Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find $g:[0,1]\rightarrow [0,1]$ such that $g(f(x))+g(x)=1$ Let $f:[0,1]\rightarrow [0,1]$ be a continuous function such that $f_\alpha= \frac{1-x}{1+\alpha x}$ , $\alpha \gt-1$.
Can you help me to find a function $g:[0,1]\rightarrow [0,1]$ such that $g(f(x))+g(x)=1$
| Consider a transform $g$ of the form $g(x) = \frac{a+bx}{c+dx}$. Then we must find $a, b, c, d$ satisfying
\begin{align*}
\frac{a + b\frac{1-x}{1+\alpha x}}{c + d\frac{1-x}{1+\alpha x}} + \frac{a + bx}{c + dx} = 1
\end{align*}
Or in other words
\begin{align*}
\frac{(a+b) + (a\alpha-b)x}{(c+d) + (c\alpha-d)x} = \frac{(c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1955071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find integral solutions for surd problem
Problem. Find all integral $x$ and $y$ satisfying $$2\sqrt{6} + 5\sqrt{10} = \sqrt{x} + \sqrt{y}.$$
Squaring both sides, we get $$274 - x - y = 2\sqrt{xy} - 20\sqrt{60}.$$ Squaring again, we get $$-160\sqrt{15xy} \in \mathbb{Z},$$
which gives us $xy = 15k^2$. We know $2\sqrt{6... | (1) Keeping in mind the linear independance of irrational quadratics we have
$$x=6X^2, y=10Y^2\Rightarrow 2\sqrt6+5\sqrt{10}=X\sqrt6+Y\sqrt{10}\Rightarrow X=2,Y=5$$ Hence $$(x,y)=(24,250)$$
(2) Applying conjugates in quadratic fields one has the system
$$\begin{cases}2\sqrt{6} + 5\sqrt{10} = \sqrt{x} + \sqrt{y}\\2\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1956295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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} |
Limit of $\frac{x}{x^2-1}$ Question
$$\lim_{x \rightarrow 1-}\frac{x}{x^2-1}$$
My attempt
$$\lim_{x \rightarrow 1-}\frac{x}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{1}{x^2-1}=\lim_{x \rightarrow 1-}x\lim_{x \rightarrow 1-}\frac{x^2(\frac{1}{x^2})}{x^2(1-\frac{1}{x^2})}=\lim_{x \rightarrow 1-}x\lim_... | Hint:
We can do partial fraction decomposition to show that
$$\frac x{x^2-1}=\frac12\left(\frac1{x-1}+\frac1{x+1}\right)$$
Then do the limit on each individual piece.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1956525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Solve $a_n - 4a_{n-1} + 4a_{n-2} = 2^n$ Solve $a_n - 4a_{n-1} + 4a_{n-2} = 2^n$
given that $a_0 = 0$, and $a_1 = 3$
My Attempt:
Get the characteristic equation and solve it.
For homogeneous equation
$x^2 -4x + 4 = 0$
$x = 2 $ or $ x = 2$
Hence, $a_n^h = (A+Bn)\cdot2^n $
Guess a particular solution: $n^22^nC$
$n^22^nC... | No, $n^22^{n^2}C$ does not work. Try with $n^22^{n}C$. Then
$$2^n=a_n - 4a_{n-1} + 4a_{n-2} = n^22^{n}C-4(n-1)^22^{n-1}C+4(n-2)^22^{n-2}C=2^{n}\cdot 2C$$
which implies that $C=1/2$.
In general, if the r.h.s. is $r^n$ and $r$ is a solution of the characteristic polynomial of multiplicity $m$, then a particular solution ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1957951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\frac{x^2y}{z}+\frac{y^2z}{x}+\frac{z^2x}{y} \geq x^2+y^2+z^2$
Let $x,y,z \in \mathbb{R}$ such that $x \geq y \geq z > 0$. Prove that $$\frac{x^2y}{z}+\frac{y^2z}{x}+\frac{z^2x}{y} \geq x^2+y^2+z^2.$$
I rearranged the inequality to get $$xy(x^2y)+yz(y^2z)+xz(z^2x) \geq xyz(x^2+y^2+z^2).$$ I then thought a... | We need to prove that $\sum\limits_{cyc}(x^3y^2-x^3yz)\geq0$ or
$$\sum\limits_{cyc}(z^3x^2+z^3y^2-2z^3xy)\geq\sum\limits_{cyc}(x^3z^2-x^3y^2)$$ or
$$\sum\limits_{cyc}z^3(x-y)^2\geq(xy+xz+yz)(x-y)(y-z)(z-x)$$
which is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1958954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Incorrect method to find a tilted asymptote Suppose I want to find the slanted asymptote for the graph of $\displaystyle y=\frac{x^2+x-6}{x+2}$.
Using division, we have $\displaystyle y=x-1-\frac{4}{x+2};\;$ so $y=x-1$ is the slanted asymptote.
I would like to find out, though, what is wrong with the following incorre... | The trouble is that the numerator in
$$\frac{x+1-\frac{6}{x}}{1+\frac{2}{x}}$$
stays large, so a small error in the denominator is magnified. Better to write
$$ \frac{1}{1+\frac{2}{x}} = 1 - \frac{2}{x} + \frac{4}{x^2} - \frac{8}{x^3} + \frac{16}{x^4} + \cdots $$
and multiply; the infinite series indicated converges f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1959652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
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Describe all natural numbers $n$ for which $3^{n}-2^{n}$ is divisible by $5$. I need to describe all natural numbers $n$ for which $3^{n}-2^{n}$ is divisible by $5$.
After testing out some $n \in \mathbb{N}$, I came to the conclusion that $3^{n} - 2^{n}$ is divisible by $5$ iff $n$ is even - i.e., if $n$ is of the for... | HINT: $$3^{2k+1} - 2^{2k+1} = 3\cdot 3^{2k} - 2 \cdot 2^{2k} = 3(3^{2k} - 2^{2k}) + 2^{2k}$$
Now can you see why the first summand is divisible by 5 and the second isn't?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1961469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove this trigonometry equation: $\sin 40^\circ \cdot \sin 50^\circ$ is equal to $\frac{1}{2} \cos 10^\circ$.
Prove that $\sin 40^\circ \cdot \sin 50^\circ$ is equal to $\frac{1}{2} \cos 10^\circ$.
I've tried writing $\sin 40^\circ$ as $\sin(40^\circ+10^\circ)$, then wrote $\sin(50^\circ+10^\circ)$ as $\sin 40^\circ... | As $\sin50^\circ=\cos(90-50)^\circ$
and $2\sin40^\circ\cdot\cos40^\circ=\sin(2\cdot40)^\circ$
Finally $\sin80^\circ=\cos(90-80)^\circ=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1963056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Solve the differential equation $4y=x^2+(y')^2$
Solve the differential equation $4y=x^2+(y')^2$.
My try:
$$4 y'=2x+2y'y''$$ hence $2y'=x+y'y''$. Let $y'=vx$, then $y''=v+xv'$, hence
$$2vx=x+vx(v+xv')$$
$$-\int {v\over (v-1)^2} dv=\int {1\over x} dx$$
For R.H.S. Let $u=v-1$
So :
$${1\over u}-\ln u=\ln x+c$$
$${1\over ... | $4y=x^2+(y')^2$
$\left(\dfrac{dy}{dx}\right)^2=4y-x^2$
$\dfrac{dy}{dx}=\pm\sqrt{4y-x^2}$
Let $u=\pm\sqrt{4y-x^2}$ ,
Then $y=\dfrac{u^2+x^2}{4}$
$\dfrac{dy}{dx}=\dfrac{u}{2}\dfrac{du}{dx}+\dfrac{x}{2}$
$\therefore\dfrac{u}{2}\dfrac{du}{dx}+\dfrac{x}{2}=u$
$\dfrac{u}{2}\dfrac{du}{dx}=\dfrac{2u-x}{2}$
$\dfrac{du}{dx}=2-\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1963899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Two spheres, triple integration, not their intersection
Two spheres, one of radius $1$ and one of radius $\sqrt{2}$, have centres that are $1$ unit apart. Find the volume of the smaller region that is outside one sphere and inside the other.
Can use either spherical or cylindrical coordinates. The $2$ spheres can be... | $x^2 + y^2 + z^2 = 2\\
x^2 + y^2 + (z-1)^2 = 1$
Would be an equation for two spheres with centers at (0,0,0) and (0,0,1) (1 unit apart), one of radius $\sqrt2$, and one of radius 1
Expand and subtract one from the other to find the the points of intersection.
$2z - 1 = 1\\
z = 1$
$x^2 +y^2 = 1$
Would you rather do this... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1964024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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How can I find the number of whole number solutions for $ x_1 + x_2 + x_3 + x_4 = 11 $? Find the number of whole number solutions for this equation
$ x_1 + x_2 + x_3 + x_4 = 11 $
where
$ x_1 \ge -2, x_2 \ge -1, x_3 \ge 0, x_4 \ge 5 $
| *$ x_1 + x_2 + x_3 + ...+x_k = n \to \left(\begin{array}{c}n+k-1\\ k-1\end{array}\right) $
where
$ x_1 \ge -2, x_2 \ge -1, x_3 \ge 0, x_4 \ge 5 $
$$ x_1 \ge -2 \to y_1=x_1 +2\ge 0\\
x_2 \ge -1 \to y_2=x_2 +1\ge 0\\
x_3 \ge 0\\
x_4 \ge 5 \to y_4=x_4 -5\ge 0$$
$$\color{red} {x_1+x_2+x_3+x_4=11\\(x_1+2)+(x_2+1)+(x_3)+(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1965452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Second partial derivative with $f(x,y)=x^3+5x^2y+y^3$ I have this problem:
I found $F_x=3x^2+10yx$ and $F_y=5x^2+3y^2$, then $D_uf=\frac{9}{5}x^2+6yx+4x^2$
$F_{x2}= \frac{58}{5}x + 6y, F_{y2}=6x+{24}{5}y$
$D_uf_2=\frac{174}{25}x+\frac{18}{5}y+\frac{32}{5}x+\frac{96}{25}y=>at(2,1)=>534/25$
The answer however is $\frac... | Your mistake was at the first derivative, you forgot to add $\frac{12}5y^2$. See this full calculation:
$$f(x,y) = x^3+5x^2y+y^3$$
$$\nabla f = \left(3x^2+10xy \atop 5x^2+3y^2 \right) $$
$$\nabla f \cdot u = \left(3x^2+10xy \atop 5x^2+3y^2 \right)\cdot \left(\frac 3 5 \atop \frac 4 5\right) =\frac {29} 5x^2+6xy+ \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1965997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why $(x-2)^2$ cannot be divided by $(x-2)(x-3)$ I know the question may seem ridiculous. My question is:
Why can't $(x-2)^2$ be divided by $(x-2)(x-3)$?
I know the answer is obvious if we do the division by hand. However, we also know that if remainder of $\frac{P(x)}{Q(x)}$ is zero, then $Q(x)$ is a divisor of $P(x)... | Let us have a polynomial $f(X)$. I claim that $X-a$ divides $f$ if and only if $a$ is a root of $f$.
From Euclidean division, we get that $f(X) = q(X)(X-a) + r(X)$, where $\deg r <\deg (X-a) = 1$ or $r = 0$.
Assume that $X-a$ divides $f$. Then $r = 0$, and $f(a)=q(a)(a-a) = 0$.
Now assume that $a$ is root of $f$. The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1966847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Formula for monic quadratic polynomials over $\mathbb{Z}$ Let $p(x)$ be a monic quadratic polynomial over $\mathbb{Z}$. Show that,for any integer $n$, there exists an integer $k$ such that
$(p(n))(p(n+1))=p(k)$
| Let P(x)= x^2 + bx + c
Therefore, P(n)= n^2 + bn + c, for an arbitrary integer n
P(n+1) = (n+1)^2 + b(n+1) + c
= n^2 + 2n + 1 + bn + b + c
= P(n) + 2n + b + 1
Therefore, P(n)P(n+1) = P(n)*{P(n) + 2n + b + 1}
= {P(n)}^2 + 2nP(n) + bP(n) + P(n)
= {P(n)}^2 + 2nP(n) + n^2 + bn + bP(n) +c
= (P(n) + n}^2 + b{P(n) + n} + c
=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1968371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Determine the elements of a triangle knowing relationships between lengths and angles A triangle's sides' lengths are three successive numbers (meaning b=a+1, c=a+2). The smallest angle is a half of the triangle's biggest angle.
Find the area of this triangle and its angles.
Solution:
Sides are 4, 5 and 6.
Area: $15\... | Using the law of sines,
\begin{align}
\dfrac{\sin \theta}{a} &= \dfrac{\sin 2\theta}{a+2} \\
\dfrac{\sin \theta}{a} &= \dfrac{2 \sin(\theta) \cos(\theta)}{a+2} \\
\cos \theta &= \dfrac{a+2}{2a} \\
\end{align}
Using the law of cosines,
\begin{align}
a^2 &= (a+1)^2 +(a+2)^2 - 2(a+1)(a+2) \cos \theta \\
\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1973819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find the values of $n$ such that $(1+\sqrt3i)^n$ is a real number
Find the values of $n$ such that $z^n=(1+\sqrt3i)^n$ is a real number.
My reasoning: The power will be real iff $\sin\arg z=0$. Since $\sin 0,\sin\pm\pi,\sin\pm2\pi,\dots=0$, $3\mid n$. Is it correct?
$$z=re^{ia}=2e^{i\pi/3}$$
$$z^n=2^n\left(\cos\frac{... | An alternate view for $a_{n} = (1+ \sqrt{3} i)^{n}$ is:
\begin{align}
a_{0} &= 1 \\
a_{1} &= 1 + \sqrt{3} i \\
a_{2} &= -2 + 2 \sqrt{3} i \\
a_{3} &= -8.
\end{align}
It is determined that $n \in {0, 3, \cdots}$. Consider $n \to 3n$ then
\begin{align}
a_{3n} &= (1 + \sqrt{3} i)^{3n} = [(1 + \sqrt{3} i)^{3}]^{n} = (-8)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1975223",
"timestamp": "2023-03-29T00:00:00",
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Show that if $x + \frac1x = 1$, then $x^5 + \frac1{x^5} = 1$. Suppose $x + \frac{1}{x} = 1$.
Without first working out what $x$ is, show that
$x^5 + \frac{1}{x^5} = 1$ as well.
| $x+\frac 1x=1\\
(x+\frac 1x)^2=1\\
x^2 + 2 + (\frac 1x)^2 = 1\\
x^2 + (\frac 1x)^2 = -1\\
x^4 + (\frac 1x)^4 = -1\\$
$x^5+(\frac 1x)^5 = (x+\frac 1x)(x^4 - x^3(\frac 1x) + x^2(\frac 1x)^2 - x(\frac 1x)^3 + (\frac 1x)^4)\\
(x+\frac 1x)(x^4 + (\frac 1x)^4 - x^2 - (\frac 1x)^2 + 1 )\\
( 1 )((-1) - (-1) + 1) = 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1976629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How can I calculate the following trigonometric integrals with the methods of contour integrals?
How can I calculate the integrals $$\int_0^\pi \frac{dx}{a+b\sin(x)}$$
and $$\int_0^\pi \frac{dx}{a+b\cos(x)}$$
using contour integrals ?
I know the residue theorem , and my idea is to choose the line from $0$ to $\pi... | The integral over the sine is a bit tricky because the integration region cannot be extended to $2 \pi$, so if you want to used the residue theorem, a unit circle is out. Rather, we use a semicircle. Thus consider the complex integral
$$\oint_C \frac{dz}{b z^2+i 2 a z-b} $$
where $C$ is the unit semicircle in the upp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1977054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to find $a, b$ and $c$ from this system of linear equations? Find $a$, $b$ and $c$ from:
$$
\begin{align}
a+c &= 3 \\
b+a &= 2 \\
c+b &= -1 \\
\end{align}
$$
I tried the following way, the answer comes wrong:
$$(a+c) + (b+a) = 3+2$$
| $$(a+c)+(c+b)+(b+a)=3+2-1$$
$$2(a+b+c)=4$$
$$a+b+c=2$$
Then if $a+b+c=2$ and $a+c=3$ then $b=-1$
Then if $a+b+c=2$ and $a+b=2$ then $c=0$
Then if $a+b+c=2$ and $b+c=-1$ then $a=3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1977561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Simplifying logarithm of a product My textbook reads (without explanation, naturally):
$\log \prod_{y=0}^{9}{(1+\frac{1}{10x+y})} = \log{(1+\frac{1}{x})}$
Wondering how this was achieved...
| Notice that $1+\frac{1}{10x+y}=\frac{10x+y+1}{10x+y}$, so the product is: $$\prod_{y=0}^{9}{\left(1+\frac{1}{10x+y}\right)}=\frac{10x+1}{10x}\cdot\frac{10x+2}{10x+1}\cdot\frac{10x+3}{10x+2}\dots\frac{10x+10}{10x+9}.$$ Everything cancels out except the first denominator and the last numerator, giving $$\frac{10x+10}{10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1979464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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How many integers $x$ satisfy: $-4How many integers are there in the set $\{x\in\Bbb R\mid-4<x^2+5x<14\}$?
The correct answer is four, but I only find $0$ and $1$ as integers in the set.
Is the answer wrong?
| This looks like a homework problem with the following intended solution:
Consider $-4<x^2+5x<14$ one inequality at a time.
(In each case, it may be helpful to draw a graph of the corresponding parabola.)
First, $x^2 + 5x > -4$ iff $(x+1)(x+4) = x^2 + 5x + 4 > 0$.
So, we have the points $x < -4$ or $x > -1$.
Second, $x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1980932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What are the conditions on $a, b, c$ so that $x^3+ax^2+bx+c$ is bijective? I would like to find the conditions on $a$, $b$, $c$ so that function $$f(x)=x^3+ax^2+bx+c$$ is bijective.
I thought about resolving the equation
$$x^3+ax^2+bx+c=y$$
but I didn't succeed. And our math teacher told us that we cannot prove that a... | Disclaimer: I started from a simple geometrical idea, but, unfortunately, it turned out messier than expected. This answer, therefore, just serves the purpose of showing that this can be done. Other answers are by far more elegant and are recommended over this one.
Let $f(x) = x^3+ax^2 + bx + c$. Since $y\mapsto y + c$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1984080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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Find the value of the fraction
What common fraction is equivalent to $$\frac{\dfrac{1}{1^3}+\dfrac{1}{3^3}+\dfrac{1}{5^3}+\dfrac{1}{7^3}+\cdots}{\dfrac{1}{1^3}+\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+\cdots} \text{ ?}$$
I didn't see how to relate the numerator to the denominator. We can't find the value of the n... | \begin{align}
& \frac{\dfrac{1}{1^3}+\dfrac{1}{3^3}+\dfrac{1}{5^3}+\dfrac{1}{7^3}+\cdots}{\dfrac{1}{1^3}+\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+\cdots} \\[12pt] = {} & \frac{ \left( \dfrac{1}{1^3}+\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+\cdots\right) - \left( \dfrac 1 {2^3} + \dfrac 1 {4^3} + \dfrac 1 {6^3} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1985223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A line L is tangent to two curves $x^2 + 1$ and $-2x^2 - 3$, find L's equation Some things I'd like to clarify beforehand:
From my understanding, for this situation to occur these conditions need to be fulfilled:
1) $f(a) = g(a)$
2) $f'(a) = g'(a)$
However, both of these conditions are not fulfilled.
What I first did:
... | $f(x) = x^2 +1\\
g(x) = -2x^2 - 3$
both are tangent to the line $L: y = mx+b$
If the curve is tangent, when we subtract one from the other, we have a root of multiplicity. The discriminant in the quadratic formula is zero.
$f(x) = x^2 -mx +1-b\\
m^2 - 4(1-b) = 0$
$g(x) = -2x^2 -mx -3-b\\
m^2 - (-2)(-3-b)(4) = 0$
$4-4b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1987845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Proof of sequence formula I need help solving the following exercise:
The sequence $(a_n)_{n\in \mathbb N}$ is given by
$$a_0 = a_1 = 1 \quad \text{and} \quad a_n=2a_{n-1}+4a_{n-2} \quad \forall n \geq 2$$
Proof the explicit formula
$$a_n = \frac{1}{2}((1+\sqrt{5})^n + (1-\sqrt{5})^n) \quad \forall n\in \mathbb... | The method presented by MrYouMath is called characteristic polynomial.
Another method is by using generating functions. E.g. $$f(x)=\sum_{n=0}^{\infty } a_nx^n=a_0+a_1x+\sum_{n=2}^{\infty } (2a_{n-1}+4a_{n-2})x^n=$$
$$a_0+a_1x+\sum_{n=2}^{\infty } 2a_{n-1}x^n + \sum_{n=2}^{\infty } 4a_{n-2}x^n=$$
$$a_0+a_1x+2x\sum_{n=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1988241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Find the radius of convergence and convergence interval Find the radius of convergence and convergence interval
$$\sum^\infty_{n=1}\frac{n^5\;(x+8)^n}{9^n\; n^\frac{17}{3}}$$
My attempt:
I'm computing $\lim_{n\to\infty}|\frac{a_{n+1}}{a_n}|$ and
I got the answer $|9(x+8)|$. So the radius of convergence $R=\frac{1}{9}... | The series is a power series around $x = -8$, that is, it is of the form
$$ \sum_{n=1}^{\infty} a_n (x - (-8))^n $$
where $a_n = \frac{n^5}{9^n n^{\frac{17}{3}}} = \frac{1}{9^n n^{\frac{2}{3}}}$. The radius of convergence is given by $R = \lim_{n \to \infty} \frac{a_n}{a_{n+1}}$ (if it exists) and not by $\lim_{n \to \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1990953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
I want to find a pair of integers $X$, $Y$ which satisfy $X^2 - 2Y^2=1$ such that $X > Y > 50$
Find a pair of integers $X, Y$, which satisfy $X^2 - 2Y^2 = 1$, such that $X > Y > 50$.
I have started by finding a pair of much smaller integers that work: $X(1) = 3$ and $Y(1) = 2$.
When I looked up a solution it was as... | $$X_{i+1}^2-2Y_{i+1}^2 = (3X_i + 4Y_i)^2 - 2(2X_i + 3Y_i)^2 =$$ $$ = (9-2\cdot4)X_i ^ 2 + (12-12)X_i \cdot Y_i + (16-2\cdot9)Y_i ^ 2= $$ $$= X_{i}^2-2Y_{i}^2 = 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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${\sqrt{2x+1}=1+\sqrt{x}}$ — I dont know if the solution is correct. Help?
${\sqrt{2x+1}=1+\sqrt{x}}$
${2x+1=1+2\sqrt{x}+x}$
${x=2\sqrt{x}}$
${x*\frac{1}{x^{1/2}}=2}$
${\sqrt{x}=2}$
${x=4}$
| You are mostly correct but there are two conditions you didn't take into account which you should have.
$\sqrt{2x + 1}=1+\sqrt{x}$ Note, this implies $2x + 1 \ge 0$ i.e $x \ge -\frac 12$.
$2x +1 = 1 + 2\sqrt{x} + x$ Now, we have "lost" that assumption. It is possible that we will end up with some extraneous answers ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 3
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Prove $\int_{0}^{1} \frac{\sin^{-1}(x)}{x} dx = \frac{\pi}{2}\ln2$ I stumbled upon the interesting definite integral
\begin{equation}
\int\limits_0^1 \frac{\sin^{-1}(x)}{x} dx = \frac{\pi}{2}\ln2
\end{equation}
Here is my proof of this result.
Let $u=\sin^{-1}(x)$ then integrate by parts,
\begin{align}
\int \frac{\sin... | $$\int_{0}^{1} \frac{\sin^{-1}x}{x} dx
= \int_{0}^{1} \int_{0}^{1} \frac{\sqrt{1-x^2}}{1-x^2+x^2y^2} dy\ dx \overset{x=\sin t}
=\int_0^1 \frac{\pi}{2(1+y)}dy=\frac{\pi}{2}\ln2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1992462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 5
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Show that $x^2+x+4$ is irreducible over $\mathbb Z_{11}$.
Show that $x^2+x+4$ is irreducible over $\mathbb Z_{11}$.
The problem can be solved by putting every element from $\mathbb{Z}_{11}$ and checking if there is any root.
$0^2 + 0 + 4 = 4 \\
1^2 + 1 + 4 = 6 \\
2^2 + 2 + 4 = 10 \\
3^2 + 3 + 4 = 5 \\
4^2 + 4 + 4 = ... | ${\rm mod}\ 11\!:\,\ 0\equiv x^2\!+\!x\!+\!4\equiv x^2\!+\!12x\!+\!4\equiv (x\!+\!6)^2\!+1\,\Rightarrow\,(x\!+\!6)^2\equiv -1.$
But $\,-1\,$ is not a square, else by little Fermat: $\left[ a^2\equiv -1\right]^5\Rightarrow\, 1\equiv -1.$
Remark $ $ We implicitly applied Euler's Criterion to show $-1$ is not a square $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1993023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solutions to $a^2+b^2+(ab)^2=c^2$ In a comment to the question $n^2+(n+1)^2+n^2\cdot(n+1)^2$ is a perfect square it is proved that:
$(1)\quad b=a+1$ give integer solutions to
$a^2+b^2+(ab)^2=c^2$ for all $a\in\mathbb N$.
In the answers to the question $\forall m\in\mathbb N\exists n>m+1\exists N\in\mathbb N:m^2+n^2+(m... | The Pell type equation $$ x^2 - (1 + A^2) y^2 = A^2 $$ has an infinite set of solutions $(x,y).$ The first three predictable types are
$$
\left(
\begin{array}{c}
A \\
0
\end{array}
\right),
$$
$$
\left(
\begin{array}{c}
A^2 - A + 1 \\
A - 1
\end{array}
\right),
$$
$$
\left(
\begin{array}{c}
A^2 + A + 1 \\
A + 1
\end{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1996487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Congruence involving prime numbers Given the function $f(x)= x^{\frac{p(p-1)}{2}}-1.$ Also let $p$ be an odd prime number. If $\epsilon$ is a number $\pmod{p}$ such that
$f(\epsilon) \equiv 0 \pmod{p}$. Then how do I prove ( prove because Hardy and Wright use it but don't prove it ) that
$$ f(\epsilon)\equiv 0 \pmod{p... | Let $f(x) = x^{p(p-1)/2} - 1$. Suppose we have an integer $a$ such that $f(a) \equiv 0 \pmod{p}$.
First, suppose $p \mid a$. Then, $f(a) = a^{p(p - 1)/2} - 1 \equiv -1 \pmod{p}$, a contradiction.
Then, suppose $p \nmid a$. By Euler's Theorem, $a^{\phi(p^2)} = a^{p(p - 1)} \equiv 1 \pmod{p^2}$. Treating this as a quadr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1997731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Combinatoric Identities Ho to prove the following two identities?
I cannot see the trick:
(a) $\binom{-x}{k}=(-1)^k\binom{x+k-1}{k}$
(b) $\binom{k+x}{2k+1}=-\binom{k-x}{2k+1}$
| Note that $(2k+1)-(k+x)-1=k-x$, the upper number of $\binom{k-x}{2k+1}$. Thus, you can negate the upper index:
$$\binom{k+x}{2k+1}=(-1)^{2k+1}\binom{(2k+1)-(k+x)-1}{2k+1}\;,$$
Therefore:
${k+x \choose 2k + 1}=-{k-x \choose 2k + 1} $
$=(-1)^{2k+1}\binom{(2k+1)-(k+x)-1}{2k+1}\;$
$=(-1)^{2k+1}\binom{2k-(k+x)}{2k+1}$
$=(-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1998306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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simplify $\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}$ simplify $\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}$.
1.$90^{\frac{3}{2}}$
2.$106\sqrt{41}$
3.$4\sqrt{41}$
4.$504$
5.$508$
My attempt:I do like this but I didn't get any of those five.
$\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}={\sqrt{45+4\s... | Considering all positive values, $45\pm4\sqrt{41}$ can be written as $45\pm4\sqrt{41}=(\sqrt{41}\pm2)^2$, thus, the given expression can be simplified as follows: $$\sqrt{(45+4\sqrt{41})^3}-\sqrt{(45-4\sqrt{41})^3}=\left(\sqrt{45+4\sqrt{41}}\right)^3-\left(\sqrt{45-4\sqrt{41}}\right)^3=\left(\sqrt{\left(\sqrt{41}+2\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2002201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Evaluate the integral $\int \:\frac{\sqrt{x^2+2\cdot\:\:x+2}}{x}dx$ $$\int \:\frac{\sqrt{x^2+2\cdot\:\:x+2}}{x}dx$$
I have tried to form a square above i also tried to get the x below under the root
but got nothing
| $z=\tan(\theta/2)$, $\tan(\theta)=\frac{2z}{1-z^2}$, $\mathrm{d}\theta=\frac{2\,\mathrm{d}z}{1+z^2}$, $\cos(\theta)=\frac{1-z^2}{1+z^2}$, $\sin(\theta)=\frac{2z}{1+z^2}$
$z=\frac{\sec(\theta)-1}{\tan(\theta)}=\frac{-1+\sqrt{x^2+2x+2}}{x+1}$
$$
\begin{align}
\int\frac{\sqrt{x^2+2x+2}}{x}\,\mathrm{d}x
&=\int\frac{\sec^3(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2002821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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How to find the value of $\lim_\limits{x \to 0}\frac{e^{x\cos{x^2}}-e^{x}}{x^5}$ How to find the exact value of the following limit?
$\lim_\limits{x \to 0}\frac{e^{x\cos{x^2}}-e^{x}}{x^5}$
| $$\cos x^2=1-\frac{x^4}{2}+o(x^4)$$
so
$$e^{x\cos x^2} = e^{x-\frac{x^5}{2}+o(x^5)} = e^xe^{-\frac{x^5}{2}+o(x^5)}$$
therefore
$$\frac{e^{x\cos x^2}-e^x}{x^5} = e^x\frac{e^{-\frac{x^5}{2}+o(x^5)}-1}{x^5} \sim e^x\frac{-\frac{x^5}{2}}{x^5} \xrightarrow[x\to0]{}-\frac{1}{2}$$
using $e^u-1\sim u$ when $u\to0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2008243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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On the closed form for $\sum_{m=0}^\infty \prod_{n=1}^m\frac{n}{4n-1}$ We have
$$\sum_{m=0}^\infty \prod_{n=1}^m\frac{n}{3n-1}=\frac{3}{2}+\frac{\ln(\sqrt[3]{2}-1)}{2^{7/3}}+\frac{\sqrt{3}}{2^{4/3}}\arctan\frac{\sqrt{3}}{2\sqrt[3]{2}-1}\tag1$$
$$\sum_{m=0}^\infty \prod_{n=1}^m\frac{n}{3n-2}=\frac{3}{2}-\frac{\ln(\sqrt[... | Yes. The logarithms have the same denominator. So factoring this out gives a simple difference of logs: namely the simple log of a quotient, which itself can be simplified to $$2\ln\left(\frac12+\frac{\surd3}2-\sqrt{2\surd3}\right)$$(if my calculation is right). A similar method will simplify the inverse cotangents to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2009343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
$\lim_{x \to 1}\frac{3}{1-x^{1/2}} - \frac{3}{1-x^{1/3}}$ - my answer is wrong (why?) Need to find
$$\lim_{x \to 1}\frac{3}{1-x^{1/2}} - \frac{3}{1-x^{1/3}}$$
One thing I use is
$$\lim_{x \to 1}\frac{x^{1/m}-1}{x^{1/n} - 1} = n/m$$
$$ \lim_{x \to 1}\frac{3}{1-x^{1/2}} - \frac{3}{1-x^{1/3}} = 3\lim_{x \to 1}\frac{x^{1/... | It's likely there was a typo, and the intended problem was
$$\lim_{x\to1}\left({3\over1-x^{1/2}}-{2\over1-x^{1/3}} \right)$$
In general, if you are taking a limit of ${A\over1-x^{1/2}}-{B\over1-x^{1/3}}$, Abdallah Hammam's trick of letting $x=t^6$ is a good way to go:
$${A\over1-x^{1/2}}-{B\over1-x^{1/3}}={A\over1-t^3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2009665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to estimate this sum? I encounter a problem and I just can't figure out a useful way to estimate it.I have tried taylor expansion but it doesn't work.
I want to know the big O-estimate of $n$ in this formula.
$$\sum_{k=2}^{n}\frac{\cos\frac{\pi}{n+1}}{\cos\frac{\pi}{n+1}-\cos\frac{k\pi}{n+1}}$$
| Since $\cos(x)=1-x^2/2+O\!\left(x^4\right)$
$$
\begin{align}
\frac{\cos\left(\frac\pi{n+1}\right)}{\cos\left(\frac\pi{n+1}\right)-\cos\left(\frac{k\pi}{n+1}\right)}
&=\frac{1-\frac{\pi^2}{2(n+1)^2}+O\left(\frac1{n^4}\right)}{\left(1-\frac{\pi^2}{2(n+1)^2}\right)-\left(1-\frac{k^2\pi^2}{2(n+1)^2}\right)+O\!\left(\frac{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2010498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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How to find x and y coordinates based on the given distance? The problem says: Find the point (coordinates $(x,y)=~?$) which is symmetrical to the point $(4,-2) $ considering the given equation $y=2x-3$
I have found the perpendicular line-slope $y=-~\frac{1}{2}x$ and the
intersection point which is shown in the graph... | Alternative rapid way: calling $x_0$ and $y_0$ the coordinates of the symmetric point, the simplest method is to note that the differences in the $x $- and $y $- coordinates between the searched point and $( \frac{6}{5}, -\frac{3}{5})$ must equal those between this last point and $(4,-2) $. Thus:
$$ 4-\frac{6}{5} = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2010926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Find the value of $\lim\limits_{n \to \infty}n\left(\left(\int_0^1\frac{1}{1+x^n}\,\mathrm{d}x\right)^n-\frac{1}{2}\right)$
Find the value of the limit $$\lim_{n \to \infty}n\left(\left(\int_0^1 \frac{1}{1+x^n}\,\mathrm{d}x\right)^n-\frac{1}{2}\right)$$
I can't solve the integral $\int_0^1 \mathrm{\frac{1}{1+x^n}}\,\... | Let $I(n)$ be given by the integral
$$\begin{align}
I(n)&=\int_0^1 \frac{1}{1+x^n}\,dx \tag 1\\\\
\end{align}$$
Then, expanding the integrand of the integral on the right-hand side of $(1)$ in the Taylor series $ \frac{1}{1+x^n}=\sum_{k=0}^\infty (-1)^kx^{nk}$ reveals
$$\begin{align}
I(n)&=\sum_{k=0}^\infty \frac{(-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2015233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Fraction Decomposition I have the following problem:
Suppose $x + y + z = 0$. Show that $$\frac{x^5 + y^5 + z^5}{5}= \frac{x^3 + y^3 + z^3}{3} \times \frac{x^2 + y^2 + z^2}{2}$$ and $$\frac{x^7 + y^7 + z^7}{7}= \frac{x^2 + y^2 + z^2}{2} \times \frac{x^5 + y^5 + z^5}{5}$$
I thought this was a fun problem to tackle, bu... | The first identity does not involve much algebra, if you write $z = -(x+y)$:
$$ z^5 + (x^5+y^5) = -5x^4y-10x^3y^2-10x^2y^3-5xy^4$$
where the $(-x^5-y^5)$ cancels out in the binomial exapnsion.
A similar thing happends in the cubes term.
So the identity reads
$$
-x^4y-2x^3y^2-2x^2y^3-xy^4 = (-x^2y-xy^2)\frac{(x^2+y^2+x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2015863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Determine the last digit of the following number. Problem: Determine the last digit of the following number $$\underbrace{7^{7^{7^{7...}}}}_{1001\text{ }7's}.$$
My Attempt: By Euler's theorem $7^4\equiv 1\pmod {10}\Rightarrow 7^7\equiv 3\pmod {10}.$ After this I am clueless.
Edit $1$: So I tried taking even and odd num... | As you observed, by Euler's theorem, the exponent works modulo $4$. And $7$ happens to be congruent to $-1 (\bmod{4})$. The exponent is a tower of $1000$ $7$'s, which, modulo $4$, is $-1$ raised to an odd power. So modulo $10$, your tower of $1001$ $7's$ is $\equiv 7^{-1} \equiv 3 (\bmod{10}).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2021202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the number of solutions $(x,y)$ in non-negative integers such that $ax+by\leq ab$, where $a$ and $b$ are positive integers. Problem: Find the number of solutions $(x,y)$ in nonnegative integers such that $ax+by\leq ab$, where $a$ and $b$ are positive integers.
My Attempt: I observed that this is equivalent to coun... | So far, you should agree that the number of solutions $(x,y)$ is
$$
\frac{(a+1)(b+1) + B}{2}, \tag{1}
$$
where $B$ is the number of points on the boundary $ax + by = ab$. So we need to count $B$.
Write $a = da'$, $b = db'$, where $d = \gcd(a,b)$, so that $\gcd(a',b') = 1$.
Our equation becomes
\begin{align*}
(da') x + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2021843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is there an identity for $\arctan(x+y)$? Ive tried looking on the internet and I can't seem to find any identities for $\arctan(x+y)$. I was wondering if anyone knows any?
Thanks!
| $$\arctan(x+y) + \arctan(x-y) = \arctan\left(\frac{2 x}{1-x^2+y^2}\right) \tag 1 $$
is valid because take tan on both sides using
$$ \tan( x+y)=\frac{\tan x + \tan y}{1- \tan x \tan y} $$
(1) holds. Similarly
$$\arctan(x+y) - \arctan(x-y) = \arctan\left(\frac{2 y}{1+x^2-y^2}\right) \tag 2 $$
Taking sum
$$2 \arctan(x+y)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2022782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Help with question on limits It is given that $\lim _{ x\rightarrow 0 }{ \frac { f(x) }{ { x }^{ 2 } } =a } $ and $\lim _{ x\rightarrow 0 }{ \frac { f(1-\cos x) }{ g(x)\sin^2x } = b }$, where $b \neq 0$, then find $\lim _{ x\rightarrow 0 }{ \frac { g(1-\cos2x) }{ x^4 } }$.
My Approach:
Since $\sin^2x + \cos^2x = 1 \imp... | You are on the right track.
Method 1.
One may recall that, by the Taylor series expansion, as $u \to 0$, $$ \cos u=1-\frac{u^2}2+o(u^2) $$ giving, as $x \to 0$, $$ \cos 2x=1-2x^2+o(x^2) $$ and $$ \cos (1-\cos 2x)=1-2x^4+o(x^4). \tag1 $$ Now you have obtained, as $u \to 0$, $$ \frac{(1-\cos u)}{g(u)(1+\cos u)} \to \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2024202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can this be solved without resorting to graphical method? I need to find the points of intersection of a circle with radius $2$ and centre at $(0,0)$ and a rectangular hyperbola with equation $xy=1$. As per the topic statement is there any way to solve this without the graphical method. I have tried setting the $y$ val... | The equation of the circle is $x^2 + y^2 = 4$ and the equation of the hyperbola is $xy=1$
So the point of intersection would be a common solution to
$xy =1$
$x^2 + y^2 = 4$
so
$y = 1/x$
$x^2 + \frac 1{x^2} = 4$
$x^4 +1 = 4x^2$
$x^4 - 4x^2 + 1 = 0$
$x^2 = \frac {4 \pm \sqrt {12}}2$
$x^2 = 2 \pm \sqrt 3$
$x = \pm \sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2024468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 1
} |
maximum value of $xy+yz+zx.$ given $x+2y+z=4$ if $x+2y+z=4$ and $x,y,z$ are real number. then find maximum value of $xy+yz+zx$
putting $x+z=4-2z$ in $y(x+z)+zx = y(4-2z)+zx = 4y-2yz+zx$
i wan,t be able to go further,could some help me with this
| $$xy+yz+zx+y^2 = (x+y)(y+z) \le \frac{(x+2y+z)^2}{4} = 4$$
$$\Rightarrow xy+yz+zx \le 4-y^2 \le 4$$
Equality occurs when $y=0$, $x=z=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2025356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Taylor expansion for $\frac{1}{x^2 + 2x + 2}$ I'm trying to find the Taylor expansion of the function:
$$ f(x) = \frac{1}{x^2 + 2x + 2} $$
about the point $ x = 0 $. I have worked out the terms up to the fourth derivative, which was very tedious. I found:
$$ f(x) = \frac{1}{2} - \frac{1}{2} x + \frac{1}{4} x^2 + 0 x^3 ... | There is a much simpler way involving no contrived derivatives, binomial expansions, etc., that unfortunately does not easily generalize to any other quadratic reciprocal.
If we notice that $x^2+2x+2$ occurs in the factorization of a certain binomial:
$$ x^8 - 16 = \left( x^4 - 4 \right) \left( x^4 + 4 \right) = \left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2025920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 7
} |
About the diophantine equation $y^3=8x^6+2x^3y-y^2$
How I can solve the equation $y^3=8x^6+2x^3y-y^2$ in integers?
I made the substition $x^3=z$ and got the equation $8z^2+2yz-y^3-y^2=0$. So I decided to apply the general formula for quadratic equations and thus I got $$z=\frac{-2y\pm \sqrt{32y^3+36y^2}}{16}=\frac{-y... | Well, after a couple of days thinking I think the diophantine equation can be solved in a different way. First of all, asumme that $y=0$, so the equation becomes $8x^6=0$, then $x=0$. Now, let's consider $y\neq 0$, we note that the LHS is divisible by $y$ and thus the RHS also has to be divisible by $y$, so we get that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2028099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that at least one of the expressions does not exceed $\sqrt[3]{3}$
Let $m,n > 1$ be positive integers. Prove that at least one of the numbers $$\sqrt[n]{m} \ , \ \sqrt[m]{n}$$ does not exceed $\sqrt[3]{3}$.
I thought about doing a proof by contradiction. That is, assume this is not the case and so both exceed $... | Without loss of generality, assume that $m\ge n$. Then $$\sqrt[m]{n} \le \sqrt[n]{n}$$
We just want to show that $$\sqrt[n]{n} \le \sqrt[3]{3}, $$
which is equivalent to
$$n^3 \le 3^n.$$
This is not hard by, for example, proof by induction.
When $n=2$ or $3$, this is true, and when $n > 3$, $$n^3 = (n-1)^3 + 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2030563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Prove that $a^n-b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots + b^{n-1})$. Exercise
Prove that $a^n-b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots + b^{n-1})$.
I've posted my solution below. In case someone has a more clever solution, feel free to post it!
(TBH, I was surprised that there was no question on Math.SE regarding thi... | The result is essentially the fact that $1+x+ \cdots + x^{n-1} = {1-x^n \over 1-x}$ (for $x \neq 1$), or equivalently that
$(1-x)(1+x+ \cdots + x^{n-1}) = 1 -x ^n$ for any $x$.
If $a=b$, or $a=0$ then the result is true.
Suppose $a \neq 0$ and $a \neq b$. Then letting $x={b \over a}$ in the above
and multiplying the re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2031059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Calculating $\lim_{x\to\infty}\left(x e^{\frac{1}{x}} - \sqrt{x^2+x+1} \right)$ I've managed to solve it by rewriting the expression as
$$\frac{1 - \frac{\sqrt{x^2 +x + 1}}{x e^{\frac{1}{x}}} }{ \frac{1}{x e^{\frac{1}{x}}} }$$
then applying L'Hospital's rule.
This took up one whole page and was very hairy, even after s... | Using the standard limit
$$
\lim_{t\to 0}\frac{e^t-1}{t}=1
$$
we have
$$
1=\lim_{x\to\infty}\frac{e^{1/x}-1}{1/x}=\lim_{x\to\infty}(xe^{1/x}-x).
$$
Now rewrite your limit as
$$
\lim_{x\to\infty}(xe^{1/x}-x+x-\sqrt{x^2+x+1})=1+\lim_{x\to\infty}(x-\sqrt{x^2+x+1}).
$$
To calculate the last limit we rewrite again
\begin{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2032212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Constructing a cubic given four points
Question: Is there an easier way to solve this problem?
Suppose the polynomial $f(x)$ is of degree $3$ and satisfies $f(3)=2$, $f(4)=4$, $f(5)=-3$, and $f(6)=8$. Determine the value of $f(0)$.
My Attempt: I started off with the general cubic $ax^3+bx^2+cx+d=f(x)$ and manuall... | You can directly find out the polynomial $f$ by considering it according as the $x$-values available:
Let $f(x)=a_0+a_1(x-3)+a_2(x-3)(x-4)+a_3(x-3)(x-4)(x-5)$ for real constants $a_0,a_1,a_2,a_3$. Note that we don't need to take into account the value $x=6$ as this is already a cubic polynomial.
Then, $f(3)=2\Rightarr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2032335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
n such that $|\sin (\sqrt{n+1})-\sin \sqrt n|< \lambda$
I broke $|\sin (\sqrt{n+1})-\sin \sqrt n|$ as $|2cos(\frac{\sqrt{n+1}+\sqrt n}{2})sin(\frac{\sqrt{n+1}-\sqrt n}{2})|$.I am facing trouble in proving that it is less than some number for some n.Please help me in his regard.Thanks.
| Note that for $0< x < \pi/2$, $\sin x < x$.
Also, for all $n > 0$, we have:
$$
\frac{1}{4n} + 1 + n > n+1 \\
\left( \frac{1}{2\sqrt{n}}+\sqrt{n}\right)^2 > n+1 > 0 \\
\frac{1}{2\sqrt{n}} +\sqrt{n}> \sqrt{n+1} \\
\sqrt{n+1}-\sqrt{n}<\frac{1}{2\sqrt{n}}
$$
Also it is obvious that
$$
\left|\cos \left( \frac{\sqrt{n+1}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2033066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find Power Series representation of the function $f(x) = {x\over 2x^2 + 1}$?
Find Power Series representation of the function $f(x) = \dfrac x{2x^2 + 1}$?
I'm not sure how to tackle this...I'm supposed to find interval of convergence.
| A variation: From the geometric series
\begin{align*}
\frac{1}{1-x}=\sum_{n=0}^\infty x^n\qquad\qquad\qquad |x|<1
\end{align*}
we obtain
\begin{align*}
\frac{x}{2x^2+1}&=\frac{x}{1-\left(-2x^2\right)}\\
&=x\sum_{n=0}^\infty (-2x^2)^n\qquad\qquad &|-2x^2|&<1\\
&=\sum_{n=0}^\infty (-2)^nx^{2n+1} &|x|&<\frac{1}{\sqrt{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2036617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$n^{\text{th}}$ term of The Maclaurin Expansion of $\dfrac{1}{(1-x)^3(1+x)(1+x+x^2)}$? I am trying to find the coefficient of $n^{\text{th}}$ term of the Maclaurin series of $$\dfrac{1}{(1-x)^3(1+x)(1+x+x^2)}.$$
How can I find the coefficient of $n^{\text{th}}$ term of this function?
| Using the hint of Mark Bennet you can write
$$ \dfrac{1}{(1-x)^3(1+x)(1+x+x^2)} = \dfrac{1}{(1-x)(1-x)(1+x)(1-x)(1+x+x^2)} = \dfrac{1}{(1-x)(1-x^2)(1-x^3)}. $$
Now, notice that for each fixed $j = 1, 2, 3$ we have:
$$ \frac{1}{1 - x^j} = 1 + x^j + x^{2j} + \ldots$$
Denote by $c_n$ the $n$-th coeficient of your rational... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2037990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
If $x+\frac{1}{x}=\frac{1+\sqrt{5}}{2}$ then $x^{2000}+\frac{1}{x^{2000}}= $? If $x+\frac{1}{x}=\frac{1+\sqrt{5}}{2}$ then
$$x^{2000}+\frac{1}{x^{2000}}=?$$
My try:
$$\left(x^{1000}\right)^2+\left(\frac{1}{x^{1000}}\right)^2=\left(x^{1000}+\frac{1}{x^{1000}}\right)^2-2$$
Continuation ?
| Here's another approach for the record.
*
*The equation $x+ \frac{1}{x}=\alpha$ where $\alpha \in [-2,2]$ can be solved as follows. Identify $\alpha$ is $2\cos (\theta)$, and observe that letting $z=e^{i\theta}$,from the definition of $\cos (\theta)$, we have
$$z+ \frac{1}{z}=2\cos(\theta)=\alpha.$$
*Also, from t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2039286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 1
} |
Evaluate $\sum_{n=1}^{\infty}{1\over n^5}$ up to the second decimal place I am trying to evaluate
$$\sum_{n=1}^{\infty}{1\over n^5}$$
up to the second decimal place. While the series is convergent, I have no idea how to construct such a bound, preferably using basic properties of series and sequences. Any hints?
| Let $N$ be such that $\sum \limits _{k \ge 2^N} ^\infty \frac 1 {n^5} \le \epsilon$. Notice that
$$\sum _{k \ge 2^N} ^\infty \frac 1 {n^5} \le \sum _{m \ge 0} \ \sum _{k = 2^{N+m}} ^{2^{N+m+1}-1} \frac 1 {n^5} \le \sum _{m \ge 0} \ \sum _{k = 2^{N+m}} ^{2^{N+m+1}-1} \frac 1 {(2^{N+m})^5} = \sum _{m \ge 0} \frac {2^{N+m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2045461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
} |
Lagrange multiplier on unit sphere for $f(x,y,z)=x^2+2y^2+3z^2$
Find the maximum and minimum value of $f(x,y,z)=x^2+2y^2+3z^2$ in the region $D=\{(x,y,z)\in \mathbb R^3| x^2+y^2+z^2=1\}$. And find a unit vector at which the maximum and minimum are attained respectively.
Attempt:
I know I need to proceed by Lagrange m... | We have: $x^2 = 1 - y^2 - z^2 \implies f(x,y,z) = 1 - y^2 - z^2 + 2y^2 + 3z^2= 1+y^2+2z^2\ge 1$, and this is the minimum of $f$ which is achieved when $y = z = 0, x = \pm 1$. For the max, we have $f_y = 2y, f_z = 4z$, and $f_y = 0 = f_z \implies y = 0 = z$ which is the only critical point in the domain $D = [-1,1]\time... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2047014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
how to show $a_{n}=[\frac{(2n)!!}{(2n-1)!!}]^2 \frac{1}{2n+1}$ converges? Question: $\displaystyle{a_{n} =
\left[{\left(2n\right)!! \over \left(2n - 1\right)!!}\,\right]^{2}
{1 \over 2n + 1}\,,\quad\mbox{prove}\ a_{n}}$ converges.
My thought: I want to prove {$a_{n}$} is an increasing sequence and it has an upper bound... | $$
\begin{align}
& a_m=\left(\frac{(2m)!!}{(2m-1)!!}\right)^2 \frac{1}{2m+1} = \prod_{n=1}^{m}\left(\frac{2n}{2n-1}\cdot\frac{2n}{2n+1}\right) \Rightarrow \\
& \log(a_m)=\sum_{n=1}^{m}\left[2\log(2n)-\log(2n-1)-\log(2n+1)\right] \\
& \text{Let}\space f(x) = -\log(x) \space\colon x\gt1 \space\Rightarrow f'(x) = -\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2047721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Solving an integral How to obtain an explicite formula for the following integral?
$$f_m(x_m)= \int_0^\infty\cdots \int_0^\infty x_1(x_1+x_2) \cdots (x_1+x_2+\cdots+x_m) e^{-\frac{(x_1+x_2+\cdots+x_m)^2}{2}} dx_1 dx_2 \cdots d x_{m-1},$$
where $m\geq 2$. We know that $f_2(x_2)= \int_{x_2}^\infty e^{-\frac{y^2}{2}} d y$... | It is convenient to consider the function $g_n(x) = f_{n+2}(x)$ with $n \geq 0$. Before the computation, we introduce some notations:
*
*$\Delta^n(a) = \{ (x_1, \cdots, x_n) \in \Bbb{R}^n : 0 < x_1 < \cdots < x_n < a \}$.
*$I_n(x)$ for $n \geq 0$ is defined recursively by $I_0 (x) = 1$ and $I_{n+1}(x) = \int_{0}^{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2048021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\sum\limits_{n=1}^\infty \frac{1}{(n+2)(n+4)^2}$
Calculate the sum of the series $$\sum_{n=1}^\infty \frac{1}{(n+2)(n+4)^2}$$
I have tried partial fraction decomposition.
$$\sum_{n=1}^\infty\frac{1}{4(n+2)}- \sum_{n=1}^\infty\frac{1}{4(n+4)}-\sum_{n=1}^\infty\frac{1}{2(n+4)^2}$$
Is this correct? What is t... | First note that your series has the same convergence as $\sum \frac{1}{n^3}$ by the limit comparison test. And the latter series converges absolutely by the $p$-series test or the integral test, so therefore so does your series.
Next, to find the value the sum converges to, trying partial fractions, we get
$$\frac{1}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2048838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve a system of equations for real $x,y,z$ Solve the following system of equations in $x,y,z \in \Bbb R$:
$$ (x+y)^3 = z $$
$$ (x+z)^3 = y$$
$$ (y+z)^3 = x$$
I found the solutions $(0,0,0)$, $(\frac{1}{2\sqrt{2}}, \frac{1}{2\sqrt{2}}, \frac{1}{2\sqrt{2}})$ and $(-\frac{1}{2\sqrt{2}}, -\frac{1}{2\sqrt{2}}, -\frac{1}{2... | This problem is very pretty:
Let's suppose that $x\geq y\geq z$ (any order will let you the same, you can check that). Because of this,
$$x+y\geq x+z\geq z+y$$
$$\Rightarrow (x+y)^3\geq (x+z)^3\geq (z+y)^3$$
But that means that:
$$z\geq y\geq x$$
So from that $x=y=z$ then for any equation you will have $8x^3=x$. From ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2049285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
limit of $\frac{x^2}{x^2+y^2}$ To prove $\frac{x^2}{x^2+y^2}$ has no limit at $(0,0)$ we can take $y=kx$ and therefore:
$$\lim_{(x,kx)\to (0,0)}\frac{x^2}{x^2+y^2}=\lim_{(x,kx)\to (0,0)}\frac{x^2}{x^2+k^2x^2}=\lim_{(x,kx)\to (0,0)}\frac{x^2}{x^2(1+k^2)}=\lim_{(x,kx)\to (0,0)}\frac{1}{(1+k^2)}$$
For different $k$ we wil... | The error is that you assume $y=kx\to3$, which is only true for $k=1$. This is not the case for the limit to $3$, but it is true when $(x,y)\to(0,0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2050138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Evalute $\int\frac{x^5}{(36x^2+1)^{3/2}}dx$ by trig sub? I'm stuck trying to evaluate the indefinite integral $\int\frac{x^5}{(36x^2+1)^{3/2}}dx$. It looks like it might be solvable by trig substitution, where $tan^2\theta+1=sec^2\theta$. That strategy seemed to payoff until I eliminated the square root. When I've work... | Hint: Change $$\frac{1}{6^6}\int\frac{\sin^5(\theta)}{\cos^4(\theta)}d\theta$$ to $$\frac{1}{6^6}\int\sin(\theta)\tan^4(\theta)d\theta$$ then go to
$$\frac{1}{6^6}\int\sin(\theta)(\sec^2(\theta) - 1)^2d\theta$$ then use a u substitution of $u = \cos\theta$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2050611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to solve system of equations involving square roots How to solve the following system of equations? I've tried some basic techniques like adding/substracting and squaring but with no effect.
$$
\left\{
\begin{array}{c}
\sqrt{1 + x_1} + \sqrt{1 + x_2} + \sqrt{1 + x_3} + \sqrt{1 + x_4} = 2\sqrt{5} \\
\sqrt{1 - x_1}... | Denote
$$\sqrt{1+x_i}=a_i,~~~\sqrt{1-x_i}=b_i, ~~~(i=1,2,3,4)$$
It's clear that $$a_i,b_i \geq 0,$$and$$a_i^2+b_i^2=2.\tag1$$
Moreover, the system of equations could be rewritten as $$\sum_{i=1}^4a_i=2\sqrt{5},~~~~\sum_{i=1}^4b_i=2\sqrt{3}.\tag2$$
From $(1)$, we have $$\sum_{i=1}^4a_i^2+\sum_{i=1}^4b_i^2=8.$$
From $(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2051270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
An interesting result in ratio and proportions If
$$
\frac{a}{b}=\frac{c}{d}=k
$$
then
$$
\frac{a+c}{b+d}=k
$$
Also
$$
\frac{a^2}{b^2}=\frac{c^2}{d^2}=k^2
$$
And
$$
\left( \frac{a+c}{b+d} \right)^2=k^2
$$
Also
$$
\frac{a^2 + c ^2}{b^2+d^2}=k^2
$$
Hence
$$
\frac{a^2 + c ^2}{b^2+d^2}= \left( \frac{a+c}{b+d} \right)^2
... | It is clear that the truth of $$ \frac{a^2+c^2}{b^2+d^2}=\frac{(a+c)^2}{(b+d)^2}$$
does not change if we swap $b\leftrightarrow d$, but the truth of $\frac ab=\frac cd$ may well change
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2051361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find all the possible values of $(a,b,c,d)$. Find all quadruples of real numbers $(a,b,c,d)$ satisfying the system of equations
$(b+c+d)^{2010}=3a$
$(a+c+d)^{2010}=3b$
$(a+b+d)^{2010}=3c$
$(a+b+c)^{2010}=3d$
I tried to find the solutions using hit and trial and by using some logic also I find two solutions which are $(... | If $(a,b,c,d)$ satisfies the equations, then we may well assume that $a\leq b\leq c\leq d$.These are non-negative because an even power of a real number is always non-negative. It follows that $$ b+c+d\geq a+c+d\geq a+b+d\geq a+b+c$$ since $x$ and hence $x^{2010}$ is increasing for $x\geq 0$, we have that $$3a=(b+c+d)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2053864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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The limit of a function with sum of two roots. I need to find the limit of the following function:
$$\lim_{x\to-\infty} \left(\sqrt{x^2+2x}+\sqrt[3]{x^3+x^2}\right).$$
I derived it to the form of:
$$\lim_{x\to-\infty}\frac{(x^2+2x)^3-(x^3+x^2)^2}{\left(\sqrt{x^2+2x}-\sqrt[3]{x^3+x^2}\right)\left((x^2+2x)^2+(x^2+2x)\sqr... | Let me change the sign on $x$, I find it easier to think about.
the expression is then
$$\sqrt{x^2-2x}+\sqrt[3]{x^2-x^3}$$
Now investigate the two limits
$$\sqrt{x^2-2x}-x\rightarrow -1$$
and
$$x+\sqrt[3]{x^2-x^3}$$ and this latter is equal
to
$$\frac{x^2}{x^2-x\sqrt[3]{x^2-x^3}+(\sqrt[3]{x^2-x^3})^2}$$
for which I ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}$ show that $a^a \cdot b^b\cdot c^c=1$ .
If $\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}$ show that $a^a \cdot b^b\cdot c^c=1$ .
My Working:
$\frac{\log a}{b-c}= \frac{\log b}{c-a}$
$ (c-a)\log a=(b-c) \log b$
$ \log a^{c-a}=\log b^{b-c}$
$ \fra... | It is given that $$\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}\tag1$$
Now $$\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log a+\log b}{b-c+c-a} \,\,\,\,\,\,\,\text {(by Addendo)}$$
$$=\frac{\log ab}{b-a} \tag2$$
So from $(1)$ and $(2)$, we get that $$\frac{\log c}{a-b}=\frac{\log ab}{b-a}$$
$$\implies ab ... | {
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"url": "https://math.stackexchange.com/questions/2055283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How can I derive the polar representation of Bernoulli's lemniscate, $r^2 = a^2\cos 2\theta$? If I start with the formula $(x^2+y^2)^2 -a(x^2-y^2) = 0$,
Converting to polar coordinates gives
$r^2 - a(r^2\cos^2 \theta - r^2 \sin^2 \theta) = 0$.
Applying the identity $ \cos^2 \theta - \sin^2 \theta = \cos 2\theta$,
I ... | You forgot to square the first term
$$
x^2+y^2 = r^2\implies (x^2+y^2)^2 = (r^2)^2 = r^4
$$
| {
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"source": "stackexchange",
"question_score": "1",
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Find the integral part of the product $\frac{2}{1} \cdot \frac{4}{3} \cdot \frac{6}{5} \cdot \frac{8}{7} \cdots \frac{2016}{2015}.$
Find the integral part of the following number $$T = \dfrac{2}{1} \cdot \dfrac{4}{3} \cdot \dfrac{6}{5} \cdot \dfrac{8}{7} \cdots \dfrac{2016}{2015}.$$
We can show that $T = 2017\int_{0}... | We may prove the inequality mentioned by achille hui in the comments without resorting to Stirling's approximation. For large values of $n$, we have:
$$ \frac{(2n)!!}{(2n-1)!!} = \prod_{k=1}^{n}\left(1-\frac{1}{2k}\right)^{-1} \tag{1} =2\prod_{k=2}^{n}\left(1-\frac{1}{2k}\right)^{-1}$$
and since $\left(1-\frac{1}{2k}\r... | {
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"source": "stackexchange",
"question_score": "5",
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Computing an integral by making it into a complex integral. I am computing an integral
$$\int_0^{2\pi} \frac{\sin^2(\theta)}{5-3\cos(\theta)}$$
I introduced the substitution $z=e^{i \theta}$ which implies $\frac{dz}{zi}=d\theta$. Using the complex analysis for $\sin$ and $\cos$ and this substitution we can write them a... | Your approach is reasonable, but there were a few careless/typographical errors along the way. For example, since $\sin(z)=\frac{z-z^{-1}}{2i}$, then $\sin^2(z)=\frac{z^2+z^{-2}-2}{-4}\ne \frac{z^2-z^{-2}}{-4}$.
METHODOLOGY $1$: MODIFY THE FORM OF THE INTEGRAND
However, there is a much more efficient way forward. P... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving no integer solution for $x^3+y^3+4=z^3$ I've been trying to shoe that there are no integer solutions for the equation:
$$x^3+y^3+4=z^3$$
What I've tried out so far is to prove by contradiction and to get to a contradiction using modulus.
Let us assume that there are $x,y,z\in\mathbb{Z}$ so that they solve the e... | Hint: Try the equation mod $9$. The cubes mod $9$ are $0,\pm1$.
Solution:
Write the equation as $x^3+y^3+(-z)^3=-4$. The range of the LHS mod $9$ is $-3,\dots,3$ and so is never $-4$.
| {
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"url": "https://math.stackexchange.com/questions/2058971",
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"source": "stackexchange",
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$\frac{2}{3}$ of the people on Weird Island tell the truth all the time and the rest lie all the time
$\frac{2}{3}$ of the people on Weird Island tell the truth all the
time and the rest lie all the time. You are sitting in a room with no
windows and two people come in from outside.
Person 1 says: "It is raining o... | According to Bayes theorem.
$P(E1 = truth) = \frac{1}{2}$
$P(E2 = lie) = 1- \frac{1}{2}$
$P(E2 = lie) = \frac{1}{2}$
$P\left(\frac{A}{E1}\right) = $people speaks truth $= \frac{2}{3}$
$P\left(\frac{A}{E2}\right) = $people speaks lie $= 1- \frac{2}{3}$
$P\left(\frac{A}{E2}\right) = $people speaks lie $= \frac{1}{3}$... | {
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What is the highest power of $18$ contained in $\frac{50!}{25!(50-25)!}$? What is the highest power of $18$ contained in $\frac{50!}{25!(50-25)!}$?
How will I be able to find the answer to such questions? Is there any special technique to find the answer to such problems? Thank you.
| $18 = 2\cdot 3^2$. We can find the power of a small prime in a large factorial by successive division to find base divisibility, then divisibility by squares, etc. So the multiplicity of powers of $2$ in $50!$, $v_2(50!),$ is
$$ v_2(50!) = \left\lfloor\frac{50}{2}\right\rfloor + \left\lfloor\frac{50}{4}\right\rfloor + ... | {
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Evaluate $\frac{1}{zx+y-1}+\frac{1}{zy+x-1}+\frac{1}{xy+z-1}$ if $x+y+z=2,x^2+y^2+z^2=3,xyz=4$ Evaluate $\frac{1}{zx+y-1}+\frac{1}{zy+x-1}+\frac{1}{xy+z-1}$ if $x+y+z=2,x^2+y^2+z^2=3, xyz=4$
The first thing that I notice is that it is symetric to $a,b,c$ but it can't help me .The other idea is finding the numbers but g... | Edit : Since the OP changes some signs, this answer also changes the signs.
You have already noticed that
$$\frac{1}{zx+y-1}+\frac{1}{yz+x-1}+\frac{1}{xy+z-1}=\frac{y}{y^2-y+4}+\frac{x}{x^2-x+4}+\frac{z}{z^2-z+4}$$
This answer shows that using that $x,y,z$ are the solutions of $t^3-2t^2+\frac 12t-4=0$ enables us to ha... | {
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"url": "https://math.stackexchange.com/questions/2062458",
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"source": "stackexchange",
"question_score": "6",
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Problem with an identity involving fractions. In a proof concerning normal distributions, the following step is a little difficult to understand:
$$\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}\sigma_Y} e^{-{(z-x-\mu_Y)^2 \over 2\sigma_Y^2}} \frac{1}{\sqrt{2\pi}\sigma_X} e^{-{(x-\mu_X)^2 \over 2\sigma_X^2}} \, dx$$
$$= \i... | To prove that
$$
-{(z-x-\mu_Y)^2 \over 2\sigma_Y^2}-{(x-\mu_X)^2 \over 2\sigma_X^2} = - { (z-(\mu_X+\mu_Y))^2 \over 2(\sigma_X^2+\sigma_Y^2) } - \frac{\left(x-\frac{\sigma_X^2(z-\mu_Y)+\sigma_Y^2\mu_X}{\sigma_X^2+\sigma_Y^2}\right)^2}{2\left(\frac{\sigma_X\sigma_Y}{\sqrt{\sigma_X^2+\sigma_Y^2}}\right)^2}
$$
we could fi... | {
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"url": "https://math.stackexchange.com/questions/2062850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the number of integer solutions of $2{x_{1}}+2{x_{2}}+{x_{3}}+{x_{4}}={12}$? What is the number of non negative integer solutions of $2{x_{1}}+2{x_{2}}+{x_{3}}+{x_{4}}={12}$ ?
I tried it as :
${x_{3}}+{x_{4}}={12} - 2{x_{1}}-2{x_{2}}$
Now, finding the solutions of ${x_{1}}+{x_{2}}$
*
*${x_{1}}+{x_{2}} = 0 ... | Here is another solution, using generating formulas:
Let the power of $x$ represent the value of $x_i$.
Then we have $$(1+x^2+x^4+\cdots)\times (1+x^2+x^4+\cdots)\times(1+x+x^2+\cdots)\times (1+x+x^2+\cdots)$$
$$=\left(\frac{1}{1-x^2}\right)^2\cdot\left(\frac{1}{1-x}\right)^2$$
$$=\frac{1}{(1-x^2)^2(1-x)^2}$$
$$=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2065465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solutions for diophantine equation $3^a+1=2^b$ I am looking for solutions for the diophantine equation
$3^a+1=2^b$
where $a\in \Bbb N$ and $b\in \Bbb N$.
Is there a power of $3$ that gives a power of $2$ when you add $1$?
Two solutions are easy to find:
*
*$3^0+1=2^1 \rightarrow 1+1=2$
*$3^1+1=2^2 \rightarrow ... | Another solution: consider the powers of $3$$\pmod 8$. We have $3^1\equiv 3\pmod 8$ and $3^2\equiv 1\pmod 8$. So by induction it follows that $3^{2k+1}\equiv 3\pmod 8$ and $3^{2k}\equiv 1\pmod 8$, for $ k\in\Bbb {N} $. So $3^a+1$ would be congruent with $4$ or $2$$\pmod 8$. Therefore we deduce that if $3^a+1=2^b $, th... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to calculate the determinant of a $4 \times 4$ matrix with multiple variables? What is the determinant: $$ \begin{vmatrix}1& a & a^2 & a^4 \\ 1 & b & b^2 & b^4 \\ 1 & c & c^2 & c^4 \\1 & d & d^2 &d^4 \end{vmatrix} $$
Someone gave me the following hint
Replace $d$ by a variable $x$; make use of the fact that the s... | As you already noted the determinant is a polynomial of 4th degree in $d$.
This polynomial is zero if you replace $d$ by $a$, $b$, or $c$ (we have two identical rows).
Moreover the coefficient of $d^3$ is zero which implies that the sum of the roots is zero. Hence the fourth root is $-(a+b+c)$.
Finally, by Vandermond... | {
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"source": "stackexchange",
"question_score": "26",
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Prove that $\ln{\left({A^6\sqrt{\pi}\over 2^{7\over6}e}\right)}=\sum\limits_{n=1}^{\infty}{(-1)^{n+1}\over n(n+1)}\eta(n)$ Show that
$$\ln{\left({A^6\sqrt{\pi}\over 2^{7\over6}e}\right)}=\sum_{n=1}^{\infty}{(-1)^{n+1}\over n(n+1)}\eta(n)$$
(where $\eta(n)$ is the Dirichlet eta function, and A is the Glaisher-Kinkelin c... | Another approach is to use a common integral representation of the Dirichlet eta function, along with an integral representation of the Hurwitz zeta function.
Specifically, we can use $$\eta(s) = \frac{1}{\Gamma(s)}\int_{0}^{\infty} \frac{x^{s-1}}{1+e^{x}} \, dx , \quad \text{Re}(s) >0,\tag{1}$$ and $$\zeta(s,z) = \fr... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Prove $\frac {x^{2a}}{x^{2a}+x^{m-b}+x^{m-c}} + \frac{x^{2b}}{x^{2b}+x^{m-c}+x^{m-a}}+ \frac {x^{2c}}{x^{2a}+x^{m-a}+x^{m-b}}=1$ for $a+b+c=m$
If $a+b+c=m$, prove that:
$$\frac {x^{2a}}{x^{2a}+x^{m-b}+x^{m-c}} + \frac {x^{2b}}{x^{2b}+x^{m-c}+x^{m-a}}+ \frac {x^{2c}}{x^{2a}+x^{m-a}+x^{m-b}}=1.$$
My Attempt:
$a+b+c=m... | Hint:
$$
\frac {x^{2a}}{x^{2a}+x^{m-b}+x^{m-c}} = \frac {x^{a}}{x^{a}+x^{m-b-a}+x^{m-c-a}} = \frac {x^{a}}{x^{a}+x^{b}+x^{c}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How prove $\sum\limits_{cyc}\sqrt{a^2+b^2-c^2}\sqrt{a^2-b^2+c^2}\le ab+bc+ca$ Let acute-angled triangle $ABC$,and $AB=c,BC=a,AC=b$,show that
$$\sum_{cyc}\sqrt{a^2+b^2-c^2}\sqrt{a^2-b^2+c^2}\le ab+bc+ca$$
I try use AM-GM
$$\sum_{cyc}\sqrt{a^2+b^2-c^2}\sqrt{a^2-b^2+c^2}\le\sum_{cyc}\dfrac{(a^2+b^2-c^2)+(a^2-b^2+c^2)}{2}=... | Let $a^2+b^2-c^2=z^2$, $a^2+c^2-b^2=y^2$ and $b^2+c^2-a^2=x^2$, where $x$, $y$ and $z$ are positives.
Hence, we need to prove that $\sum\limits_{cyc}\sqrt{(x^2+y^2)(x^2+z^2)}\geq2(xy+cz+yz)$,
which after using C-S $\sqrt{(x^2+y^2)(x^2+z^2)}\geq x^2+yz$
gives $\sum\limits_{cyc}(x^2-xy)$, which is $\sum\limits_{cyc}(x-y... | {
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"source": "stackexchange",
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How to quickly solve $y=\int_{-\pi/4 }^{\pi/4 } \left[\cos x + \sqrt{1+x^2}\sin^3x\cos^3x\right]dx$? I'm currently trying to solve $$y=\int_{-\pi/4 }^{\pi/4 } \left[\cos x + \sqrt{1+x^2}\sin^3x\cos^3x\right]dx,$$ which is a GRE math subject test problem. I was able to get the answer ($\sqrt{2}$) by breaking up the inte... | If you noticed that $\sqrt{1+x^2}\sin^3 x \cos^3 x$ was an odd function, you would immediately get that $$\displaystyle\int_{-\pi/4}^{\pi/4}\sqrt{1+x^2}\sin^3 x \cos^3 x\,dx = 0,$$
and thus, the given integral simplifies to just $y = \displaystyle\int_{-\pi/4}^{\pi/4}\cos x\,dx$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2074230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Showing that $f(x) = x^3$ is injective? This is my attempt. Is it right? Is there a simpler way? Is there a way which relies on less background knowledge?
A function is injective iff $f(x) = f(y) \implies x = y$. This is equivalent to $x \not = y \implies f(x) \not = f(y)$. We will prove this latter statement by contr... | Is there a way which relies on less background knowledge?
Yes, there is (probably the most elementary one, which is based essentially in the monotonicity of multiplication):
Claim 1. If $x\neq 0$, $y\neq 0$ and $x^3=y^3$ then $x$ and $y$ have the same sign.
Proof: Assume that $x^3=y^3$ with $x>0$ and $y<0$. Then
$$0<... | {
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"source": "stackexchange",
"question_score": "8",
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Find the remainder of $51!$ when divided by $61$? Find the remainder of $51!$ when divided by $61$ ?
My Try :
By Wilson's theorem $60! ≡ −1\pmod{61}$
Then, I can write
$(60)(59)(58)(57)(56)(55)(54)(53)(52)51!≡−1\pmod{61}$
$(−1)(−2)(−3)(−4)(−5)(−6)(−7)(−8)(−9)51!≡−1\pmod{61}$
$(362880)51!\equiv1\pmod{61}$
How can I p... | Continued from your working without evaluating $9!$ explicitly:
$$51!9!\equiv 1 \mod 61$$
$$51! 2(3)(4)(5)(6)(7)(8)(9) \equiv 1 \mod 61$$
Since $2(5)(6)=60$,
$$51! (60)(3)(4)(7)(8)(9) \equiv 1 \mod 61$$
$$51! (-1)(3)(4)(7)(8)(9) \equiv 1 \mod 61$$
$$51! (3)(4)(7)(8)(9) \equiv -1 \mod 61$$
Since $9(7)=63$,
$$51! (3)(4)(... | {
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"source": "stackexchange",
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How to find $3^{3^{3^{\dots}}}\pmod{100}$? I can show that $3^{3^{3^n}}\equiv7\pmod{10}$ since
$3^1\equiv3\pmod{10}$
$3^2\equiv9\pmod{10}$
$3^3\equiv7\pmod{10}$
$3^4\equiv1\pmod{10}$
Thus, it reduces to $3^{(3^{3^n}\mod4)}$. I can then notice that
$3^1\equiv3\pmod4$
$3^2\equiv1\pmod4$
Reducing it down to $3^{(3^{(3^n\... | We know that
$$3^2 \equiv 1 \pmod{4} \\
3^{20} \equiv 1 \pmod{25}$$
with the last following from Euler Theorem. Therefore
$$3^{20} \equiv 1 \pmod{100}$$
The problem then reduces to finding the powers of $3 \pmod{20}$.
Again
$$3^2 \equiv 1 \pmod{4} \\
3^4 \equiv 1 \pmod{5} \\$$
Therefore $3^4 \equiv 1 \pmod{20}$.
We t... | {
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"url": "https://math.stackexchange.com/questions/2077664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Mathematical Induction on Fibonacci numbers I have ben stuck on this for a while.
Let $F(N)$ be the Fibonacci numbers with $F(1)=F(2)=1$. Show that $4(-1)^n + 5(F(N))^2$ is a square for all integers $N$.
| This doesn't prove it inductively, so if you specifically need an inductive proof, this wouldn't work.
Instead, this uses the closed form for the Fibonacci sequence, which is that $F(N)=\dfrac{\alpha^N-\beta^N}{\sqrt{5}}$, where $\alpha=\frac{1+\sqrt{5}}{2}$ and $\beta=\frac{1-\sqrt{5}}{2}=\frac{-1}{\alpha}$.
The expr... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral involving $\sqrt{\tan x}$ $$\int \dfrac{1}{1+\sqrt{\tan x}}\quad dx$$
$$\int \dfrac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}\quad dx$$
It is difficult for me to solve this integration.
| $\displaystyle \mathcal{I} = \int\frac{1}{1+\sqrt{\tan x}}dx = \int \frac{\sqrt{\cot x}}{1+\sqrt{\cot x}}dx$
substitute $\cot x= t^2$ and $\displaystyle dx = -\frac{1}{1+t^4}dt$
$\displaystyle \mathcal{I}= -\int\frac{t}{(1+t)(1+t^4)}dt = -\frac{1}{2}\int\frac{\bigg((1+t^4)+(1-t^4)\bigg)t}{(1+t)(1+t^4)}dt$
$\displaysty... | {
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"source": "stackexchange",
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Prove something similar to a variant of Cauchy-Shwarz inequality Cauchy-Shwarz Inequality is:
$$(a_1b_1 + a_2b_2 + \cdots + a_nb_n)^2 \leq (a_1^2 + a_2^2 + \cdots + a_n^2)(b_1^2 + \cdots + b_n^2)$$
However, it can be manipulated as:
$$\sqrt{(a_1-b_1)^2 + (a_2-b_2)^2 + \cdots + (a_n-b_n)^2} \leq \sqrt{a_1^2 + a_2^2 + \c... | Hint:
Replacing $b_i$ by $-b_i$ one can transform
$$
\sqrt{(a_1-b_1)^2+\cdots+(a_n-b_n)^2}\leq \sqrt{a_1^2+\cdots+a_n^2}+\sqrt{b_1^2+\cdots+b_n^2}
$$
into
$$
\sqrt{(a_1+b_1)^2+\cdots+(a_n+b_n)^2}\leq \sqrt{a_1^2+\cdots+a_n^2}+\sqrt{b_1^2+\cdots+b_n^2}
$$
Then,
\begin{align*}
\sqrt{(a_1+b_1+c_1)^2+\cdots+(a_n+b_n+c_n)^... | {
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"url": "https://math.stackexchange.com/questions/2083990",
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Circumradius of triangle ABC where H is orthocentre and $(AH)(BH)(CH) = 3 $ and $(AH)^2 + (BH)^2 + (CH)^2 = 7 $ Let ABC is an acute angled triangle with orthocentre H. D, E, F are feet of perpendicular from A, B, C on opposite sides. Let R is circumradius of ΔABC. Given $(AH)(BH)(CH) = 3$
and $ (AH)^2 + (BH)^2 + (CH)^... | As you wrote, we have
$$AH=2R\cos A,\quad BH=2R\cos B,\quad CH=2R\cos C$$
Now using that
$$\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1\tag1$$
(the proof for $(1)$ is written at the end of this answer)
we get
$$\frac{7}{4R^2}+2\cdot\frac{3}{8R^3}=1,$$
i.e.
$$(R+1)(2R-3)(2R+1)=0$$
to have
$$\color{red}{R=\frac 32}$$
L... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2084344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the power series of rational function How to find the power series of rational functions of type, e.g.
$$\frac{1}{1-6x+12x^2},\frac{6x}{1-6x+12x^2}$$
where denominator can't be factorized over the real domain.
Is there a way to use the method of undetermined coefficients (by completing the square), or is it necess... | Here is an approach based upon the geometric series expansion (see the comment from @YvesDaoust).
We obtain
\begin{align*}
\frac{1}{1-6x+12x^2}&=\frac{1}{1-6x(1-2x)}\\
&=\sum_{k=0}^\infty (6x)^k(1-2x)^k\\
&=\sum_{k=0}^\infty (6x)^k\sum_{j=0}^k\binom{k}{j}(-2x)^j\\
&=1+6x+24x^2+72x^3+144x^4\\
&\qquad+\color{grey}{0}x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2086397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the last two digits in the decimal representation (base 10) of 17^20 My attempt is in this image. But I want the answer to this question by using modulo congruence method since the method used by me involves tedious calculations.
| The units digit of $17^1$ is equal to $7$.
The units digit of $17^2$ is equal to $9$.
The units digit of $17^3$ is equal to $3$.
The units digit of $17^4$ is equal to $1$.
Calculating,
$\quad 17^4 = (17^2)(17^2) = (289)(289) \equiv (-11)(-11) \equiv 21 \pmod{100}$
Continuing,
$\quad 21^2 \equiv 41 \pmod{100}$
$\quad 21... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
There exists $n$ such that there exists an n-digit number M for which $M(10^n + 1)$ is a perfect square. Show that there exists a positive integer $n$ greater than 3 such that there exists an $n$-digit positive integer $M$ for which $M(10^n + 1)$ is a perfect square.
| As I explained in the comments, this is impossible if $10^n+1$ is squarefree. On the other hand, if $p^2$ divides $10^n+1$ for some prime $p$, then $M_0=\frac{10^n+1} {p^2}$ is an integer, and $M_0(10^n+1)=\left(\frac{10^n+1} {p}\right)^2$ is a perfect square. $M_0$ does not necessarily have $n$ digits; but we can alwa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solutions for $a^2+b^2+c^2=d^2$ I have to find an infinite set for $a,b,c,d$ such that $a^2+b^2+c^2=d^2$ and $a,b,c,d\in\mathbb N$
I have found one possible set which is $x,2x,2x,3x$ and $x\in\mathbb N$
But I want another infinite sets and how to reach them because I have found my solution by hit and trial .
| Neater solution. (not obvious where I got formula from)
Note that: $(b^2-a^2)^2+(2ab)^2+(2cd)^2-(c^2+d^2)^2=(a^2+b^2+c^2-d^2)(a^2+b^2-c^2+d^2)$.
So if you have a solution $(a,b,c,d)$ to $a^2+b^2+c^2-d^2=0$ then another solution is:
$$\bigg(b^2-a^2,2ab,2cd,c^2+d^2\bigg)$$
Applying this to your solution $(1,2,2,3)$ gives... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Find the limit of $6^n(2-x_n)$ where $x_n=\sqrt[3]{6+\sqrt[3]{6+\dots+\sqrt[3]{6}}}$ with $n$ roots
Let $x_n=\sqrt[3]{6+\sqrt[3]{6+\dots+\sqrt[3]{6}}}$ where the expression in the RHS has $n$ roots.
Find the following limit: $\lim \limits_{n\to \infty}6^n(2-x_n)$
My approach: I had two approaches. The first one was t... | takes $f(x)= \sqrt[3]{x+6}$ then see that $x_{n+1} = f(x_n)$
and $$|f'(x) |= \frac{1}{3\sqrt[3]{(x+6)^2}} < \frac{1}{3\sqrt[3]{36}} <\frac{1}{3\sqrt[3]{2^3\times 3}}=\frac{1}{6\sqrt[3]{3}}$$
so that
for any $x,y $ one has
$$|f(x)-f(y)| = |\int_x^yf'(s)ds|\le \frac{1}{6\sqrt[3]{3}}|x-y|$$
show then by induction using ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Finding the Smith Normal Form of an integer matrix I want to put the following integer matrix into Smith Normal Form:
$$\begin{pmatrix} -9 & 6 \\ 5 & -2 \\ 6 & 3 \end{pmatrix}$$
I have done this and found the answer to be $$\begin{pmatrix} 1 & 0 \\ 0 & 3 \\ 0 & 0\end{pmatrix}$$
Could someone verify whether this is corr... | $$\begin{pmatrix}-9 & 6 \\ 5 & -2 \\ 6 & 3\end{pmatrix}\to\begin{pmatrix}2 & -5 \\ 6 & -9 \\ 3 & 6\end{pmatrix}\to \begin{pmatrix}2 & -5 \\ 0 & 6 \\ 1 & 11\end{pmatrix}\to\begin{pmatrix}1 & 11 \\ 0 & 6 \\ 0 & -27\end{pmatrix}\to\begin{pmatrix}1 & 0 \\ 0 & 6 \\ 0 & 3\end{pmatrix}\to\begin{pmatrix}1 & 0 \\ 0 & 3 \\ 0 & 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2092186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Rotation Matrix of rotation around a point other than the origin In homogeneous coordinates, a rotation matrix around the origin can be described as
$R = \begin{bmatrix}\cos(\theta) & -\sin(\theta) & 0\\\sin(\theta) & \cos(\theta) & 0 \\ 0&0&1\end{bmatrix}$
with the angle $\theta$ and the rotation being counter-clockwi... | Your first formula is correct. Remember, the point to which this is applied appears on the RIGHT:
$$
T(x,y) * R * T(-x,-y) (P)
$$
So to evaluate the expression above, we first translate $P$ by $(-x, -y)$, then rotate the result, then translate back. Let's see what happens when $P$ is the point $(x, y, 1)$. That amounts... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2093314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "48",
"answer_count": 3,
"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.