Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find the inverse of $5+4\sqrt[3]{2}+3\sqrt[3]{4}$? I'm trying to find the inverse of $5+4\sqrt[3]{2}+3\sqrt[3]{4}$ in $\mathbb{Z}[\sqrt[3]{2}]$. I know it is a unit, so there is an inverse, but I feel like I may be doing too much work in the wrong direction. Here's what I have so far:
Let $\alpha = 5+4\sqrt[3]{2}+3\sqrt[3]{4}$ and $\alpha^{-1} = a+b\sqrt[3]{2}+c\sqrt[3]{4}$ for some $a,b,c \in \mathbb{Z}$.
$(a+b\sqrt[3]{2}+c\sqrt[3]{4})(5+4\sqrt[3]{2}+3\sqrt[3]{4})=1$
$=5a+6b+8c+4a\sqrt[3]{2}+5b\sqrt[3]{2}+6c\sqrt[3]{2}+3a\sqrt[3]{4}+4b\sqrt[3]{4}+5c\sqrt[3]{4}=1$
$=a(5+4\sqrt[3]{2}+3\sqrt[3]{4})+b(6+5\sqrt[3]{2}+4\sqrt[3]{4})+c(8+6\sqrt[3]{2}+5\sqrt[3]{4})=1$
and trying to solve for a,b, and c, but I don't know how?
Edit: Regrouping to $(5a+6b+8c)+(4a+5b+6c)\sqrt[3]{2}+(3a+4b+5c)\sqrt[3]{4}=1$
| $$
\left(
\begin{array}{ccc|c}
5 & 6 & 8 & 1 \\
4 & 5 & 6 & 0 \\
3 & 4 & 5 & 0
\end{array}
\right)
$$
$$
\left(
\begin{array}{ccc|c}
1 & 1 & 2 & 1 \\
4 & 5 & 6 & 0 \\
3 & 4 & 5 & 0
\end{array}
\right)
$$
$$
\left(
\begin{array}{ccc|c}
1 & 1 & 2 & 1 \\
1 & 1 & 1 & 0 \\
3 & 4 & 5 & 0
\end{array}
\right)
$$
$$
\left(
\begin{array}{ccc|c}
0 & 0 & 1 & 1 \\
1 & 1 & 1 & 0 \\
3 & 4 & 5 & 0
\end{array}
\right)
$$
$$
\left(
\begin{array}{ccc|c}
0 & 0 & 1 & 1 \\
1 & 1 & 1 & 0 \\
0 & 1 & 2 & 0
\end{array}
\right)
$$
$$
\left(
\begin{array}{ccc|c}
0 & 0 & 1 & 1 \\
1 & 1 & 1 & 0 \\
0 & 1 & 0 & -2
\end{array}
\right)
$$
$$
\left(
\begin{array}{ccc|c}
0 & 0 & 1 & 1 \\
1 & 0 & 1 & 2 \\
0 & 1 & 0 & -2
\end{array}
\right)
$$
$$
\left(
\begin{array}{ccc|c}
0 & 0 & 1 & 1 \\
1 & 0 & 0 & 1 \\
0 & 1 & 0 & -2
\end{array}
\right)
$$
Apparently
$$ a = 1, \; \; b = -2, \; \; c = 1 \; \; . $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Diophantine Equation on squares and cubes Find all integer solutions to: $(a^2+b)(b^2+a)=(a-b)^3$
I've found some of the trivial cases, just finding difficulty proving the existence (or not existence) of others. Perhaps taking $mod5$ or something?
| (EDITED)
$$(a^2+b)(b^2+a)-(a-b)^3 = b(a^2 b+3 a^2-3 a b+2 b^2+a)$$
The factor $b$ means $a=arbitrary, b=0$ are solutions.
The curve $a^2 b+3 a^2-3 a b+2 b^2+a=0$ has genus $0$ and rational parametrization
$$ a = {\frac {-2{s}^{2}}{ \left( 2\,s+1 \right) \left( s+1 \right) }}, b = -{
\frac { \left( 4\,s+1 \right) s}{(2s+1)^2}}
$$
If $s = S/T$ with $S, T$ coprime integers, we get
$$ \eqalign{a &= -1 + \frac{2T}{S+T} - \frac{T}{2S+T}\cr
b &= -1 + \frac{3T}{2(2S+T)} - \frac{T^2}{2(2S+T)^2} \cr}$$
If $2S+T$ is divisible by any prime, this can't be an integer. So the only possible cases are:
$2S+T=\pm 1$, which leads to the integer solutions $(a,b) = (-1,-1), (0,0), (8,-10), (9,-21)$.
EDIT: Oops, you could also have $2S+T = \pm 2$ if $T$ is divisible by $4$.
This leads to the additional solution $(9,-6)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given $a + b$ is prime and $a < -1$, show $a^2 + ab + b < 2$ I want to show $a^2 + ab + b < 2$ given $a, b \in \mathbb{Z}$, $a + b$ is prime and finally $a < -1$.
I've reduced the bounds a little. As $a + b$ is prime, $a + b \ge 2$. Since $a \in \mathbb{Z}$ the first integer less than $-1$ is $-2$, and thus $a \le -2$.
Starting with $a + b \ge 2$ I've done the following steps.
$$a + b \ge 2 \implies b \ge 2 - a$$
$$\implies ab \le a(2 - a)$$
$$\implies a^2 + ab + b \le a^2 + a(2 - a) + b$$
$$\implies a^2 + ab + b \le a^2 + 2a - a^2 + b \implies a^2 + ab + b \le 2a + b$$
However this is nowhere near a tight enough upper bound to show the result I am looking for. How should I continue?
| Let $a+b = p$. $p$ prime so $p\geq 2$. Then $a^2+ab+b = (p-1)a+p$. What can you conclude now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2736169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluation of a certain polynomial on a matrix. Given $A=\left[\begin{matrix}\dfrac{-1+i\sqrt{3}}{2i}&\dfrac{-1-i\sqrt{3}}{2i}\\\dfrac{1+i\sqrt{3}}{2i}&\dfrac{1-i\sqrt{3}}{2i}\end{matrix}\right]$ where $i=\sqrt{-1}$.
Also Given$f(x)=x^2+2.$
Find $f(A)$
My approach:
A can be written as
$A=i\left[\begin{matrix}w&w^2\\w^2&w\end{matrix}\right]$ Here $1+w+w^2=0$ i.e $w,w^2 \in {cube \space roots\space of \space \space unity }$
$\therefore \space f(A)=-((\left[\begin{matrix}w&w^2\\w^2&w\end{matrix}\right])^2-2)$
$=2-\left[\begin{matrix}-1&2\\2&-1\end{matrix}\right]$
Another approach:
$A=\left[\begin{matrix}\dfrac{-1+i\sqrt{3}}{2i}&\dfrac{-1-i\sqrt{3}}{2i}\\\dfrac{1+i\sqrt{3}}{2i}&\dfrac{1-i\sqrt{3}}{2i}\end{matrix}\right]$
$=i\left[\begin{matrix} cos(\dfrac{2\pi}{3})+isin(\dfrac{2\pi}{3})&cos(\dfrac{2\pi}{3})-isin(\dfrac{2\pi}{3})\\-cos(\dfrac{2\pi}{3})+isin(\dfrac{2\pi}{3})&cos(\dfrac{2\pi}{3})-isin(\dfrac{2\pi}{3})\end{matrix}\right] $
Now using DeMoivres' Theorem I was thinking what to be done. Let me think...
Now the options given are:
a. $\left[\begin{matrix}1&0\\0&1\end{matrix}\right]$
b. $\dfrac{3-i\sqrt{3}}{2} \cdot \left[\begin{matrix}1&0\\0&1\end{matrix}\right]$
c. $\dfrac{5-i\sqrt{3}}{2} \cdot \left[\begin{matrix}1&0\\0&1\end{matrix}\right]$
d. $(2+i\sqrt{3}) \cdot \left[\begin{matrix}1&0\\0&1\end{matrix}\right]$
But my answer does not come close to any of the given options. Could someone selflessly help me out please!
| Your expression for $A$ is not correct. You have
$$
A=-i\begin{bmatrix} w&w^2\\ -w^2&-w\end{bmatrix}=-iw\begin{bmatrix} 1&w\\-w&-1\end{bmatrix}
$$
Then
$$
A^2+2I=-w^2\begin{bmatrix} 1-w^2&0\\0&1-w^2\end{bmatrix}+2I=\begin{bmatrix} w-w^2+2&0\\0&w-w^2+2\end{bmatrix}=(2+i\sqrt 3)I,
$$
since
$$
w-w^2+2=-\frac12+\frac i2\sqrt3+\frac12+\frac i2\sqrt3+2=2+i\sqrt3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2736930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that if the matrix $\begin{bmatrix}a & b\\b & c\end{bmatrix}$ is nonnegative definite, then it has a factorization $LL^{T}$ Prove that if the matrix $\begin{bmatrix}a & b\\b & c\end{bmatrix}$ is nonnegative definite, then it has a factorization $LL^{T}$ in which $L$ is lower triangular.
I am using the following theorem:
Theorem: If $A$ is real, symmetric, and pisitive definite matrix, then it has a unique factorization, $A=LL^{T}$, in which $L$ is lower triangular with a positive diagonal.
Clearly $A$ is real, is symmetric since $\begin{bmatrix}a & b\\b & c\end{bmatrix}^{T}=\begin{bmatrix}a & b\\b & c\end{bmatrix}$ and is positive definite or is of the form P, di is positive definite is of the form $\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}$ for Find the precise conditions on $a,b,c$ so that $\left[\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right]$ is nonnegative definite. but
P has a $LL^{T}$ factorization trivially.
Is this reasoning right? Thank you very much.
| The (semi)definedness of the matrix tells you that $a\ge0$, $c\ge0$ and $ac-b^2\ge0$.
In particular, if $a=0$, then also $b=0$, so the matrix is
$$
\begin{bmatrix} 0 & 0 \\ 0 & c\end{bmatrix}=
\begin{bmatrix} 0 & 0 \\ 0 & \sqrt{c}\end{bmatrix}
\begin{bmatrix} 0 & 0 \\ 0 & \sqrt{c}\end{bmatrix}^T
$$
Suppose $a>0$. We can go with Gauss-Jordan elimination; let's first assume $d=ac-b^2>0$:
\begin{align}
\begin{bmatrix} a & b \\ b & c \end{bmatrix}
&\to
\begin{bmatrix} 1 & b/a \\ b & c \end{bmatrix}
&& R_1\gets\tfrac{1}{a}R_1
\\[6px]&\to
\begin{bmatrix} 1 & b/a \\ 0 & d/a \end{bmatrix}
&& R_2\gets R_2-bR_1
\\[6px]&\to
\begin{bmatrix} 1 & b/a \\ 0 & d \end{bmatrix}
&& R_2\gets aR_2
\end{align}
This means that
\begin{align}
\begin{bmatrix} a & b \\ b & c \end{bmatrix}
&=\begin{bmatrix} a & 0 \\ b & a^{-1} \end{bmatrix}
\begin{bmatrix} 1 & b/a \\ 0 & d \end{bmatrix} \\[6px]
&=\begin{bmatrix} \sqrt{a} & 0 \\ b/\sqrt{a} & \sqrt{d}/\sqrt{a} \end{bmatrix}
\begin{bmatrix} \sqrt{a} & 0 \\ 0 & 1/\sqrt{ad} \end{bmatrix}
\begin{bmatrix} 1 & b/a \\ 0 & d \end{bmatrix} \\[6px]
&=\begin{bmatrix} \sqrt{a} & 0 \\ b/\sqrt{a} & \sqrt{d}/\sqrt{a} \end{bmatrix}
\begin{bmatrix} \sqrt{a} & b/\sqrt{a} \\ 0 & \sqrt{d}/\sqrt{a} \end{bmatrix}
\end{align}
If $d=0$ the above decomposition is good as well:
$$
\begin{bmatrix} \sqrt{a} & 0 \\ b/\sqrt{a} & 0 \end{bmatrix}
\begin{bmatrix} \sqrt{a} & b/\sqrt{a} \\ 0 & 0 \end{bmatrix}
=\begin{bmatrix} a & b \\ b & c\end{bmatrix}
$$
because $b^2/a=c$.
| {
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"url": "https://math.stackexchange.com/questions/2737020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Alternate solutions to similar integrals? So, when I started this problem I couldn't see any reasonable trig substitutions that would make it simpler, and I don't know of a better method, so I attempted to solve it using repeated use of integration by parts. Is there a much easier, better, alternate solution to integrals of this nature in general? I'd also like to know where I went wrong with this solution.
$$
\begin{align*}
\int\sin^4x\,\mathrm{d}x&\rightarrow\\
u_1&=\sin^4x\\
u_1'&=4\cos x\sin^3x\\
v_1&=x\\
v_1'&=1\\
\leftarrow\int\sin^4x\,\mathrm{d}x&=x\sin^4x-4\int x\cos x\sin^3x\,\mathrm{d}x\\
\rightarrow\int x\cos x\sin^3x\,\mathrm{d}x&\rightarrow\\
u_2&=\sin^3x\\
u_2'&=3\cos x\sin^2x\\
v_2&=\int x\cos x\,\mathrm{d}x\\
v_2'&=x\cos x\\
\rightarrow\int x\cos x\,\mathrm{d}x\rightarrow\\
u_3&=x\\
u_3'&=1\\
v_3&=\sin x\\
v_3'&=\cos x\\
\leftarrow\int x\cos x\,\mathrm{d}x&=x\sin x-\int\sin x\,\mathrm{d}x\\
&=x\sin x+\cos x\\
\leftarrow\int x\cos x\sin^3x\,\mathrm{d}x&=\left(\sin^3x\right)\left(x\sin x+\cos x\right)-\int\left(3\cos x\sin^2x\right)\left(x\sin x+\cos x\right)\,\mathrm{d}x\\
&=x\sin^4x+\cos x\sin^3x-3\int x\cos x\sin^3 x\,\mathrm{d}x+3\int\cos^2
x\sin^2x\,\mathrm{d}x\\
4\int x\cos x\sin^3x\,\mathrm{d}x&=x\sin^4
x+\cos x\sin^3 x+3\int\cos^2x\sin^2x\,\mathrm{d}x\\
&=x\sin^4
x+\cos x\sin^3 x+3\int\left(1-\sin^2x\right)\sin^2x\,\mathrm{d}x\\
&=x\sin^4
x+\cos x\sin^3 x+3\int\sin^2x\,\mathrm{d}x-\int\sin^4x\,\mathrm{d}x\\
\rightarrow\int\sin^2x\,\mathrm{d}x\rightarrow\\
\cos2x&=1-2\sin^2x\\
\sin^2x&=\frac{1}{2}(1-\cos2x)\\
\leftarrow\frac{1}{2}\int1-\cos2x\,\mathrm{d}x&=\frac{1}{2}x-\frac{1}{4}\sin{2x}\\
\leftarrow4\int x\cos x\sin^3x\,\mathrm{d}x&=x\sin^4
x+\cos x\sin^3 x+3\left[\frac{1}{2}x-\frac{1}{4}\sin{2x}\right]-\int\sin^4x\,\mathrm{d}x\\
&=x\sin^4 x+\cos x\sin^3 x+\frac{3}{2}x-\frac{3}{4}\sin{2x}-\int\sin^4x\,\mathrm{d}x\\
\leftarrow\int\sin^4x\,\mathrm{d}x&=x\sin^4x-\left[x\sin^4 x+\cos x\sin^3 x+\frac{3}{2}x-\frac{3}{4}\sin{2x}-\int\sin^4x\,\mathrm{d}x\right]\\
&=x\sin^4x-x\sin^4x-\cos x\sin^3x-\frac{3}{2}x+\frac{3}{4}\sin2x+\int\sin^4x\,\mathrm{d}x\\
\end{align*}
$$
As you can see, I can't continue because the integral cancels itself.
| $\sin^2 x = \frac 12 (1 - \cos 2x)\\
\sin^4 x = \frac 14 (1 - \cos 2x)^2 = \frac 14 (1 - 2\cos 2x + \cos^2 2x)\\
\cos^2 2x = \frac 12 (1+\cos 4x)\\
\sin^4 x = \frac 14 (1 - \cos 2x)^2 = \frac 14 (1 - 2\cos 2x + \frac 12(1+\cos 4x))$
and simplify.
This is a nifty trick you can use if you know Euler's identity.
$e^{ix} = \cos x + i\sin x\\
\cos x = \frac {e^{ix} + e^{-ix}}{2}, \sin x = \frac {e^{ix} - e^{-ix}}{2i}\\
\sin^4 x = \frac {e^{4ix} + 4e^{2ix} + 6 + 4e^{-2ix} + e^{-4ix}}{2^4}\\
\sin^4 x = \frac {\cos 4x + 4\cos 2x + 3}{8}\\
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2740586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Flux of a Sphere without using Divergence Theorem I have solved the problem below. my answer is $2\pi a^3$ but in the text book the answer is $4\pi a^3$. Please I would appreciate if someone can tell me where i made a mistake.
PROBLEM
Find the Flux of F= $xi+yj+zk$ outward across the sphere $x^2+y^2+z^2=a^2$
SOLUTION
Sphere: $$x^2+y^2+z^2=a^2$$
So $$z^2=a^2-x^2-y^2$$
$$\hat n dS = (i\frac{\partial z}{\partial x}+ j\frac{\partial z}{\partial y} +k)dA = \frac{x}{z}i+ \frac{y}{z}j +k)dA = \frac{1}{z}(xi+yj +zk)dA $$
$$F.\hat n dS = (xi+yj+zk). \frac{1}{z}(xi+yj +zk)dA = \frac{x^2+y^2+z^2}{z}dA= \frac{a^2}{\sqrt{a^2-x^2-y^2}}dA$$
$$F.\hat n dS = \frac{a^2}{\sqrt{a^2-r^2}}dA $$
I used the substitution u= $a^2 -r^2$
$$\iint_S F.\hat n\mathrm dS = \int_0^{2\pi} \int_0^{a} \frac{a^2}{\sqrt{a^2-r^2}}r\mathrm dr\mathrm d\theta = 2\pi a^3$$
| You are considering the flux over an hemisphere, indeed we have
$$2\int_0^{2\pi} \int_0^{a} \frac{a^2}{\sqrt{a^2-r^2}}r\mathrm dr\mathrm d\theta = 4\pi\left[-a^2(a^2-r^2)^\frac12\right]_0^a =4\pi a^3$$
Note that $4\pi a^3$ is the correct result since the $|F|=a$ and $F$ is parallel to $n$ thus the flux is $4\pi a^2\cdot a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2743221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to factor $a^{3} + b^{3} + c^{3} - 3abc$ into a product of polynomials The question is in the title.
This question is from "Algebra" by Gelfand.
My initial thought is that if $a$, $b$ and $c$ are $1$ or $-1$, then the polynomial evaluates to $0.$ So, maybe two of the factors will be $(a + b + c - 3)$ and $(a + b + c + 3)$. An alternative option that combines these two might be $a^{2} + b^{2} + c^{2} - 3$.
*
*Is the thought process correct here, and would trial and error be a good way to decide between the linear and the quadratic options I described above?
*As you can tell, I am largely doing guess work here. Is there a more systematic way of deciding what terms to add and subtract in orders to factor the polynomial?
Note: The factoring need not be done all the way to linear factors. All that is needed is a product of polynomials.
| I suggests that you use $(a+b)^3=a^3+b^3+3ab(a+b)\Rightarrow a^3+b^3=(a+b)^3-3ab(a+b)$ instead, you will need to use it twice like this:
$a^3+b^3+c^3-3abc$
$=(a+b)^3+c^3-3ab(a+b)-3abc$
$=(a+b+c)^3-(3c(a+b)^2+3(a+b)c^2)-3ab(a+b+c)$
$=(a+b+c)^3-3c(a+b)(a+b+c)-3ab(a+b+c)$
$=(a+b+c)^3-(a+b+c)(3ab+3bc+3ca)$
$=(a+b+c)(a^2+b^2+c^2+2ab+2bc+2ca)-(a+b+c)(3ab+3bc+3ca)$
$=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2746204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
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Lines tangent to graph of equation So I have the following problem and can't seem to arrive at the official answer:
Consider the curve whose equation is $$(2-x)y^2 = x^3$$
Obtain the equations of the lines tangent to the graph of the curve at the points where $x=3/2$.
My attempt
My first thought is that you can isolate y in this equation and then have $y$ as a function of $x$ by the function $$f(x) = \sqrt{(x^3/(-x+2)^2)}$$
But when you turn the equation into a function then obviously you can't have a single $x$ with two outputs, therefore you can't have more than one line tangent to the curve at $x=3/2$. I find this confusing. You can have the equation become a function but you can still have the equation with two outputs for one x.
So I decided to just use implicit differentiation here. Then the problem boils down to finding $$y - y(3/2) = y'(3/2)(x - 3/2)$$
So after a fair amount of work you find $y(3/2)$ and $y'(3/2)$. I would share what I got but really I'd like to see if other people come up with other results.
Official answers are $$y = 3\sqrt{3}x - 3\sqrt{3}$$ and $$y = -3\sqrt{3}x + 3\sqrt{3}$$
| Note that fo $x\ne 2$
$$(2-x)y^2 = x^3\implies y^2=\frac{x^3}{2-x}\implies 2ydy=-\frac{2(x-3)x^2}{(2-x)^2}dx \implies\frac{dy}{dx}=-\frac{2(x-3)x^2}{2y(2-x)^2}$$
and for $x=\frac32$
$$y^2=\frac{x^3}{2-x}=\frac{\frac{27}{8}}{\frac{1}{2}}=\frac{27}{4}\implies y={\pm} \frac{3\sqrt{3}}2 $$
and we obtain
*
*$(x,y)=\left(\frac32,\frac{3\sqrt{3}}2\right)\implies \frac{dy}{dx}=3\sqrt 3$
*$(x,y)=\left(\frac32,-\frac{3\sqrt{3}}2\right)\implies \frac{dy}{dx}=-3\sqrt 3$
and finally
*
*$\left(y-\frac{3\sqrt{3}}2\right)=3\sqrt 3\left(x-\frac{3}2\right)\implies y=3\sqrt 3x-\frac{9\sqrt{3}}2+\frac{3\sqrt{3}}2\implies y=3\sqrt 3x-3\sqrt{3}$
*$\left(y+\frac{3\sqrt{3}}2\right)=-3\sqrt 3\left(x-\frac{3}2\right)\implies y=-3\sqrt 3x+\frac{9\sqrt{3}}2-\frac{3\sqrt{3}}2\implies y=-3\sqrt 3x+3\sqrt{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2746510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Computing determinant of a specific matrix: This question might seem very easy to some. But I am having a very tough time solving it.
$$
A = \begin{pmatrix}
1 + x^2 - y^2 - z^2& 2(xy + z) & 2(zx-y) \\
2(xy - z) & 1 + y^2 - z^2 - x^2 & 2(yz + x) \\
2(zx + y) & 2(yz - x) & 1 + z^2 - x^2 - y^2
\end{pmatrix}
$$
Then $ \det A$ is:
(a) $(1 + xy + yz + zx)^3$
(b) $(xy + yz + zx)^3$
(c) $(1 + x^2 + y^2 + z^2)^3$
(d) $(1 + x^3 + y^3 + z^3)^2$
I am interested in an approach other than the brute-force approach of expanding the $3 \times 3$ determinant. I tried row and column operations, but I wasn't able to see a simplification. I couldn't write A as product of 2 matrices either, so I am clueless. (Also I want a rigorous proof, not methods like put $(x,y,z) = (0,1,2) $ etc. and eliminate 3 options to get the right answer.)
| Let $B = \begin{bmatrix}0 & z & -y\\ -z & 0 & x \\ y & -x & 0\end{bmatrix}$ be half of the anti-symmetric part of $A$. Since its square
$$B^2 = \begin{bmatrix}-y^2-z^2 & xy & xz\\ xy & -x^2-z^2 & yz\\xz & yz & - x^2-y^2\end{bmatrix}$$
has same off-diagonal elements as half of the symmetry part of $A$, we obtain following decomposition of $A$.
$$A = (1+r^2) I_3 + 2B(I+B)$$
It is sort of well known $B$ has eigenvalues $0, \pm ir$ where $r = \sqrt{x^2+y^2+z^2}$.
This means $A$ has eigenvalues $$1+r^2 \quad\text{and}\quad 1+r^2 \pm 2ir(1 \pm ir) = 1-r^2 \pm 2ir = (1 \pm ir)^2$$
and hence
$$\det(A) = (1+r^2)(1+ir)^2(1-ir)^2 = (1+r^2)^3 = (1+x^2+y^2+z^2)^3$$
| {
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Generalizing $ \frac{a^2+b^2+c^2}{2} \times \frac{a^3+b^3+c^3}{3} = \frac{a^5+b^5+c^5}{5}$ If $a+b+c=0$ as discussed in this, this, and this post, then,
$$ \frac{a^2+b^2+c^2}{2} \times \frac{a^3+b^3+c^3}{3} = \frac{a^5+b^5+c^5}{5}\tag1 $$
$$ \frac{a^3+b^3+c^3}{3} \times \frac{a^4+b^4+c^4}{2} = \frac{a^7+b^7+c^7}{7}\tag2 $$
$$ \frac{a^2+b^2+c^2}{2} \times \frac{a^5+b^5+c^5}{5} = \frac{a^7+b^7+c^7}{7}\tag3 $$
Using these basic identities, we can prove the nice squared identities here,
$$ \frac{a^3+b^3+c^3}{3}\times \frac{a^7+b^7+c^7}{7} = \left(\frac{a^5+b^5+c^5}{5}\right)^2 $$
and here for $(a^7+b^7+c^7)^2$. Some investigation shows that,
$$\frac{(a^5+b^5+c^5)}{5\times18}\times\big(9(a^6+b^6+c^6) -(a^3+b^3+c^3)^2\big)= \frac{a^{11}+b^{11}+c^{11}}{11}\tag4$$
$$\frac{(a^7+b^7+c^7)}{7\times18}\times\big(9(a^6+b^6+c^6) +(a^3+b^3+c^3)^2\big)= \frac{a^{13}+b^{13}+c^{13}}{13}\tag5$$
Q: What would be the corresponding identities, as concise as possible, for $p=17$ and $p=19$?
| When denote
$$
s_k = \dfrac{a^k+b^k+c^k}{k},
$$
then some expressions for $s_{17}$ and $s_{19}$:
$$
s_5 \left(6s_{12}-7s_5s_7-\frac{1}{2}{s_3^4}\right) = s_{17},
$$
$$
s_7 \left(6s_{12}-5s_5s_7+\frac{3}{2}{s_3^4}\right) = s_{19}.
$$
Or
$$
s_5 \left(s_{12}+3s_6^2+4s_4s_8\right) = s_{17},
$$
$$
s_7 \left(3s_{12}+9s_6^2-4s_4s_8\right) = s_{19}.
$$
| {
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Find $U+V$ given $U\alpha^{-5} + V\beta^{-5} = 22$
Find $U+V$ given $U\alpha^{-5} + V\beta^{-5} = 22$, when $\alpha, \beta$ are roots
of $x^2-x-1$ and $U,V$ are integers.
For simplicity I only show the U term:
$$U+V = U\alpha^{-5}\alpha^{5}+ ... \\ = U \alpha^{-5}(\alpha^4 \alpha)\\
= U \alpha^{-5}((\alpha+1)^2 \alpha)+ ...\\
=U \alpha^{-5}((\alpha +1) \alpha + 2(\alpha+1) + \alpha)+ ... \\
= U \alpha^{-5}(\alpha +1+\alpha + 2 \alpha + 2 + \alpha)+ ...\\ = U \alpha^{-5 }(5\alpha + 3) + ... \\
= 5 U \alpha^{-4}+5V\beta^{-4} + 3 \cdot 22 $$
Now How to get rid of $\alpha^{-4}, \beta^{-4}$? Please help me, do give hints only.
Thanks a lot!
| Start with $\alpha^2 = \alpha + 1$ and $\beta^2 = \beta + 1$ since these satisfy $x^2 = x+1$. Clearly, both $\alpha$ and $\beta$ are non-zero.
Now, $\alpha^{-1} = \alpha - 1$. Square both sides : $\alpha^{-2} = \alpha^2 - 2\alpha + 1 = 2 - \alpha$. Squaring again, $\alpha^{-4} = (2-\alpha)^2 =$ $ 4 - 4 \alpha + \alpha^2 = 5 - 3 \alpha$. Another $\alpha$ multiplication gives $\alpha^{-5} = 5\alpha^{-1} - 3 = 5\alpha - 8$. Can you see the Fibonacci sequence here?
The exact same applies for $\beta$.
Therefore, $U\alpha^{-5} + V \beta^{-5} = U(5 \alpha - 8) + V(5 \beta- 8) = 22$. Hence, $5\alpha U + 5\beta V = 22 + 8(U+V)$.
Hence, $5(1 + \sqrt 5)U + 5(1 - \sqrt 5)V = 44 + 16(U+V)$, rewrite to $5\sqrt{5}(U-V) = 44 + 11(U+V)$. Hence, we get that $U-V = 0$ otherwise the LHS will be irrational and hence $44 = -11(U+V)$ so $U+V = -4$. Of course, we get $U= V = -2$ from here.
Key idea : if $\alpha$ is the root of a quadratic equation, higher powers of $\alpha$ can be expressed as $c \alpha + d$ for some $c,d$.
| {
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Figure out all positive integers n with consecutive + integers a,b,c. When $2018^n$ = $a^4$ + $b^4$ + $({b^2+c^2})^2$,
then what is the possible positive integers n be?
| Since we have consecutive integers $a,b,c$, take $a=b-1$ and $c=b+1$. Then, we have $2^n 547^n=(b-1)^4+b^4+(b^2+(b+1)^2)^2=(b-1)^4+2b^4+2b^2(b+1)^2+(b+1)^4=(b-1)^4+(b+1)^4+4b^4+4b^3+2b^2$. If $b=2k$, then expanding the braces, we get the RHS is divisible by $2$, not $4$. This means $n$ is at most 1.
If $b=2k+1$, then again the RHS is divisible by 2, not 4. So $n \leq 1$.
We check the case $n=1$. If $b=4$, the RHS is $2018$. So, we conclude that $n=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How do I derive the formula for the reciprocal of a hypotenuse? Given $a^2 + b^2 = c^2$,
Why is it that the equation below yields the reciprocal of the hypotenuse c, ($\frac{1}c$)?
$\sqrt{(\frac{a}{a^2+b^2})^2 + (\frac{b}{a^2+b^2})^2}$
Worked example:
$3^2 + 4^2 = c^2$
$c = 5$
$\sqrt{(\frac{3}{3^2+4^2})^2 + (\frac{4}{3^2+4^2})^2}$
= $\sqrt{(\frac{3}{25})^2 + (\frac{4}{25})^2}$
= $\sqrt{(.0144) + (.0256)}$
= $\sqrt{.04}$
= .2 or $1/5$
| Observe that, for $ab\ne0$,
$$
\left(\frac{a}{a^2+b^2}\right)^2 + \left(\frac{b}{a^2+b^2}\right)^2=\frac{a^2}{(a^2+b^2)^2}+\frac{b^2}{(a^2+b^2)^2}=\frac{a^2+b^2}{(a^2+b^2)^2}=\frac{1}{a^2+b^2}.
$$ Hope you can take it from here.
| {
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Limits of a product $N$ is a positive integer.
I want to calculate this limit, but i couldn't get anywhere when i tried.
$$P[n]=\left(1+\frac {1}{n^2}\right)\left(1+\frac {2}{n^2}\right)\cdots\left(1+\frac {n-1}{n^2}\right)$$ as $n\to \infty$
I tried to apply $\ln()$ at both sides to transform into a sum etc..
tried to use functions to limit the superior and inferior intervals of the function like this:
$$\exp\left(\left(n-1\right)\ln\left(1+\frac1{n^2}\right)\right)<P[n]<\exp\left((n-1)\ln\left(1+\frac{n-1}{n^2}\right)\right)$$
The only thing I get is that:
$1<\lim(P[n])<e$
Can anyone help me to solve this?
| First realize that, $$\prod_{i=1}^{n} 1+ \frac i{n^2} = \prod_{i=1}^n \frac {n^2+i}{n^2} = \prod_{i=n^2+1}^{n^2+n} \frac {i}{n^2} $$
Then,
$$\prod_{i=n^2+1}^{n^2+n} \frac {i}{n^2} = \exp \left ( \sum_{i=n^2+1}^{n^2+n} \log(i) - 2n\log(n) \right )$$
Then we use a Riemann inequality since $\log$ is an increasing function,
$$\int_{n^2}^{n^2+n} \log(x)dx \leq \sum_{i=n^2+1}^{n^2+n} \log(i) \leq \int_{n^2+1}^{n^2+n+1} \log(x)dx \mathrm{\ \ (Riemann \ inequality)} $$
$$\begin{align} \int_{n^2}^{n^2+n} \log(x)dx -2n\log(n) & = [xlog(x)-x]_{n^2}^{n^2+n} -n\log(n^2) \\
& = (n^2+n) \log(n^2+n) -n^2 - n - n^2\log(n^2) +n^2 -n \log(n^2)\\
& = (n^2 +n) \log(1+\frac 1 n) - n \\
& = (n^2 +n)(\frac 1 n - \frac 1 {2n^2} + O(n^{-3})) - n \underset{n\to \infty}\to \frac 12 \end{align} $$
Likewise
$$\begin{align} \int_{n^2+1}^{n^2+n+1} \log(x)dx -2n\log(n) & = [x\log(x)-x]_{n^2+1}^{n^2+n+1} -n\log(n^2) \\
& = (n^2+1)\log(1+\frac{n}{n^2+1})+n\log(1+\frac{n+1}{n^2})-n \underset{n\to \infty}\to \frac 12 \end{align} $$
Therefore $\log(P_n) \underset{n\to \infty}\to \frac 12$ and $P_n \underset{n\to \infty}\to \sqrt e$
The only result I used is $x\mathcal \in V(0), \ \log(1+x)=x-\frac 12 x^2+O(x^3)$ to take the limits. Thanks to @achille hui for the correction.
| {
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How prove $\cos(\frac{2\pi}{17}) + \cos(\frac{18\pi}{17})+\cos(\frac{26\pi}{17})+\cos(\frac{30\pi}{17}) = \frac{\sqrt{17}-1}{4}$
Prove that $\cos(\frac{2\pi}{17}) + \cos(\frac{18\pi}{17})+\cos(\frac{26\pi}{17})+\cos(\frac{30\pi}{17}) = \frac{\sqrt{17}-1}{4}$
Regards that value of $\cos(2\pi/17)$, I can't find the easy way to solve that expression.
Even if I had time, I wouldn't try that method to find the all roots others cosines expressions. IMHO
| Let $p$ be an odd prime number. Then
$$g_p=\sum_{k=0}^{p-1}\exp(2\pi i k^2/p)$$
is a quadratic Gauss sum. Gauss proved that $g_p=\sqrt p$ or
$i\sqrt p$ according to whether $p\equiv1$ or $p\equiv3\pmod 4$.
It is quite easy to prove this up to sign, but hard to prove the sign.
So $g_{17}=\sqrt{17}$. Therefore
\begin{align}
\sqrt{17}&=1+2\exp(2\pi i/17)+2\exp(8\pi i/17)+2\exp(18\pi i/17)
+2\exp(32\pi i/17)\\
&+2\exp(16\pi i/17)+2\exp(4\pi i/17)+2\exp(30\pi i/17)
+2\exp(26\pi i/17)\\
&=1+4\cos(2\pi/17)+4\cos(18\pi/17)+4\cos(26\pi/17)+4\cos(30\pi/17).
\end{align}
| {
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Simplification of $\frac{\sqrt{n}}{\sqrt{4n+\sqrt{n}} + 2\sqrt{n}}$ Can somebody explain how pass to this : $\sqrt{n} / \left(\sqrt{4n+\sqrt{n}} + 2*\sqrt{n}\right)$
to this :
$1 \big/ \left(2 + 2\sqrt{1+ \frac{\sqrt{n}}{4n}}\right)$
Thanks in advance
| \begin{align}
\frac{\sqrt{n} }{ \sqrt{4n+\sqrt{n}} + 2\sqrt{n}}
=&\;\frac{\frac{\sqrt{n}}{\sqrt{n}} }{ \frac{\sqrt{4n+\sqrt{n}}}{\sqrt{n}} + 2\frac{\sqrt{n}}{\sqrt{n}}}\\[2ex]
=&\;\frac{1 }{ \sqrt{\frac{4n+\sqrt{n}}{n}} + 2}\\[2ex]
=&\;\frac{1 }{ \sqrt{4+\frac{\sqrt{n}}{n}} + 2}\\[2ex]
=&\;\frac{1 }{ \sqrt{4\left(1+\frac{\sqrt{n}}{4n}\right)} + 2}\\[2ex]
=&\;\frac{1 }{ 2\sqrt{1+\frac{\sqrt{n}}{4n}} + 2}\\[2ex]
=&\;\frac{1}{2 + 2\sqrt{1+ \frac{\sqrt{n}}{4n}}}
\end{align}
| {
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Removing a discontinuity? How would you "remove the discontinuity" of $f$ ? In other words, how would you define $f(4)$ in order to make $f$ continuous at $x=4$?
$$f(x) = \dfrac{x^2-x-12}{x-4}$$
| if $ x \neq 4 $
$f(x)= \dfrac{x^2-x-12}{x-4}= \dfrac{(x-4)(x+3)}{x-4}=\dfrac{1(x+3)}{1}= x+ 3$
so we have
f(x)=x+3
(for $ x \neq 4$)
$\lim_{x\to4}f(x) = \lim_{x\to4}(x+3) $= 7
we want to make f continuous for this reason
the ammount of function at 4 should be equall to $$\lim_{x\to4}f(x)=7$$
so we define a function that in x=4 it equall to $\lim_{x\to4}f(x)=7$
$$g(x)=
\begin{cases}
\dfrac{x^2-x-12}{x-4}= x+3, & x\neq 4 \\
7, & x=4
\end{cases}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof check modulo arithmetic what is wrong with this proof,
so I am supposed to show $7x^2-15y^2=1$ has no integer solutions...
so since $7x^2=1+3(5y^2)$ so $7x^2\equiv 1 \pmod{3}$
hence in mod $3$, $x\equiv 0, 1$ or $2$ so $x^2\equiv 0, 1$ and $7x^2\equiv0$ or $7$
and since $7$ is congruent to $1 \bmod3$ I end up with solutions.
I also tried it with mod $5$ since $15y^2=5(3y^2)$ and I ended up with no solution so why not in case above?
| There are also no integer solutions to $$ 13 x^2 - 17 y^2 = 1. $$
We know that there are rational solutions, therefore $p$-adic, as we can solve
$$ 13 x^2 - 17 y^2 = 25 $$
in integers.
$$ 13 \cdot 7^2 - 17 \cdot 6^2 = 637 - 612 = 25 $$
| {
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Number of terms in a Polynomial Expansion For a binomial $(a + b)^n$, the number of terms is n + 1.
For a trinomial $(a + b + c)^n$, the number of terms is $\frac{(n+1)(n+2)}{(2)}$.
For a multinomial $(a + b + c +d)^n$, the number of terms is $\frac{(n+1)(n+2)(n+3)}{(6)}$.
I'm guessing that for $(a + b + c + d + e)^n$, the number of terms formula would include $(n+1)(n+2)(n+3)(n+4)$ on the numerator but I don't know what should be its denominator.
Question:
*
*What is the number of terms for $(a + b + c + d + e)^n$?
*Do we have a general formula for the number of terms of a polynomial expansion?
*What if the given is $(a^2 + a + b)^n$, can I still use the formula $\frac{(n+1)(n+2)}{(2)}$.when 2 terms in the expansion has the same variable?
*What if the given is $(a + b + Constant)^n$, would the constant affect the number of terms?
| for (a+b+c+d+e)^n it would be [(n+4)(n+3)(n+2)(n+1)]/24
| {
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Whenever $a+b+c=1$, $\frac{bc+a+1}{a^2+1} + \frac{ac+b+1}{b^2+1} + \frac{ab+c+1}{c^2+1} \le \frac{39}{10}$ Prove that for a + b + c =1 and a,b,c are positive real numbers,
then
$$\frac{bc+a+1}{a^2+1} + \frac{ac+b+1}{b^2+1} + \frac{ab+c+1}{c^2+1} \le \frac{39}{10}$$
My try:
if one term is proven to be $\le \frac{13}{10} $ then we prove the inequality.
sub in $a =1-b-c$
then we have
$$\frac{bc+b+c+2}{b^2+c^2+2bc-2b-2c+2}\le\frac{13}{10}$$
further simplifying, then we have $$0\le13b^2+13c^2+16bc-16b-16c+6$$
since $b^2+c^2\ge 2bc$ then we can substitute it? (not sure about this)
then we have
$$0\le42bc-16b-16c+6$$ since $b+c\ge2\sqrt{bc}$ then we can sub it again? (also not sure about this)
then we get a quadratic inequality
$$21bc -16\sqrt{bc} +6$$
but it is not true for all positive real numbers a, b and c.
Is it impossible to use this approach and get the answer (is there a better way to solve it) or did i do something wrong?
| This is wrong. If $a=1$ and $b=c=0$ then the left-hand side is $3$, which is less than $\frac{39}{10}$.
| {
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Maximum power of $2$ which divides $3^{1024}-1$ What is the maximum power of $2$ which completely divides $3^{1024}-1$?
I proceeded thus:
$\phi(2^n)=2^{n-1}$ for all $n\ge1$
$$3^{1024}=3^{2^{10}}\equiv1\pmod {2^{11}}$$
$$3^{1024}-1\equiv0\pmod {2^{11}}$$
Since $\phi(2^{11})=2^{10}$. So, maximum power of $2$ must be $11$.
But the answer says it is $12$.
Where am I wrong and how to solve it correctly?
| As requested in the comments:
We start by factoring the polynomial $x^{1024}-1$. To do that, we note that $1$ is a root, as is every $2^k-$root of $1$ for $k=0,\cdots 10$. For $k>0$ such a root of unity is also a $2^{k-1}-$st root of $-1$ so our polynomial is divisible by $$(x-1)\times \prod_{k=0}^9(x^{2^k}+1)$$
Comparing the lead terms shows that this is in fact equal to our polynomial.
Now, let $x=3$. We remark that $$3\equiv -1 \pmod 4 \implies 3^{2i}+1\equiv 2 \mod 4$$ so most of the terms in the product are divisible by $2$ but not by $4$. $3-1=2$, of course, and $3+1=4$ is the only term in the product divisible by a higher power of $2$. $(3-1)(3+1)$ then gives us a factor of $2^3$ and the other nine terms in the product each give us exactly one factor of $2$, making the answer $12$ as desired.
| {
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Probability of exactly three of a kind in a roll of 5 dice The answer according to edx course HarvardX: FC1x Fat Chance: Probability from the Ground Up
is $$\frac{6*5*5* \left(^5_2\right)}{6^5} =\frac{1500}{6^5}$$
The remaining two dice can be same say this is valid favourable outcome 4,4,4,5,5.
But according to my reasoning the probability is much more than that,
My reasoning:
6 options for 3 of a kind
5 options for 4th dice
4 options for 5th dice(let us consider only the cases where 4th and 5th dice are different, for sake of showing that even with excluding certain favourable outcomes, namely in which 4th and 5th dice are same, my probability is higher than the course answer.)
therefore total number of ways
$$ \frac{ 6*5*4*5!}{3!} = 2400$$
5! ways of arranging 5 items, divided by 3!, since 3 are of a kind.
$$ p=\frac{2400}{6^5} $$
Which is greater than the probability calculated by the edx course, and I have not even considered the case when we allow 4th and 5th dice to have same number.
what is wrong with my reasoning?
| Assume the dice are distinguishable. Then there are $6^5$ possible outcomes since there are six possible outcomes for each of the five dice.
Three of a kind: There are $\binom{5}{3}$ ways for three of the five dice to display the same outcome and six possible outcomes those three dice could display. There are $\binom{5}{2}2!$ ways for the two remaining dice to display two of the five other possible values (as there are $\binom{5}{2}$ ways to select two of the remaining five values and $2!$ ways to arrange those values on the remaining two distinct dice), giving
$$\binom{5}{3}\binom{6}{1}\binom{5}{2}2! = 1200$$
favorable outcomes.
Thus, the probability that three of a kind is obtained is
$$\Pr(\text{three of a kind}) = \frac{1200}{6^5}$$
Edit: Evidently, we were also supposed to consider all cases in which exactly three of the dice show the same outcome, so we must add the results for a full house.
Full house: There are $\binom{5}{3}$ ways for three of the five dice to display the same outcome and six outcomes those dice could display. There are five possible outcomes the other two dice could both display. Hence, there are
$$\binom{5}{3}\binom{6}{1}\binom{5}{1} = 300$$
ways to obtain a full house. Thus, the probability of obtaining a full house is
$$\frac{300}{6^5}$$
Total: The probability that exactly three of the dice display the same number is found by adding the probabilities for three of a kind and a full house, which yields
$$\frac{1200}{6^5} + \frac{300}{6^5} = \frac{1500}{6^5}$$
as the given answer states.
Check: We know that the total number of outcomes is $6^5 = 7776$.
All different: There are $\binom{6}{5}$ ways of selecting five different outcomes and $5!$ arrangements of those outcomes on the dice. Thus, there are
$$\binom{6}{5}5! = 720$$
ways to obtain five different numbers.
One pair: There are $\binom{5}{2}$ ways for two of the dice to display the same outcome and six possible outcomes those two dice could display. There are $\binom{5}{3}3!$ ways for the three remaining dice to display three of the remaining five values. Hence, there are
$$\binom{5}{2}\binom{6}{1}\binom{5}{3}3! = 3600$$
ways to obtain a pair.
Two pairs: There are $\binom{6}{2}$ possible outcomes for the pairs. There are $\binom{5}{2}$ ways for two of the five dice to show the smaller of those outcomes and $\binom{3}{2}$ ways for two of the other three dice to show the larger of those outcomes. There are four possible outcomes for the remaining die. Hence, there are
$$\binom{6}{2}\binom{5}{2}\binom{3}{2}\binom{4}{1} = 1800$$
ways to obtain two pairs.
Three of a kind: We showed above that there are $1200$ ways to obtain three of a kind.
Full house: We showed above that there are $300$ ways to obtain a full house.
Four of a kind: There are $\binom{5}{4}$ ways for four of the five dice to display the same outcome and six outcomes those dice could display. There are five possible outcomes for the remaining die. Hence, there are
$$\binom{5}{4}\binom{6}{1}\binom{5}{1} = 150$$
ways to obtain four of a kind.
Five of a kind: All the dice must show the same outcome. There are six possible outcomes. Hence, there are $6$ ways to obtain five of a kind.
Total: The above cases are mutually exclusive and exhaustive. Observe that
$$720 + 3600 + 1800 + 1200 + 300 + 150 + 6 = 7776 = 6^5$$
| {
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Sum of $4$ dice rolls greater than the product Let us roll a fair die $4$ independent times, and denote the outcomes as $X_1, X_2, X_3$ and $X_4$.
What is the probability of $X_1+X_2+ X_3+X_4 > X_1X_2 X_3X_4$?
My try:
I could get the answer for $2$ rolling case by enumerating possibilities, but couldn't get this larger problem. Can someone help me with this?
Thanks in advance for any help!
| A solution to the equation has at least one $1$.
When you roll all 2s, the equation is false.
$2 + 2 + 2 + 2 > 2 \times 2 \times 2 \times 2$
$8 > 16 \implies false$
Adding one to any of the numbers always maintains the falsehood of this equation, or the truth of the reverse. Assuming the opposite equation:
$x_1 + x_2 + x_3 + x_4 \le x_1x_2x_3x_4$
Adding $1$ to one of the die results in the product results in a value that is $x_2x_3x_4$ higher.
$(x_1 + 1)x_2x_3x_4 = x_1x_2x_3x_4 + x_2x_3x_4$
Adding $1$ to one of the die results in the sum makes the sum $1$ higher, and $x_2x_3x_4 >= 1$, so we can say:
*
*If $x_1 + x_2 + x_3 + x_4 \le x_1x_2x_3x_4$, then $(x_1 + 1) + x_2 + x_3 + x_4 \le (x_1 + 1)x_2x_3x_4$.
Therefore, any solution to $x_1 + x_2 + x_3 + x_4 > x_1x_2x_3x_4$ must have at least one $1$.
The lowest possible dice roll with exactly one $1$ goes like this:
$1 + 2 + 2 + 2 > 1 \times 2 \times 2 \times 2$
$7 > 8 \implies false$
Therefore, the solution must require at least two $1$s.
Three or more $1$s.
If there are at least 3 $1$s, the equation is always true. Note this includes the case of all four $1$s.
$1 + 1 + 1 + x > 1 \times 1 \times 1 \times x$
$x + 3 > x \implies true$
Exactly two $1$s.
With exactly two $1$s, let's explore the other two values.
If one of the other values is a $2$, then that restricts the other value to $3$ or less.
$1 + 1 + 2 + 2 > 1 \times 1 \times 2 \times 2$
$6 > 4 \implies true$
$1 + 1 + 2 + 3 > 1 \times 1 \times 2 \times 3$
$7 > 6 \implies true$
$1 + 1 + 2 + 4 > 1 \times 1 \times 2 \times 4$
$8 > 8 \implies false$
If both the other values are at least $3$, then we don't have a solution.
$1 + 1 + 3 + 3 > 1 \times 1 \times 3 \times 3$
$8 > 9 \implies false$
All solutions
*
*All four $1$s. 1 solution.
*Three $1$s and we don't care what the last value is (this excludes the four $1$s case): 5 possible choices for the last value, 4 choices as to on which die it occurs: 20 solutions
*Exactly two $1$s: The other two values are both $2$. There are $4\choose2$ or 6 choices to distribute the two $2$s.
*Exactly two $1$s: The other two values are $2$ and $3$. There are four choices for the $2$, leaving three choices for the $3$, for a total of 12 solutions here.
I count 39 possible solutions, out of $6^4$ possibilities.
$39\div1296 = 13\div432 \approx0.0301 $
| {
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If I draw 4 cards from a deck of 52 cards, what is the probability that the cards are 2 of one suit and 2 of another? 4 cards are drawn from a stack of 52, what is the probability that 2 cards are of one suit and the other 2 cards are of a different suit?
I.e. 2 Diamonds and 2 Clubs or 2 Spades and 2 Diamonds etc..
I think it might be:
$$2\left(\frac{13C2 \cdot 13C2}{52C4}\right)$$
However, I'm not entirely sure that multiplying it by 2 to account for all possible suits is the correct approach.
| In my opinion "2 cards are of one suit and the other 2 cards are of a different suit" is a bit ambiguous. Let us consider the following two possible interpretations.
2 different suits: two cards of one suit and two of another suit
The total number of hands of $4$ cards is $\binom{52}{4}.$
We have $\binom{4}{2}$ ways to choose the suits. Then we have $\binom{13}{2}$ ways for the values of the suit with two cards and $\binom{13}{2}$ for the other two.
So the probability should be
$$\frac{\binom{4}{2}\cdot\binom{13}{2}^2}{\binom{52}4}=\frac{6\cdot\binom{13}{2}^2}{\binom{52}4}.$$
3 different suites: two cards of one suit and two of the other suits
The total number of hands of $4$ cards is $\binom{52}{4}.$
We have $4$ ways to choose the suit with two cards and $\binom{3}{2}$ to choose the other two suits. Then we have $\binom{13}{2}$ ways for the values of the suit with two cards and $13\cdot 13$ for the other two.
So the probability should be
$$\frac{4\cdot\binom{3}{2}\cdot\binom{13}2\cdot 13\cdot 13}{\binom{52}4}=\frac{26\cdot\binom{13}{2}^2}{\binom{52}4}.$$
| {
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} |
Logarithmic expression how to simplify $$ \log_{3}24 - 3\log_{3}5\times \log_{5}2$$
What I can get is:
$$ \log_3{24} - \log_3{5^3} \times \log_{5}2$$
Change of base rule to get it all in base 3:
$$ \log_5{2} = \frac{\log_3{2}}{\log_3{5}} $$
Now I have:
$$\log_3{24} - \frac{\log_3{5^3}\times \log_3{2}}{\log_3{5}}$$
How to continue from here?
| Continuing from what you have,
\begin{align}
\log_3{24} - \frac{\log_3{5^3}\times \log_3{2}}{\log_3{5}} &= \log_3{24} - \frac{3\cdot\log_3{5}\cdot \log_3{2}}{\log_3{5}} \\
&=(\log_3{3} + \log_3{8}) - 3\cdot \log_3{2} \\
&= 1 + 3 \log_3 2 - 3 \log_3 2 \\
&= 1.
\end{align}
In questions like these, it's often a good idea to reduce exponents as much as possible: you started by converting $3 \log_3 5$ to $\log_3 (5^3)$, but the former expression is simpler to work with.
You can see this further in how expanding $\log_3 24$ was a useful step to finish.
| {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Expected number of die rolls - conditional probability The aim here is to roll a fair die until a 3 is rolled twice in a row, where the number of rolls that this takes (including the final two where threes are consecutive) is given by the random variable X. We are also told of the random variable Y, and this is the number of rolls of this die until anything other than a three occurs.
We need to find the mean and variance of X, and I realise we can set up E[X]=E[E[X|Y]], and the corresponding equation for the variance. I'm having trouble setting up E[X|Y] and E[X^2|Y], all we know is that E[X] is finite, but no more. Y itself I think is a geometric random variable with parameter 5/6, but am confused as to how to proceed.
| In fact, you may compute the entire distribution of $X$. Let's denote $a_n = \mathbb{P}(X = n)$, where $n=1,2,...$ . We have $a_1 = 0$ and $a_2 = 1/6^2$.
Let also $X_1$ and $X_2$ be the first and the second rolls of the die.
Then, using the law of total probability, for $n>2$ we have
$$
a_n = \frac 56 \mathbb{P}(X = n |X_1 \neq 3) + \frac 16 \mathbb{P}(X = n |X_1 = 3) = \\
\frac 56 \mathbb{P}(X = n-1) + \frac {1}{6} \frac{5}{6} \mathbb{P}(X = n | X_1 = 3, X_2 \neq 3) + \frac {1}{6} \frac{1}{6} \mathbb{P}(X = n | X_1 = 3, X_2 = 3) = \\
\frac 56 a_{n-1} + \frac{5}{6^2} a_{n-2} + \frac{1}{6^2} \chi_{n=2}.
$$
The idea is that if you miss 3 on the first step and need to wait for exactly $n$ steps to complete $3,3$ then you are now waiting for $3,3$ in exactly $n-1$ steps .
We thus have
$$
a_1 = 0, a_2 = \frac{1}{6^2}, \ \ a_n = \frac 56 a_{n-1} + \frac{5}{6^2} a_{n-2}, \text{ for } n >2.
$$
From this
$$
\mathbb{E} X = \sum_{1}^\infty n a_n = \frac{2}{6^2} + \sum_{3}^\infty n a_n =
\frac{2}{6^2} + \frac{5}{6}\sum_{2}^\infty (n+1)a_n + \frac{5}{6^2} \sum_{1}^\infty ( n + 2 ) a_n = \\
\frac{2}{6^2} + \frac 56 + \frac{10}{6^2} + \frac 56 \mathbb{E}X + \frac{5}{6^2} \mathbb{E} X,
$$
where we used the fact that $\sum a_n = 1$. Hence,
$$
\mathbb{E} X = 42.
$$
You may also compute the variance easily, using the recurrence relation for $a_n$, and playing with the corresponding sum.
| {
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} |
Binomial Inequality in which Binomial coefficient is in square root.
$$1\cdot \sqrt{C_{1}}+2\cdot \sqrt{C_{2}}+\cdots \cdots +100\cdot \sqrt{C_{100}}\leq \frac{1}{2}\cdot (2^{100}-1)+\frac{20301}{12}$$
where $\displaystyle C_{r}=\binom{n}{r}$
Try: Using Cauchy Schwarz Inequaity
$$\bigg(1^2+2^2+\cdots \cdots +100^2\bigg)\bigg(C_{1}+C_{2}+\cdots \cdots +C_{100}\bigg)\geq \bigg(1\cdot \sqrt{C_{1}}+2\cdot \sqrt{C_{2}}+\cdots \cdots +100\cdot \sqrt{C_{100}}\bigg)^2$$
$$\bigg(1\cdot \sqrt{C_{1}}+2\cdot \sqrt{C_{2}}+\cdots \cdots +100\cdot \sqrt{C_{100}}\bigg)\leq \bigg[\frac{100\cdot 101\cdot 201}{6}\cdot (2^{100}-1)\bigg]^{\frac{1}{2}}$$
i am not understand how can i prove my original inequality,
could some help me , Thanks
| By C-S,
as you have done,
$\begin{array}\\
\left(\sum_{k=1}^n k\sqrt{\binom{n}{k}}\right)^2
&\le \sum_{k=1}^n k^2\sum_{k=1}^n\binom{n}{k}\\
&=\dfrac{n(n+1)(2n+1)}{6}(2^n-1)\\
&\lt\dfrac{2(n+1)^3}{6}2^n\\
&=\dfrac{(n+1)^3}{3}2^n\\
\text{so}\\
\sum_{k=1}^n k\sqrt{\binom{n}{k}}
&\lt\dfrac{(n+1)^{3/2}}{\sqrt{3}}2^{n/2}\\
\end{array}
$
and this is less than
$2^{n-1}$
for $n \ge 12$
according to Wolfy.
So it is much less
for your case of
$n = 100$.
| {
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} |
$2^{x}+ 5^{y}= 7^{z}$ Help me solve this integers exponents equation:
$$2^{x}+ 5^{y}= 7^{z}$$
We have: $z= \frac{i(2\,\pi\,n-i\,\log(2^{x}+ 5^{y}))}{\log(7)}$
I ask because I read that a slightly complication of the equation occuring in Fermat's last theorem can lead to an undecidable case and I wonder whether the given form already is sufficient to achieve this.
| Only $x=y=z=1$, corresponding to $2+5=7$, can work. Most of this proof rehashes the comments, the real purpose here is to consolidate all the steps into one place.
1) $x$ is odd, or else $2^x \equiv 7^z \equiv 1 \bmod 3$ and $5^y$ can't be divisible by $3$.
2) $y$ is odd, or else $5^y \equiv 7^z \equiv 1 \bmod 3$ and $2^x$ can't be divisible by $3$.
3) If $x$ and $y$ are both odd and $x\ge 3$, then $2^x+5^y \equiv 0+5=5 \bmod 8$ but $7^z \in \{1,7\} \bmod 8$. This forces $x=1$.
4) If $x=1$ and $y\ge 2$, then $2^x+5^y \equiv 2+0=2 \bmod 25$ but $7^z \in \{1,7,18,24\} \bmod 25$. This forces $y=1$ leaving $2+5=7$ as the only possible equality.
| {
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"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
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} |
Complex Number Series If
$$
z + 1/z = -1
$$
where $z$ is a complex number then value of the sum from r = 1 to r = 99 of $$
( z^r + 1/z^r)^2
$$
is equal to
A) 198
B) 3
C) 99
D) 0
I tried by putting z = x + iy but that gives, |z| = 1
| HINT: observe that
$$
\left(z^r+\frac1{z^r}\right)^2=\left(z^{2}\right)^r+\left(\frac1{z^2}\right)^r+2
$$
then split the sum and using the geometric series
$$
\sum_{n=1}^Nq^n=\frac{q-q^{N+1}}{1-q}\;,\;\;\;\;\;q\in\Bbb C
$$
we can evaluate the first two sums:
\begin{align*}
\sum_{r=1}^{99}\left(z^r+\frac1{z^r}\right)^2
=&\sum_{r=1}^{99}\left(z^{2}\right)^r+\sum_{r=1}^{99}+\left(\frac1{z^2}\right)^r+2\cdot99\\
=&\frac{z^2-(z^2)^{100}}{1-z^2}+\frac{z^{-2}-(z^{-2})^{100}}{1-z^{-2}}+198\\
=&z^2\frac{1-z^{198}}{1-z^2}+\frac1{z^{198}}\frac{1-z^{198}}{1-z^{2}}+198\\
\end{align*}
but $z+1/z=-1$ holds iff $z^2+z+1=0$, that is $z=e^
{\pm i\frac23\pi}$, thus $z^3=1$ (you can see this also as follows: supposing $z\neq1$, you can multiply this last one by $z-1$ to get $z^3-1=0$).
Hence since $3|198$ we have that $1-z^{198}=0$ thus the first two summands of the last line are thrown away and the only survivor os $198$.
| {
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"answer_count": 2,
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} |
Find $a\in\mathbb{R}$ such that the image of $f(x)=\frac {x^2+ax+1}{x^2+x+1}$ is included in $[0,2]$.
Find $a\in\mathbb{R}$ such that the image of $f(x)=\frac {x^2+ax+1}{x^2+x+1}$ is included in $[0,2]$.
My attempt:
We have: $f'(x)=-\frac{(a-1)(x^2-1)}{(x^2+x+1)^2}\implies x = 1$ and $x = -1$ points of extrema.
then for $a\geq 1$:
so then $$2-a=0\implies a=2$$ and $$\frac {a+2}3=2\implies a=4.$$
and for $a\leq1:$
so then $$2-a=3\implies a=-1$$ and $$\frac {a+2}3=0\implies a=-2.$$
Now my answers are in the type of interval. How do I know which interval to choose?
| The derivative of $f(x)$ is equal to $\dfrac{(a-1)(x^2-1)}{(x^2+x+1)^2}$ so the extremes are independent of a and taken at $x=\pm 1$. This extremes are $\dfrac{2\pm a}{3}$ It follows $0\le a\le 2$
| {
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$\sum_{m=0}^{\infty}x^m\sum_{k=0}^{\infty}W_{m,k}f_k=\sum_{n=1}^{\infty}\frac{1}{n(n+1)}\sum_{k=0}^{\infty}\left(\frac{n+x}{n(n+1)}\right)^k f_k$? How do we solve
$\sum_{m = 0}^{\infty} x^m \sum_{k = 0}^{\infty} W_{m, k} f_k = \sum_{n =
1}^{\infty} \frac{1}{n (n + 1)} \sum_{k = 0}^{\infty} \left( \frac{n + x}{n (n
+ 1)} \right)^k f_k$
for $W_{m,k}$ ?
Here is what I have attempted so far:
$\begin{array}{l}
\left\{ W_{m, k} : \sum_{m = 0}^{\infty} x^m \sum_{k = 0}^{\infty} W_{m, k}
\frac{f^{(k)} (0)}{k!} = \sum_{n = 1}^{\infty} \frac{1}{n (n + 1)} \sum_{k =
0}^{\infty} \frac{f^{(k)} (0)}{k!} \left( \frac{n + x}{n (n + 1)} \right)^k
\right\}\\
\left\{ W_{m, k} : \sum_{m = 0}^{\infty} \sum_{k = 0}^{\infty} x^m W_{m, k}
\frac{f^{(k)} (0)}{k!} = \sum_{n = 1}^{\infty} \sum_{k = 0}^{\infty}
\frac{\frac{f^{(k)} (0)}{k!} \left( \frac{n + x}{n (n + 1)} \right)^k}{n (n
+ 1)} \right\}\\
\left\{ W_{m, k} : \sum_{m = 0}^{\infty} \sum_{k = 0}^{\infty} x^m W_{m, k}
\frac{f^{(k)} (0)}{k!} = \sum_{m = 1}^{\infty} \sum_{k = 0}^{\infty}
\frac{\left( \frac{m + x}{m (m + 1)} \right)^k}{m (m + 1)} \frac{f^{(k)}
(0)}{k!} \right\}\\
\left\{ W_{m, k} : \sum_{m = 0}^{\infty} \sum_{k = 0}^{\infty} x^m W_{m, k}
\frac{f^{(k)} (0)}{k!} = \sum_{m = 0}^{\infty} \sum_{k = 0}^{\infty}
\frac{\left( \frac{m + 1 + x}{(m + 1) (m + 2)} \right)^k}{(m + 1) (m + 2)}
\frac{f^{(k)} (0)}{k!} \right\}\\
\left\{ W_{m, k} : \sum_{m = 0}^{\infty} \sum_{k = 0}^{\infty} x^m W_{m, k}
\frac{f^{(k)} (0)}{k!} = \sum_{m = 0}^{\infty} \sum_{k = 0}^{\infty}
\frac{(m + 1 + x)^k}{((m + 1) (m + 2))^k (m + 1) (m + 2)} \frac{f^{(k)}
(0)}{k!} \right\}
\end{array}$
| $\sum_{m = 0}^{\infty} x^m \sum_{k = 0}^{\infty} W_{m, k} f_k = \sum_{n =
1}^{\infty} \frac{1}{n (n + 1)} \sum_{k = 0}^{\infty} \left( \frac{n + x}{n (n
+ 1)} \right)^k f_k
$
I'll naively
expand the right side,
not worrying about convergence.
$\begin{array}\\
\sum_{n=1}^{\infty} \frac{1}{n (n + 1)} \sum_{k = 0}^{\infty} \left( \frac{n + x}{n(n+1)} \right)^k f_k
&=\sum_{n=1}^{\infty} \frac{1}{n (n + 1)} \sum_{k = 0}^{\infty} \left( \frac{1}{n(n+1)} \right)^k f_k(n+x)^k\\
&=\sum_{n=1}^{\infty} \frac{1}{n (n + 1)} \sum_{k = 0}^{\infty} \left( \frac{1}{n(n+1)} \right)^k f_k\sum_{m=0}^k \binom{k}{m}x^mn^{k-m}\\
&=\sum_{n=1}^{\infty} \frac{1}{n (n + 1)} \sum_{m=0}^{\infty}\sum_{k=m}^{\infty} \left( \frac{1}{n(n+1)} \right)^k f_k \binom{k}{m}x^mn^{k-m}\\
&=\sum_{m=0}^{\infty}x^m\sum_{n=1}^{\infty} \frac{1}{n (n + 1)} \sum_{k=m}^{\infty} \left( \frac{1}{n(n+1)} \right)^k f_k \binom{k}{m}n^{k-m}\\
&=\sum_{m=0}^{\infty}x^m\sum_{k=m}^{\infty}f_k\sum_{n=1}^{\infty} \frac{1}{n (n + 1)} \left( \frac{1}{n(n+1)} \right)^k \binom{k}{m}n^{k-m}\\
&=\sum_{m=0}^{\infty}x^m\sum_{k=m}^{\infty}f_k\binom{k}{m}\sum_{n=1}^{\infty} \frac{1}{n (n + 1)} \left( \frac{n}{n(n+1)} \right)^k n^{-m}\\
&=\sum_{m=0}^{\infty}x^m\sum_{k=m}^{\infty}f_k\binom{k}{m}\sum_{n=1}^{\infty} \frac{1}{n (n + 1)} \left( \frac{1}{n+1} \right)^k\frac1{n^{m}}\\
&=\sum_{m=0}^{\infty}x^m\sum_{k=m}^{\infty}f_k\binom{k}{m}\sum_{n=1}^{\infty} \left( \frac{1}{n+1} \right)^{k+1}\frac1{n^{m+1}}\\
\end{array}
$
so it looks like
$ W_{m, k}
=\binom{k}{m}\sum_{n=1}^{\infty} \left( \frac{1}{n+1} \right)^{k+1}\frac1{n^{m+1}}
$
and
$ W_{m, k}
=0$
for
$k < m$.
| {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find $C\in\mathbb R$ such that $(2,3,5)$ be in $\text{Im}(F)$ Let $$\color{red}{\underbrace{F:\mathbb R^3\to\mathbb R^3\;|}_{\text{Added by me}}}\; F(x,y,z)=(x+Cz,\;x+y+2z,\;x+Cy)$$ a linear transformation. Find $C\!\!\!\!\!\color{red}{\underbrace{\in\mathbb R}_{\text{Added by me}}}\!\!\!\!\!\!$ such that $(2,3,5)$ be in $\text{Im}(F)$.
I don't know hot to interpet the statement. I think I need to solve the equation $$F(x,y,z)=(2,3,5)\qquad\text{i.e.}\qquad(x+Cz,\;x+y+2z,\;x+Cy)=(2,3,5),$$ but this is a system equations of $3\times 4$. Is correct my reasoning? If so, how can it solve to find $C$?
Thank you!
| $$(x+Cz, x+y+2z, x+Cy)=(2,3,5)$$
$$\implies x+Cz=2;~~~ x+y+2z=3;~~~~ x+Cy=5$$
$$\begin{bmatrix}1&0&C\\1&1&2\\1&C&0\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}2\\3\\5\end{bmatrix}$$
applying row transformation, $R_2\leftarrow R_2-R_1$, $R_3\leftarrow R_3-R_1$
$$\begin{bmatrix}1&0&C\\0&1&2-C\\0&C&-C\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}2\\1\\3\end{bmatrix}$$
$R_3\leftarrow R_3-C R_2$
$$\begin{bmatrix}1&0&C\\0&1&2-C\\0&0&C^2-3C\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}2\\1\\3-C\end{bmatrix}$$
For unique solution $(x,y,z)$, we take $C$ different from $3$.
NOTE:- for $C=0$, rank of coefficient matrix is not equal to rank of augmented matrix. Hence, for $C=0$ solution does not exist.
Take $C=1$
$$\begin{bmatrix}1&0&1\\0&1&1\\0&0&-2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}2\\1\\2\end{bmatrix}$$
$R_3\leftarrow \frac{-1}{2}R_3$ followed by $R_2\leftarrow R_2-R_3$, $R_1\leftarrow R_1-R_3$
$$\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}3\\2\\-1\end{bmatrix}$$
$$x=3;~~y=2;~~z=-1.$$
$$(x+Cz, x+y+2z, x+Cy)=(3+1(-1),3+2+2(-1),3+1(2))=(2,3,5)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2785445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Suppose $a,b,c\geq 0$ and $ab+bc+ca=1$, prove that $a\sqrt{a^2+1}+b\sqrt{b^2+1}+c\sqrt{c^2+1}\geq 2$. Suppose $a,b,c\geq 0$ and $ab+bc+ca=1$, prove that
$$a\sqrt{a^2+1}+b\sqrt{b^2+1}+c\sqrt{c^2+1}\geq 2.$$
In my opinion, I do this problem by th following:
$$\sum_{cyc}a\sqrt{a^2+1}=\sum_{cyc}a\sqrt{(a+b)(c+a)},$$
but I do not know how to continue following this way.
It seems easy to enlarge this quantity, but how to reduce this quantity.
Any help and hint will welcome!!
| By C-S, Schur and AM-GM we obtain:
$$\sum_{cyc}a\sqrt{a^2+1}=\sum_{cyc}a\sqrt{(a+b)(a+c)}\geq\sum_{cyc}a(a+\sqrt{bc})=$$
$$=\sum_{cyc}(a^2-\sqrt{a^3b}-\sqrt{a^3c}+a\sqrt{bc})+\sum_{cyc}(\sqrt{a^3b}+\sqrt{ab^3})\geq2\sum_{cyc}ab=2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2785890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $\binom{2n}{n}\geq2^n$ for all $n\in\mathbb{N}_0$. Show that $\binom{2n}{n}\geq2^n$ for all $n\in\mathbb{N}_0$.
First, observe that $\binom{2n}{n} = \frac{(2n)!}{n!(2n-n)!}=\frac{(2n)!}{(n!)^2}=\frac{1\cdot 2\cdot 3\cdot...\cdot 2n-1\cdot 2n}{(n!)^2}=\frac{(1\cdot 3\cdot 5\cdot ... \cdot 2n-1)\cdot(2\cdot 4\cdot 6\cdot ... \cdot 2n)}{(n!)^2}=\frac{(1\cdot 3\cdot 5\cdot ... \cdot 2n-1)\cdot(2\cdot 1\cdot 2\cdot 2\cdot 2\cdot 3\cdot ... \cdot 2n)}{(n!)^2}=\frac{(1\cdot 3\cdot 5\cdot ... \cdot 2n-1)\cdot 2^n\cdot n!}{(n!)^2}=2^n\cdot\frac{1\cdot 3\cdot 5\cdot ... \cdot 2n-1}{(n!)}.$
For $n=0$ we have:
$\frac{(2*0)!}{(0!)^2}=1=2^0.$
Now to our inductive step $n+1$:
\begin{align*}
& \frac{(2(n+1))!}{((n+1)!)^2}\geq2^{n+1}\\
\Leftrightarrow\,\, & \frac{(2n+2)!}{((n+1)!)^2}\geq2^{n+1}\\
\Leftrightarrow\,\, & \frac{1\cdot 2\cdot 3\cdot...\cdot 2n+1 \cdot 2n+2}{((n+1)!)^2}\geq2^{n+1}\\
\Leftrightarrow\,\, & \frac{(1\cdot 3\cdot 5\cdot ... \cdot 2n+1)\cdot(2\cdot 4\cdot 6\cdot ... \cdot 2n \cdot 2n+2)}{((n+1)!)^2}\geq2^{n+1}\\
\Leftrightarrow\,\, & \frac{(1\cdot 3\cdot 5\cdot ... \cdot 2n+1)\cdot 2^{n+1} \cdot (n+1)!}{((n+1)!)^2}\geq2^{n+1}\\
\Leftrightarrow\,\, & 2^{n+1}\cdot \frac{(1\cdot 3\cdot 5\cdot ... \cdot 2n+1)}{((n+1)!)}\geq2^{n+1}\\
\end{align*}
Is this proof correct/sufficient? Was induction even needed here or should I have argued instead that $\frac{1\cdot 3\cdot 5\cdot ... \cdot 2n-1}{(n!)}>0$? Is this step necessary even for the induction?
| Another approach: the expression equals
$$\frac{2n(2n-1)\cdots (2n-(n-1))}{n!}=\frac{2n}{n}\frac{2n-1}{n-1}\frac{2n-2}{n-2}\cdots \frac{2n-(n-1)}{n-(n-1)}.$$
Since $2n-k \ge 2(n-k),$ $ k=0,\dots ,n-1,$ each fraction on the right is at least $2.$ Since there are $n$ fractions, we have the result.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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This function has two different integrals? $f(x)=∫\frac{1}{x^2}dx$
Integrating by u-substitution:
$u=x^2$
$du=2dx$
$\frac{1}{2}du = dx$
$∫\frac{1}{x^2}dx=$ $∫\frac{1}{u}\times\frac{1}{2}du$
$\frac{1}{2}$∫ $\frac{1}{u}du$
$=\frac{1}{2}ln u+c$
$=\frac{1}{2}ln x^2+c$
$=lnx+c$
Another way:
$∫\frac{1}{x^2}dx=∫x^{-2}dx $
$∫x^{-2}dx$
$=\frac{x^{-1}}{-1} + c$
$=-\frac{1}{x} + c$
Where have I gone wrong?
| It’s not $du=2dx$, it’s $du=2xdx$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Indefinite integral of $\sqrt{x^2-x}$ i was trying to compute the indefinite integral:
$$
\int\sqrt{x^2-x}dx
$$
but i got stuck: after a few (unsuccessful) attempts for some $u$-substitution, i tried integration by parts:
$$
\int\sqrt{x^2-x} \ dx=\int(x)'\sqrt{x^2-x} \ dx= \\ x\sqrt{x^2-x}-\int x(\sqrt{x^2-x})'dx= \\
=x\sqrt{x^2-x}-\frac{1}{2}\int x\frac{2x-1}{\sqrt{x^2-x}}dx= \\
=x\sqrt{x^2-x}-\int \frac{x^2}{\sqrt{x^2-x}}dx+\frac{1}{2}\int \frac{x}{\sqrt{x^2-x}}dx=...
$$
and now what? Can anybody help?
| hint
$$\sqrt {x^2-x}=\frac {1}{2}\sqrt {(2x-1)^2-1} $$
then put $$2x-1=\cosh (t) $$
and use
$$\cosh^2(t)-1=\sinh^2(t)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluating definite integral $\int_0^{2\pi} \frac{1}{13-5\sin\theta}\,\mathrm{d}\theta$ Question: $$\int_0^{2\pi} \frac{1}{13-5\sin\theta}\,\mathrm{d}\theta$$ is equals to
(a) $-\frac\pi6$
(b) $-\frac{\pi}{12}$
(c) $\frac\pi{12}$
(d) $\frac\pi6$
My attempt: Denoting given integral by $I$ and letting $z=e^{iθ}$ then given integral becomes,
\begin{align*}
I&=\int_C\frac{1}{13-5(\frac{z-\bar{z}}{2i})}\frac{\mathrm dz}{iz}\\
&=\frac{1}{i}\int_C\frac{2i}{26iz-5z^2+5|z|^2}\mathrm dz\\
&=2\int_C\frac{\mathrm dz}{-5z^2+26iz+5}\hspace{0.5in}\text{As }C: |z|=1\\
&=2\int_C \frac{\mathrm dz}{(z-5i)(z-i/5)}\\
&=2\left(\frac{5}{24i}\int_C\frac{1}{z-5i}-\frac{5}{24i}\int_C \frac{1}{z-i/5}\right)
\end{align*}
Now as point $z=5i$ lies outside $C$ so it's integral evaluates to $0$ and by Cauchy integral formula, above becomes,
$$I=0-2\frac{5}{24i}2\pi i = -\frac{5\pi}{6}$$
But none of the given answer matches with mine. So is am i incorrect? Please help me..stuck on this from hours...
| Use periodicity to rewrite your integral
$$I=\int_{-\pi}^{\pi} \frac{1}{13-5\sin\theta}\,\mathrm{d}\theta$$
With the change of variable $t=\tan\frac\theta2$,
$$I=\int_{-\infty}^{+\infty}\frac{2\mathrm dt}{(1+t^2)\left(13-5\dfrac{2t}{1+t^2}\right)}=\int_{-\infty}^{+\infty}\frac{2\mathrm dt}{13t^2-10t+13}$$
Then
$$I=\dfrac2{13}\int_{-\infty}^{+\infty}\frac{\mathrm dt}{\left(t-\dfrac5{13}\right)^2+\left(\dfrac{12}{13}\right)^2}=\dfrac{2}{13}\left(\dfrac{13}{12}\right)^2\int_{-\infty}^{+\infty}\frac{\mathrm dt}{\left(\dfrac{t-\frac5{13}}{\frac{12}{13}}\right)^2+1}$$
Now with the change of variable $u=\dfrac{t-\frac5{13}}{\frac{12}{13}}$,
$$I=\dfrac{2}{13}\left(\dfrac{13}{12}\right)^2\dfrac{12}{13}\int_{-\infty}^{+\infty}\dfrac{\mathrm du}{1+u^2}=\dfrac{2}{13}\left(\dfrac{13}{12}\right)^2\dfrac{12}{13}\pi=\frac{\pi}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2792437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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How to solve the following trigonometrical equation? I have the following equations
\begin{align*}
R_1\cos(\omega T_1-\phi_{1})& =Q_1-R_2\cos(\omega T_2-\phi_{2})\\
R_1\sin(\omega T_1-\phi_{1})& =-Q_2-R_2\sin(\omega T_2-\phi_{2}).
\end{align*}
From these equations how can I obtain the following solution
$$\omega T_2=\pm \arccos\left(\frac{Q_1^2+Q_2^2-R_1^2+R_2^2}{2R_2\sqrt{Q_1^2+Q_1^2}}\right)-\arctan\left(\frac{Q_2}{Q_1}\right)+\phi_2+2k\pi .$$
| From $$a\sin\theta+b\cos\theta+c=0,$$
$$a^2\sin^2\theta=a^2(1-\cos^2\theta)=(b\cos\theta+c)^2$$
and
$$(a^2+b^2)\cos^2\theta+2bc\cos\theta+c^2-a^2=0.$$
You solve the quadratic equation for $\cos\theta$, and $\sin\theta$ follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2792861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Show that the annulus $1 \leq x^2 + y^2 \leq 3$ is positively invariant
Given the planar dynamical system
$$\dot x = x - y - x^3, \\ \dot y = x + y - y^3$$
show this is positively invariant in the annulus $1 \leq x^2 + y^2 \leq 3$. Hints:
$$x^4 + y^4 = (x^2 + y^2)^2 − 2x^2y^2$$
$$x^4 + y^4 = \frac 12 (x^2 + y^2)^2 + \frac 12 (x^2 - y^2)^2$$
My attempt:
I am really unsure whether what I am doing is correct or not, but I seem to get the right answer.
$$V:= x^2 + y^2$$
Take the derivative,
$$\frac{dV}{dt} = \frac{dV}{dx} \frac{dx}{dt}$$
which gives
$$\frac{dV}{dt} = \pmatrix{2x \\ 2y} \cdot \pmatrix{x-y-x^3 \\ x+y-y^3} = 2x^2 - 2xy - 2x^4 + 2xy + 2y^2 - 2y^4 \\ = 2(x^2 + y^2 -x^4 - y^4)$$
By using the 2 hints given I get that
$$2(x^2 + y^2 -x^4 - y^4) = 2(x^2 + y^2) - 2(x^2 + y^2)^2 + 4x^2y^2$$
and
$$2(x^2 + y^2 -x^4 - y^4) = 2(x^2+y^2) - (x^2+y^2)^2 - (x^2-y^2)$$
Converting these into polar coordinates will give
$$2r^2(1-r^2) + 4x^2y^2 \geq 0$$
when $r^2=1$, and
$$2r^2 \left( 1-\frac{r^2}{2} \right) - (x^2-y^2)^2 < 0$$
when $r^2=3$. Therefore can conclude from this that they are both pointing inwards.
Even though I am getting the right answer, I don't know if my methodology is correct. The place where I am not confident is when I call $V := x^2+y^2$ and then take like the orbital derivative.
| You found that
$$
2V(1-V)\le \dot V\le V(2-V)
$$
Treating these differential inequalities in the Bernoulli fashion now divide by $V^2$ to get for $U=1/V$
$$
2U-2\le -\dot U\le 2U-1\\
1\le\dot U+2U\le 2\\
U_0e^{-2t}+\frac12(1-e^{-2t})\le U(t)\le U_0e^{-2t}+(1-e^{-2t})\\
\frac{1}{1+(1/V_0-1)e^{-2t}}\le V(t)\le \frac2{1+(2/V_0-1)e^{-2t}}
$$
which shows that from any position (except the origin) the solution curve will move to the annulus $1\le V\le 2$.
| {
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"source": "stackexchange",
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How many solutions are there to the sum $x+y+z+w = 12$ when $x \geq 4$ and $y \leq 3$? I am currently trying to work out how many integer solutions there are to the sum
$$x+y+z+w = 12$$
when $x \geq 4$ and $y \leq 3$ (and $z,w \geq 0$).
I have worked out that
*
*With $x,y,z,w \geq 0$ the number of solutions is $455$.
*With $x \geq 4$ and $y,z,w \geq 0$ the number of solutions is $165$.
*With $y \leq 3$ and $x,z,w \geq 0$ the number of solutions is $290$.
but I have been unable to proceed from here. Can anyone help me to deduce the answer to this one?
| We wish to solve the equation
$$x + y + z + w = 12 \tag{1}$$
in the nonnegative integers subject to the restrictions $x \geq 4$ and $y \leq 3$.
To handle the restriction $x \geq 4$, let $x' = x - 4$. Then $x'$ is a nonnegative integer. Substituting $x' + 4$ for $x$ in equation 1 yields
\begin{align*}
x' + 4 + y + z + w & = 12\\
x' + y + z + w & = 8 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers. The number of nonnegative integers solutions of equation 2 subject to the restriction $y \leq 3$ is equal to the number of nonnegative integer solutions of equation 1 subject to the restrictions $x \geq 4$, $y \leq 3$.
If we ignore the restriction $y \leq 3$ for the moment, a particular solution of equation 2 corresponds to the placement of three addition signs in a row of eight ones. For instance,
$$1 1 1 + 1 1 + + 1 1 1$$
corresponds to the solution $x' = 3$ ($x = 7$), $y = 2$, $z = 0$, $w = 3$. The number of such solutions is the number of ways we can select which three of the eleven positions required for eight ones and three addition signs will be filled with addition signs, which is
$$\binom{8 + 3}{3} = \binom{11}{3}$$
From these, we must subtract those cases in which the restriction $y \leq 3$ is violated.
Suppose $y > 3$. Then $y' - 4$ is a nonnegative integer. Substituting $y' + 4$ for $y$ in equation 2 yields
\begin{align*}
x' + y' + 4 + z + w & = 8\\
x' + y' + z + w & = 4 \tag{3}
\end{align*}
Equation 3 is an equation in the nonnegative integers. Since a particular solution corresponds to the placement of three addition signs in a row of four ones, equation 3 has
$$\binom{4 + 3}{3} = \binom{7}{3}$$
solutions in the nonnegative integers.
Thus, the number of solutions of equation 1 in the nonnegative integers that satisfy the restrictions $x \geq 4$ and $y \leq 3$ is
$$\binom{11}{3} - \binom{7}{3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\int\limits_{x}^{+\infty}\frac{ds}{1+s^2}\geq \frac{x}{1+x^2},~x\geq 0.$
Prove that for all $x\geq 0$:
$$\int\limits_{x}^{+\infty}\frac{ds}{1+s^2}\geq \frac{x}{1+x^2}.$$
Attempt. I have tried algebra, by using $f(x)=\int\limits_{x}^{+\infty}\frac{ds}{1+s^2}- \frac{x}{1+x^2}$, where $$f'(x)=-\frac{1}{1+x^2}-\bigg(\frac{x}{1+x^2}\bigg)'=\ldots=-\frac{2}{(1+x^2)^2}<0,$$
so $f$ is decreasing and: $$f(x)\leq f(0)=\frac{\pi}{2},$$
but i didn't get anywhere. Geometrically, I tried bu using the area under the graph of $\frac{1}{1+x^2}$ after $x$ and the rectangle over $[0,x]$, of height $\frac{1}{1+x^2}$, but i couldn't prove the desired inequality.
Thanks in advance.
| You can express the right hand side as a definite integral:
$$\int\limits_{x}^{+\infty}\frac{ds}{1+s^2}\geq \frac{x}{1+x^2} \iff \int\limits_{x}^{+\infty}\frac{1}{1+s^2}ds\geq \int\limits_{x}^{+\infty}\frac{s^2-1}{(1+s^2)^2}ds \iff \\
\int\limits_{x}^{+\infty}\frac{1}{1+s^2}-\frac{s^2-1}{(1+s^2)^2}ds\geq 0 \iff \int\limits_{x}^{+\infty}\frac{2}{(1+s^2)^2}ds\geq 0. \qquad \checkmark$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can we say that $\frac{a+b}{c+d} \leq \frac{a}{c} + \frac{b}{d} $ If $a,b,c,d$ are positive real numbers, can we say that $\frac{a+b}{c+d} \leq \frac{a}{c} + \frac{b}{d} $ is always true? If no, can you please give insignts on under which conditions this might be true.
Any references to a similar type of inequalities are also welcome
Thank you,
| Solution to new version:
Since $$\frac{a}{c+d} \leq \frac{a}{c}$$
and
$$\frac{b}{c+d} \leq \frac{b}{d}$$
we have
$$\frac{a+b}{c+d} = \frac{a}{c+d} + \frac{b}{c+d} \leq \frac{a}{c} + \frac{b}{d} $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Find primes $p$, for which the $x^2 \equiv 7 \pmod{p}$ has a solution
I want to determine for which primes $p$ the following congruence has a solution.
$$x^2 \equiv 7 \pmod{p}.$$
I have thought the following.
We are looking for the primes $p$ for which the Legendre symbol $\left( \frac{7}{p}\right)=1$. Then $(7,p)=1$.
If $p$ is odd, we have from Euler's theorem that
$$\left( \frac{7}{p}\right)=7^{\frac{p-1}{2}} \pmod{p}$$
So we would want that $7^{\frac{p-1}{2}} \pmod{p}=1$.
But does this help in order to find the desired primes $p$ ?
| It's a little longer. From law of quadratic reciprocity, we have
$$\biggl(\frac7p\biggr)=\biggl(\frac p7\biggr)(-1)^{\tfrac{p-1}2},$$
\begin{align}&\text{On the other hand, }&&\begin{cases}(-1)^{\tfrac{p-1}2}=1&\text{if }\;p\equiv 1\mod 4,\\(-1)^{\tfrac{p-1}2}=-1&\text{if }\;p\equiv 3\;(\text{or }-1)\mod 4.
\end{cases}&&\hspace{10em}\end{align}
Also the list of non-zero squares mod. $7$ is$\enspace\bigl\{1,\,2,\,4\bigr\}$.
So we have to solve the systems of congruences:
\begin{align}
&\begin{cases}p\equiv 1\mod4,\\p\equiv 1,\,2,\,4\mod 7,
\end{cases}
& &\qquad\begin{cases}p\equiv -1\mod4,\\p\equiv 3,\,5,\,6\mod 7,
\end{cases}
\end{align}
To solve these systems, we use the explicit formula for the inverse isomorphism of the abstract form of the Chinese remainder theorem.
More precisely, start from a Bézout's relation between $4$ and $7$, e.g.
$\;4\cdot 4-1\cdot 7=1$. Then
$$\begin{cases}p\equiv a\mod4,\\p\equiv b\mod 7
\end{cases}\iff p\equiv 4\cdot 4\, b- 7a \mod 28$$
In all, you should find, if I'm not mistaken:
$$p\equiv \pm 1,\pm3,\pm 9\mod 28. $$
Added:
As @OscarLanzi observed, this list can be simplified to
$$x^2 \equiv 7 \;\text{ has a solution mod. }p \iff p \equiv 3^k \mod 28 \;\text{ for some }k.$$
This is because $3$ has order $6\bmod 28$. Indeed the successive power of $3$ mod. $28$ are:
$$ 1,\;3,\;9,\;3^3\equiv -1,\;3^4\equiv -3,\;3^5\equiv -9, \;3^6\equiv -27\equiv 1 \mod 28.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Showing $f(x,y):=(y^2-x^2)(y^2-2x^2)$ has no extrema how can I show that $f(x,y):=(y^2-x^2)(y^2-2x^2)$ has no extrema?
I already computed the first derivatives for $x$ and $y$ but that doesn't really help me either
$f_{1_x}=8x^3-6xy^2$
$f_{2_x}=24x^2-6y^2$
$f_{1_y}=-6x^2y+4y^3$
$f_{2_y}=-6x^2+12y^2$
| We need to find the critical points by
*
*$f_x=-2x(y^2-2x^2)-4x(y^2-x^2)=0$
*$f_y=2y(y^2-2x^2)+2y(y^2-x^2)=0$
which has solution for $x=y=0$ and for $x,y\neq 0$ we have
*
*$f_x=-2x(y^2-2x^2)-4x(y^2-x^2)=0\\\implies -2(y^2-2x^2)-4(y^2-x^2)=-6y^2+8x^2=0\implies3y^2=4x^2$
*$f_y=2y(y^2-2x^2)+2y(y^2-x^2)=0\implies (y^2-2x^2)+(y^2-x^2)=2y^2-3x^2=0$
which has not solutions.
Then note that
$$f(x,y)=(y^2-x^2)(y^2-2x^2)=y^4-3x^2y^2+2x^4=(y^2-\frac32x^2)^2-\frac14x^4$$
and around$(x,y)=(0,0)$ with $f(0,0)=0$ we have that
*
*$f(0,y)=y^4\ge 0$
*$f(x,\sqrt{\frac32}\,x)=-\frac14x^4\le 0$
| {
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"url": "https://math.stackexchange.com/questions/2799690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find $x^5+x^{-5}$ given the value of $x^2+x^{-2}$.
Find $x^5+\dfrac1{x^5}$ in its simplest form given that $x^2+\dfrac1{x^2}=a$ for $a,x>0$.
Attempt:
We write $$x^2+\frac1{x^2}=a\implies x^4-ax^2+1=0\implies x^5=ax^3-x$$ and $$\frac1{x^2}=a-x^2\implies \frac1{x^4}=a^2-2ax^2+x^4\implies \frac1{x^5}=\frac{a^2}x-2ax+x^3$$ so $$\begin{align}x^5+\frac1{x^5}&=(1+a)x^3-(1+2a)x+a^2\cdot\frac1x\\&=(1+a)x^3-(1+2a)x+a^2(ax-x^3)\\&=(1+a-a^2)x^3-(1+2a-a^3)x\\\implies x^5+\frac1{x^5}&=(a^2-a-1)x(a+1-x^2)\end{align}$$ But is this in its simplest form?
| With $S_n:=x^n+x^{-n}$, you can establish the recurrence
$$S_{n+1}=S_nS_1-S_{n-1}.$$
From this, $$S_2=S_1^2-2,
\\S_3=S_1(S_2-1),
\\S_4=S_2^2-2,
\\S_5=S_1(S_2^2-S_2-1).
$$
You draw $S_1=\sqrt{S_2+2}$ from the first identity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 7
} |
Prove or disprove that, for any $n \in \mathbb{N_+}$, there exist $a,b \in \mathbb{N_+} $ such that $\frac{a^2+b}{a+b^2}=n.$ Problem
Prove or disprove that, for any $n \in \mathbb{N_+}$,
there exist $a,b \in \mathbb{N_+} $ such that $$\frac{a^2+b}{a+b^2}=n.$$
My Thought
Assume that the statement is ture. Then, the equality is equivalent to that
$$a^2-na+b-nb^2=0.$$
Regard it as a quadratic equation with respect of $a$.Then $$a=\dfrac{n \pm \sqrt{n^2+4nb^2-4b}}{2}.$$ Thus, $n^2+4nb^2-4b$ must be a square number. Let
$$n^2+4nb^2-4b=k^2,k \in \mathbb{N_+}.$$
How to go on with this? May it work?
P.S.
The statement seems to be true. Here are parts of verification examples:
\begin{array}{r|r|r}
n&a&b \\ \hline
1&1&1\\
2&5&3\\
3&5&2\\
4&10&4\\
5&27&11\\
6&69&27\\
\vdots&\vdots&\vdots
\end{array}
Besides, the equation could be rewritten as
$$n(2a-n)^2-(2nb-1)^2=n^3-1,$$
which is a $\textbf{ Pell-like equation}$. This will help?
| Solution for non-square $n$ is provided in @Oldboy's answer and in linked questions. This answer handles the case for square $n$.
Case 1: $n=k^2,k \equiv 0 \pmod {2}$
Choose
\begin{align}
a=\frac{k^2(k^3+2)}{4},
b=\frac{k^4}{4}.
\end{align}
Conditions imply that $k^2 \equiv 0 \pmod {4}$ and so both $a$ and $b$ are integers. By algebraic manipulation we can show that $(a^2+b)/(b^2+a)=k^2=n$ (it is quite technical).
Case 2: $n=k^2,k \equiv 1 \pmod {2}$
Choose
\begin{align}
a=\frac{(k^2+1)(k^2-k+2)}{4}, b=\frac{(k-1)(k^2+1)}{4}.
\end{align}
Here $2 \mid k^2+1$ and $2 \mid k^2-k+2$ implies $a$ is an integer and similarly $2 \mid k-1$, $2 \mid k^2+1$ for $b$. Again it can be verified that $(a^2+b)/(b^2+a)=k^2=n$.
This result is obtained by mindless following of solution of quadratic diophantine equation on https://www.alpertron.com.ar/QUAD.HTM. Basically for square $n$ and our equation the site instructs us to find $(X-\sqrt{n}Y)(X+\sqrt{n}Y)=4n(n^3-1)$ such that $4n \mid Y+2$ ($2$ being calculated there as $\beta$ and $4n$ being a determinant). So the problem is essentially to look at divisors $d$ of $4n(n^3-1)$ that satisfy above divisibility criteria. For $n=k^2$ the factorization is $2\cdot2\cdot(k-1)k^2(k+1)(k^2-k+1)(k^2+k+1)$ (not into primes, but fortunately this is enough). So by testing combinations of these factors (using Maple e.g.), it turns out that choices of $d=2k$ and $d=2k(k+1)$ work (for even and odd $k$ cases respectively, that is). Those choices when substituting all the way back simplify to the cases described above, but it is too long/technical to get there...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2802933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 1,
"answer_id": 0
} |
Solve the initial value problem $ (x+1)^2 dx+(2xy+x^2-1)dy =0 , \ y(0)=1 \ $ Solve the initial value problem $$ (x+1)^2 dx+(2xy+x^2-1)dy =0 \ , \ \ y(0)=1 $$
Answer:
Consider the above equation with $ \ M(x,y)dx+N(x,y)dy=0 \ $ , then we get
$ M(x,y)=(x+1)^2, \ N(x,y)=2xy+x^2-1 \ $
Since $ \ \frac{\partial M}{\partial {y}}=0 \neq 2y+2x=\frac{\partial N}{\partial {x}} \ $ , the equation is not exact.
Let $ \ \mu =x^a y^b \ $ is an integrating factor.
Multiplying the differential equation by $ \ \mu=x^a y^b \ $ , we get
$$ (x^{a+2}y^b+2x^{a+1}y^b+x^ay^b)dx+(2x^{a+1} y^{b+1}+x^{a+2} y^b-x^ay^b)dy=0 \ $$ Consider it with $ M'dx+N'dy=0 \ $ , we gte
$ M'= x^{a+2}y^b+2x^{a+1}y^b+x^ay^b , \ N'=2x^{a+1} y^{b+1}+x^{a+2} y^b-x^ay^b \ $
Since the new equation is exact , we have
$ \frac{\partial M'}{\partial {y}}=\frac{\partial N'}{\partial {x}} \\ \Rightarrow bx^{a+2}y^{b-1}+2bx^{a+1} y^{b-1}+bx^a y^{b-1}=(a+1)2x^{a} y^{b+1}+(a+2)x^{a+1} y^{b}-ax^{a-1} y^{b} $
From this I have to find out the values $ \ a, \ b \ $.
But I can not find $ a,b \ $
Help me
| $$ (x+1)^2 dx+(2xy+x^2-1)dy =0 \ , \ \ y(0)=1$$
Yes, the integrating factor is $\mu=x^ay^b$. There's a formula for this case
$$\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}=a\frac{N(x,y)}{x}-b\frac{M(x,y)}{y}\\ -2x-2y=a\left(\frac{x^2+2xy-1}{x}\right)-b\left(\frac{x^2+2x+1}{y}\right)\\ -2x-2y=ax+2ay-ax^{-1}-b(x^2+2x+1)y^{-1}$$Here everything its right except for $a$ because can't have two values$\ a=-2\ \&\ a=0$. Then the ODE isn't exact.
Try using Maclaurin Series to aproximate the solution $\ y(x)=\sum_{n=0}^\infty \frac{y^{(n)}(0)}{n!}x^n$
$$ (x+1)^2 dx+(2xy+x^2-1)dy =0 \ , \ \ y(0)=1\\ y'(x)=-\frac{(x+1)^2}{x^2+2xy-1}$$
We have $y(0)=1$ that means $x=0$ and $y=1$ substitute to find $y'(0)$. I'll expand the solution to four terms so do the same for $y''(0)$ and $y'''(0)$
$$y''(x)=-\frac{(x^2+2xy-1)(2x+2)-(x+1)^2[2x+2y+2xy'(x)]}{(x^2+2xy-1)^2}$$
$$y''(x)=2\left[\frac{(x+1)^2[x+y+xy'(x)]}{(x^2+2xy-1)^2}-\frac{x+1}{x^2+2xy-1}\right]$$
$y'''(x)=2\left\{\frac{4(x+1)[x+y+xy'(x)]}{(x^2+2xy-1)^2}+\frac{(x+1)^2[1+2y'(x)+xy''(x)]}{(x^2+2xy-1)^2}-\frac{4(x+1)^2[x+y+xy'(x)]^2}{(x^2+2xy-1)^3}-\frac{1}{x^2+2xy-1}\right\}$
Then $\quad y'(0)=1$,$\quad y''(0)=4$,$\quad y'''(0)=24$
$$y(x)\approx1+x+2x^2+4x^3+\ldots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2803725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding $f(\sqrt 3 )$
Let $$x^2 + \left( f(x)-2 \right) x + 2\sqrt 3 - 3 - \sqrt 3 f(x) = 0$$ for all $x \in \mathbb R$ and let $f(x)$ be continuous for all $x \in \mathbb R$. Find $f(\sqrt 3)$.
When I substituted $x= \sqrt 3$, I got $0=0$. How else do I solve this problem? I don't think there are many more obvious constraints.
The given answer is:
$2(1-\sqrt 3) $
| If $x\ne \sqrt{3}$ then you can write $f$ explicitly:
$$f(x)= {-x^2+2x+3-2\sqrt{3}\over x - \sqrt 3 }$$
So you can only calculate $$\lim_{x\to \sqrt{3}}f(x) =\lim_{x\to \sqrt{3}} {-x^2+2x+3-\color{red}{2}\sqrt{3}\over x - \sqrt 3 }$$ $$= \lim_{x\to \sqrt{3}} {(\sqrt{3}-x)(\sqrt{3}+x)+2(x-\sqrt{3})\over x - \sqrt 3 }=$$
$$= \lim_{x\to \sqrt{3}} (-(\sqrt{3}+x)+2)=2-2\sqrt{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2803952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Trigonometric equation: $2\arcsin \left(\frac{2x}{1+x^2}\right)- \pi x^3 = 0$
The number of solutions of the equation $$2\arcsin \left(\dfrac{2x}{1+x^2}\right)- \pi x^3 = 0$$ is?
Let $x= \tan \theta$
$\implies \sin 2\theta = \sin(\dfrac \pi 2 \tan^3\theta)$
I had to delete the rest of my attempt because it was totally wrong. What are the methods to solve this problem?
| Consider
$$
f(x)=\arcsin\frac{2x}{1+x^2}
$$
Then
$$
f'(x)=\frac{1}{\sqrt{1-\dfrac{4x^2}{(1+x^2)^2}}}\frac{2(1+x^2)-4x^2}{(1+x^2)^2}
=\frac{1+x^2}{|1-x^2|}\frac{2(1-x^2)}{(1+x^2)^2}
$$
Therefore
$$
f'(x)=\begin{cases}
\dfrac{2}{1+x^2} & |x|<1 \\[6px]
-\dfrac{2}{1+x^2} & |x|>1
\end{cases}
$$
which implies
$$
f(x)=\begin{cases}
c_--2\arctan x & x<-1 \\[6px]
c_0+2\arctan x & -1\le x\le 1 \\[6px]
c_+-2\arctan x & x>1
\end{cases}
$$
and it's easy to see that $c_0=0$, $c_-=-\pi$ and $c_+=\pi$.
Thus the equation can be split into three cases:
$$
\begin{cases}
-2\pi-4\arctan x-\pi x^3=0 \\[4px]
x<-1
\end{cases}
\qquad
\begin{cases}
4\arctan x-\pi x^3=0 \\[4px]
-1\le x\le 1
\end{cases}
\qquad
\begin{cases}
2\pi-4\arctan x-\pi x^3=0 \\[4px]
x>1
\end{cases}
$$
The graph suggests that the solutions are only $-1$, $0$ and $1$ (which are indeed solutions).
Let's consider $\alpha(x)=-2\pi-4\arctan x-\pi x^3$ for $x\le-1$; then
$$
\alpha'(x)=-\frac{4}{1+x^2}-3\pi x^2=-\frac{4+3\pi x^2+3\pi x^4}{1+x^2}
$$
Note that the discriminant of $3\pi t^2+3\pi t+4$ is $9\pi^2-48\pi=3\pi(3\pi-16)<0$ so $\alpha$ is decreasing. As $\alpha(-1)=0$, there are no solutions of your equation for $x<-1$. Similarly, there are no solutions for $x>1$.
In the case $-1\le x\le 1$, the function $\beta(x)=4\arctan x-\pi x^3$ is odd, so we can study it over $[0,1]$. We have
$$
\beta'(x)=\frac{4}{1+x^2}-3\pi x^2=-\frac{3\pi x^4+3\pi x^2-4}{1+x^2}
$$
The derivative vanishes only for
$$
x=\sqrt{\frac{-3\pi+\sqrt{9\pi^2+48\pi}}{6\pi}}\approx 0.5668
$$
Thus $0$ and $1$ are the only solutions on the interval $[0,1]$; by symmetry, $0$ and $-1$ are the only solutions on the interval $[-1,0]$.
It's not needed to look for an approximation: the following set of inequalities are equivalent to each other:
\begin{gather}
\sqrt{\frac{-3\pi+\sqrt{9\pi^2+48\pi}}{6\pi}}<1 \\[6px]
\frac{-3\pi+\sqrt{9\pi^2+48\pi}}{6\pi}<1 \\[6px]
-3\pi+\sqrt{9\pi^2+48\pi}<6\pi \\[6px]
\sqrt{9\pi^2+48\pi}<9\pi \\[6px]
9\pi^2+48\pi<81\pi^2 \\[6px]
48<72\pi \\[6px]
2<3\pi
\end{gather}
and the last one is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2804754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
$\sum_{n=1}^{\infty}\frac{4n^2+26n+34}{n^4+10n^3+35n^2+50n+24}$ convergent? serie is convergent?
$$\sum_{n=1}^{\infty}\frac{4n^2+26n+34}{n^4+10n^3+35n^2+50n+24}$$
Factor: $$\frac{4n^2+26n+34}{(n+1)(n+2)(n+3)(n+4)}$$
What test can i Use?
| $$
\frac{4 n^2+26 n+34}{(n+1) (n+2) (n+3) (n+4)} = \frac{1}{n+2}-\frac{4}{n+3}+\frac{1}{n+4}+\frac{2}{n+1}
$$
Then
$$
\sum_{n=1}^{\infty}\frac{4 n^2+26 n+34}{(n+1) (n+2) (n+3) (n+4)} = \lim_{n\to\infty}\left(2\sum_{k=2}^{n}\frac{1}{k}+\sum_{k=3}^{n}\frac{1}{k}-4\sum_{k=4}^{n}\frac{1}{k}+\sum_{k=5}^{n}\frac{1}{k}\right)=2\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)-4\frac{1}{4} = \frac{7}{4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2810694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to integrate the product of two or more polynomials raised to some powers, not necessarily integral This question is inspired by my own answer to a question which I tried to answer and got stuck at one point.
The question was:
HI DARLING.
USE MY ATM CARD, TAKE ANY AMOUNT OUT, GO SHOPPING AND TAKE YOUR FRIENDS FOR LUNCH.
PIN CODE: $\displaystyle \int_{0}^{1} \frac{3x^3 - x^2 + 2x - 4}{\sqrt{x^2 - 3x + 2}} \, dx $
I LOVE YOU HONEY.
Anyone knows? Are we gonna get an integer number?
My attempt:
Does this help?
$$\frac{3x^3-x^2+2x-4}{x-1}=3x^2+2x+4$$
(long division)
\begin{align*}
I&=\int\frac{3x^3-x^2+2x-4}{[(x-1)(x-2)]^{1/2}} dx = \\
&=\int\frac{(3x^2+2x+4)(x-1)^{1/2}}{(x-2)^{1/2}} dx = \\
&=\int 3(u^4-4u^2-4)(u^2+1)^{1/2}du \times 2
\end{align*}
after the substitution
\begin{gather*}
(x-2)^{1/2}=u\\
du=\frac1{2(x-2)^{1/2}}dx\\
u^2=x-2\\
(x-1)^{1/2}=(u^2+1)^{1/2}
\end{gather*}
Update: This may help us proceed.
I tried to proceed:
$$6\int (u^4-4u^2-4)(u^2+1)^{1/2} du = 6\int ((t-3)^2-8)t \frac{dt}{2u}$$
after $u^2+1=t$ and $dt=2udu$
\begin{align*}
u^4-4u^2-4
&= (u^2+1)^2-(6u^2+5) \\
&= (u^2+1)^2-6(u^2+1)+1 \\
&= ((u^2+1)-3)^2-8
\end{align*}
I wonder whether this question can be solved from here?
Update:
This has been getting a lot of views, and I think most people came for the sort of problem mentioned in the title (where I got stuck) rather than the original problem itself.
Keepin this in mind, I'm reopening the question and here's the kind of answers I expect — Solutions to the original problem are good, but I'd prefer solutions that continue from the part where I got stuck — the polynomial in $u$ — that's the sort of problem mentioned in the title.
| $\require{begingroup}\begingroup$This should help to get closer to the final result (if you want to calculate this manually):
$$\newcommand{\dd}{\; \mathrm{d}} I=\int_0^1 \frac{3x^3 - x^2 + 2x - 4}{\sqrt{x^2 - 3x + 2}} \dd x =
\int_0^1 \frac{3x^3 - x^2 + 2x - 4}{\sqrt{(x-\frac32)^2 -\frac14}} \dd x$$
It will take some computing but we can get that
$3x^3 - x^2 + 2x - 4 = 3(x-\frac32)^3+\frac{25}2(x-\frac32)^2+\frac{77}4(x-\frac32)+\frac{55}8$.
So we have
$$I=
\int_{-3/2}^{-1/2} \frac{3u^3+\frac{25}2u^2+\frac{77}4u+\frac{55}8}{\sqrt{u^2-\frac14}} \dd u =
\begin{vmatrix}
t=2u & u=\frac t2 \\
\dd t=2\dd u & \dd u = \frac12\dd t
\end{vmatrix} =
\frac12 \int_{-3}^{-1} \frac{\frac{3t^3}8+\frac{25}8t^2+\frac{77}8t+\frac{55}8}{\sqrt{\frac{t^2}4-\frac14}} \dd t =
\frac18 \int_{-3}^{-1} \frac{3t^3+25t^2+77t+55}{\sqrt{t^2-1}} \dd t =
\frac18 \int_{-3}^{-1} \frac{3t(t^2-1)+25(t^2-1)+80t+80}{\sqrt{t^2-1}} \dd t =
\frac18 \int_{-3}^{-1} (3t+25)\sqrt{t^2-1} +80 \frac{t+1}{\sqrt{t^2-1}} \dd t
$$
You can check that Wolfram Alpha returns the same value for the original integral and this integral. (To be honest, I am not sure how I am supposed to get a PIN number from the result.)
Now you could divide up this into separate integrals which should be not too difficult:
*
*How to calculate this integral with square roots: $\int\frac{ \sqrt{x+1} }{ \sqrt{ x-1 }} \, dx$
*Indefinite integral of $\int\sqrt{x^2-1}dx$
*For $\int t\sqrt{t^2-1} \dd t$ the substitution $s=t^2-1$ seems reasonable.
$\endgroup$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2814179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
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Prove $\sin^2 a + \sin^2 b + \sin^2 c = 2(1-\cos a \cos b \cos c)$ when $a=b+c$ I want to prove this identity:
$$\sin^2 a + \sin^2 b + \sin^2 c = 2(1-\cos a \cos b \cos c) \qquad\text{when}\;a=b+c$$
Can somebody give a hint in the easiest way possible? I am debugging this for hours and can't get the left side to be the right.
| If $a = b+ c$ then
$\cos a = \cos b\cos c - \sin b\sin c$
$\sin a = \sin b\cos c + \cos b\sin c$
So solving $\sin^2 a + \sin^2 b + \sin^2 c = 2(1-\cos a \cos b \cos c)$
involves solving
$\sin^2 b + \sin^2 c + \sin^2 b\cos^2 c + 2\cos b\sin b\cos c\sin c + \cos^2 b \sin^2 c = 2(1 - \cos b^2 \cos^2 c + \cos b\cos c\sin b\sin 2)$
$\sin^2 b + \sin^2 c + \sin^2 b\cos^2 c +\cos^2 b \sin^2 c +\cos^2 b \cos^2 c+\cos^2 b \cos^2 c= 2$
$\sin^2 b + \sin^2 c + \cos^2 c(\sin^2 b +\cos^2 b)+\cos^2 b(\sin^2 c + \cos^2 c) = 2$
$\sin^2 b +\sin^2 c +\cos^2 c + \cos^2 b = 2$
which is always true.
So all $a = b+c$ will solve the equation.
...
So I have to wonder if the problem wasn't supposed to be not that we were told $a = b+c$ but we were suppose to solve the equation to find that $a = b+c$ is a solution.
Well, no point restarting from scratch. If we *don't know that $a = b+c$ ...
Let $d = b+c$.
So
$\sin^2 a + \sin^2 b + \sin^2 c = 2(1-\cos a \cos b \cos c)$
$\sin^2 a + \sin^2 d + \sin^2 b + \sin^2 c = 2(1-\cos a \cos b \cos c) +\sin^2 d$
$\sin^2 a + 2(1 - \cos d\cos b \cos c) = 2(1-\cos a \cos b \cos c) +\sin^2 d$ (from above)
$1 - \cos^2 a - 1 + \cos^2 d = 2\cos b\cos c(\cos d - \cos a)$
$(\cos d - \cos a)(\cos d + \cos a) = 2\cos b\cos c(\cos d - \cos a)$.
So If $\cos d = \cos a$ then $a = \pm d = \pm (b + c)$ are one system of solutions.
Other wise
$\cos d + \cos a = 2\cos b\cos c$
$\cos b\cos c - \sin b\sin c + \cos a = 2\cos b\cos c$
$\cos a = \cos b\cos c + \sin b\sin c = \cos (b-c)$ so
$a = \pm (b-c)$.
So (modulo $2\pi$) all solution are $a = \pm b \pm c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2815136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Convergence of $\sum_{n=1}^{\infty} 3^n \sin(\frac{1}{4^nx})$ I wish to prove the convergence of:
$$\sum_{n=1}^\infty 3^n \sin\left(\frac 1 {4^nx}\right)$$
for $1\le x \lt \infty$, using Cauchy's criterion.
Here is what I tried:
\begin{align}
|S_{n+p}-S_n| & = \left| 3^{n+1} \sin\left(\frac 1 {4^{n+1}x}\right)
+ \cdots+3^{n+p} \sin\left(\frac 1 {4^{n+p}x}\right)\right| \\[10pt]
& \le \left|4^{n+1} \sin\left(\frac 1 {4^{n+1}x}\right) \cdots 4^{n+p} \sin\left(\frac 1 {4^{n+p}x}\right) \sin\left(\frac 1 {4^{n+1}x}\right)(4^{n+1}+\cdots+4^{n+p}) \right|
\end{align}
I tried using geometric series sum from here but came empty handed.
how can I show that $|S_{n+p}-S_n|\lt \varepsilon$?
| Note that
$$3^n \sin\left(\frac{1}{4^nx}\right)\sim \frac{3^n}{4^nx}$$
and by ratio test $\sum \frac{3^n}{4^nx}$ converges $\forall x\neq 0$, therefore the given series converges by limit comparison test with $\sum \frac{3^n}{4^nx}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2815627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $ \lim_{x \to -0.5^{-}} \left\lfloor\frac{1}{x} \left\lfloor \frac{-1}{x} \right\rfloor\right\rfloor$ Evaluate $$L=\lim_{x \to -0.5^{-}} \left\lfloor\frac{1}{x} \left\lfloor \frac{-1}{x} \right\rfloor\right\rfloor $$
My try:
Let $t=\frac{1}{x}$ Now when $ t \to -0.5^{-}$ we have $t \to -2^{+}$ we get
$$L=\lim_{t \to -2^{+}} \left\lfloor t \left\lfloor -t \right\rfloor \right\rfloor =\lim_{h \to 0}\left\lfloor (-2+h) \left\lfloor (2-h) \right\rfloor \right\rfloor$$
How can we proceed now since we cannot take limit inside greatest integer function?
| Assume $x=-0.5-\epsilon$ with $\epsilon >0$, then we have that
$$\frac1x=\frac1{-0.5-\epsilon}=-\frac{2}{1+2\epsilon}=-2(1-2\epsilon)=-2+4\epsilon$$
therefore for $\epsilon$ sufficiently small we have
$$\left\lfloor\frac{1}{x} \left\lfloor \frac{-1}{x} \right\rfloor\right\rfloor=\left\lfloor (-2+4\epsilon) \left\lfloor 2-4\epsilon \right\rfloor\right\rfloor=\left\lfloor (-2+4\epsilon) \right\rfloor=-2$$
| {
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"url": "https://math.stackexchange.com/questions/2815938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Linear equation help $$\frac{1-x}{4} + \frac{5x+1}{2} = 3 - \frac{2(x+1)}{8}$$
I got x=1 but the book says x=4/5 and I don't understand how to get to that, I tried working backwards too but I just can't figure it out. Any help would be appreciated, thanks.
Here's my work
using 8 as the lcm
$$
\begin{align}
\frac 81(\frac {(1-x)}4) + \frac 81(\frac {(5x+1)}2) &= \frac 81(3) - \frac 81\frac {2(x+1)}8 \\
2(1-x) + 4(5x+1) &= 24 - 2x + 2 \\
2 - 2x + 20x + 4 &= 24 - 2x + 2 \\
6 + 18x &= 26 -2x \\
20x &= 20 \\
x &= 1 \\
\end{align}
$$
| You have a sign mistake here
$$2(1-x) + 4(5x+1) = 24 - 2x \color{red}{+ 2}$$
It should be
$$2(1-x) + 4(5x+1) = 24 - 2x \color{blue}{- 2}$$
Because
$$-\frac {2(x+1)}8 =-\frac x4-\frac 14$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int_0^{\pi/2} \ln \left(a^2\cos^2\theta +b^2\sin^2\theta\right) \,d\theta$ where $a,b$ are finite natural numbers
Evaluate $$\int_0^{\pi/2} \ln \left(a^2\cos^2\theta +b^2\sin^2\theta\right) \,d\theta$$ where $a,b$ are finite natural numbers
I have spent about a day thinking over this problem. I tried integration by parts, differentiating under integral sign (Feynman's trick, with respect to $a, b$), using some trigonometric and logarithmic properties like changing $\cos^2\theta$ to $\cos2\theta$ and hereafter some logarithmic properties, etc., but failed miserably.
Also tried to use the property that $$\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx$$ in-between, but still to no avail. I also tried to find similar questions on MSE but did not get a related one.
Can someone please help me to solve this integral?
Edit
My try (Feynman's trick) :
Let $$I(a)=\int_0^{\pi/2} \ln \left(a^2\cos^2\theta +b^2\sin^2\theta\right) d\theta$$
Hence $$I'(a) =\frac 1a \int_0^{\pi/2} \frac {2a^2\cos^2\theta d\theta}{a^2\cos^2\theta +b^2\sin^2\theta}$$ $$=\frac 1a\left[ \frac {\pi}{2}+\int_0^{\pi/2} \frac {a^2\cos^2\theta -b^2\sin^2\theta}{a^2\cos^2\theta +b^2\sin^2\theta}\right]$$
Wherein between I broke $2a^2\cos^2\theta=a^2\cos^2\theta +b^2\sin^2\theta+a^2\cos^2\theta -b^2\sin^2\theta$
But now how do I continue further
| Let $I(a,b)=\int_0^{\pi/2} \log(a^2\cos^2(\theta)+b^2\sin^2(\theta))\,d\theta$. Differentiating under the integral with respect to $a^2$ reveals
$$\begin{align}
\frac{\partial I(a,b)}{\partial (a^2)}&=\int_0^{\pi/2}\frac{1}{a^2+b^2\tan^2(\theta)}\,d\theta\\\\
&=\frac{\pi/2}{a(a+b)}\tag1
\end{align}$$
Integrating $(1)$ with respect to $a^2$, we obtain
$$I(a,b)=\pi \log(a+b)+C$$
For $a=b$, $I(a,a)=\pi \log(a)$ from which we find that $C=-\pi\log(2)$.
Putting it all together yields
$$I(a,b)=\pi \log\left(\frac{a+b}2\right)$$
NOTE:
We see from symmetry that $I(a,b)=\pi \log\left(\frac{|a|+|b|}2\right)$ $\forall (a,b)\in \mathbb{R}^2\setminus (0,0)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplify the polynomial $xy(x+y)+(x+y)+(x+y)^2=13xy$ to the form $y^2=4x^3-g_{2}x-g_{3}$
I try to simplify a polynomial to the form: $y^2=4x^3-g_{2}x-g_{3}$, which is the elliptic curves. And the polynomial is $xy(x+y)+(x+y)+(x+y)^2=13xy$.
I try to let the $u=x+y$ and $v=x-y$, then I get the $u^3-uv^2+4u-9u^2+13v^2=0$. But how to get ahead?
| You made a good start but there was a complication you did not anticipate.
The homogeneous version of your equation is:
$\, 0 = -W X Y Z + (X+Y)(X+Z)(Y+Z), \,$ where $\, W=13 \,$ is a constant. Now substitute $\, X = 1 + c_3 x + \sqrt{c_1} y, $
$\, Y = 1 + c_3 x - \sqrt{c_1} y, $ $\, Z = 2 x + c_2, \,$
where $\,c_1,c_2,c_3 \,$ depend on $\,W.\,$ After the substitutions, we eliminate the $\, x y^2 \,$ term with $\, c_3 = W. \,$
We eliminate the $\,x^2\,$ term with
$\, c_2 = (2W^2 +16W + 8)/(W^3 - 4W^2 - 8W). \,$ Now the coefficients of $\,y^2\,$ and of $\,x^3\,$ needs adjustment to get the final form. Let $\, c_1 = (-W^3 + 4W^2 + 8W)/12 .\,$
The equation now is
$\, 0 = 2W(1+W)( -y^2 + 4x^3 - g_2x - g_3), \,$ where
$$\, g_2 = \frac{12(W^4 - 8W^3 + 16W + 16)}{W^2(W^2 - 4W - 8)^2},
\quad g_3 = \frac{8(W^4 - 8W^3 - 8W - 8)}{W^3(W^2 - 4W - 8)^2}. $$
In our case of $\, W=13 \,$ these invariants become
$\, g_2 = 134508/2007889, \, g_3 = 86984/26102557. \,$
There is a $2$-torsion point $\, (-1/13,0). \,$ There is a generator point
$\, (-97/1417, 168 \sqrt{-3}/1417^{3/2}). \,$
After noticing the $1417$ in the denominators, we can scale $\,x,y\,$ to simplify the equation. After scaling it is
$\, 0 = -y^2 + 4x^3 - 134508x - 9481256, \,$ with generator point
$\, (-97, 168 \sqrt{-3}). \,$
This is the elliptic curve with LMFDB label 8190.bw4 which has no rational generator.
| {
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Determining rational functions that simplify to a simple expression
Question: Is there a way to find, if possible, another way to write $(\sqrt[3]2-1)^2$ and $(\sqrt[3]4-1)^2$, in the form of $\frac {a+b}{c+d}$?
What I meant was that, let's take the second expression $(\sqrt[3]4-1)^2$ as an example. The expression is also equal to$$(\sqrt[3]4-1)^2=3\left(\frac {2-\sqrt[3]4}{2+\sqrt[3]4}\right)$$
You can check this with Wolfram Alpha. However, I'm not sure how to do that with $(\sqrt[3]2-1)^2$ and what the work is. I feel like this should be such a basic thing to do, but I'm struggling with where to begin.
| As for $(\sqrt[3]{2}-1)^2$:
$$
\dfrac{4-3\sqrt[3]{2}}{2+\sqrt[3]{2}} = (\sqrt[3]{2}-1)^2. \tag{1}
$$
We can easily check it: denote $x = \sqrt[3]{2}$, then $$(x-1)^2(2+x)=(x^2-2x+1)(2+x)= \\ 2x^2-4x+2+x^3-2x^2+x=\\ x^3-3x+2 = \\ 4-3x.$$
Let's try to find $(x-1)^2$ in the form
$$
\dfrac{a+bx}{c+dx}, \qquad a,b,c,d\in\mathbb{Z}\tag{2}$$ manually; and prove this way that $(1)$ is unique representation for $(x-1)^2$ (where $x=\sqrt[3]{2}$) in the form $(2)$ (of course, if avoid scaling of $a,b,c,d$).
Then
$$
(x-1)^2(c+dx)=a+bx; \\
(x^2-2x+1)(c+dx)=a+bx; \\
dx^3+(c-2d)x^2+(d-2c)x+c=bx+a; \\
(c-2d)x^2+(d-2c)x+(c+2d)=bx+a.
$$
Since $1,\sqrt[3]{2}, \sqrt[3]{4}$ are linearly independent over $\mathbb{Q}$, we have system of equations:
$$
c-2d=0, \\
b=d-2c, \\
a=c+2d.
$$
So, $c=2d$, $b=-3d$, $a=4d$. Setting $d=1$, we'll get $(1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the value of $\frac{4x}{y+1} + \frac{16y}{z+1} + \frac{64z}{x+1}$
Let $x,y,z$ be positive real numbers such that $x+y+z = 1$ and $xy+yz+zx = \frac{1}{3}$. Find the vlaue of $$\frac{4x}{y+1} + \frac{16y}{z+1} + \frac{64z}{x+1}.$$
So far using the fact that $(x+y+z)^2 = x^2+y^2+z^2 + 2(xy+yz+zx)$, I was able to get $x^2+y^2+z^2 = \frac{1}{3}$. Then I tried finding the common denominator of $\frac{4x}{y+1} + \frac{16y}{z+1} + \frac{64z}{x+1}$, which got very messy quickly and I think there's a nicer way to do this problem, but for now I am stuck.
| The point is that $x=y=z=1/3$.
Too see this consider
$$0=(x+y+z)^2-3(xy+xz+yz)=\frac{(x-y)^2+(x-z)^2+(y-z)^2}{2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Functional Equation satisfying f(2x)=f(x) Determine all contimouse functions $f: \mathbb{R} \to \mathbb{R}$ satisfying $f\left(2x\right)=f\left(x\right)$
Attempt
Let $a$ and $b$ be two distinct points on $\mathbb{R}$.
Then consider the sequences of real numbers
$$\{x_n\}=\frac{a}{2},\frac{a}{2^2},\frac{a}{2^3},\frac{a}{2^4} \dots\frac{a}{2^n}$$
$$\{y_n\}=\frac{b}{2},\frac{b}{2^2},\frac{b}{2^3},\frac{b}{2^4} \dots\frac{b}{2^n}$$
We now have
$$f\left(a\right)=f\left(\frac{a}{2}\right)=f\left(\frac{a}{2^2}\right)=f\left(\frac{a}{2^3}\right) \dots f\left(\frac{a}{2^n}\right) \dots= f\left(0\right)$$
Simmilarly
$$f\left(b\right)=f\left(\frac{b}{2}\right)=f\left(\frac{b}{2^2}\right)=f\left(\frac{b}{2^3}\right) \dots f\left(\frac{b}{2^n}\right) \dots= f\left(0\right)$$
Hence $f\left(a\right)=f\left(b\right)=f\left(0\right)$
Hence $f$ is constant function.
Note we use the continuity to establish that $$\lim_{n \to \infty} f\left(x_n\right)=f\left(\lim_{n \to \infty}x\right)=f\left(0\right)$$
| Let $x\ne 0$. for $n\in\mathbb N,$
$$f (x)=f (\frac {x}{2^n} )$$
$f $ is continuous at $0$ and $$\lim_{n\to+\infty}\frac {x}{2^n}=0$$
$$\implies \lim_{n\to+\infty}f (\frac {x}{2^n})=f (0) $$
thus
$$(\forall x\in\mathbb R) \;\;f (x)=f (0) $$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find $y$ in $-\frac12+\frac y2=1$ without combining $-\frac12+\frac y2$? I am trying to find the value of a variable. The problem given was$$
\frac{1}{2}(-1+y) = 1.
$$
I understand that $\dfrac{1}{2}(-1+y) = 1$ is the same as $-\dfrac{1}{2} + \dfrac{y}{2} =1$. How do I get $y$ by itself starting from $-\dfrac{1}{2} + \dfrac{y}{2} =1$? I tried to multiply $-\dfrac{1}{2}$ by $-2$ and then multiply $1$ by $-2$:$$
-\frac{1}{2} × (-2) + \frac{y}{2} = 1 × (-2).
$$
That leaves me with $\dfrac{y}{2} = -2$. Then I multiplied $\dfrac{y}{2}$ by $2$ and also multiplied $-2$ by $2$:
$$
\frac{y}{2} × 2 = (-2) × 2.
$$
That leaves me with $y = -4$.
But this is not correct. The answer should be $3$. What step am I missing or getting wrong?
| As Ethan Bolker pointed out in the comments, without having to divide each summand by $2$ in the expression $$\frac{y-1}{2}=1,$$
one can multiply both sides by two to get $$\require{cancel}\frac{y-1}{\cancel{2}}\cdot\cancel{2}=1\cdot 2\iff y-1=2\iff y=3.$$
Note: I use the symbol $\iff$ to mean “if and only if,” or “this is equivalent to.”
Regarding your work, you forgot to multiply the $y/2$ term in the step when you multiplied both sides by $-2$: the expression $$-\frac{1}{2}\cdot -2 +\color{red}{\frac{y}{2}}=1\cdot -2,$$
should be $$-2\left(-\frac{1}{2}+\frac{y}{2}\right)=1\cdot -2\iff -\frac{1}{2}\cdot -2 +\color{blue}{\frac{y}{2}\cdot -2}=-2\iff 1-y=-2$$
Can you take it from here?
| {
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How to find the finite limit of this function?
Let $f(x) = \dfrac{1-\cos \{x\}}{(x^4 + ax^3 +bx^2 +cx)^2}$. If $l= \lim_{x\to 1^+}f(x), m = \lim_{x\to 2^+}f(x) $ and $n= \lim_{x\to 3^+}f(x),$ where $l,m$ and $n$ non-zero finite then:
$a+b+c=? $
$l+m+n=?$
$\lim_{x\to 0^+}f(x)=? $
where {} denotes the fractional part function.
The trouble I am facing with this question is that for $n^+$ for $n \in \mathbb N$, the numerator is turning out to be $0$ as $\{x\}= 0$ and $\cos 0 =1$ and the deominator is finite.
I even tried the taylor expansions of $\cos \{x\}$ but that didn't help. I don't need the full solution, just want a hint to be able to proceed.
| Try breaking the fraction part function into different intervals, like $(1,2)$ etc. For $x\in (1,2) $ we have $\{x\} = x-1$ and so
$$\begin{align}
l=\lim_{x\to1^+} \frac{1-\cos\{x\}}{(x^4+ax^3+bx^2+cx)^2} &= \lim_{x\to1^+}\frac{2\sin^{2}(\frac{x-1}{2})}{(\frac{x-1}{2})^2}\cdot \frac{(\frac{x-1}{2})^2}{(x^4+ax^3+bx^2+cx)^2}\\
&= \lim_{x\to1^+}\frac{(x-1)^2}{2(x^4+ax^3+bx^2+cx)^2}
\end{align}$$
For the limit to be nonzero finite, we need $0/0$ form, so we need $(x=1)$ to be root of denominator quartic. Similar analysis for limits $m,n$ gives us three roots of quartic on denominator, which are $x=1,2,3$.
The fourth root is obviously $x=0$. Call the denominator quartic as $P(x)=x^4+ax^3+bx^2+cx$. We have shown $P(x) = x(x-1)(x-2)(x-3)$. So now finding limits $l,m,n$ is simple:
*
*$l= \frac{1}{2 \cdot 1\cdot 4\cdot 1} = 1/8$
*$m= \frac{1}{2 \cdot 1\cdot 1\cdot 4} = 1/8$
*$n= \frac{1}{2 \cdot 1\cdot 4\cdot 9} = 1/72$
From here we get $l+m+n = \frac{19}{72}$
For finding $a,b,c$, we have $P(x)=x^4+ax^3+bx^2+cx = x(x-1)(x-2)(x-3)$ so $P(1) = a+b+c+1=0$ or $a+b+c=-1$.
| {
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The ring $\mathbb{Z}[\sqrt{2}]=\{a+b\sqrt{2}:a,b\in\mathbb{Z}\}$ is a UFD. Contradiction? Considering the ring $\mathbb{Z}[\sqrt{2}]=\{a+b\sqrt{2}:a,b\in\mathbb{Z}\}$, I know that this ring is an euclidian domain and therefore a unique factorization domain.
Now, $23=(3+4\sqrt{2})(3-4\sqrt{2})=(11+7\sqrt{2})(11-7\sqrt{2})$.
$(3+4\sqrt{2}),(3-4\sqrt{2}),(11+7\sqrt{2}),(11-7\sqrt{2})$ are irreducibles in $\mathbb{Z}[\sqrt{2}]$, since for example: $$N(3+4\sqrt{2})=(3+4\sqrt{2})(3-4\sqrt{2})=-23$$Where $N$ is the norm $N(a+b\sqrt{d})=(a+b\sqrt{d})(a-b\sqrt{d})$. Since $23$ is prime, $(3+4\sqrt{2})$ is irreducible.
Therefore $23$ is not expressed by a unique product of irreducibles, doesn't this contradicts the fact that $\mathbb{Z}[\sqrt{2}]$ is UFD?
| There is no contradiction because $3+4\sqrt{2}$ and $11+7\sqrt{2}$ are associates:
$$
\frac{11+7\sqrt{2}}{3+4\sqrt{2}} = 1+\sqrt{2}
$$
and $1+\sqrt{2}$ is a unit, since it has norm $-1$. Or note that
$$
\frac{3+4\sqrt{2}}{11+7\sqrt{2}} = -1+\sqrt{2}
$$
| {
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Want to Confirm Answer for the Sum of Fractions with Triangular Number Sequence as the Denominators I am trying to compute the sum of these fractions:
$$\frac{3}{1}+\frac{3}{1+2}+\frac{3}{1+2+3} + \dots + \frac{3}{1+2+3+\dots+100}.$$
I believe the denominators are a triangular number sequence, therefore the expression should convert to
$$3\Bigl(\frac{2}{1×2}+\frac{2}{2×3}+\frac{2}{3×4} + \dots +\frac{2}{100×101}\Bigr)$$
then, this can then be converted to
$$6\biggl[\Bigl( \frac{1}{1}-\frac{1}{2}\Bigr) + \Bigl(\frac{1}{2}-\frac{1}{3}\Bigr) + \Bigl(\frac{1}{3}-\frac{1}{4}\Bigr) + \dots + \Bigl(\frac{1}{100}-\frac{1}{101}\Bigr)\biggr]$$
then, this simplifies to $6\Bigl(\frac{1}{1}$ - $\frac{1}{101}\Bigr)$
with the final answer being $\mathbf{6\frac{100}{101}}$
However, the answer key we obtained specifies the answer as $5\frac{95}{101}$
I would like to have some help on finding out if I made any errors with my solution hence our answer not being the same as the $5\frac{95}{101}$ given by the answer key? Thank you in advance.
| Call the $n$th Triangular number $T_n$. Then, noting $T_{100}=5050$, the reciprocal sum of the first $100$ triangular numbers is:
\begin{align*}
\frac{1}{1}&+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\dotsb+\frac{1}{5050}\\
&=2\left[\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\dotsb+\frac{1}{10100}\right]
\end{align*}
The right hand side transforms as
\begin{align*}
2&\left[\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+\dotsb+ \left(\frac{1}{100}-\frac{1}{101}\right)\right]\\
&=2\left(\frac{1}{1}-\frac{1}{101}\right)
\end{align*}
Now multiply by $3$ to get your desired answer: $6\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{600}{101}$
| {
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Minimizing $c(x-2)(y-2)(z-2)$ subject to $xyz =V$ Given constants $c > 0$ and $V > 0$, how can I solve the following optimization problem?
$$\begin{array}{ll}
\text{minimize} & c \, (x-2) (y-2) (z-2)\\
\text{subject to} & xyz = V\\
& x,y,z > 2
\end{array}$$
My attempt
Since $V=xyz$, we get $z = \frac{V}{xy}$
Let $S = c(x-2)(y-2)(z-2) = c(x-2)(y-2)(\frac{V}{xy}-2)$
Simplifying we get,
$$S = c(4x+4y-2xy-2\frac{V}{x}-2\frac{V}{y}+4\frac{V}{xy}+V-8)$$
Partially differentiate $S$ w.r.t. $x$ and $y$ and equate to $0$.
$\frac{\partial S}{\partial x} = c (4 -2y + \frac{2V}{x^{2}} - 4\frac{V}{x^{2}y}) = 0$ -------- (A)
$\frac{\partial S}{\partial y} = c (4 -2x + \frac{2V}{y^{2}} - 4\frac{V}{y^{2}x}) = 0$ -------- (B)
If we subtract (B) from (A), then one solution is $x=y$.
Is it correct till this point ?
| Let$f(x,y,z):=c (x - 2) (y - 2) (z - 2)$, and $g(x,y,z)=x y z - V$. Then solve for $x,y,z,\lambda$ in:
$$\frac{\partial f(x,y,z)}{\partial x}=\lambda \frac{\partial g(x,y,z)}{\partial x}\\
\frac{\partial f(x,y,z)}{\partial y}=\lambda \frac{\partial g(x,y,z)}{\partial y}\\\frac{\partial f(x,y,z)}{\partial z}=\lambda \frac{\partial g(x,y,z)}{\partial z}\\g(x,y,z)=0$$
The complete solution set should be:
$$\left\{x\to 2,y\to 2,z\to \frac{V}{4},\lambda \to 0\right\},\left\{x\to 2,y\to \frac{V}{4},z\to 2,\lambda \to 0\right\},\color{red}{\left\{x\to \sqrt[3]{V},y\to \sqrt[3]{V},z\to \sqrt[3]{V},\lambda \to \frac{c \left(\sqrt[3]{V}-2\right)^2}{V^{2/3}}\right\}}\\
\left\{x\to -\sqrt[3]{-1} \sqrt[3]{V},y\to -\sqrt[3]{-1} \sqrt[3]{V},z\to -\sqrt[3]{-1} \sqrt[3]{V},\lambda \to c \left(1-\frac{4 \left((-1)^{2/3} \sqrt[3]{V}+\sqrt[3]{-1}\right)}{V^{2/3}}\right)\right\},\left\{x\to (-1)^{2/3} \sqrt[3]{V},y\to (-1)^{2/3} \sqrt[3]{V},z\to (-1)^{2/3} \sqrt[3]{V},\lambda \to \frac{c \left(\sqrt[3]{V}+2 \sqrt[3]{-1}\right)^2}{V^{2/3}}\right\},\left\{x\to \frac{V}{4},y\to 2,z\to 2,\lambda \to 0\right\}$$
Since the $x,y,z>2$ and $V,c>0$, then only the answer in red is meaningful.
Following your approach for $z=\frac V{x y}$,for $h(x)=c (x-2) (y-2) \left(\frac{V}{x y}-2\right)$:
$$\frac{\partial h(x,y)}{\partial x}=0\\
\frac{\partial h(x,y)}{\partial y}=0
$$
Your complete solution set should be:
$$\left\{\left\{x\to 2,y\to \frac{V}{4}\right\},\color{red}{\left\{x\to \sqrt[3]{V},y\to \sqrt[3]{V}\right\}},\left\{x\to -\sqrt[3]{-1} \sqrt[3]{V},y\to -\sqrt[3]{-1} \sqrt[3]{V}\right\},\left\{x\to (-1)^{2/3} \sqrt[3]{V},y\to (-1)^{2/3} \sqrt[3]{V}\right\},\{x\to 2,y\to 2\},\left\{x\to \frac{V}{4},y\to 2\right\}\right\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2833562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Rationalize this I'm having trouble rationalizing the denominator of this fraction. Would you kindly explain this to a fellow self-learning math student?
$$\frac{10}{\sqrt[4]{3}-1}$$ knowing that $a^n-b^n=\left(a-b\right)\left(a^{n-1}+a^{n-2}\cdot b+a^{n-3}\cdot b^2+...+a\cdot b^{n-2}+b^{n-1}\right)$
$a,b\in \mathbb{R}$ and $n\in \mathbb{N}$
| $$\dfrac{10}{\sqrt[4]{3}-1} = \dfrac{5\cdot (\sqrt[4]{3^4}-1^4)}{\sqrt[4]{3}-1} = 5\left(\sqrt[4]{3^3}\cdot 1^0+\sqrt[4]{3^2}\cdot 1^1+\sqrt[4]{3^1}\cdot 1^2+\sqrt[4]{3^0}\cdot 1^3\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2833654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Cube root of numbers such as $2+11i$ How can I find powers and roots of complex numbers with ugly argument such as cube root of 2+11i? I saw Find the solutions to $z^3 = 2 + 11i$., but the answer includes some guessing and I would like to have some algorithm for such a tasks, which I can program. I would also like to solve it symbolically and thus precisely. Of course I know the formula: $a^b=e^{b \ln(a)}$ But when I try to use it:
$\sqrt[3]{2+11i}=(2+11i)^\frac{1}{3}=e^{\frac{1}{3} \ln(2+11i)}=e^{\frac{1}{3} \big(\ln(\sqrt{125})+i \arctan(\frac{11}{2}) \big)}=$
$=\sqrt{5} \big(\cos(\frac{1}{3} \arctan(\frac{11}{2}))+i \sin (\frac{1}{3} \arctan(\frac{11}{2})) \big)$
And I cannot solve this further symbolically, of course when I solve it in floating point numbers I get the right solution $2+i$.
| Note that $\tan 3\alpha=\dfrac {t(3-t^2)}{1-3t^2}$, where $t=\tan\alpha$.
Putting $\theta=3\alpha$ where $\tan\theta=\tan3\alpha=\dfrac {11}2$, we have
$$\begin{align}
\frac {11}2&=\frac {t(3-t^2)}{1-3t^2}\\
11(1-3t^2)&=2t(3-t^2)\\
2t^3-33t^2-6t+11&=0\end{align}$$
By inspection, putting $t=\frac 12$ gives $\text{LHS}=0=\text {RHS}$, hence $t=\tan\alpha=\tan\frac \theta 3=\frac 12$ is a solution.
$$\large\begin{align}
\left[re^{i(2n\pi+\theta)}\right]^{\frac 13}&=r^{\frac 13}e^{i\left(\frac {2n\pi+\theta}3\right)}\\&=\big(\sqrt{125}\big)^\frac 13e^{i\left(\frac {2n\pi}3+\frac\theta 3\right)}
&&\scriptsize{\theta=\arctan(\frac {11}2)}\\
&=\sqrt{5}e^{i\left(\frac {2n\pi}3+\frac\theta 3\right)}\\
&=\sqrt{5}e^{i\left(\frac {2n\pi}3+\arctan\frac 12\right)}\end{align}$$
Hence the principal solution ($n=0)$ is
$$\begin{align}
\large\sqrt{5}e^{i\cdot\arctan\frac 12}
&=\sqrt{5} \left(\cos (\arctan \left(\tfrac 12\right)+i\sin(\arctan\left(\tfrac 12\right)\right)\\
&=\sqrt{5}\left(\frac 2{\sqrt5}+i\frac 1{\sqrt{5}}\right)\\
&=2+i\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2836570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Eliminating $\theta$ from $a\cos(\theta-\alpha)=x$ and $b\cos(\theta-\beta)=y$ Eliminate $\theta$ in following equations
$$\begin{align}
a \cos(\theta-\alpha) &= x \\
b \cos(\theta- \beta) &=y
\end{align}$$
I am trying to solve this problem but still I am unable to get the perfect answer
I added both the equations but it transformed it to
$2 \cos(\theta+(\alpha + \beta)/2)$
| For simplicity, define $u:=x/a$ and $v:=y/b$, so that we have
$$\begin{align}
u &= \cos(\theta-\alpha) = \cos\theta \cos\alpha + \sin\theta\sin\alpha \\
v &= \cos(\theta-\beta) = \cos\theta\cos\beta + \sin\theta\sin\beta
\end{align}$$
This is a linear system in $\cos\theta$ and $\sin\theta$. Solving, we obtain
$$\cos\theta = \frac{v\sin\alpha - u \sin\beta}{
\sin\alpha \cos\beta - \cos\alpha\sin\beta} = \frac{v\sin\alpha-u\sin\beta}{\sin(\alpha-\beta)} \qquad\qquad \sin\theta = \frac{u \cos\beta - v \cos\alpha}{\sin(\alpha-\beta)}$$
Then, because $\cos^2\theta + \sin^2\theta = 1$, we can write
$$\frac{\left(v\sin\alpha-u\sin\beta\right)^2}{\sin^2(\alpha-\beta)} + \frac{\left(u\cos\beta-v\cos\alpha\right)^2}{\sin^2(\alpha-\beta)} = 1$$
so that
$$u^2 + v^2 - 2 u v (\sin\alpha\sin\beta+\cos\alpha\cos\beta) = \sin^2(\alpha-\beta)$$
which becomes
$$u^2 + v^2 - 2 u v \cos(\alpha-\beta) = \sin^2(\alpha-\beta)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2838356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding the determinant of a $2\times 2$ matrix Let $R$ be a commutative ring with 1. Show with the Leibniz signature formula that $\det\biggl(\begin{bmatrix}
\lambda & A_{12}\\
0 & A_{22}\\
\end{bmatrix}\biggl) = \lambda \cdot \det(A_{22})$ applies whereby $\lambda \in R \land A_{22} \in R^{n,n}$.
My idea:
$\sigma_{1} = \operatorname{id} =
\begin{pmatrix}
\lambda & A_{12}\\
0 & A_{22}\\
\end{pmatrix}
$ and $\sigma_{2} = \begin{pmatrix}
0 & A_{22}\\
\lambda & A_{12}\\
\end{pmatrix}
$
I know that if the number of transpositions are equal then I will get a positive mathematical operator otherwise I will get a negative one. But I think this approach is only useful if I have actual numbers in the matrix. I got the following out of this:
$\det A = \lambda \cdot A_{22} - 0 \cdot A_{12}$
Just to make clear what I mean here is an example where it works:
$A = \begin{pmatrix}
1 & 3&4\\
2&1&7\\
6&7&8
\end{pmatrix}
$ $\sigma_{1} = \operatorname{id} =
\begin{pmatrix}
1 & 3&4\\
2&1&7\\
6&7&8
\end{pmatrix}
$ $\Rightarrow$ $\sigma_{2} =
\begin{pmatrix}
2 & 1&7\\
1&3&4\\
6&7&8
\end{pmatrix}
$ $\Rightarrow$ $\sigma_{3} =
\begin{pmatrix}
1 & 3&4\\
6&7&8\\
2&1&7
\end{pmatrix}
$
$\Rightarrow$ $\sigma_{4} =
\begin{pmatrix}
6 & 7&8\\
2&1&7\\
1&3&4
\end{pmatrix}
$ $\Rightarrow$ $\sigma_{5} =
\begin{pmatrix}
6 & 7&8\\
1&3&4\\
2&1&7
\end{pmatrix}
$ $\Rightarrow$ $\sigma_{6} =
\begin{pmatrix}
2 & 1&7\\
6&7&8\\
1&3&4
\end{pmatrix}
$
so that $\det A = 8 -48 -49 -24 +126 +56 = 69$.
Additional information:
Leibniz signature formula:
$$\det: R^{n,n} \to R, A = [a_{ij}] \mapsto \det(A):= \sum_{\sigma \in S_{n}}sgn(\sigma) \prod_{i=1}^n a_{i},_{\sigma(i)}$$
How do I show now that $\det\biggl(\begin{bmatrix}
\lambda & A_{12}\\
0 & A_{22}\\
\end{bmatrix}\biggl) = \lambda \cdot \det(A_{22})$?
Could someone show me how to solve this proof in a proper way?
| While you are right to be concerned that it's a little more tricky if $A_{22}$ is a larger matrix, since $A$ is only a $2\times 2$ matrix, $A_{22}$ is just a number and your work is correct!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2838630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Remainder on division with $22$ What is the remainder obtained when $14^{16}$ is divided with $22$?
Is there a general method for this, without using number theory? I wish to solve this question using binomial theorem only - maybe expressing the numerator as a summation in which most terms are divisible by $22$, except the remainder?
How should I proceed?
| Using the Euclidean Algorithm
Note that
$$
\begin{align}
14^{16}&\equiv0&\pmod2\tag1
\end{align}
$$
Reducing mod $11$ and using Fermat's Little Theorem, we get
$$
\begin{align}
14^{16}
&\equiv3^6&\pmod{11}\\
&\equiv3&\pmod{11}\tag2
\end{align}
$$
Solving $11a+2b=1$ using the Euclidean Algorithm as implemented in this answer,
$$
\begin{array}{r}
&&5&2\\\hline
1&0&1&-2\\
0&1&-5&11\\
11&2&1&0\\
\end{array}\tag3
$$
we have
$$
11(1)+2(-5)=1\tag4
$$
from which we get
$$
\begin{align}
-10&\equiv0&\pmod2\\
-10&\equiv1&\pmod{11}
\end{align}\tag5
$$
Multiplying $(5)$ by $3$ gives
$$
\begin{align}
-30&\equiv0&\pmod2\\
-30&\equiv3&\pmod{11}
\end{align}\tag6
$$
Using the Chinese Remainder Theorem
The Chinese Remainder Theorem says that the solution to $(6)$ is unique mod ${22}$. Thus, we get the smallest positive solution to be
$$
\begin{align}
14&\equiv0&\pmod2\\
14&\equiv3&\pmod{11}
\end{align}\tag7
$$
Thus, $(1)$, $(2)$, and $(7)$ yields
$$
\begin{align}
14^{16}&\equiv14&\pmod{22}\tag8
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2839477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
Minimum $e$ where $a,b,c,d,e$ are reals such that $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$ I have a question about this 1978 USAMO problem:
Given that $a,b,c,d,e$ are real numbers such that $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$, find the maximum value that $e$ can attain.
I had the following solution:
Let $a+b+c+d=x$. Then $x+e=8\implies e=8-x$. Also, $\frac{a^2+1}{2}+\frac{b^2+1}{2}+\frac{c^2+1}{2}+\frac{d^2+1}{2}\geq (a+b+c+d)=x$ by AM-GM inequality. Hence, $a^2+b^2+c^2+d^2\geq 2x-4$.
Now we have $2x-4+e^2\leq a^2+b^2+c^2+d^2+e^2=16$. Substituting $x=8-e$, we get $e^2-2e-4\leq 0$. We can easily calculate that the lowest value that $e$ can attain is $1 -\sqrt{5}$.
However, the answer given on the internet is $\frac{16}{5}$. Where am I going wrong?
EDIT $1$ -- Is this a case of how the value $1-\sqrt{5}$ can never be attained by $e$, although the inequality is true?
EDIT $2$ -- It seems that we need to find the maximum. By my method, I’ve found the maximum to be $1+\sqrt{5}$. This is greater than $\frac{16}{5}$. Have I found a sharper inequality?
| Given $a+b+c+d+e=8$ and $a+b+c+d=8-e$
$$(8-e)^2=(a+b+c+d)^2$$
Now expand $(a+b+c+d)^2$
$$=a^2+b^2+c^2+d^2+2(ab+ac+ad+bc+bd+cd)$$
$$\le4(a^2+b^2+c^2+d^2)$$
And Now, $a^2+b^2+c^2+d^2+e^2=16$
$a^2+b^2+c^2+d^2=16-e^2$ $$(16-e^2)=a^2+b^2+c^2+d^2$$$$4(a^2+b^2+c^2+d^2)\ge(a+b+c+d)^2$$$$4(16-e^2)\ge(8-e)^2$$$$64-4e^2\ge64-16e^2+e^2$$$$5e^2-16e\le0$$$$e(5e-16)\le0$$$$e\le\dfrac{16}{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
prove that the polynomial $x^8 -x^5 +x^2 -x +1$ is positive for all real values of x is there any factorization possible of the above expression or can it be shown that is a a sum of two or three squares?
i tried various factorizations but none of them were conclusive.
| $$x^8-x^5+x^2-x+1 = \frac{1}{2}\bigg[2x^8-2x^5+2x^2-2x+2\bigg]$$
So $$ = \frac{1}{2}\bigg[x^8+(x^8-2x^5+x^2)+(x^2-2x+1)+1\bigg]$$
So $$ = \frac{1}{2}\bigg[x^8+(x^4-x)^2+(x-1)^2+1\bigg]>0\forall x \in \mathbb{R}.$$
Added:: For $x=0,x^8-x^5+x^2-x+1>0$
Using Arithmetic Geometric Inequality $(x\neq 0)$
$$\frac{x^8}{2}+\frac{x^2}{2}\geq |x^5|\geq x^5$$
$$\frac{x^2}{2}+\frac{1}{2}\geq |x|\geq x$$
So $$\frac{x^8}{2}+\frac{x^2}{2}-x^5+\frac{x^2}{2}-x+\frac{1}{2}+\frac{1}{2}>0\forall x \in \mathbb{R}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding the inverse of $A$
Find the inverse of $$A =\left[\begin{matrix}0 & 1 & 0 & 0\\ 0& 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ a & b & c & d\end{matrix}\right]$$
My attempts:
$$A^{-1} = \frac {\operatorname{adj}A}{\det A}$$
As I can find the $\det A$ that is $\det A = -b$. Here, how can I find the inverse? Is there any easy method/ or easy procedure to find the inverse of $A$?
| You're wrong: expanding by the first column, you obtain at once $\;\det A=-a$. We'll suppose $a\ne0$.
To find the inverse you can perform column reduction:
\begin{align}
\begin{bmatrix}
\begin{array}{cccc|cccc}
0&1&0&0&1&0&0&0 \\
0&0&1&0 &0&1&0&0\\
0&0&0&1&0&0&1&0\\
a&b&c&d&0&0&0&1
\end{array}
\end{bmatrix}&\xrightarrow[\scriptstyle\to\,\text{(c$_2$,c$_3$,c$_4$, c$_1$)}]{\text{(c$_1$,c$_2$,c$_3$,c$_4$)}}
\begin{bmatrix}
\begin{array}{cccc|cccc}
1&0&0&0&0&0&0&1 \\
0&1&0&0 &1&0&0&0\\
0&0&1&0&0&1&0&0\\
b&c&d&a&0&0&1&0
\end{array}
\end{bmatrix}\xrightarrow{\scriptstyle\tfrac1a\text{c}_4\,\to\,\text{c}_4}\\[1ex]
\begin{bmatrix}
\begin{array}{cccc|cccc}
1&0&0&0&0&0&0&\frac1a \\
0&1&0&0 &1&0&0&0\\
0&0&1&0&0&1&0&0\\
b&c&d&1&0&0&1&0
\end{array}
\end{bmatrix}&\xrightarrow[\scriptstyle\text{(c}_2-c\,\text{c}_4,\text{c}_3-d\,\text{c}_4)]{(\text{c}_1,\text{c}_2,\text{c}_3)\,\to\,(\text{c}_1-b\,\text{c}_4,\:}
\begin{bmatrix}
\begin{array}{cccc|cccc}
\!\!1&0 &0&0&-\frac ba&-\frac ca& -\frac da&\frac1a \\
0&1&0&0 &1&0&0&0\\
0&0&1&0&0&1&0&0\\
0&0&0&1&0&0&1&0
\end{array}
\end{bmatrix}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2845376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Series Convergence of Harmonic Means
Let $\{x_n\}$ be a sequence of real numbers such that $0< x_1 <x_2$. If $$x_n= \frac{2}{\frac{1}{x_{n-1}}+\frac{1}{x_{n-2}}}, $$then show that
$$\lim_{n\to\infty}x_n=\frac{3x_1x_2}{2x_1+x_2}.$$
| Note that
\begin{align}
& x_n = \frac{2}{\frac{1}{x_{n-1}} + \frac{1}{x_{n-2}}} \\
\iff & \frac{1}{x_{n-1}} + \frac{1}{x_{n-2}} = \frac{2}{x_{n}} \\
\iff & a_n = \frac{1}{2}a_{n-1} + \frac{1}{2}a_{n-2},
\end{align}
where $a_i = \frac{1}{x_i}$.
It's easy to verify that $$a_n = \alpha + \beta\left(-\frac{1}{2}\right)^n.$$
Consequently, $$\lim_{n\to \infty}a_n = \alpha.$$
Using $$a_1 = \frac{1}{x_1} = \alpha-\frac{\beta}{2},$$
and $$a_2 = \frac{1}{x_2} = \alpha+\frac{\beta}{4},$$
one gets $$\lim_{n\to\infty}x_n = \frac{1}{\alpha} = \frac{3x_1 x_2}{2x_1 + x_2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2847198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Asymptotes of an implicit curve Per the method described at How to find asymptotes of implicit function? , I proceeded to find the asymptotes of
$$
x^3 + 3x^2y - 4y^3 - x + y + 3 = 0
$$
Whilst, it correctly generates the asymptote $$ y=x $$ , the remaining asymptote(s):-
$$ 2y + x = ±1 $$ can't be deduced.
Instead another erroneous line $ y = 0 $ is outputted.
Some help will be appreciated.Alternate methods of solution would work too:)
| Giving
$$
f(x,y) = x^3 + 3 x^2 y - 4 y^3 - x + y - 3 = 0
$$
the asymptotic directions can be explored by substituting $ y = a x + b\;\; $ into $f(x,y)$ giving
$$
f(x,ax+b) = 3 x^2 (a x+b)-4 (a x+b)^3+a x+b+x^3-x-3 = (1+3a-4a^3)x^3+3b(1-4a^2)x^2 +(a(1-12b^2)-1)x+b(1-4b^2)-3
$$
Now the conditions for $f(x,ax+b)$ to have line behavior are
$$
\left\{
\begin{array}{rcl}
3 b-12 a^2 b=0 \\
-4 a^3+3 a+1=0 \\
\end{array}
\right.
$$
and solving we find two solutions:
$$
\{a =-\frac 12, \forall b\} \cup \{a = 1, b = 0\}
$$
so we have
$$
L_1\to y = x\\
L_2\to y = -\frac x2 + b
$$
Considering $f(x,y)$ at their asymptotic values it's value should be the same or
$$
\lim_{x\to 0}f(x,-\frac 12 x+b)=\lim_{x\to 0}f(x,x) = -3
$$
so this can be solved by choosing $b$ such that $1-4b^2 = 0\;\;$ giving
$$
L_1\to y = x\\
L_2\to y = -\frac x2 \pm\frac 12
$$
Attached a plot showing in blue $f(x,y) = 0$ and in red $L_1, L_2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2849494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
} |
Reference request: Transformations under which the discriminant is invariant I had an occasion to think about this quadratic equation:
$$
ax^2 +bx(1-x) + c(1-x)^2 = 0.
$$
Its solution is
$$
x = \frac{2c-b\pm\sqrt{b^2-4ac\,}}{2(a-b+c)}.
$$
The thing under the radical is the same thing we were all taught in eighth or ninth grade. The mapping $(a,b,c)\mapsto (a-b+c, b-2c, c)$ doesn't change it. Is this an instance of the conclusion of some well known result?
| Let $P(x)=ax^2+bx(1-x)+c(1-x)^2$. Then
\begin{align}
P(x)&=ax^2+bx(1-x)+c(1-x)^2\\
&=x^2\big [a+b(\frac{1}{x}-1)+c(\frac{1}{x}-1)^2\big ]
\end{align}
Hence, $P(x)$ is obtained from $a+bx+cx^2$ by a projective transformation. Also,
\begin{align}
a+bx+cx^2 &= x^2\big [a\frac{1}{x^2}+b\frac{1}{x}+c\big ]\\
\end{align}
and so $a+bx+cx^2$ can be obtained from $Q(x)=ax^2+bx+c$ by a projective transformation. It's well known that the discriminant is invariant under these projective transformations and so we have that the discriminant of $P(x)$ is equal to the discriminant of $Q(x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2849891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
What is the probability that both tubes are good?
This a conditional probability where we calculate:
Probability that both are good knowing that one of them is good= prob that both are good/ probability that one of them is good
Which gives me: (1/3) / (6/10)=5/9
But the correct answer is: 5/13
How should I proceed?
| You presumably calculated the probability both are good as $$\frac{6}{10}\times\frac{5}{9}=\frac{1}{3}$$
Similarly the probability both are bad is $$\frac{4}{10}\times\frac{3}{9}=\frac{2}{15}$$
And so the probability at least one is good $$1-\frac{2}{15} = \frac{13}{15}$$
which you could alternatively calculate as $\frac{6}{10}+\frac{4}{10}\times\frac{6}{9}$ (since if the first is bad then the second might be good) or as $\frac{6}{10}\times\frac{4}{9}+\frac{4}{10}\times\frac{6}{9}+\frac{6}{10}\times\frac{5}{9}$ (looking at all the possibilities for both: good & bad, or bad & good, or good & good)
so the answer to the question is $$\frac{\frac13}{\,\frac{13}{15}\,}=\frac{5}{13}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2852031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the value of this product: $\prod_{n=1}^\infty \;\frac{3}{1+2 \cos(\frac{\pi}{3^n})} \;=\;? $ emphasized text$$\left[\, \prod_{n=1}^\infty \;\frac{3}{1+2 \cos\left(\frac{\pi}{3^n}\right)}\, \right]\, =\;? $$
Where $\;[\, .]\;$ denotes the integral part function.
$\mathbf {My Attempt}$
I tried to confine the n-th term to get
$$\frac12 \le \cos\left(\frac{\pi}{3^n}\right) \lt 1$$
$$ 1 \lt \;\frac{3}{1+2 \cos\left(\frac{\pi}{3^n}\right)}\; \le \frac32 $$
But this didn't help as the product now is bounded below but unbounded above $\rightarrow \infty$
Using Wolfram Alpha, the product approaches 1.5708. So, the floor is 1.
I tried also to bound product from above using
$$\left(\frac{\pi}{3^n}\right) \lt \left(\frac{\pi}{2^n}\right)$$
$$\cos\left(\frac{\pi}{3^n}\right) \gt \cos\left(\frac{\pi}{2^n}\right) \quad(\cos x \text { is decreasing on } ]0, \frac{\pi}{3}]
)$$
And use the Telescoping product
$$\cos\left(\frac{\pi}{2^n}\right)=\frac{\sin\left(\frac{\pi}{2^{n-1}}\right)}{2\sin\left(\frac{\pi}{2^n}\right)}$$
But this doesn't help much.
Any hint?
| $$1+2\cos2x=1+2(1-2\sin^2x)=\dfrac{\sin3x}{\sin x}=\dfrac{f(n+1)}{f(n)}$$
where $f(m)=\sin(3^mx)$
Here $3^mx=?$
Related:$\cos x(2\cos2x-1)=\cos3x$
| {
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"url": "https://math.stackexchange.com/questions/2852712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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An inequality with substitution If $a,b,c$ positive real numbers, then I have to prove $ \frac {1}{18} \sum\limits_{cycl}^{} \frac{a^2}{b^2} + \sum\limits_{cycl}^{} \frac {a}{2a+b+c} \ge \frac {11}{12}$
We have that $\frac {1}{18} \sum\limits_{cycl}^{} \frac{a^2}{b^2} \ge \frac {3}{18} = \frac {1}{6}$
If we assume that $x=2a+b+c,y=a+2b+c,z=a+b+2c,$ then $a=\frac{3x-y-z}{4},b=\frac{3y-x-z}{4},c=\frac{3z-x-y}{4},$ and
$\sum\limits_{cycl}^{} \frac {a}{2a+b+c} = \frac{1}{4} [9 - \sum\limits_{cycl}^{} (\frac {x}{y}+\frac {y}{x})]$
Since
$\sum\limits_{cycl}^{} (\frac {x}{y}+\frac {y}{x}) \ge 6$
so we have
$ \frac{3}{4} \ge \sum\limits_{cycl}^{} \frac {a}{2a+b+c} = \frac{1}{4} [9 - \sum\limits_{cycl}^{} (\frac {x}{y}+\frac {y}{x})]$
and I stuck there. Thank you
| Let $c=\max\{a,b,c\}$.
Since by C-S $$\sum_{cyc}\frac{a}{2a+b+c}=\sum_{cyc}\frac{a^2}{2a^2+ab+ac}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(2a^2+ab+ac)}=\frac{(a+b+c)^2}{2\sum\limits_{cyc}(a^2+ab)},$$
it's enough to prove that
$$\frac{1}{18}\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)+\frac{(a+b+c)^2}{2\sum\limits_{cyc}(a^2+ab)}\geq\frac{11}{12}$$ or
$$\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}-3\geq\frac{27}{2}-\frac{9(a+b+c)^2}{\sum\limits_{cyc}(a^2+ab)}$$ or
$$\frac{a^2}{b^2}+\frac{b^2}{a^2}-2+\frac{b^2}{c^2}+\frac{c^2}{a^2}-\frac{b^2}{a^2}-1\geq\frac{9\sum\limits_{cyc}(a^2-ab)}{2\sum\limits_{cyc}(a^2+ab)}$$ or
$$\frac{(a^2-b^2)^2}{a^2b^2}+\frac{(c^2-a^2)(c^2-b^2)}{a^2c^2}\geq\frac{9((a-b)^2+(c-a)(c-b))}{2\sum\limits_{cyc}(a^2+ab)}.$$
Id est, it's enough to prove that
$$\frac{(a+b)^2}{a^2b^2}\geq\frac{9}{2\sum\limits_{cyc}(a^2+ab)}$$ and
$$\frac{(a+c)(b+c)}{a^2c^2}\geq\frac{9}{2\sum\limits_{cyc}(a^2+ab)}.$$
Both these inequalities we can prove by AM-GM.
Indeed, $$2(a+b)^2\sum_{cyc}(a^2+ab)\geq2(a+b)^2(a^2+ab+b^2)\geq2\cdot4\cdot3a^2b^2>9a^2b^2 $$ and
$$2(a+c)(b+c)\sum_{cyc}(a^2+ab)\geq2(a+c)c(a^2+ac+c^2)\geq2c(a+a)\cdot3ac=12a^2c^2>9a^2c^2.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $\frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}+\frac{1}{(x+4)(x+5)}=0.8$ Solve:
$$\frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}+\frac{1}{(x+4)(x+5)}=0.8$$
This is taken from one of the TAU entry tests (I have one in 2 weeks :) )
So, I don't really recognize anything speical here except that there's some similiraty between some pairs of denominators, my intuation is telling me to just multipy everything and try to solve but it looks like a big dead polynomial.
Is there an elegent way to solve this?
Solution:
$$\frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}+\frac{1}{(x+4)(x+5)}=0.8$$
$$\frac{1}{x+1}-\frac{1}{x+5}=0.8$$
$$4=0.8(x+1)(x+5)$$
$$4x^2+24x=0$$
$$4x(x+6)=0$$
Solution: $x_1=0$ and $x_2=-6$
Definately very elegent! :)
| Hint: your equation is equivalent to $$\frac{4x(6+x)}{5(1+x)(5+x)}=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How do I prove the following inequality? How do I proceed to solve the inequality $$\frac{(a^2+b^2)}{(a+b)} + \frac {(b^2+c^2)}{(b+c)} + \frac{(a^2+c^2)}{(a+c)} \geq (a+b+c)$$ where $a , b , c > 0$
I have thought of taking the terms on the $LHS$ and converting them to $AM$ and then use the $AM-GM$ theorem , but I cant figure out how to convert $\frac {(a^2+b^2)}{(a+b)}$ to $AM$.I have tried finding $a_1$ and $a_2$ by doing
$ (\frac{(a_1+a_2)}{2}\bigr)^2 \geq$ $\frac{(a^2+b^2)}{(a+b)}$
| Using Cauchy Schwarz in Engel form, we have
$$\!\!\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{a+c}+
\frac{b^2}{a+b}+\frac{c^2}{b+c}+\frac{a^2}{a+c}\geq \frac{(a+b+c)^2}{2(a+b+c)}+\frac{(b+c+a)^2}{2(a+b+c)}= a+b+c$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Conic Section Equation from Michael Spivak's Book So i've been reading Michael Spivak's Calculus lately and now i feel im stuck in his conic section equation, page 81.
What i dont understand is, how can the first equation becomes the second? After squaring it, i got different coefficient especially in the x² part.
| There is indeed a typo. From:
$$M (\alpha x + \beta) + B = \pm C \sqrt{(\alpha x + \beta)^2 + y^2}$$
square both sides:
$$\left [M (\alpha x + \beta) + B \right ]^2 = \left [\pm C \sqrt{(\alpha x + \beta)^2 + y^2} \right]^2$$
Now on the left use $(a + b)^2 = a^2 + 2 a b + b^2$:
$$M^2(\alpha x + \beta)^2 + 2 B M (\alpha x + \beta) + B^2 = C^2 \left [(\alpha x + \beta)^2 + y^2 \right ]$$
Then subtract the right-hand side:
$$M^2(\alpha x + \beta)^2 + 2 B M (\alpha x + \beta) + B^2 - C^2 (\alpha x + \beta)^2 - C^2 y^2 = 0$$
Regroup the terms according to the powers of $(\alpha x + \beta)$:
$$(M^2 - C^2) (\alpha x + \beta)^2 + 2BM(\alpha x + \beta) + B^2 - C^2 y^2 = 0$$
Now we look at the coefficients of $x^2$ and $y^2$.
The coefficient of $x^2$ must come from the first term, because it's the only one in which a term containing $x$ is squared. Therefore it's $\alpha^2 (M^2 - C^2)$. Clearly, the coefficient of $y^2$ is $-C^2$. Therefore we can write
$$-C^2 y^2 + \alpha^2(M^2 - C^2) x^2 + Ex + F = 0$$
for some $E, F$ (notice that there are no terms with $y$). Changing signs, we get:
$$C^2 y^2 - \alpha^2 (M^2 - C^2) x^2 + E x + F = 0$$
or equivalently,
$$C^2 y^2 + \alpha^2 (C^2 - M^2) x^2 + E x + F = 0.$$
For the last equation, instead of dividing by $\alpha^2$, we can write:
$$C^2 y^2 + (C^2 - M^2) (\alpha x)^2 + \frac E \alpha (\alpha x) + F = 0$$
and so if we substitute $\alpha x$ with $x$ and let $G = \frac E \alpha$ and $H = F$, we get
$$C^2 y^2 + (C^2 - M^2) x^2 + G x + H = 0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the remainder when $x^{10}+1$ is divided by $(x^2+1)(x^2+x+1)$ Find the remainder when $x^{10}+1$ is divided by $(x^2+1)(x^2+x+1)$
I have done it until the the divisor is of second degree.
But here the degree of the remainder is $4$
This means the remainder will be of the form. : $ax^3+bx^2+cx+d$ which makes it difficult to solve. Is it possible to solve $4$ variants with $4$ equations?
| Note that:
$$\frac{x^{10}+1}{(x^2+1)(x^2+x+1)}=\frac{(x^2+1)(x^8-x^6+x^4-x^2+1)}{(\color{green}{x^2+1})(x^2+x+1)}$$
Using long division:
$$
\require{enclose}
\begin{array}{r}
\color{blue}{x^6-x^5-x^4+2x^3-2x+1} \\[-3pt]
x^2+x+1 \enclose{longdiv}{x^8-x^6+x^4-x^2+1} \\[-3pt]
\underline{x^8+x^7+x^6}\phantom{2} \\[-3pt]
-x^7-2x^6+x^4-x^2+1 \\[-3pt]
\underline{-x^7-x^6-x^5} \\[-3pt]
-x^6+x^5+x^4-x^2+1 \\[-3pt]
\underline{-x^6-x^5-x^4} \\[-3pt]
2x^5+2x^4-x^2+1 \\[-3pt]
\underline{2x^5+2x^4+2x^3} \\[-3pt]
-2x^3-x^2+1 \\[-3pt]
\underline{-2x^3-2x^2-2x} \\[-3pt]
x^2+2x+1 \\[-3pt]
\underline{x^2+x+1} \\[-3pt]
\color{red}x \\[-3pt]
\end{array}
$$
Hence:
$$x^{10}+1=(x^2+1)(x^2+x+1)(\color{blue}{x^6-x^5-x^4+2x^3-2x+1})+\color{red}x(\color{green}{x^2+1}).$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $B$ if $B=A-{{1}\over{2}} A^2+{{1}\over{3}} A^3 -{{1}\over{4}} A^4+...$
Let $$
\ A=\begin{bmatrix} 0 & a & a^2 & a^3 \\ 0 & 0 & a & a^2 \\ 0 & 0 & 0 & a \\ 0 & 0 & 0 & 0 \end{bmatrix}
$$ and
$B=A-{{1}\over{2}} A^2+{{1}\over{3}} A^3 -{{1}\over{4}} A^4+...$
$i)$ Find the matrix $B$
$ii)$ Prove that $A=B+ {{1}\over{2!}} B^2+ {{1}\over{3!}} B^3+...$
My attempt:
$i)$
I calculated $A^2$ by multiplying $A$ by itself, then foundnd $A^3$ by multiplying $A$ by $A^2$, ans so on.
Then I noted that $A^n=0$ for $n\geq 4$
$A^2= A.A=\begin{bmatrix} 0 & 0 & a^2 & 2a^3 \\ 0 & 0 & 0 & a^2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}
$
$A^3= A^2.A=\begin{bmatrix} 0 & 0 & 0 & a^3 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}
$
$A^4=A^3.A =\begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}
$
$A^n=0$ for every $n\geq 4$, so:
$B= \begin{bmatrix} 0 & a & {{1}\over{2}} a^2 & {{1}\over {3}}a^3 \\ 0 & 0 & a & {{1}\over{2}} a^2 \\ 0 & 0 & 0 & a \\ 0 & 0 & 0 & 0 \end{bmatrix}
$
But is there any way easier than my way ?
And what about $(ii)$ ?
| You are on the right track. For (ii), just compute $B^2$, $B^3$, $B^4$, and observe that $B^4 = O$. Afterwards, verify that
$$A = B + \frac{1}{2!}B^2 + \frac{1}{3!} B^3.$$
For your information,
$$B^2 = \begin{bmatrix}0 & 0 & a^2 & a^3 \\ 0 & 0 & 0 & a^2 \\0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{bmatrix},$$ and
$$B^3 = \begin{bmatrix}0 & 0 & 0 & a^3 \\ 0 & 0 & 0 & 0 \\0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\end{bmatrix}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Sequences and Sums There is a list of numbers $a_{1} , a_{2} , …, a_{2010}$ . For $1 \leq n \leq 2010$, where $n$ is positive
integer, let $a_1+a_2+ \ldots +a_n = S_n$ . If $a_1 = 2010$ and $S_n = a_nn^2$ for all n, what is
the value of $a_{2010}$ ?
I've been trying to manipulate the formula but I cant seem to find a good relationship between $a_1$ and $a_{2010}$ like
$$
a_{2010} = \frac{a_1 +a_2 ... +a_{2010}}{2010^2}
$$
Then tried to use the definition $S_n = a_nn^2 $ over and over again but I cant find a good formula.
| We have $a_n\cdot n^2=a_1+a_2+\ldots+a_n=S_n$. Then
$$
\begin{array}{cc}
\begin{array}{rlrl}
\frac{\quad}{\quad}a_1\cdot 1^2=&2010\\
\frac{\quad}{\quad}a_2\cdot 2^2=&a_1+a_2\\
\frac{\quad}{\quad}a_3\cdot 3^2=&a_1+a_2+a_3\\
\frac{\quad}{\quad}a_4\cdot 4^2=&a_1+a_2+a_3+a_4\\
\end{array}
&
\begin{array}{rl}
\implies a_1=&2010\\
\implies a_2 =& \frac{1}{2^2-1}2010\\
\implies a_3=&\frac{1}{3^2-1}2010+\frac{1}{3^2-1}\frac{1}{2^2-1}2010\\
\implies a_4=&\frac{1}{4^2-1}2010+\frac{1}{4^2-1}\frac{1}{3^2-1}2010+\frac{1}{4^2-1}\frac{1}{3^2-1}\frac{1}{2^2-1}2010\\
\end{array}
\end{array}
$$
Now is easy to see
$$
a_n=2010\left(\frac{1}{n^2-1}+\frac{1}{n^2-1}\cdot\frac{1}{(n-1)^2-1}+\ldots+\frac{1}{n^2-1}\cdot\frac{1}{(n-1)^2-1}\cdot\ldots\cdot\frac{1}{2^2-1}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2861332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 5
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Find the area of the polygon whose vertices are the solutions in the complex plane of the equation $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$ If the area of the polygon whose vertices are the solutions in the complex plane of the equation $x^7+x^6+x^5+x^4+x^3+x^2+x+1=0$ can be expressed as $\frac{a\sqrt b+c}{d}$.Find $a+b+c+d$
The polynomial is $\frac{x^8-1}{x-1}$ has roots $\operatorname{cis}(2\pi k/8)$ for $k \in \{1, \ldots, 7\}$.
Thus the value is the area of the regular octagon minus the area of a triangle formed by two adjacent sides.
The area of an octagon (by splitting into triangles) with radius $1$ is $8 \cdot \frac{1}{2} \cdot \frac{\sqrt{2}}{2} = 2\sqrt{2}$.
I am stuck here.The answer for $a+b+c+d=10$
| $$\frac{a\sqrt{b}+c}{d}~=~2\sqrt{2}~~~~\text{with}~~~~a+b+c+d~=~10$$
The numerator of the $abcd$-fraction contains one square root plus a number. Hence the sum of the four variable equals an integer the easiest way to get rid of $c$ is by setting it $0$. From thereone we not got $a+b+d=10$. Now $b$ has to be $2$ so that we can arrive at the square root of $2$. Now we got
$$a+d=8~~\text{and}~~\frac{a\sqrt{2}}{d}=2\sqrt{2}$$
So $a$ has to be a multiply of $d$, to be exact $a=2d$. Plugging this into $a+d=8$ leads us to $a=\frac{16}3$ and $d=\frac83$. Verifying by setting this values in the first equations:
$$\begin{align}
\frac{\frac{16}{3}\sqrt{2}+0}{\frac83}~=~2\sqrt{2}~~~~&\text{and}~~~~\frac{16}3+2+0+\frac83~=~10\\
\frac{16\sqrt{2}}{8}~=~2\sqrt{2}~~~~&\text{and}~~~~\frac{16+8+6}{3}~=~10\\
2\sqrt{2}~=~2\sqrt{2}~~~~&\text{and}~~~~10~=~10
\end{align}$$
Therefore $a=\frac{16}3,b=2,c=0,\frac83$ but this only holds for $a,b,c,d \in \mathbb{Q}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
The sides of a triangle are in Arithmetic progression If the sides of a triangle are in Arithmetic progression and the greatest and smallest angles are $X$ and $Y$, then show that
$$4(1- \cos X)(1-\cos Y) = \cos X + \cos Y$$
I tried using sine rule but can't solve it.
| The law of sines helps!
From the given we obtain
$$\sin X+\sin Y=2\sin(X+Y)$$ or
$$2\sin\frac{X+Y}{2}\cos\frac{X-Y}{2}=4\sin\frac{X+Y}{2}\cos\frac{X+Y}{2}$$ or
$$\cos\frac{X-Y}{2}=2\cos\frac{X+Y}{2}$$ or
$$\cos\frac{X}{2}\cos\frac{Y}{2}=3\sin\frac{X}{2}\sin\frac{Y}{2}$$ or
$$\cos\frac{X-Y}{2}=4\sin\frac{X}{2}\sin\frac{Y}{2}$$ and
$$\cos\frac{X+Y}{2}=2\sin\frac{X}{2}\sin\frac{Y}{2}$$
We need to prove that
$$16\sin^2\frac{X}{2}\sin^2\frac{Y}{2}=2\cos\frac{X+Y}{2}\cos\frac{X-Y}{2},$$
which is obvious now.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Help summing the telescoping series $\sum_{n=2}^{\infty}\frac{1}{n^3-n}$. I know a priori that the series $$\sum_{n=2}^{\infty}\frac{1}{n^3-n}$$ converges. However, I am tasked with summing the series by treating it as a telescoping series.
By partial fraction decomposition, the series can be written as: $$\sum_{n=2}^{\infty}\left(\frac{1}{2(n+1)}+\frac{1}{2(n-1)}-\frac{1}{n}\right)$$
$s_n=\sum_{n=2}^{\infty}\left(\frac{1}{2(n+1)}+\frac{1}{2(n-1)}-\frac{1}{n}\right)=\left(\frac{1}{6}+\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{8}+\frac{1}{4}-\frac{1}{3}\right)+...+\left(\frac{1}{2(n+1)}+\frac{1}{2(n-1)}-\frac{1}{n}\right)$
I then grouped the terms by their position in each partial sum,
First terms: $\frac{1}{6},\frac{1}{8},\frac{1}{10},\frac{1}{12}...$
Second terms: $\require\cancel{\cancel{\frac{1}{2}}},\cancel{\frac{1}{4}},\cancel{\frac{1}{6}},\cancel{\frac{1}{8}}...$
Third terms: $\cancel{-\frac{1}{2}},-\frac{1}{3},\cancel{-\frac{1}{4}},-\frac{1}{5}...$
Cancelling leaves the series: $$\sum_{n=3}^{\infty}\frac{1}{2n}-\sum_{n=1}^{\infty}\frac{1}{2n+1}$$
However I'm stuck here since I see a divergent harmonic series summed with another harmonic series but I know the original series in question is convergent to $\frac{1}{4}$. What can I do? I suspect my error can be fixed somehow by adjusting the bounds of the sums...? Thanks in advance.
| Your partial fraction decomposition is fine, but you need not to break up into separate positive and negative sums. Instead, you need to cancel terms within the summation.
$$\sum_{n=2}^{\infty}\bigg(\frac{1}{2(n+1)}+\frac{1}{2(n-1)}-\frac{1}{n}\bigg)=\\\frac 12\sum_{n=2}^{\infty}\bigg(\frac{1}{(n+1)}+\frac{1}{(n-1)}-\frac{2}{n}\bigg)=\\\frac 12\sum_{n=2}^{\infty}\left(\frac{1}{n-1}-\frac{1}{n}\right)-\left(\frac{1}{n}-\frac{1}{n+1}\right)$$
Now notice that the negative part of the $n$ term of the sum cancels with the positive part of the $n+1$ term of the sum, so all the terms disappear except the positive part of the $n=2$ term. Our sum then equals
$$\frac 12\left(\frac 11-\frac 12\right)=\frac 14$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2865699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 3,
"answer_id": 0
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For a given $D\in \mathbb{N}$ are there infinite solutions to $D=b^2-4ac$ with $a,b,c \in \mathbb{N}$ I am wondering if for given $D\in \mathbb{N}$ we can find an infinite amount of solutions to $D=b^2-4ac$. Obvious $D$ is the discriminant of a binary quadratic form or a polynomial of degree $2$. Maybe the answer can be given with the use of those, but I do not know. Thanks for your answers.
|
I am wondering if for given $D\in \mathbb{N}$ we can find an infinite amount of solutions to $D=b^2-4ac$
No, not in general.
Consider $D=1$. Then $b^2-1=4ac$. Hence $(b+1)(b-1)=4ac$.
This obviously has four immediate solutions. $b=\pm 1$ and $a=0\vee c=0$.
Since $4|(b+1)(b-1)$, $b$ has to be odd. Which gives $b=2k+1$ for some $k\in\mathbb{N}$ and we get:
$(2k+2)2k=4k^2+4k=4k(k+1)$
Hence we have an infinte amount of solutions. With $k=a$ and $k+1=c$
Consider $D=3$. Then $3=b^2-4ac$. If $a=c$ we get $3=b^2-4a^2=(b-2a)(b+2a)$ but $3$ is prime. Hence there are only (at most) four solutions, for that case, which result from $b-2a=1 \wedge b+2a=3$ and $b-2a=3\wedge b+2a=1$.
If $a\neq c$.
We get, after adding $1$ to the equation:
$b^2+1=4ac+4=4(ac+1)$
Hence $4|b^2+1$ as above $b$ has to be odd.
$(2k+1)^2+1=4k^2+4k+1+1=4k^2+4k+2=2(2k^2+2k+1)$
$2k^2+2k+1=2(ac+1)$
But $2\nmid 2(k^2+k)+1$
As Lord Shark the Unknwon suggested in the comments, it is enough to view the problem $\mod 4$. Because we can always just factor out the 4 of $D$ and reduce to $0, 1,2, 3$ that way.
The case $D\equiv 0\mod 4$ is trivial and has infinite solutions, since then $b^2=4ac$ and if $b=2k$ we get $4k^2=4ac$. Now $a=c=k$ gives infinite solutions.
The last case is $D\equiv 2\mod 4$.
This fails for a similar argument like the shown $D=3$ case.
$2=b^2-4ac$.
$b^2=4ac+2=2(2ac+1)$
Yet again, $b$ has to be even. This gives us:
$4k^2=2(2ac+1)\Leftrightarrow 2k^2=2ac+1$
But now the LHS is even, while the RHS is odd.
| {
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"source": "stackexchange",
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} |
How to solve $\sqrt{49-x^2}-\sqrt{25-x^2}=3$? I recognize the two difference of squares: $49-x^2$ and $25-x^2$.
I squared the equation to get:
${49-x^2}-2(\sqrt{(49-x^2)(25-x^2)})+{25-x^2}=9$
However, I can't quite figure out how to remove the root in the middle. Any help is appreciated.
| First express one root: $$\sqrt{49-x^2}=3+\sqrt{25-x^2}$$
Now square:
$$ 49-x^2=9+6\sqrt{25-x^2}+ 25-x^2\implies \boxed{5=2\sqrt{25-x^2}}$$
and square again...
| {
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} |
Evaluating $\lim_{x\to0} \frac{\cos x - \cos 3x}{\sin 3x^2 - \sin x^2}$
$$ \lim_{x\to0}\frac{\cos x-\cos (3x)}{\sin (3x^2)-\sin (x^2)} $$
Is there a simple way of finding the limit?
I know the long one: rewrite it as
$$ -\lim_{x\to 0}\frac{\cos x-\cos(3x)}{\sin(3x^2)}\cdot\frac{1}{1-\dfrac{\sin(3x^2)}{\sin(x^2)}} $$
and then find both limits in separately applying L'Hospital's rule several times. The answer is $2$.
| I would use Taylor polynomials at order $2$.
$$\frac{\cos x-\cos (3x)}{\sin (3x^2)-\sin (x^2)}=\frac{1-\frac{x^2}{2}-\left(1-\frac{(3x)^2}{2}\right)+o(x^2)}{3x^2-x^2+o(x^2)}=\frac{4x^2+o(x^2)}{2x^2+o(x^2)}=2+o(1)
$$
| {
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Maximum minus minimum of $c$ where $a+b+c=2$ and $a^2+b^2+c^2=12$ Let $a,b,$ and $c$ be real numbers such that
$a+b+c=2 \text{ and } a^2+b^2+c^2=12.$
What is the difference between the maximum and minimum possible values of $c$?
$\text{(A) }2\qquad \text{ (B) }\frac{10}{3}\qquad \text{ (C) }4 \qquad \text{ (D) }\frac{16}{3}\qquad \text{ (E) }\frac{20}{3}$
As I was reading the solution for this problem, I noticed that it said to use Cauchy–Schwarz inequality. I know what this inequality is (dot product of two vectors < vector 1*vector 2), but I don't understand how it can be applied in this situation. Thanks!
https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12B_Problems/Problem_17
| Hint:
$$(2-c)^2 = (a+b)^2 = (a \cdot 1 + b \cdot 1)^2 \le (a^2 + b^2) (1^2 + 1^2) = 2(12-c^2)$$
Finding $c$ satisfying this inequality amounts to solving a quadratic.
$$c^2 - 4c + 4 \le 24 - 2 c^2$$ $$3 c^2 - 4 c - 20 \le 0$$ So $c$ is between the two roots $-2$ and $10/3$. Note that you should check that for each of these values of $c$, there exist valid choices of $a$ and $b$ that satisfy the two original equalities. Specifically, $a=b=2$ and $c=-2$ works, as well as $a=b=-2/3$ and $c=10/3$.
| {
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$\frac{1}{n+1}+\frac{1}{(n+1)^2}...=\frac{1}{n}$? On a problem book solution I was faced with the following step:
$$\frac{1}{n+1}+\frac{1}{(n+1)^2}...=\frac{1}{n}$$
I identified $\frac{1}{n+1}+\frac{1}{(n+1)^2}...$ as a geometric series so the sum would be $\frac{1}{1-r}$ so that $\frac{1}{1-\frac{1}{n+1}}=1+\frac{1}{n}$. I do not understand what I am doing worng.
Question:
What am I doing wrong?
Thanks in advance!
| If you know that $a+ar+ar^2+ar^3+\cdots = \dfrac{a}{1-r}$ for $|r|\lt 1$
then here you have $a= \dfrac1{n+1}$ and $r= \dfrac1{n+1}$
so the sum is $\dfrac{\frac1{n+1}}{{1-\frac1{n+1}}}=\dfrac{1}{n+1-1}=\dfrac{1}{n}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2868522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
last 2 digits of a sequence $x+\frac{1}{x} = 3$, what are the last 2 digits of $x^{2^{2013}}+\frac{1}{x^{2^{2013}}}$?
Getting the next value, we have to square then subtract by 2, I am clueless in getting to the next step
| Definitely you see the pattern : $x^{2m} + \frac 1{x^{2m}} = \left(x^m + \frac 1{x^m}\right)^2 - 2$ for all $m \geq 1$.
So, if the value at $m = 1$ is $3$, then the value at $m=2$ is $3^2 - 2 = 7$, at $m=4$ is $7^2 - 2 = 47$ etc.
Since we are only looking at the last two digits of the number, it is also enough to track the last two digits of the number at each point of time, and see if there is any cycle.
However, a cycle is already visible : at $m = 8$, we get $47^2 = 2209 - 2 = 22\color{blue}{07}$, so the last two digits are $07$. For $m=16$, we will again get $(07)^2 - 2 = 47$.
Therefore, the last two digits of the number $x^{2^n} + \frac 1{x^{2^n}}$ for $n = 1,2,3,...$ looks like : $03,07,47,07,47,07,47,...$
So now, it should be obvious which term appears at $n=2013$.
| {
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How to find the limit of the sequence $a_n=\frac{n^n}{3^n\cdot n!}$ as $n$ tends to infinity.
Let $a_n=\dfrac{n^n}{3^n\cdot n!}.$ Show that $a_n\to0$ as $n\to\infty$.
I know that I can use the ratio test for sequences, and $n^n$ increases faster than $3^n \cdot n!$ so it will tend towards infinity so I invert the sequence so I have to show that $\frac{3^n \cdot n!}{n^n}$ tends towards $0$.
I divide $a_{n+1}$ by $a_n$, I get $$\frac{3^{n+1} \cdot (n+1)! \cdot n^n}{(n+1)^{n+1} \cdot 3^n \cdot n!}.$$ I can get simplify up to get $\frac{3n^n}{(n+1)^{n+1}}$ but I do not know how to simplify any further so I can find the limit as $n$ tends to infinity.
| A Third Proof
Since $$\frac{a_{n}}{a_{n-1}}=\frac{1}{3}\left(1+\frac{1}{n-1}\right)^{n-1}$$ for $n=2,3\cdots$
then $$a_{n}=a_1\cdot\prod_{k=2}^{k=n}\frac{a_{k}}{a_{k-1}}=a_1\cdot \prod_{k=2}^{k=n}\frac{1}{3}\left(1+\frac{1}{k-1}\right)^{k-1}.$$
Notice that $$\left(1+\frac{1}{k-1}\right)^{k-1}<e$$ for $k=2,3,\cdots$
Thus, $$0<a_{n}<\frac{1}{3} \cdot \left(\frac{e}{3}\right)^{n-1}$$for $n=2,3\cdots.$
Let $n \to \infty$. Since $0<\dfrac{e}{3}<1$,then $\left(\dfrac{e}{3}\right)^{n-1} \to 0$. By the squeeze theorem, we may claim $$\lim_{n \to \infty}a_n=0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 5,
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} |
Evaluate $\int_{0}^{\infty}\frac{dx}{(x+\sqrt{1+x^2})^{2n}}\cdot \frac{1}{1+x^2}$ $$F(n)=\large \int_{0}^{\infty}\frac{\mathrm dx}{(x+\sqrt{1+x^2})^{2n}}\cdot \frac{1}{1+x^2}$$
$\large x=\tan u$
$\large \mathrm dx=\sec^2 u\mathrm du$
$$F(n)=\large \int_{0}^{\pi/2}\frac{\mathrm du}{(\tan u+\sec u)^{2n}}$$
$\large \tan u+\sec u=\frac{2\tan(u/2)}{1-\tan^2(u/2)}+\frac{1+\tan^2(u/2)}{1-\tan^2(u/2)}$
$$F(n)=\large \int_{0}^{\pi/2}\frac{\mathrm du}{\left(\frac{2\tan(u/2)}{1-\tan^2(u/2)}+\frac{1+\tan^2(u/2)}{1-\tan^2(u/2)}\right)^{2n}}$$
$\large t=\tan(u/2)$
$\large \mathrm du=\frac{2}{\sec^2(u/2)}\mathrm dt=\frac{2}{1+t^2}\mathrm dt
$
$$F(n)=\large 2\int_{0}^{1}\left(\frac{t-1}{t+1}\right)^{2n}\cdot \frac{\mathrm dt}{1+t^2}$$
I trying to evaluate $F(n)$, but I got stuck, how can I continue?
| Alternatively:
$$F(n)=\int_{0}^{\infty}\frac{dx}{(x+\sqrt{1+x^2})^{2n}}\cdot \frac{1}{1+x^2}=
\int_{0}^{\infty}\frac{(x-\sqrt{1+x^2})^{2n}}{1+x^2}dx$$
Change: $x-\sqrt{1+x^2}=t \Rightarrow x=\frac{1-t^2}{2t}, dx=-\frac{1+t^2}{2t^2}dt$, then:
$$F(n)= \int_{-1}^{0}\frac{t^{2n}}{\frac{(1+t^2)^2}{4t^2}}\cdot \left(-\frac{1+t^2}{2t^2}\right)dt=-2 \int_{-1}^{0}\frac{t^{2n}}{1+t^2}dt=\\
-2 \int_{-1}^{0}\frac{t^{2n}+t^{2n-2}-t^{2n-2}}{1+t^2}dt=\\
-2\int_{-1}^{0}t^{2n-2}dt+2\int_{-1}^{0}\frac{t^{2n-2}}{1+t^2}dt=\\
-\frac{2t^{2n-1}}{2n-1}\big{|}_{-1}^0-F(n-1).$$
| {
"language": "en",
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2 ways to find a Laurent series? A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 8.17,19,36 --> These exercises involve possibility of computing multiple Laurent series.
-
(Q1) For Exer 8.17, do we obtain multiple Laurent series depending on how we rewrite $\frac1{z+1} = \frac1{2+z-1}$?
I did similarly for Exer 8.18, and it seems that that's the point based on Exer 8.32 where I obtained 4 Laurent series (the 4th being convergent on the $\emptyset$!)
For (Q1)
*
*Take out $z-1$
$$\frac1{z+1} = \frac1{2+z-1} = \frac1{(z-1)(\frac{2}{z-1}+1)} \to \ \text{a Laurent series for} \ |z-1| > 2$$
*Take out $2$
$$\frac1{z+1} = \frac1{2+z-1} = \frac1{(2)(1+\frac{z-1}{2})} \to \ \text{a Laurent series for} \ |z-1| < 2$$
-
(Q2) For Exer 8.19, is there only one Laurent series unlike in Exer 8.17,18?
For (Q2)
$$\frac{z-2}{z+1} = 1 + \frac{-3}{z+1} \ \text{only}?$$
(Q3) For Exer 8.36, I came up with 2 Laurent series. Are they both valid?
Rewrite $\frac1{(z^2-4)(z-2)} = \frac1{(z-2)^2(z+2)}$. Rewrite $\frac1{z+2} = \frac1{z-2+4}$:
*
*Take out $z-2$
$$\frac1{z+2} = \frac1{z-2+4} = \frac1{(z-2)(\frac{4}{z-2}+1)} \to \ \text{a Laurent series for} \ |z-2| > 4$$
--> This is not the book's answer, and it doesn't seem to have a $c_{-1}$. It looks like the integral will be 0.
(Q3.1) What's wrong with this Laurent series? I guess something like $C[2,1] \subsetneq \{|z-2| > 4\}$, so it doesn't apply or something.
*Take out $4$
$$\frac1{z+2} = \frac1{z-2+4} = \frac1{(4)(1+\frac{z-2}{4})} \to \ \text{a Laurent series for} \ |z-2| < 4$$
--> This is the book's answer (apart from the region), and it gives the same answer as with Cauchy Integral Formula 5.1 (and later Residue Theorem 9.10) namely $\frac{- \pi i}{8}$
(Q3.2) Book says that for $\frac1{(4)(1+\frac{z-2}{4})}$, the region of convergence is $\color{red}{0 <} |z-2| < 4$. Why $0 <$?
| Note that isolated singularities of a rational function specify different regions of convergence when expanding the function around a point in a Laurent series. So, for each of these different regions there is a specific representation of $f$ as Laurent series.
Ad Q.1: Yes, we obtain two different Laurent series expansions of $f$ around $z=1$, one for each region of convergence. Since there are simple poles at $z=1$ and $z=-1$ we have to distinguish two regions of convergence when expanding around the pole $z=1$.
\begin{align*}
D_1:&\quad 0< |z-1|<2\\
D_2:&\quad |z-1|>2
\end{align*}
*
*The first region $D_1$ is a punctured disc with center $z=1$, radius $2$ and the pole at $z=-1$ at the boundary of the disc. It admits for the fraction with pole at $z=1$ a representation as principal part of a Laurent series and for the fraction with pole at $z=-1$ a power series.
*The region $D_2$ contains all points outside the disc with center $z=1$ and radius $2$. It admits for both fractions a representation as principal part of a Laurent series.
Note that in $D_1$ we have to exclude $0$, since $f$ is not defined there.
We have in $D_1$ the representation
\begin{align*}
\frac{1}{(z-1)(z+1)}&=\frac{1}{z-1}\cdot\frac{1}{2+(z-1)}\\
&=\frac{1}{z-1}\cdot\frac{1}{2\left(1+\frac{z-1}{2}\right)}\\
&=\frac{1}{2}\cdot\frac{1}{z-1}\sum_{j=0}^\infty(-1)^j\left(\frac{z-1}{2}\right)^j\\
&=-\sum_{j=-1}^\infty\left(-\frac{1}{2}\right)^j(z-1)^j
\end{align*}
and in $D_2$ we have the representation
\begin{align*}
\frac{1}{(z-1)(z+1)}&=\frac{1}{z-1}\cdot\frac{1}{2+(z-1)}\\
&=\frac{1}{(z-1)^2}\cdot\frac{1}{1+\frac{2}{z-1}}\\
&=\frac{1}{(z-1)^2}\sum_{j=0}^\infty(-1)^j\left(\frac{2}{z-1}\right)^j\\
&=\frac{1}{4}\sum_{j=-\infty}^{-2}\left(-\frac{1}{2}\right)^j(z-1)^j
\end{align*}
Ad Q.2: Yes, there is only one region of convergence, since we have only one simple pole at $z=-1$. The region of convergence is $|z-1|>0$ and the Laurent-series expansion is according to your calculation.
Ad Q.3,Q3.1: We obtain for $|z-2|>4$ the Laurent series expansion
\begin{align*}
\frac{1}{(z^2-4)(z-2)}&=\frac{1}{(z-2)^2}\cdot\frac{1}{z+2}\\
&=\frac{1}{(z-2)^3}\cdot\frac{1}{1+\frac{4}{z-2}}\\
&=\frac{1}{(z-2)^3}\sum_{j=0}^\infty(-4)^j\frac{1}{(z-2)^j}\tag{1}\\
&=\cdots
\end{align*}
Ad Q.3.2: The function under consideration
\begin{align*}
\frac{1}{(z^2-4)(z-2)}
\end{align*}
has a pole at $z=2$. Since this function is not defined at $z=2$ we have to exclude $0$ from the region and get $0<|z-2|<4$.
Hint: This answer with some more detailed considerations might be helpful.
| {
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symmetric polynomial recursion to solve the system, $x^5+y^5=33$, $x+y=3$ I was just reading on symmetric polynomials and was given the system of equations$$x^5+y^5=33 \text{ , } x+y=3$$
In the text they said to denote $\sigma_1=x+y$ and $\sigma_2=xy$, and to use recursion. I understand how the formula below works
$$S_5=x^5+y^5=\sigma_1 S_4-\sigma_2 S_3$$
But doing the recursion takes a long time. Is there a shorter way to do the recursion to solve this specific problem?
Thanks
| Let $x+y=3=a$, $xy=b$, $x^5+y^5=33=c$.
\begin{align}
x^5+y^5&=(x+y)(x^4-x^3y+x^2y^2-xy^3+y^4)
\\
c&=
a(x^4+y^4-x^2xy+(xy)^2-xyy^2)
\\
c&=
x^4+y^4-b(x^2+y^2)+b^2
,\\
x^4+y^4&=(x^2+y^2)^2-2x^2y^2=(x^2+y^2)^2-2b^2
,\\
c&=
a((x^2+y^2)^2-2b^2-b(x^2+y^2)+b^2)
=
a(x^2+y^2)(x^2+y^2-b)-b^2)
,\\
x^2+y^2&=(x+y)^2-2xy=a^2-2b
,\\
c&=
a(a^2-2b)(a^2-2b-b)-ab^2
,
\end{align}
thus we have a quadratic equation in terms of $b$:
\begin{align}
5ab^2-5a^3b-c+a^5&=0
,
\end{align}
\begin{align}
b&=\frac{(5a^3\pm\sqrt{5a^6+20ac}}{10a}
,\\
b_1&=2
,\\
b_2&=7
.
\end{align}
So, we need to consider two cases: $xy=2$ and $xy=7$.
In both cases we also have $x+y=3$.
Case 1 $xy=2$.
The quadratic equation with roots $x,y$ is
\begin{align}
t^2-3t+2&=0
,\\
x,y&=1,2
.
\end{align}
Indeed, $1^5+2^5=33$, $1+2=3$,
so in Case 1 we have two real solutions, $x=1,y=2$ and $x=2,y=1$.
Case 2 $xy=7$.
The quadratic equation with roots $x,y$ is
\begin{align}
t^2-3t+7&=0
,\\
x,y&=\tfrac32\pm\tfrac{\sqrt{19}}2\cdot i
,
\end{align}
so in Case 2 we have
two complex conjugate solutions,
\begin{align}
x&=\tfrac32+\tfrac{\sqrt{19}}2\cdot i
,\\
y&=\tfrac32-\tfrac{\sqrt{19}}2\cdot i
\end{align}
and
\begin{align}
x&=\tfrac32-\tfrac{\sqrt{19}}2\cdot i
,\\
y&=\tfrac32+\tfrac{\sqrt{19}}2\cdot i
.
\end{align}
| {
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Integral $\int_0^\frac{\pi}{2} \arcsin(\sqrt{\sin x}) dx$ I am trying to calculate $$I=\int_0^\frac{\pi}{2} \arcsin(\sqrt{\sin x}) dx$$ So far I have done the following. First I tried to let $\sin x= t^2$ then:
$$I=2\int_0^1 \frac{x\arcsin x}{\sqrt{1-x^4}}dx =\int_0^1 (\arcsin^2 x)'\frac{x}{\sqrt{1+x^2}}dx $$
$$=\frac{\pi^2}{8}-\int_0^1 \frac{\arcsin^2 x}{(1+x^2)^{3/2}}dx$$ We can expand into power series the integral, we have: $\arcsin^2z=\sum\limits_{n\geq1}\frac {2^{2n-1}z^{2n}}{n^2\binom {2n}n}$ and using the binomial series for $(1+x^2)^{-3/2}$ will result in: $$\sum_{n\geq1}\frac{2^{2n-1}x^{2n}}{n^2\binom {2n}n}\sum_{k\ge 0}\binom{-3/2}{k}x^{2k}$$ But I dont know how to simplify this. I tried one more thing, letting $\sin x= \sin^2 t$ gives:
$$I=2\int_0^\frac{\pi}{2}\frac{x\sin x}{\sqrt{1+\sin^2 x}}dx$$ Since $\int \frac{\sin x}{\sqrt{1+\sin^2x}}dx=-\arcsin\left(\frac{\cos x}{\sqrt 2} \right)+C$ we can integrate by parts to obtain: $$I=2\int_0^\frac{\pi}{2}\arcsin\left(\frac{\cos x}{\sqrt 2}\right)dx=2\int_0^\frac{\pi}{2}\arcsin\left(\frac{\sin x}{\sqrt 2}\right)dx$$ But I am stuck, so I would appreciate some help.
Edit: By letting $\frac{\sin x}{\sqrt 2} =t $ We get: $$I=2\int_0^\frac1{\sqrt{2}} \frac{\arcsin x}{\sqrt{\frac12-x^2}}dx=2\text{Li}_2\left(\frac1{\sqrt 2}\right)-\frac{\pi^2}{24}+\frac{\ln^2 2}{4}$$ Where the latter integral was evaluated with wolfram. I would love to see a proof for that.
| Mathematica gives:
$$\frac{1}{24} \left(-6 \text{Li}_2\left(\frac{1}{2}-\frac{1}{\sqrt{2}}\right)+6
\text{Li}_2\left(3-2 \sqrt{2}\right)+4 \pi ^2-3 \log ^2(2)+3 \log
^2\left(\sqrt{2}-1\right)+3 \log ^2\left(3+2 \sqrt{2}\right)+ \log (64) \log
\left(\sqrt{2}-1\right)+6 \sinh ^{-1}(1)^2-12 \log \left(2
\left(1+\sqrt{2}\right)\right) \sinh ^{-1}(1)-6 i \pi \left(2 \sinh ^{-1}(1)-\log
\left(3+2 \sqrt{2}\right)\right)\right)$$
which strongly suggests that hand calculation will be extremely difficult and error prone. Incidentally, the numerical value of that expression is $1.5122$, and the value of the numerical evaluation of the original is also $1.5122$, which strongly suggests the answer is correct.
Here is a graph of the function, and shaded value of the integral, which suggests there isn't a major error:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2876690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
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.