ The sign of the inequality reverses if $f''(x) < 0, \forall x \in D$.
Consider $f(x)=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3272376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Radical equation solve $\sqrt{3x+7}-\sqrt{x+2}=1$. Cannot arrive at solution $x=-2$ I am to solve $\sqrt{3x+7}-\sqrt{x+2}=1$ and the solution is provided as -2.
Since this is a radical equation with 2 radicals, I followed suggested textbook steps of isolating each radical and squaring:
$\sqrt{3x+7}-\sqrt{x+2}=1$
$(3x+... | Let $\sqrt{3x+7}=a,\sqrt{x+2}=b$
$\implies a,b\ge0$ and $a-b=1$
and $a^2-3b^2=1$
Or $(b+1)^2-3b^2=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3273876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 1
} |
Proofs involving strict inequalities let $a,b \in R$
Prove that if $3 \lt a \lt 5$ and $b= 2 + \sqrt{a-2}$ then, $3 \lt b \lt a$
My approach was simply to start with the first inequality and transform it into b and see what happens.
Subtracting 2 gives: $1 \lt a-2 \lt 3$
Taking the square root gives: $1 \lt \sqrt{a-2} ... | You are asked to prove that if $3<a<5$ and $b=2+\sqrt{a-2}$ then $3<b<a$.
You proved $3<b$ (and $b<\sqrt3+2$), but you didn't prove $b<a$.
$b<a$ means $\sqrt{a-2}+2<a$ or $\sqrt{a-2}<a-2$.
This suggests letting $x=a-2$ and proving $\sqrt x<x$.
Well, if $x>1$, which is true since $a>3$, then $x^2>x>0$ so $x>\sqrt x$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3274468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solve the limit $\lim\limits_{x\to 0}\left(\frac {e^x}{x}-\frac {1}{\arctan{x}}\right)$ without using L'Hopital (with my attempts) Help me solve this limit, using simple limit work: $$\lim_{x\to 0}\left(\frac {e^x}{x}-\frac {1}{\arctan{x}}\right)$$
I tried extracting $\frac {e^x}{x}$ but that was dead end, then $\frac ... | Everything is locally a power series, so,
as $x \to 0$,
$\begin{array}\\
\left(\frac {e^x}{x}-\frac {1}{\arctan{x}}\right)
&=\left(\frac {1+x+O(x^2)}{x}-\frac {1}{x-x^3/3+O(x^5)}\right)\\
&=\left(\frac {(1+x+O(x^2))(1-x^2/3+O(x^4))-1}{x(1-x^2/3+O(x^4))}\right)\\
&=\left(\frac {(1+x+O(x^2))-1}{x(1+O(x^2))}\right)\\
&=\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3278120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Integrating quadratics in denominator I'm following a book on Calculus that introduces partial fraction expansion. They discuss common outcomes of the partial fraction expansion, for example that we are left with an integral of the form:
$$
\int \frac{dx}{x^2+bx+c}
$$
And then we can use complete the square and $u$-sub... | Check your book properly. For them to make the statement
...this is possible because $4c-b^2>0$
must mean that they have said somewhere above that they are considering this method subject to that restriction.
When such discussions are made, they're usually split into cases.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to factor a fourth degree polynomial I'm working on a math problem but I am having a hard time figuring out the method used by my textbook to make this factorization:
$$x^4 + 10x^3 + 39x^2 + 70x + 50 = (x^2 + 4x + 5)(x^2 + 6x + 10)$$
I've tried to see if this equation can be factored by grouping or by long division... | Let
$$f (x) = x^4 + 10 x^3 + 39 x^2 + 70 x + 50$$
Converting to a depressed quartic see here, we see that the $x$ term drops out as well
$$f\left(x-\frac{5}{2}\right)=x^4+\frac{3 x^2}{2}+\frac{25}{16}=\left(x^2+\frac{5}{4}\right)^2-x^2=\left(x^2+\frac{5}{4}+x\right)\left(x^2+\frac{5}{4}-x\right)$$
The factorization of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3282632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 7,
"answer_id": 3
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Prove that the diophantine equation $2x^2-5y^2=7$ has no integer solutions. My attempt: I rewrote it as $2x^2=5y^2+7. 2x^2$ is always even, so in order for the RHS to be even, this means that $5y^2$ must be odd since an odd number plus $7$ is even.
If I evaluate when y is odd, so if $y=2k+1$ for some integer $k$, I ge... | We can write the equation in this form too:
$$2(x^2+y^2) = 7(y^2+1)$$
This means that $7 | x^2+y^2$, also its easy to show that $\gcd(x, y) = 1$, becouse if $\gcd(x, y) = d > 1$, then $d | 7$. Now suppose that $d = 7$, then for $x=7a$ and $y=7b$:
$$2x^2 - 5y^2 = 2\times 49 \times a^2 - 5 \times 49 \times b^2 = 49(2a^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3283559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Ratio of area of intersection of two circles and enclosing rectangle Consider two circles with radius $r$ with distance $d < 2r$ from their centres. The area of their intersection is given by:
$A = r^2\cos^{-1}\left(\frac{d^2} {2dr}\right) + r^2\cos^{-1}\left(\frac{d^2} {2dr}\right) -
\frac{1}{2}\sqrt{(-d+2r)(d^2)(d+2... | The area of a circular sector is given by:
$$A_{\text{circular sector}} = \frac{r^2}{2} (\theta - \sin(\theta))$$
The intersection of two circles is just twice this:
$$A_{\text{intersection}} = r^2 (\theta - \sin(\theta))$$
The height of the rectangle can be found using simple trigonometry:
$$h = 2r \sin\left(\frac{\th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3283658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
A finite product : $\prod_{k=0}^{n-1}(1-\frac{1}{n-1+k})$ Find the maximum and minimum of the following products :
$A)$ $\prod_{k=0}^{n-1}(1-\frac{1}{n-1+k})$
$B)$ $\prod_{k=0}^{n-1}(1-\frac{1}{n+1-k})$
My idea is :
$n-1+k>k$ then : $\frac{1}{n-1+k}<\frac{1}{k}$
We obtain :
$\prod_{k=0}^{n-1}(1-\frac{1}{k})$
But I... | Let us note that for $a,b\in\mathbb N$, $2\le a \le b$ we have $$ \prod_{m=a}^b m = \frac{\prod_{m=1}^b m}{\prod_{m=1}^{a-1} m} = \frac{b!}{(a-1)!}$$
For the product A) we have
$$ \prod_{k=0}^{n-1} \left(1-\frac{1}{n-1+k}\right) = \prod_{k=0}^{n-1}\frac{n-2+k}{n-1+k} = \frac{\prod_{k=0}^{n-1}(n-2+k)}{\prod_{k=0}^{n-1}(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3285778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Shortest distance around a pyramid
Transcript:
The diagram shows a square based pyramid with base PQRS and vertex O. All the edges are length 20 meters. Find the shortest distance, in meters, along the outer surface of the pyramid from P to the midpoint of OR.
The only way I have been able to solve this question is ... | Draw the black and blue lines as shown below:
$\hspace{3cm}$
The length of blue line (using Cosine theorem):
$$c+d=\sqrt{20^2+y^2-2\cdot 20y\cos 60^\circ}+\sqrt{10^2+y^2-2\cdot 10y\cos 60^\circ}=\\
\sqrt{y^2-20y+400}+\sqrt{y^2-10y+100}$$
Set its derivative to zero:
$$(c+d)'=\frac{y-10}{\sqrt{y^2-20y+400}}+\frac{y-5}{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3289765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 5,
"answer_id": 4
} |
Unsure about solution to $\int_0^\frac{\pi}{4} \frac{\ln\left| 1 + \tan(x)\right|}{\left( 1 + \tan(x)\right)^n}\:dx$ Spurred on by a question posed on MSE, I was hoping to resolve the following definite integral:
\begin{equation}
I_n = \int_0^\frac{\pi}{4} \frac{\ln\left| 1 + \tan(x)\right|}{\left( 1 + \tan(x)\right)^n... | This is not an answer but a different method to find a closed form of ${{I}_{n}}$
For $x={{\tan }^{-1}}\left( u \right)$ we have:
$$
{{I}_{n}}=\int_{0}^{1}{\frac{\ln \left( 1+u \right)}{{{\left( 1+u \right)}^{n}}\left( 1+{{u}^{2}} \right)}du}
$$
Now using this result (thanks to Sangchul Lee) :
$$
\frac{1}{{{(1+x)}^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3290624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
minimum value of of $(5+x)(5+y)$
If $x^2+xy+y^2=3$ and $x,y\in \mathbb{R}.$
Then find the minimum value of $(5+x)(5+y)$
What I try
$$(5+x)(5+y)=25+5(x+y)+xy$$
$x^2+xy+y^2=3\Rightarrow (x+y)^2-3=xy$
I am finding $f(x,y)=22+5(x+y)+(x+y)^2$
How do I solve it? Help me please
| The feasible set is a (compact) ellipse $E$ in the $(x,y)$-plane. We have to consider the Lagrangian
$$\Phi:=(x+5)(y+5)-\lambda(x^2+xy+y^2-3)\ .$$
The equations
$$\Phi_x=y+5-\lambda(2x+y)=0,\qquad\Phi_y=x+5-\lambda(x+2y)=0\ ,$$
or
$$\left\{\eqalign{2\lambda x+(\lambda-1)y&=5 \cr (\lambda-1)x+2\lambda y&=5\cr}\right.$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3291916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
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Let $x, y, z$ be real numbers. If $x + y + z = 1$ and $x^2 + y^2 + z^2 = 1$, then what is the minimum value of $x^3 + y^3 + z^3$? Through manipulation, I got as far as $x^3 +y^3 + z^3 = 1 + 3xyz$
I tried solving it using AM - GM inequality but it only gave me the maximum value.
GM - HM does not to help either.
| Since you got that $x^3+y^3+z^3=1+3xyz$, easy to understand that you meant $x^2+y^2+z^2=1,$ which gives $xy+xz+yz=0.$
Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, $u=\frac{1}{3},$ $v^2=0$ and we obtain: $$0\leq\prod_{cyc}(x-y)^2=27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)=27\left(-\frac{4}{27}w^3-w^6\right),$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3292578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Proving an integral equation
Question:
I have had a hard time solving and understanding the solution to this problem.
Given Solution:
Questions like why (and how) this step:
Or why $5I_{3}$ = $8I_{1}$
Have been puzzling me. Can anyone please give a clear answer to why and how each step is done. Thanks!
| The point is that, since
$$
\frac{d}{dx}\left[x^n(4-x^2)^{3/2}\right] = 4n x^{n-1}\sqrt{4-x^2} - (n+3)x^{n+1}\sqrt{4-x^2},
$$
and the fudamental theorem of calculus tells us that
$$
\int_0^2\frac{d}{dx}\left[x^n(4-x^2)^{3/2}\right]dx = 2^n(4-2^2)^{3/2}-0^n(4-2^2)^{3/2} = 0,
$$
we must have that
$$
\int_0^2\left[4n x^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3293448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find integers $x$ and $y$ such that $|5^x - 2^y| = 1$.
Find integers $x$ and $y$ such that $$\large |5^x - 2^y| = 1$$
Below is a graph of the equation $|5^x - 2^y| = 1$. As can clearly be seen, $(0, 1)$ and $(1, 2)$ are the solutions. I don't know the other answers, maybe there could be none.
| $$5^x - 2^y = 1$$
If $y=1$, then there is no integer $x$ such that $5^x=3$.
If $y=2$, then $5^x=5$ implies $x=1$.
If $y\ge 3$, then$$5^x\equiv 1\pmod 8$$
So, we see that $x$ has to be even.
Then, we have
$$1-(-1)^y\equiv 1\pmod 3\implies (-1)^y\equiv 0\pmod 3$$
which is impossible.
$$5^x - 2^y = -1$$
If $y=1$, then th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3293892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solving the following limit without L'Hospital's rule: $\lim_{x\to 0} \frac{\sin(x^2+2)-\sin(x+2)}{x} $
I have been trying to solve the following limit $$\lim_{x\to 0} \frac{\sin(x^2+2)-\sin(x+2)}{x}.$$
I came across the right answer as shown by the steps below, but I would to check if the steps are correct or if so... | First notice that
$$\lim_{x\to 0} \frac{\cos x - 1}{x} = \lim_{x\to 0} \frac{2\sin^2\frac{x}2}{x} = \lim_{x\to 0} \left(\frac{\sin^2\frac{x}2}{\frac{x}2}\right)\cdot\sin\frac{x}2 = 1 \cdot 0 = 0$$
so
$$\lim_{x\to 0}\frac{ \cos{x^2}-\cos{x}}{x} = \lim_{x\to 0} \left(\frac{\cos x^2-1}{x^2}\right)x + \lim_{x\to 0} \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3294127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
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$ \int_0^\frac{\pi}{2}\ln^n\left(\tan(x)\right)\:dx$ I'm currently working on a definite integral and am hoping to find alternative methods to evaluate. Here I will to address the integral:
\begin{equation}
I_n = \int_0^\frac{\pi}{2}\ln^n\left(\tan(x)\right)\:dx
\end{equation}
Where $n \in \mathbb{N}$. We first observe... | $$\begin{split}
I_{2k} &= \int_0^\infty\frac{\ln^{2k}u}{u^2 + 1}du \\
&= \int_0^1\frac{\ln^{2k}u}{u^2 + 1}du +\int_1^{+\infty}\frac{\ln^{2k}\left(u\right)}{u^2 + 1}du \\
&=\int_0^1\frac{\ln^{2k}u}{u^2 + 1}du +\int_0^{1}\frac{\ln^{2k}t}{t^2 + 1}dt \,\,\,\left(\text{by } u\rightarrow \frac
1 t\right)\\
&=2\int_0^1\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3294446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 0
} |
Given three positive numbers $a,b,c\in R_{+}^{*}$ prove the following inequality
If $\ abc=1$ then prove that
$$\sum_{cyc}\frac{1}{3-a+a^{6}}≤1$$
where $a,b,c>0$
I think this inequality can be proved by holder ?
My attempt using $\ am-gm$
$$3-a+a^{6}≥3-a \quad(etc)$$
$$\sum_{cyc}\frac{1}{3-a+a^{6}}≤\displayst... | $$\sum_{cyc}\frac{1}{a^6-a+3}\leq1$$ it's
$$\sum_{cyc}\left(\frac{1}{3}-\frac{1}{a^6-a+3}\right)\geq0$$ or
$$\sum_{cyc}\left(\frac{1}{3}-\frac{1}{a^6-a+3}-\frac{5}{9}\ln{a}\right)\geq0.$$
Let $f(x)=\frac{1}{3}-\frac{1}{x^6-x+3}-\frac{5}{9}\ln{x}.$
Thus, $$f'(x)=\tfrac{(1-x)(5x^{11}+5x^{10}+5x^9+5x^8+5x^7-5x^6-29x^5-29x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3295529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Given positive $a, b, c$, prove that $(a^2 + b^2 + c^2)^3 \ge (a^3 + b^3 + c^3)(ab + bc + ca)(a + b + c)$.
Given positive $a, b, c$, prove that $$\large (a^2 + b^2 + c^2)^3 \ge (a^3 + b^3 + c^3)(ab + bc + ca)(a + b + c)$$
As a starting point,
$$(a^2 + b^2 + c^2)^3 \ge (a^3 + b^3 + c^3)(ab + bc + ca)(a + b + c)$$
$$\i... | Your SOS works!
Indeed, by your work we obtain:
$$(a^2+b^2+c^2)^3-(a^3+b^3+c^3)(ab+ac+bc)(a+b+c)=$$
$$=(a^2+b^2+c^2)^2(ab+ac+bc)\left(\frac{a^2+b^2+c^2}{ab+ac+bc}-\frac{(a^3+b^3+c^3)(a+b+c)}{(a^2+b^2+c^2)^2}\right)=$$
$$=\frac{1}{2}\sum_{cyc}(a-b)^2((a^2+b^2+c^2)^2-2ab(ab+ac+bc))\geq$$
$$\geq\frac{1}{2}\sum_{cyc}(a-b)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3295819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Sum of an Infinite Sequence
If $$ S = \frac{1}{1\cdot3\cdot5} + \frac{1}{3\cdot 5 \cdot 7} +
\frac{1}{5\cdot 7 \cdot 9} \cdots $$
$$S =\, ? $$
My Attempt
Let the general term be $ a_n $.
Then, $$ a_n = \frac{1}{(2n-1)(2n+1)(2n+3)} $$
$$ a_n = \frac{1}{4} \left[\frac{(2n+3)-(2n-1)}{(2n-1)(2n+1)(2n+3)} \right]$$
$$... | You are right but more methodically:
For the series $$T_n=\frac{1}{(2n-1)(2n+1)(2n+3)}$$
Let $$V_n=\frac{1}{(2n-1)(2n+1)},$$ then $$T_n=\frac{1}{4}[V_{n}-V_{n+1}].$$ By telescopic summing we get $$S_N=\sum_{n=1}^{N}=\frac{1}{4} [V_1-V_{N+1}]=\frac{1}{4} \left[\frac{1}{1.3}-\frac{1}{(2N+1)(2N+3)}\right].$$ Hence $$S_{\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3297033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Determine polynomial of $3$ degree real coefficients $f(x)$ such that $f(x) \vdots (x-2)$ and $f(x)$ dividing by $x^2-1$ have remainder $2x$ I need help in this problem.
Problem: Determine polynomial of $3$ degree real coefficients $f(x)$ such that $f(x) \vdots (x-2)$ and $f(x)$ divided by $x^2-1$ has remainder $2x$ a... | Let $f(x) = ax^3 + bx^2 + cx + d$
and also $f(x) = (px+q)(x^2-1) + 2x$
Comparing the coefficients of $f(x)$, we have
$a = p$, $b = q$, $c = 2 - p$, $d = -q$
Also $f(2) = 0 \implies 8a + 4b + 2c + d = 0$
Substituting the values of $a, b, c$ in this equation and simplifying
$6p + 3q + 4 = 0$
You can find infinite number ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find modulus and argument of $\zeta = {\frac {1-\cos4(\theta) + i \sin4(\theta)} {\sin2(\theta)+ 2i\cos^2(\theta)}}$ A guideline on which identity to use would be greatly appreciated, as $1-\cos2A=2\sin^2A$ identity isn't giving me the correct answer I think.
Given that:
$$\zeta = {\frac {1-\cos4(\theta) + i \sin4(\th... | Note that\begin{align}\bigl(1-\cos(4\theta)\bigr)^2+\sin^2(4\theta)&=2-2\cos(4\theta)\\&=2-2\bigl(1-2\sin^2(2\theta)\bigr)\\&=4\sin^2(2\theta)\\&=16\sin^2(\theta)\cos^2(\theta)\end{align}and that\begin{align}\sin^2(2\theta)+4\cos^4(\theta)&=4\sin^2(\theta)\cos^2(\theta)+4\bigl(1-\sin^2(\theta)\bigr)^2\\&=4\sin^2(\theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3298918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Finding a closed form for recurrence relations $a_n=na_{n-1}+1$ and $a_n=na_{n-1}+n$ Consider the sequence defined by
$$
\begin{cases}
a_0=1\\
a_n=n\cdot a_{n-1}+1 & \text{if }n\ge 1
\end{cases}
$$
Find a closed form for $a_n$.
Second case is as follows:
$$
\begin{cases}
a_0=1\\
a_n=n\cdot a_{n-1}+n & \text{if }n\ge 1
... | Try exponential generating functions (and another link here). For example, let's look at the first recurrence
$$f(x)=\sum\limits_{n=0}\color{red}{a_n}\frac{x^n}{n!}=
1+\sum\limits_{n=1}a_n\frac{x^n}{n!}=\\
1+\sum\limits_{n=1}(n\cdot a_{n-1} + 1)\frac{x^n}{n!}=
1+\sum\limits_{n=1}a_{n-1} \frac{x^n}{(n-1)!} + \sum\limits... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve inequality Given positive numbers $a,b,c$ satisfying $a^2+b^2+c^2=1$, prove the following inequality
$$\frac{a}{\sqrt{1-bc}} + \frac{b}{\sqrt{1-ac}}+\frac{c}{\sqrt{1-ab}}\le\frac{3}{\sqrt{2}}$$
Thanks
I have tried using CS, try to make use of $a+b+c\leq\sqrt3$, $abc\leq\frac{1}{3\sqrt3}$, but got nowhere –
| By C-S
$$\sum_{cyc}\frac{a}{\sqrt{1-bc}}\leq\sqrt{\sum_{cyc}a\sum_{cyc}\frac{a}{1-bc}}.$$
Thus, it's enough to prove that
$$\sum_{cyc}\frac{a}{1-bc}\leq\frac{9}{2(a+b+c)},$$ which is true by SOS:
$$\frac{9}{2(a+b+c)}-\sum_{cyc}\frac{a}{1-bc}=\sum_{cyc}\left(\frac{3}{2(a+b+c)}-\frac{a}{1-bc}\right)=$$
$$=\frac{1}{2(a+b+... | {
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"url": "https://math.stackexchange.com/questions/3301091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Number of trees $T = (V,E) \;$ such that $V = \left \{ 1,2,3,4,5,6,7 \right \} \;$, $\deg(1) = 2$ , $\deg(t)<5 \;\forall\; 2\leq t\leq 7$ I tried summing all of the cases (I found 4), but somehow i'm getting the wrong answer.
For ex. the number of trees on 7 vertices such that the following sequence is the sequence of ... | The unlabelled trees with $7$ vertices are
We have to rule out the last three which have a vertex of degree $\geq 5$ or no vertex of degree $2$.
So for each of the $8$ cases we compute the number of labelled trees such that
$\deg(1) = 2$ and $\deg(t)<5 \;\forall\; 2\leq t\leq 7$ and we add them all together:
$$5\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3302554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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$\omega = \frac{1 + \sqrt3 i}{2}$ , $ \omega^5 = ? $
$\omega = \frac{1 + \sqrt3 i}{2} $, $ \omega^5 = ? $
$\omega^3 = 1$ by definition?
So, $\omega^5 = \omega^2$
But why do i get wrong answer?
| Not by definition, just check that
$$\omega=\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i=(\cos(\frac{\pi}{3})+i\sin(\frac{\pi}{3}))=e^{i\frac{\pi}{3}}$$
If you make $\omega^3$, then you have $e^{i\frac{\pi}{3}\cdot 3}=e^{i\pi}=-1$.
So $\omega^5=\omega^3\omega^2=-\omega^2$ is your mistake
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3303770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Minimizing $9 \sec^2{x} + 16 \csc^2{x}$ Find the minimum value of $$9 \sec^2{x} + 16 \csc^2{x}$$
My turn :
Using AM-GM
$$9\sec^2{x} + 16\csc^2{x} \geq 2 \sqrt{144 \sec^2{x} \csc^2{x}}$$
$$9 \sec^2{x} + 16\csc^2{x} \geq 24 \sec{x} \csc{x} $$
But the equality sign holds iff
$$9 \sec^2{x} = 16\csc^2{x}$$
Then $$ \tan{x} =... | Your mistake in the following.
You proved that $$\frac{9}{\cos^2x}+\frac{16}{\sin^2x}\geq\frac{24}{|\sin{x}\cos{x}|}=\frac{48}{|\sin2x|},$$
but you did not find a minimal value.
After your first step we see that $$\frac{48}{|\sin2x|}\geq48,$$ but it does not give a minimal value because the value $48$ does not occur.
O... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding a function with given conditions I want to find the equation of a function $f(x)$ such that the following conditions are met:
$(i)\ f(3) = 0, f(0) =1$.
$(ii)\ f(x)$ is an even function.
$(iii)\ f(x)$ has vertical asymptodes at $x=\pm 4$.
$(iv)\ f(x)$ has horizontal asymptode at $ y=2$.
Now since the function is... | I am assuming you want to find a rather than all functions that satisfy that property. You can approach this with a sort of an ad hoc method.
First lets find some function $f(x)$ that satisfies $(i)$. This should be easy enough, the polynomial $x(x-3)$ does the job. (This i incorrect as I misread the condition to be $f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3305372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Maximizing area of rectangle inscribed in circle sector of radius 2 Question:
A rectangle is inscribed in a circle sector. The top two corners of the rectangle lies on the radius of the circle sector and the bottom two corners lie on the arc of the circle sector.
The radius is 2 and angle is $\frac{2\pi}{3}$.
Find the... | I got $$A(\alpha)=\frac{\sin(2\alpha)}{2}-\frac{\sin^2(\alpha)}{\tan(60^{\circ})}$$ so $$A'(\alpha)=8\left(\cos(2\alpha)-\frac{\sin(2\alpha)}{\tan(60^{\circ})}\right)$$
and
$$A'(\alpha)=0$$ if $$\tan(2\alpha)=\tan(60^{\circ})$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given the sum and the product of two variables; what's the sum of their reciprocals? I'm sorry if it's a simple problem, but I need an explanation..
If
$$ x + y = 2 $$
$$ xy = 3$$
Then:
$${1\over x} + {1\over y} = z $$
$${z\in \{ \Bbb R}\}$$
My attempt was as simple like that:
$$x=2-y$$
$$(2-y)y=3$$
$$y^2 -2y + 3 = 0... | Your approach is correct up to where you solved for the values of $y$ and $x$. The sum $\frac{1}{x} + \frac{1}{y}$ will always be equal to $\frac{1}{1 - i \sqrt 2} + \frac{1}{1 + i \sqrt 2}$. Cross multiplying gives you $\frac{1 + i \sqrt 2 + 1 - i \sqrt 2}{(1 - i \sqrt 2)(1 + i \sqrt 2)} = \frac{2}{1 + 2} = \frac{2}{3... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Number theory congruence
Prove that $n^{12} - a^{12}$ is divisible by $91$ if $n$ and $a$ are co prime to $91$
What I tried:
*
*$91$ divides $n^{12}-a^{12}$
*Therefore $n^{12}-a^{12}=91k$
*But $91=13\times7$
*$n^{12}-a^{12}=(13\times7)k$
*$n^{12}=(13\times7)k+a^{12}$
*$n^{12}$ divides $91$
*Therefore $n$ is ... | Two things should leap to mind. And a third thing comes with practice.
1) $91 = 13*7$ so to prove $91|M$ it will work to prove a) $7|M$ and b) $13|M$. If you can prove a) and b) you are done.
2) $n^{12} - a^{12} = (n^6 - a^6)(n^6 + a^6) = (n^3 - a^3)(n^3 + a^3)(n^6 + a^6)=....$. The expression $n^{12}-a^{12}$ is "ve... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3314191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Value of a and P(x) when P(x) is a rational number when satisfies a certain equation This is the question as I still don't have permission to post picture, but it is basically when P(x) = x^3+x^2+ax+1, when a is a rational number, P(X) is also rational number for every x that satisfy x^2+2x-2=0
Consider the integral e... | Just another way without explicit root finding: $x^2=2-2x\implies x^3=2x-2x^2=2x-2(2-2x)=6x-4$, so in this case, $P(x)=(6x-4)+(2-2x)+ax+1=(a+4)x-1$.
As $x^2+2x-2$ has no rational roots ($\pm1,\pm2$ are not roots); to have $P$ rational, we must have $a_0=-4 \implies P_0=-1$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\int_{-1}^{1}\frac{\log(1+x)}{1+x^2}dx = \frac{\pi}{4}\log(2)-\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}$ I have to prove that
$$
I=\int_{-1}^{1}\frac{\log(1+x)}{1+x^2}dx=\frac{\pi}{4}\log(2)-\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}
$$
I know that
$$ I_{+}=\int_{0}^{1}\frac{\log(1+x)}{1+x^2}dx=\frac{\pi}... | \begin{align}
I&=\int_{-1}^1\frac{\ln(1+x)}{1+x^2}\ dx\overset{x=\frac{1-y}{1+y}}{=}\int_0^\infty\frac{\ln2-\ln(1+y)}{1+y^2}\ dy\\
&=\frac{\pi}{2}\ln2-\int_0^\infty\frac{\ln(1+y)}{1+y^2}\ dy\\
&=\frac{\pi}{2}\ln2-\left(\int_0^1\frac{\ln(1+y)}{1+y^2}\ dy+\underbrace{\int_1^\infty\frac{\ln(1+y)}{1+y^2}\ dy}_{\large y\map... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3317286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Help me out with the sum $\sum_{n= 0}^{N-1} \frac{ \left(a-b \cos{\left(\frac{2 \pi n}{N} \right)} \right)^2}{a^2 + b^2 -2ab\cos{\frac{2\pi n}{N}}}$ I am trying to find an analytical expression for the summation below
$$
\sum_{n= 0}^{N-1} \frac{ \left(a-b \cos{\left(\frac{2 \pi n}{N} \right)} \right)^2}{a^2 + b^2 -2ab\... | My method gets the wrong answer, which may be because I approximate it at the start by an integral.
$$\int_0^N \frac{(a-b\cos(2\pi n/N))^2}{a^2+b^2-2ab\cos(2\pi n/N)}dn$$
Substitute $z=\exp(2\pi i n/N)$ to get
$$\frac N{2\pi i}\oint\frac{(a-(z+z^{-1})b/2)^2} {a^2+b^2-ab(z+z^{-1})}\frac{dz}z\\
=\frac N{8\pi i}\oint\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3320890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
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let $x > -1$ prove for all $n \geq 1$, $(1+x)^n \geq 1+nx$ I tried to prove this by induction:
Base case for n = 1 satisfies $\because$ $1+x$ = $1+x$
I.H: $(1+x)^k \geq 1 + kx$
Inductive Step for $k+1$:
$(1+x)^{k+1} \geq 1 + (k+1)x$
$(1+x)^k (1+x) \geq 1 + kx + x$
$(1+kx) (1+x) \geq 1 + kx + x$ by I.H.
$kx^2 + kx + x +... | You can save your proof by added "is implied by" ($\Leftarrow$) marks between lines to indicate the desired direction:
Inductive Step for k+1
:
$(1+x)^{k+1}≥1+(k+1)x\Leftarrow$
$(1+x)^k(1+x)≥1+kx+x\Leftarrow$ by I.H.
$(1+kx)(1+x)≥1+kx+x\Leftarrow$
$kx^2+kx+x+1≥1+kx+x\Leftarrow$
$kx^2≥0$
which is true as $k \ge 1 > 0$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3322144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Equation of a line after being reflected in another line using the transformation matrix For a specific case, consider the line $ y = 3x + 1 $. How can I find the equation of the new line when this is reflected in the line $ y = 2x $ ? I would like to solve this using solely matrices and not considering angles of line... | There is a much faster way to make the computations: remember that a reflection is its own inverse, so if $\mathbf r'$ has coordinates $(x',y')$, we have
$$\mathbf r=T\mathbf r'\iff \begin{pmatrix}x\\y\end{pmatrix}=T\begin{pmatrix}x'\\y'\end{pmatrix}=\tfrac 15 \begin{pmatrix}-3x'+4y'\\4x'+3y'\end{pmatrix}$$
and pluggin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3322528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the sum of all values of $x$ which satisfy the following systems of equations: Systems of equations are as follows:
$$\left\lbrace\begin{aligned}
y&=x^2-5x+5\\
z&=x^2-12x+35\\
y^z&=1\\
x,&y,z \in\mathbb{R}\\
\end{aligned}\right.$$
I got the obvious $x=5,7,1$ and $4$ but apparently there are other answers as the su... | There are infinitely many values of $x$ satisfying this system. For if you write the second equation as $z=x^2-5x+5-7x+30,$ and substitute for $x^2-5x+5$ using the first equation, we get $$z=y-7x+30.$$ Finally, the last equation implies $z=0$ or $y=1.$
If $z=0,$ then $y=7x-30$ and if $y=1,$ then $z=31-7x.$ Thus we have... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3322742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Evaluating $\lim_{x \to -\infty} \frac{1}{-\sqrt{\frac{1}{x^6}}\sqrt{x^6+4}}$ Problem:
$$\lim_{x \to -\infty} \frac{1}{-\sqrt{\frac{1}{x^6}}\sqrt{x^6+4}}$$
$$\lim_{x \to -\infty} \sqrt{\frac{1}{x^6}}=0$$ so...
$$\lim_{x \to -\infty} \frac{1}{-\sqrt{\frac{1}{x^6}}\sqrt{x^6+4}}=\frac{1}{0}$$
The answer is $-1$ and I kn... | It is $$\lim_{x\to \infty}\frac{1}{-\sqrt{1+\frac{4}{x^6}}}=…$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3323085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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If $a+b+c = 4, a^2+b^2+c^2=7, a^3+b^3+c^3=28$ find $a^4+b^4+c^4$ and $a^5+b^5+c^5$ I have tried to solve it but cannot find any approach which would lead me to the answer
| Maple:
simplify(a^4+b^4+c^4, {a+b+c = 4, a^2+b^2+c^2 = 7, a^3+b^3+c^3 = 28}) = 209/2
simplify(a^5+b^5+c^5, {a+b+c = 4, a^2+b^2+c^2 = 7, a^3+b^3+c^3 = 28}) = 334
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3325068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $9 \mathrel| (4^n+6n-1)$ by induction I know that this question was already answered, but I would like to know if the second step of induction its okay the way I did it.
This question is different from Induction proof for $n\in\mathbb N$, $9 \mathrel| (4^n+6n-1)$ and Let $n ∈ N, n ≥ 1$. Prove that $4^n + 6n - 1$ ... | If $n \equiv 0 \pmod 3$, then for an integer $k$:
$$4^n + 6n-1 \equiv 4^{3k} + 18k - 1 \equiv 1^k + 0 - 1 \equiv 0 \pmod 9.$$
If $n \equiv 1 \pmod 3$, then:
$$4^n + 6n-1 \equiv 4 \cdot4^{3k} + 18k + 6- 1 \equiv 4+0+6-1 \equiv 0 \pmod 9.$$
Finally, if $n \equiv 2 \pmod 3$, we have:
$$4^n + 6n-1 \equiv 16 \cdot4^{3k} + 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3326477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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What is the general solution of this equation :$2^x 3^y+1=7^z$ with $x, y , z$ are integers? I have got these triplet solution $(x,y,z)=(1,1,1),(4,1,2)$ for this equation:
$$2^x 3^y+1=7^z$$
with $x, y , z$ are integers, But i can't get general solution of it, I have attempted to use Gausse theorem for the solution... | As Aqua notes, we must have $z\gt0$, so let's write $z=c+1$. Then
$$2^x3^y=7^{c+1}-1=(7-1)(1+7+7^2+\cdots+7^c)$$
It follows that $x,y\gt0$ as well, so let's write $x=a+1$ and $y=b+1$. We now have
$$2^a3^b=1+7+7^2+\cdots+7^c$$
If $b\gt0$, then, since $7\equiv1$ mod $3$, the number of terms on the right hand side must be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3327032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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tangent inequality in triangle Let $a$, $b$ and $c$ be the measures of angles of a triangle (in radians).
It is asked to prove that
$$\tan^2\left(\dfrac{\pi-a}{4}\right)+\tan^2\left(\dfrac{\pi-b}{4}\right)+\tan^2\left(\dfrac{\pi-c}{4}\right) \ge 1$$
When does equality occur ?
My try :
Letting $u:= \tan\left(\dfrac{\pi... | Using the inequality for concave upward graph of $y\,=({tan\,(x)})^2$.
That is $\frac{(\sum_{}^{}{tan}^2{\frac{180-A}{4})}}{3} \;$ $ \ge\,$ ${tan}^2(({\frac{(180-A)+(180-B)+(180-C)}{4}} )/3)$ =1/3
Hence $(\sum_{}^{}{tan}^2{\frac{180-A}{4}})$ $\ge\,1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3327429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Inverse of rational function $y= \frac{3-x}{1+x^2}$ I have the function $$y= \frac{3-x}{1+x^2}$$ and I want to find the inverse of this function.
I know that
$$x= \frac{1 \pm \sqrt{1-4y(3-y)}}{2y}$$
My question is how do I find the domain where the function is
$$x= \frac{1- \sqrt{1-4y(3-y)}}{2y}$$
and
$$x= \frac{1... | I guess you got the idea, you just need to be more careful with the signs.
To find the inverse of:
$$inverse\:\frac{3-x}{1-x^2}$$
We'll need to find the inverse function in the form $y=g(x)$, given the above $f(x)$.
First, we need to get the quadratic equation solution general formula:
given:
$$Ax^2+Bx+C=0$$
$y_1(x)$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3328423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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The Riccatti equation for The Cox-Ingerson-Ross Model. (My Question)
I went through the calculations halfway, but I cannot find out how to calculate the following Riccatti equation. Please tell me how to calculate this The Riccatti equation with its computation processes. If you have other solutions, please let me know... | I solved by myself. The following is this solution.
*
*Let $T-t=s$, one reaches the following equation.
\begin{eqnarray}
B'(s) + \beta B(s) + \frac{1}{2} \sigma^2 B(s)^2 =1
\end{eqnarray}
*One finds out it is the Riccatti equation because of $A(s)=0$. Therefore, one reaches the following equation.
\begin{eqn... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3333207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Eliminating parameter $\beta$ from $x=\cos 3 \beta + \sin 3 \beta$, $y = \cos \beta - \sin \beta$
Based on the given parametric equations:
$$\begin{align}
x &=\cos 3 \beta + \sin 3 \beta \\
y &= \cos \beta \phantom{3}- \sin \beta
\end{align}$$
Eliminate the parameter $\beta$ to prove that $x-3y+2y^3=0$.
What I got so... | Hint:
$$\cos3\beta+\sin3\beta=4(\cos^3\beta-\sin^3\beta)-3(\cos\beta-\sin\beta)$$
$$(\cos\beta-\sin\beta)^3=\cos^3\beta-\sin^3\beta-3\cos\beta\sin\beta(\cos\beta-\sin\beta)$$
$$y^2=?$$
Replace the values of $\cos\beta\sin\beta,\cos\beta-\sin\beta$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3333626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to compare $\pi, e\cdot 2^{1/3}, \frac{1+\sqrt{2}}{\sqrt{3}-1}$ This is in the GRE exam where we are supposed to answer fast so I think there might be some trick behind this to allow us to do that. But so far the best I can do is to write $\frac{1+\sqrt{2}}{\sqrt{3}-1}=\frac{1+\sqrt{6}+\sqrt{2}+\sqrt{3}}{2}$ and co... | If you raise them to the third power you get
$\pi^3 \approx (3+\frac 17)^3 \approx 3^3 + 3*3^2*\frac 17 \approx 27*(1\frac 17)$
$e^3*2 \approx 2(3-0.29)^3 \approx 2(3^3 - 3*3^2*0.29)\approx 27*2*(0.71)\approx 27*1.42$ so $e*2^{\frac 13} > \pi$. (And as a weird unexpected bonus I get that $e^3*2 \approx 27*\sqrt 2$.)
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3334376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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$24$ is the largest integer divisible by all integers less than its square root
Show that $24$ is the largest integer divisible by all integers less
than its square root.
This is what I have done :
Let $m$ be the greatest integer such that $m^2\leq n$, so $i\mid n$ for all i $\in \{1,2,\cdots m-1,m\}$ so lcm$(1,2... | Here's a hint / observation:
The number $25$ is not divisible by $2, 3,$ or $4$, so $25$ doesn't work.
If $n > 25$, the number must be divisible by $2^2 \cdot 3 \cdot 5 = 60$. But $\sqrt{60} > 7$, so the number would need a factor of $7$ also. Now we're looking at $2^2 \cdot 3 \cdot 5 \cdot 7 = 420$, which means we nee... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3335171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Show that $\int_0^1 \frac{x \text{csch}(a x)}{\sqrt{\cosh (2 a)-\cosh (2 a x)}} \mathrm dx=\frac{\pi \sin ^{-1}(\tanh (a))}{2 \sqrt{2} a^2 \sinh (a)}$ Gradshteyn & Ryzhik $3.535$ states that
$$\int_0^1 \frac{x \operatorname{csch}(a x)}{\sqrt{\cosh (2 a)-\cosh (2 a x)}} \mathrm dx=\frac{\pi \sin ^{-1}(\tanh (a))}{2 \s... | Let $f \colon (0,\infty) \to (0,\infty)$,
\begin{align}
f(a) &= \int \limits_0^1 \frac{x \operatorname{csch}(a x)}{\sqrt{\cosh(2a) - \cosh(2ax)}} \, \mathrm{d} x = \int \limits_0^1 \frac{x \operatorname{csch}(a x)}{\sqrt{2 \left[\frac{1}{1-\tanh^2(a)} - \frac{1}{1-\tanh^2(ax)}\right]}} \, \mathrm{d} x \\
&=\frac{1}{\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3335701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Question about the particular part in a non homogeneous recurrence Question about the particular part in the following non homogeneous recurrence :
$$a_n - 6a_{n-1} + 9a_{n-2} = n * 3^n $$
I have the following particual part : $$ a_n = n * 3^n$$
Now the solution of the homogenous part is $$x_1 = 3, x_2 = 3$$ and is of... | Another, general, take is using generating functions. Define $A(z) = \sum_{n \ge 0} a_n z^n$, multiply the recurrence by $z^n$, sum over $n \ge 0$ and recognize some sums:
$\begin{align*}
\sum_{n \ge 0} a_{n + 2} z^n
- 6 \sum_{n \ge 0} a_{n + 1} z^n
+ 9 \sum_{n \ge 0} a_n z^n
&= \sum_{n \ge 0} n \cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3335815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Can i do this to an infinite series? $$let , Y=\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2...}}}}}$$
$$Then , Y= \sqrt{2Y}$$
$$Y^2 = 2Y$$
$$Y^2 - 2Y = 0$$
$$Y = 0 , Y =2$$
Now y can't be zero so ,
$$Y = 2$$
Is This Correct?
| If you assume the number exists and has a value then that is true.
But it could be that the number doesn't exist.
Consider $1 + 2 + 4 + 8 + 16+..... = M$ then $2M = 2+4 + 8 + 32 + ....$ and $2M + 1 = 1 + 2 + 4+ 8+16 + 32 + .... = M$ so $2M + 1 = M$ and $M = -1$.
That's obviously not true.
ANd this fails because $\lim\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3336396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How would I go about computing this finite sum? How would I go about computing the sum
$$
\sum_{k=1}^{n} \dfrac{(-k^2+2k+1)2^k}{(k(k+1))^2}.
$$
I have tried partial fractions but have gotten stuck trying to find the coefficients. I decomposed it like this:
$$
\dfrac{2^k(-k^2+2k+1)}{(k(k+1))^2} = \frac{a_0}{k} + \frac{a... | Hint:
Without the need for solving the coefficients:
$\dfrac{-k^2+2k+1}{k^2(k+1)^2} =\dfrac{k^2+2k+1-2k^2}{k^2(k+1)^2}=\dfrac{(k+1)^2}{k^2(k+1)^2}-\dfrac{2}{(k+1)^2}=\dfrac{1}{k^2}-\dfrac{2}{(k+1)^2}.$
Thus, the sum is reduced to
$$\sum_{k=1}^{n}\dfrac{2^k}{k^2}-\sum_{k=1}^{n}\dfrac{2^{k+1}}{(k+1)^2}.$$
And then note t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3342041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Does an elementary indefinite integral for 2^sin(x) exist? Wolframalpha says no, and I've had little luck elsewhere, maybe because search engines aren't the best at making sense of the "^" symbol.
And if not, are any numerical methods known for solving this integral?
| $\int 2^{\sin x}~dx$
$=\int e^{\ln2\sin x}~dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{\ln^{2n}2\sin^{2n}x}{(2n)!}~dx+\int\sum\limits_{n=0}^\infty\dfrac{\ln^{2n+1}2\sin^{2n+1}x}{(2n+1)!}~dx$
$=\int\left(1+\sum\limits_{n=1}^\infty\dfrac{\ln^{2n}2\sin^{2n}x}{(2n)!}\right)~dx+\int\sum\limits_{n=0}^\infty\dfrac{\ln^{2n+1}2\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3343239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
An inequality involving $|a|^p$: how can I prove this? How can I prove the following lemma?
Let $1<p<\infty$ and let $\epsilon >0$. Then there exists a constant $C \geq 0$ (may depend on $p$ and $\epsilon$) such that for all $a, b \in \mathbb{C}$, $||a+b|^p - |b|^p| \leq \epsilon |b|^p +C|a|^p $.
I am suffering since... | Let
$$x =|a+b|,\quad y = |b|,$$
then the issue inequality takes the form of
$$|x^p-y^p| \le \varepsilon y^p + C|x+e^{i\varphi} y|^p,\tag1$$
where the phase $\varphi$ depends of the phases of $(a+b)$ and $b.$
Since for the arbitrary $u,v\in\mathbb C$
\begin{cases}
||u|-|v|| \le |u+v|\\
||u|-|v|| \le |u-v|,
\end{cases}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3344903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Is $SU(3)$ generated by the exponentials of the Gell-Mann matrices? It is well known that the lie algebra of $SU(3)$ is spanned by the (half) Gell-Mann matrices $T_i=\lambda_i/2$, and that these generate $SU(3)$ via the map $\theta_iT_i \rightarrow e^{\theta_iT_i}$.
However, the above map involves taking a linear comb... | For SU$(3)$, the following set $F_m$ and its algebraic relations differ from that of the eight Gell Mann generators $g_m$.
The $F_m$ have no simple relation to the $g_m$. For example, in the usual representation, $7$ of the $8$ generators $g_m$ have a diagonal element that is zero;
Is the set of the eight finite SU(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3346177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Is there any simpler way to find $\sin 2 y$ from $\cos(x+y)=\tfrac13$ and $\cos(x-y)=\tfrac15$? Is there any simpler way to find $\sin 2 y$ from $\cos(x+y)=\tfrac13$ and $\cos(x-y)=\tfrac15$? Note: $x$ and $y$ are obtuse angles.
My attempt that is not simple is as follows.
Expand both known constraints, so we have
\be... | You might note that $7+4\sqrt3=(2+\sqrt3)^2$ and $2(2+\sqrt3)=(1+\sqrt3)^2$, and that your $\cos^2y$ has opposite sign to $\sin^2y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3346306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Calculate $\sum_{k=1}^n k2^{n-k}$ I want to Calculate $\sum_{k=1}^n k2^{n-k}$. Here's my attempt:
$$\begin{align}
&\sum_{k=1}^n k2^{n-k} = \sum_{k=1}^n k2^{n+1}2^{-k-1} =2^{n+1}\sum_{k=1}^n k2^{-k-1} & &\text{(1)} \\
&k2^{-k-1} = -\frac{d}{dk}2^{-k} & &\text{(2)}
\end{align}$$
Plugging $(2)$ to $(1)$,
$$\begin{align... | Actually, the answer is quite clean. It is $2^{n+1}-n-2$.
You can prove this recursively.
Start with, let's say $n=2$.
This sum is equal to $1(2)+2(1)=2^3-2-2$.
Now, for $n=3$, the sum is equal to $1(4)+2(2)+3(1)$.
How do we get there? First we multiply $1(2)+2(1)$ by $2$, to get $1(4)+2(2)$, and then we add $3$ to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3348824",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Prove that $\int_0^1 \frac{x^2}{\sqrt{x^4+1}} \, dx=\frac{\sqrt{2}}{2}-\frac{\pi ^{3/2}}{\Gamma \left(\frac{1}{4}\right)^2}$ How to show
$$\int_0^1 \frac{x^2}{\sqrt{x^4+1}} \, dx=\frac{\sqrt{2}}{2}-\frac{\pi ^{3/2}}{\Gamma \left(\frac{1}{4}\right)^2}$$
I tried hypergeometric expansion, yielding $\, _2F_1\left(\frac{1}{... | An alternative approach. By termwise integration of a Maclaurin series
$$ I=\int_{0}^{1}\frac{x^2}{\sqrt{1+x^4}}\,dx = \sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]\frac{(-1)^n}{4n+3} $$
and due to the fact that $P_{2n}(0)=\left[\frac{1}{4^n}\binom{2n}{n}\right](-1)^n$, the RHS of the previous line is the valu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3349364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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The sum of infinite series $(1/2)(1/5)^2 + (2/3)(1/5)^3 + (3/4)(1/5)^4..................$ Initially, I broke this series as $1/2 = 1-1/2, 2/3 = 1-1/3, 3/4 = 1-1/4$, then I got two series as it is
$$= (1/5)^2 + (1/5)^3 + (1/5)^4 +.......-[1/2(1/5)^2 + 1/3(1/5)^3 + 1/4(1/5)^4 +.....]$$
After solving the first part, we re... | This second sum can be written as $$\sum_{n=2}^{\infty}\frac{1}{n}\left(\frac{1}{5}\right)^n = \frac{1}{25}\sum_{n=0}^{\infty}\frac{1}{n+2}\left(\frac{1}{5}\right)^n$$
This is the Lerch Transendent, so $$=\frac{1}{25}\phi\left(\frac{1}{5}, 1,2\right)$$
By the identities $\phi(z,s,a) = \frac{1}{z}\left(\phi(z,s,a-1) - \... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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Find all $x\in \mathbb{Z}$ with $x \equiv 2 \mod 4$ , $x \equiv 8\mod9$ and $x \equiv 1\mod5$
Find all $x\in \mathbb{Z}$ with $x \equiv 2 \mod 4$ , $x \equiv 8\mod9$ and $x \equiv 1\mod5$
My attempt:
$\gcd\left(9,5,4\right) = 1 = 9r+5s+4\times0$
$9 = 1\times5+4$
$5 = 1\times4+1$
so $1 = 5-(9-5) = 2\times5 - 1\times9$... | From $x\equiv2\mod4$, let $x=4a+2$ that $a$ is an integer. Then sub it into the second equation,$$4a+2\equiv8\mod9\\4a\equiv6\mod9\\2a\equiv3\mod9\\2a\equiv12\mod9\\a\equiv6\mod9$$ After that, we do the same thing: let $a=9b+6$ that $b$ is an integer, $x=4\left(9b+6\right)+2=36b+26$. Then sub it into the third equation... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3354136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
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Proving $n^2(n^2-1)(n^2+1)=60\lambda$ such that $\lambda\in\mathbb{Z}^{+}$ I'm supposed to prove that the product of three successive natural numbers the middle of which is the square of a natural number is divisible by $60$. Here's my attempt.
My Attempt:
$$\text{P}=(n^2-1)n^2(n^2+1)=n(n-1)(n+1)[n(n^2+1)]$$
It is now ... | The product of three consecutive numbers is divisible by $3$.
If $n$ is even, then $4\mid n^2$. Otherwise $n^2-1$ and $n^2 + 1$ are both even.
Now the hard part:
\begin{array}{c|c}
n \pmod 5 & n^2 \pmod 5\\ \hline
0 & 0 \\
1 & 1 \\
2 & 4 \\
3 & 4 \\
4 & 1
\end{array}
or, without modular notation,
\begin{array}{c|c}
n &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3356512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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Tree diagram for Conditional probability problem of two sons being disease carriers The question : A woman has a 50% chance of carrying hemophilia. She also has two sons If she is a carrier, each son independently has 0.5 probability of having the disease. If she is not a carrier, her sons will independently be normal ... | Let $C$ denote the event that the woman is a carrier.
Let $N_{1}$ denote the event that the first son is normal.
Let $N_{2}$ denote the event that the second son is normal.
$\begin{aligned}P\left(N_{1}\right) & =P\left(N_{1}\mid C\right)P\left(C\right)+P\left(N_{1}\mid C^{\complement}\right)P\left(C^{\complement}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3359530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
jacobian of a dot product of 2 functions take the function $f: \mathbb R^3 \rightarrow \mathbb R^2$ an a function $g: \mathbb R^2 \rightarrow \mathbb R^2 $ defined via $$ f
\begin{pmatrix}
x\\
y\\
z
\end{pmatrix} =
\begin{pmatrix}
x+z^3\\
xyz
\end{pmatrix}
$$
and $$ g
\begin{pmatrix}
s\\
t
\end{pmatrix} =
\begin{pma... | I can see that you have no trouble bar silly mistakes in computing the Jacobian. More precisely, we have :
$$
Df(x,y,z) = \begin{pmatrix}
1 & 0 & 3z^2 \\
zy&xz&xy
\end{pmatrix} ;
Dg(s,t) = \begin{pmatrix}
2s & 1 \\
1&1
\end{pmatrix}
$$
Now, the chain rule is the following : (with variables $a,b,c$ to avoid confusion)
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3360562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $P^{-1}(D + A)P$ is a diagonal matrix, with $D$ a diagonal matrix and $D+A$ hermitian, is $P^{-1}AP$ also diagonal? Suppose $D+A$ is hermitian matrix, and $D$ is a diagonal matrix. Then $A$ is a hermitian matrix as well.
Since $D+A$ is a hermitian matrix, there exists $P$ such that $P^{-1}(D+A)P$ is a diagonal matri... | Here is an counter-example. Put \begin{align*} A = \begin{bmatrix} 3 & i \\ -i & 1 \end{bmatrix}, D= \begin{bmatrix} -2 & 0 \\ 0 & 0 \end{bmatrix} \end{align*}
We can find an unitary matrix $U = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ i & -i \end{bmatrix} $ such that $$U^{-1}(D+A)U = \frac{1}{2}
\begin{bmatrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3360663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding the equation for image of a map Consider the morphism $\mathbb A^1 \to \mathbb A^2$ given by $f(t) = (t^n,(1-t)^n)$. What is the defining equation of the image? Is there a systematic way to do this?
| Too long for a comment.
In M2
R=QQ[t,x,y,MonomialOrder=>Eliminate 1]
I1=ideal(x-t,y-(1-t))
toString (gens gb I1)_0_0
-- x+y-1
I2=ideal(x-t^2,y-(1-t)^2)
toString (gens gb I2)_0_0
-- x^2-2*x*y+y^2-2*x-2*y+1
I3=ideal(x-t^3,y-(1-t)^3)
toString (gens gb I3)_0_0
-- x^3+3*x^2*y+3*x*y^2+y^3-3*x^2+21*x*y-3*y^2+3*x+3*y-1
I4=i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3362578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find all natural numbers $n$ such that $n+1$ divides $3n+11$ Following the example of my teacher:
Find all natural numbers $n$ such that $n-2$ divides $n+5$.
$$n+5 = n-2+7$$
As $n-2|n-2$, $n-2$ will divide $n+5$ if and only if $n-2|7$. Yet, $7$ has the divisors $-7$, $-1$, $1$, $7$ hence the equations:
*
*$n-... | Set $n+1=m$
$3n+11=3(m-1)+11=3m+8\equiv8\pmod m$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3364097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Apostol Theorem 1.35 (Every nonnegative real number has a unique nonnegative square root)
In the above theorem I am not able to understand inequalities in the two marked boxes.
To be specific in the first inequality $0<c<b$, I don't understand why we have $c<b$ when $c=\frac{1}{2}(b+\frac{a}{b})$. In the second inequ... | At the beginning of that paragraph $c=b - \frac{(b^2-a)}{2b}$ which is clearly $<b$ as we substract from $b$ a positive number (as $b^2-a>0$ in this subcase)
and that $b - \frac{(b^2-a)}{2b}$ equals $\frac{1}{2}(b + \frac{a}{b})$ is a simple fraction manipulation:
$$c= b - \frac{(b^2-a)}{2b}= \frac{2b^2}{2b} - \frac{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3365590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Factorizing $x^5+1$ as a product of linear and quadratic polynomials. I am encountering some trouble with this question:
Factorize $$x^5+1$$ as a product of real linear and quadratic polynomials.
I know that if we subtract 1 from $x^5+1$, we get that $x^5 = -1$, but I am unsure where to go from here.
Can anyone help wi... | Let $$s_n(x)=1+x+x^2+\dots +x^{n-1}.$$
Then $s_{n+1}(x)=x^n+s_n(x)$ and $xs_n(x)+1=s_{n+1}(x)$. Thus
$$x^n+s_n(x)=xs_n(x)+1,$$
that is, $$x^n-1=(x-1)s_n(x).$$
We let $p_n(x)=s_n(-x)=1-x+x^2-x^3+\dots+(-1)^{n-1}x^{n-1}.$ We then have
$$(x+1)p_n(x)=1+(-1)^{n-1}x^n.$$
Set $n=5$ so that $(-1)^{5-1}=(-1)^4=1$ and
$$\begin{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3371200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 5
} |
Given three positive numbers $a,b,c$. Prove that $\sum\limits_{sym}\frac{a+b}{c}\geqq 2\sqrt{(\sum\limits_{sym}a)(\sum\limits_{sym}\frac{a}{bc}})$ .
(A problem due to Mr. Le Khanh Sy). Given three positive numbers $a, b, c$. Prove that
$$\sum\limits_{sym}\frac{a+ b}{c}\geqq 2\sqrt{(\sum\limits_{sym}a)(\sum\limits_{s... | Let $c\neq {\rm mid}\{a,b,c\}$ and $c'={\rm mid}\{a,b,c\}.$
Now, let another variables be $a'$ and $b'$.
Thus, since our inequality does not depend on any permutations of $a'$, $b'$ and $c'$, we need to prove that:
$$\frac{a'+b'}{c'}+\frac{b'+c'}{a'}+\frac{c'+a'}{b'}\geq2\sqrt{(a'+b'+c')\left(\frac{a'}{b'c'}+\frac{b'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3373080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Find the limit of a sequence $p_n=\frac{1}{2^n} \sum_{j=0}^{\frac{a\sqrt{n}-1}{2}}\binom{n}{j}$ Given sequence $p_n=\frac{1}{2^n} \sum_{j=0}^{\frac{a\sqrt{n}-1}{2}}\binom{n}{j}$ where a is some natural number.
Show $\lim_{n \to \infty}p_n=0$
| This is a direct consequence of the central limit theorem!
EDITED: With probabilistic intuition, $p_n\rightarrow 0$ is obvious. It is just about tail probability.
Detailed proofs
(Proof using the Chebyshev inequality)
$p_n$ is the probability
\begin{align}
\mathbb{P}\left(X \leq \frac{a\sqrt{n}-1}{2}\right)
\end{align}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Demonstrate that $(x-(\frac{x^2-2}{2*x})\bigr)^2 \gt 2$ for the given conditions. I am stuck on a proof where I need to demonstrate that $(x-(\frac{x^2-2}{2*x})\bigr)^2 \gt 2$.
The proof provides me with the information that $x^2\gt 2$ and $x>0$.
I've taken the following steps to simplify the algebra...to the point whe... | Note that
$$\frac{\frac{1}{4}\cdot x^2}{\frac{1}{x^2}} \gt 1 \implies \frac{1}{4}\cdot x^2-\frac{1}{x^2}>0$$
and then
$$\frac{1}{4}\cdot x^2+\frac{1}{x^2}>\frac{2}{x^2}$$
For the solution we can proceed as follows
$$\left(x-\left(\frac{x^2-2}{2 x}\right)\right)^2 \gt 2$$
$$\left(x^2-2x\left(\frac{x^2-2}{2x}\right)+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3380864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find all five digit number $\overline{abcde}$ such that $\overline{abcde} = \overline{(ace})^2$ Find all five digit number $\overline{abcde}$ such that $$\overline{abcde} = \overline{(ace})^2$$
This question popped in my mind while solving other elementary numbers and I have been trying to solve it ever since but witho... | We can minimize trial and error with some clever use of modular arithmetic.
Let $N=100a+10c+e$ be the square root. Thus $N^2\equiv e^2$ and we require also $N^2\equiv e\bmod 10$. Therefore $e^2\equiv e$ forcing $e\in\{0,1,5,6\}$.
We also know that $(100a)^2<10000(a+1)$ or $a^2<a+1$ forcing $a=1$. Then $N^2<20000$ bu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3381453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
prove that ${3^{3n}} + 3^{2n} + 3^{n } + 1$ is divided by $4$. by induction I tried to take the 3 out but it is not helping me much.
| Let $a_n=3^{3n}+3^{2n}+3^{n} + 1 = (3^3)^n+(3^2)^n+(3^1)^n + (3^0)^n$.
Since
$$(x-3^3)(x-3^2)(x-3^1)(x-3^0)= x^4 - 40 x^3 + 390 x^2 - 1080 x + 729$$
we have
$$
a_{n+4} =40 a_{n+3} - 390 a_{n+2} + 1080 a_{n+1} - 729a_{n}
$$
The claim that $a_n$ is a multiple of $4$ for all $n$ follows by induction since it is true for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3381956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Find the value of the periodic continued fraction with given terms
Find the value of the periodic continued fraction with the terms
$1, 3, 4, 3, 2, 3, 4, 3, 2, 3, 4, 3, 2, . . .$
We see that it starts to be periodic after $1$, i.e, $3,4,3,2$ then $3,4,3,2$ etc...
I know that $x= \frac{A_{k+1}}{B_{k+1}}$ = $\frac{A... | Yes, you're right. We have $5x^2-14x-7=0$. This is the answer.
\begin{eqnarray}
x&=&3+\dfrac{1}{4+\dfrac{1}{3+\dfrac{1}{2+\dfrac{1}{x}}}}\\
&=&3+\dfrac{1}{4+\dfrac{1}{3+\dfrac{1}{\dfrac{2x+1}{x}}}}\\
&=&3+\dfrac{1}{4+\dfrac{1}{3+\dfrac{x}{2x+1}}}\\
&=&3+\dfrac{1}{4+\dfrac{1}{\dfrac{6x+3}{2x+1}+\dfrac{x}{2x+1}}}\\
&=&3+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3382551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find $\sup_{a a^* + b b^* = 1} | a^2x + aby |$ for fixed $x,y \in \mathbb{C}$, $|x| = 1$, $y \ne 0$. While trying to come up with an answer to this question without using the theorem used in the existing answer, whose proof is non-trivial, I tried to find
$$ \tag{1}
\sup_{a a^* + b b^* = 1} | a^2x + aby |
$$
for some f... | Here is the solution to Christian Blatter's maximization problem, writing $p$ for $\psi$:
We have
$$f(p)
=\cos^2(p)+\cos(p)\sin(p)y
=\frac12(\cos(2p)+1)+\frac12\sin(2p) y
$$
so maximizing this is the same as
$g(q)
:=\cos(q)+\sin(q)y
$
with $p = q/2$.
Let
$\tan(r) = y$,
so
$\sin(r)
=\frac{y}{\sqrt{1+y^2}}
$
and
$\cos(r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3385667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Does the fractional equation $\frac{1}{x-5} +\frac{1}{x+5}=\frac{2x+1}{x^2-25}$ have any solutions? We have a partial fraction equation:
$$\frac{1}{x-5} +\frac{1}{x+5}=\frac{2x+1}{x^2-25}$$
I multiplied the equation by the common denominator $(x+5)(x-5)$ and got $0=1$. Is this correct?
| Rearrange the equation as
$$\frac{2x}{x^2-25}-\frac{1}{x-5}-\frac{1}{x+5}
=\frac{2x+1}{x^2-25}-\frac{2x}{x^2-25}= \frac{1}{x^2-25}=0$$
which leads to the solutions
$$x=\pm \infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3392171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.