Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
Trig substitution for $\sqrt{9-x^2}$ I have an integral that trig substitution could be used to simplify.
$$ \int\frac{x^3dx}{\sqrt{9-x^2}} $$
The first step is where I'm not certain I have it correct. I know that, say, $\sin \theta = \sqrt{1-cos^2 \theta}$, but is it correct in this case $3\sin \theta = \sqrt{9 - (3\cos \theta)^2}$?
Setting then $x = 3\cos \theta; dx = -3\sin \theta d\theta$
$$-\int \frac{(3\cos\theta)^3}{3\sin\theta}3\sin\theta d\theta$$
$$-27\int\cos^3\theta d\theta$$
$$-27\int(1-\sin ^2\theta)\cos \theta d\theta$$
Substituting again, $u=\sin \theta; du=\cos \theta d\theta$
$$-27\int(1-u^2)du $$
$$-27u + 9u^3 + C$$
$$-27\sin \theta + 9 \sin^3 \theta + C$$
$$-9\sqrt{9-x^2} + 3\sin\theta\cdot 3\sin\theta\cdot \sin \theta + C$$
$$-9\sqrt{9-x^2} + (\sqrt{9-x^2})^2 \cdot \frac{\sqrt{9-x^2}}{3} + C$$
$$-9\sqrt{9-x^2} + \frac{1}{3}(9-x^2)(9-x^2)^{\frac{1}{2}} + C$$
$$-9\sqrt{9-x^2} + \frac{1}{3}(9-x^2)^\frac{3}{2} + C $$
I guess I have more doubts that I've done the algebra correctly than the substitution, but in any case I'm not getting the correct answer. Have I calculated correctly? Is the answer simplified completely?
EDIT
Answer needed to be simplified further:
$$-9\sqrt{9-x^2} + \frac{1}{3}(\sqrt{9-x^2}^2 \sqrt{9-x^2}) + C$$
$$-9\sqrt{9-x^2} + \frac{1}{3}((9-x^2)\sqrt{9-x^2}) + C$$
$$\sqrt{9-x^2} \left (-9 + \frac{1}{3}(9-x^2) \right ) + C$$
$$\sqrt{9-x^2} \left (-6 - \frac{x^2}{3} \right ) + C$$
$$ \bbox[5px,border:2px solid red]
{
- \left ( 6+ \frac{x^2}{3} \right ) \sqrt{9-x^2}
}
$$
This is the answer the assignment was looking for.
| alternative is to use $u=\sqrt{9-x^{2}}$
$$
\begin{aligned}
\int{\frac{x^{3}}{\sqrt{9-x^{2}}}\ dx}&=\int{(u^{2}-9)\ du}\\
&=\frac{u^{3}}{3}-9u+c\\
&=\frac{\left(\sqrt{9-x^{2}}\right)^{3}}{3}-9\sqrt{9-x^{2}}+c
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Simplify homogeneous polynomial of degree 3 with all coefficients equal 1 I want to simplify this homogeneous polynomials:
$$x^3 + y^3 + z^3 + x^2 y + x y^2 + x^2 z + x z^2 + z^2 y + z y^2 + x y z$$ to a simpler form.
For example, the homogeneous polynomials of degree 2 can be simplified as follows:
$$x^2 + y^2 + z^2 + x y + x z + y z = \frac{1}{2}((x + y)^2 + (x + z)^2 + (y + z)^2)$$. So I wonder whether degree of 3 could also be simplified. I know this kind of "simplification" is not very well-defined and very dependent on experience. I have tried mathematica but in vain. Is there any simpler form?
| Maybe $$(x^2+y^2+z^2)(x+y+z)+xyz$$ or $$\frac{x^5y+y^5z+z^5x-x^5z-y^5x-z^5y}{(x-y)(x-z)(y-z)},$$ where $\prod\limits_{cyc}(x-y)\neq0?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3587227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving $a^2 + b^2 + c^2 \geqslant ab + bc + ca$
Let $a, b, c \in \mathbb{R}$. Show that $a^2 + b^2 + c^2 \geqslant ab + bc + ca$.
My reasoning went as follows and I would like to know if it's correct.
$a^2 + b^2 + c^2 \geqslant ab + bc + ca$
$\Leftrightarrow a^2 + b^2 + c^2 -ab - bc - ca \geqslant 0$
$\Leftrightarrow 2a^2 + 2b^2 + 2c^2 -2ab - 2bc - 2ca \geqslant 0$
$\Leftrightarrow a^2 + a^2 + b^2 + b^2 + c^2 + c^2 -2ab - 2bc - 2ca \geqslant 0$
$\Leftrightarrow (a^2 -2ab + b^2) + (b^2 - 2bc + c^2)+ (c^2 - 2ca + a^2) \geqslant 0$
$\Leftrightarrow (a -b)^2 + (b-c)^2 + (c-a)^2 \geqslant 0$.
And the last inequality holds since squares are non-negative.
I've seen couple of different ways of doing this and they seem to divide by $2$ and I didn't really understand it so instead can I just use the fact that $2a^2 = a^2+a^2$?
| Yes, you can.
But I think it's better to release this way and the fact $2a^2=a^2+a^2$ by the cyclic sum:
$$a^2+b^2+c^2-ab-ac-bc=\sum_{cyc}(a^2-ab)=\frac{1}{2}\sum_{cyc}(2a^2-2ab)=$$
$$=\frac{1}{2}\sum_{cyc}(a^2-2ab+b^2)=\frac{1}{2}\sum_{cyc}(a-b)^2\geq0.$$
Another way:
$$a^2+b^2+c^2-ab-ac-bc=a^2-(b+c)a+b^2-bc+c^2=$$
$$=\left(a-\frac{b+c}{2}\right)^2+b^2-bc+c^2-\frac{(b+c)^2}{4}=\left(a-\frac{b+c}{2}\right)^2+\frac{3}{4}(b-c)^2\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3587457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\frac1{a(1+b)}+\frac1{b(1+c)}+\frac1{c(1+a)}\ge\frac3{1+abc}$ I tried doing it with CS-Engel to get $$
\frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)} \geq \frac{9}{a+b+c+ a b+b c+a c}
$$
I thought that maybe proof that $$
\frac{1}{a+b+c+a b+b c+a c} \geq \frac{1}{3(1+a b c)}
$$ or $$
3+3 a b c \geq a+b+c+a b+b c+a c
$$, but I don't know how
| As in Michael's solution, this uses the same idea of "+1" to split up the fraction.
Written up this way, it might seem like a more natural approach.
WTS
$$ \sum \frac{ 1 + abc } { a (1+b)} \geq 3 $$
$$\sum \frac{ 1 + abc + a + ab } { a (1+b) } \geq 6$$
$$ \sum \frac{ ab ( 1 + c) + (1+a ) } { a (1+b) } \geq 6$$
$$ \sum \frac{ ab (1+c) } { a(1+b) } + \frac{ (1+b) } { b (1 + c) } \geq 6$$
Applying AM-GM to all 6 terms, the result follow.
From the conditions, we can easily deduce that the equality case is only $ a = b = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3588912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Unable to prove an assertion with induction I need to prove:
$ \displaystyle \sum_{k=1}^n\frac{1}{(5k + 1) (5k + 6)} = \frac{1}{30} - \frac{1}{5(5n + 6)} $ with mathematical induction for all $n \in \mathbb{N}$.
After successfully proving it for n = 1, I try to prove it in the Induction-Step for n + 1:
$ \displaystyle \sum_{k=1}^{n+1}\frac{1}{(5k + 1) (5k + 6)} = \frac{1}{30} - \frac{1}{5(5(n+1) + 6)} $
which can be summarised to:
$ \displaystyle \sum_{k=1}^{n}\frac{1}{(5k + 1) (5k + 6)} + \frac{1}{(5(n+1) + 1) (5(n+1) + 6)} = \frac{1}{30} - \frac{1}{5(5(n+1) + 6)} $
Using the Induction Hyptohesesis this turns to:
$ ( \frac{1}{30} - \frac{1}{5(5n + 6)} ) + \frac{1}{(5n + 6 ) (5n+6+5)} $
when we let s = 5n +6 we get:
$ \frac{1}{30} - \frac{1}{5s} + \frac{1}{(s) (s+5)} = \frac{1}{30} - \frac{1}{5(s + 5)} $
substracting both sides by $\frac{1}{30}$ yields:
$- \frac{1}{5s} + \frac{1}{(s) (s+5)} = - \frac{1}{5(s + 5)}$ and this is where I'm stuck. I'm working on this for more than 6h but my mathematical foundation is too weak to get a viable solution. Please help me.
| Your proof by induction is complete as soon as you compute the difference $$\frac{1}{5s} - \frac{1}{5(s + 5)}=\frac{(s+5)-s}{5s(s + 5)}=\frac{1}{s(s + 5)}.$$
There is also a direct proof: the sum is telescopic
$$ \sum_{k=1}^n\frac{1}{(5k + 1) (5k + 6)}=\frac{1}{5}\sum_{k=1}^n\left(\frac{1}{5k + 1}-\frac{1}{5(k+1) + 1}\right)=\frac{1}{5}\sum_{k=1}^n\frac{1}{5k + 1}-\frac{1}{5}\sum_{k'=2}^{n+1}\frac{1}{5k' + 1}\\=\frac{1}{5}\sum_{k=1}^n\frac{1}{5\cdot 1 + 1}
-\frac{1}{5}\sum_{k=1}^n\frac{1}{5\cdot (n+1) + 1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3589245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Would my method of solving this symmetric difference set be correct and is my answer correct? Problem: If A = {1, 2, 4, 6, 7} and B = {1, 2, 4, 6} and C = {1, 3, 4, 7}, what is A β B β C?
1) A β B = {1, 2, 4, 6, 7} β {1, 2, 4, 6} = {1, 2, 4, 6, 7} β {1, 2, 4, 6} = {7}
2) B β A = {1, 2, 4, 6} β {1, 2, 4, 6, 7} = {1, 2, 4, 6} β {1, 2, 4, 6, 7} = {}
3) A β B = (A β B) βͺ (B β A) = {7} βͺ {} ={7}
4) (A β B) β C = {7} β {1, 3, 4, 7} = {7} β {1, 3, 4, 7} = {}
5) C β (A β B) = {1, 3, 4, 7} β {7} = {1, 3, 4, 7} β {7}= {1, 3, 4}
6) (A β B) β C = ((A β B) β C) βͺ (C β (A β B)) = {} βͺ {1, 3, 4} = {1, 3, 4}
| Your answer and working appear to be correct. A noteowrthy fact: for a family of sets $A_1,\cdots,A_n$, we can show that
$$\bigoplus_{i=1}^n A_i = \{ \text{all elements belonging to an odd number of the } A_n \}$$
You can use this fact to quickly verify your answer in a case like this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3589677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integral $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x}{1+\sin^4 x}{\rm d}x$ Someone gives a solution as follows:
\begin{align*} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x}{1+\sin^4 x}{\rm d}x&=2\int_0^{\frac{\pi}{2}}\frac{\sin^2 x}{1+\sin^4 x}{\rm d}x\\ &=2\int_0^{\frac{\pi}{2}}\frac{\cos^2 x}{1+\cos^4 x}{\rm d}x\\ &=2\int_0^{\frac{\pi}{2}}\frac{\sec^2 x}{\sec^4 x+1}{\rm d}x\\ &=2\int_0^{\frac{\pi}{2}}\frac{{\rm d}(\tan x)}{(1+\tan^2 x)^2+1}\\ &=2\int_0^{+\infty} \frac{{\rm d}t}{(1+t^2)^2+1}\\ &\color{red}{=\frac{1}{2}\int_{-\infty}^{+\infty}\frac{{\rm d}u}{\left(u-\frac{1}{u\sqrt{2}}\right)^2+(1+\sqrt{2})}}\\ &=\frac{1}{2}\int_{-\infty}^{+\infty}\frac{{\rm d}u}{u^2+(1+\sqrt{2})}\\ &=\frac{\pi}{2} \sqrt{\sqrt{2}-1} \end{align*}
I wonder how he obtains the sixth equality colored red.
| $$I=2\int_{0}^{\pi/2}\frac{\sin^2 x}{1+\sin^4 x} dx=2\int_{0}^{\pi/2} \frac{\tan^2 t \sec^2 x}{(1+\tan^2x)^2+\tan^4 x}dx=2\int_{0}^{\infty} \frac{t^2 dt}{2t^4+2t^2+1}$$
$$\implies I=2\int_{0}^{\infty} \frac{dt}{2t^2+1/t^2+2}=\frac{1}{\sqrt{2}}\int_{0}^{\infty}
\frac{(\sqrt{2}-1/t^2)+(\sqrt{2}+1/t^2)}{2t^2+1/t^2+2}dt$$
$$\implies I=\frac{1}{\sqrt{2}}\left( \int_{0}^{\infty} \frac{(\sqrt{2}-1/t^2)dt}{(t\sqrt{2}+1/t)^2-2(\sqrt{2}-1)}+\int_{0}^{\infty} \frac{(\sqrt{2}+1/t^2)dt}{(t\sqrt{2}-1/t)^2+2(\sqrt{2}+1)}\right)$$
Let the twin transformations $u=t\sqrt{2}+1/t$ and $v=t\sqrt{2}-1/t$ in the two integrals above, then
the first integral vanishes because both upper and lower limits become the same
$$I=\frac{2}{\sqrt{2}}\int_{0}^{\infty} \frac{du}{u^2+2(\sqrt{2}+1)}=\frac{\pi}{2\sqrt{(\sqrt{2}+1})}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3598456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What method should I use to solve this system of polynomial equations? What method should I use to solve this system of polynomial equations?
$$\begin{cases} 3x^3 - 3y^3 + z^3 - xyz - 3 &= 0\\
3y^3 - x^3 - z^3 - xyz + 5 &= 0\\
x^3 - y^3 + z^3 - xyz - 2 &= 0\end{cases}$$
I've run out of ideas. Tried adding, subtracting but it gave me nothing. Maybe there's a good substitution I don't see?
| Hint. By summing the first two equations we obtain
$$xyz=x^3+1.$$
Then solve the linear system with respect to $y^3$ and $z^3$:
$$\begin{cases}
3x^3 - 3y^3 + z^3 - (x^3+1) - 3 = 0\\
x^3-y^3+z^3-(x^3+1)-2=0
\end{cases}$$
and we find
$$y^3=x^3-\frac{1}{2},\quad z^3=x^3+\frac{5}{2}.$$
Going back to the first equation cubed we get
$$x^3\left(x^3-\frac{1}{2}\right)\left(
x^3+\frac{5}{2}\right)=(x^3+1)^3.$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3599880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
if $\omega$ is a primitive cube root of unity then $-\omega$ is a primitive sixth root of unity. I was given a statement that if $\omega$ is a primitive cube root of unity then $-\omega$ is a primitive sixth root of unity.
The roots of $x^nβ1$ in $\mathbb C$ which are not also roots of $x^m β1$ for some $1 β€ m β€ n$ are called the primitive nβth complex roots of unity
So if this example works with $n=6$ I would get the following:
We have $x^6β1 = (xβ1)(x^5+x^4+x^3+x^2+x+1)$, $x^6β1 = (x^2β1)(x^4+x^2+1)$ and $x^6 β1 = (x^3 β1)(x^3 + 1)$. The roots of $xβ1$, $x^2 β1$, $x^3 β1$ are $1,β1,\frac {β1} 2 Β± \frac{{\sqrt3}} 2 i$. Thus the remaining two roots of $x^6β1$, namely, $Ο^1 = \frac {1} 2 + \frac{{\sqrt3}} 2 i$ and $Ο^5 = \frac {1} 2 - \frac{{\sqrt3}} 2 i $ are the primitive 6βth complex roots of unity.
Is it correct to prove the above statement by just pointing out that if we set $\omega = \frac {β1} 2 Β± \frac{{\sqrt3}} 2 i$ that $-\omega = \frac {1} 2 - \frac{{\sqrt3}} 2 i$ and $-\omega = \frac {1} 2 + \frac{{\sqrt3}} 2 i$ which are both sixth roots of unity.
Thanks in advance.
| Yes, that is correct.
Another way, without computing anything is this:
Let $\zeta$ be a primitive 3rd root of unity. Then $\zeta^3=1,\zeta\neq 1$.
Now a 6th root of unity $\omega$ must satisfy $\omega^6=1$. Primitive means that $\omega^2\neq 1$ and $\omega^3\neq 1$. (It has order 6 and not a strict divisor of 6).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3604101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the points on the curve $x^4+y^4+3xy=2$ closest and farthest to the origin I need to find the points on the curve $x^4+y^4+3xy=2$ that are closest and farthest to the origin.
I believe this might be a Lagrange multiplier problem, but I am not sure. I was thinking that maybe minimizing/maximizing the function would be the way to go.
| I saw people recommending to use lagrange multiplier, but I know none of that. Instead, I tried this.
Firstly, you want to maximise/minimise (sqrt) $x^2+y^2$ with $x^4+y^4+3xy-2=0$. Consider $x^2+y^2=t$ and the intersection point $(u, v)$. i.e.
$$u^2+v^2=t, u^4+v^4+3uv-2=0$$
If at the point $(u, v)$, we can get both of the $\frac{dy}{dx}$ to be equal, then it means the curve $x^2+y^2-t=0$ and $x^4+y^4+3xy-2=0$ are tangent to each other.
Finding $\frac{dy}{dx}$ of both functions
Implicit differentiation gives
$$x^2+y^2=t \implies \frac{dy}{dx}=-\frac{x}{y}$$
$$x^4+y^4+3xy-2=0 \implies \frac{dy}{dx}=-\frac{4x^3+3y}{4y^3+3x}$$
We want to equal those. After simplifying
$$4xy^3+3x^2=4x^3y+3y^2$$
$$3(x^2-y^2)=4xy(x^2-y^2)$$
$$xy=\frac{3}{4} \textrm{ or } x=\pm y$$
If $xy=\frac{3}{4}$, $x^4+y^4+3(\frac{3}{4})-2=0 \implies x^4+y^4=-\frac{1}{4}$ which is impossible since LHS is non-negative.
Therefore, $x=\pm y$
**If ** $x=y$, $2x^4+3x^2-2=0\implies(2x^2-1)(x^2+2)=0\implies x=\pm \frac{1}{\sqrt{2}}, y=\pm \frac{1}{\sqrt{2}}, \sqrt{x^2+y^2}=1$
**If ** $x=-y$, $2x^4-3x^2-2=0\implies(2x^2+1)(x^2-2)=0\implies x=\pm \sqrt{2}, y=\mp \sqrt{2}, \sqrt{x^2+y^2}=2$
$$\therefore 1<\sqrt{x^2+y^2}<2$$
Side note: To the people doing lagrange multiplier, stop and calm down :) As mentioned in the comments, (apparently) this is the foundation of lagrange multiplier - Langrange multiplier is basically a generalisation of this to any constraint-equations and more variables :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3604466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Proving existence of a limit I try to prove that the folowing limit does not exist.
$$\displaystyle\lim_{(x,y)\to(0,0)} \frac{x\sin (ax^2+by^2)}{\sqrt{x^2+y^2}}, a,b>0, a\neq b$$
I make the assumption that this limit exist. So, I try to write the limit $$\displaystyle\lim_{(x,y)\to(0,0)} \frac{x}{\sqrt{x^2+y^2}}$$ as a limit of the function $ \dfrac{x\sin (ax^2+by^2)}{\sqrt{x^2+y^2}}$ and another function, that the limits of those functions exist, and thus from the limits function algebra, the $\displaystyle\lim_{(x,y)\to(0,0)} \frac{x}{\sqrt{x^2+y^2}}$ exists which is a contradiction, because this limit obviously does not exist. Maybe the limit $\displaystyle\lim_{(x,y)\to (0,0)} \dfrac{\sin (ax^2+by^2)}{ax^2+by^2}=1$ can help in someway.
Any Ideas?? Thank you
| As Peter Foreman's question comment indicates, the limit does exist and is equal to $0$. One way to see this is with
$$\begin{equation}\begin{aligned}
\lim_{(x,y)\to(0,0)} \frac{x\sin (ax^2+by^2)}{\sqrt{x^2+y^2}} & = \lim_{(x,y)\to(0,0)} \frac{\sin(ax^2+by^2)(ax^2+by^2)(x)}{(ax^2+by^2)\sqrt{x^2+y^2}} \\
& = \lim_{(x,y)\to(0,0)}\left(\frac{\sin(ax^2+by^2)}{(ax^2+by^2)}\right)\left(\frac{(ax^2+by^2)(x)}{\sqrt{x^2+y^2}}\right)
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
The first factor's limit is $1$, as you've already noted in your question text. For the second factor, using polar coordinates so $x = r\cos(\theta)$ and $y = r\sin(\theta)$, you get
$$\begin{equation}\begin{aligned}
\lim_{(x,y)\to(0,0)}\frac{(ax^2+by^2)(x)}{\sqrt{x^2+y^2}} & = \lim_{r \to 0}\frac{(ar^2\cos^2(\theta) + br^2\sin^2(\theta))(r\cos(\theta))}{r} \\
& = \lim_{r \to 0}r^2(a\cos^2(\theta) + b\sin^2(\theta))\cos(\theta) \\
& = 0
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Thus, the product of the limits of the $2$ factors in \eqref{eq1A} is $0$, showing the limit is $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3606145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
For $n\ge 3$ determine all real solutions of the system of $n$ equations.
Question: For $n\ge 3$ determine all real solutions of the system of $n$ equations: $$x_1+x_2+\cdots+x_{n-1}=\frac{1}{x_n}\\ \cdots \\ x_1+x_2+\cdots+x_{i-1}+x_{i+1}+\cdots+x_n=\frac{1}{x_i}\\ \cdots \\ x_2+\cdots+x_{n-1}+x_n=\frac{1}{x_1}.$$
My approach: It is given that $$x_1+x_2+\cdots+x_{n-1}=\frac{1}{x_n}\\ \cdots \\ x_1+x_2+\cdots+x_{i-1}+x_{i+1}+\cdots+x_n=\frac{1}{x_i}\\ \cdots \\ x_2+\cdots+x_{n-1}+x_n=\frac{1}{x_1}.$$
Define $$S_n:=x_1+x_2+\cdots+x_n.$$
This implies that $$x_1+x_2+\cdots+x_{n-1}+x_n=x_n+\frac{1}{x_n}\\ \cdots \\ x_1+x_2+\cdots+x_{i-1}++x_i+x_{i+1}+\cdots+x_n=x_i+\frac{1}{x_i}\\ \cdots \\ x_1+x_2+\cdots+x_{n-1}+x_n=x_1+\frac{1}{x_1}.$$
Therefore, we have $$x_j+\frac{1}{x_j}=S_n, \forall 1\le j\le n.$$
Now for any $1\le j\le n,$ we have $$x_j+\frac{1}{x_j}=x_n+\frac{1}{x_n}\\ \implies \frac{1}{x_j}-\frac{1}{x_n}=x_n-x_j\\ \implies \frac{x_n-x_j}{x_nx_j}=x_n-x_j\\ \implies (x_n-x_j)\left(\frac{1}{x_nx_j}-1\right)=0\\ \implies x_j=x_n \text{ or } x_j=\frac{1}{x_n}.$$
Now let us have $i (0\le i\le n-1)$ of the $(n-1)$ numbers $x_j, 1\le j\le n-1$ such that $$x_j=x_n$$ and the rest $(n-1-i)$ numbers such that $$x_j=\frac{1}{x_n}.$$
Therefore, since $$x_1+x_2+\cdots+x_{n-1}=\frac{1}{x_n}\\\implies i.x_n+(n-1-i).\frac{1}{x_n}=\frac{1}{x_n}\\ \implies i.x_n^2+(n-2-i)=0\\\implies i.x_n^2=2+i-n.$$
Now since $x_n\in\mathbb{R}$ and $x_n\neq 0\implies x_n^2>0.$ Now since $i\ge 0\implies i.x_n^2\ge 0 \implies 2+i-n\ge 0\implies i\ge n-2.$ Therefore $i=n-2,n-1$.
Now when $i=n-2$, we have $(n-2)x_n^2=0.$ Now since it is given that $n\ge 3\implies n-2\ge 1>0$. Therefore $x_n^2=0\implies x_n=0$. But $x_n\neq 0$. Therefore $i\neq n-2$.
Now when $i=n-1$, we have $(n-1)x_n^2=1\implies x_n^2=\frac{1}{n-1}\implies x_n=\pm\frac{1}{\sqrt{n-1}}.$ Therefore we have $$x_1=x_2=\cdots=x_n=\pm\frac{1}{\sqrt{n-1}}.$$
Therefore the required set of solutions are $$(x_1,x_2,\cdots,x_n)=\left(\frac{1}{\sqrt{n-1}},\frac{1}{\sqrt{n-1}},\cdots,\frac{1}{\sqrt{n-1}}\right),\left(-\frac{1}{\sqrt{n-1}},-\frac{1}{\sqrt{n-1}},\cdots,-\frac{1}{\sqrt{n-1}}\right).$$
Can someone check if my solution is correct or not? And if correct, is there a more better and efficient solution than this?
| I get the same result.
Here's my work.
I abhore "..."s,
so I'll write the equations like this:
For $i=1$ to $n$,
$\sum_{k=1, k\ne i}^n x_k
=\dfrac1{x_i}
$.
Filling in the missing term,
$\sum_{k=1}^n x_k
=\dfrac1{x_i}+x_i
$.
Letting
$S = \sum_{k=1}^n x_k$,
we have
$S
= \dfrac1{x_i}+x_i
$
so
$x_i^2-Sx_i+1 = 0$
or
$x_i
= \dfrac{S\pm\sqrt{S^2-4}}{2}
$.
Also,
for $i \ne j$,
$\dfrac1{x_i}+x_i
= \dfrac1{x_j}+x_j
$
or,
multiplying by $x_ix_j$,
$x_j-x_i^2x_j
=x_i-x_ix_j^2
$
or
$x_i-x_j
=x_ix_j^2-x_i^2x_j
=x_ix_j(x_j-x_i)
$.
If $x_i \ne x_j$
then
$x_ix_j = -1$
so
$x_j = -\dfrac1{x_i}
$.
Set $i = n$
so the other values
are either
$x_n$ or
$-\dfrac1{x_n}
$.
Suppose $m$ of them
are $x_n$.
Then
$S
=mx_n-(n-m)\dfrac1{x_n}
$
so
$x_n+\dfrac1{x_n}
=mx_n-(n-m)\dfrac1{x_n}
$
or,
writing $x$ for $x_n$,
$(m-1)x=(n-m+1)\dfrac1{x}
$.
We can't have
$m=1$
for then $\dfrac1{x} = 0$.
Therefore
$m \ge 2$
so that
$(m-1)x^2=(n-m+1)
$
or
$x
=\pm\sqrt{\dfrac{n-m+1}{m-1}}
$.
I will choose
$x
=\sqrt{\dfrac{n-m+1}{m-1}}
$
for now.
Then
$\begin{array}\\
S
&=m\sqrt{\dfrac{n-m+1}{m-1}}-(m-n)\sqrt{\dfrac{m-1}{n-m+1}}\\
&=x+\dfrac1{x}\\
&=\sqrt{\dfrac{n-m+1}{m-1}}+\sqrt{\dfrac{m-1}{n-m+1}}\\
\end{array}
$
so,
multiplying by
$\sqrt{(n-m+1)(m-1)}
$,
$m(n-m+1)-(m-n)(m-1)
=(n-m+1)+(m-1)
$
or
$n
=mn-m^2+m-(m^2-(n+1)m+n)
=2m(n+1)-2m^2-n
$
or
$0
=2(m(n+1)-m^2-n)
$
or
$0
= m^2-(n+1)m+n
$
or
$0
=(m-1)(m-n)
$.
Since
$m > 1$,
we have
$m = n$
so all the
$x_i$ are the same,
so
$S
= \dfrac{n}{\sqrt{n-1}}
$
and each
$x_i
=\dfrac1{\sqrt{n-1}}
$.
Note that
$\dfrac1{\sqrt{n-1}}
+\sqrt{n-1}
=\dfrac{1+n-1}{\sqrt{n-1}}
=\dfrac{n}{\sqrt{n-1}}
= S
$.
If
$x
=-\sqrt{\dfrac{n-m+1}{m-1}}
$
then all the signs
are reversed,
so the final result is the same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3608386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find stability function of Runge Kutta Method starting from a 3x4 Butcher Tableau Given the scalar IVP $\dot{y}(t) = f(t, y(t))$ , $y(0)=y_0$,
and the butcher Tableau
$$
\begin{array}
{c|cccc}
0 & 0 & 0 & 0\\
\frac{1}{2} & \frac{1}{2} & 0 & 0\\
1& -1& 2& 0\\
\hline
& \frac{1}{6} &\frac{2}{3} &\frac{1}{6}
\end{array}
$$
We're asked to find the stability function of that Runge Kutta Method.
| $$k_1=f(t_j, y_j) = \lambda y_j$$
$$k_2= f(t_j + \frac{1}{2}, y_j + \frac{1}{2}hk_1) = \lambda(y_j + \frac{1}{2}hk_1)$$
$$k_3= f(t_j + 1, y_j - hk_1 + 2hk_2) = \lambda (y_j - hk_1 + 2hk_2)$$
$$y_{j+1}=y_j +\frac{h}{6}(k_1 + 4k_2 + k_3)$$
Thus
$$y_{j+1}=y_j +\frac{h}{6}(\lambda y_j + 4\lambda(y_j + \frac{1}{2}h(\lambda y_j)) + \lambda (y_j - h(\lambda y_j) + 2h(\lambda(y_j + \frac{1}{2}h(\lambda y_j)))))
$$
Which simplifies to
$$
y_{j+1}= y_j(1 + h\lambda + \frac{(h \lambda)^2}{2} + \frac{(h \lambda)^3}{6})
$$
If we let $z := h\lambda$, then our stability function is $$g(z) = 1 + z + \frac{z^2}{2} + \frac{z^3}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3610699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Different solutions with different results for an inequality Find m such that the following inequality:
$$\left|4x-2m-\frac{1}{2}\right| > -x^2 +2x + \frac{1}{2} - m$$
is always true for $\forall x \in R$.
1st solution:
1st case
$$4x-2m-\frac{1}{2} > -x^2 + 2x +\frac{1}{2} -m$$
$$<=>x^2+2x-m-1>0$$
$$\Leftrightarrow 1^2+(m+1)< 0$$
$$\Leftrightarrow m<- 2$$
2nd case
$$4x-2m-\frac{1}{2}< -(-x^2 + 2x +\frac{1}{2} -m)$$
$$\Leftrightarrow x^2-6x+3m>0$$
$$\Leftrightarrow 3^2-3m<0$$
$$\Leftrightarrow m>3$$
2nd solution:
The inequality is the same as:
$$(x-1)^2+|4x-2m-\frac{1}{2}|>\frac{3}{2}-m$$
Since the left-hand side is always positive, in order for the inequality to be always true, $\frac{3}{2}-m$ has to be negative, or $m > \frac{3}{2}$
The 2 solutions give different answers, so I was quite confused
But I get more confused as Wolfram Alpha gives me the solution:
$$m > \sqrt{3} - \frac{1}{4} \text{ or } m < -\sqrt{3} - \frac{1}{4} $$
There's a high chance that Wolfram Alpha's solution is correct (after testing out some $m$ value). How do I approach their solution? (Or maybe if you believe that solution is wrong, then what's the exact solution to the problem?)
| Your two solutions are mathematically correct, but you when you reach $m<-2$, you have calculated for which values of $m$, we have $m+2>0$ and you have not shown that $\forall m>-2, (x+1)^2>m+2, \forall x \in R$. This explain the error in the first method. For the second method, you have shown that the inequality:
$$(x-1)^2+|4x-2m-\frac{1}{2}|>\frac{3}{2}-m$$
have solutions when $m<\frac{3}{2}$, but this not explain for which $m$, we have:
$$\left|4x-2m-\frac{1}{2}\right| > -x^2 +2x + \frac{1}{2} - m$$
In general, when you have to solve this type of inequality, you have to solve the system:
$$\left\{\begin{matrix}
4x-2m-\frac{1}{2}\geq0
\\ -x^2 +2x + \frac{1}{2} - m\geq0
\\ 4x-2m-\frac{1}{2} > -x^2 +2x + \frac{1}{2} - m
\end{matrix}\right.
\vee
\left\{\begin{matrix}
4x-2m-\frac{1}{2}<0
\\ -x^2 +2x + \frac{1}{2} - m<0
\\ -(4x-2m-\frac{1}{2}) > -x^2 +2x + \frac{1}{2} - m
\end{matrix}\right.
$$
These system are quite complicated, and I will give directly the solutions:
$$\left\{\begin{matrix}
x>\sqrt{m+2}-1
\\ x<-\sqrt3\sqrt{3-m}+3
\end{matrix}\right.$$
When $m<-2$ and $m>3$ the square roots are not defined, so these values of $m$ are correct.
Also, we want the two expression to be the same because we want a single solution, so:
$$\sqrt{m+2}-1=-\sqrt3\sqrt{3-m}+3$$
The solutions are:
$$m=\sqrt3-\frac{1}{4} \vee m=-\sqrt3-\frac{1}{4}$$
When $-\sqrt3-\frac{1}{4}<m<\sqrt3-\frac{1}{4}$ there are solutions, while when:
$$m > \sqrt{3} - \frac{1}{4} \lor m < -\sqrt{3} - \frac{1}{4}$$
there aren't.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3612123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to use prove this $p^4\equiv p\pmod {13}$
Let a prime number $p$, and $n$ a positive integer such $$p\mid n^4+n^3+2n^2-4n+3.$$
Show that $$p^4\equiv p\pmod {13}.$$
A friend of mine suggested that I might be able to use the results problem.
| We will have modulo $13$ all the time. Notice that $$N:= n^4+n^3+2n^2-4n+3 \equiv (n^2-6n-4)^2 \equiv (n-3)^4$$
Clearly $13\mid N\iff 13\mid n-3$, so we assume $p\ne 13$ and thus $13$ does not divide $n-3$. Say prime divisor $p$ of $N$ is good if $p^4{\equiv}p$.
Suppose the statment is not true, so there exists a prime $q$ which is not good.
Also we can assume that there is no good prime divisor of $N$: if $p\ne 13$ is good than we can divide $N$ by $p\equiv p^4$ modulo $13$ and then observe $N' = N/p^4$ modulo $13$. Also we can reduce $N$ with all divisors of $N$ of form $d^4$. So if we make now a prime factorisation for $N$ modulo $13$ we have $$N\equiv 2^a4^b5^c6^d7^e8^f10^g11^h12^i$$
(clearly there is no $0,1,3$ and $9$) where all the exponents are nonnegative and less than $4$ and at least one is positive. Of course, we can reduce this even more:
$$N\equiv 2^{a+2b+d+3f+g+2i} 3^{d+i} 5^{c+g}7^e11^h$$
or $$N\equiv 2^x 3^y 5^z(-6)^e(-2)^h$$
where or exponents are nonegative and less than $4.$ Again we can assume if $y=0$ (else we divide $N$ by $3$ modulo $13$). So we can write:$$ N\equiv (-1)^{3z+e+h} 2^{x+3z+e+h} = (-1)^t2^{x+t}$$
where $t= 3z+e+h$. So $$N\equiv -1\pm2,\pm4,\pm8$$
Now this can not be of the form $(...)^4$ and we have a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3612556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
} |
An A level question about partial differentiation
The equation of a curve is $2x^4+xy^3+y^4=10$. Show that $$\frac{dy}{dx}=-\frac{8x^3+y^3}{3xy^2+4y^3}.$$
I understand that you are to work out:
\begin{align}
\frac{dz}{dx} = 8x^3 + y^3\\
\frac{dz}{dy} = 3xy^2 + 4y^3
\end{align}
and therefore, $\dfrac{dy}{dx}$.
My answer almost matches the requirement, but I don't understand how the negative symbol got there.
| $$2x^4+xy^3+y^4=10$$
D.w.r.t.$x$
we get $$ 8x^
3+y^3+3xy^2y'+4y^3 y'=0 \implies \frac{dy}{dx}=y'= -\frac{8x^3+y^3}{3xy^2+4y^3}$$
next you have actually found partial derivatives $\frac{\partial z }{\partial x}$ and
$\frac{\partial z }{\partial y}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3614462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
How does $\frac {1} {a^n}$ compare to $\frac {1} {b^n}$ when $a>b$ and $n>0$? To put it briefly, my question is : suppose $a>b$ and $n>0$, how does $\frac {1} {a^n}$ compare to $\frac {1} {b^n}$ ?
I have considered various cases without arriving at finding a general rule.
In view of deriving the order relation between $\frac {1}{a^n}$ and $\frac {1}{b^n}$ in each case, I use this rule : let a given ordering relation ( greater than/ less than) hold between $N$ and $M$, then
*
*if $N$ and $M$ have the same sign, their ( multiplicative) inverses have the reverse order
*if $N$ and $M$ have opposite signs, then, the ( multiplicative ) inverses preserve the order.
I apply this rule to the $n$th power of $a$ and of $b$, previously ordered in each case.
My "strategy" was as follows: (1) first determining the order relation of the $n$th powers, and then (1) deriving from this the order relation of the inverses of the $n$th powers. But finally, what I end up with is a mess.
I managed to find a sort of rule for the $n$th powers, but not for their inverses. The rule for $n$th powers was as follows :
"In case a> b , and n > 0 , then $n$th-powers conserve the order, that is , $a^n > b^n$, except when $n$ is even and either (1) $a$ and $b$ are both negative , or (2) $a$ and $b$ have different signs and $a$ is smaller than $b$ in absolute value."
If there a way to find a general rule for the cases distinguished below.
| If $n>0$ is an integer, this is $n\in\mathbb{Z}^+$, and $a>b$ are any two real numbers, then we have even nine different cases:
*
*$0<b<a$;
*$b<0<a$, $|b|<|a|$ and $n$ is even;
*$b<0<a$, $|b|<|a|$ and $n$ is odd;
*$b<0<a$, $|b|=|a|$ and $n$ is even;
*$b<0<a$, $|b|=|a|$ and $n$ is odd;
*$b<0<a$, $|b|>|a|$ and $n$ is even;
*$b<0<a$, $|b|>|a|$ and $n$ is odd;
*$b<a<0$ and $n$ is even;
*$b<a<0$ and $n$ is odd.
What happens then?
*
*$\Rightarrow\ {0<b^n<a^n}\ \Rightarrow\ {\frac{1}{b^n}>\frac{1}{a^n}>0}$.
*$\Rightarrow\ {0<b^n<a^n}\ \Rightarrow\ {\frac{1}{b^n}>\frac{1}{a^n}>0}$.
*$\Rightarrow\ {b^n<0<a^n}\ \Rightarrow\ {\frac{1}{b^n}<0<\frac{1}{a^n}}$.
*$\Rightarrow\ {b^n=a^n>0}\ \Rightarrow\ {\frac{1}{b^n}=\frac{1}{a^n}>0}$.
*$\Rightarrow\ {b^n<0<a^n}\ \Rightarrow\ {\frac{1}{b^n}<0<\frac{1}{a^n}}$.
*$\Rightarrow\ {b^n>a^n>0}\ \Rightarrow\ {0<\frac{1}{b^n}<\frac{1}{a^n}}$.
*$\Rightarrow\ {b^n<0<a^n}\ \Rightarrow\ {\frac{1}{b^n}<0<\frac{1}{a^n}}$.
*$\Rightarrow\ {b^n>a^n>0}\ \Rightarrow\ {0<\frac{1}{b^n}<\frac{1}{a^n}}$.
*$\Rightarrow\ {b^n<a^n<0}\ \Rightarrow\ {0>\frac{1}{b^n}>\frac{1}{a^n}}$.
So be careful because
*
*$n$ even or odd matters only if at least one between $a$ and $b$ is smaller than $0$;
*the signs of $a$ and $b$ matter and, if they are different ($b<0<a$), the absolute values also matter!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3614621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Trouble with $4\times4$ matrix determinant $$
\begin{vmatrix}
1 & -6 & 7 & 5 \\
0 & 0 & 3 & 0 \\
3 & -2 & -8 & 6\\
2 & 0 & 5 & 4\\
\end{vmatrix}
$$
Clearly I want to expand along the second row yielding:
$((-1)^5)3$ times the following matrix
$$
\begin{vmatrix}
1 & -6 & 5 \\
3 & -2 & 6 \\
2 & 0 & 4 \\
\end{vmatrix}
$$
and then breaks down into several smaller matrices:
2 times
$$
\begin{vmatrix}
-6 & 5 \\
-2 & 6 \\
\end{vmatrix}
$$
and 4 times
$$
\begin{vmatrix}
1 & -6 \\
3 & -2 \\
\end{vmatrix}
$$
which should come out to be $-3[(2(-36+10))+(4(-2+18))]$
$-3[(2(-16))+(4(16))]$
$-3(-32+64)=32 \times -3$
but the answer is -36 I don't know what went wrong?
| Your computations are almost correct. But it turns out that $-36+10=-26$, not $-16$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3614807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Question about the basis of an intersection of subspaces
Let $ U=\left\langle\left(\begin{array}{c}-5 \\ 2 \\ 0\end{array}\right),\left(\begin{array}{c}-2 \\ 2 \\ -1\end{array}\right)\right\rangle $ and $ W=\left\langle\left(\begin{array}{c}-3 \\ 0 \\ 1\end{array}\right),\left(\begin{array}{c}-1 \\ -2 \\ -2\end{array}\right)\right\rangle
$ be vector subspaces of $ \mathbb{R}^{3} $.
a) Determine the basis and dimension of $U, W$ and $U\cap W$.
...
Solution:
$ B_{U}=\left\{\left(\begin{array}{c}-5 \\ 2 \\ 0\end{array}\right),\left(\begin{array}{c}-2 \\ 2 \\ -1\end{array}\right)\right\} \implies \dim(U)=2$
$ B_{W}=\left\{\left(\begin{array}{c}-3 \\ 0 \\ 1\end{array}\right),\left(\begin{array}{c}-1 \\ -2 \\ -2\end{array}\right)\right\} \implies \dim(W)=2$
$ B_{U\cap W}=\left\{\left(\begin{array}{c}-3 \\ 0 \\ 1\end{array}\right)\right\} \implies \dim(U\cap W)=1$
My question: how do I get to $B_{U\cap W}$? I do not understand the solution. Could someone please explain?
| Hint :
You write the equation :
$$\lambda_{1}\begin{pmatrix}-5 \\ 2 \\ 0\end{pmatrix}+\lambda_{2}\begin{pmatrix}-2 \\ 2 \\ -1\end{pmatrix}
= \gamma_{1}\begin{pmatrix}-3 \\ 0 \\ 1\end{pmatrix} +\gamma_{2}\begin{pmatrix}-1 \\ -2 \\ -2\end{pmatrix}$$
And after you get a very simple linear system
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3616968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Efficient computation of $\sum_{i=1}^{i=\left \lfloor {\sqrt{N}} \right \rfloor}\left \lfloor \frac{N}{i^{2}} \right \rfloor$ I have tried to find a closed form but did not succeed but is there an efficient way to calculate the following expression
$\sum_{i=1}^{i=\left \lfloor {\sqrt{N}} \right \rfloor}\left \lfloor \frac{N}{i^{2}} \right \rfloor$
So Far I have noticed the following
$\sum_{i=1}^{i=\left \lfloor {\sqrt{N}} \right \rfloor}\left \lfloor \frac{N}{i^{2}} \right \rfloor = \sum_{i=1}^{i=\left \lfloor {\sqrt{N}} \right \rfloor}\left \lfloor \frac{N-N \mod i^{2}}{i^{2}} \right \rfloor = N *\left \lfloor {\sqrt{N}} \right \rfloor - \sum_{i=1}^{i=\left \lfloor {\sqrt{N}} \right \rfloor} N \mod i^{2}$
I want to solve it in either log(N) or as a closed form
| One approach to improving the efficiency of calculating this would be to take
$$\sum_{i=1}^{\infty}\left\lfloor\frac{N}{i^2}\right\rfloor$$ and ask how many times is the summand a $1$? How many times is it a $2$? And so on. Keep reading to the end, and this reduces the computation from $O(n^{1/2})$ time to $O(n^{1/3})$ time.
$\left\lfloor\frac{N}{i^2}\right\rfloor=1$ whenever $1\leq\frac{N}{i^2}<2$. So whenever $\sqrt{N}\geq i>\sqrt{\frac{N}{2}}$. There are $\left\lfloor\sqrt{N}\right\rfloor-\left\lfloor\sqrt{\frac{N}{2}}\right\rfloor$ such values of $i$.
So the sum is the same as $$\sum_{i=1}^{\lfloor\sqrt{N/2}\rfloor}\left\lfloor\frac{N}{i^2}\right\rfloor+1\cdot\left(\left\lfloor\sqrt{N}\right\rfloor-\left\lfloor\sqrt{\frac{N}{2}}\right\rfloor\right)$$
The original expression has $\lfloor\sqrt{N}\rfloor$ nonzero terms. Now it is written with $\lfloor\sqrt{N/2}\rfloor+2$ terms, which is an improvement if $N$ is at least $64$. You could continue like this, counting how many times $2$ appears in the original sum.
$\left\lfloor\frac{N}{i^2}\right\rfloor=2$ whenever $2\leq\frac{N}{i^2}<3$. So whenever $\sqrt{\frac{N}{2}}\geq i>\sqrt{\frac{N}{3}}$. There are $\left\lfloor\sqrt{\frac{N}{2}}\right\rfloor-\left\lfloor\sqrt{\frac{N}{3}}\right\rfloor$ such values of $i$.
So the sum is the same as $$\sum_{i=1}^{\lfloor\sqrt{N/3}\rfloor}\left\lfloor\frac{N}{i^2}\right\rfloor+1\cdot\left(\left\lfloor\sqrt{N}\right\rfloor-\left\lfloor\sqrt{\frac{N}{2}}\right\rfloor\right)+2\cdot\left(\left\lfloor\sqrt{\frac{N}{2}}\right\rfloor-\left\lfloor\sqrt{\frac{N}{3}}\right\rfloor\right)$$
$$=\sum_{i=1}^{\lfloor\sqrt{N/3}\rfloor}\left\lfloor\frac{N}{i^2}\right\rfloor+\left\lfloor\sqrt{N}\right\rfloor+\left\lfloor\sqrt{\frac{N}{2}}\right\rfloor-2\left\lfloor\sqrt{\frac{N}{3}}\right\rfloor$$
Now there are $\lfloor\sqrt{N/3}\rfloor+3$ terms, which is an improvement over the previous version if $N$ is at least $72$. Keep going like this $M$ iterations, and the sum is equal to $$\sum_{i=1}^{\left\lfloor\sqrt{N/(M+1)}\right\rfloor}\left\lfloor\frac{N}{i^2}\right\rfloor+\sum_{j=1}^M\left\lfloor\sqrt{\frac{N}{j}}\right\rfloor-M\left\lfloor\sqrt{\frac{N}{M+1}}\right\rfloor$$ which is a sum with $\left\lfloor\sqrt{\frac{N}{M+1}}\right\rfloor+M+1$ terms, each of which has roughly the same computational complexity as the terms in the original sum. For a given $N$, there is an $M$ that minimizes this count of summands. If we ignore the floor function, calculus optimization leads us to $M\approx(N/4)^{1/3}$. And using that value for $M$, the number of terms in the summation is $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{4}}\right)N^{1/3}$. That would be a noteworthy improvement over the original summand count of $\sqrt{N}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3618219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Sum of squared Fresnel sine integral I'm trying to find the following sum:
$$ \sum_{n=0}^{\infty} \frac{S\left(\sqrt{2n}\right)^2}{n^3}$$
where $S(n)$ is the fresnel sine integral, however, I think I made a mistake somewhere.
To start, I considered using parseval's identity:
$$ 2\pi\sum_{n=-\infty}^{\infty} |c_n|^2 = \int_{-\pi}^{\pi} |f(x)|^2 \space dx$$
where $f(x)$ is:
$$ f(x) = \sqrt{\frac{ix}{2}} + \sqrt{\frac{-ix}{2}} $$
$c_n$ becomes:
$$c_n = \frac{1}{2\pi}\int_{-\pi}^{\pi} \left(\sqrt{\frac{ix}{2}} + \sqrt{\frac{-ix}{2}} \right ) e^{-inx}\space dx $$
This integral is complicated, so I plugged it into wolfram alpha and found that
$$ c_n = \frac{1}{2\pi} \left(-\sqrt{2\pi} \cdot \frac{S\left(\sqrt{2n}\right)}{n^{3/2}} \right)$$
so,
$$ |c_n|^2 = \frac{1}{4\pi^2} \left(2\pi \cdot \frac{S\left(\sqrt{2n}\right)^2}{n^{3}} \right) = \frac{1}{2\pi} \left(\frac{S\left(\sqrt{2n}\right)^2}{n^{3}} \right)$$
I think $|c_n|^2$ is finite for all n and is an even function of n. If this is true, then parseval's identity gives:
$$ 2\pi\sum_{n=-\infty}^{\infty} \frac{1}{2\pi} \left(\frac{S\left(\sqrt{2n}\right)^2}{n^{3}} \right) = \int_{-\pi}^{\pi} \left|\sqrt{\frac{ix}{2}} + \sqrt{\frac{-ix}{2}}\right|^2 \space dx$$
and if $|c_n|^2$ is even then this expression becomes:
$$ 2\sum_{n=0}^{\infty} \left(\frac{S\left(\sqrt{2n}\right)^2}{n^{3}} \right) = \int_{-\pi}^{\pi} \left|\sqrt{\frac{ix}{2}} + \sqrt{\frac{-ix}{2}}\right|^2 \space dx$$
I believe that
$$ \int_{-\pi}^{\pi} \left|\sqrt{\frac{ix}{2}} + \sqrt{\frac{-ix}{2}}\right|^2 \space dx = \int_{-\pi}^{\pi} \left(\sqrt{\frac{ix}{2}} + \sqrt{\frac{-ix}{2}}\right)^2 \space dx$$
and if I plug in the second integral into wolfram alpha again, I find that (EDIT user Claude Leibovici correctly found that):
$$ \int_{-\pi}^{\pi} \left(\sqrt{\frac{ix}{2}} + \sqrt{\frac{-ix}{2}}\right)^2 \space dx = \pi^2$$
So, in total I have:
$$ 2\sum_{n=0}^{\infty} \left(\frac{S\left(\sqrt{2n}\right)^2}{n^{3}} \right) = \pi^2$$
or
$$ \sum_{n=0}^{\infty} \left(\frac{S\left(\sqrt{2n}\right)^2}{n^{3}} \right) = \frac{\pi^2}{2}$$
The problem is that wolfram alpha suggests that the sum approaches .549, but my answer is ~4.93. Where did I make a mistake?
| With help from reddit user GamblingTheory the solution is as follows:
let
$$ f(x) = -\sqrt{2\pi} * \left( \sqrt{\frac{ix}{2}} + \sqrt{-\frac{ix}{2}} - \sqrt{\pi}\right)$$
thus,
$$ c_n = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) * e^{-inx} \space dx = \frac{S({\sqrt{2n}})}{n^{3/2}}$$
Using Parseval's Identity:
$$\sum_{-\infty}^{\infty}|c_n|^2 = \sum_{-\infty}^{\infty} \frac{\left(S({\sqrt{2n}})\right)^2}{n^{3}} = \int_{-\pi}^{\pi} \left|\sqrt{\frac{ix}{2}} + \sqrt{\frac{-ix}{2}} -\sqrt{\pi}\right|^2 \space dx = \int_{-\pi}^{\pi} \left(\sqrt{\frac{ix}{2}} + \sqrt{\frac{-ix}{2}} - \sqrt{\pi}\right)^2 \space dx = \frac{\pi^{2}}{3}$$
We rewrite the sum like so:
$$\sum_{-\infty}^{\infty} \frac{\left(S({\sqrt{2n}})\right)^2}{n^{3}} = 2\sum_{1}^{\infty} \frac{\left(S({\sqrt{2n}})\right)^2}{n^{3}} + \frac{2\pi^2}{9}$$
thus
$$2\sum_{1}^{\infty} \frac{\left(S({\sqrt{2n}})\right)^2}{n^{3}} + \frac{2\pi^2}{9} = \frac{\pi^{2}}{3}$$
$$\sum_{1}^{\infty} \frac{\left(S({\sqrt{2n}})\right)^2}{n^{3}} = \frac{\pi^{2}}{6} - \frac{\pi^2}{9} = \frac{\pi^2}{18}$$
I'm not sure if this sum is useful in anyway, but I thought it was a fun sum to compute especially because it is reminiscent of:
$$ \sum_{1}^{\infty} \frac{1}{n^{3}} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3621209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to make sense of a weird Summation? I have been toying with some sums, as I saw that $123456789\div987654312 = \frac{1}{8}$. Which led me to wonder if this can be generalized to other bases. I.E. in base 4 it would be $(1*4^{2}+2*4^{1}+3*4^{0})\div(3*4^{2}+1*4^{1}+2*4^{0})=\frac{1}{2}$. So this led me to the generalization of, if B is base $$\left(\sum^{B-1}_{i=0}(B-1-i)*B^{i}\right)\div\left(\left(\sum^{B-2}_{i=2}(i+1)*B^{i}\right)+B + 2\right)=\frac{1}{B-2}$$
I've tested many values on my calculator and I've confirmed for B = 3 to B = 52. If anyone smarted than me could either explain this to me or at least point me a direction to understand.
| (This has hung around unanswered for more than long enough.)
Iβll rewrite the equation like this
$$\frac{\sum\limits_{i=0}^{b-1}(b-1-i)b^i}{b+2+\sum\limits_{i=2}^{b-2}(i+1)b^i}=\frac1{b-2}$$
and (as MPW suggested in the comments) get rid of the fractions to produce the following equation, which is equivalent for $b\ne 2$:
$$(b-2)\sum_{i=0}^{b-1}(b-1-i)b^i=b+2+\sum_{i=2}^{b-2}(i+1)b^i\,.\tag{1}$$
We can rewrite the lefthand side as
$$\begin{align*}
(b-2)\sum_{i=0}^{b-1}(b-1-i)b^i&=(b-2)\sum_{i=1}^bb^i-(b-2)\sum_{i=0}^{b-1}(i+1)b^i\\
&=(b-2)\frac{b^{b+1}-b}{b-1}-(b-2)\sum_{i=0}^{b-1}(i+1)b^i\\
&=\frac{b(b-2)(b^b-1)}{b-1}-(b-2)\sum_{i=0}^{b-1}(i+1)b^i
\end{align*}$$
and the righthand side as
$$\begin{align*}
b+2+\sum_{i=2}^{b-2}(i+1)b^i&=b+2+\sum_{i=0}^{b-1}(i+1)b^i-(1+2b+b^b)\\
&=1-b-b^b+\sum_{i=0}^{b-1}(i+1)b^i\,,
\end{align*}$$
to make the summations match. That will allow us to combine them on one side of the equation, transforming $(1)$ into the equivalent equation
$$\begin{align*}
\frac{b(b-2)(b^b-1)}{b-1}&=1-b-b^b+\sum_{i=0}^{b-1}(i+1)b^i+(b-2)\sum_{i=0}^{b-1}(i+1)b^i\\
&=1-b-b^b+(b-1)\sum_{i=0}^{b-1}(i+1)b^i\,.\tag{2}
\end{align*}$$
There are a number of ways to evaluate that summation; here is a very elementary one that requires only knowing how to reverse the order of a double summation:
$$\begin{align*}
\sum_{i=0}^{b-1}(i+1)b^i&=\sum_{i=0}^{b-1}\left(\sum_{j=0}^i1\right)b^i\\
&=\sum_{i=0}^{b-1}\sum_{j=0}^ib^i\\
&=\sum_{j=0}^{b-1}\sum_{i=j}^{b-1}b^i\\
&=\sum_{j=0}^{b-1}\frac{b^b-b^j}{b-1}\\
&=\frac1{b-1}\left(b^{b+1}-\sum_{j=0}^{b-1}b^j\right)\\
&=\frac1{b-1}\left(b^{b+1}-\frac{b^b-1}{b-1}\right)\,.
\end{align*}$$
We can now simplify $(2)$ to
$$\frac{b(b-2)(b^b-1)}{b-1}=1-b-b^b+b^{b+1}-\frac{b^b-1}{b-1}$$
and thence, after multiplication by $b-1$, to
$$b(b-2)(b^b-1)=2b-b^2-2b^{b+1}+b^{b+2}\,;$$
this is still equivalent to $(1)$, since $b\ne 0$, and is easily verified.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3623567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Finding det(A) with standard basis vectors For example, let $e_1 = [1 \quad 0]$ and $e_2 = [0 \quad 1] $be standard basis vectors. A is a $2 \times 2$ matrix. $Ae_1 = [-3\quad 7]$ and $Ae2 = [3 \quad 5] $
How do I find the $\det(A)$?
| Let $a,b,c,d \in \mathbb{R}$, then
\begin{align}
&A = \begin{bmatrix}
a&b \\ c&d
\end{bmatrix}\begin{bmatrix}
1 \\ 0
\end{bmatrix} = \begin{bmatrix}
-3 \\ 7
\end{bmatrix} \implies a = -3, \quad \text{and} \quad c = 7 \\
&A = \begin{bmatrix}
a&b \\ c&d
\end{bmatrix}\begin{bmatrix}
0 \\ 1
\end{bmatrix} = \begin{bmatrix}
3 \\ 5
\end{bmatrix} \implies b = 3, \quad \text{and} \quad d = 5.\\
& \qquad\text{therefore, }\det \begin{vmatrix}
A
\end{vmatrix} = \det\begin{vmatrix}
-3 & 3 \\ 7 & 5
\end{vmatrix} = -15-21 = -36.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3624245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Explicit formula for orthonormalized vectors (using Gram-Schmidt) For $n\in\mathbb{N}$, let
$$
B(n):=\{b_1,b_2,\ldots,b_n\}:=\left\{\begin{pmatrix}1\\2\\3\\4\\\vdots\\n\end{pmatrix},\begin{pmatrix}2\\-1\\0\\0\\\vdots\\0\end{pmatrix},\begin{pmatrix}3\\0\\-1\\0\\\vdots\\0\end{pmatrix},\begin{pmatrix}4\\0\\0\\-1\\\vdots\\0\end{pmatrix},\ldots,\begin{pmatrix}n\\0\\0\\0\\\vdots\\-1\end{pmatrix}\right\}.
$$
Now I would like to orthogonalize (by Gram-Schmidt) and then normalize $B(n)$.
Question: Is it possible to determine the orthonormalized vectors explicitly?
I write the vectors $b_k\in B(n)~(k=1,2,\ldots,n)$ as
$$
b_k=(b_{k,1},b_{k,2},\ldots,b_{k,k},b_{k,k+1},\ldots,b_{k,n})^T.
$$
where $b_{k,1}=k$ and $b_{k,k}=-1$ while $b_{k,i}=0$ otherwise. Hence, for $k,j\geq 2$ and $k\neq j$, I note that $b_k\cdot b_j= kj$ and $b_k\cdot b_k=k^2+1$.
Moreoever, since $b_k\cdot b_1=0$ for all $k\geq 2$, the orthogonalized vectors, which I denote by $C(n):=\{c_1,c_2,\ldots,c_n\}$, are
$$
\begin{align*}
c_1&:=b_1,\\
c_2&:=b_2,\\
c_k&:=b_k-\frac{2k}{5}b_2-\sum_{i=3}^{k-1}\frac{c_i\cdot b_k}{c_i\cdot c_i}c_i,\quad k\geq 3
\end{align*}
$$
And if we normalize, $d_k:=\frac{c_k}{\sqrt{c_k\cdot c_k}}$ one has the desired set $D(n):=\{d_1,d_2,\ldots,d_n\}$ of orthonormalized vectors.
For the first three vectors, what I get is:
$$
\begin{align*}
d_1&=\frac{b_1}{\sqrt{b_1\cdot b_1}}=\frac{b_1}{\sqrt{\frac{1}{6}(n(n+1)(2n+1))}}=\frac{1}{\sqrt{\frac{n(n+1)(2n+1)}{6}}}\begin{pmatrix}1\\2\\3\\\vdots\\n\end{pmatrix}\\
d_2&=\frac{b_2}{\sqrt{b_2\cdot b_2}}=\frac{b_2}{\sqrt{5}}=\begin{pmatrix}\frac{2}{\sqrt{5}}\\-\frac{1}{\sqrt{5}}\\0\\\vdots\\0\end{pmatrix}\\
d_3&=\frac{5}{\sqrt{70}}c_3,\quad c_3=b_3-\frac{6}{5}b_2=\begin{pmatrix}\frac{3}{5}\\\frac{6}{5}\\-1\\0\\\vdots\\0\end{pmatrix}\\
d_4&=\frac{c_4}{\sqrt{c_4\cdot c_4}},\quad c_4=b_4-\frac{8}{5}b_2-\frac{c_3\cdot b_4}{c_3\cdot c_3}c_3=?
\end{align*}
$$
But I don't see a way to reach explicit formulae for $d_j~(j=1,2,3,\ldots,n)$ in general.
| It seems that
$$v_k=(1,2,3,\ldots,k-1,-x_k,0,\ldots,0)$$
works for constants $x_k$ that are easy to work out by dotting with $b_1$. Then normalize each $v_k$. If $v_k$ is normal to $v_1$ then it is normal to $v_h$ for all $h\gt k$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3624886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\lim\limits_{n\to\infty}\frac1{n^4}Β·\left[1Β·\sum\limits_{k=1}^nk+2Β·\sum\limits_{k=1}^{n-1}k+3Β·\sum\limits_{k=1}^{n-2}k+\cdots+nΒ·1\right]$
Find the value of$$L=\lim_{n\to\infty}\frac1{n^4}\cdot\left[1\cdot\sum_{k=1}^nk+2\cdot\sum_{k=1}^{n-1}k+3\cdot\sum_{k=1}^{n-2}k+\cdots+n\cdot1\right].$$
My attempt is as follows:
$$S=\sum_{i=1}^{n}i\sum_{j=1}^{n-i+1}j$$
$$S=\sum_{i=1}^{n}i\cdot\dfrac{(n-i+1)(n-i+2)}{2}$$
$$S=\dfrac{1}{2}\sum_{i=1}^{n}i\cdot(n^2-ni+2n-in+i^2-2i+n-i+2)$$
$$S=\dfrac{1}{2}\sum_{i=1}^{n}i\cdot(n^2+i^2-2ni-3i+3n+2)$$
$$S=\dfrac{1}{2}\left(\dfrac{n^3(n+1)}{2}+\dfrac{n^2(n+1)^2}{4}-\dfrac{n(n+1)(2n+1)(2n+3)}{6}+\dfrac{n(n+1)(3n+2)}{2}\right)$$
$$S=\dfrac{n(n+1)}{4}\left(n^2+\dfrac{n(n+1)}{2}-\dfrac{(2n+1)(2n+3)}{3}+3n+2\right)$$
$$S=\dfrac{n(n+1)}{4}\left(\dfrac{6n^2+3n(n+1)-2(2n+1)(2n+3)+6(3n+2)}{6}\right)$$
$$S=\dfrac{n(n+1)}{4}\left(\dfrac{9n^2+3n-2(4n^2+8n+3)+18n+12}{6}\right)$$
$$S=\dfrac{n(n+1)}{4}\left(\dfrac{n^2+5n+6}{6}\right)$$
$$S=\dfrac{n(n+1)(n+2)(n+3)}{24}$$
$$L=\lim_{n\to\infty}\frac1{24}{\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{2}{n}\right)\left(1+\dfrac{3}{n}\right)}$$
hence answer is $\dfrac{1}{24}$.
But is there any shorter way to do this as in the last we got $S=\dfrac1{24}n(n+1)(n+2)(n+3)$, which is a nice closed expression.
| Let $$S_n=1\cdot\sum_{k=1}^nk+2\cdot\sum_{k=1}^{n-1}k+3\cdot\sum_{k=1}^{n-2}k+\cdots+n\cdot1.$$
$$S_1=1=\frac{1\cdot2\cdot3\cdot4}{24}.$$
Now, by the assumption of the induction we obtain:
$$S_{n+1}=S_{n}+(n+1)+2\cdot n+3\cdot(n-1)+...+(n-1)\cdot3+n\cdot2+(n+1)\cdot1=$$
$$=\frac{n(n+1)(n+2)(n+3)}{24}+\sum_{k=1}^{n+1}(n+2-k)k=$$
$$=\frac{n(n+1)(n+2)(n+3)}{24}+(n+2)\cdot\frac{(n+1)(n+2)}{2}-\frac{(n+1)(n+2)(2n+3)}{6}=$$
$$=\frac{(n+1)(n+2)}{24}(n(n+3)+12(n+2)-4(2n+3))=$$
$$=\frac{(n+1)(n+2)(n+3)(n+4)}{24}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3626127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Shortest distance from circle to a line
Let $C$ be a circle with center $(2, 1)$ and radius $2$. Find the shortest distance from the line $3y=4x+20$.
This should be very simple, but I seem to end up with no real solutions.
The shortest distance would be from the center of the circle perpendicular to the line right?
Solving the line for $y$ we get $y=\frac{4}{3}x+\frac{20}{3}$
Substituting this to the equation of the circle we get $(x-2)^2+(\frac{4}{3}x+\frac{20}{3}-1)^2=2^2$, but solving this for $x$ ended up with no real roots. What am I missing here?
| The perpendicular line to the given line has slope $=\frac{-1}{4/3}=-\frac{3}{4}$
Also the perpendicular should pass through the centre of the circle so that the shortest distance between the given line and the given circle is the distance between the given line and the point of intersection of the perpendicular with the circle. SEE THE GRAPH BELOW.
The perpendicular line is $y=-\frac{4}{3}x+\frac{5}{2}$, this intersects with the circle at the point $(\frac{2}{5},\frac{11}{5})$.
The distance between the given line and $(\frac{2}{5},\frac{11}{5})$ is given by:
$$d=\frac{|[3y-4x-20]_{x=2/5,y=11/5}|}{\sqrt{3^2+(-4)^2}}$$
$$=\frac{|\frac{33}{5}-\frac{8}{5}-20|}{\sqrt{25}}=\frac{|-15|}{5}=\frac{15}{5}=3 \text{ units}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3626410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 3
} |
Find linear mapping in basis Let A:
\begin{bmatrix}0&0&0\\0&0&1\\1&2&3\end{bmatrix}
be a matrix associated with the linear mapping T in the basis B={(1,1,1), (0,2,2), (0,0,3)}.
Find the standard matrix for T.
If I'm not misunderstanding:
We want to find the transformation T in the standard basis. I.e. what the transformation would be with the following basis vectors:
\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}
Currently the transformation T has done a transformation T on A under the basis B.
I'm not sure how think about it.
| Here's how to start thinking about it. I'll use $\mathcal B$ to denote the coordinate representation in the basis $\mathcal B = \left\{(1,1,1),(0,2,2),(0,0,3) \right\}$ and $\mathcal S$ to denote the coordinate representation in the standard basis. The transformation matrix for the basis $\mathcal B$ is given by
$$A = \left[ \begin{array}{ccc} 0&0&0 \\ 0&0&1 \\ 1&2&3 \end{array} \right].$$
This tells you that the first basis vector in $\mathcal B$ is mapped to the third basis vector in $\mathcal B$, the second is mapped to two times the third, and so on. That is,
$\left[ \begin{array}{ccc} 0&0&0 \\ 0&0&1 \\ 1&2&3 \end{array} \right]\left[\begin{array}{c}1\\0\\0 \end{array} \right]_{\mathcal B} = \left[\begin{array}{c}0\\0\\1 \end{array} \right]_{\mathcal B} = (1)\left[\begin{array}{c}0\\0\\3 \end{array} \right]_{\mathcal S},$
$\left[ \begin{array}{ccc} 0&0&0 \\ 0&0&1 \\ 1&2&3 \end{array} \right]\left[\begin{array}{c}0\\1\\0 \end{array} \right]_{\mathcal B} = \left[\begin{array}{c}0\\0\\2 \end{array} \right]_{\mathcal B} = (2)\left[\begin{array}{c}0\\0\\3 \end{array} \right]_{\mathcal S}, \text{ and }$
$\left[ \begin{array}{ccc} 0&0&0 \\ 0&0&1 \\ 1&2&3 \end{array} \right]\left[\begin{array}{c}0\\0\\1 \end{array} \right]_{\mathcal B} = \left[\begin{array}{c}0\\1\\3 \end{array} \right]_{\mathcal B} = (1)\left[\begin{array}{c}0\\2\\2 \end{array} \right]_{\mathcal S} + (3)\left[\begin{array}{c}0\\0\\3 \end{array} \right]_{\mathcal S}.$
So you see that in the standard basis
\begin{equation}
\begin{split}
T \text{ maps } & (1,1,1) \text{ to } (0,0,3), \\
T \text{ maps } & (0,2,2) \text{ to } (0,0,6), \\
T \text{ maps } & (0,0,3) \text{ to } (0,2,11).
\end{split}
\end{equation}
Now see if you can use this information to construct the matrix for $T$ in the standard basis.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3627805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding a line integral of a vector field I'm working on the following:
For $k \in \mathbb{N}$ let the oriented piecewise $C^1$ curve $\gamma_k$ be parametrised by
$$\gamma_k(\theta):=\left ((k+1)\cos\left(\frac{\theta}{k+1}\right)β\cos(\theta),(k+1)\sin\left (\frac{\theta}{k+1}\right)β\sin(\theta) \right )$$
with $ \theta \in[0, 2(k+1)\pi]$. Then let
$$g := \left(\frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2}\right)$$
and find $$\int_{\gamma_k}g\cdot ds$$
for all $k \in \mathbb{N}$.
I tried letting $x(\theta) = (k+1)cos(\frac{ΞΈ}{k+1})βcos(ΞΈ)$, giving a $dx = sin(\theta) - sin(\frac{\theta}{k+1})$ and something similar, but with cosine for $y(\theta)$. Then I plugged them into $g$ and dotted the resulting vector with $\langle dx, dy \rangle$, but that results in a pretty ugly integrand. Is there a better way to approach this?
| This is a hefty computation and it needs to be done step-by-step. First notice that in evaluating $g(\gamma(\theta))$ you need to compute $x^2 + y^2$:
\begin{eqnarray*}
x^2 + y^2 & = & (k+1)^2\cos^2\left (\frac{\theta}{k+1}\right ) - 2(k+1)\cos\theta\cos\left (\frac{\theta}{k+1}\right ) + \cos^2\theta \\
&& + (k+1)^2\sin^2\left (\frac{\theta}{k+1}\right ) - 2(k+1)\sin\theta\sin\left (\frac{\theta}{k+1}\right ) + \sin^2\theta \\
& = & 1 + (k+1)^2 - 2(k+1)\left [\cos\theta\cos\left (\frac{\theta}{k+1}\right ) + \sin\theta\sin\left (\frac{\theta}{k+1}\right )\right ] \\
& = & 1 + (k+1)^2 - 2(k+1)\cos\left (\frac{k\theta}{k+1}\right )
\end{eqnarray*}
where we use the identity $\cos(\alpha\pm \beta) = \cos\alpha\cos\beta \mp \sin\alpha\sin\beta$. This means that
$$
g(\gamma(\theta)) \;\; =\;\; \frac{\left (\sin\theta - (k+1)\sin\left (\frac{\theta}{k+1}\right ), \; (k+1)\cos\left (\frac{\theta}{k+1}\right ) - \cos\theta \right )}{1 + (k+1)^2 - 2(k+1)\cos\left (\frac{k\theta}{k+1}\right )}.
$$
Similarly, we can compute $d\gamma$ as
$$
d\gamma(\theta) \;\; =\;\; \left (\sin\theta - \sin\left (\frac{\theta}{k+1}\right ), \; \cos\left (\frac{\theta}{k+1}\right ) - \cos\theta\right ).
$$
Computing the line integral goes as follows:
\begin{eqnarray*}
\int_\gamma g\cdot ds & = & \int g(\gamma(\theta))\cdot \gamma'(\theta)d\theta \\
& = & \int_0^{2(k+1)\pi} \frac{\sin^2\theta + (k+1)\sin^2\left (\frac{\theta}{k+1}\right ) + \cos^2\theta + (k+1)\cos^2\left (\frac{\theta}{k+1}\right ) - 2(k+1)\left [\cos\theta\cos\left (\frac{\theta}{k+1}\right ) + \sin\theta\sin\left (\frac{\theta}{k+1}\right ) \right ]}{1 + (k+1)^2 - 2(k+1)\cos\left (\frac{k\theta}{k+1}\right )} d\theta \\
& = & \int_0^{2(k+1)\pi} \frac{1 + (k+1) - 2(k+1)\cos\left (\frac{k\theta}{k+1}\right )}{1 + (k+1)^2 - 2(k+1)\cos\left (\frac{k\theta}{k+1}\right )} d\theta \\
& = & \int_0^{2(k+1)\pi} \frac{-k + 2(k+1)\left [ 1 - \cos\left (\frac{k\theta}{k+1}\right ) \right ]}{k^2 + 2(k+1)\left [1 - \cos\left (\frac{k\theta}{k+1}\right )\right ]} d\theta \\
& = & \int_0^{2(k+1)\pi} 1 - \frac{k(k+1)}{k^2 + 2(k+1)\left [1 - \cos\left (\frac{k\theta}{k+1}\right )\right ]} d\theta \\
& = & 2(k+1)\pi - \int_0^{2(k+1)\pi} \frac{k(k+1)}{k^2 + 2(k+1)\left [1 - \cos\left (\frac{k\theta}{k+1}\right )\right ]} d\theta.
\end{eqnarray*}
I'm not going to bother computing this integral by hand. But using Wolfram we see that we obtain an integral of the type:
$$
\int \frac{1}{k^2 + 2(k+1)\left (1-\cos \left (\frac{k\theta}{k+1}\right ) \right )}d\theta \;\; =\;\; \frac{2(k+1)\tan^{-1}\left (\frac{(k+2)\tan\left (\frac{k\theta}{2(k+1)}\right ) }{k} \right ) }{k^2(k+2)} \\
$$
While evaluating this integral at the endpoints would yield zero, this would not be correct. What we would like to do instead is to observe the behavior of the function and see if we can find another way of computing the integral. Observe the graph of the function for $k=1$ and also the lines at $x = 4\pi$ and $x = 2\pi$.
Notice that by the periodicity of the graph, we can bypass evaluating at the endpoints by taking $2k$-times the abstract expression and instead integrate between $0$ and $\frac{(k+1)\pi}{k}$ (thankfully, this behavior scales with $k$, hence it will work for all values). What we compute is the following:
\begin{eqnarray*}
\int_0^{2(k+1)\pi} \frac{1}{k^2 + 2(k+1)(1-\cos u)}du & = & \int_0^{\frac{(k+1)\pi}{k}} \frac{2k}{k^2 + 2(k+1)(1-\cos u)}du \\
& = & \left . \frac{4k(k+1)\tan^{-1}\left (\frac{(k+2)\tan\left (\frac{kx}{2(k+1)}\right ) }{k} \right ) }{k^2(k+2)} \right |_0^{\frac{(k+1)\pi}{k}} \\
& = & \frac{4(k+1)\tan^{-1}\left (\frac{(k+2)\tan\left (\frac{\pi}{2}\right ) }{k} \right ) }{k(k+2)} \\
& = & \frac{4(k+1)}{k(k+2)}\cdot \frac{\pi}{2} \\
& = & \frac{2(k+1)\pi}{k(k+2)}.
\end{eqnarray*}
Finally, we arrive at the final answer which is
\begin{eqnarray*}
\int_{\gamma_k}g\cdot ds & = & 2(k+1)\pi - \frac{2(k+1)\pi}{k(k+2)} \\
& = & 2(k+1)\pi \left [1 - \frac{1}{k(k+2)} \right ] \\
& = & 2(k+1)\pi\left [ \frac{k^2+2k-1}{k(k+2)}\right ]
\end{eqnarray*}
and finally finally:
$$
\int_{\gamma_k}g(s)\cdot ds \;\; =\;\; \frac{2\pi(k+1)(k^2+2k-1)}{k(k+2)}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3637737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Showing that $\sum_{n=1}^\infty n^{-1}\left(1+\frac{1}{2}+...\frac{1}{n}\right)^{-1}$ is divergent
How to show that $\sum_{n=1}^\infty \frac{1}{n\left(1+\frac{1}{2}+...\frac{1}{n}\right)}$ diverges?
I used Ratio test for this problem and this is the result: $$\lim_{n\to\infty} \left(1+\frac{1}{n}\right)\left(1-\frac{1}{(n+1)+\frac{(n+1)}{2}+...+1}\right)= 1$$
Then I thought using abel or dirichlet test. But I couldn't solve it.
| First, We need a lemma.
Lemma:
For $n>1$,
$$1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^n-1} < n$$
This could be checked easily with induction.
To prove the question we use Cauchy Condensation Test, and see that its equal to show $$\sum_{k=1}^{\infty}1+\frac{1}{2}+\frac{1}{3}+ ... +\frac{1}{2^k}$$ is divergent.
It can be checked easily that:
$$\frac{1}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^k}} > \frac{1}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^{k+1}-1}}$$
Now we apply our lemma for $k\geq 2$ and see:
$$\frac{1}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^k}} > \frac{1}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^k}+...+\frac{1}{2^{k+1}-1}} > k+1$$
hence:
$$\sum_{k=1}^{\infty}{\frac{1}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^k}}} = \frac{1}{1+\frac{1}{2}} + \sum_{k=2}^{\infty}{\frac{1}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^k}}} > \sum_{k=3}^{\infty}{k}$$
Since the left hand side of inequality is divergent, the other side is so. Finally by Cauchy Condensation Test we see that the we have proved the divergence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3640050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to integrate with a function in the limit $\int ^a _0 (\int_0 ^ \sqrt {a^2 - x^2} (a^2 - x^2 - y^2)dy)dx = 303$
I'm trying to solve the above question where the integral has a function as the limit.
Seeing the similarity between the limit and the variable integration, I substituted the $\sqrt {a^2 - x^2}$ with $u$.
Hence I end up with:
$\int ^a _0 (\int_0 ^ u (u^2 - y^2)dy)dx = 303$
After that, I'm not too sure how to proceed as there isn't much to go on as far as integrating $y$ with respect to $u$ is concerned.
EDIT:
Building onto the advice given:
$\int ^a _0 (\int_0 ^u (u^2 - y^2)dy)dx = 303$
$\int ^a _0 [u^2y - \cfrac{y^3}{3}]^u _0 dx = 303$
$\int ^a _0 [u^3 - \cfrac{u^3}{3}] dx = 303$
$[u^3x - \cfrac{u^3x}{3}]^a_0$ = 303
$((a^2 - a^2)^{1.5}a - \cfrac{(a^2 - a^2)^{1.5}a}{3} = 303$
EDIT1:
$\int ^a _0 (\int_0 ^\sqrt{a^2 - x^2} (a^2 - x^2 - y^2)dy)dx = 303$
$\int ^a _0 [a^2y - x^2y - \cfrac{y^3}{3}]^\sqrt{a^2 - x^2}_0 dx = 303$
$\int ^a _0 [a^2\sqrt{a^2 - x^2} - x^2\sqrt{a^2 - x^2} - \cfrac{((a^2 - x^2)^{1.5})}{3}dx = 303$
$\int ^a _0 [(a^2 - x^2)\sqrt{a^2 - x^2} - \cfrac{((a^2 - x^2)^{1.5})}{3}dx = 303$
$\int ^a _0 [(a^2 - x^2)^{1.5} - \cfrac{((a^2 - x^2)^{1.5})}{3}dx = 303$
$\int ^a _0 [\cfrac{2}{3}(a^2 - x^2)^{1.5}dx = 303$
| As a general rule, whenever possible, try to draw a picture of the region over which you are integrating. In this case, the limits of integration describe a quarter circle in the first quadrant of radius $a$, centered at the origin. This suggests a change of variable to polar coordinates, which turns the integral into
$$\int_0^{\pi/2}\int_0^a(a^2-r^2)r\,dr\,d\theta={\pi\over2}\int_0^a(a^2-r^2)r\,dr={\pi\over2}a^4\int_0^1(1-u^2)u\,du={\pi\over2}a^4\left({1\over2}-{1\over4}\right)={\pi\over8}a^4$$
Remark: It strikes me as a little odd that the problem would set the integral equal to $303$. I would have expected $32\pi$ instead (for which the solution would simply be $a=4$). But so be it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3642799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Two nested radicals with similar result Sorry for my last post it's my bad .So I ask to this (the true ^^)nested radical :
$$\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{\cdots }}}}}}}}=\frac{1+\sqrt{5}+\sqrt{30-6\sqrt{5}}}{4}$$
The period is $4$ and the related equation is :
$$\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+x}}}}=x$$
There is a big similarity with this How to prove $\sqrt{5+\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5+\sqrt{5+\sqrt{5-\cdots}}}}}}} = \frac{2+\sqrt 5 +\sqrt{15-6\sqrt 5}}{2}$
Following the answer of Tito Piezas III can solve this nested radical (with $2$).
My question:
Can someone explain these similarities between these two nested radicals ?
Any helps is highly appreciated .
Thanks a lot .
| We can solve the equation $\sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2+x}}}}=x$ as follows.
Since $2+x\ge 0$, we have $x\ge -2$. Since $\sqrt{2+x}\le 2$, $x\le 2$. So we can put $x=2\cos 32\alpha$ for some $0\le\alpha\le \tfrac{\pi}{32}$. Then
$$\sqrt{2+x}=\sqrt{2+2\cos 32\alpha}=2\cos 16\alpha$$
$$\sqrt{2-2\cos 16\alpha}=2\sin 8\alpha$$
$$\sqrt{2-2\sin 8\alpha}=\sqrt{2-2\cos\left(\frac{\pi}2- 8\alpha\right)}=2\sin\left(\frac{\pi}4- 4\alpha\right)$$
$$\sqrt{2+2\sin\left(\frac{\pi}4-4\alpha\right)}=\sqrt{2+2\cos\left(\frac{\pi}4+4\alpha\right)}=2\cos\left(\frac{\pi}8+2\alpha\right)$$
So $$2\cos\left(\frac{\pi}8+2\alpha\right)=2\cos 32\alpha $$
$$\sin\left(17\alpha+\frac{\pi}{16}\right)\sin\left(15\alpha-\frac{\pi}{16}\right)=0$$
$$17\alpha+\frac{\pi}{16}=\pi n\mbox{ or }15\alpha-\frac{\pi}{16}=\pi n,\,\, n\in\Bbb Z$$
For the range $0\le\alpha\le \tfrac{\pi}{32}$ fits only $15\alpha-\frac{\pi}{16}=0$, so $\alpha=\frac{\pi}{15\cdot 16}$ and
$$x=2\cos 32\alpha=2\cos\frac{2\pi}{15}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3642922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If $f : \mathbb R \rightarrow \mathbb R $ such that $f(x^2+x)+2f(x^2-3x+2) = 9x^2-15x$. Find $f(2016)$.
Determine all $f : \mathbb R \rightarrow \mathbb R $ such that $$f(x^2+x)+2f(x^2-3x+2) = 9x^2-15x$$ for all $x$. Find $f(2016)$.
Similar problem appeared on this site before: $f(x^2 + x)+2f(x^2 - 3x + 2)=9x^2 - 15x$ then find $f(2016)$. (The question is now deleted.) The same problem with finding $2011$ (instead of $2016$) appeared in 2011 Singapore Mathematical Olympiad as problem 17 (Wayback Machine).
Iβve tried put $x=0,1$ and got \begin{align*}
f(0)+2f(2)&=0\\
f(2)+2f(0)&=-6
\end{align*}
which gives me $f(0)=-4$, $f(2)=2$.
Similarly, if we notice that $x^2+x=x^2-3x+2$ holds for $x=\frac12$, we can find the value at the point $\frac34=\left(\frac12\right)^2+\frac12$.
But the above doesnβt seem to help for other values.
Thank you very much for helping.
| First (observation):
Note that we can determine $f(0), f(2)$ easily:
$$
x=0 \qquad \rightarrow \qquad f(0)+2f(2)=0;\\
x=1 \qquad \rightarrow \qquad f(2)+2f(0)=-6;
$$
so
$$
f(0)=-4,\quad f(2)=2.
$$
Same way we can determine $f(6), f(20)$ (substituting $x=-3, x=4$).
Same way we can determine $f(56), f(30)$ (substituting $x=-6, x=7$).
...
Second (solution):
Let's focus on $x=-a, x=a+1$, where $a\in\mathbb{R}$:
$$
x=-a \qquad \rightarrow \qquad f(a^2-a)+2f(a^2+3a+2) = 9a^2+15a; \\
x=a+1 \qquad \rightarrow \qquad f(a^2+3a+2)+2f(a^2-a) = 9a^2+3a-6;
$$
so (when denote $A=f(a^2-a)$, $B=f(a^2+3a+2)$):
$$
\left\{ \begin{array}{l}A+2B = 9a^2+15a; \\ B+2A = 9a^2+3a-6;\end{array} \right.$$
$$
\left\{ \begin{array}{l}B+A = 6a^2+6a-2;\\ B-A = 12a+6;\end{array} \right.
$$
and
$$
\left\{ \begin{array}{l}f(a^2-a) = A = 3a^2-3a-4; \\
f(a^2+3a+2) = B = 3a^2+9a+2. \end{array}\right.\tag{1}
$$
From $(1)$ we conclude that for each $z$ which can be written in the form
$$
z = a^2-a, \qquad a \in\mathbb{R} \tag{2}
$$
(in fact, for $z\ge -\frac{1}{4}$)
we have
$$
f(z) = 3z-4.
$$
Therefore $f(z)$ is linear function for $z\ge -\frac{1}{4}$.
Since $z=2016$ admits representation $(2)$, then $f(2016)=3\cdot 2016-4 = 6044.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3644316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 0
} |
Computing $\int_0^1\frac{1-2x}{2x^2-2x+1}\ln(x)\text{Li}_2(x)dx$ Any idea how ot approach
$$I=\int_0^1\frac{1-2x}{2x^2-2x+1}\ln(x)\text{Li}_2(x)dx\ ?$$
I came across this integral while I was trying to find a different solution for $\Re\ \text{Li}_4(1+i)$ posted here.
here is how I came across it;
using the identity
$$\int_0^1\frac{\ln(x)\text{Li}_2(x)}{1-ax}dx=\frac{\text{Li}_2^2(a)}{2a}+3\frac{\text{Li}_4(a)}{a}-2\zeta(2)\frac{\text{Li}_2(a)}{a}$$
multiply both sides by $\frac{a}{3}$ then replace $a$ by $1+i$ and consider the the real parts of both sides we have
$$\Re\ \text{Li}_4(1+i)=-\frac16\Re\ \text{Li}_2^2(1+i)+\frac23\zeta(2)\Re\ \text{Li}_2(1+i)+\frac13\Re \int_0^1\frac{(1+i)}{1-(1+i)x}\ln(x)\text{Li}_2(x)dx$$
For the integral, use $\Re\frac{1+i}{1-(1+i)x}=\frac{1-2x}{2x^2-2x+1}$ which gives $I$.
What I tried is subbing $1-2x=y$ which gives
$$I=\int_{-1}^1\frac{-y}{1+y^2}\ln\left(\frac{1+y}{2}\right)\text{Li}_2\left(\frac{1+y}{2}\right)dy=\int_{-1}^1 f(y)dy=\underbrace{\int_{-1}^0 f(y)dy}_{y\to\ -y}+\int_{0}^1 f(y)dy$$
$$=\int_0^1\frac{y}{1+y^2}\ln\left(\frac{1-y}{2}\right)\text{Li}_2\left(\frac{1-y}{2}\right)dy-\int_0^1\frac{y}{1+y^2}\ln\left(\frac{1+y}{2}\right)\text{Li}_2\left(\frac{1+y}{2}\right)dy$$
I think I made it more complicated. Any help would be appriciated.
| $$I=\int_0^1\frac{y}{1+y^2}\ln\left(\frac{1-y^2}{4}\right)\text{Li}_2\left(\frac{1-y}{2}\right)dy+\frac{7}{32} \zeta (3) \log (2)+\frac{ \text{Li}_4\left(\frac{1}{2}\right)}{8}-\frac{157 \pi ^4}{46080}-\frac{ 11\log ^4(2)}{48}+\frac{19}{384} \pi ^2 \log ^2(2)$$
Put $$y=\frac{1-x}{1+x}$$
$$\int_0^1\frac{y}{1+y^2}\ln\left(\frac{1-y^2}{4}\right)\text{Li}_2\left(\frac{1-y}{2}\right)dy=\int_0^1\frac{\ln(x)}{1+x}\text{Li}_2\left(\frac{x}{1+x}\right)\ dx-2\int_0^1\frac{\ln(1+x)}{1+x}\text{Li}_2\left(\frac{x}{1+x}\right)\ dx-\int_0^1\frac{x\ln(x)}{1+x^2}\text{Li}_2\left(\frac{x}{1+x}\right)\ dx+2\int_0^1\frac{x\ln(1+x)}{1+x^2}\text{Li}_2\left(\frac{x}{1+x}\right)\ dx$$
be continued
$$\int_0^1\frac{\ln(1+x)}{1+x}\text{Li}_2\left(\frac{x}{1+x}\right)\ dx=3\operatorname{Li}_4\left(\frac12\right)-\frac{\pi^4}{30}+\frac{21}8\ln2\zeta(3)-\frac{\pi^2}{12}\ln^22$$
$$\int_0^1\frac{x\ln(x)}{1+x^2}\text{Li}_2\left(\frac{x}{1+x}\right)\ dx=\frac{C^2}{2}+\frac{15 \text{Li}_4\left(\frac{1}{2}\right)}{16}-\frac{701 \pi ^4}{46080}+\frac{7}8\ln2\zeta(3)+\frac{5 \log ^4(2)}{128}-\frac{3}{64} \pi ^2 \log ^2(2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3655021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Find $a,b \in \mathbb{N}$ with $\mathrm{lcm}(a,b)=12\gcd(a,b)\,$ and $\,a\bmod b = 5$
Find $a,b \in\mathbb{N}$ such that:
Remainder of $a$ divided by $b$ equals $5$
$\mathrm{lcm}(a,b)=12\gcd(a,b)$
I don't know how or where to start. The most similar problem I found was Find $a, b \in \mathbb{N}$ that verify $a + b = 1271, lcm(a, b) = 330.gcd(a, b).$
Still I can't figoure out how to solve my problem, any help will be appreciated, thanks.
| Remember $lcm(a,b) = \frac {ab}{\gcd(a,b)}$.
So $\frac {ab}{\gcd(a,b)}= 12\gcd(a,b)$ and
$\frac {a}{\gcd(a,b)}\frac {b}{\gcd(a,b)}= 12$.
Let $a' = \frac a{\gcd(a,b)}; b' =\frac b{\gcd(a,b)} = 12$ so
$(a',b') = \{(1,12), (2,6), (3,4),(4,3), (6,2), (12,1)\}$.
But $\gcd(a',b') = 1$ so $(a',b') \ne (2,6)$ or $(6,2)$.
Now $a= mb + 5$ for some integer $m$
So we have $a'\gcd(a,b) = mb'\gcd(a,b) + 5$. Let $d = \gcd(a,b)$ and we have:
$a'd =md + 5$
$(a'-mb')d = 5$
$d = \frac {5}{a'-mb'}$
$5$ is prime so $a'-mb' = 1,5$.
We have 4 cases $(a',b') = (1,12), (3,4), (4,3), (12,1)$.
Case 1: $a'=1;b'=12$ and $1-12m = \pm 1, \pm 5$. That can only be $m=0$ and $d=5$ and $a=1*5=5; b=12*5= 60$. Then $5 = 0*60 + 5$ and $lcm(5,60)= 60=12*5=12\gcd(5,60)$.
Case 2: $a'=3;b'=4$ and $3-4m =1,5$. That's impossible.
Case 3: $a'=4;b'=3$ and $4-3m=1,5$. That can only be $m=1$ and $d=5$ and $a=4*5=20$ and $b=3*5 =15$. and $20 = 1*15+5$ and $lcm(20,15) =60=12*5=12\gcd(20,15)$
Case 4: $a'=12; b'=1$ and $12-3m = 1,5$. That's impossible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3658753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How can I approach this inequality? Let $a, b$ and $c$ be three non-zero positive numbers. Show that:
$$\sqrt{\frac{2a}{a + b}} + \sqrt{\frac{2b}{b + c}} + \sqrt{\frac{2c}{a + c}} \leq 3$$
I know the triangular inequality would help here, but I don't know how to approach it.
I started by $a+bβ₯a$ then that gives $\frac{1}{a+b}β€\frac{1}{a}$ by muliplyting both sides by $2a$ we get $\frac{2a}{a+b}β€\frac{2a}{a}$ which leads eventually to $\frac{2a}{a+b}β€2$ and by adding the square root to both sides we get $\sqrt{\frac{2a}{a+b}}\leq\sqrt2$ and doing the same thing to the other terms we get $\sqrt{\frac{2a}{a+b}}+\sqrt{\frac{2b}{b+c}}+\sqrt{\frac{2c}{c+a}}\leq3\sqrt2$ beyond that I don't have any idea if that would lead to anything useful or not.
| It's not a clean solution but it works. Consider the change of variable $0<x:=\frac{b}{a}, 0<y:=\frac{c}{b}, 0 < x :=\frac{a}{c} $. Observe that $xyz=1$. Now the inequality reads
$$f(x,y,z)=\sqrt{\frac{2}{1+x}}+\sqrt{\frac{2}{1+y}}+\sqrt{\frac{2}{1+z}}\leq 3$$
That means we have to find de maxima of $f$ subject to restriction $g(x,y,z):=xyz=1$. Now we calculate the gradients
$$\nabla f(x,y,z)= -\sqrt{2}\left(\frac{1}{(1+x)^{3/2}},\frac{1}{(1+y)^{3/2}},\frac{1}{(1+z)^{3/2}} \right).$$
$$ \nabla g(x,y,z) = \lambda (yz,xz,xy)$$
Then we solve $\nabla f= \lambda \nabla g$. Now we solve a system of equations.
\begin{align}
\frac{1}{(1+x)^{3/2}} & = \lambda yz \quad (1) \\
\frac{1}{(1+y)^{3/2}} & = \lambda xz \quad (2)\\
\frac{1}{(1+z)^{3/2}} & = \lambda xy \quad (3)\\
1 & = xyz \quad (4) \\
\end{align}
Observe by right side of these equations $\lambda,x,y,z \neq 0 $. Dividing (1) with (2) and $(1)$ with $(3)$, in addition with some calculations we get
\begin{align}
h(x) &= h(y) \\
h(x) &= h(z). \\
\end{align}
Where $h(x) := \frac{x}{(1+x)^{3/2}}$ which satisfies $h''(x)\leq 0$ in particular is concave in $[0,\infty[$ so it doesn't take the same value more than two times,that means two of the $x,y,z$ are equal. Hence, without lost of generality $x=y$. With this
\begin{align*}
\lambda &= \frac{x}{(1+x)^{3/2}} \\
\frac{1}{(1+z)^{3/2}} &= \lambda x^2 \\
z & = 1/x^2
\end{align*}
doing the substitutions and algebraic manipulations we get $x^2=1$. So $x=y=z=1$,
Hence the maximum is $f(1,1,1)=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3659032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Prove that $ a^2+b^2+c^2 \le a^3 +b^3 +c^3 $ If $ a,b,c $ are three positive real numbers and $ abc=1 $ then prove that $a^2+b^2+c^2 \le a^3 +b^3 +c^3 $
I got $a^2+b^2+c^2\ge 3$ which can be proved $ a^2 +b^2+c^2\ge a+b+c $. From here how can I proceed to the results? Please help me to proceed. Thanks in advance.
| This is the same as proving that
$$(a^2+b^2+c^2)(abc)^{1/3}\le a^3+b^3+c^3$$
for all positive $a$, $b$, $c$. This is AM/GM. We get
$$a^{7/3}b^{1/3}c^{1/3}\le\frac{7a^3+b^3+c^3}9$$
applying AM/GM to seven copies of $a^{3}$ and one each of $b^{3}$ and $c^{3}$. Cyclicly
permute and add.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3659533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Solving a first order differential equation by finding an integrating factor : Edit Problem:
Solve the following differential equations by first finding an integrating factor.
$$ (y^2(x+1) + y ) \, dx + ( 2xy + 1 ) \, dy = 0 $$
Answer:
\begin{align*}
M_y &= 2(x+1)y + 1 = 2xy + 2y + 1 \\
N_x &= 2y \\
\frac{1}{N(x,y)} \left[ \frac{\partial M(x,y)}{\partial y} - \frac{\partial N(x,y)}{\partial x} \right] &=
\frac{ 2xy + 2y + 1 - 2y } { 2xy + 1 } \\
\frac{1}{N(x,y)} \left[ \frac{\partial M(x,y)}{\partial y} - \frac{\partial N(x,y)}{\partial x} \right] &=
\frac{ 2xy + 1 } { 2xy + 1 } = 1 \\
\end{align*}
This means that:
$$ e ^ { \int \frac{1}{N(x,y)} \left[ \frac{\partial M(x,y)}{\partial y} - \frac{\partial N(x,y)}{\partial x} \right] \, dx } $$
is the integrating factor we seek. Call this integrating factor $I$.
\begin{align*}
I &= e ^ { \int 1 \, dx } = e^x \\
(y^2(x+1) + y ) e^x \, dx + ( 2xy + 1 ) e^x \, dy &= 0
\end{align*}
Now we have:
\begin{align*}
M &= (y^2(x+1) + y ) e^x \\
M_y &= ( 2(x+1)y + 1 )e^x = ( 2xy + 2y + 1)e^x \\
N &= ( 2xy + 1 ) e^x \\
N_x &= ( 2xy + 1 ) e^x
\end{align*}
As I understand it, I was suppose to get $M_y = N_x$. That is, the de should have been exact. What did I do wrong?
Now, I have an updated answer. However, It is still wrong. I feel I am much closer to the right answer. Here is my updated answer:
\begin{align*}
M_y &= 2(x+1)y + 1 = 2xy + 2y + 1 \\
N_x &= 2y \\
\frac{1}{N(x,y)} \left[ \frac{\partial M(x,y)}{\partial y} - \frac{\partial N(x,y)}{\partial x} \right] &=
\frac{ 2xy + 2y + 1 - 2y } { 2xy + 1 } \\
\frac{1}{N(x,y)} \left[ \frac{\partial M(x,y)}{\partial y} - \frac{\partial N(x,y)}{\partial x} \right] &=
\frac{ 2xy + 1 } { 2xy + 1 } = 1 \\
\end{align*}
This means that:
$$ e ^ { \int \frac{1}{N(x,y)} \left[ \frac{\partial M(x,y)}{\partial y} - \frac{\partial N(x,y)}{\partial x} \right] \, dx } $$
is the integrating factor we seek. Call this integrating factor $I$.
\begin{align*}
I &= e ^ { \int 1 \, dx } = e^x \\
(y^2(x+1) + y ) e^x \, dx + ( 2xy + 1 ) e^x \, dy &= 0
\end{align*}
Now we have:
\begin{align*}
M &= (y^2(x+1) + y ) e^x \\
M_y &= ( 2(x+1)y + 1 )e^x = ( 2xy + 2y + 1)e^x \\
N &= ( 2xy + 1 ) e^x \\
N_x &= ( 2xy + 1 ) e^x + (2y)e^x = (2xy + 2y + 2)e^x1
\end{align*}
Hence the differential equation is exact. We have:
\begin{align*}
F_x &= (y^2(x+1) + y ) e^x \\
F &= \int (y^2(x+1) + y ) e^x \, dx = \int (x y^2 + y^2 + 1 ) e^x \, dx
\end{align*}
Recall that:
$$ \int x e^x \, dx = x e^x - e^x + C $$
\begin{align*}
F &= y^2 \int xe^x \, dx + (y^2+1) \int e^x \, dx \\
F &= y^2 ( xe^x - e^x) + (y^2 + 1)e^x + \phi(y) \\
F &= y^2 xe^x - y^2 e^x + y^2 e^x + e^x + \phi(y) \\
F &= y^2 xe^x + e^x + \phi(y) \\
F_y &= 2xy e^x + \phi'(y) \\
2xy e^x + \phi'(y) &= ( 2xy + 1 ) e^x \\
\phi'(y) &= e^x \\
\phi(y) &= ye^x + c \\
F &= y^2 xe^x + e^x + ye^x + c
\end{align*}
However, the book gets:
$$ x y^2 e^x + y e^x = c $$
Where did I go wrong?
Problem:
Solve the following differential equations first finding an integrating factor.
$$ ( 5xy + 4y^2 + 1 ) \, dx + ( x^2 + 2xy ) \, dy = 0 $$
Answer:
Now, I try $x^3$ as an integrating factor. This gives me:
$$ ( 5x^4 y + 4 x^3 y^2 + x^3 ) \, dx + ( x^5 + 2x^4 y ) \, dy = 0 $$
Now, we see if it is exact.
\begin{align*}
M_y &= 5x^4 + 8 x^3 y \\
N_x &= 5x^4 + 8 x^3 y
\end{align*}
The equation is exact. Let $F$ be the solution we seek:
\begin{align*}
F_x &= 5x^4 y + 4 x^3 y^2 + x^3 \\
F &= x^5 y + x^4 y^2 + \frac{x^4}{4} + \phi(y) \\
F_y &= 5x^4 + 2x^4 y + \phi'(y) = x^5 + 2x^4 y \\
\phi'(y) &= 0 \\
\phi(y &= C
\end{align*}
Hence the solution we seek is:
$$ 4x^5 y + 4x^4 y^2 + x^4 + C = 0 $$
Where did I go wrong?
| Everything looks good up to the last formula, you missed there one factor in the product rule for $N_x$:
$$
N_x=(2xy+1)_xe^x+(2xy+1)(e^x)_x=(2y)e^x+(2xy+1)e^x,
$$
restoring the needed equality $M_y=N_x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3663339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to improve $\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin(nx)}{\sin(x)}\right)^{4}dx<\frac{\pi^{2}n^{2}}{4}$ I have proved this inequality $\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin(nx)}{\sin(x)}\right)^{4}dx<\frac{\pi^{2}n^{2}}{4}$.
Using $\left|\sin(nx)\right|\leq n\left|\sin(x)\right|$ on $[0,\frac{\pi}{2n}]$ and $\frac{\left|\sin(nx)\right|}{\left|\sin(x)\right|}\leq\frac{\pi}{2x}$ on $[\frac{\pi}{2n},\frac{\pi}{2}]$,we can have
$$\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin(nx)}{\sin(x)}\right)^{4}dx<\frac{\pi^{2}n^{2}}{8}+\frac{\pi^{2}}{8}\left(n^{2}-1\right)<\frac{\pi^{2}n^{2}}{4}.$$
But using mathematica I found this inequality can still be improved.
And after calculating some terms I found it seems that when $n\geq 2$ we can have
$$\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin(nx)}{\sin(x)}\right)^{4}dx<\frac{\pi^{2}n^{2}}{8}.$$
But I cannot prove this.So is there any method to improve my result?Any help will be thanked.
| Alternative solution:
When $n = 2, 3, 4$, the inequality is verified directly.
In the following, assume that $n\ge 5$.
Let
$$I_n = \int_0^{\pi/2} \frac{x}{n^2}\left(\frac{\sin n x}{\sin x}\right)^4\mathrm{d} x.$$
We have
\begin{align}
I_n &= \underbrace{\int_0^{\pi/n} \frac{x}{n^2}\left(\frac{\sin n x}{\sin x}\right)^4\mathrm{d} x}_{I_{n,1}}
+ \underbrace{\int_{\pi/n}^{\pi/2} \frac{x}{n^2}\left(\frac{\sin n x}{\sin x}\right)^4\mathrm{d} x}_{I_{n,2}}.
\end{align}
First, we have
\begin{align}
I_{n,1} &\le \int_0^{\pi/n} \frac{x}{n^2}(\sin nx)^4 \frac{1}{x^4} \left(\frac{\frac{\pi}{n}}{\sin\frac{\pi}{n}}\right)^4\mathrm{d} x \\
&= \frac{1}{n^2}\left(\frac{\frac{\pi}{n}}{\sin\frac{\pi}{n}}\right)^4
\int_0^{\pi/n} \frac{(\sin nx)^4}{x^3} \mathrm{d} x \\
&= \left(\frac{\frac{\pi}{n}}{\sin\frac{\pi}{n}}\right)^4
\int_0^{\pi} \frac{(\sin y)^4}{y^3} \mathrm{d} y\\
&\le \left(\frac{\frac{\pi}{5}}{\sin\frac{\pi}{5}}\right)^4
\int_0^{\pi} \frac{(\sin y)^4}{y^3} \mathrm{d} y
\end{align}
where we have used: i) $\frac{\sin x}{x} \ge \frac{\sin \frac{\pi}{n}}{\frac{\pi}{n}}$
on $0 \le x \le \frac{\pi}{n}$; ii) the substitution $y = nx$; iii) $\frac{\frac{\pi}{n}}{\sin\frac{\pi}{n}}$ is non-increasing for $n\ge 2$.
Second, we have
\begin{align}
I_{n, 2} &= \int_{\pi/n}^{\pi/2} \frac{x}{n^2}\left(\frac{\sin n x}{\sin x}\right)^4\mathrm{d} x\\
&\le \int_{\pi/n}^{\pi/2} \frac{x}{n^2}\left(\frac{\pi}{2x}\right)^4\mathrm{d} x \\
&= -\frac{\pi^2}{8n^2} + \frac{\pi^2}{32}\\
&\le \frac{\pi^2}{32}
\end{align}
where we have used $\sin x \ge \frac{2}{\pi}x$ for $0 \le x \le \frac{\pi}{2}$.
Thus, we have
$$I_n \le \left(\frac{\frac{\pi}{5}}{\sin\frac{\pi}{5}}\right)^4
\int_0^{\pi} \frac{(\sin y)^4}{y^3} \mathrm{d} y
+ \frac{\pi^2}{32} < \frac{\pi^2}{8}.$$
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3664974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
how is this solved? $\lim_{xβ0} \frac{1}{x^2}\left(\left(\tan(x+\frac{\pi}{4})\right)^{\frac{1}{x}}βe^2\right)$ I know how to solve the trigonometric part of the limit, i.e $\left(\tan(x+\frac{\pi}{4})\right)^{1/x}$, which I think is $e^2$, I do not however, know how to carry forward the question with that.
Also, I am sincerely sorry for the upsetting format, i have no clue how to use mathjax.
| $$\frac{\left(\tan\left(x+\frac{\pi}{4}\right)\right)^{\frac{1}{x}}-e^2}{x^2}=\frac{\left(\frac{1+\tan{x}}{1-\tan{x}}\right)^{\frac{1}{x}}-e^2}{x^2}=$$
$$=\frac{\left(1+\frac{2\tan{x}}{1-\tan{x}}\right)^{\frac{1}{x}}-e^2}{x^2}=\frac{\left(1+\frac{2\tan{x}}{1-\tan{x}}\right)^{\frac{1-\tan{x}}{2\tan{x}}\cdot\frac{2\tan{x}}{x(1-\tan{x})}}-e^2}{x^2}=$$
$$=\frac{e^{\frac{\ln\left(1+\frac{2\tan{x}}{1-\tan{x}}\right)}{\frac{2\tan{x}}{1-\tan{x}}}\cdot\frac{2\tan{x}}{x(1-\tan{x})}}-e^2}{x^2}=$$
$$=\frac{\left(e^{\frac{\ln\left(1+\frac{2\tan{x}}{1-\tan{x}}\right)}{\frac{2\tan{x}}{1-\tan{x}}}\cdot\frac{\tan{x}}{x(1-\tan{x})}}-e\right)\left(e^{\frac{\ln\left(1+\frac{2\tan{x}}{1-\tan{x}}\right)}{\frac{2\tan{x}}{1-\tan{x}}}\cdot\frac{\tan{x}}{x(1-\tan{x})}}+e\right)}{x^2}\rightarrow$$
$$\rightarrow\frac{2e^2\left(e^{\frac{\ln\left(1+\frac{2\tan{x}}{1-\tan{x}}\right)}{\frac{2\tan{x}}{1-\tan{x}}}\cdot\frac{\tan{x}}{x(1-\tan{x})}-1}-1\right)}{x^2}=$$
$$=\frac{2e^2\left(e^{\frac{\ln\left(1+\frac{2\tan{x}}{1-\tan{x}}\right)}{\frac{2\tan{x}}{1-\tan{x}}}\cdot\frac{\tan{x}}{x(1-\tan{x})}-1}-1\right)}{ \frac{\ln\left(1+\frac{2\tan{x}}{1-\tan{x}}\right)}{\frac{2\tan{x}}{1-\tan{x}}}\cdot\frac{\tan{x}}{x(1-\tan{x})}-1}\cdot\frac{\frac{\ln\left(1+\frac{2\tan{x}}{1-\tan{x}}\right)}{\frac{2\tan{x}}{1-\tan{x}}}\cdot\frac{\tan{x}}{x(1-\tan{x})}-1}{x^2}\rightarrow $$
$$\rightarrow2e^2\cdot \frac{\frac{\ln\left(1+\frac{2\tan{x}}{1-\tan{x}}\right)}{\frac{2\tan{x}}{1-\tan{x}}}\cdot\frac{\tan{x}}{x(1-\tan{x})}-1}{x^2}=\frac{e^2\left(\ln\left(1+\frac{2\tan{x}}{1-\tan{x}}\right)-2x\right)}{x^3}=$$
$$=\frac{e^2(\ln(1+\tan{x})-\ln(1-\tan{x})-2x)}{x^3}\rightarrow$$
$$\rightarrow\frac{e^2\left(\frac{\frac{1}{\cos^2x}}{1+\tan{x}}+\frac{\frac{1}{\cos^2x}}{1-\tan{x}}-2\right)}{3x^2}=$$
$$=\frac{e^2\left(\frac{2}{\cos^2x(1-\tan^2{x})}-2\right)}{3x^2}=\frac{2e^2(1-\cos2x)}{3x^2\cos2x}=\frac{4e^2\sin^2x}{3x^2\cos2x}\rightarrow\frac{4e^2}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3672397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that $x^2+y^2-\sqrt{2}xy < 4$ provided $1\le x,y \le \frac{5}{2}$. If possible I would like an elegant solution to the following problem:
Let $x,y\in\mathbb{R}$ such that $1\le x \le \frac{5}{2}$ and $1\le y \le \frac{5}{2}$. Prove that
$x^2+y^2-\sqrt{2}xy < 4$
I'm aware that you can use Lagrange multipliers but I want an elementary solution using elementary inequalities that are available up to the high school level, or inequalities used in olympiads.
Remarks
I've substituted the obvious specific values for $x,y$ (we can assume without loss of generality that $x\le y$) below
\begin{array}{|c|c|c|}
\hline
x & y &x^2+y^2-\sqrt{2}xy \text{ (rounded to 3 d.p.)}\\ \hline
1 & 1 & \approx 0.587\\ \hline
1 & \frac{5}{2} & \approx 3.714\\ \hline
\frac{5}{2} & \frac{5}{2} & \approx 3.661\\ \hline
\end{array}
Some thoughts:
*
*I've tried solving it in different equivalent forms. Using the following identity
$x^4+y^4 = \left(x^2+y^2+\sqrt{2}xy \right) \left(x^2+y^2-\sqrt{2}xy \right)$
I attempted to prove $x^4+y^4 < 4 \left(x^2+y^2 + \sqrt{2}xy \right)$ (though I'm not sure if it really helps) but to no avail.
*Also, crudely approximating via $x^2+y^2 -\sqrt{2}xy < \left(\frac{5}{2}\right)^2 + \left(\frac{5}{2}\right)^2 - \sqrt{2} \cdot 1 \cdot 1 $ doesn't work since the RHS is clearly larger than 4
| Hint. Make the substitution $$x=t+u,\\y=\sqrt 2 u.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3674487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
diophantine question parity and sign If I have an equation say $$3(1+x+x^2)(1+y+y^2)=4x^2y^2-1 \quad (1)$$ and I rewrite it as
$$4+3(x+y+xy+x^2+y^2+xy^2+x^2y)-x^2y^2=0$$ and I find a positive integer solution where both x and y are even, can I show that no odd positive integer solutions exist?
| Since the equation is symmetric in $x$ and $y$, we only need to consider the case $x\le y$.
Dividing the both sides of
$$3(1+x+x^2)(1+y+y^2)=4x^2y^2-1\tag1$$
by $x^2y^2$ gives
$$3\bigg(1+\frac 1x+\frac{1}{x^2}\bigg)\bigg(1+\frac 1y+\frac{1}{y^2}\bigg)=4-\frac{1}{x^2y^2}\tag2$$
Suppose here that $x\ge 8$.
Then, we have
$$ \text{(LHS of $(2)$)}\le 3\bigg(1+\frac 18+\frac{1}{8^2}\bigg)\bigg(1+\frac 18+\frac{1}{8^2}\bigg)=\frac{15987}{4096}$$
and
$$\text{(RHS of $(2)$)}\ge 4-\frac{1}{8^4}=\frac{16383}{4096}$$
This is a contradiction. So, we have $x\le 7$.
Now, $(1)$ can be written as
$$(3+3x-x^2)y^2+3(1+x+x^2)y+4+3x+3x^2=0$$
Seeing this as a quadratic equation on $y$ and considering the discriminant, we get
$$D=9(1+x+x^2)^2-4(3+3x-x^2)(4+3x+3x^2)$$
It is necessary that $D$ is a square number.
For $x=1$, we get $D=-119$ which is not a square number.
For $x=3$, we get $D=1041$ which is not a square number.
For $x=5$, we get $D=11281$ which is not a square number.
For $x=7$, we get $D=46441$ which is not a square number.
Therefore, $(1)$ has no solution $(x,y)$ such that both $x$ and $y$ are positive odd integers.
Added :
For $x=2$, $D$ is a square number, but $y=-2$ is not positive.
For $x=4$, $D$ is a square number, and $y=-1,64$.
For $x=6$, $D$ is not a square number.
Therefore, $$\color{red}{(x,y)=(4,64),(64,4)}$$ are the only positive integer solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3675459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding $\lim_{n\to\infty} \left(\frac{\sqrt{n^2+n}-1}{n}\right)^{2\sqrt{n^2+n}-1}$
Find the following limit without using the L'Hopital rule:
$$\lim_{n\to\infty} \left(\dfrac{\sqrt{n^2+n}-1}{n}\right)^{2\sqrt{n^2+n}-1}$$
Answer: $e^{-1}$
My attempt: Since the limit is of the form $1^{\infty}$, I decided to use the standard formula:
$$\lim_{x\to a} f^g = e^{\lim_\limits{x\to a}(f-1)g}$$
(See link)
Let $l=(f-1)g$.
We have,
$$l=\left(\dfrac{\sqrt{n^2+n}-(1+n)}{n}\right)(2\sqrt{n^2+n}-1)$$
This on solving boils down to
$$l=2n+3-\sqrt{1+\frac 1n}(2n+3)+\frac 1n$$
Now if I tend $n$ towards infinity, then $l\to 0$ and the limit i.e. $e^l$, is equal to $1$, which contradicts the given answer.
Please help. Thanks!
Edit:
A proof for the "standard formula" I have used.
Edit 2: just noticed a typo in the power which I have now fixed.
| $$a_n=\left(\frac{\sqrt{n^2+n}-1}{n}\right)^{2\sqrt{n^2+n}-1}\implies\log(a_n)=\left({2\sqrt{n^2+n}-1}\right)\log\left(\frac{\sqrt{n^2+n}-1}{n}\right)$$ Now, using Taylor expansions
$$\log\left(\frac{\sqrt{n^2+n}-1}{n}\right)=-\frac{1}{2 n}-\frac{1}{4 n^2}-\frac{1}{24
n^3}+O\left(\frac{1}{n^4}\right)$$
$${2\sqrt{n^2+n}-1}=2 n-\frac{1}{4 n}+\frac{1}{8 n^2}-\frac{5}{64
n^3}+O\left(\frac{1}{n^4}\right)$$
$$\log(a_n)=-1-\frac{1}{2 n}+\frac{1}{24 n^2}+O\left(\frac{1}{n^3}\right)$$
$$a_n=e^{\log(a_n)}=\frac 1e \left(1-\frac{1}{2 n}+\frac{1}{6 n^2} \right)+O\left(\frac{1}{n^3}\right)$$ which shows the limit and how it is approached.
Moreover, this gives a shortcut estimation of $a_n$. Suppose $n=10$
$$a_{10}=\left(\frac{10}{\sqrt{110}-1}\right)^{1-2 \sqrt{110}}\approx 0.350040$$ while the truncated expansion gives $\frac{571}{600 e} \approx 0.350099$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3677279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
If $(1 + \operatorname{cis}x)(1 + \operatorname{cis} 2x) = a + bi$, prove that $a^2 + b^2 = (4\cos x \cos\frac x2)^2$ Could anyone help me with this? I'm stuck.
If $$(1 + \operatorname{cis}x)(1 + \operatorname{cis} 2x) = a + bi$$ prove that $$a^2 + b^2 = \left(4\cos x \cos\frac x2\right)^2$$
For reference, $\operatorname{cis}x = \cos x + i\sin x$.
I found that
$$a = 1 + \cos x + \cos2x + \cos3x \quad\text{and}\quad
b = \sin x + \sin2x + \sin3x$$
| Hint:
$$1+\text{cis}2y=1+\cos2y+i\sin2y=2\cos y(\cos y+i\sin y)$$
$$|1+\text{cis}2y|=|2\cos y|\cdot|\cos y+i\sin y|=2|\cos y|$$
$$|1+\text{cis}2y|^2=4|\cos y|^2=4\cos^2y$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3679659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Cevianas AD, BE and CF of ABC compete at P. Show $\frac{S_{DEF}}{2S_{ABC}}=\frac{PD .PE.PF}{PA.PB.PC}$ In an ABC triangle, the cevianas AD, BE and CF compete in P. Show that
$\frac{S_{DEF}}{2S_{ABC}}=\frac{PD .PE.PF}{PA.PB.PC}$
Using areas relation, I found
$\frac{3S_{ABC}-\overbrace{(S_{PAB}+S_{PAC}+S_{PBC})}^{S_{ABC}}}{S_{ABC}}=\frac{PA}{PD}+\frac{PB}{PE}+\frac{PC}{PF}\implies
2S_{ABC}=S_{ABC}\left(\frac{PA}{PD}+\frac{PB}{PE}+\frac{PC}{PF}\right)\\
\frac{S_{PDF}}{S_{PAC}}=\frac{PDΓPF}{PAΓPC}\\
\frac{S_{PDF}}{S_{PAB}}=\frac{PDΓPE}{PAΓPB}\\
\frac{S_{PEF}}{S_{PBC}}=\frac{PEΓPF}{PBΓPC}\\
S_{PDF}+S_{PED}+S_{PFE}=S_{DEF}$
Can someone help me to finish this proof?
Thanks for antetion!
|
Let $a=\frac{a_1}{a_2}$, $b=\frac{b_1}{b_2}$, $c=\frac{c_1}{c_2}$. Per the Ceva's theorem, $abc = 1$. Evaluate
\begin{align}
\frac{AP}{PD} &= \frac{S_{ABE}}{S_{DBE}}
= \frac{\frac{b_2}{b_1+b_2}S_{ABC} }{ \frac{a_1}{a_1+a_2}\frac{b_1}{b_1+b_2}S_{ABC} }
= \frac1b(1 + \frac1a )=c(1+a)\\
\end{align}
Likewise, $\frac{BP}{PE} = a(1+b)$ and $\frac{CP}{PF} = b(1+c)$. Then,
$$\frac{PA}{PD}\frac{PB}{PE}\frac{PC}{PF}
=(1+a)(1+b)(1+c)\tag1
$$
Also,
$$\frac{S_{AEF}}{S_{ABC}} = \frac{b_2}{b_1+b_2}\frac{c_1}{c_1+c_2} =\frac c{(1+c)(1+b)} \\
\frac{S_{BDF}}{S_{ABC}}= \frac a{(1+a)(1+c)} ,\>\>\>\frac{S_{CDE}}{S_{ABC}}= \frac b{(1+b)(1+a)} $$
and
\begin{align}
\frac{S_{DEF}}{S_{ABC}} & = \frac{S_{ABC}- S_{AEF}- S_{BDF} - S_{CDE}}{S_{ABC}} \\
&= 1-\frac c{(1+c)(1+b)} - \frac a{(1+a)(1+c)} - \frac b{(1+b)(1+a)}\\
&= \frac {abc+1}{(1+a)(1+b)(1+c)}\tag2\\
\end{align}
With $abc=1$, (1) and (2) lead to
$$\frac{S_{DEF}}{2S_{ABC}}=\frac{PD}{PA}\frac{PE}{PB}\frac{PF}{PC}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3687310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluation of $\int \:\frac{1}{\sqrt[3]{\left(1+x\right)^2\left(1-x\right)^4}}dx$ using trig substitution Recently I came accross this integral :
$$ \int \:\frac{1}{\sqrt[3]{\left(1+x\right)^2\left(1-x\right)^4}}dx $$
I would evaluate it like this, first start by the substitution:
$$ x=\cos(2u) $$
$$ dx=-2\sin(2u)du $$
Our integral now becomes:
$$\int \:\frac{-2\sin \left(2u\right)du}{\sqrt[3]{\left(\cos \left(2u\right)+1\right)^2\left(\cos \:\left(2u\right)-1\right)^4}}$$
$$\cos(2u)=\cos(u)^2-\sin(u)^2$$
Thus:
$$\cos(2u)+1=2\cos(u)^2$$
$$\cos(2u)-1=-2\sin(u)^2$$
Thus our integral now becomes:
$$\int \:\frac{-\sin \left(2u\right)du}{\sqrt[3]{4\cos \left(u\right)^416\sin \left(u\right)^8}}=\frac{1}{2}\int \:\frac{-\sin \left(2u\right)du}{\sqrt[3]{\cos \left(u\right)^4\sin \left(u\right)^8}}$$
we know:
$$\sin \left(u\right)=\cos \left(u\right)\tan \left(u\right)$$
Thus our integral becomes:
$$\int \:\frac{-\tan \left(u\right)\cos \left(u\right)^2du}{\cos \:\left(u\right)^4\sqrt[3]{\tan \left(u\right)^8}}=\int \frac{-\tan \:\left(u\right)\sec \left(u\right)^2du}{\sqrt[3]{\tan \:\left(u\right)^8}}\:$$
By letting $$v=\tan \:\left(u\right)$$
$$dv=\sec \left(u\right)^2du$$
Our integral now becomes:
$$\int -v\:^{1-\frac{8}{3}}dv=-\frac{v^{2-\frac{8}{3}}}{2-\frac{8}{3}}+C=\frac{3}{2\sqrt[3]{v^2}}+C$$
Undoing all our substitutions:
$$\frac{3}{2\sqrt[3]{\tan \left(u\right)^2}}+C$$
$$\tan \:\left(u\right)^2=\frac{1}{\cos \left(u\right)^2}-1=\frac{2}{1+\cos \left(2u\right)}-1=\frac{2}{1+x}-1$$
Our integral therefore:
$$\frac{3}{2\sqrt[3]{\frac{2}{1+x}-1}}+C$$
However, online integral give me anti-derivative of $$\frac{-3\sqrt[3]{\frac{2}{x-1}+1}}{2}+C$$ so I want to know where I went wrong
| Another way (more complex than @Quanto's one)
$$I=\int\dfrac{dx}{\sqrt[3]{\left(1-x\right)^4\left(x+1\right)^2}}=\int\dfrac{dx}{\left(x-1\right)^\frac{4}{3}\left(x+1\right)^\frac{2}{3}}$$
$$u=\dfrac{1}{\sqrt[3]{x-1}}\implies I=-3\int\dfrac{du}{\left(\frac{1}{u^3}+2\right)^\frac{2}{3}}$$
$$v=\sqrt[3]{2u^3+1}\implies I=\frac 12 \int dv=\frac v 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3687521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How can I find the conditional entropy? I was given the joint pmf of 3 variables :
$$P_{X,Y,Z}(x,y,z) =
\begin{cases}
{\frac{3}{16}} &\quad\text{if }(x,y,z)\in \{001,111\}\\
{\frac{1}{8}} &\quad\text{if }(x,y,z)\in \{000,010,100,110\}\\
{\frac{1}{16}} &\quad\text{if } (x,y,z)\in \{011,101\}
\\
\end{cases}$$
and the pmf of each variable:
$$P_{X}(x) =
\begin{cases}
{\frac{1}{2}} &\quad\text{if }x\in \{0\}\\
{\frac{1}{2}} &\quad\text{if } x\in \{1\}
\\
\end{cases}$$
$$P_{Y}(y) =
\begin{cases}
{\frac{1}{2}} &\quad\text{if }y\in \{0\}\\
{\frac{1}{2}} &\quad\text{if } y\in \{1\}
\\
\end{cases}$$
$$P_{Z}(z) =
\begin{cases}
{\frac{1}{2}} &\quad\text{if }z\in \{0\}\\
{\frac{1}{2}} &\quad\text{if } z\in \{1\}
\\
\end{cases}$$
How can I find the conditional entropy $H(X,Y|Z=0)$?
| Note that simply $P_{X,Y|Z}(x,y|z=0)=\frac{P_{X,Y,Z}(x,y,z=0)}{p(z=0)}$, which is equal to $\frac{1}{4}$, in the given distribution. Hence
$$H(X,Y|Z=0)=2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3688703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $x$ and $y$ are positive integers, and $p$ is a prime, find all triples ($x$, $y$, $p$) such that $x^5 + x^4 + 1 = p^y$ If $x$ and $y$ are positive integers, and $p$ is a prime, find all triples ($x$, $y$, $p$) such that $x^5 + x^4 + 1 = p^y$. (Titu Andreescu)
My attempt:
I factorised the LHS to get $(x^3 - x +1)(x^2 + x + 1)$ = $p^y$, and then I tried to solve the 2 equations $(x^3 - x + 1) = p^m$ and $(x^2 + x + 1) = p^n$ where $m + n = y$ and they are integers, but I could not get anywhere. I also took the difference between $(x^3 - x + 1)$ and $(x^2 + x + 1)$, which was $x(x-2)(x+1)$, and I know this is divisible by $p$, but this didnt work as well. Am I going in the right direction with my first step (the factorisation), or must I try something different?
Please give some hints, but not the solution. (I want to solve it, not let others tell me the answer) Thanks a lot.
| Note that $x^5 + x^4 + 1 = (x^3 - x + 1)(x^2 + x + 1)$, and
$$x^3 - x + 1 \geq x^2 + x + 1 \iff x^3 - x^2 - 2x \geq 0 \iff x(x^2-x-2) \geq 0 \iff x \geq 2.$$
Hence, for $x \geq 2$, we know if $x^3 - x + 1 = p^m$ and $x^2 + x + 1 = p^n$, then $x^2 + x + 1 \mid x^3 - x + 1$. We have
$$\begin{aligned}
x^2 + x + 1 &\mid x^3 - x + 1\\
x^2 + x + 1 &\mid x^3 + x^2 + x\\
\implies x^2 + x + 1 &\mid x^2 + 2x - 1\\
\implies x^2 + x + 1 &\mid x - 2\\
\end{aligned}$$
If $x = 2$, then this is trivially true. Else, $x > 2$, so $x^2 + x + 1 \leq x - 2 \implies x^2 + 3 \leq 0$, which is a contradiction. Therefore, $x = 1, 2$ only, of which we check to find both work: $(x, y, p) = (1, 1, 3), (2, 2, 7)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3690560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Show inequality $a\sqrt{b - 1} + b\sqrt{a - 1} \le ab$ Given numbers $a$ and $b$; $a, b \ge 1.$
I'm trying to prove
$$a\sqrt{b-1} +b\sqrt{a - 1} \le ab.$$
Also conditions for turning it to equality.
I tried to use AM-GM to the $(a - 1)(b - 1)$, which less than ab, but got nothing.
Applying AM-GM to $a\sqrt{b-1}$ and $b\sqrt{a - 1}$ also didn't give me some result. How can i do it?
| Set $x=a-1,y=b-1.$
$$\begin{aligned}LHS&=(x+1)\sqrt y +(y+1)\sqrt x\\\\ &=\frac{(x+1)(y+1)}{2}\cdot \left[\frac{2}{\sqrt y + \frac{1}{\sqrt y}}+\frac{2}{\sqrt x + \frac{1}{\sqrt x}}\right]\\\\&\leq(x+1)(y+1)=RHS,\end{aligned}$$
because $t+{1\over t} \geq 2\;$ for all $t>0.$
Equality occurs when $t=1$ or, in terms of $a$ and $b,$ when $a=b=2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3692552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to evaluate limit of a sequence $ \lim_{n \to \infty} \frac{2 \cdot 3^{2n - 1} - \left( -2 \right)^n}{2 \cdot 3^n - 3 \cdot 2^{2n + 1}} $ I need a help with evaluating a limit of a sequence
$$ \lim_{n \to \infty} \frac{2 \cdot 3^{2n - 1} - \left( -2 \right)^n}{2 \cdot 3^n - 3 \cdot 2^{2n + 1}}. $$
The problem is that $\lim_{n \to \infty} \left( -2 \right)^n$ does not exist. What can we even tell from this when there's a part oscilating between $+\infty$ and $- \infty$. Wolfram says it should equal to $ - \infty$, but how to get there? The only thing I know might help is to factor out the fastest growing terms.
$$
\lim_{n \to \infty} \frac{2 \cdot 3^{2n - 1} - \left( -2 \right)^n}{2 \cdot 3^n - 3 \cdot 2^{2n + 1}}
=
\frac{1}{3}\lim_{n \to \infty} \frac {2 \cdot 3^{2n} - \left( -2 \right)^n}{2 \cdot 3^n - 6 \cdot 2^{2n}}
=
\frac{1}{3} \lim_{n \to \infty} \left( \frac {3}{2} \right)^{2n} \cdot \frac {2 - \left( - \frac {2} {9} \right)^n}{2 \cdot \left( \frac{3}{4} \right)^n - 6}
$$
Now we can see that $\left( \frac {3} {4} \right)^n $ goes to zero. What about $\left(- \frac{2}{9} \right)^n $? I guess that despite the fact that the values oscilate between $+$ and $-$, the overall fraction has to go to zero, hence giving
$$\left| \frac{1}{3} \cdot \infty \cdot \frac{2}{-6} \right| = \left| - \frac{\infty}{9} \right| = -\infty$$
Does that make sense?
| Divide by $3^{2n}$, numerator becomes $\frac{2}{3} - (-1)^n (\frac{2}{3})^n$, so the second term converges to 0. In he denominator you also get term that converges to $0$ - another term that converges to $0$, so in total the expression has limit of the type $\frac{1-0}{0-0}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3692904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
For angles $A$ and $B$ in a triangle, is $\cos\frac B2-\cos \frac A2=\cos B-\cos A$ enough to conclude that $A=B$? Brief enquiry:
$$\cos\frac B2-\cos \frac A2=\cos B-\cos A$$
Optionally
$$\sqrt\frac{1+\cos B}{2}-\cos B=\sqrt\frac{1+\cos A}{2}-\cos A$$
Is above equality sufficient to prove that it implies $A=B$?
Detailed explanation and motivation for this question:
Consider a triangle with bisectors of equal length:
By definition:
$\lvert AE\rvert = \lvert BD\rvert = D\\\frac A2+\frac A2 = A,\space\space\space\frac B2 + \frac B2 = B$
By cosine law:
$$\lvert AE\rvert^2=y^2+a^2-2ya\cos B\\\lvert BD\rvert^2=x^2+a^2-2xa\cos A\\x^2=\lvert BD\rvert^2+a^2-2\lvert BD\rvert a\cos \frac B2\\y^2=\lvert AE\rvert^2+a^2-2\lvert AE\rvert a\cos \frac A2$$
By sine law:
$$\frac{x}{D}=\frac{\sin \frac B2}{\sin A};\space\space\space \frac{y}{D}=\frac{\sin \frac A2}{\sin B}$$
$$\bigl[D=\lvert AE\rvert = \lvert BD\rvert\bigr]$$
Since bisectors are equal:
$$x^2-2xa\cos A=y^2-2ya\cos B\implies \lvert BD\rvert \cos \frac B2 +x\cos A = \lvert AE\rvert \cos \frac A2 +y\cos B $$
$$D (\cos\frac B2-\cos \frac A2)=y\cos B-x\cos A $$
Dividing by D and substituting y and x we obtain:
$$\cos\frac B2-\cos \frac A2=\frac{\sin \frac A2\cos B}{\sin B}-\frac{\sin \frac B2\cos A}{\sin A}$$
Consider triangles $\Delta$ABE and $\Delta$BAD
Area of triangle $\Delta$ABE:
$$A = \frac{aD}{2}\sin\frac A2 = \frac{aD}{2}\sin B \implies \sin\frac A2 = \sin B$$
Similarly for triangle $\Delta$BAD
$$A=\frac{aD}{2}\sin\frac B2 = \frac{aD}{2}\sin A\implies \sin\frac B2 = \sin A$$
Therefore:
$$\cos\frac B2-\cos \frac A2=\cos B-\cos A$$
Optionally
$$\sqrt\frac{1+\cos B}{2}-\cos B=\sqrt\frac{1+\cos A}{2}-\cos A$$
Is above equality sufficient to prove that it implies A = B?
| $$\cos\dfrac A2-\cos\dfrac B2=2\left(\cos^2\dfrac A2-\cos^2\dfrac B2\right)$$
What if $\cos\dfrac A2-\cos\dfrac B2=0?$
else
$$\dfrac12=\cos\dfrac A2+\cos\dfrac B2$$
which is clearly possible as $0<\cos\dfrac A2,\cos\dfrac B2<1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3697195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Galois group of $x^6-2x^4+2x^2-2$ over $\mathbb{Q}$
Find Galois group of $x^6-2x^4+2x^2-2$ over $\mathbb{Q}$ and describe an extension corresponding to any of it's proper subgroups of maximal order.
I know that the roots are $$\sqrt{\frac{1}{3}\left(2 - \frac{2}{\sqrt[3]{17 + 3\sqrt{33}}} + \sqrt[3]{17 + 3 \sqrt{33}}\right)}$$ and $$\sqrt{\frac{2}{3}+\frac{1\mp3i}{3\sqrt[3]{17 + 3\sqrt{33}}}-\frac{1}{6}(1\pm i\sqrt3)\sqrt[3]{17 + 3\sqrt{33}}}$$
This looks a bit too complex to me. Can you please help me?
UPD: I know that Galois group of $x^3-2x^2+2x-2$ is $S_3$.
| Hint: Write $x^6-2x^4+2x^2-2=g(x^2)$ and start by considering the spitting field of $g$ and its Galois group.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3698677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Inequality involving AM-GM but its wierd Let a, b, c be positive real numbers. Prove that
$ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+ \frac{3\cdot\sqrt[3]{abc}}{a+b+c} \geq 4$
Ohk now i know using AM-GM that
$ \frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq 3 \cdot \sqrt[3] {\frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{a}} = 3 \cdot 1=3 $
Now if i could have shown that the other term $\geq 1$. I would be done.
But the problem is that (again using AM-GM)
$\sqrt[3]{abc} \leq \frac{a+b+c}{3}
\Rightarrow 3 \cdot \sqrt[3]{abc} \leq a+b+c \Rightarrow \frac{3 \cdot \sqrt[3]{abc}}{a+b+c} \leq 1$
So if first part is $\geq 3$ and second part is $\leq 1$, How will i show that it is greater than $4$?
Is my approach correct? Or is there something wrong with the question?
Thanks.
(Source: https://web.williams.edu/Mathematics/sjmiller/public_html/161/articles/Riasat_BasicsOlympiadInequalities.pdf ,pg-$13$ exercise $1.3.4.a$ )
| Your steps are right, but they are don't give a proof.
You proved that $$A=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq3$$ and $$B=\frac{3\sqrt[3]{abc}}{a+b+c}\leq1.$$
We need prove that $$A+B\geq4,$$ which is impossible by your work.
For example, for $A=3.1$ and $B=0.8$ we obtain $$A+B\geq4$$ is wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3703357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Limitations of Arithmetic-Geometric Inequality? I have to find the range of function $f(x) = x+(1/x) +1$, where $x$ is positive. Now I did it with two ways which we can see below, in equations $(1)$ and $(2)$, by using the AM-GM inequality.
$$\frac 1 3 \left( x + \frac 1 x + 1 \right) \ge \sqrt[3]{x \cdot \frac 1 x \cdot 1} \implies x + \frac 1 x + 1 \ge 3 \tag 1$$
$$\frac 1 4 \left( x + \frac 1 x + \frac 1 2 + \frac 1 2 \right) \ge \sqrt[4]{x \cdot \frac 1 x \cdot \frac 1 2 \cdot \frac 1 2} \implies x + \frac 1 x + 1 \ge 4 \cdot (1/4)^4 \tag 2$$
So now after seeing these my question is why is my answer different in each case? Are there any limitations of this inequality?
| It is clear that $f(x)$ is unlimited from above. For a lover bound, the following is useful:
For $x>0\;$ is $x+\frac{1}{x}\geq 2,$ with equality at $x=1.$
Therefore, the range of the function $f(x)=x+\frac{1}{x}+1\;$ is $\;[3, \infty).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3704956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Tried to apply the ratio test to determine the convergence interval, but get the limit as a constant. So here is the series: $\Sigma_{n=1}^{\infty} \frac{x^{2n}}{1+x^{4n}}$
$$\left| \frac{x^{2(n+1)} (1+x^{4n})}{x^{2n}(1+x^{4(n+1)})}\right| = \left|\frac{x(1+x^{4n})}{1+x^{4n+4}} \right| \overset{\text{ n } \rightarrow \infty }{\rightarrow}\left[\frac{\infty}{\infty} \right]= \frac{x+x^{4n+1}}{1+x^{4n+4}} = \lim_{n \to \infty} \frac{x}{1+x^{4n+4}} +\lim_{ n\to \infty} \frac{x^{4n+1}}{1+x^{4n+4}} = \\ 0 + \lim_{ n\to \infty} \frac{x^{4n+1}}{1+x^{4n+4}} = \frac{\infty}{\infty} =\left[\text{ apply L'Hospital rule (4n+1)th times } \right] = \frac{1}{(4n+4)(4n+3)(4n+2)x^3} = 0 $$
Where is my mistake? Maybe some other technique would be useful for these series?
$$\lim_{n \to \infty} \frac{x}{1+x^{4n+4}} +\lim_{ n\to \infty} \frac{x^{4n+1}}{1+x^{4n+4}} \\ \text{ if } x = 0 \text{ then limit equals } 0 \\ \text{if x=1 then limit equals to 1 } \\ \text{if } 0<x<1 \text{ then limit is equal to x [ the same goes for } -1<x<0 \\ \text{ if } x> 1 \text{ then limit is equal to} \frac{1}{x^3} \text{ the same goes for x<1} $$
| Without ratio test...
For $\vert x \vert \lt 1$ you have
$$0 \le \left\vert \frac{x^{2n}}{1+x^{4n}} \right\vert \lt \vert x \vert^{2n}$$ hence the series converges as $\sum x^{2n}$ converges.
For $\vert x \vert \gt 1$ you have
$$0 \le \left\vert \frac{x^{2n}}{1+x^{4n}} \right\vert \lt 1/\vert x \vert^{2n}$$ hence the series converges for the same reason.
If $x=1,-1$ the series diverges as its term is constant and equal to $1/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3705297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
1st order linear differential equation $y'+\frac{xy}{1+x^2} =x$ Can anyone help me with this one task. I need to resolve 1st order linear equation of this equation.
$$y'+\frac{xy}{1+x^2} = x.$$
I stopped when this result came out
$$e^{\ln|y|}=e^{-\frac{1}{2}\ln|1+x^2|}\cdot e^C.$$
I try solve this by wolfram
$$y=\frac{1}{\sqrt{x^2+1}}\cdot C$$
But when I try to calculate $y'$, I get a strange equation. I think I had to be wrong somewhere. I will be grateful for your help.
| We multiply the ODE by the integrating factor $e^{\int{\frac{x}{x^2+1}}dx}=e^{\frac{ln(1+x^2)}{2}}=\sqrt{1+x^2}$ to obtain $\frac{d}{dx}(y\sqrt{1+x^2})=x\sqrt{1+x^2}$. Then integrating gives $y\sqrt{1+x^2}=\int x\sqrt{1+x^2}dx= \frac{(1+x^2)^{\frac{3}{2}}}{3}+C$ and finally we have $y(x)=\frac{C}{\sqrt{1+x^2}}+\frac{x^2+1}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3705414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
A tangent to ellipse $E_1$ intersects ellipse $E_2$ at $P$ and $Q$. Prove that the tangents to $E_2$ at $P$ and $Q$ are perpendicular to each other. A tangent to the ellipse $x^2 + 4y^2 = 4$ meets the ellipse $x^2 + 2y^2 = 6$ at P and Q. Prove that tangents at $P$ and $Q$ of ellipse $x^2 + 2y^2 = 6$ are perpendicular to each other.
Let $E_1: \frac{x^2}{4} + \frac{y^2}{1} = 1$ and $E_2: \frac{x^2}{6} + \frac{y^2}{3} = 1$ be the equation of the ellipses.
Let $R(h, k)$ be the point of tangent on $E_1$
Then let $L_1: y - k = m(x - h)$ be a line passing through two points $P$ and $Q$ on the outer ellipse. This line is tangential to $E_1$ at $R$.
To find the slope of the tangent at $R$ we take the derivative of $E_1$ at $R$.
$$m = \frac{-h}{4k}$$
$$y - k = \frac{-h}{4k}(x - h)$$
$$ h^2 + 4k^2 = 4ky + hx \tag{1}$$
We know that $R(h, k)$ satisfies $E_1$. Hence, $$h^2 + 4k^2 = 4 \tag{2}$$
$(1)$ and $(2)$ implies,
$$4ky + hx = 4 \tag{3}$$
$(3)$ is the equation of the tangent to $E_1$ at $R$.
I don't know how to proceed further.
I can think of various approaches, but can't work it out.
*
*Proving that the point of intersection of the two tangents lies on a circle of diameter $PQ$. (We know that angle in a semicircle is $90$ degrees).
*We can prove that product of slopes of the two tangents to $E_2$ is $-1$.
| Let $A=(2\cos \theta, \sin \theta)$ be the point of tangency on $E_1$. The slope of this tangent line will be
$$t_A=-\frac{\cot \theta}{2}.$$
The equation of the tangent line $\ell$ at $A$ on ellipse $E_1$ is given by:
$$\ell: x\cos \theta+2y \sin \theta=2. \tag{1}$$
Let $P=(x_1,y_1)$ and $Q=(x_2,y_2)$ be the points on $E_2$ where the tangent $\ell$ intersects $E_2$.
Consider the intersection of $E_2$ ($x^2+2y^2=6$) and $\ell$ (equ (1)):
*
*We can get a quadratic equation in $x$ which will be of the form:
$$x^2(1+\sin^2\theta)-4x \sin \theta +4(1-3\sin^2\theta)=0.$$
This should have roots $x_1,x_2$ (the $x-$coordinates of $P$ and $Q$). Using Viete's relation,
$$\color{red}{x_1x_2 = \frac{4(1-3\sin^2 \theta)}{1+\sin^2\theta}=\frac{-4(2-3\cos^2 \theta)}{1+\sin^2\theta}}.$$
*We can get a quadratic equation in $y$ which will be of the form:
$$y^2(1+\sin^2\theta)-4y \cos \theta +(2-3\cos^2\theta)=0.$$
This should have roots $y_1,y_2$ (the $y-$coordinates of $P$ and $Q$). Using Viete's relation,
$$\color{blue}{y_1y_2 = \frac{(2-3\cos^2 \theta)}{1+\sin^2\theta}}=\color{green}{\frac{-x_1x_2}{4}}.$$
Slope of tangents at $P$ and $Q$ are
$$t_P=\frac{-x_1}{2y_1} \quad \text{ and } \quad t_Q=\frac{-x_2}{2y_2}.$$
For these tangents to be perpendicular, we want to show that
$$t_p \cdot t_Q=-1 \implies \boxed{\frac{x_1x_2}{4y_1y_2}=-1}.$$
Plug in the expressions for $x_1x_2$ and $y_1y_2$ and you get the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3705940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Parametrization of Parabola After seeing recent curve I wonder if a parametrization of red curve
of equation $ \sqrt{x}+ \sqrt{y}=1 $ can be found for extended domain/range. Parametrization $ ( x= \cos^4 t, y=\sin^4 t \; )$ is bounded $\pm1$ for $(x,y).$
This interesting curve is a parabola, intersection of a cone touching three coordinate planes and another plane $z=1$.
3D equation of this cone with vertex at origin and touching the three orthogonal planes can be factored: ( actually I back calculated)
$$ x^2+y^2+z^2-2 xy-2 yz-2 zx=0 $$
$$ (x^2+y^2+z^2-2 xy +2 xz-2 ay)- 4 a x =0 $$
$$ (y-x-z)^2 - 4 ax =0 $$
$$y=x+z -2 \sqrt{zx} = ( \sqrt{z} -\sqrt{x})^2 $$
$$ \sqrt{y}= \sqrt{z} -\sqrt{x} $$
So, combination of signs there are 8 cones with their 24 parabola intersections that can be packed around the origin touching the 3 orthogonal planes along contact lines at $45^\circ$ to the axes.
$$ \pm \sqrt{x} \pm \sqrt{y} \pm \sqrt{z} =0 $$
Taking for the present case
$$z=1 \rightarrow \sqrt{x} +\sqrt{y} =1\;$$
Intersection of cones with planes parallel to generators result in parabolic arc intersections. The cones touch the coordinate planes. Hence all the parameter lines on surface are parabolas.
| One approach is to convert the equation to the standard bivariate polynomial form for a conic section. Start with the given equation $\sqrt{x}+\sqrt{y}=1$ and square both sides:
$x+2\sqrt{xy}+y=1$
$2\sqrt{xy}=1-(x+y)$, square again:
$4xy=1-2(x+y)+(x+y)^2$
Using the quarter-square multiplication formula $4xy=(x+y)^2-(x-y)^2$ we get
$1-2(x+y)+(x-y)^2=0$
Note that the variable terms are a linear term involving one combination of $x$ and $y$ and a squared linear term involving an independent linear combination of $x$ and $y$. This combination guarantees a parabola.
The derived equation lends itself to the identification
$x-y=t$, whereupon
$x+y=(1+t^2)/2$
By taking appropriate linear combinations we solve for $x$ and $y$:
$x=(1+2t+t^2)/4=(1+t)^2/4$
$y=(1-2t+t^2)/4=(1-t)^2/4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3709258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
For $\pi<\alpha<\frac{3\pi}{2}$ what is the value of $\sqrt{4\sin^4\alpha + \sin^2 2\alpha} + 4\cos^2\left(\frac{\pi}{4} - \frac{\alpha}{2}\right)$? I solved part of it this way:
$$\sqrt{4\sin^4\alpha + \sin^2 2\alpha} + 4\cos^2\left(\frac{\pi}{4} - \frac{\alpha}{2}\right)$$
$$= \sqrt{4\sin^4\alpha + 4\sin^2\alpha \cos^2\alpha} + 2\cos\left(\frac{\pi}{2} - \alpha\right) + 2$$
because $\sin 2\alpha = 2\sin\alpha\cos\alpha$ and $\cos2\alpha = \cos^2\alpha - \sin^2\alpha = 2\cos^2\alpha - 1$.
The expression then reduces to:
$$\sqrt{4\sin^2\alpha(\sin^2 \alpha + \cos^2\alpha)} + 2\sin\alpha + 2$$
since $\cos\left(\frac{\pi}{2} - \alpha\right) = \sin\alpha$. Now, because $\sin^2 \alpha + \cos^2\alpha = 1$, the expression becomes:
$$\sqrt{4\sin^2\alpha} + 2\sin\alpha + 2$$
Keep in mind that $\pi < \alpha < \frac{3\pi}{2}$. How do you proceed? I always seem to be left with extraneous sines of $\alpha$, while according to my textbook, the answer is $2$.
| Now, since $\pi<\alpha<\frac{3\pi}{2}$, it's $$-2\sin\alpha+2\sin\alpha+2=2$$ because for $x<0$ we have:
$$\sqrt{x^2}=-x.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3710752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find solution set of $\dfrac{8^x+27^x}{12^x+18^x}=\dfrac{14}{12}$ What I've done is factoring it.
$$\dfrac{2^{3x}+3^{3x}}{2^{2x}\cdot 3^{x}+3^{2x}\cdot{2^{x}}}=\dfrac{7}{2\cdot 3}$$
This looks like it can be factored more but it doesn't work from my attempts.
| Rearrange the equation
$$6(8^x+27^x)-7(12^x+18^x)=0$$
and factorize
$$(2^x+3^x)(3\cdot 3^x - 2\cdot 2^x)(2\cdot 3^x - 3\cdot 2^x)=0
$$
which leads to the solutions $x=\pm1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3711923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Finding $\lim_{n\to \infty} \sum_{r=1}^n \frac{6n}{9n^2-r^2}$ This is a question from an entrance exam for a college here in India.
We have to find $$ \lim_{n\to\infty}\sum_{r=1}^n \frac{6n}{9n^2-r^2}$$
So far I have tried:
$$ S(n)= \sum_{r=1}^n \frac{6n}{9n^2-r^2}$$
$$ S(n)= \sum_{r=1}^n \Bigl(\frac{1}{3n+r} + \frac{1}{3n-r}\Bigr)$$
from what I can notice as $n$ gets larger $\frac{1}{3n}$ gets closer to $0$ and so does the neighborhood of summation if I may call it, $\frac{1}{2n}$ and $\frac{1}{4n}$. So shouldn't it be closer to $0$. The answer key suggests that its $\log{2}$. I'm confused after the second step. Thanks for the answers.
| $S(n) = \underbrace{\sum\limits_{1 \le k \le n} \frac{1}{3n - k }}_{A} + \underbrace{\sum\limits_{1 \le k \le n} \frac{1}{3n + k}}_{B}$
Let's denote $H_{n} = \sum\limits_{1 \le k \le n} \frac{1}{k}$ - harmonic series (we assume, that it's a known series).
Than A = substitution $\{ k = 3n - k \} \rightarrow \sum\limits_{1 \le 3n-k \le n} \frac{1}{3n - (3n - k)} = \sum\limits_{2n \le k \le 3n - 1} \frac{1}{k} = \sum\limits_{1 \le k \le 3n-1}\frac{1}{k} - \sum\limits_{1 \le k \le 2n-1} \frac{1}{k} = H_{3n-1} - H_{2n-1}$.
B = substitution $\{ k = -3n + k \} \rightarrow \sum\limits_{1 \le -3n + k \le n} \frac{1}{k} = \sum\limits_{3n+1 \le k \le 4n} \frac{1}{k} = \sum\limits_{1 \le k \le 4n} \frac{1}{k} - \sum\limits_{1 \le k \le 3n} \frac{1}{k} = H_{4n} - H_{3n}$
So: A + B = $H_{4n} - H_{3n} + H_{3n-1} - H_{2n-1} = H_{4n} - H_{3n-1} - \frac{1}{3n} + H_{3n-1} - H_{2n-1} = \boxed{H_{4n} - H_{2n-1} - \frac{1}{3n}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3713885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Find the domain and range of $f(x) = \frac{x+2}{x^2+2x+1}$: The domain is: $\forall x \in \mathbb{R}\smallsetminus\{-1\}$
The range is: first we find the inverse of $f$:
$$x=\frac{y+2}{y^2+2y+1} $$
$$x\cdot(y+1)^2-1=y+2$$
$$x\cdot(y+1)^2-y=3 $$
$$y\left(\frac{(y+1)^2}{y}-\frac{1}{x}\right)=\frac{3}{x} $$
I can't find the inverse... my idea is to find the domain of the inverse, and that would then be the range of the function. How to show otherwise what is the range here?
| Notice that
\begin{align*}
f(x) = \frac{x+2}{x^{2} + 2x + 1} = \frac{(x+1) + 1}{(x+1)^{2}} = \frac{1}{x+1} + \frac{1}{(x+1)^{2}}
\end{align*}
If we set $y = 1/(x+1)$ and consider $c\in\textbf{R}$, we arrive at the following equation
\begin{align*}
y + y^{2} = c \Longleftrightarrow y^{2} + y - c = 0\Longleftrightarrow y = \frac{-1\pm\sqrt{1+4c}}{2}
\end{align*}
which has real roots if and only if $c\geq-1/4$.
So the answer is $\text{Im}(f) = [-1/4,\infty)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3715987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Closed form for $\int_0^1 \frac{\mathrm{Li}_3(-x)\mathrm{Li}_2(x)}{x}\ \mathrm{d}x$ I am looking for a closed form for:
$$\int_0^1 \frac{\mathrm{Li}_3(-x)\mathrm{Li}_2(x)}{x}\ \mathrm{d}x.$$
I am assuming integration by parts multiple times but I can't get anywhere with it. Any help/ hint would be greatly appreciated. Thanks!
| A conjectural evaluation for the remaining alternating sum is given in equation (19) of https://arxiv.org/abs/1908.04770, namely
$$
\sum_{n=1}^{\infty} \frac{(-1)^n H_{n-1}}{n^5}
= \frac{1}{13} \left( \frac{1}{3}\operatorname{Li}_6\left(-\frac{1}{8}\right) - 162 \operatorname{Li}_6\left(-\frac{1}{2}\right) - 126 \operatorname{Li}_6\left(\frac{1}{2}\right) \right)
-\frac{1787}{624} \zeta(6) + \frac{3}{8} \zeta(3)^2
\\
+\frac{31}{16}\zeta(5) \log(2) - \frac{15}{26} \zeta(4) \log^2(2) + \frac{3}{104} \zeta(2) \log^4(2) - \frac{1}{208} \log^6(2)
$$
Subtract this from $-\frac{9}{16} \zeta(6)$ to get the answer for the original question. Note the relation
$$
\int_0^1 \frac{\operatorname{Li}_3(-x)\operatorname{Li}_2(x)}{x} \mathrm{d} x
= -\frac{9}{16} \zeta(6) - \sum_{n=1}^{\infty} \frac{(-1)^n H_{n-1}}{n^5}
= -\frac{49}{32} \zeta(6) - \sum_{n=1}^{\infty} \frac{(-1)^n H_{n}}{n^5}
$$
where the denominator in the coefficient of $\zeta(6)$ in the above answer is corrected, the $-\frac{49}{28} \zeta(6)$ there should read $-\frac{49}{32}\zeta(6)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3716117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
verifying the result $\int_{B^4}e^{x^2+y^2-z^2-w^2}\,dx\,dy\,dz\,dw=\pi^2(\sinh(1)+1-\frac{1}{e})$ the problem is:
$$I=\int_{B^4} e^{x^2+y^2-z^2-w^2} \,dx\,dy\,dz\,dw $$
where $B^4$ is the unit 4 ball, explicitly: $$ B^4=\{(x,y,z,w)\in\Bbb{R}^4 \mid x^2+y^2+z^2+w^2\leq 1\}$$
here is my solution:
using fubini's theorem we obtain:
$$I=\int_{B^2}e^{-z^2-w^2}\left(\int_{{B^2}_{\sqrt{1-z^2-w^2}}} e^{x^2+y^2} \, dx\, dy\right)\,dz\,dw$$
where $B^2$ is the unit 2 ball and ${B^2}_{\sqrt{1-z^2-w^2}}$ is the 2 ball centred at the origin with radius $\sqrt{1-z^2-w^2}$.
next we calculate $$\int_{{B^2}_{\sqrt{1-z^2-w^2}}}e^{x^2+y^2}dxdy$$
using polar cordinates, so:
$$\int_{{B^2}_{\sqrt{1-z^2-w^2}}}e^{x^2+y^2}dxdy=\int_{0}^{2\pi}\int_{0}^{\sqrt{1-z^2-w^2}}e^{r^2}rdrd\theta=\pi(e^{1-z^2-w^2}-1)$$
plugging this result into the outer integral we get:
$$I=\int_{B^2}e^{-z^2-w^2}(\pi(e^{1-z^2-w^2}-1))dzdw=\pi e \int_{B^2}e^{-2(z^2+w^2)}dzdw-\pi \int_{B^2}e^{-z^2-w^2}dzdw$$
again using polar cordinates we receive:
$$\int_{B^2}e^{-2(z^2+w^2)}dzdw=\int_{0}^{2\pi}\int_{0}^{1}e^{-2r^2}rdrd\theta=\frac{\pi}{2}(1-e^{-2})$$
and:
$$\int_{B^2}e^{-z^2-w^2}dzdw=\int_0^{2\pi}\int_0^1re^{-r^2}drd\theta=\pi(1-\frac{1}{e})$$
plugging this in we get:
$$I=\pi^2(\sinh(1)+1-\frac{1}{e})$$
is this solution correct or have I made a mistake along the way?
| could you verify using two sets of polar coordinates e.g.:
$$r_1=\sqrt{x^2+y^2}$$
$$r_2=\sqrt{z^2+w^2}$$
$$x=r_1\cos\theta_1,y=r_1\sin\theta_1,z=r_2\cos\theta_2,w=r_2\sin\theta_2$$
$$dxdydzdx=r_1r_2dr_1dr_2d\theta_1d\theta_2$$
and we know that $r_1^2+r_2^2\le1$
and our integral would become:
$$\iiiint_{\Omega^4}e^{r_1^2-r_2^2}r_1r_2dr_1d\theta_1dr_2d\theta_2$$
Then we would just need to find a condition for $\theta_1,\theta_2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3717304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Finding the image of an absolute value function I am trying to find the image of the following function:
$$f:\mathbb{R}\rightarrow\mathbb{R}, \\f(x)=|1-x^2|.$$
I know trivially that the image is in ${[0,\infty[}$, but I'm looking for a way to algebraically solve it.
So I have tried taking a $y\in im(f)$
$\Rightarrow$ there must be a $x\in dom(f)$ such that $f(x)=y$
$\Rightarrow |1-x^2|=y$.
And at this point I do not know what to do. Do I separate into cases? What is the next logical algebraic reasoning?
Any help is much appreciated.
| Please note that there are usually three ways to find the range of a function. I show how they work by finding the range of the given function through these ways.
First Way
We can find the range of a function by starting from the known functions, of which we know the range, and then trying to transform them into the given function.
As we know, the range of the function $g(x)=x^2$ is $R_g=[0, \infty )$. So we have$$-x^2 \le 0 \quad \Rightarrow \quad 1-x^2 \le 1.$$Now, according to the definition of the absolute value function$$|u|= \begin{cases}u, & \text{if } u \ge 0 \\ -u, & \text{if } u \lt 0 \end{cases}.$$Since $1-x^2\le 1$,$$0\le 1-x^2 \le 1 \qquad \text{ or } \qquad 1-x^2 \lt 0.$$So we have$$\begin{cases} 0 \le 1-x^2 \lt 1 \\ 1-x^2 \lt 0 \end{cases}$$ $$\Rightarrow \quad |1-x^2|= \begin{cases}1-x^2, & \text{if } 0 \le 1-x^2 \le 1 \\ -(1-x^2), & \text{if } 1-x^2 \lt 0 \end{cases}$$ $$\Rightarrow \quad |1-x^2|= \begin{cases}1-x^2 \ge 0, & \text{if } 0 \le 1-x^2 \le 1 \\ -(1-x^2) \gt 0, & \text{if } 1-x^2 \lt 0 \end{cases}.$$Thus, we conclude that the range of the function $f(x)=|1-x^2|$ is$$R_f=[0, \infty ).$$
Second Way
We can find the range of a function by finding the inverse map of the function; the range of the function is the domain of its inverse map.
So, let us find the inverse map of the function $f(x)=|1-x^2|$, by using the definition of the absolute value function mentioned above, as follows.$$y=|1-x^2| \quad \Rightarrow \quad y=\begin{cases} 1-x^2, & \text{if } 1-x^2 \ge 0 \\-(1-x^2), & \text{if } 1-x^2 \lt 0 \end{cases}$$ $$\Rightarrow \quad \begin{cases} y=1-x^2, & \text{if } x^2 \le 1 \\y=-1+x^2, & \text{if } x^2 \gt 1 \end{cases}$$ $$\Rightarrow \quad \begin{cases} x=\pm \sqrt{1-y}, & \text{if } -1\le x \le 1 \\ x=\pm \sqrt{1+y}, & \text{if } x \in (-\infty , -1) \cup (1, \infty ) \end{cases}.$$ The domain of the radical function in the first case is $1-y \ge 0$, that is, $y \le 1$. But, since $x$ is restricted to $(-1, 1)$, $y$ can be never less than $0$. So the domain of the first case is $y \in [0, 1]$.
The domain of the radical function in the second case is $1+y \ge 0$, that is, $y \ge -1$. But, since $x$ is restricted to $(-\infty , -1) \cup (1, \infty )$, $y$ can be never less than or equal to $0$. So the domain of the first case is $y \in (0, \infty )$.
Please note that the domain of a piecewise-defined function equals the union of the domain of the pieces. So the domain of the inverse map is$$[0,1] \cup (0, \infty ) = [0, \infty ).$$Thus, we conclude that the range of the function $f(x)=|1-x^2|$ is$$R_f=[0, \infty ).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3723717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Power series approximation for $\ln((1+x)^{(1+x)}) + \ln((1-x)^{(1-x)})$ to calculate $ \sum_{n=1}^\infty \frac{1}{n(2n+1)} $ Problem
Approximate $f(x) = \ln((1+x)^{(1+x)}) + \ln((1-x)^{(1-x)})$ and then calculate $ \sum_{n=1}^\infty \frac{1}{n(2n+1)} $
My attempt
Let
$$f(x) = \ln((1+x)^{(1+x)}) + \ln((1-x)^{(1-x)}) \iff $$
$$f(x) = (1+x)\ln(1+x) + (1-x)\ln(1-x) \quad $$
We know that the basic Taylor series for $\ln(1+x)$ is
$$ \ln(1+x) = \sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1} \quad (1)$$
As far as $\ln(1-x)$ is concerned
$$y(x) = \ln(1-x) \iff y'(x) = \frac{-1}{1-x} = - \sum_{n=0}^\infty x^n \text{ (geometric series)} \iff$$
$$y(x) = \int -\sum_{n=0}^\infty x^n = - \sum_{n=0}^\infty \frac{x^{n+1}}{n+1} \quad (2)$$
Therefore from $f(x), (1), (2)$ we have:
$$ f(x) = (1+x)\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1} - (1-x)\sum_{n=0}^\infty \frac{x^{n+1}}{n+1} \iff$$
$$ f(x) = \sum_{n=0}^\infty \frac{2x^{n+2} + (-1)^n x^{n+1} - x^{n+1} }{n+1} $$
Why I hesitate
It all makes sense to me up to this point. But the exercise has a follow up sub-question that requires to find:
$$ \sum_{n=1}^\infty \frac{1}{n(2n+1)} $$
I am pretty sure that this sum is somehow connected with the previous power series that we've found, but I can't find a way to calculate it, so I assume that I have made a mistake.
Any ideas?
| Summing $$S=\sum_{n=1}^{\infty} \frac{1}{n(2n+1)}=2\sum_{k=1}^{\infty} \left( \frac{1}{2n}-\frac{1}{2n+1}\right)=2 [1/2-1/3+ 1/4-1/5_....]$$ $$\implies S=2(1-\ln 2).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3724442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Writing a system in Polar form and finding the equilibrium points
Consider the system
\begin{align}
\dot{x}&= -y+x(1-2x^2-3y^2)\\
\dot{y}&=x+y(1-2x^2-3y^2)
\end{align}
(a)Find the equilibrium points and determine their stability
(b)Rewrite the equations in polar coordinates
My Attempt:
(a) As far as I know, an equilibrium point will be a $(x,y)$ coordinate so that the Right Hand Side of the above two equations is equal to zero. But when I try to compute them directly, the calculation becomes very messy. So I would like to know whether there is a better way of doing that.
And for the stability.. should I have to consider the linearization of the system and obtain their Eigen values? (to do that I will first need the equilibrium points anyway)
(b)By using $x=r\cos\theta$ and $y=\sin\theta$, We get
\begin{align}
\dot{x}&= \dot{r}\cos\theta-r\sin\theta\times\dot{\theta}\\
\dot{y}&=\dot{r}\sin\theta+r\cos\theta.\times\dot{\theta}
\end{align}
Thus
\begin{align}
\dot{r}&= r-2r^3-r^3\sin^2\theta\\
\dot{\theta}&=1
\end{align}
I would like to know whether there is a way to get rid of that $\sin^2\theta$ term
Appreciate your help
| We have the system
$$\begin{align}
\dot{x}&= -y+x(1-2x^2-3y^2)\\
\dot{y}&=x+y(1-2x^2-3y^2)
\end{align}$$
We can solve $x' = y' = 0$ and find a single critical point of
$$(x, y) = (0, 0)$$
The Jacobian is
$$J(x, y) = \begin{bmatrix}
\dfrac{\partial x'}{\partial x}& \dfrac{\partial x'}{\partial y}\\
\dfrac{\partial y'}{\partial x}& \dfrac{\partial y'}{\partial y}
\end{bmatrix} = \begin{bmatrix} -6 x^2-3 y^2+1 & -6 x y-1 \\
1-4 x y & -2 x^2-9 y^2+1 \end{bmatrix}$$
Evaluating the Jacobian at the critical point, $J(0,0)$, we have the eigenvalues
$$\lambda_{1, 2} = 1 \pm i$$
This gives us a phase portrait (unstable spiral since we have a positive real part)
To convert to polar, using $x = r \cos \theta, y = r \sin \theta$, we have
$$\begin{align}r' &= \dfrac{xx' +yy'}{r} = \dfrac{1}{2} r \left(r^2 \cos (2 t)-5 r^2+2\right) \\\theta' &= \dfrac{y' x -x' y}{r^2} = 1 \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3730343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove Stirling's formula given that for $I_n = \int_0^{\pi/2} \sin^n\theta \, d\theta$ we have $I_{2n+1}/I_{2n} \rightarrow 1$ For $I_n = \int_0^{\pi/2} \sin^n\theta \, d\theta$ it is possible to show (using integration by parts and $\sin^2\theta = 1-\cos^2\theta$) that: $nI_n = (n-1)I_{n-2}$
We can then repeatedly apply this relation to $\frac{I_{2n+1}}{I_{2n}}$ to show:
$$\frac{I_{2n+1}}{I_{2n}} = \frac{2^{4n+1}(n!)^4}{\pi (2n)!(2n+1)!}$$
This ratio converges to $1$ since $I_{2n+2}/I_{2n} \rightarrow 1$ (easy to show with the above recursive relation) which is always larger than $I_{2n+1}/I_{2n}$. Since $I_{2n}/I_{2n} = 1$, we can sandwich the desired ratio.
I also know that (from Show that $n!e^n/n^{n+1/2} \leq e^{1/(4n)}C$) that: $r_n = n!e^n/n^{n+1/2} \leq e^{1/(4n)}C$ for all $n$ and for $C = \lim_{n\rightarrow \infty} n!e^n/n^{n+1/2}$. Now I want to show Stirling's formula:
$$n! \sim \sqrt{2\pi}n^{n+1/2}/e^n$$
What I've tried
Using the upper bound on $r_n$ it is clear that all that remains is to show that $C = \sqrt{2\pi}$. I have seen trick on related problems involving expressing the integral $I_n$ in terms of the beta function, but I think another trick is needed as well (perhaps one involving introducing a term $e^{i\theta}$ but I am struggling to get anything more specific than that.
| Note that you know a little bit more from the linked question - that $n!/(n^{n+1/2} e^{-n})$ decreases to $C$.
Using this and the upper bound you have, note that $$ \frac{I_{2n+1}}{I_{2n}} \le \frac{2^{4n + 1} C^4 n^{4n + 2} e^{-4n} e^{1/n}}{\pi \cdot C(2n)^{2n+1/2}e^{-2n} \cdot C(2n+1)^{2n + 3/2} e^{-2n - 1}} \\ = \frac{C^2}{2\pi} \frac{e^{1+1/n}}{(1 + 1/2n)^{2n + 3/2}} =: a_n$$
Since $I_{2n+1}/I_{2n} \to 1,$ we can conclude that $\liminf a_n \ge 1$. But $\liminf a_n = \frac{C^2}{2\pi}$, thus telling us that $C \ge \sqrt{2\pi}$.
Similarly, we can develop the lower bound $$ \frac{I_{2n+1}}{I_{2n}} \ge \frac{C^2}{2\pi} \frac{e}{e^{1/8n + 1/8n + 4} (1+1/2n)^{2n + 3/2}} =: b_n, $$
and argue that $\frac{C^2}{2\pi} = \limsup b_n \le 1,$ ergo $C \le \sqrt{2\pi}$.
Thus we have shown that $\sqrt{2\pi} \le C \le \sqrt{2\pi},$ which of course implies that $C = \sqrt{2\pi}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3730786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Integrate $\int\frac{2\cos x-\sin x}{3\sin x+5\cos x }\,dx$ How can I evaluate this integral $$\int\dfrac{2\cos x-\sin x}{3\sin x+5\cos x } \mathop{dx}$$
I tried using the half angle formula
$$\sin x=\dfrac{2\tan\dfrac x2}{1+\tan^2\dfrac x2}, \cos x=\dfrac{1-\tan^2\dfrac x2}{1+\tan^2\dfrac x2}$$
substituted and simplified I got
$$\int\dfrac{2-2\tan^2\dfrac{x}{2}-2\tan\dfrac{x}{2}}{5\tan^2\dfrac{x}{2}+6\tan\dfrac{x}{2}-5}dx$$
substituted $\tan^2\dfrac x2=\sec^2\dfrac x2-1$
$$\int\dfrac{4-2\sec^2\dfrac{x}{2}-2\tan\dfrac{x}{2}}{5\left(\tan\dfrac{x}{2}+\dfrac{3}{5}\right)^2-\dfrac{34}{5}}dx$$
I can't eliminate $\tan\frac x2$ term in numerator. I think I am not in right direction. your help to solve this integral is appreciated. thank in advance
| Bioche's rules say you should use the substitution
$$t=\tan x,\qquad\mathrm dx=\frac{\mathrm dt}{1+t^2},$$
to obtain the integral of a rational function in $t$, which you can compute via partial fractions decomposition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3733361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Parametrization the curve of intersection of sphere $x^2+y^2+z^2=5$ and cylinder $x^2+\left(y-\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^2$. I've seen similar posts here but none of the answers helped me. I am trying to parametrize a curve of intersection of a (top half $z>0$) sphere $x^2+y^2+z^2=5$ and cylinder $x^2+\left(y-\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^2$.
I tried $$x=\frac{1}{2}\cos t $$ $$y=\frac{1}{2}+\frac{1}{2}\sin t$$ $$z=\sqrt{5-\left(\frac{1}{2}\cos t\right)^2-\left(\frac{1}{2}+\frac{1}{2}\sin t\right)^2}=\sqrt{\frac{9}{2}-\frac{1}{2}\sin t}$$ for $t \in (0,2\pi)$, but I don't think it is correct. Even if it was, is there a better (more simple) approach?
Note: I need to find the circulation of a field $F=(y+z,x-z,0)$ over this curve so I need a good parametrizatian so that I was able to integrate it.
| Use the fact that the right-circular cylinder of radius $r$ centered at $(h, k)$ in the $xy$-plane can be parametrized by $\mathbf G(t, z) = \langle h + r \cos t, k + r \sin t, z \rangle$ by the Pythagorean Identity, i.e., $\cos^2 t + \sin^2 t = 1.$
Considering that we are in the intersection of the right-circular cylinder $(x - 0)^2 + \bigl(y - \tfrac 1 2 \big)^2 = \tfrac 1 4$ and the sphere $x^2 + y^2 + z^2 = 5,$ both of these equations hold. Using the parametrization $\mathbf G,$ we have that $x = \tfrac 1 2 \cos t$ and $y = \tfrac 1 2 + \tfrac 1 2 \sin t,$ hence plugging these identities into the equation $x^2 + y^2 + z^2 = 5$ gives rise to the equation $\tfrac 1 4 \cos^2 t + \tfrac 1 4 + \sin t + \frac 1 4 \sin^2 t + z^2 = 5.$ Of course, we can now solve for $z$ as a function $z(t)$ of $t.$
Our original parametriation $\mathbf G(t, z)$ can now be viewed as a parametrization $\mathbf G(t) = \langle \tfrac 1 2 \cos t, \tfrac 1 2 + \tfrac 1 2 \sin t, z(t) \rangle.$
One other way to parametrize the intersection of these surfaces is by spherical coordinates $x = \sqrt 5 \sin \phi \cos \theta,$ $y = \sqrt 5 \sin \phi \sin \theta,$ and $z = \sqrt 5 \cos \theta.$ By plugging these identities into the equation of the right-circular cylinder, we obtain the equation $$5 \sin^2 \phi \cos^2 \phi + 5 \sin^2 \phi \sin^2 \theta - \sqrt 5 \sin \phi \sin \theta + \frac 1 4 = \frac 1 4.$$ By the Pythagorean Identity, this simplifies to $5 \sin^2 \phi = \sqrt 5 \sin \phi \sin \theta.$ Considering that $0 \leq \phi \leq \pi,$ and $\sin \phi > 0$ whenever $0 \leq \phi < \pi,$ we can solve for $\sin \theta = \sqrt 5 \sin \phi$ whenever $0 < \phi < \pi.$ But this implies that $\theta = \arcsin(\sqrt 5 \sin \phi),$ and this is not desirable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Geometric representation of range and kernel of a linear transformation Let $T$ be a linear operator on $\Bbb{R}^3$ defined by $TX$ = $AX$ where $X$ is a $3$$\times$$1$ column vector and $A$ is
$$\begin{bmatrix}1 & 3 & 4\\3 & 4 & 7\\-2 & 2 & 0\end{bmatrix}$$
I need to show that the range and kernel of $T$ are a plane and a line passing through the origin respectively.
I have found that the dimension of range and kernel are $2$ and $1$ respectively. So all elements in the kernel are of the form $ax$ where $a$ is a real number and $x$ is a fixed element in $\Bbb{R}^3$.From here how can I conclude the above.
| First, X is not a "3 x 3 column vector"! "3 x 3" would mean a matrix with three rows and three columns. X is a 3 dimensional column vector.
The most direct way (not necessarily the simplest) to solve this is to actually find the kernel and range.
A vector $v= \begin{bmatrix}a \\ b \\ c \end{bmatrix}$ is in the range of A if and only if there exist a vector $u= \begin{bmatrix}x \\ y \\ z \end{bmatrix}$ such that Au= v.
That is $\begin{bmatrix} 1 & 3 & 4 \\ 3 & 4 & 7 \\ -2 & 2 & 0 \end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}x+ 3y+ 4z \\ 3x+ 4y+ 7z \\ -2x+ 2y\end{bmatrix}= \begin{bmatrix}a \\ b \\ c \end{bmatrix}$.
Given any $a, b, c$ there must exist $x, y, z$ such that $x+ 3y+ 4z= a, 3x+ 4y+ 7z= b$, and $=-2x+ 2y= c$. We can determine for which a, b, c that is true by finding $x, y, z$ in terms of $a, b, c$. From the last equation, $-2x+ 2y= c, y= x+ c/2$. Replacing y with that in the first two equations, $x+ 3(x+ c/2)+ 4z= 4x+ c/2+ 4z= a$ or $4x+ 4z= a- c/2$, and $3x+ 4(x+ c/2)+ 7z= 7x+ 2c+ 7z= b$ or $7x+ 7z= b- 2c$. Dividing the first of those two equations by $ =4, x+ z= a/4- c/8$. Dividing the second of those two equations by $7, x+ z= b/7- 2c/7$. Since they are both equal to $x+ z$, we must have $a/4- c/8= b/7- 2c/7$. So $a/4= b/7- 2c/7+ c/8= b/7- 9c/56, a= 4b/7- 9c/14= (8b- 9c)/14$. So any vector in the range must be of the form (assuming my arithmetic is correct) $\begin{bmatrix}\frac{8b- 9c}{14} \\ b \\ c\end{bmatrix}$. That is a two dimensional subspace (i.e. a plane). Taking $b= 7, c= 0$ we have $\begin{bmatrix} 4 \\ 7 \\ 0\end{bmatrix}$. Taking b= 0, c= 14 we have $\begin{bmatrix}-9 \\ 0 \\ 14\end{bmatrix}$. Those two vectors can be taken as basis vectors for the range.
The kernel is actually easier to find. It is the set of all vectors, u, such that Au= 0. If we take $u= \begin{bmatrix}x \\ y \\ z \end{bmatrix}$ then we must have
$Av= \begin{bmatrix} 1 & 3 & 4 \\ 3 & 4 & 7 \\ -2 & 2 & 0 \end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix}x+ 3y+ 4z \\ 3x+ 4y+ 7z \\ -2x+ 2y\end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}$.
So now we have the three equations $x+ 3y+ 4z= 0, 3x+ 4y+ 7z= 0$, and $-2x+ 2y= 0$. The last equation immediately gives y= x. Replacing y with x in the first two equations we get $4x+ 4z= 0 and 7x+ 7z= 0$. Both of those give z= -x. So every vector in the kernel is of the form $\begin{bmatrix} x \\ x \\ -x\end{bmatrix}$ for x any number. Taking x= 1 the single vector $\begin{bmatrix}1 \\ 1 \\ -1 \end{bmatrix}$ spans that one dimensional subspace.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3736378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\sqrt{a+b+\sqrt{\left(2ab+b^2\right)}}$ Evaluate $\sqrt{a+b+\sqrt{\left(2ab+b^2\right)}}$
My attempt:
Let $\sqrt{a+b+\sqrt{\left(2ab+b^2\right)}}=\sqrt{x}+\sqrt{y}$
Square both sides: $a+b+\sqrt{\left(2ab+b^2\right)}=x+2\sqrt{xy}+y$
Rearrange: $\sqrt{\left(2ab+b^2\right)}-2\sqrt{xy}=x+y-a-b$
That's where my lights go off.
Any leads? Thanks in advance.
| After taking squares, you can proceed as follows
$$(a+b)+\sqrt{\left(2ab+b^2\right)}=(x+y)+2\sqrt{xy}$$
Compare corresponding (conjugate) parts on both the sides of above equation, we get
$$x+y=a+b\tag 1$$
$$2\sqrt{xy}=\sqrt{2ab+b^2}\iff 4xy=2ab+b^2\tag 2$$
$$x-y=\pm\sqrt{(x+y)^2-4xy}=\pm\sqrt{(a+b)^2-(2ab+b^2)}=\pm a\tag3$$
Solving (1) & (3) we get $x$ & $y$ as follows
$$x=a+\frac{b}{2}, \ y=\frac b2\ \ \ \text{OR}\ \ \ x=\frac b2, \ y=a+\frac{b}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3738201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Partition of $\Bbb R^2$ arising from matrix multiplication. Describe the partition of $\Bbb R^2$ arising from the action, by matrix multiplication, of the subgroup $H=\left\{\begin{pmatrix}
1 & 0\\
a & 1
\end{pmatrix}:a\in\Bbb R\right\}$ of $GL_2(\Bbb R)$.
First of all I don't think I really understand the question itself. Does it means I pick elements of $H$ and apply matrix multiplication to $\begin{pmatrix}
x\\
y
\end{pmatrix}$ where $x,y\in\Bbb R$? I don't know how to describe the partition of a cartesian in terms of matrix multiplication And don't know what partition to describe I am confused. And why is the answer like this:"The lines $x=c$ for $c\neq0$ and the points $(0,b)$"?
| Let's first prove that indeed $H\le\operatorname{GL}_2(\mathbb{R})$, and that we really have a $H$-action on $\mathbb{R}^2$.
*
*Subgroup - closure:
$\space\begin{pmatrix}
1 & 0 \\
a & 1 \\
\end{pmatrix}
\begin{pmatrix}
1 & 0 \\
b & 1 \\
\end{pmatrix}=
\begin{pmatrix}
1 & 0 \\
a+b & 1 \\
\end{pmatrix}\in H$;
*Subgroup - "closure by inverses":
$\space\begin{pmatrix}
1 & 0 \\
a & 1 \\
\end{pmatrix}^{-1}=
\begin{pmatrix}
1 & 0 \\
-a & 1 \\
\end{pmatrix}\in H$;
*Action - property #0:
$\space\begin{pmatrix}
1 & 0 \\
a & 1 \\
\end{pmatrix}
\begin{pmatrix}
x \\
y \\
\end{pmatrix}
=
\begin{pmatrix}
x \\
ax+y \\
\end{pmatrix} \in \mathbb{R}^2
$
*Action - property #1:
$\space\begin{pmatrix}
1 & 0 \\
0 & 1 \\
\end{pmatrix}
\begin{pmatrix}
x \\
y \\
\end{pmatrix}
=
\begin{pmatrix}
x \\
y \\
\end{pmatrix}
,\space\space\forall \begin{pmatrix}
x \\
y \\
\end{pmatrix}\in \mathbb{R}^2$
*Action - property #2:
$
\space\biggl(\begin{pmatrix}
1 & 0 \\
a & 1 \\
\end{pmatrix}
\begin{pmatrix}
1 & 0 \\
b & 1 \\
\end{pmatrix}
\biggr)
\begin{pmatrix}
x \\
y \\
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 \\
a+b & 1 \\
\end{pmatrix}
\begin{pmatrix}
x \\
y \\
\end{pmatrix}
=
\begin{pmatrix}
x \\
(a+b)x+y \\
\end{pmatrix}
=
\begin{pmatrix}
x \\
ax+bx+y \\
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 \\
a & 1 \\
\end{pmatrix}
\begin{pmatrix}
x \\
bx+y
\end{pmatrix}
=
\begin{pmatrix}
1 & 0 \\
a & 1 \\
\end{pmatrix}
\biggl(
\begin{pmatrix}
1 & 0 \\
b & 1 \\
\end{pmatrix}
\begin{pmatrix}
x \\
y \\
\end{pmatrix}
\biggr)
,\space\space
\forall \begin{pmatrix}
1 & 0 \\
a & 1 \\
\end{pmatrix}, \begin{pmatrix}
1 & 0 \\
b & 1 \\
\end{pmatrix}\in H, \space\space\forall \begin{pmatrix}
x \\
y \\
\end{pmatrix}\in \mathbb{R}^2
$
From the general action theory, we know that the "acted upon" set, i.e. here $\mathbb{R}^2$, is partitioned into "orbits", given by (here $X$ is the generic point of $\mathbb{R}^2$):
$$O(X)=\{AX, A\in H\}=\biggl\{\begin{pmatrix}
x \\
ax+y \\
\end{pmatrix}, \space \space a\in \mathbb{R}\biggr\}\tag 1$$
By $(1)$, the orbits by $x=0$ points "don't feel" the action of $H$ (via the parameter $a$): the points of the $y$-axis ($x=0$) constitue each a "degenerate" orbit:
$$O(X_{x=0})=O(y)=\biggl\{\begin{pmatrix}
0 \\
y \\
\end{pmatrix}, \space \space a\in \mathbb{R}\biggr\}=\biggl\{\begin{pmatrix}
0 \\
y \\
\end{pmatrix}\biggr\}$$
Off $y$-axis points ($x\ne 0$), instead, are moved all along the "vertical" direction by the point $(x,y)$. Summarizing, the partition of $\mathbb{R}^2$ induced by this action is made of single-point orbits (one for each point of the $y$-axis) and "vertical", straight lines for the rest of the real plane.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3738666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Limit of a fraction involving square and third roots. $$\lim_{n\to\infty} \frac{\sqrt{n^2+1} - \sqrt{n^2+n}}{\sqrt[3]{n^3+1} - \sqrt[3]{n^3+n^2+1}}$$
Is it a good idea to substitute the numerator and denominator using that $a-b=\frac{a^2-b^2}{a+b}$ and $a-b=\frac{a^3-b^3}{a^2+ab+b^2}$, since Iβll end up having to multiplicate a polynomial by square and third roots?
| Instead, write $\sqrt{n^2+1}= |n|\sqrt{1+\frac{1}{n^2}}$, and similar for other expression in the numerator. For the denominator, multiply both numerator and denominator. Then in the denominator you will have $n^2$, and in the numerator the term
$$
(n^3+1)^{\frac{2}{3}} + ((n^3+1)(n^3+n^2+1))^{\frac{2}{3}} + (n^3+n^2+1)^{\frac{2}{3}}
$$
Then, you can perform the same operation on them, e.g. on the first one it is $|n|(1+\frac{1}{n^3})^{\frac{1}{3}}$, and a lot of terms will cancel out
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3738751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
If $a=7!$, $b=_{13}P_k$, $\frac{ab}{\operatorname{lcm}(a,b)}=120$, find the $k$
Question:
$a=7!$, $b=_{13}P_k$, $\dfrac{ab}{\operatorname{lcm}(a,b)}=120$, then find the $k$.
My attempts:
$$\frac{ab}{\operatorname{lcm}(a,b)}=\gcd{(a,b)}=120$$
$\gcd{(7!, _{13}P_k)}=120 \Longrightarrow \begin{cases} \dfrac{13!}{(13-k)!Γ5!}\in \mathbb {Z^+} \\ \dfrac{13!}{(13-k)!Γ5!} \mod 2β 0 \\
\dfrac{13!}{(13-k)!Γ5!} \mod 3β 0 \\
\dfrac{13!}{(13-k)!Γ5!} \mod 7β 0 \end{cases} \Longrightarrow \begin{cases}
\dfrac{13Γ12Γ\cdots (13-k+1)!}{120} \in \mathbb {Z^+} \\
\dfrac{13!}{(13-k)!Γ5!} \mod 2β 0 \\
\dfrac{13!}{(13-k)!Γ5!} \mod 3β 0 \\
\dfrac{13!}{(13-k)!Γ5!} \mod 7β 0 \end{cases} \Longrightarrow 13-k+1=10 \Longrightarrow k=4$
Is my solution correct? Do I have any missing or unnecessary/unneeded steps?
| This was based on the original post
The problem as stated is incorrect or there is a typo. If $\frac{ab}{\gcd(a,b)}=120$, then that would mean $\text{lcm}(a,b)=120$. But $120=\text{lcm}(a,b) \geq a=7!$. This is not possible.
After the original post got modified:
Perhaps, the $\gcd(a,b)=120$. In which case we can do the following:
\begin{align*}
\gcd(a,b)&=120\\
\gcd(7!,b)&=120\\
\gcd\left(42,\frac{b}{120}\right)&=1
\end{align*}
This means $\frac{b}{120}$ is (at least) not divisible by $2,3$ and $7$, where $b=\frac{13!}{(13-k)!}$. We know that $13!=2^{10} \cdot 5^2 \cdot 7^{1} \dotsb$. So
$$\frac{b}{120}=\frac{13!}{(2^3\cdot 3 \cdot 5) \, (13-k)!}=\frac{2^{7} \cdot 3^4\cdot 5^1 \cdot 7^{1} \dotsb}{(13-k)!}$$
This means our $k$ should be such that the prime factorization of $(13-k)!$ should also have exactly these powers of the primes $2,3$ and $7$.
Since $2$ should only appear with exponent $7$ in the prime factorization of $(13-k)!$, so $13-k \geq 8 \implies k \leq 5$ and $13-k \leq 9 \implies k \geq 4$.
Since $3$ should only appear with exponent $4$ in the prime factorization of $(13-k)!$, so $13-k \geq 9 \implies k \leq 4$.
Thus $\color{red}{k=4}$ will satisfy the gcd condition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3739141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Show that if $|z| < 1$, then $\displaystyle\frac{z}{1+z} + \frac{2z^{2}}{1+z^{2}} + \frac{4z^{4}}{1+z^{4}} + \frac{8z^{8}}{1+z^{8}} + \ldots$ Show that if $|z| < 1$, then
\begin{align*}
\frac{z}{1+z} + \frac{2z^{2}}{1+z^{2}} + \frac{4z^{4}}{1+z^{4}} + \frac{8z^{8}}{1+z^{8}} + \ldots
\end{align*}
converges
MY ATTEMPT
Let us start by noticing that
\begin{align*}
\frac{z}{1-z} - \frac{z}{1-z} + \frac{z}{1+z} = \frac{z}{1-z} + \frac{z - z^{2} -z - z^{2}}{1-z^{2}} = \frac{z}{1-z} - \frac{2z^{2}}{1-z^{2}}
\end{align*}
Similarly, we have that
\begin{align*}
\frac{z}{1-z} - \frac{2z^{2}}{1-z^{2}} + \frac{2z^{2}}{1+z^{2}} = \frac{z}{1-z} + \frac{2z^{2} - 2z^{4} - 2z^{2} - 2z^{4}}{1-z^{4}} = \frac{z}{1-z} - \frac{4z^{4}}{1-z^{4}}
\end{align*}
Based on this pattern, it can be proven by induction the given series is the same as
\begin{align*}
\frac{z}{1-z} = z + z^{2} + z^{3} + \ldots = \sum_{n=1}^{\infty}z^{n}
\end{align*}
which converges, since it is the geometric series inside the circle $|z| < 1$
(we may apply the ratio test, for instance).
Could someone tell me if my approach is correct? Any contribution is appreciated.
| Your approach is correct! It can be verified in the following way too.
Consider the partial sums $$S_n=\sum_{j=0}^{n}\frac{2^jz^{2^j}}{1+z^{2^j}}$$
Then $$\sum_{m=0}^{\infty}\frac{2^mz^{2^m}}{1+z^{2^m}}=\lim_{n\rightarrow\infty}S_n$$
Observe (prove by induction on $n$) that $$S_n+\frac{z}{z-1}=\frac{2^{n+1}z^{2^{n+1}}}{z^{2^{n+1}}-1}$$
Now $$\frac{z}{z-1}+\sum_{m=0}^{\infty}\frac{2^mz^{2^m}}{1+z^{2^m}}=\lim_{n\rightarrow\infty}\frac{2^{n+1}z^{2^{n+1}}}{z^{2^{n+1}}-1}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3740757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\lim\limits_{x \to 1} \left( \frac{1}{x-1}-\frac{3}{1-x^3} \right)$ Evaluate $$\lim_{x \to 1} \left( \frac{1}{x-1}-\frac{3}{1-x^3} \right)$$
My attempt: $$\lim_{x \to 1} \left( \frac{1}{x-1}-\frac{3}{1-x^3} \right) = \lim_{x \to 1} \frac{x+2}{x^2+x+1}=1$$
According to the answer key, this limit does not exist. I turned that into one fraction, then I factored the polynomial on the numerator as $-(x-1)(x+2)$ and the one on the denominator as $(x+1)(-x^2-x-1)$. What did I do wrong?
| $$\frac1{x-1}-\frac3{1-x^3}=\frac1{x-1}+\frac3{x^3-1}\to\infty$$
as the terms do not cancel each other (they have the same sign).
Assuming a typo,
$$\frac1{x-1}+\frac3{1-x^3}=\frac1{x-1}-\frac3{x^3-1}=\frac{x^2+x-2}{x^3-1}=\frac{x+2}{x^2+x+1}\to1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3745531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Polynomial $f(x) = ax^2 β bx + c $ (where $a$, $b$ & $c$ are positive integers) Let the polynomial $f(x) = ax^2 β bx + c $ (where $a$, $b$ & $c$ are positive integers). If $f(p) = f(q) = 0$, where $ 0 < p < q < 1$, then find the minimum possible value of $a$.
The vertex is $-\frac{-b}{2a}=\frac{b}{2a}>0$ and lies between $0$ & $1$.
$f(0)>0$ and also $f(1)>0$, hence $c>0$ and $a-b+c>0$, also $b^2-4ac>0$. Even after proceeding up to these steps I am not able to find the minimum value of $a$.
| Let the quadratic be $a(x-u)^2-v$, where $u=(p+q)/2$ and $-v=f(u)\lt0$.
Then $f(0)=au^2-v\ge1$ so $au^2\gt1$ and likewise $a(1-u)^2\gt1$. Either $u\le\frac12$ or $1-u\le\frac12$ so $\frac a4\gt1$, and $a\ge5$.
An example is $f(x)=5x^2-5x+1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3746177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find dV/dA in terms of r. The volume of a sphere, $V$ cmΒ³, of radius $r$ is given by the formula $V = \frac{4}{3} \pi r^3$. The surface area of a sphere $A$ cmΒ² of radius $r$ cm is given by the formula $A=4\pi r^2$. Find $dV/dA$ in terms of $r$.
Here's my workings to the question:
$$V= \frac{\frac{4}{3}\pi r^{3}}{4\pi r^2}A = \frac{1}{3}rA$$
So, $$\frac{dV}{dA} = \frac{1}{3}r =\frac{r}{3}.$$
I am not sure about this answer, so it would help to know if anyone got the same answer. Thank you!
| Given $V= \frac{4\pi}3 r^3$ and $A=4\pi r^2$, we have
$$V= \frac{4\pi}3\left(\frac{A}{4\pi}\right)^{3/2}= \frac{A^{3/2}}{3\sqrt{4\pi}}
$$
Then,
$$\frac{dV}{dA}= \frac12\frac{\sqrt A}{\sqrt{4\pi}}= \frac r2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3751080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
How can I integrate $\int \frac{u^3}{(u^2+1)^3}du?$ How to integrate following
$$\int\frac{u^3}{(u^2+1)^3}du\,?$$
What I did is here:
Used partial fractions
$$\dfrac{u^3}{(u^2+1)^3}=\dfrac{Au+B}{(u^2+1)}+\dfrac{Cu+D}{(u^2+1)^2}+\dfrac{Au+B}{(u^2+1)^3}$$
After solving I got
$A=0, B=0, C=1, D=0, E=-1, F=0$
$$\dfrac{u^3}{(u^2+1)^3}=\dfrac{u}{(u^2+1)^2}-\dfrac{u}{(u^2+1)^3}$$
Substitute $u^2+1=t$, $2u\ du=dt$, $u\ du=dt/2$
$$\int\frac{u^3}{(u^2+1)^3}du=\int \frac{dt/2}{t^2}-\int \frac{dt/2}{t^3}$$
$$=\frac12\dfrac{-1}{t}-\frac{1}{2}\dfrac{-1}{2t^2}$$
$$=-\dfrac{1}{2t}+\dfrac{1}{4t^2}$$
$$=-\dfrac{1}{2(u^2+1)}+\dfrac{1}{4(u^2+1)^2}+c$$
My question: Can I integrate this with suitable substitution? Thank you
| Substitute $u=\tan\theta\implies du=\sec^2\theta \ d\theta$
$$\int \frac{u^3}{(u^2+1)^3}du=\int \frac{\tan^3\theta}{(\tan^2\theta+1)^3}\sec^2\theta\ d \theta$$
$$=\int \frac{\tan^3\theta\sec^2\theta}{\sec^6\theta}\ d\theta$$
$$=\int\sin^3\theta\cos\theta d\theta$$
$$=\int\sin^3\theta\ d(\sin\theta)$$
$$=\frac{\sin^4\theta}{4}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3753883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 10,
"answer_id": 0
} |
When the function equation $f(x)f(y)=axy+b$ is solvable
Assume $a,b$ are constants. The question is whether there is a continuous function $f$ defined on $\mathbb R$ or $\mathbb C$ so that
$$
f(x)f(y)=axy+b
$$
Of course, such a function $f$ exists if $b=0$ by taking $$f(x)=\sqrt{a}x\,.$$ Likewise if $a=0$ then $f$ exists by taking $$f(x)=\sqrt{b}\,.$$ But I don't know whether the condition $a=0$ or $b=0$ is also necessary for the solvability of this function equation.
| If $x$ and $y$ are independent variables, you only have to set $y = x$, to get $f(x)^2 = ax^2 + b$ and therefore $f(x) = \sqrt{ax^2 + b}$. Now,
$$f(x)f(y) = \sqrt{(ax^2 + b)(ay^2 + b)} = \sqrt{a^2x^2y^2 + abx^2 + aby^2 + b^2}$$
Needs to be equal to $axy + b$, i.e. taking the squares of both expressions
$$a^2x^2y^2 + abx^2 + aby^2 + b^2 = (axy + b)^2 = a^2x^2y^2 + 2abxy + b^2$$
which entails $a = 0$ or $b = 0$. Thus it is either $f(x) = x\sqrt{a}$ or $f(x) = \sqrt{b}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find range of $f(x)=\frac{5}{\sin^2x-6\sin x\cos x+3\cos^2x}$ Find range of $f(x)=\frac{5}{\sin^2x-6\sin x\cos x+3\cos^2x}$
My attempt :
\begin{align*}
f(x)&=\dfrac{5}{9\cos^2x-6\sin x\cos x+\sin^2x-6\cos^2x}\\
&= \dfrac{5}{(3\cos x+\sin x)^2-6\cos^2x}
\end{align*}
The problem is if I'm going to use
$$-1\leqslant\sin x\leqslant1\;\text{and}-1\leqslant\cos x\leqslant1$$
I think I need to have only one term.
Edit : I have made some more progress
$$-3\leqslant 3\cos x\leqslant 3$$
$$\therefore -4\leqslant 3\cos x+\sin x\leqslant 4$$
$$ 0\leqslant (3\cos x+\sin x)^2\leqslant 16$$
| this $f(x)=\frac{5}{\sin^2x-6\sin x\cos x+3\cos^2x}$ can be simplified more neatly, if you use:
$\sin^2(\theta)+ \cos^2(\theta)=1$
$\cos(2\theta)=2\cos^2(\theta)-1$
$\sin(2\theta)=2\sin(\theta)\cos(\theta)$
And the simplified denominator would be in the form:
$$ a\cos(2x)+b\sin(2x)+c$$
Then you can use $$-\sqrt{a^2+b^2} \le a\cos(x) \pm b\sin(x) \le \sqrt{a^2+b^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3755235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Rotational Volume of $y^2 = \frac{x^3}{2a-x}$ around asymptote $x = 2a$ My approach is to use a shell method over the range $[0, 2a]$. One cylinder will be $C = 2\pi xy \ dx$. Let us only work with the positive quadrant and multiply by two for ease:
https://www.desmos.com/calculator/pvuquz8orz
then the volume is: $$V = 2 \int_{0}^{2a} 2 \pi (2a-x)\sqrt{\frac{x^3}{2a-x}} \ dx$$
One question I have is that when $x \rightarrow 2a$ it goes to infinity, would this mean I need to treat the definite integral as:
$$V = 2 \lim_{b\to2a}\int_{0}^{b} 2 \pi (2a-x)\sqrt{\frac{x^3}{2a-x}} \ dx = 4 \pi \lim_{b\to2a}\int_{0}^{b} (2a-x)\sqrt{\frac{x^3}{2a-x}} \ dx$$
I am not sure how to evaluate this integral either. Is this setup at least conceptually correct?
| You're on the right track. As long as $x>0$ then
$$ 4\pi\int_0^{2a} (2a-x)x \sqrt{\frac{x}{2a-x}}\:dx = 4\pi\int_0^{2a} x \sqrt{x(2a-x)}\:dx$$
Then use the substitution $x=2a\sin^2\theta$:
$$= 64\pi a^3 \int_0^{\frac{\pi}{2}} \sin^4\theta\cos^2\theta\:d\theta$$
Denote $I$ as just the integral part without the constants. Under the variable interchange $\theta\leftrightarrow \frac{\pi}{2} -\theta$ we get that
$$2I = \int_0^{\frac{\pi}{2}}\sin^4\theta\cos^2\theta+\cos^4\theta\sin^2\theta\:d\theta = \frac{1}{8}\int_0^{\frac{\pi}{2}}\sin^2 2\theta\:2d\theta$$
Using the fact that $\int_a^b \sin^2 t \:dt = \frac{b-a}{2}$ whenever $a$ and $b$ are integer multiples of $\frac{\pi}{2}$, we can simplify this down to
$$I = \frac{\pi}{32}$$
which means our final answer is
$$2\pi^2 a^3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3757285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the greatest integer less than $3^\sqrt{3}$ without using a calculator and prove the answer is correct. Find the greatest integer less than $3^\sqrt{3}$ without using a calculator and prove the answer is correct.
I'm puzzled on how to solve this problem, any help is appreciated. There was hints about turning the exponents into fractions and picking fractions between : $3^x < 3^\sqrt3 <3^y$
Then I simplified: $x< \sqrt3<y$
$x^2< 3<y^2$
$\sqrt2^2<3<\sqrt4^2$
So $x=\sqrt2$ and $y=\sqrt4=2$
$3^\sqrt2 < 3^\sqrt3 <3^2$
| Since $3 = \frac{48}{16} <\frac{49}{16}$, you have $\sqrt{3}<\frac{7}{4}.$ So you might try to take $y=\frac{7}{4}.$ It's easy to calculate $3^7 = 2187$ which is close to $2401 = 7^4.$ So $3^7<7^4$ and you have $3^{7/4}<7.$ So your answer is $6$ or less.
Try $x=5/3$ to get the lower bound.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3757754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Integral: $\int \dfrac{dx}{(x^2-4x+13)^2}$?
How can I integrate $$\int \dfrac{dx}{(x^2-4x+13)^2}?$$
Here is my attempt:
$$\int \dfrac{dx}{(x^2-4x+13)^2}=\int \dfrac{dx}{((x-2)^2+9)^2}$$
Substitute $x-2=3\tan\theta$, $\ dx=3\sec^2\theta d\theta$
\begin{align*}
&=\int \dfrac{3\sec^2\theta d\theta}{(9\tan^2\theta+9)^2}\\
&=\int \dfrac{3\sec^2\theta d\theta}{81\sec^4\theta}\\
&=\dfrac{1}{27}\int \cos^2\theta d\theta\\
&=\dfrac{1}{27}\int \frac{1+\cos2\theta}{2} d\theta\\
&=\dfrac{1}{54}\left(\theta+\frac{\sin2\theta}{2}\right)+C
\end{align*}
This is where I got stuck. How can I get the answer in terms of $x$?
Can I solve it by other methods?
| Since you selected $$x - 2 = 3 \tan \theta$$ as your substitution, it follows that $$\tan \theta = \frac{x-2}{3},$$ and by considering a right triangle with legs $3$ and $x-2$ with hypotenuse $\sqrt{3^2 + (x-2)^2}$ via the Pythagorean theorem, we obtain $$\sin \theta = \frac{x-2}{\sqrt{3^2 + (x-2)^2}}, \\ \cos \theta = \frac{3}{\sqrt{3^2 + (x-2)^2}}.$$ Therefore, $$\frac{1}{2} \sin 2\theta = \sin \theta \cos \theta = \frac{3(x-2)}{3^2 + (x-2)^2} = \frac{3(x-2)}{x^2 - 4x + 13}.$$ We also easily have $$\theta = \tan^{-1} \frac{x-2}{3}.$$ Therefore $$\frac{1}{54}\left( \theta + \frac{1}{2} \sin 2\theta \right) = \frac{1}{54} \left( \tan^{-1} \frac{x-2}{3} + \frac{3(x-2)}{x^2 - 4x + 13} \right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3761986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 9,
"answer_id": 0
} |
How to evaluate $\int x^5 (1+x^2)^{\frac{2}{3}}\ dx$? I am trying to evaluate $\int x^5 (1+x^2)^{\frac{2}{3}} dx$
This is apparently a binomial integral of the form $\int x^m (a+bx^k)^ndx$. Therefore, we can use Euler's substitutions in order to evaluate it. Since $\dfrac{m+1}{k} = \dfrac{5+1}{2} = 3 \in \mathbb{Z}$ we will use the substitution: $$ a+bx^k = u^{\frac{1}{n}}$$
Therefore,
$$ u^3 = 1 + x^2 \iff x = \sqrt{u^3 +1} \text{ (Mistake Here. Check the comments) } $$ $$\iff dx = \frac{3u^2}{2\sqrt{u^3}+1}$$
So the new integral is,
$$ \int x^5 (1+x^2)^{\frac{2}{3}} dx = \frac{3}{2} \int u^4 (u^3+1)^{\frac{7}{6}} du$$
Instead of simplifying the integral, the substitution did nothing by keeping it at the same form, with different values on the variables $m,k,n$.
I tried to substitute once again and it doesn't seem to lead in any known paths, anytime soon.
Any ideas on how this could be evaluated?
| Take $x^5 \times (1 + x^2)^{\frac{2}{3}} = (x^{15} \times (1+x^2)^2)^{\frac{1}{3}}$ or some other form of that sort and integrate.
Thus, you'll be integrating $(x^{19} + 2x^{17} + x^{15})^{\frac{1}{3}}$
Applying substitution method here may work.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3762071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Check and comment my proof of $a+b \geq 2 \sqrt{ab}$ I want to prove
$$
a+b \geq 2 \sqrt{ab}, \quad \text{for} \quad a, b>0 \tag 1
$$
First question:
Isn't this wrong? Shouldn't it be
$$
a+b \geq 2 \sqrt{ab}, \quad \text{for} \quad a, b\geq0 \tag 2
$$
Or
$$
a+b > 2 \sqrt{ab}, \quad \text{for} \quad a, b>0 \tag 3
$$
And the proof:
I start backwards.
\begin{align}
a+b-2&\sqrt{ab}\geq 0 \tag 4 \\
(\sqrt a)^2+(\sqrt b)^2-2&\sqrt{ab}\geq 0 \tag 5 \\
(\sqrt a)^2+(\sqrt b)^2-2&\sqrt{a}\cdot \sqrt{b}\geq 0 \tag 6
\end{align}
Let $x= \sqrt a$ and $y=\sqrt b$, so we see that
\begin{align}
x^2+y^2-2xy&\geq 0 \tag 7\\
(x-y)^2\geq 0 \tag 8
\end{align}
So finally we have
$$
(\sqrt a- \sqrt b)^2 \geq 0 \tag 9
$$
If $a=b=0$ we have $0\geq 0$ which is true for $0=0$ ($0>0$ is always false).
For $a>0$ and $b>0$ the square $(\sqrt a- \sqrt b)^2$ is always strictly positive.
And also, $a<0$ and $b<0$ is not valid because $\sqrt a$ and $\sqrt b$ are only defined for $a\geq 0$ and $b\geq 0$.
| I would just want to point out that in the third equation and the proof,
\begin{align} \text{for }~ a ,~ b > 0 , \ ( \sqrt{a} - \sqrt{b} )^2 \end{align}
is non-negative, since both of them could be equal.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3762541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to solve polynomial rational relations for $y$ (e.g $\sqrt{4-3y-y^2} = x(y+4)$)? From time to time, I struggle to solve polynomial relations for $y$.
A trivial example is :
$$ \frac{y}{x} = x \iff y = x^2$$
Easy.
But consider this relation:
$$ \sqrt{4-3y-y^2} = x(y+4)$$
No matter how much I mess around it, seems impossible to bring it in $y = f(x)$ form.
*
*$ \frac{\sqrt{4-3y-y^2}}{(y+4)}= x $
*$ 4 - 3y - y^2 = x^2(y^2 + 8y + 16) \iff (x^2-1)y^2+y(-8x-3)+4(1-4x) = 0$
Is there a trivial methodology that I am missing or is it indeed impossible to inverse some relations?
| For $-4<y\le 1, x>0$
We have $$\sqrt{(y+4)(1-y)}=x (y+4) \implies (1-y)(y+4)=x^2(y+4)^2 \implies \frac{1-y}{y+4}=x^2$$
$$\implies y(x)=\frac{1-4x^2}{1+x^2}.$$
See the graph of $y(x)$ below:
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3764098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Solving for positive reals: $abcd=1$, $a+b+c+d=28$, $ac+bc+cd+da+ac+bd=82/3$ $$a,b,c,d \in \mathbb{R}^{+}$$
$$ a+b+c+d=28$$
$$ ab+bc+cd+da+ac+bd=\frac{82}{3} $$
$$ abcd = 1 $$
One can also look for the roots of polynomial
$$\begin{align}
f(x) &= (x-a)(x-b)(x-c)(x-d) \\[4pt]
&= x^4 - 28x^3 + \frac{82}{3}x^2 - (abc+abd+acd+bcd)x + 1
\end{align}$$
and $f(x)$ has no negative roots... but how else do I proceed?
There is a trivial solution $\frac{1}{3}, \frac{1}{3}, \frac{1}{3}, 27$. We just need to prove it's unique.
| The hint.
By your work $$\frac{x^4-28x^3+\frac{82}{3}x^2+1}{x}-\frac{244}{27}=\frac{(3x-1)^3(x-27)}{27x}$$ because by Rolle $$\left(\frac{x^4-28x^3+\frac{82}{3}x^2+1}{x}\right)'=\frac{(3x-1)^2(x^2-18x-3)}{3x^2}$$ has three positive roots and one of them must be $\frac{1}{3},$ which is also an element of $\{a,b,c,d\}$.
Indeed, let $0<a\leq b\leq c\leq d$.
Thus, $f'$ has positive roots on $[a,b]$ on $[b,c]$ and on $[c,d]$ and we know that one of these roots it's double $\frac{1}{3}$.
Let $\frac{1}{3}\in[a,b]$ and $\frac{1}{3}\in[b,c].$
Thus, $b=\frac{1}{3}$, which says $$abc+abd+acd+bcd=\left(\frac{x^4-28x^3+\frac{82}{3}x^2+1}{x}\right)_{x=\frac{1}{3}}=\frac{244}{27}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3764736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Help finding a centre of a circle I have been trying to work out this problem for hours but I cannot get it. A circle has its centre on the positive $x$-axis at $(a,0)$. The radius of this circle is $4$. Lines $y=2x$ and $y=-2x$ are tangent to the circle. I have to find the $x$-coordinate $a$.
So I get that the equation for this circle should be $(x-a)^2+y^2=16$, and then I figure that I have to substitute $2x$ in this equation for the $y$, so I would get $(x-a)^2 + 4x^2=16$. After this, I don't know what to do to get $a$. Any help? Thanks
| Why not just solve the quadratic? Collect like terms in $x$ as follows. Move everything to one side to get $$(x-a)^2 + 4x^2 - 16 = 0.$$ Then expand the square:
$$\begin{align}
0 &= x^2 - 2ax + a^2 + 4x^2 - 16 \\
&= 5x^2 - 2ax + a^2 - 16.
\end{align}$$
Then using the quadratic formula gives you $$x = \frac{2a \pm \sqrt{(-2a)^2 - 4(5)(a^2-16)}}{2(5)} = \frac{a \pm 2\sqrt{20-a^2}}{5}.$$ This means the intersection of the line and the circle will have two solutions if $a^2 - 20 > 0$, no solutions if $a^2 - 20 < 0$, and exactly one solution if $a^2 - 20 = 0$. Since the line is supposed to be tangent, $a$ must satisfy the last case, hence $a = 2\sqrt{5}$. We take the positive value for $a$ since we know the circle's center is on the positive $x$-axis.
Alternatively, we can exploit the geometry of similar right triangles. Let $P = (x_0, 2x_0)$ be the point of tangency of the line $y = 2x$ with the circle, and let the origin be $O = (0,0)$ and the center of the circle $A = (a,0)$. Let $Q = (x_0, 0)$ be the foot of the perpendicular from $P$ to the $x$-axis. Then we have $$\triangle PQO \sim \triangle APO,$$ hence $$\frac{AP}{PO} = \frac{PQ}{QO} = \frac{2x_0}{x_0} = 2.$$ Since $AP = 4$, the radius of the circle, it follows that $PO = 2$, and by the Pythagorean theorem, $$a = AO = \sqrt{AP^2 + PO^2} = \sqrt{4^2 + 2^2} = \sqrt{20} = 2 \sqrt{5}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3765329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Maximum of function abs Let function $f(x)=|2x^3-15x+m-5|+9x$ for $x\in\left[0,3\right]$ and $m\in R$. Given that $\max f(x) =60$ with $x\in\left[0,3\right]$, find $m$.
I know how to solve this kind of problem for $g(x)=|2x^3β15x+mβ5|$. However, the $+9x$ is confusing me.
| Case 1:
Let $2x^3-15x+m-5\gt0$ at the point where maximum occurs.
Then our $f(x)$ becomes $2x^3-6x+m-5$
note that $f(x)$ decreases for $(0,1)$ and then increases for $x\gt1$ so the maximum of $f(x)$ is at $x=3$(as $x\in\left[0,3\right]$)
Plugging in the $f(3)=60$ we get $m = 29$.
We can put in $x=3$ and $m = 29$ in $2x^3-6x+m-5$ and verify that it is positive.
Case 2 :
Let $2x^3-15x+m-5\lt0$ at the point where maximum occurs
Then our $f(x)$ becomes $-2x^3+24x-m+5$
This function increases in $(0,2)$ and then decreases for $x\gt2$ so maximum for $f(x)$ is at $x=2$
Putting $f(2)=60$ we get $m=-23$.
We can put in $x=2$ and $m = -23$ in $-2x^3+24x-m+5$ and verify that it is negative.
So we get $m= \left({29\ \ and\ -23}\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3766297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $\prod_{n\ge0}^{ }\frac{\left(n+a\right)\left(n+b\right)}{\left(n+c\right)\left(n+d\right)}=\frac{\Gamma(c)\Gamma(d) }{ \Gamma(a)\Gamma(b)}$ How to prove the following identity?
$$\prod_{n\ge0}^{ }\frac{\left(n+a\right)\left(n+b\right)}{\left(n+c\right)\left(n+d\right)}\tag{$a,b,c,d \in \mathbb R$}=\frac{\Gamma(c)\Gamma(d) }{ \Gamma(a)\Gamma(b)}$$
For $a+b=c+d$.
The product can be rewritten as :
$$\prod_{n\ge0}^{ }\frac{\left(n+a\right)\left(n+b\right)}{\left(n+c\right)\left(n+d\right)}=\prod_{n\ge0}^{ }\frac{n^{2}+\left(a+b\right)n+ab}{n^{2}+\left(c+d\right)n+cd}$$$$=\prod_{n\ge0}^{ }\frac{n^{2}+\left(a+b\right)n+ab}{n^{2}+\left(a+b\right)n+cd}=\prod_{n\ge0}^{ }\left(1+\frac{ab-cd}{n^{2}+\left(a+b\right)n+cd}\right)$$
But I think this is useless.
Or we can say:
$$\prod_{n\ge0}^{ }\frac{\left(n+a\right)\left(n+b\right)}{\left(n+c\right)\left(n+d\right)}$$$$=\lim_{N \to \infty} \frac{ab}{cd}\cdot\frac{\left(a+1\right)\left(b+1\right)}{\left(c+1\right)\left(d+1\right)}\cdot...\cdot\frac{\left(a+N\right)\left(b+N\right)}{\left(c+N\right)\left(d+N\right)}$$$$=\lim_{N \to \infty}\frac{\left(a+N\right)!\left(b+N\right)!}{\left(c+N\right)!\left(d+N\right)!}\cdot\frac{\left(c-1\right)!\left(d-1\right)!}{\left(a-1\right)!\left(b-1\right)!}$$$$=\frac{ \Gamma(c)\Gamma(d)}{ \Gamma(a)\Gamma(b)}\lim_{N \to \infty}\frac{\left(a+N\right)!\left(b+N\right)!}{\left(c+N\right)!\left(d+N\right)!}$$
How to finish?
| You can finish the proof by using the asymptotics $$
\mathop {\lim }\limits_{N \to + \infty } \frac{{(N + a)!}}{{N^{N + a + 1/2} e^{ - N} \sqrt {2\pi } }} = 1,
$$ which can be derived from Stirling's formula ($a$ is an arbitrary fixed complex number).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3766409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int \frac{2-x^3}{(1+x^3)^{3/2}} dx$
Evaluate:
$$\int \frac{2-x^3}{(1+x^3)^{3/2}} dx$$
I could find the integral by setting it equal to $$\frac{ax+b}{(1+x^3)^{1/2}}$$
and differentiating both sides w.r.t.$x$ as
$$\frac{2-x^3}{(1+x^3)^{3/2}}=\frac{a(1+x^3)^{3/2}-(1/2)(ax+b)3x^2(1+x^3)^{-1/2}}{(1+x^3)}$$$$=\frac{a-ax^3/2-3bx^2}{(1+x^3)^{3/2}}$$
Finally by setting $a=2,b=0$, we get $$I(x)=\frac{2x}{(1+x^3)^{1/2}}+C$$
The question is: How to do it otherswise?
| $$\int \frac{2-x^3}{(1+x^3)^{3/2}} dx=\int \frac{2x^{-3}-1}{(x^{-2}+x)^{3/2}} dx$$
Now substitute $t=x^{-2}+x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3768479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Show that if $a$ and $b$ have the same sign then $|a + b| = |a| + |b| $, and if $a$ and $b$ have opposite signs then $|a+b| < |a| + |b|$ I'm considering the different cases for $a$ and $b$
Case 1) $ a\geq 0$ and $ b\geq 0$
Given both terms are positive, $ a + b \geq 0 $
$$
|a+b| = a + b = |a| + |b|\\
$$
Case 2) $ a< 0$ and $ b< 0$
Given both terms are negative, $ a + b < 0 $
$$
|a+b| = -(a + b) = -a -b = |a| + |b|\\
$$
Case 3) $ a> 0$ and $ b< 0$
Case 3a) $ a + b \geq 0 $
Knowing that $ b < -b $ because $ b < 0 < -b $:
$$
|a+b| = a + b < a - b = |a| + |b|
$$
Case 3b) $ a + b \leq 0 $
Knowing that $ -a < a $ because $ -a < 0 < a $:
$$
|a+b| = -(a + b) = -a - b < a - b = |a| + |b|
$$
Case 4) $ a< 0$ and $ b> 0$
Case 4a) $ a + b \geq 0 $
Knowing that $ a < -a $ because $ a < 0 < -a $:
$$
|a+b| = a + b < -a + b = |a| + |b|
$$
Case 4b) $ a + b \leq 0 $
Knowing that $ -b < b $ because $ -b < 0 < b $:
$$
|a+b| = -(a + b) = -a -b < -a + b = |a| + |b|
$$
Is my proof correct?
I feel like it's somewhat incomplete as I'm not considering when either $a$ or $b$ are equal to zero in the cases 3 and 4.
| I'll suggest some techniques for being briefer, which helps you know if you've covered all bases.
Since $|a+b|-|a|-|b|$ is unchanged when multiplying $a,\,b$ by $-1$, for the first problem we can assume without loss of generality that their common sign means they're $\ge0$, reducing the equation to the trivial $a+b=a+b$. The same invariance-under-changing-sign logic tells us the second problem only requires us to consider the case $a\ge 0>b$, whence$$(|a|+|b|)^2-|a+b|^2=a^2+b^2+2a|b|-a^2-b^2-2ab=4a|b|>0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3769795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
I need a hash function that will give all values from $0$ to $2^n - 1$. Will $f(x) = \frac{x^2 + x}{2} \; \text{mod} \; 2^n$ do the job? I need a hash function that will give all values from $0$ to $2^n - 1$. I want to know if this function does what is needed?
$$f(x) = \frac{x^2 + x}{2} \; \text{mod} \; 2^n$$
the values of $x$ are integers in the range $[0, 2^{n} - 1]$.
| Consider if there are any $x, y \in [0, 2^n - 1]$ where $x \neq y$ and $f(x) = f(y)$. If so, then there's a $k \in \mathbb{Z}$ where
$$\begin{equation}\begin{aligned}
\frac{x^2 + x}{2} - \frac{y^2 + y}{2} & = k\left(2^n\right) \\
x^2 + x - (y^2 + y) & = k\left(2^{n+1}\right) \\
x^2 - y^2 + x - y & = k\left(2^{n+1}\right) \\
(x - y)(x + y) + (x - y) & = k\left(2^{n+1}\right) \\
(x - y)(x + y + 1) & = k\left(2^{n+1}\right)
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Now, if $x - y$ is even, then $x + y + 1$ is odd, and vice versa. This means all of the $n + 1$ or more factors of $2$ must be in just one of those $2$ factors on the left, i.e.,
$$x - y \equiv 0 \pmod{2^{n+1}} \; \; \text{ or } \; \; x + y + 1 \equiv 0 \pmod{2^{n+1}} \tag{2}\label{eq2A}$$
The first condition doesn't occur since it was stated $x \neq y$, and you have $|x - y| \lt 2^{n+1}$. The second one can't be true either since $x + y + 1 \gt 0$ and, with $x, y \le 2^n - 1$, you have $x + y + 1 \le (2^n - 1) + (2^n - 1) + 1 = 2^{n+1} - 1$.
This shows all of the values generated will be unique.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$f\left( x \right) = {x^3} + x$, then $\int\limits_1^2 {f\left( x \right)dx} + 2\int\limits_1^5 {{f^{ - 1}}\left( {2x} \right)dx} $
If $f\left( x \right) = {x^3} + x$, then
$$\int\limits_1^2 {f\left( x \right)dx} + 2\int\limits_1^5 {{f^{ - 1}}\left( {2x} \right)dx} $$
is________.
My approach is as follows:
$$g = {f^{ - 1}} \Rightarrow g\left( x \right) = {f^{ - 1}}\left( x \right)$$
$$g\left( {2x} \right) = {f^{ - 1}}\left( {2x} \right)$$
$$2y = {8x^3} + 2x$$
$${f^{ - 1}}\left( {{x^3} + x} \right) = x$$
$$\int\limits_1^2 {f\left( x \right)dx} + 2\int\limits_1^5 {{f^{ - 1}}\left( {2x} \right)dx} $$
I am not able to proceed further.
| After the substitution $u=2x$ in the second integral, your expresion becomes
$$\int\limits_1^2 {f\left( x \right)dx} + 2\int\limits_1^5 {{f^{ - 1}}\left( {2x} \right)dx}= \int\limits_1^2 {f\left( x \right)dx} + \int\limits_{f(1)=2}^{f(2)=10} {{f^{ - 1}}\left( {u} \right)du}$$
Now, you can use an integration rule for the inverse function and obtain
$$\int\limits_1^2 {f\left( x \right)dx} + \int\limits_{f(1)=2}^{f(2)=10} {{f^{ - 1}}\left( {u} \right)du} = 2\cdot f(2) - 1\cdot f(1) =18$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Finding $\lim_{n\to\infty}\frac1{n^3}\sum_{k=1}^{n-1}\frac{\sin\frac{(2k-1)\pi}{2n}}{\cos^2\frac{(k-1)\pi}{2n}\cos^2\frac{k\pi}{2n}}$
For all $n\ge 1$, let
$$
a_{n}=\sum_{k=1}^{n-1} \frac{\sin \left(\frac{(2 k-1) \pi}{2 n}\right)}{\cos ^{2}\left(\frac{(k-1) \pi}{2 n}\right) \cos ^{2}\left(\frac{k \pi}{2 n}\right)}
$$
Find $\displaystyle\lim_{n\to\infty}\frac{a_n}{n^3}$.
This is a problem from the 2019 Putnam competition.
The official solutions use two different strategies; one reduces the expression as a telescoping sum and the other uses the asymptotic property of sine near zero. The two community-wiki answers below essentially elaborate on the official solutions.
Remark. I am curious if one can write $\displaystyle\frac{a_n}{n^3}$ as a Riemann sum so that one can write the limit as an integral. Naively, the fraction looks very much like $\frac{1}{n}\sum_{k=1}^{n-1}\cdots$, which is in the setting of Riemann sums. (This only serves as a comment, not a requirement for solving the problem.)
| This is based on one of the official solutions for the problem.
We first write $a_n$ as a telescoping sum. Notice that
$$
\frac{1}{AB} = \left(\frac{1}{A}-\frac{1}{B}\right)\cdot \frac{1}{B-A}\,.
$$
It follows that the summand of $a_n$ can be written as
$$
\left(\frac{1}{\cos ^{2}\left(\frac{(k-1) \pi}{2 n}\right)}-\frac{1}{\cos ^{2}\left(\frac{k \pi}{2 n}\right)}\right)\cdot
\frac{\sin \left(\frac{(2 k-1) \pi}{2 n}\right)}{\cos ^{2}\left(\frac{k \pi}{2 n}\right)-\cos ^{2}\left(\frac{(k-1) \pi}{2 n}\right)}\tag{1}
$$
If we can show that the quantity
$$
\frac{\sin \left(\frac{(2 k-1) \pi}{2 n}\right)}{\cos ^{2}\left(\frac{k \pi}{2 n}\right)-\cos ^{2}\left(\frac{(k-1) \pi}{2 n}\right)}\tag{2}
$$
is independent of $k$, then we have a telescoping sum.
By the double angle and sum-product identities for cosine, we have
\begin{align}
&\phantom{=}2 \cos ^{2}
\left(\frac{(k-1) \pi}{2 n}\right)
-2 \cos ^{2}\left(\frac{k \pi}{2 n}\right) \\
&=\cos \left(\frac{(k-1) \pi}{n}\right)-\cos \left(\frac{k \pi}{n}\right)
\quad &(2\cos^2x = \cos 2x - 1)
\\
&=2 \sin \left(\frac{(2 k-1) \pi}{2 n}\right) \sin \left(\frac{\pi}{2 n}\right)
\quad &(\cos \theta-\cos \varphi=-2 \sin \left(\frac{\theta+\varphi}{2}\right) \sin \left(\frac{\theta-\varphi}{2}\right))
\end{align}
and it follows that the summand in $a_n$ can be written as
$$
\frac{1}{\sin \left(\frac{\pi}{2 n}\right)}\left(-\frac{1}{\cos ^{2}\left(\frac{(k-1) \pi}{2 n}\right)}+\frac{1}{\cos ^{2}\left(\frac{k \pi}{2 n}\right)}\right)
$$
Thus the sum telescopes and we find that
$$
a_{n}=\frac{1}{\sin \left(\frac{\pi}{2 n}\right)}\left(-1+\frac{1}{\cos ^{2}\left(\frac{(n-1) \pi}{2 n}\right)}\right)=-\frac{1}{\sin \left(\frac{\pi}{2 n}\right)}+\frac{1}{\sin ^{3}\left(\frac{\pi}{2 n}\right)}
$$
Finally, since $\lim_{x\to 0}\frac{\sin x}{x}=1$, we have
$$
\lim_{n\to\infty} n\sin\frac{\pi}{2n} = \frac{\pi}{2}
$$
and thus
$$
\lim_{n\to\infty}\frac{a_n}{n^3} = \frac{8}{\pi^3}\;.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3772724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to evaluate $\sum_{n=1}^{\infty}\:\frac{2n+1}{2n(n+1)^2}$? Note: Similar questions have been asked here and here, but this is quite different.
I am trying to evaluate
$$\sum_{n=1}^{\infty}\:\frac{2n+1}{2n(n+1)^2} \quad (1)$$
I re-wrote the fraction as $$ \frac{2n+1}{2n(n+1)^2} = \frac1{2(n+1)} \cdot \frac{2n+1}{n(n+1)}= \frac1{2(n+1)} \left( \frac1n + \frac1{n+1} \right) = \frac1{2} \left( \frac1{n(n+1)} + \frac1{(n+1)^2} \right) = \frac12 \left( \left( \frac1n -\frac1{(n+1)} \right) + \frac1{(n+1)^2} \right)$$
Hence
$$(1) = \frac12 \sum_{n=1}^{\infty}\: \left( \frac1n -\frac1{(n+1)} \right) + \frac12 \sum_{n=1}^{\infty}\:\frac1{(n+1)^2} = \frac12\lim_{n \to \infty}1-\frac1{n+1}+\frac{\pi^2}{12} = \frac{\pi^2}{12} + \frac12$$
Therefore
$$\sum_{n=1}^{\infty}\:\frac{2n+1}{2n(n+1)^2} = \frac{\pi^2}{12} + \frac12$$
I am unsure about $$ \sum_{n=1}^{\infty}\:\frac1{(n+1)^2} = \frac{\pi^2}{12} $$
We know the basic p-series $$ \sum_{n=1}^{\infty}\:\frac1{n^2} = \frac{\pi^2}{6} $$
Is this solution correct?
| As has been already pointed out in the comments,
$$\sum_{n = 1}^\infty \frac{1}{(n+1)^2} = \frac{1}{4} + \frac 19 + ... = \frac{\pi^2}{6}-1$$
Therefore, your solution would be
$$\frac{1}{2}\left( \sum_{n=1}^\infty \frac 1n - \frac {1}{n+1} + \sum_{n=1}^\infty\frac{1}{(n+1)^2}\right) = \frac 12 \left( 1 + \frac{\pi^2}{6} - 1\right) = \boxed{\frac{\pi^2}{12}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3773202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Need help with finishing integration I have the following integral: $$y=\int \frac{1}{1-2\sqrt{x}} \, dx$$
I first got $u=2\sqrt{x}$ which gives us $x=\frac{u^2}{4}$. Plugging this in I got:
$$y=\int \frac{1}{1-2\sqrt{\frac{u^2}{4}}} \, du=\int \frac{1}{1-2(\frac{u}{2})} \, du=\int \frac{1}{1-u} \, du$$
After this I am not sure where to go from there. Even though there might be easier methods I'd prefer to stick with this method so please relate your responses to my correct or incorrect work shown above.
| We have $$ \int \frac{1}{1-2\sqrt{x}}dx$$
Substitute $u = 2\sqrt{x}$ so that $x = \frac{u^2}{4}$ and $dx = \frac{u}{2}du$. We get $$ \int \frac{u}{2(1-u)}du $$
Now, it can be done from here, but since you're learning, make another substitution which might help you seeing "typical" integral here.
Let $t = 1-u$ so that $u= 1-t$ and $du = -dt$. We get
$$ \int \frac{1-t}{2t}(-dt) = -\frac{1}{2}\int \frac{1}{t} - \frac{t}{t} dt = -\frac{1}{2} \Big ( \int \frac{1}{t} dt - \int 1 dt \Big) $$
You should know those integrals. Preciselly:
$$ \int \frac{1}{t} dt = \ln|t| + C_1 $$
and
$$ \int 1 dt = t + C_2$$
So that taking $C=-\frac{1}{2}(C_1 + C_2) \in \mathbb R$ we have:
$$ \int \frac{1}{1-2\sqrt{x}}dx = -\frac{1}{2} \big ( \ln|t| - t \big) + C $$ and now go back to variable $x$:
$$\int \frac{1}{1-2\sqrt{x}}dx= -\frac{1}{2}\ln|1-u| + \frac{1}{2}(1-u) + C = -\frac{1}{2}\ln|1-2\sqrt{x}| + \frac{1}{2}(1-2\sqrt{x}) + C $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3774975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
What is the integer part of the following fraction: $\frac{2012^{2013}+2013^{2014}}{2012^{2012}+2013^{2013}}$
What is the integer part of the following fraction: $\dfrac{2012^{2013}+2013^{2014}}{2012^{2012}+2013^{2013}}$.
This is a competition problem for 7th grade students.
The answer to this question is $2012$.
Is there any way to simplify/evaluate it so we can see the integer part clearly? Or can we just estimate?
| Observe that for any $a,b > 0,$ we have $\dfrac{a^{a+1}+b^{b+1}}{a^a + b^b} = \dfrac{a(a^a + b^b)+(b-a)b^{b}}{a^a + b^b} = a + (b-a)\dfrac{b^b}{a^a+b^b}.$ Clearly, $b^b < a^a + b^b.$ So in order for $\dfrac{b^b}{a^a+b^b}$ to be a fraction, we need $\dfrac{b^b}{a^a+b^b} < \frac{1}{b-a}.$ Here we have $a = 2012, b = 2013.$ Hence $(b-a)\dfrac{b^b}{a^a+b^b} = \dfrac{b^b}{a^a+b^b} < 1,$ so the integer part is $a = 2012.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3775785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Finding the asymptotes of the polar curve $ r=\frac{ 5\cos^2 \theta+3 }{ 5\cos^2 \theta-1 }$ Find all asymptotes ( their equations in polar or cartesian form) for the polar curve
$$ r(\theta)=\dfrac{ 5\cos^2 \theta+3 }{ 5\cos^2 \theta-1 }$$
When denominator goes to zero we have one set, but what is the other set?
| You are right, values are given by
$$5\cos^2 \theta-1=0 \implies \cos \theta = \pm\frac1{\sqrt 5}\implies \theta_{1,2}=\pm \arccos \left(\frac1{\sqrt 5}\right),\: \theta_{3,4}=\pm \arccos \left(-\frac1{\sqrt 5}\right)$$
To find the asymptotes equations let consider for any direction $\theta_0$ (wlog assume $ \theta_0=\theta_{1}$)
$$y=\tan \theta_0 \, x+q \iff q=r\cos \theta (\tan \theta-\tan \theta_0)$$
then
$$q=\lim_{\theta\to \theta_0} (5\cos^2 \theta+3)\cos\theta \frac{\tan \theta-\tan \theta_0}{5\cos^2 \theta-1}=\sqrt 5$$
indeed
$$ (5\cos^2 \theta+3)\cos(\theta)\to \frac4{\sqrt 5}$$
and by lβHopital
$$q=\lim_{\theta\to \theta_0}\frac{\tan \theta-\tan \theta_0}{5\cos^2 \theta-1}= \lim_{\theta\to \theta_0}-\frac{1}{10\cos^3\theta\sin \theta}=\frac 54$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3776144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.