Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Function for cosine transformation around $\pi/2$ Given the cosine of an angle $x$ relatively close to $\pi/2$, is there a function $f$ such as: $f(cos(x))=cos((x+\pi/2)/2)$ ?
| It seems like you just want a function $f$ so that $f(\cos(x))=\cos((x+\pi/2)/2)$ for $x$ close to $\pi/2$. First do the following using trigonometric identities:
$$
f(\cos(x)) = \cos\left(\frac{x}{2} + \frac{\pi}{4}\right) = \cos\left(\frac{x}{2}\right)\cos\left(\frac{\pi}{4}\right) - \sin\left(\frac{x}{2}\right)\sin\left(\frac{\pi}{4}\right) \\
f(\cos(x)) = \frac{1}{\sqrt{2}}\cos\left(\frac{x}{2}\right) - \frac{1}{\sqrt{2}}\sin\left(\frac{x}{2}\right)
$$
So if we can write $\cos(x/2)$ and $\sin(x/2)$ in terms of $\cos(x)$, then we can write $f$ explicitly. But this is simple remembering the half angle formula (or we can just derive it using the cosine double angle formula since I don't remember the half angle one hehe) For $\cos$, we have
$$
\cos(2x) = 2\cos^2(x) - 1 \implies \cos(x) = 2\cos^2\left(\frac{x}{2}\right) - 1 \implies \cos\left(\frac{x}{2}\right) = \sqrt{\frac{1 + \cos(x)}{2}}
$$ taking the positive root since $\cos(x/2)$ is positive assuming $x$ is close to $\pi/2$. For $\sin$, we have
$$
\cos(2x) = 1- 2\sin^2(x) \implies \cos(x) = 1 - 2\sin^2\left(\frac{x}{2}\right) \implies \sin\left(\frac{x}{2}\right) = \sqrt{\frac{1 -\cos(x)}{2}}
$$ since $\sin(x/2)$ is positive for $x$ near $\pi/2$. So, going back to the actual function, we have
$$
f(\cos(x)) = \frac{1}{\sqrt{2}}\sqrt{\frac{1 + \cos(x)}{2}} - \frac{1}{\sqrt{2}}\sqrt{\frac{1 -\cos(x)}{2}} = \frac{1}{2}\sqrt{1 + \cos(x)} - \frac{1}{2}\sqrt{1 - \cos(x)}
$$
So now we can just replace $\cos(x)$ with $u$ to get
$$
f(u) = \frac{1}{2}\sqrt{1 + u} - \frac{1}{2}\sqrt{1 - u}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/467834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
4-dim. generalization of $ab+ac+bc=0$ The equation $ab+ac+bc=0$ can be parameterized by $(a,b,c)=\lambda(-pq, p(p+q), q(p+q))$.
Is there a (similar) parameterization for $ab+ac+ad+bc+bd+cd=0$?
What about the 5-dimensional case?
Edit:
Proof of the parametrization in the 3-dim. case:
Write $x=a+b, y=a-b$.
Then $ab+ac+bc=0 <=> (x-y)(x+y) +2(x+y)c+2(x-y)c=0 <=> (x+2c)^2-y^2-4c^2=0$. Then use the pythogorean triple parametrization to get $x+2c=\lambda(m^2+n^2), y=\lambda(m^2-n^2), c=\lambda mn$.
Now $a=(x+y)/2= \lambda(m^2+n^2-2mn+m^2-n^2)/2=m(m-n), b=(x-y)/2=\lambda n(n-m), c=\lambda mn$. Now set $m=p$ and $n=q+p$ and the result follows (with $b$ and $c$ switched)
Edit2: yes, I am looking for a parameterization in the integers. Sorry, that I didn't make myself clear.
| Solution of above inquiry regarding parametric form for equations (given below) for four and five variables is shown:
$ab+ac+ad+bc+bd+cd=0$
$ab+ac+bc+bd+ad+cd+ea+eb+ec+ed=0$
For $ab+ac+ad+bc+bd+cd=0$
$(a,b,c,d)=((p^2+2p+1),(2p+2),-(2p+2),(4))$
$p=2$ we get $(a,b,c,d)=(9,6,-6,4)$
For $ab+ac+bc+bd+ad+cd+ea+eb+ec+ed=0$
$(a,b,c,d,e)=((3p+7),-3(p-3),(3p-5),-(3p+5),(3p^2-p+12))$
For $p=0$ we get: $(a,b,c,d,e)=(7,9,-5,-5,12)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/468433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proof on an equilateral triangle with a cevian extended to its circumcircle Consider the following figure with equilateral triangle $ABC$ and a cevian $AQ$ extended to $P$ on its circumcircle.
We are required to prove that:
$\frac{1}{PB} + \frac{1}{PC} = \frac{1}{PQ}$
Let $\angle PAC = \alpha$ and let length of $AB = s$
By the Law of Sines,
$\frac{PC}{\sin\alpha} = \frac{s}{\sin\angle CPA} = \frac{s}{\sin60^{\circ}} \implies PC = \frac{s\sin\alpha}{\sin60^{\circ}}$
Similarly, $PB = \frac{s\sin(60^{\circ} - \alpha)}{\sin60^{\circ}}$
$\frac{1}{PB} + \frac{1}{PC} = \frac{\sin60}{s}\left(\frac{\sin(60 - \alpha) + \sin\alpha}{\sin\alpha\sin(60 - \alpha)}\right)$
It remains to be proven that:
$PQ = \frac{\sin60\sin(60 - \alpha)\sin\alpha}{\sin60\sin(60 - \alpha) + \sin60\sin\alpha}$
I'm utterly lost from here.
| I have read that this problem appeared in South African mathematical competition in years 1994 and 1997. The proof I know is based on the areas of triangles.
As $\angle APB=\angle APC=60^\circ$ and $\angle BPC=120^\circ$, the areas of triangles $BPQ,$ $QPC$ and $BPC$ are $\frac{1}{2}BP\cdot PQ\cdot \sin 60^\circ,$ $\frac{1}{2}PQ\cdot PC\cdot \sin 60^\circ$
and $\frac{1}{2}BP\cdot PC\cdot \sin 120^\circ$ respectively. As $\sin 60^\circ=\sin 120^\circ$ and the sum of areas of triangles $BPQ$ and $QPC$ is the area
of $BPC,$ we have that $BP\cdot PQ+PQ\cdot PC=BP\cdot PC.$ Now divide this equation by $BP\cdot PQ\cdot PC.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/469637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How prove this $\tan{x}>\frac{3x}{2+\sqrt{1-x^2}}$ let $0<x<1$,prove that
$$\tan{x}>\dfrac{3x}{2+\sqrt{1-x^2}}$$
This problem have nice solution?
my idea:
let $$f(x)=\tan{x}-\dfrac{3x}{2+\sqrt{1-x^2}}=\tan{x}-3x\dfrac{2-\sqrt{1-x^2}}{3+x^2}$$
and other idea:
let $x=\cos{t}$,then
$$\tan{(\cos{x})}>\dfrac{3\cos{t}}{2+\sin{t}}$$
other (2) idea:
$$\tan{x}>x+\dfrac{1}{3}x^3$$
| Your idea works equally well too.
We need to prove that $$ \cos{t} + \frac{\cos{t}^3}{3} > \frac{3\cos{t}}{2+\sin{t}} $$
Since $x$ varies between $(0,1)$, both sin(t) and cos(t) vary from (0,1).
That reduces our inequality to:
$$ 1 + \frac{\cos^2{t}}{3} > \frac{3}{2+\sin{t}} $$
Expressing cos in terms of sin, and further simplifying we get:
$$ (2+\sin{t})^2(2-\sin{t}) > 3 $$
We substitute $y = \sin{t}$, where $y \in (0,1)$:
$$ -y^3 -2y^2 + 4y + 8 > 3 $$
Analyzing the derivative of the LHS, we can easily find that the minimum value of the polynomial in $(0,1)$ is $8$, and therefore the inequality is true in the given interval.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/470172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
integrating $\int \sqrt{2-2\cos(x)} \, dx$ So i am having some trouble getting the solution to the integral:
$\int \sqrt{2-2\cos(x)} \, dx$
i made my first substitution
$u = 2-2\cos(x)$
$u' = 2\sin(x) \, dx$
then...
$\int \dfrac{1}{\sqrt{4-u}} \, du$
then the next sub of
$s = 4-u$
$s' = -du$
then...
$\int \dfrac{1}{\sqrt{s}} \, ds$
which gave me...
$2\sqrt{s}$
subbing back $s$...
$2\sqrt{4-u}$
subbing back $u$...
$2\sqrt{2\cos(x)+2}$
but i believe this is incorrect, can someone tell me where i went wrong?
thanks :)
| $$
\begin{aligned}
\int \sqrt{2-2 \cos x} d x=&\int \sqrt{2(1-\cos x)} d x \\
&=\int \sqrt{2 \cdot 2 \sin ^2 \frac{x}{2}} d x \\
&=2 \operatorname{sgn}\left(\sin \frac{x}{2}\right) \int \sin \frac{x}{2} d x \\
&=-4 \operatorname{sgn}\left(\sin \frac{x}{2}\right) \cos \frac{x}{2}+C
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/472024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Least square solutions when $Ax=b$ has no solution I have this linear alegbra question that I don't know how to start.
The question:
$$A = \left( \begin{matrix}1 & -2 \\3 & 1 \\2 & 3 \\\end{matrix}\right ) $$
$$b= \left (\begin{matrix}
0 \\
1 \\
0 \\
\end{matrix}\right )$$
$$r = \left ( \begin{matrix}
x \\
y \\
\end{matrix} \right )$$
The matrix equation of $Ar-b$ has no solution because it is inconsistent. If we define the square error to be $\Delta^2 = \delta^T\delta$ where δ=Ar−b, we can still solve the least square sense.
Explain why $\Delta^2$ is a function of x and y and that $\Delta^2 $ is minimised by the solution of $$A^TAr = A^Tb$$
Need some guidance in solving this two parts.
| We have
$$A^T A = \left( \begin{matrix}1&3&2\\-2&1&3 \\\end{matrix}\right ) \left( \begin{matrix}1 & -2 \\3 & 1 \\2 & 3 \\\end{matrix}\right ) = \left ( \begin{matrix} 14 & 7\\ 7 & 14 \end{matrix} \right )$$
and
$$A^T b = \left( \begin{matrix}1&3&2\\-2&1&3 \\\end{matrix}\right ) \left ( \begin{matrix} 0\\1\\0 \end{matrix} \right )= \left( \begin{matrix} 3\\1 \end{matrix} \right )$$
So we result in a $2\times 2$ linear equation system with an unique solution:
$$\left ( \begin{matrix} 14 & 7\\ 7 & 14 \end{matrix} \right ) r = \left( \begin{matrix} 3\\1 \end{matrix} \right)$$
So far for the solution. Now why is $\Delta^2$ a function of $r = (x,y)^T$? Well, because the definition says
$$\Delta^2 = (Ar-b)^T (Ar-b) = (Ar)^T(Ar) - (Ar)^T b - b^T Ar + b^Tb = r^T A^TA r - r^T A^Tb - b^T Ar + b^Tb = r^T (A^TAr-2A^Tb) + b^Tb$$
with constant $A, b$. Thus, $\Delta^2 = \Delta^2(r) = \Delta^2(x,y)$. Using the latter expanded form, minimizing $\Delta^2$ is the same as minimizing $r^T A^TAr - 2r^TA^Tb$. The rest is basic calculus.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/473019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the sum of all real solutions for $x$ to the equation $(x^2 + 2x + 3)^{(x^2+2x+3)^{(x^2+2x+3)}} = 2012.$ Find the sum of all real solutions for $x$ to the equation $(x^2 + 2x + 3)^{(x^2+2x+3)^{(x^2+2x+3)}} = 2012.$
I just know $x^{x^x}$ is increasing in $x$ and hence the equation has a unique solution, nut then I dont know how to move on, I also know viete' formula but I dont know if it helps here, thanks in advance.
| The sum is $-2$. Can you see why? Hint: I have not computed the solutions.
Details: As you observed, there is a unique positive $b$ such that $b^{(b^b)}=2012$. Moreover, this $b$ is in the interval $(2,3)$, by the Intermediate Value Theorem, since $2^{(2^2)}$ is too small and $3^{(3^3)}$ is too big.
Note that $x^2+2x+3=(x+1)^2+2$, so $x^2+2x+3$ attains a minimum value of $2$. Thus the equation $x^2+2x+3-b=0$ has two real solutions. The sum of these is the negative of the coefficient of $x$, that is, $-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/473825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 3,
"answer_id": 0
} |
How do you solve this differential equation using variation of parameters? $\color{green}{question}$:
How do you solve this differential equation using variation of parameters?
$$y"-\frac{2x}{x^2+1}y'+\frac{2}{x^2+1}y=6(x^2+1)$$
$\color{green}{I~tried}$ . . .
$using~the~\color{blue}{Laplace~transform}~method$ . . .
$$L[\int_{0}^{\infty }\frac{sinxt}{1+t^{2}}dt]$$
$$=\int_{0}^{\infty }e^{-sx}(\int_{0}^{\infty }\frac{sinxt}{1+t^{2}}dt)$$
$$=\int_{0}^{\infty }\frac{1}{1+t^{2}}(\int_{0}^{\infty }e^{-px}sinxtdx)dt\\\\\\=\int_{0}^{\infty }\frac{1}{1+t^{2}}\frac{t}{s^{2}+t^{2}}dt$$
$$=\int_{0 }^{\infty }\frac{1}{s^{2}-1}(\frac{t}{1+t^{2}}-\frac{t}{s^{2}+t^{2}})$$
$$=\frac{Lns}{s^{2}-1}$$
Is my solution correct?
Should I use the inverse Laplace?
How can I get a complete and correct answer?
Thanks for any hint.
| I will map out the steps for you and have you fill in the details.
We are asked to solve this using Variation of Parameters (VoP), given:
$$\tag 1 y''-\dfrac{2x}{x^2+1}y'+\dfrac{2}{x^2+1}y=6(x^2+1)$$
Step 1
Find the homogeneous solution to $(1)$, so we have:
$$\tag 2 y''-\dfrac{2x}{x^2+1}y'+\dfrac{2}{x^2+1}y=0$$
This yields:
$$y_h = c_1(x^2-1) + c_2 x$$
Step 2
We are now going to make use of VoP, so we set: $y_1 = x^2-1$ and $y_2 = x$ from $y_h$ and $f = 6(x^2+1)$ from $(1)$.
We calculate the Wronskian of $y_1$ and $y_2$, yielding $W(x^2-1, x) = -x^2-1$.
Using VoP, we have:
$$u_1 = \int \dfrac{-y_2 f}{W(x^2-1, x)} dx = \int \dfrac{-x 6(x^2+1)}{-x^2-1} dx = 3x^2$$
$$u_2 = \int \dfrac{y_1 f}{W(x^2-1, x)} dx = \int \dfrac{(x^2-1)6(x^2+1)}{-x^2-1} dx = 6x-2x^3$$
Now, $y_p$ is given by:
$$y_p = y_1 u_1 + y_2 u_2 = (x^2-1)(3x^2) + (-2x^3+6x)(x) = x^4 + 3x^2$$
Step 3
Our final solution is given by:
$$y(x) = y_h(x) + y_p(x) = c_1(x^2-1) + c_2 x + x^4 + 3x^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/474830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int \frac{ \mathrm dy}{(y^2+z^2)\left(\sqrt{\frac{a^2}{2}+z^2+y^2}\right)}$ How do you evaluate this integral?
$$\int \frac{ \mathrm dy}{(y^2+z^2)\left(\sqrt{\frac{a^2}{2}+z^2+y^2}\right)}$$
Thanks for any help.
| $$
? \equiv \int \frac{ \mathrm dy}{(y^2+z^2)\left(\sqrt{\frac{a^2}{2}+z^2+y^2}\right)}
$$
With $\quad\displaystyle{y = \sqrt{{a^{2} \over 2} + z^{2}\,}\ \tan\left(\theta\right)}$
\begin{align}
?
&=
\int
{\cos\left(\theta\right)
\over
a^{2}\sin^{2}\left(\theta\right)/2 + z^{2}}\,{\rm d}\theta
=
{1 \over z^{2}}\ {z \over 2^{-1/2}a}\int
{2^{-1/2}a\cos\left(\theta\right)/z
\over
\left\lbrack 2^{-1/2}a\sin\left(\theta\right)/z \right\rbrack^{2} + 1}\,{\rm d}\theta
\\[3mm]&=
{\sqrt{2} \over az}\,\arctan\left(a\sin\left(\theta\right) \over \sqrt{2}z\right)
=
{\sqrt{2} \over az}\,
\arctan\left({a \over \sqrt{2}z}\,
{\tan\left(\theta\right) \over \sqrt{\tan^{2}\left(\theta\right) + 1}}\right)
\\[3mm]&={\large
{\sqrt{2} \over az}\,
\arctan\left({a \over \sqrt{2}z}\,
{y \over \sqrt{y^{2} + a^{2}/2 + z^{2}}}\right)}
\end{align}
$+$ some constant !!!.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/474887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Performing operations and simplification
The equation is $$2a-7b- \dfrac{4(a^2-16b^2)}{2a-3b}$$
I need to simplify the expression.
| Perhaps the main objective here is simply finding a common denominator, expand, and simplify:
$$\begin{align} 2a-7b- \dfrac{4(a^2-16b^2)}{2a-3b} & = \dfrac{(2a - 7b)(2a - 3b) - 4a^2 + 64 b^2 }{2a - 3b} \tag{1}\\ \\ & = \dfrac{4a^2 -20ab+ 21 b^2 - 4a^2 + 64b^2}{2a - 3b}\tag{2} \\ \\ &= \dfrac{85b^2 - 20ab}{2a-3b} \tag{3}\\ \\ & = \dfrac{5b(17b - 4a)}{2a - 3b}\tag{4}\end{align}$$
$(1)\quad $ Finding the common denominator and subtracting.
$(2)\quad$ Expanding the product in the numerator: $(2a - 7b)(2a-3b)$
$(3)\quad$ Adding/subtracting like terms in numerator.
$(4)\quad$ Factoring out common factor in numerator.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/476473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\tan A + \tan B + \tan C = \tan A\tan B\tan C\,$ in a triangle I want to prove this (where each angle may be negative or greater than $180^\circ$):
When $A+B+C = 180^\circ$ \begin{equation*} \tan A + \tan B + \tan C = \tan A\:\tan B\:\tan C. \end{equation*}
We know that
\begin{equation*}\tan(A+B) = \frac{\tan A+\tan B}{1-\tan A\tan B}\end{equation*} and that \begin{equation*}\text{and that}~A+B = 180^\circ-C.\end{equation*}
Therefore $\tan(A+B) = -\tan C.$
From here, I got stuck.
| Use $\tan(A+B)=\tan(180^\circ-C)$:
$$\frac{(\tan A + \tan B)}{(1-\tan A \tan B)} = \frac{(\tan 180^\circ- \tan C)}{(1-\tan 180^\circ \tan C)}$$
Since $\tan 180^\circ=0$,
$$\frac{ (\tan A +\tan B)}{(1-\tan A\tan B) }= \frac{-\tan C}{1}$$
Therefore,
$$\tan A + \tan B = -\tan C + \tan A \tan B \tan C$$
Hence, the result
$$\tan A + \tan B + \tan C= \tan A \tan B \tan C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/477364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 11,
"answer_id": 3
} |
Generalization of $\lfloor \sqrt n+\sqrt {n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\rfloor=\lfloor\sqrt {25n+49}\rfloor$ I have been asking the following question at MSE with an answer:
$\lfloor \sqrt n+\sqrt {n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\rfloor=\lfloor\sqrt {25n+49}\rfloor$ is true?
I found this relational expression by using computer. Then, I got interested in the generalization of this expression. After receiving an answer, I decided to start an easy example.
First, I got the following with a proof:
$$\lfloor\sqrt n+\sqrt {n+1}\rfloor=\lfloor\sqrt {4n+1}\rfloor=\lfloor\sqrt{4n+2}\rfloor.$$
Note that $\lfloor x \rfloor$ is the largest integer not greater than $x$.
Proof: By the AM-GM inequality, we get $n\lt\sqrt{n(n+1)}\lt n+1/2$. Then, we get
$$4n+1\lt\left(\sqrt n+\sqrt{n+1}\right)^2=4n+2-2\left(n+1/2-\sqrt{n(n+1)}\right)\lt4n+2.
$$
By the way, there are no perfect squares in $\left(4n+1, 4n+2\right].$ (since $k^2\equiv0, 1(mod 4)$ for any natural number $k$, there is no natural number $k$ such that $k^2=4n+2$.) Now, the proof is completed.
Second, I've just got new relational expressions by using computer:
$$\lfloor\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}\rfloor=\lfloor\sqrt{9n+8}\rfloor,$$
$$\lfloor\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}\rfloor=\lfloor\sqrt{16n+20}\rfloor$$
for any natural numbre $n$.
I won't ask these expression because I'm going to prove these by myself.
Third, I haven't got any expression about 'six terms expression'.
Then, I expect the following theorem would be proven true:
Theorem: For any real number $c$, there exists a natural number $n$ such that
$$\lfloor\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}+\sqrt{n+5}\rfloor\not=\lfloor\sqrt{36n+c}\rfloor.$$
Then, here is my question.
Question: The theorem I wrote is true? If it is true, please show me how to prove it. If it is not true, please give me a counterexample.
Any help would be appriciated.
| Let $g(n)=\sqrt n+\sqrt{n+1}+\ldots+\sqrt{n+5} $ and $f(n)=\lfloor g(n)\rfloor$.
When searching for a $c$ such that
$$f(n) = \lfloor \sqrt{36n+c}\rfloor $$
for all $n$,
each $n$ gives us some conditions on $c$, namely that
$$f(n)\le\sqrt{36n+c}<f(n)+1 $$
that is
$$\tag1f(n)^2-36n\le c<(f(n)+1)^2-36n.$$
This way, $n=1$ gives us $64\le c<85$ and $n=11$ gives us $88\le c<133$. Since these inequalites contradict each other, no $c$ works for all $n$, thus your theorem is true.
We can find suitable $c$ if we are allowed to ignore the first few values of $n$:
Note that for $k>0$ $$\sqrt{n+k}-\sqrt{n}=\frac{k}{\sqrt{n+k}+\sqrt n}\in\left(\frac{k}{2\sqrt{n+k}},\frac{k}{2\sqrt{n}}\right),$$
hence
$$ 6\sqrt n+\frac{15}{2\sqrt{n+5}}<g(n) <6\sqrt n+\frac{15}{2\sqrt n}$$
and
$$\tag236n+90\sqrt{\frac{n}{n+5}}+\frac{225}{4(n+5)}< g(n)^2<36n+90+\frac{225}{4n}.$$
Assume $n\ge20$, so that $4(n+5)\le 5n$.
From $\sqrt{1+\frac 5n}<1+\frac{5}{2n}$ we have $\sqrt{\frac{n}{n+5}}>\frac1{1+\frac5{2n}}>1-\frac5{2n}$, so the left hand side of (2) becomes
$$ g(n)^2>36n+90\left(1-\frac5{2n}\right)+\frac{45}{n}=36n+90-\frac{180}{n}.$$
Thus for $n>180$ we have $\sqrt{36n+89}<g(n)<\sqrt{36n+91}$ and hence
$$ f(n)\in\bigl\{\lfloor\sqrt{36n+89} \rfloor,\lfloor\sqrt{36n+90} \rfloor,\lfloor\sqrt{36n+91} \rfloor\bigr\}.$$
Since $36n+90$ and $36n+91$ cannot be perfect squares ($x^2\equiv 18\pmod{36}$ and $x^2\equiv 19\pmod{36}$ have no solutions), we conclude
$$ f(n)=\lfloor\sqrt{36n+89} \rfloor=\lfloor\sqrt{36n+90} \rfloor=\lfloor\sqrt{36n+91} \rfloor$$
at least for $n>180$, and by inspection for all $n>3$.
Remark: The above computations also suggest which value of $c$ we shold check in the general case:
Similar as above we find that
$$ \sqrt n+\ldots +\sqrt{n+k}\approx (k+1)\sqrt n+\frac{0+1+\ldots +k)}{2\sqrt n}=(k+1)\sqrt n+\frac{k(k+1)}{4\sqrt n}$$
and therefore
$$ \left(\sqrt n+\ldots +\sqrt{n+k}\right)^2\approx (k+1)^2n+\frac{k(k+1)^2}{2}.$$
Further details depend on whether the integers near $\frac{k(k+1)^2}{2}$ are squares modulo $(k+1)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/477429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 1,
"answer_id": 0
} |
Find all values of parameters to make a certain function continuous. $$f(x) = \begin{cases}
\frac{\sin(x^2 - 4)}{x-2} & \text{if $x \in (-\infty, 2)$} \\
ax+b & \text{if $x \in [2,5]$} \\
{\sqrt{4(\ln(x-5))^2 + 1} \over \ln(x-5)} & \text{if $x \in (5, \infty)$}
\end{cases}$$
My process is equating the limit as $x$ approaches $2$ from the left and from the right:
\begin{gather*}
\lim_{x\to 2^-} \frac{\sin(x^2 - 4)}{x-2} = L = 4\\
\lim_{x\to 2^+} ax+b = L = 2a+b
\end{gather*}
Then $2a+b = 4 \implies b = 4 - 2a$.
\begin{gather*}
\lim_{x\to 5^-} ax+b = -5a+b \\
\lim_{x\to 5^+} {\sqrt{4(\ln(x-5))^2 + 1} \over \ln(x-5)} = -2
\end{gather*}
Then $-5a+b = -2$
$\therefore -5a+4-2a = -2 $
$\therefore -7a = -6$
$\therefore a = 6/7$
Is this right for getting the $a$ part?
| Looks pretty good! You just made a small sign mistake. You should have:
$$
\lim_{x\to 5^-} ax+b = \color{red}{+5}a+b
$$
Hence, by equating the left hand side and right hand side limits and substituting, we obtain:
$$
-2 = 5a + b = 5a + (4-2a) = 3a + 4 \iff -6 = 3a \iff \boxed{a = -2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/478853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Don't know how to find all the roots So i got this problem :
Find all the roots of $r^{3}=(-1)$
i can only think to use :
$\sqrt[n]{z} =\sqrt[n]{r}\left[\cos \left(\dfrac{\theta + 2\pi{k}}{n}\right) + i \sin\left(\dfrac{\theta + 2\pi k}{n}\right)\right] $
i know that $\theta = \tan^{-1}(\dfrac{b}{a}) $as in from $ z=a + bi$
$r=\sqrt[]{a^2 + b^2}$ , $k=0,1,2,3,4,.. n-1$
so anyone could explain this to me ?
lets say that $r^{3}=(-1) == z^{3}=(-1)$ a since i dont like it and it messes me up using the formula
so far :
$ z^{3} = (-1) $ so $z = \sqrt[3]{(-1 + 0 \cdot i)}$
$ r= \sqrt[]{(-1)^{2} + 0^{2}}=1$
$\tan^{-1}(\dfrac{b}{a})=\tan^{-1}(\dfrac{0}{-1})=\tan^{-1}(0)=0$ <- not sure about his part
k=0;
$ z^{3}=\sqrt[3]{1}(\cos(\dfrac{\pi + 2\pi \cdot 0}{3})+i \cdot \sin(\dfrac{\pi + 2\pi \cdot 0}{3})$
$ =\cos(\dfrac{\pi}{3}) +i\cdot \sin(\dfrac{\pi}{3})= \dfrac{\sqrt[]{3}}{2} +\dfrac{i}{2}$
k=1;
$ \sqrt[3]{1}(\cos(\dfrac{\pi+2\pi \cdot 1}{3})+i \cdot \sin(\dfrac{\pi+2\pi \cdot 1}{3})$
$=\cos(\pi) + i \cdot \sin(\pi)=(-1) + 0 = -1$
k=2;
$\sqrt[3]{1}(\cos(\dfrac{\pi + 2\pi \cdot 2}{3})+i \cdot \sin(\dfrac{\pi + 2\pi \cdot 2}{3})$
$=\cos(\dfrac{5\pi}{3})+i \cdot \sin(\dfrac{5\pi}{3})=-\dfrac{\sqrt[]{2}}{2}-i\cdot\dfrac{\sqrt[]{2}}{2}$
so is this correctet or im missing something here?
Roots $(\dfrac{\sqrt[]{3}}{2} +\dfrac{i}{2};-1;-\dfrac{\sqrt[]{2}}{2}-i\cdot\dfrac{\sqrt[]{2}}{2})$
| Just to hop and skip around some of the theory hovering in the background of some of these answers...
O.K., firstly by the Fundamental Theorem of Algebra $r^3=-1$ has at most three (distinct) solutions. This means that if you can find three then you are done (this is why in the '$k$' solutions they only need $k=0,1,2$).
Obviously $-1$ is one solution. Now what we want to do is find another (complex) number that when cubed gives you $-1$. Now how do complex numbers multiply? What does it look like? Multiplication for complex numbers is better handled when we write in polar form
$$z=|z|(\cos \theta+i\sin\theta),$$
where $|z|$, the magnitude, is the distance from $z$ to the origin and $\theta$, the argument $\arg(z)$, is the\an angle between the positive $x$-axis and the ray connecting $z$ to the origin.
It turns out that when you multiply complex numbers together they stretch\contract and rotate... more precisely
$$|z_1\cdot z_2|=|z_1||z_2|\text{ and }\arg(z_1\cdot z_2)=\arg(z_1)+\arg(z_2).$$
You can show this by multiplying together
$$z_1\cdot z_2=\left(|z_1|(\cos \theta_1+i\sin\theta_1)\right)\cdot\left(|z_2|(\cos \theta_2+i\sin\theta_2)\right).$$
Now $|-1|=1$ so we want $|r^3|=|r|^3=1$ and because $|r|\geq0$ we need $|r|=1$.
Next $\arg(-1)=\pi$ and $\arg(r^3)=3\arg(r)$ so we want
$$\pi=3\arg(r)\Rightarrow \arg(r)=\pi/3,$$
and indeed by DeMoivre (AKA multiplying complex numbers by themselves)
$$(\cos(\pi/3)+i\sin(\pi/3))^3=\cos \pi+i\sin \pi=-1$$
as required.
Now that is two solutions. Finally by the Conjugate Root Theorem, as the coefficients of $r^3=-1$ are all real the conjugate of this solution is also a root.
So we have the solution set
$$\{-1,\cos(\pi/3)+i\sin(\pi/3),\cos(\pi/3)-i\sin(\pi/3)\}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/479009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Given that $z=2-i$ and $z^2=3-4i$ find the roots of the equation $(z+i)^2=3-4i$ Given that $z=2-i$ and $z^2=3-4i$ find the roots of the equation $(z+i)^2=3-4i$
How do you use the given properties to find the roots? I can only obtain them the long way by working through
$$\begin{align}
&(z+i)(z+i)&=3-4i\\
&z^2+2zi+i^2&=3-4i\\
&z^2+2zi-1-3+4i&=0\\
&z^2+2zi+4i-4&=0\\
&z^2+2z-2z+2iz+4i-4&=0\\
&(z-(2-2i))(z+2)&=0\\
\end{align}$$
$$z=2-2i, z=-2$$
| \begin{align*}
(z+i)^2 &= 3 -4i\\
(z + i)^2 &= (2 -i)^2\\
(z+i)^2 - (2-i)^2 &= 0\\
[(z+i) - (2-i)][(z+i) + (2-i)] &= 0\\
[z - (2 - 2i)][z - 2] &= 0
\end{align*}
Therefore $z = 2 -2i$ or $z = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/479476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Limit of $\lim_{x\to\infty}{\frac{\cos(\frac{1}{x})-1}{\cos(\frac{2}{x})-1}}$ Find the limit of:
$$\lim_{x\to\infty}{\frac{\cos(\frac{1}{x})-1}{\cos(\frac{2}{x})-1}}$$
| $$\lim_{x \to \infty} \frac{\cos \frac1x - 1}{\cos \frac2x - 1} = \lim_{x \to \infty} \frac{\sin^2 \frac{1}{2x}}{\sin^2 \frac{1}{x}} =\lim_{x \to \infty} \frac{\sin^2 \frac{1}{2x}}{(\frac{1}{2x})^2}\frac{(\frac{1}{x})^2}{(\sin^2 \frac{1}{x})}\frac{1}{4}=1*1* \frac14=\frac14.$$
note that $$\lim_{x \to \infty} \frac{\sin \frac1x }{\frac1x}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/481421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 9,
"answer_id": 8
} |
$x^2 +y^2 + z^2$ is irreducible in $\mathbb C [x,y,z]$
Is $x^2 +y^2 + z^2$ irreducible in $\mathbb C [x,y,z]$?
As $(x^2+y^2+z^2)= (x+y+z)^2- 2(xy+yz+zx)$,
$$(x^2+y^2+z^2)=\left(x+y+z+\sqrt{2(xy+yz+zx)}\right)\left(x+y+z-\sqrt{2(xy+yz+zx)}\right).$$
But how to show that none of these factors belong to $\mathbb C [x,y,z]$?
| Let, $i$ be a square root of $-1$ in $K$.Then we have, $x^2+y^2+z^2=(x+iy)(x-iy)+z^2$
Since $x+iy$ is irreducible in $K[x,y]$ and $x+iy$ does not divide $x-iy;$ irreducibly follows from the Eienstein's Citeria in $K[x,y,z]=K[x,y][z]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/486668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 7,
"answer_id": 2
} |
proof that the inequality $n (n-1)^n > n^n$ holds for all $n \ge 4$ I am trying to prove that the inequality $n(n-1)^n > n^n$ holds for all $n\ge4$.
I tried using mathematical induction, but I really couldn´t find a way how to get past the $P(n) \implies P(n+1)$ step. I get $n(n-1)^n > n^n$ implies $n^{n+1} > (n+1)^n$ I tried expanding with binomial theorem but to no avail. I also know that in the limit the equality holds (but I want to prove it for all $n \ge 4$)
| Rearrange the inequality a little:
$$\left(\frac{n-1}{n}\right)^n > \frac1n.$$
Then you get
$$\begin{align}
\left(\frac{n}{n+1}\right)^{n+1} &= \left(\frac{n}{n+1}\right)^{n+1} \left(\frac{n}{n-1}\right)^n \left(\frac{n-1}{n}\right)^n\\
&= \frac{n}{n+1} \left(\frac{n^2}{n^2-1}\right)^n\left(\frac{n-1}{n}\right)^n\\
&> \frac{n}{n+1} \left(\frac{n^2}{n^2-1}\right)^n \frac1n\\
&= \frac{1}{n+1} \left(\frac{n^2}{n^2-1}\right)^n\\
&> \frac{1}{n+1}.
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/486948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Irrational inequalities question: $\sqrt { -3x+1 } + \sqrt {6x+1} < \sqrt {3x+4}$ and $\sqrt { -6x+10 } + \sqrt {-x+2} \gt \sqrt {4x+5}$ Consider the following inequalities:
\begin{equation*}
\sqrt { -3x+1 } + \sqrt {6x+1} \lt \sqrt {3x+4}, \\
\sqrt { -6x+10 } + \sqrt {-x+2} \gt \sqrt {4x+5}.
\end{equation*}
Attempt at a solution; after performing all the standard actions, I got: $\frac 16\lt x \le\frac 13 $ or $\frac {-1}{6}\le x \lt 0 $, for first inequality, and: $\frac {-34}{97}\lt x \lt 1 $ for the second.
Official results state however: $\frac {-1}{6}\le x \lt 0 $ for first inequality, and
$\frac {-5}{4}\le x \lt 1 $ for the second.
What was my mistake?
| The second equation became $ 97x^2-66x-31<0$
$(97x+31)(x-1)<0$
Therefore
$$-\frac{31}{97}<x<1$$
This condition satisfies all conditions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/487356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving inequality $(a+\frac{1}{a})^2 + (b+\frac{1}{b})^2 \geq \frac{25}{2}$ for $a+b=1$ If $a, b$ are positive real numbers and $a+b = 1$, prove that :
$$\left(a+\frac{1}{a}\right)^2 + \left(b+\frac{1}{b}\right)^2 \geq \frac{25}{2}$$
I can see that the value $\frac{25}2$ is attained for $a=b=\frac12$. But I do not know how to show that this is the minimal possible value.
Thank you.
| Let $f(p,q)=\left(p+\frac{1}{p}\right)^2+\left(q+\frac{1}{q}\right)^2$.
Using Lagrange multiplier's method:
$$L=\left(p+\frac{1}{p}\right)^2+\left(q+\frac{1}{q}\right)^2+\lambda (1-p-q)\\
\begin{cases}L_p=2\left(p+\frac1p\right)(1-\frac1{p^2})-\lambda=0\\
L_q=2\left(q+\frac1q\right)(1-\frac1{q^2})-\lambda=0\\
L_{\lambda}=1-p-q=0\end{cases} \Rightarrow p=q=\frac12$$
SOC:
$$L_{pp}=2\left(1-\frac1{p^2}\right)^2+2\left(p+\frac1p\right)\cdot \frac1{p^3}>0\\
L_{pq}=0\\
\Delta=L_{pp}L_{qq}-L_{pq}^2=L_{pp}^2>0$$
Hence: $f(\frac12,\frac12)=12.5$ is minimum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/487486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 10,
"answer_id": 9
} |
Having trouble proving epsilon deltas of limits with roots I am having a big issue proving this equation... Specifically at the part that involves the roots. I can't seem to get $\sqrt{2x-1}/\sqrt{x-3} - \sqrt{7}$ to look like $x-4$ in terms of epsilon. In other words I don't know how I can algebraically manipulate the root part of this proof to make it look like the $x-4$ part of it.
| Given any $\epsilon>0$, let $\delta = \min\{\epsilon/A, B\}$, where $A,B$ are some positive constants that we're going to figure out later. Then if $0<|x-4|<\delta$, observe that:
\begin{align*}
\left| \frac{\sqrt{2x-1}}{\sqrt{x-3}}-\sqrt{7} \right|
&= \left| \frac{\sqrt{2x-1}-\sqrt{7}\sqrt{x-3}} {\sqrt{x-3}} \right| \\
&= \left| \frac{\sqrt{2x-1}-\sqrt{7}\sqrt{x-3}} {\sqrt{x-3}} \cdot \frac{\sqrt{2x-1}+\sqrt{7}\sqrt{x-3}}{\sqrt{2x-1}+\sqrt{7}\sqrt{x-3}} \right| \\
&= \left| \frac{(2x-1)-7(x-3)} {\sqrt{x-3}(\sqrt{2x-1}+\sqrt{7}\sqrt{x-3})} \right| \\
&= \left| \frac{-5(x-4)} {\sqrt{x-3}(\sqrt{2x-1}+\sqrt{7}\sqrt{x-3})} \right| \\
&\leq \left| \frac{-5(x-4)} {\sqrt{x-3}(0+\sqrt{7}\sqrt{x-3})} \right| \qquad\text{since }\sqrt{2x-1}\geq0\\
&= \left| \frac{-5(x-4)} {\sqrt{7}(x-3)} \right|\\
&= \left| \frac{-5}{\sqrt 7} \right| \cdot \frac{1}{|x-3|} \cdot |x-4|\\
&= \frac{5}{\sqrt 7} \cdot \frac{1}{|x-3|} \cdot |x-4|\\
&< \frac{5}{\sqrt 7} \cdot \frac{1}{C} \cdot \frac \epsilon A \qquad \text{since } |x-4|<\delta \leq \epsilon/A\\
&= \epsilon \\
\end{align*}
The hard part is figuring out $A,B,C$. Somehow, we need to choose $B$ such that $|x-4|<\delta \leq B \implies |x-3|>C$ for some $C>0$ so that $\dfrac{1}{|x-3|} < \dfrac1C$ (this justifies the second to last step). Once we've done that, it suffices to let $A = \dfrac{5}{C\sqrt 7}$ (this justifies the last step).
With this in mind, suppose we choose $B=1/2$. Then we have:
\begin{align*}
|x-4| &< 1/2 \\
-1/2 < x-4 &< 1/2 \\
1/2 < x-3 &< 3/2 \\
\end{align*}
Hence, we have $|x-3| > 1/2$, so we obtain $C=1/2$. Thus, we may let $A = \dfrac{10}{\sqrt 7}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/487554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to find Basis of vector space of real $n \times n$ matrices $V$ is vector space of real $n \times n$ matrices and $W$ is a set of elements of $A$ with trace equal to zero. How can we find the basis of $W$?
I have seen a few questions previously posted including one that was identical with the exception that it considered only $2 \times 2$ matrices.
I have tried to use the logic derived from them and worked out the following solution:
The dimension of the basis of $W$ will be $n-1$. The basis will have $n \times n$ matrices of form
Mi = { a11 = 1, all elements except ai +1 i + 1 =0, ai +1 i + 1 = -1} . So M1 = {a11 = 1, a22 = -1, rest all elements = 0}
Please correct me if I'm wrong.
| Your answer is partially correct. For instance
$$
A=\left( {\begin{array}{*{20}{c}}
2 & 0 & 1 \\
0 & { - 1} & 0 \\
0 & 0 & { - 1} \\
\end{array}} \right) \in M_3
$$
is not in the span of your basis because $a_{1,3} \ne 0$ but $trace(A)=0$. When $n=2$ your subspace is
$$\begin{align} M_2 &=\left\{ \left( {\begin{array}{*{20}{c}}
a & b \\
c & d \\
\end{array}} \right) | a,b,c,d \in \mathbb{R},a+d=0
\right\} \\
&=\left\{ \left( {\begin{array}{*{20}{c}}
a & b \\
c & {-a} \\
\end{array}} \right) | a,b,c \in \mathbb{R}
\right\} \\
&=\left\{ a \left( {\begin{array}{*{20}{c}}
1 & 0 \\
0 & {-1} \\
\end{array}} \right) + b\left( {\begin{array}{*{20}{c}}
0 & 1 \\
0 & 0 \\
\end{array}} \right)+
c\left( {\begin{array}{*{20}{c}}
0 & 0 \\
1 & 0 \\
\end{array}} \right)| a,b,c \in \mathbb{R}
\right\}
\end{align}$$
so a a natural basis for $M_2$ is
$$B=\left\langle \left( {\begin{array}{*{20}{c}}
1 & 0 \\
0 & {-1} \\
\end{array}} \right), \left( {\begin{array}{*{20}{c}}
0 & 1 \\
0 & 0 \\
\end{array}} \right),
\left( {\begin{array}{*{20}{c}}
0 & 0 \\
1 & 0 \\
\end{array}} \right)
\right\rangle.$$
So $dim(M_2)=3=2*2-1$. To extend above idea, we have for $n \in \mathbb{N}$
$$\begin{align} M_n &=\left\{ \left( {\begin{array}{*{20}{c}}
{{a_{1,1}}} & \cdots & {{a_{1,n}}} \\
\vdots & \ddots & \vdots \\
{{a_{n,1}}} & \cdots & {{a_{n,n}}} \\
\end{array}} \right) | \sum\limits_{j=1}^{n}{a_{j,j}}=0
\right\} \\
&=\left\{ \left( {\begin{array}{*{20}{c}}
{{a_{1,1}}} & \cdots & {{a_{1,n}}} \\
\vdots & \ddots & \vdots \\
{{a_{n,1}}} & \cdots & {{a_{n,n}}} \\
\end{array}} \right) | a_{n,n}=-\sum\limits_{j=1}^{n-1}{a_{j,j}}
\right\} \\
&=\left\{ \left( {\begin{array}{*{20}{c}}
{{a_{1,1}}} & \cdots & {{a_{1,n}}} \\
\vdots & \ddots & \vdots \\
{{a_{n,1}}} & \cdots & {{\sum\limits_{j=1}^{n-1}{-a_{j,j}}}} \\
\end{array}} \right)
\right\} \\
&=\left\{ {{a_{1,1}}\left( {\begin{array}{*{20}{c}}
1 & 0 & \cdots & 0 \\
0 & \vdots & \vdots & \vdots \\
\vdots & \vdots & \vdots & 0 \\
0 & \ldots & 0 & { - 1} \\
\end{array}} \right) + {a_{1,2}}\left( {\begin{array}{*{20}{c}}
0 & 1 & 0 & 0 \\
0 & 0 & \vdots & \vdots \\
\vdots & \vdots & \vdots & 0 \\
0 & \ldots & 0 & { - 1} \\
\end{array}} \right) + \cdots + {a_{n,n - 1}}\left( {\begin{array}{*{20}{c}}
0 & 0 & \cdots & 0 \\
0 & \vdots & \vdots & \vdots \\
\vdots & \vdots & 0 & 0 \\
0 & \vdots & 1 & { - 1} \\
\end{array}} \right)} \right\}
\end{align}
$$
and a natural basis for $M_n$ is
$$B={\left\langle {\left( {\begin{array}{*{20}{c}}
1 & 0 & \cdots & 0 \\
0 & \vdots & \vdots & \vdots \\
\vdots & \vdots & \vdots & 0 \\
0 & \ldots & 0 & { - 1} \\
\end{array}} \right),\left( {\begin{array}{*{20}{c}}
0 & 1 & 0 & 0 \\
0 & 0 & \vdots & \vdots \\
\vdots & \vdots & \vdots & 0 \\
0 & \ldots & 0 & { - 1} \\
\end{array}} \right), \cdots ,\left( {\begin{array}{*{20}{c}}
0 & 0 & \cdots & 0 \\
0 & \vdots & \vdots & \vdots \\
\vdots & \vdots & 0 & 0 \\
0 & \vdots & 1 & { - 1} \\
\end{array}} \right)} \right\rangle }
$$
hence $dim(M_n)=n^2 -1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/488659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Techniques of Proof Prove If n is an integer, then $n^2+n^3$ is an even number.
I don't know if I'm just reiterating what I'm asked to prove or if my ideas are actually proving the statement.
If $n^2+n^3$ is an even number then $n^2+n^3=2n$ because $2n$ is an even number. By solving the equation by algebraic operations we get $n^2+n^3-2n=0$ and it follows that $n(n^2+n-2)=0$. By factoring we see that $n(n+2)(n-1)=0$ so $n= 0, 1, 2$ which are all integers so therefore $n$ must be an integer.
| We have two cases, because $n \in \mathbb{Z}$ is even or odd.
If $n$ is even, then there is a $k \in \mathbb{Z}$ such that $n=2k$, so
$$n^2+n^3=(2k)^2+(2k)^3=4k^2+8k^3=2(2k^2+4k^3)$$
with $q \in \mathbb{Z}$ such that $q=2k^2+4k^3$ we have $n^2+n^3=2q$, i.e. $n^2+n^3$ is even.
If $n$ is odd, then there is a $m \in \mathbb{Z}$ such that $n=2m+1$, so
$$\begin{align}n^2+n^3&=(2m+1)^2+(2m+1)^3\\&=(4m^2+4m+1)+(8m^3+12m^2+6m+1)\\
&=8m^3+16m^2+10m+2\\
&=2(4m^3+8m^2+5m+1)
\end{align}$$
with $p \in \mathbb{Z}$ such that $p=4m^3+8m^2+5m+1$ we have $n^2+n^3=2p$, i.e. $n^2+n^3$ is even.
In both cases $n^2+n^3$ is even. Therefore for all $n \in \mathbb{Z}$, $n^2+n^3$ is even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/488746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Prove the following set has a multiplicative inverse Let $F = \{x + y\sqrt{7} : x, y \in Q\}$ with the usual addition and multiplication operations. Show that $F$ has a multiplicative inverse.
So far, I have
$$\left(x_1 + y_1\sqrt{7}\right) \left(\frac{1}{x_1} + \left(\frac{1}{y_1}\right)\sqrt{7}\right)$$
But that led me to $8+2\sqrt{7}(\frac{1}{x_1} + \frac{1}{y_1}) = 1$...I don't know what goes from there.
| Suppose $(a+b\sqrt{7})(x+y \sqrt{7}) = 1$ (and $(x,y) \neq (0,0)$). This gives $(ax+7 by) + (xb+ay)\sqrt{7} = 1$, from which we get two equations: $ax+7 by = 1$, $xb+ay = 0$. Now solve for $a,b$ to get
$a = \frac{x}{x^2-7 y^2}$, $b=-\frac{y}{x^2-7 y^2}$.
To see that $x^2 \neq 7 y^2$, note that $\sqrt{7}$ is irrational.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/489189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
The maximum and minimum of a function How can I find the maximum and minimum of $$L=\sqrt{1-x^2}+\sqrt{1-y^2}+\sqrt{1-z^2},$$
given that $x^2+y^2+z^2=1$?
I worked on some basic algebra, but it doesn't get me anywhere, so I'm stuck on it.
| By using lagrange multipliers $$ f(x,y,z) = \sqrt{1 - x^2} + \sqrt{1 - y^2} + \sqrt{1 - z^2} $$ subject to $$ g(x,y,z) = x^2 + y^2 + z^2 - 1 = 0 $$
$$ F(x,y,z,\lambda) = f + \lambda g = \sqrt{1 - x^2} + \sqrt{1 - y^2} + \sqrt{1 - z^2} + \lambda( x^2 + y^2 + z^2 - 1) $$
$$ F_x = \frac{-2x}{2\sqrt{1-x^2}} + 2\lambda x = 0 $$
$$ F_y = \frac{-2y}{2\sqrt{1-y^2}} + 2\lambda y = 0 $$
$$ F_z = \frac{-2z}{2\sqrt{1-z^2}} + 2\lambda z = 0 $$
$$ F_{\lambda} = x^2 + y^2 + z^2 - 1 = 0 $$
now solve this system of equations :-)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/489463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Hypergeometric function argument simplification Let $_2 F_1 (a,b,c,z)$ be the hypergeometric function. As a result of some integration, I obtained the following expression
$$
f(x) = \frac{\Gamma(2k)\Gamma(2m)}{\Gamma(m)^2\Gamma(k)^2} \frac{\Gamma(m+k)^2}{\Gamma(2k+2m)} \frac{_2 F_1 (2k,k+m,2k+2m,1-\frac{1}{x})}{x^{k+1}}.
$$
The parameters $m$ and $k$ are positive numbers. I will have to use this result in some other integrals, namely the average bit error rate and the capacity integrals:
$$
P(E) = \int_0^{+\infty} \alpha \mathrm{erfc}(\sqrt{\beta x}) f(x) dx,
$$
and
$$
C = \frac{1}{\ln(2)}\int_0^{+\infty} \log(1+x) f(x) dx.
$$
I thought of using the Meijer G function to obtained some closed-form expression but I need to transform the argument of the hypergeometric function in $f(x)$ to be linear. Is there any relation that could help me ? and is there a relation that can simplify the product of such hypergometric function and rational function $\frac{1}{x^{k+1}}$ ?
Thank you.
| Those things are not really very hard only quite tedious since it takes a lot of time to solve them even when using Mathematica, We want to compute;
\begin{equation}
lhs_{k,m} := F_{2,1}\left[ \begin{array}{cc} 2k & k+m \\ 2k+2 m \end{array}; 1- \frac{1}{x} \right]
\end{equation}
Firstly take $m=0$ then:
\begin{eqnarray}
lhs_{k,0}&=&\sum\limits_{p=0}^\infty (k+m)^{(p)} \frac{(1-\frac{1}{x})^p}{p!}\\
&=& \sum\limits_{p=0}^\infty \binom{k+m+p-1}{p} (1-\frac{1}{x})^p \\
&=& \sum\limits_{p=0}^\infty \binom{-k-m}{p} (\frac{1}{x}-1)^p = (1+\frac{1}{x} - 1)^{-k-m}=x^{k+m}
\end{eqnarray}
Now we take $m=1$.
\begin{eqnarray}
lhs_{k,1}&=& \sum\limits_{p=0}^\infty \frac{(2k)(2k+1)}{(2k+p)(2k+p+1)} \cdot (k+m)^{(p)} \frac{(1-\frac{1}{x})^p}{p!}\\
&=& (2k)(2k+1) \sum\limits_{p=0}^\infty \left( \frac{1}{2k+p} - \frac{1}{2k+p+1} \right) \binom{-k-1}{p} (\frac{1}{x} - 1)^p \\
&=& (2k)^{(2)} \int\limits_0^1 \left( t^{2k-1} - t^{2 k} \right) (1+t (\frac{1}{x} - 1))^{-k-1} dt \\
&=& (2k)^{(2)} \frac{x^{2 k}}{(1-x)^{2k+1}} \int\limits_1^{1/x} \frac{(u-1)^{2k-1}}{u^{k+1}} (1-x u) du \\
&=&(2k)^{(2)} \frac{x^{2 k}}{(1-x)^{2k+1}} \sum\limits_{p=0}^{2k-1} \binom{2k-1}{p} (-1)^{p+1}\cdot \\
&&
\left(
\frac{(\frac{1}{x})^{p-k} - 1}{p-k} 1_{k\neq p} + \log(1/x) 1_{k=p} - x \frac{(\frac{1}{x})^{p-k+1} - 1}{p-k+1} 1_{k \neq p+1} - x \log(1/x) 1_{k=p+1}
\right)
\end{eqnarray}
From the above it is not hard to see that those calculations can be carried through to higher values of $m$ and then the generic result can be proven by induction. The result reads:
\begin{eqnarray}
lhs_{k,m}= \frac{(2k)^{(2m)}}{(2m-1)!} \sum\limits_{q=0}^{2m-1} \binom{2m-1}{q} (-1)^q \frac{x^{2k+q}}{(1-x)^{2k+q}} \sum\limits_{p=0}^{2k+q-1} \binom{2k+q-1}{p} (-1)^{q-1-p}
\left(
\frac{(\frac{1}{x})^{p-k-m+1} -1}{p-k-m+1} 1_{p-k-m \neq 1} + \log(\frac{1}{x}) 1_{p-k-m=1}
\right)
\end{eqnarray}
for $k=1,2,\cdots$ and $m=1,2,\cdots$ and $lhs_{0,m}=1$ and $lhs_{k,0}=x^k$.
In[18]:= mmax = 5;
mat = Table[
Hypergeometric2F1[2 k, k + m, 2 k + 2 m, 1 - 1/x], {k, 0,
mmax}, {m, 0, mmax}];
tmp = Simplify[
Table[Pochhammer[2 k, 2 m]/(2 m - 1)! Sum[
Binomial[2 m - 1, q] (-1)^q x^(2 k + q)/(1 - x)^(2 k + q) Sum[
Binomial[2 k + q - 1, p] (-1)^(q - 1 - p) If[
p - k - m != -1, (1/x^(p - k - m + 1) - 1)/(p - k - m + 1),
Log[1/x]], {p, 0, 2 k + q - 1}], {q, 0, 2 m - 1}], {k, 0,
mmax}, {m, 0, mmax}] - mat, Assumptions -> x > 1];
MatrixForm[tmp]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/489932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
If $\,x-\frac 1 x=k, \, k$ being any integer,then $\,\,x^5-\frac {1}{x^5}=?$ I am stuck with the following problem which one of friends gave me :
If $\,x-\frac 1 x=k, \, k$ being any integer,then $\,\,x^5-\frac {1}{x^5}=?$
The options are $\,\,k^5+4k^3+4k, \,k^5+5k^3+6k,\,k^5+5k^3+5k,\,k^5+5k^3+4k $.
We see that $x-\frac 1 x=k \implies x=\frac{k \pm \sqrt{k^2+4}}{2}$. Now putting this value to $\,\,x^5-\frac {1}{x^2}$ makes the calculation complicated.
Can anyone help? Thanks and regards to all.
EDIT: The problem contained a typo and thanks to @noam for pointing that out. Now using binomial expansion of $x^5-\frac{1}{x^5}$, we see that option 3 is the correct choice.
| Let $a=x,b=\frac 1x$ and hence $$a^5-b^5=(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4)$$ where $a-b=k,a^2+b^2=k^2+2,\,a^4+b^4=k^4+4k^2+2$. Now putting this values in the expression ,we get $a^5-b^5=k^5+5k^3+5k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/492605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Factorising a 3 x 3 determinant - What Am I doing Wrong? $$\begin{vmatrix}
1 & a & a^3 \\
1 & b & b^3 \\
1 & c & c^3 \\
\end{vmatrix}$$
subtracting the top row from the middle and bottom rows
$$ =
\begin{vmatrix}
1 & a & a^3 \\
0 & (b - a) & (b^3 - a^3)\\
0 & (c - a) & (c^3 - a^3)\\
\end{vmatrix}
$$
expanding the determinant
$$
= \begin{vmatrix}
(b - a) & (b^3 - a^3)\\
(c - a) & (c^3 - a^3)\\
\end{vmatrix}
$$
pulling out factors
$$
= (b-a)(c-a)
\begin{vmatrix}
1 & (b^2 + a^2)\\
1 & (c^2 + a^2)\\
\end{vmatrix}
$$
and evaluating gives
$$
= (b-a)(c-a)(c^2 - b^2)
$$
yet the answer in the book gives
$$(a+b+c)(a-b)(b-c)(c-a)$$
I am self studying so thanks in advance
| But $b^3-a^3 = (b-a)(b^2+ab+a^2)$ and $c^3-a^3 = (c-a)(c^2+ac+a^2)$. Hence
$$(c^2+ac+a^2) - (b^2+ab+a^2) = c^2+ac- b^2-ab = (c-b)(a+b+c).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/494270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Conic Ellipse problem For an ellipse, $9x^2+4y^2=36$, find vertices and foci.
I would first standardised the equation in form $(x)^2/a^2 + y^2/b^2=1$ thus…………(i)
divide all sides by 36,I get:
$$(9x^2)/36+(4y^2)/36=36/36$$ which is equiv. to $x^2/4+y^2/9=36/36$
based on $x^2/4+y^2/9=36/36$ we can rewrite this as:
$$x^2/2^2 +y^2/3^2 =1$$ which is in standard $x^2/a^2 +y^2/b^2 =1$
we can even pluck out our values for a,b thus: a=2 and b=3
but by pythagoras,we know $c^2=a^2-b^2$ hence $c^2=2^2-3^2$
so in my case, $c^2=4-9= -5$ and $\therefore$ $c=√(-5)$
But I reckon the values of a,b can swap depending on major axis since we can re-write standard equation as $$(x)^2/b^2 + y^2/b^2=1$$ which will bring a different value of $c^2$.
Questions: How do I make sure I get the correct value of $c^2$?
Question: How do ensure the ellipse is plotted on the correct major axis?
|
For the ellipse $9x^2 + 4y^2 = 36$ ...
This is, as you have found, equivalent to $$\dfrac{x^2}{4} + \dfrac{y^2}{9} = 1$$
Now in an ellipse, the larger denominator corresponds with $a$.
(Contrast this with a hyperbola, where the denominator of the positive variable term is $a^2$)
So we find that $a = 3, b = 2$. The major axis is vertical, as the $a$ was under the $y^2$.
Therefore, the vertices are up and down $(a = 3)$ units from the center, which is the origin in this case.
The covertices are $(b = 2)$ units left and right of center.
$c = \sqrt5$, and the foci are $c$ units up and down from the center.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/496083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding all the set of values for k such that $-2x^3+24x+8=k$ find all the set of values for k such that $-2x^3+24x+8=k$ has more than one solution.
This problem can only be solved using basic Calculus 1 knowledge.
I thought about setting the equation to zero and then computing the x value that can then be rewritten in terms of k but I am not sure that is the correct approach can someone help me?
| Let $f(x) = -2x^3+24x+8 = k$
Now Let $f(x) = -2x^3+24x+8$ and we will find the nature of cure and plot the curve in $X-Y$ plane
So $f^{'}(x) = -6x^2+24$ and for Max. and Min. $f^{'}(x) = 0$
we get $-6(x^2-4) = 0 \Rightarrow x = \pm 2$
Now $f^{''}(x) = -12x$
So at $x = -2$ , we get $f^{''}(-2) = -12 \times -2 = 24>0$
So $x = -2$ is a point of Minimum
Similarly at $x = 2$ , we get $f^{''}(2) = -12 \times 2 = -24>0$
So $x = 2$ is a point of Maximum
So $f(-2) \leq f(x) \leq f(+2)$
So $-24 \leq f(x) \leq 40$
So value of $k$ for which the equation $f(x) = -2x^3+24x+8$ has more then one real roots is
$-24 < k < 40\Rightarrow k\in \left(-24,40\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/497118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How prove this inequality $(a^3+b^3+c^3)(ab+bc+ac)\ge 6abc(a^2+b^2+c^2-ab-bc-ac)$ let $a,b,c$ are postive numbers, show that
$$(a^3+b^3+c^3)(ab+bc+ac)\ge 6abc(a^2+b^2+c^2-ab-bc-ac)$$
my try: let
$$a+b+c=p,ab+bc+ac=q,abc=r$$
and the
$$a^3+b^3+c^3=(a+b+c)^3-3(ab+bc+ac)(a+b+c)=p^3-3pq$$
Thank you
| Note that:
$$
(a^3+b^3+c^3)\geq a^3+b^3+c^3-3abc= (a+b+c)(a^2+b^2+c^2-ab-bc-ac)
$$
Therefore it is enough to prove that
$$
(a+b+c)(ab+bc+ac)\ge 6abc
$$
This is done as follows:
$$
(a+b+c)(ab+bc+ac)\ge 3(abc)^{\frac{1}{3}}\times 3(abc)^{\frac{2}{3}}=9abc \geq 6abc.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/498517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Determine the number of real roots in the equation... Determine the number of real roots in the equation $2x^3 + x^2 = 3$.
I know about finding the different roots, and solving giving that it has (for example) 2i as a root, but I'm not sure how to just find the amount of real roots.
I've found nothing that can help so far.
| HINT:
$$2x^3+x^2-3=2(x^3-1)+(x^2-1)=2\{(x-1)(x^2+x+1)\}+(x-1)(x+1)$$
$$=(x-1)\{2(x^2+x+1)+x+1\}=(x-1)(2x^2+3x+3)$$
Alternatively, using Remainder Theorem, $(x-1)|(2x^3+x^2-3)$
So, by actual division $2x^3+x^2-3=(x-1)(2x^2+3x+3)$
Do you know how to determine the nature of a Quadratic Equation?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/500346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How prove this inequality $ \frac{b^3+c^3}{a}+\frac{c^3+a^3}{b}+\frac{a^3+b^3}{c} \ge 2(a^2+b^2+c^2)+3\left((b-c)^2+(c-a)^2+(a-b)^2\right)$ let $a,b,c$ are positive numbers, show that
$$ \frac{b^3+c^3}{a}+\frac{c^3+a^3}{b}+\frac{a^3+b^3}{c} \ge 2(a^2+b^2+c^2)+3\left((b-c)^2+(c-a)^2+(a-b)^2\right)\cdots (1)$$
my try:
$$\Longleftrightarrow \frac{b^3+c^3}{a}+a^2+\frac{c^3+a^3}{b}+b^2+\frac{a^3+b^3}{c}+c^2 \ge 3(a^2+b^2+c^2)+3\left((b-c)^2+(c-a)^2+(a-b)^2\right)$$
$$\left(a^3+b^3+c^3\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge 9(a^2+b^2+c^2)-6(ab+bc+ac)$$
$$(a^3+b^3+c^3)(ab+bc+ac)\ge abc[9(a^2+b^2+c^2)-6(ab+bc+ac)]$$
some days ago,I have ask this same problem：How prove this inequality $(a^3+b^3+c^3)(ab+bc+ac)\ge 6abc(a^2+b^2+c^2-ab-bc-ac)$
then I can't prove it,Thank you
maybe $(1)$ have other nice methods?
| Hint: Read Arash's proof and tighten up his loose bounds.
Similar to the proof by Arash Beh in your other linked question, we have
$$(a^3 + b^3 + c^3) = (a+b+c) ( a^2 + b^2 +c^2 - ab - bc -ca) + 3abc $$
He showed that
$$ (a+b+c)(a^2+b^2+c^2 - ab - bc - ca)(ab+bc+ca) \geq 9abc( a^2 + b^2 + c^2 - ab - bc -ca),$$
which is obvious because
$$(a+b+c) (ab+bc+ca) \geq 9abc.$$
We then add
$$ 3abc (ab+bc+ca) = 3 abc( ab+bc+ca)$$
to both sides, and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/500437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Is this a valid proof of a Diophantine equation Problem: Find all $x,y,z\in\mathbb{Z}$ satisfying $x^2 + x = y^2 + y + z^2 + z$.
Approach: It is equivalent to solve $x^2 + x - y^2 - y = z^2 + z$ or $(x-y)(x+y+1) = z(z+1).$ Let $m=x-y$ and $x+y+1=n.$ Then $mn=z(z+1).$ We can find that $x=(m+n-1)/2,y=(n-m-1)/2.$ There are integer solutions iff $n\not\equiv m \mod 2.$
| Everything can be done much easier! Rewrite this equation a little differently.
$X(X+a)+Y(Y+a)=Z(Z+a)$
Formulas for the solution can then be written, $p,k$ - where are integers and sets us.
$X=pk$
$Y=\frac{(p^2-1)k}{2}+\frac{(p-1)a}{2}$
$Z=\frac{(p^2+1)k}{2}+\frac{(p-1)a}{2}$
If we use the solutions of Pell's equation $p^2-2s^2=1$
Then the solution can be written:
$X=2(s+p)sL+as(2s+p)$
$Y=(2s+p)pL+as(2s+p)$
$Z=(2s^2+2ps+p^2)L+2as(s+p)$
And more.
$X=2s(s-p)L+ap(s-p)$
$Y=(p-2s)pL+ap(s-p)$
$Z=(2s^2-2ps+p^2)L+ap(2s-p)$
$L$ - given by us and can be any integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/500642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that a triangle is right angled ... ... if one side is double the other and the angle opposite these sides differ by $\frac \pi 3$
It would be better if this was proven using only algebraic trigonometric identities(sine/cosine rules, etc.)
Let A,B,C be the angles and a,b,c be the respective sides opposite to them.
\begin{align} b &=2a \\
|B-A|&=\frac \pi 3 \\
\sin(B-A) &= \frac {\sqrt{3}} 2\\
\sin B \cos A-\cos B \sin A &=\frac {\sqrt{3}} 2\\
\frac b k \frac {b^2+c^2-a^2}{2bc}-\frac a k \frac {a^2+c^2-b^2}{2ac}&=\frac{\sqrt 3} 2 \\
\frac 1 {kc} (b^2-a^2)&=\frac {\sqrt 3} 2 \\
\frac {3a^2} {kc} &= \frac {\sqrt 3} 2
\end{align}
And I'm stuck at that.
Another approach
\begin{align}
\frac a {\sin A } &= \frac b {\sin B} \\
\frac {a}{\sin A} &= \frac {2a} {\sin (\frac \pi 3+A)}\\
\frac {\sqrt 3} 2\cos A-\frac 1 2 \sin A&= 2\sin A\\
\tan A&= \frac {\sqrt 3} 5\\
\end{align}
That doesn't lead anywhere too
| Let $\;\Delta ABC\;$ be our triangle, with sides $\;|AB|=x\;,\;\;|BC|=2x\;$ , and
$$\angle A=\beta\;,\;\;\angle C=\alpha\;\;,\;\;\beta-\alpha=\frac\pi3$$
Apply now the Law of Sines:
$$\frac{|AB|}{\cos\alpha}=\frac{|BC|}{\sin\beta}\iff \frac x{\sin\alpha}=\frac{2x}{\sin\left(\alpha+\frac\pi3\right)}=\frac{2x}{\frac12\sin\alpha+\frac{\sqrt3}2\cos\alpha}\implies$$
$$\implies \frac x{\sin\alpha}=\frac{4x}{\sin\alpha+\sqrt3\cos\alpha}\implies\sin\alpha+\sqrt3\cos\alpha=4\sin\alpha\implies$$
$$\implies\tan\alpha=\frac1{\sqrt3}\implies \alpha=\frac\pi6$$
and thus
$$\beta=\alpha+\frac\pi3=\frac\pi6+\frac\pi3=\frac\pi2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/503744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Integrating $\int \frac{x^2+1}{x(x^2-1)}$ How would I integrate the following.
$$\int \frac{x^2+1}{x(x^2-1)}$$
I have done the following.
$$\frac{x^2}{(x)(x+1)(x-1)}$$
$$\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x-1}$$
I then did $\quad \displaystyle A(x^2-1)+B(x-1)+C(x+1)=x^2+1$
Then $\quad Ax^2-1A+Bx-1B+Cx+C=x^2+1$
Grouping I get $$Ax^2=x^2,\quad A=1, \quad C=1,\quad Bx+Cx=0, \quad B=-1$$
Then plug back in
$$\int\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x-1}$$
But I am not sure if I did it correctly.
| A simpler solution is to note that you can calculate
$$\int \frac{x^2-1}{x(x^2-1)} \mbox{and} \int \frac{3x^2-1}{x(x^2-1)}$$
Now
$$x^2+1=(3x^2-1)-2(x^2-1)$$
Thus, your integral is
$$\int \frac{3x^2-1}{x^3-x}- 2\int \frac{x^2-1}{x(x^2-1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/505009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Integrate $\int \frac{x^2}{(x^2+4)^2}$ $$\int \frac{x^2\;dx}{(x^2+4)^2}$$I suppose you must have to use trigonometric substitution, or something, but do not even know where to begin to solve this integral, guys, please help me!!
| Integrate by parts:
$$\int\frac{x^2}{(x^2+4)^2}dx=\int x\frac{x}{(x^2+4)^2}dx=\frac{1}{2} \int x(\frac{-1}{x^2+4})'dx=\frac{1}{2}(\frac{-x}{x^2+4} +\int \frac{1}{x^2+4}dx)=\frac{1}{2}(\frac{-x}{x^2+4} +\frac{1}{2}\arctan\frac{x}{2}) +C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/506047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Proof that $n+1$ divides $\sum_{k=1}^n k^q$ for $q > 0$ Since $$\sum_{k=1}^n k = \frac{n\left(n+1\right)}{2}$$
$$\sum_{k=1}^n k^2 = \frac{n\left(n+1\right)\left(2n+1\right)}{6}$$
$$\sum_{k=1}^n k^3 = \frac{n^2\left(n+1\right)^2}{4}$$
... and so on. Is there a proof that generally $\left(n+1\right) | \sum_{k=1}^n k^q$ for $q\in\mathbb{N}$ positive?
| I think what you claim in your comments is true. We can use induction. Let's assume that all sums up to $m$ contains this term $(n+1)$.
\begin{align*}
(n+1)^{m+2} - 1&=
\sum_{k=1}^n (k+1)^{m+2} - \sum_{k=1}^n k^{m+2} \\
&= \sum_{k=1}^n \left( 1 + k + k^2 + \dots + k^{m+1} \right )\\
&= n + (n+1) \left( \text{ some junk term } \right )+\sum_{k=1}^n k^{m+1}\\
\end{align*}
which gives
$$(n+1)^{m+2} - (n+1) -(n+1) \left( \text{ some junk term } \right ) = \sum_{k=1}^n k^{m+1} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/507006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Prove, that $\sin x- a^3\cos x\leq \frac 1 3 \sqrt{1+a^6}$ Let a and $x$ be natural numbers with the property that $\sin x\leq a\cos x$.
Prove that $\sin x- a^3 \cos x\leq \frac 1 3 \sqrt{1+a^6}$.
Again, I'm looking for a second solution. I don't know how to use LaTex and my solution is hard to write here. It would be very nice to see a different solution from mine.
| Not sure if it is correct.
\begin{align*}
\sin x- a^3\cos x &= \sqrt{1 + (-a^3)^2} \left( \frac{1}{\sqrt{1 + a^6}} \sin(x) - \frac{a^3}{\sqrt{1 + a^6}} \cos(x)\right )\\
&= \sqrt{1+a^6} \left( \cos(\phi) \sin(x) - \sin(\phi) \cos(x)) \right )\\
&= \sqrt{1 +a^6} \sin \left( x - \phi \right )\\
\end{align*}
Where $\displaystyle \phi = \arctan(a^3)$. Given, constraint $\sin(x) \le a \cos(x) \implies x \le \arctan(a) \le \frac \pi 2 $
\begin{align*}
\sin(x-\arctan(a^3)) &\le \sin (\arctan(a) - \arctan(a^3)) \\
&= \sin \left(\arctan \left( \frac{a - a^3}{1 + a^4 } \right ) \right )\\
&= \frac{a - a^3}{ \sqrt{ (a -a^3)^2 + (1+a^4)}} \\
\end{align*}
From calculus, that thing seems to have maximum value of $\displaystyle \frac 1 3 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/507132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find $\lim_{x \to \pi/2} \frac{1 - \sin{x}}{(2x - \pi)^2}$ without using L'Hospital?
$$\lim_{x \to \pi/2} \frac{1 - \sin{x}}{(2x - \pi)^2}$$
I tried L'Hospital and it works like this in wolfram but if I don't use it what should I do?
| The best thing is to put $x = \frac{\pi}{2}+h$. Now observe that as $x \to \pi/2$, $h\to 0$. Also we get $2x-\pi=2h$. Then your quantity becomes
\begin{align*}
\lim_{x\to \pi/2} \frac{1-\sin(x)}{(2x-\pi)^{2}} &=\lim_{h \to 0} \frac{1-\sin\Bigl(\frac{\pi}{2}+h\Bigr)}{4h^{2}}\\ &=\lim_{h \to 0} \frac{1-\cos(h)}{4h^{2}} \\ &= \frac{1}{4} \cdot \lim_{h \to 0} \frac{2 \cdot \sin^{2}\frac{h}{2}}{h^2} \\ &= \frac{1}{4} \cdot \lim_{h \to 0} \frac{2 \cdot \sin^{2}\frac{h}{2}}{4 \cdot \displaystyle\Bigl(\frac{h}{2}\Bigr)^{2}}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/507577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
If $\tan(\pi \cos\theta) =\cot(\pi \sin\theta)$, then what is the value of $\cos(\theta -\frac{\pi}{4})$? Problem :
If $\tan(\pi \cos\theta) =\cot(\pi \sin\theta)$, then what is the
value of $\cos(\theta -\frac{\pi}{4})$?
My approach :
Solution: $\tan(\pi \cos\theta) =\cot(\pi \sin\theta)$
$\Rightarrow \tan(\pi \cos\theta) = \tan \{ \frac{\pi}{2} - (\pi \sin\theta) \} $
$\Rightarrow \pi \cos\theta = \frac{\pi}{2} - (\pi \sin\theta)$
$\Rightarrow \frac{1}{2} =\frac{1}{\sqrt{2}}[\sin\frac{\pi}{4} \cos\theta + \cos\frac{\pi}{4} \sin\theta] $
$\Rightarrow \frac{1}{\sqrt{2}} = \sin(\frac{\pi}{4} + \theta)$
$\Rightarrow \frac{\pi}{4} = \frac{\pi}{4} + \theta$
$\Rightarrow \theta = 0$
$\therefore \cos(\theta - \frac{\pi}{4})$ = $\frac{1}{\sqrt{2}}$ But this is wrong answer.. please suggest where I am wrong... thanks.
| It is given that,
$\tan(\pi \cos\theta) =\cot(\pi \sin\theta)\\
\frac {\sin(\pi \cos\theta)} {\cos(\pi \cos\theta)} = \frac {\cos(\pi \sin\theta)} {\sin(\pi \sin\theta)}\\
\sin(\pi \cos\theta)\sin(\pi \sin\theta) - \cos(\pi \sin\theta)\cos(\pi \cos\theta) = 0\\
\cos(\pi \cos\theta + \pi \sin\theta) = \cos(\frac \pi 2)$
Therefore,
$\pi \cos\theta + \pi \sin\theta = 2n\pi \pm \frac \pi 2 , [n \in Z]\\
\cos\theta + \sin\theta = 2n \pm \frac 1 2$
Since $\sin\theta + \cos\theta$ always lies between $\sqrt 2$ and $-\sqrt 2$, $n$ can only assume the value $0$. Thus,
$\cos\theta + \sin\theta = \pm \frac 1 2\\
\frac 1 {\sqrt 2} \cos\theta + \frac 1 {\sqrt 2} \sin\theta = \pm \frac 1 {2 \sqrt 2}\\
\cos\frac \pi 4 \cos\theta + \sin\frac \pi 4 \sin\theta = \pm \frac 1 {2 \sqrt2}\\
\cos(\theta - \frac \pi 4) = \pm \frac 1 {2 \sqrt 2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/509898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Induction with inequality problem Prove by induction that $2k(k+1) + 1 < 2^{k+1} - 1$ for $ k > 4$.
Can some one pls help me with this?
I reformulated like this
$ 2k(k+1) + 1 < 2^{k+1} - 1 $
$ 2k^2+2k+2<2^{k+1}$
and I tried like this
Take $k=k+1$
$ 2^{k+2} -1 > 2(k+1)(k+2) + 1 $
$2^{k+2} > 2(k+1)(k+2) + 2$
$ 2^{k+2} > 2k^2+2k+2 +4k+4$
I dont know how to proceed further
Please help me.
| You want to show $2k(k+1)+2<2^{k+1}$
For this, $k=5$ holds true (why?)
Further, $2^{k+2} = 2\cdot 2^{k+1} > 2\cdot(2k^2+2k+2) = 4k^2 + 4k + 4 = (2k^2 + 6k + 6)+2(k^2-k-1) $
Now can you recognise the first part in brackets and show that the second part is positive for $k> 4$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/510151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to find $\sqrt{1+{4\over x}+{4\over x^2} }$? If $$abx^2 = (a-b)^2(x+1)$$ then what is $$\sqrt{1+{4\over x}+{4\over x^2} }$$
(A) $a+b \over a-b$ (B)$a-b\over a+b$ (C) $a^2+ab$ (d) None
EDIT: What I've done is this:
$$abx^2=(a^2+b^2-2ab)(x+1)$$
=> $$abx^2 - (a^2+b^2+2ab)x - (a-b)^2=0$$
=> $$abx^2 - (a-b)^2x - (a-b)^2=0$$
but this is getting me nowhere.
| Ans. is $\frac{a+b}{a-b}$
As @Tattwamasi Amrutam said find the value of $x$ from
$$abx^2-(a-b)^2x-(a-b)^2=0$$then
$$x=\frac{(a-b)^2 \pm \sqrt{(a-b)^4+4ab(a-b)^2}}{2ab}$$
$$x=\frac{(a-b)^2\pm \sqrt{a^4+b^4-4a^3b-4ab^3+6a^2b^2+4a^3b+4ab^3-8a^2b^2}}{2ab}$$
(I got expansion of $(a-b)^4$ using pascal's triangle)
After the simplification,
$$x=\frac{2a^2-2ab}{2ab}$$
and there is another root that u can find.
For this root only i got
$$\sqrt{1+\frac{4}{x}+\frac{4}{x^2}}=\sqrt{\bigg(1+\frac{2}{x}\bigg)^2}$$
$$1+\frac{2}{x}=1+\bigg(\frac{4ab}{2a^2-2ab}\bigg)=\frac{a+b}{a-b}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/511189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Stuck on decomposing partial fraction I want to decompose the following, and I think got stuck in the thick of it $$ \dfrac{2x^3+3x+1}{(x+1)^2}$$
I tried like this:
OK, after advice from @Daniel Fischer and @lab bhattacharjee I decided to use division:
first separated the equation as $[\dfrac{2x^3+3x+1}{(x+1)}]\times \dfrac{1}{(x+1)} $
The I used polynomial division on $[\dfrac{2x^3+3x+1}{(x+1)}]$ to get $ [2x^2-2x+3- \dfrac{2}{x+1}] \times \dfrac{1}{(x+1)} $
with me, $ [2x^2-2x+3- \dfrac{2}{x+1}] \times \dfrac{1}{(x+1)}$ reduces to $\dfrac{2x^2-2x+3}{x-1}-2 $
Is this still on course, so far?
| As the highest power of numerator is greater that that of denominator
using Partial Fraction Decomposition,
$$ \dfrac{2x^3+3x+1}{(x+1)^2}=2x+A+\frac B{x+1}+\frac C{(x+1)^2}$$
On multiplication by $(x+1)^2$ we get $$2x^3+3x+1=(x+1)^2(2x+A)+B(x+1)+C$$
$$\implies 2x^3+3x+1=2x^3+x^2(4+A)+x(2+2A+B)+A+B+C$$
Now compare the coefficients of the different powers of $x$ to find $A,B,C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/514398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is the value of $f(0)+f(8)$? Suppose $f$ is a polynomial of degree $7$ which satisfies $f(1) =2$, $f(2)=5$, $f(3)=10$, $f(4)=17$, $f(5)=26$, $f(6)=37$ and $f(7)=50$. What is the value of $f(0)+f(8)$?
| The given values are the values of the quadratic polynomial $X^2+1$. Therefore, we have $f(X) - X^2 - 1$ a polynomial of degree $7$ with the seven zeros $1,2,3,4,5,6,7$, so
$$f(X) = c\cdot\prod_{k=1}^7 (X-k) + X^2+1.$$
with an unknown $c$ ($c\neq 0$ if the degree is exactly $7$). Then we have
$$f(8) = c\cdot 7! + 65$$
and
$$f(0) = c\cdot(-1)^7 7! + 1,$$
whence $$f(0) + f(8) = 65 + 1 + c(7! - 7!) = 66.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/514630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
How to solve the limit $\lim_{n \to \infty} \sqrt[8]{n^2+1} - \sqrt[4]{n+1}$? How do I show that this limit is zero?
$$\lim_{n \to \infty} \sqrt[8]{n^2+1} - \sqrt[4]{n+1} = 0$$
I've done the multiply by conjugates thing, which seems to lead nowhere:
$$\lim_{n \to \infty} (\sqrt[8]{n^2+1} - \sqrt[4]{n+1}) (\frac{\sqrt[8]{n^2+1} + \sqrt[4]{n+1}}{\sqrt[8]{n^2+1} + \sqrt[4]{n+1}})$$
$$=$$
$$\lim_{n \to \infty} \frac{\sqrt[4]{n^2+1} + \sqrt[2]{n+1}}{\sqrt[8]{n^2+1} + \sqrt[4]{n+1}}$$
I also considered using the squeeze theorem, as $\sqrt[8]{n^2+1} - \sqrt[4]{n+1} \leq \sqrt[8]{n^2+1} - \sqrt[4]{n}$, but I'm not sure how to bound it from below.
What's the correct approach to solving this limit?
| From identity
$$\sqrt[8]a-\sqrt[4]b=\frac{a-b^2}{(\sqrt[8]{a}+\sqrt[4]{b})(\sqrt[4]{a}+\sqrt{b})(\sqrt{a}+b)}$$
for $a=n^2+1$ and $b=n+1$ we have that
$$\lim_{n\to\infty}\sqrt[8]{n^2+1}-\sqrt[4]{n+1}=$$
$$=\lim_{n\to\infty}\frac{1}{\sqrt[8]{n^2+1}+\sqrt[4]{n+1}}\frac{1}{\sqrt[4]{n^2+1}+\sqrt{n+1}}\frac{n^2+1-{(n+1)^2}}{\sqrt{n^2+1}+n+1}=$$
$$=\lim_{n\to\infty}\frac{1}{\sqrt[8]{n^2+1}+\sqrt[4]{n+1}}\frac{1}{\sqrt[4]{n^2+1}+\sqrt{n+1}}\frac{-2}{\sqrt{1+1/n^2}+1+1/n}=0\cdot0\cdot(-2)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/515452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
How to find the general solution of $\tan \left(x+\frac{\pi }{3}\right)+3\tan \left(x-\frac{\pi }{6}\right)=0$ Find the general solution of the equation.
\begin{eqnarray} \tan \left(x+\frac{\pi }{3}\right)+3\tan \left(x-\frac{\pi }{6}\right)=0\\ \end{eqnarray}
The answers in my textbook are $n\pi $ and $n\pi +\frac{\pi }{3}$.
Previously, I compute the similar questions by using the following operations :
\begin{eqnarray}
\\\tan 3x&=&\cot 5x\\
\\\tan 3x&=&-\tan \left(\frac{\pi }{2}+5x\right)\\
\\\tan 3x&=&\tan \left(-\frac{\pi }{2}-5x\right)\\
\\3x&=&-\frac{\pi }{2}-5x\\
\\8x&=&n\pi -\frac{\pi }{2}\\
\\x&=&\frac{n\pi }{8}-\frac{\pi }{16}\\
\end{eqnarray}
But now there is a 3 in front of the second tan.
What should I do?
I do not know whether my method is correct. If you have any other methods, would you mind telling me the methods?
Thank you for your attention.
Update 1 : Found one of the answers, are my operations correct?
| The solution is given as follows.
I don't know how to use spoiler for a block of code.
\documentclass[preview,border=12pt]{standalone}
\usepackage{amsmath}
\begin{document}
\abovedisplayskip=0pt\relax
\begin{gather*}
\tan (x+\frac{\pi}{3}) + 3\tan (x-\frac{\pi}{6})=0\\
\tan (x+\frac{\pi}{3}) = 3\tan (\frac{\pi}{6}-x)\\
\frac{\tan x + \tan \frac{\pi}{3}}{1-\tan x \tan \frac{\pi}{3}} = 3\frac{\tan \frac{\pi}{6} -\tan x}{1+\tan \frac{\pi}{6}\tan x}\\
\frac{\tan x + \sqrt 3}{1-\sqrt 3\tan x } = 3\frac{\frac{\sqrt 3}{3} -\tan x}{1+\frac{\sqrt 3\tan x}{3}}\\
\frac{\tan x + \sqrt 3}{1-\sqrt 3\tan x } = \frac{ \sqrt 3 -3 \tan x}{1+\frac{\sqrt 3\tan x}{3}}\\
\frac{\tan x + \sqrt 3}{1-\sqrt 3\tan x } = \frac{ \sqrt 3 -3 \tan x}{1+\frac{\sqrt 3\tan x}{3}}\times\frac{3}{3}\\
\frac{\tan x + \sqrt 3}{1-\sqrt 3\tan x } = \frac{ 3\sqrt 3 -9 \tan x}{3+ \sqrt 3\tan x}\\
(1- \sqrt 3\tan x) (3\sqrt 3 -9 \tan x) = (\tan x + \sqrt 3)(3+ \sqrt 3\tan x)\\
3\sqrt 3 -9\tan x -9\tan x +9\sqrt 3\tan^2x = 3\tan x +\sqrt 3\tan^2 x +3\sqrt 3 +3\tan x\\
8\sqrt 3\tan^2 x -24 \tan x=0\\
\sqrt 3\tan^2 x -3\tan x=0\\
\tan x (\sqrt 3 \tan x -3)=0\\
\tan x =0 \text{ or } \sqrt 3 \tan x -3 =0\\
\tan x=0 \text{ or } \tan x =\sqrt 3\\
x=n\pi \text{ or } x=\frac{\pi}{3}+n\pi
\end{gather*}
\end{document}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/516507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to find $\lfloor 1/\sqrt{1}+1/\sqrt{2}+\dots+1/\sqrt{100}\rfloor $ without a calculator? $$
\left\lfloor\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} +\dots+ \frac{1}{\sqrt{100}}\right\rfloor =\,?
$$
I rationalized the denominator and then I think I should somehow group the numbers, but i don't know how.
Thanks in advance!
| Doing it in 9th grade math is quite a challenge. But perhaps this would come close.
For any positive number $t$, we have
$$ \dfrac{1}{\sqrt{t}} > 2 \sqrt{t+1} - 2 \sqrt{t} > \dfrac{1}{\sqrt{t+1}}$$
To see the first inequality, note that
$$ \left(\dfrac{1}{\sqrt{t}} + 2 \sqrt{t}\right)^2 = \dfrac{1}{t} + 4 + 4 t > 4 + 4 t = (2 \sqrt{t+1})^2$$
Similarly for the second, by looking at $\left(2 \sqrt{t+1} - \dfrac{1}{\sqrt{t+1}}\right)^2$.
So $$\eqalign{\dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \ldots + \dfrac{1}{\sqrt{100}}
&> (2 \sqrt{2} - 2 \sqrt{1}) + (2 \sqrt{3} - 2 \sqrt{2}) + \ldots + (2 \sqrt{101} - 2 \sqrt{100})\cr &= 2 \sqrt{101} - 2 > 2 \sqrt{100} - 2 = 18\cr}$$
while
$$\eqalign{\dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \ldots + \dfrac{1}{\sqrt{100}}
&< 1 + (2 \sqrt{2} - 2 \sqrt{1}) + \ldots + (2 \sqrt{100} - 2 \sqrt{99})\cr
& = 1 + 20 - 2 = 19\cr}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/516846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Find points on curve $y = (x^2 - 1)^3 / (x^2 + 1)^3$ when slope of tangent line is zero? As soon as a get to the quotient rule I can't simplify to solve for $x$.
This is what I have tried:
Since the derivative much be equal to zero:
$$0 = [6x(x^2 -1)^2 (x^2+1)^3] - [6x(x^2 + 1)^2(x^2-1)^3]$$
How do I simplify further and solve for $x$?
| Note that there is a common term $6x$, of $(x^2 - 1)^2$ and a common term of $(x^2 + 1)^2$, so your equation can be rewritten as
$$0 = 6x (x^2 - 1)^2 (x^2 + 1)^2 \Big((x^2 + 1) - (x^2 - 1)\Big) = 12x (x^2 - 1)^2(x^2 + 1)^2$$
Now set each term to $0$ individually.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/517268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find remainder when $F(x)$ is divided by $P(x)$ Find the remainder when $F(x)=x^{276}+12$ is divided by $P(x)=x^2+x+1$?
| You have $F(x) = (x^2+x+1) q(x) + r(x) = \frac{x^3-1}{x-1} q(x) + r(x)$, with $\partial r <2$. Let $x$ be a cube root of $1$ different from $1$, then $F(x) = r(x)$, and since $276 = 0 \mod 3$, we have $x^{276} = 1$, and so $F(x) = 13$ at both 'non one' roots.
Hence $r(x) = 13$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/518616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Interesting trigonometric equation Find all $x\in \mathbb{R}$ such that
$\arccos x = 2\arctan \sqrt{\frac{1-x}{1+x}}$
Now, here is my approach, please state anything that is not correct/fully motivated.
$\cos 2\theta =2\cos ^2\theta -1$
$\cos ^2\theta =\frac{\cos ^2\theta}{1}=\frac{\cos ^2\theta}{\cos ^2\theta+\sin^2\theta}=\frac{1}{1+\tan ^2\theta }$
$\Rightarrow \cos 2\theta = \frac{2}{1+\tan ^2\theta}-1$
$\cos \arccos x = x = \cos(2\arctan \sqrt{\frac{1-x}{1+x}} ) = \frac{2}{1+\frac{1-x}{1+x}}-1=1+x-1=x. $
Hence, we have shown that $\cos LHS=\cos RHS$ and thus the solutions are given by the intersections of the domains of the functions.
$f(x)=\arccos x \ \ \ \ \ D_{f} = \left \{ x\in \mathbb{R}: -1\leq x\leq 1 \right \}$
$g(x)=2\arctan \sqrt{\frac{1-x}{1+x}} \ \ \ \ \ D_{f} = \left \{ x\in \mathbb{R}: x\neq 1 \wedge x\leq 1 \wedge x> -1 \right \}$
Solutions are thus given by:
$D_{f}\cap D_{g}=\left \{ -1< x< 1 \right \}$
Follow up question: Is it enough to show that $\cos LHS = \cos RHS$ and proceed as I've done?
| Hint
By calculating the derivative
$$\left(2\arctan \sqrt{\frac{1-x}{1+x}}\right)'=-\frac{1}{\sqrt{1-x^2}}=(\arccos x)'$$
and verifying that the two expressions have the same value at $0$ (for example) we conclude that the equality hold for all $x\in(-1,1]$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/520712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Show that the even and odd terms of the sequence described by $x_1=\frac{1}{3}, x_{n+1}=\frac{{x_n}^2+1}{5x_n}$ are monotonic and bounded. Let $(x_n)_n$ be a sequence of real numbers such that $$x_1=\dfrac{1}{3}, x_{n+1}=\dfrac{{x_n}^2+1}{5x_n}.$$ Show that the even terms are monotone decreasing and bounded and the odd terms are monotone increasing and bounded.
$\textbf{My Attempt:}$
I first showed that $x_1<x_3$, then assumed for some positive integer $k$ that $x_{2k-1}<x_{2k+1}$. From there I attempted show that $0<x_{2k+1}-x_{2k-1}$ implies $0<x_{2k+3}-x_{2k+1}$, by substituting in $$x_{2k+3}=\dfrac{\left(\dfrac{(x_{2k+1})^2+1}{5x_{2k+1}}\right)^2+1}{5\left(\dfrac{(x_{2k+1})^2+1}{5x_{2k+1}}\right)},$$ and $$x_{2k+1}=\dfrac{\left(\dfrac{(x_{2k-1})^2+1}{5x_{2k-1}}\right)^2+1}{5\left(\dfrac{(x_{2k-1})^2+1}{5x_{2k-1}}\right)}.$$
However, the algebra became very messy, so I think there may be a better approach that I've missed. I know the sequences of even and odd terms should both be bounded by $\frac 1 2$. I again tried to show this using induction, but ran into trouble when trying to show that $x_{2k+1}\leq \frac 1 2 \implies x_{2k+3}\leq \frac 1 2$.
Any suggestions/hints would be greatly appreciated. Thanks!
| $$x_{n+1} = \frac{1}{5} ( \frac{x_n}{1} + \frac{1}{x_n})$$
Let $C = 2/5,\ D=\frac{1}{5}(C+\frac{1}{C})$
So $x_1=1/3 < D$, $C<x_2 =(3+1/3)/5=2/3<1$, $x_1 < x_3 =13/30<D$ so that $$
C<x_4 <x_2,
$$
which also implies $$ D>x_5> x_3. $$
(1) With this observation, we use induction on odd term $$D>x_{2n+3} > x_{2n+1}$$
$k=3$-case is proved. If $k=2n-1$ is satisfied, that is, $x_{2n-1}<x_{2n+1}<D$, so $$(A)\ x_{2n}=\frac{1}{5} (\frac{1}{x_{2n-1}} + x_{2n-1}) > x_{2n+2} = \frac{1}{5} (\frac{1}{x_{2n+1}} + x_{2n+1}) >C
$$ So continually we have $$x_{2n+1}=\frac{1}{5} (\frac{1}{x_{2n}} + x_{2n}) < x_{2n+3} = \frac{1}{5} (\frac{1}{x_{2n+2}} + x_{2n+2}) < D.$$
(2) So we can conclude from odd term's increasing and bound
that even term is decreasing and bounded (See (A)).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/522269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Proving by absurd that $d \nmid 4,5,10, 20$ What I'm trying to solve is the following:
Given that $(a:b) = 2$, prove that $(a^2 + 2b^2+10:20) = 2$.
So, basically, I think that what I need to do is to show that if $d = a^2 + 2b^2 + 10$, then $2\mid d$ and $d \nmid 4, d \nmid 5, d \nmid 10$ and $d \nmid 20$.
So, starting with 2 and 4 is very straightforward: knowing that $(a:b) = 2$, then we can write $a = 2 \cdot k$ and $b = 2 \cdot q$. Following this:
$2 \mid a^2, 2 \mid 2b^2, 2 \mid 10 \Rightarrow 2 \mid d$.
In the other hand, I want to prove that $4 \nmid d$. As we know:
$2 \mid a \Rightarrow 2^2 \mid a^2$, $2 \mid b \Rightarrow 2^2 \mid b^2 \Rightarrow 4 \mid 2b^2$, but $4 \nmid 10$, so it's clear that $4 \nmid d$.
Then, the following steps are proving that $5 \nmid d$ and $10 \nmid d$. But, I'm not quite sure if my statements are correct. I've say the following:
If $a^2 \equiv 0 \pmod{4} \Rightarrow a^2 \equiv 1 \pmod{5}$. And the same with $b^2$, so that it follows that $2b^2 \equiv 2 \pmod{5}$.
But.. are those statements necessarily correct? Or how should I attack this problem?
| Your argument is fine for $4\nmid a^2+2b^2+10$. It is enough to prove that $5\nmid a^2+2b^2+10$. Suppose that $5\mid a^2+2b^2+10$. First observe that $5\nmid a$. Because if $5\mid a$ then $5\mid 2b^2+10$ and necessarily $5\mid b$ and hence $5\mid (a,b)=2$ which is not true.
For $a$ such that $5\nmid a$, we can see the following:
$$
a^2\equiv -1\quad or\quad 1 \mod 5
$$
The same is true for $b^2$ and hence we have:
$$
a^2+2b^2\equiv 3\quad or \quad -1\quad or\quad 1 \quad -3 \mod 5\implies 5\nmid a^2+2b^2.
$$
Therefore $ 5\nmid a^2+2b^2+10$ which proves your result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/523085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Why does $(3\sqrt3)^2 = 27$? How does $(3\sqrt3)^2 = 27$?
I've tried to solve this using binomial expansion and using the FOIL method from which I obtain $9 + 3\sqrt3 +3\sqrt3 + 3$.
it has been a while since I've done this kind of thing so it may be something obvious that I can't see.
| If you really insist on expanding, just use the fact that $3\sqrt{3} = \sqrt{3} + \sqrt{3} + \sqrt{3}$, and therefore expand away $(\sqrt{3} + \sqrt{3} + \sqrt{3})^2$ using the trinomial expansion formula $(a+b+c)^2 = a^2 + b^2 +c^2 +2ab+2bc+2ac$.
Later edit:
And if you really insist on using FOIL, then use a little imagination to write $3\sqrt{3} = 2\sqrt{3} + \sqrt{3}$, thus your desired expression becoming $(2\sqrt{3} + \sqrt{3})(2\sqrt{3} + \sqrt{3})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/526289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 6
} |
Find the residue of $ \bigl(\frac{z}{2z+1}\bigr)^3$ Find the residue of : $\displaystyle \left(\frac{z}{2z+1}\right)^3$
In this case we have $-\frac{1}{2}$ as a pole. Now I want to use the following formula:
$\displaystyle \frac{\phi^{k-1}(z_{0})}{(k-1)!}$
Now would I let $k = 3$ in this case then?
After that I am a little unsure what to do. The answer is $\displaystyle-\frac{3}{16}$ .
| Here is a way to not use that formula. Note that $$\dfrac{z}{2z+1}=\dfrac{1}{2}\cdot\frac{z}{z+\frac12}=\frac{1}{2}\left(1-\frac{1}{2}\cdot\frac{1}{z+\frac{1}{2}}\right)=\frac12-\frac14\cdot\frac{1}{z+\frac12}.$$
You can use $(a+b)^3=a^3+3a^2b+3ab^2+b^3$ with $a=\dfrac12$ and $b=-\dfrac14\cdot\dfrac{1}{z+\frac12}$ to see that $\left(\dfrac{z}{2z+1}\right)^3 = -\dfrac{3}{16}\cdot\dfrac{1}{z+\frac12} + $ other stuff.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/527869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Showing "$30$ divides $n^5-n$ for all $n\in\Bbb N$" using induction
Prove that $(n^5 - n)$ divides by $30$ for every $ n\in N$, using induction only.
How on earth do I do that? Thing is $(n^5 - n)$ can't be opened using any known formula...
| Without using induction
$$n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=n(n^2-1)(n^2-4+5)$$
$$=\underbrace{n(n+1)(n-1)(n+2)(n-2)}_{\text{product of }5\text{ consecutive integers }}+5\underbrace{n(n+1)(n-1)}_{\text{product of }3\text{ consecutive integers }}$$
Now from this or this, the product of $n$ consecutive integers is divisible by $n!$
Alternatively, using Fermat's Little theorem,
$n^5-n$ is divisible by $5$
$n^3-n$ is divisible by $3$
$n^2-n$ is divisible by $2$
Now, $n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=(n^2+1)(n^3-n)$
and
$n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=n(n-1)(n+1)(n^2+1)=(n^2-n)(n+1)(n^2+1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/529757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Forced nonlinear oscillator - analytical methods This is an example from Kovacic & Brennan (2011). Consider the following equation of motion for a forced, non-linear oscillator (Duffing's equation):
$$
\ddot x + 2 \zeta \dot x + \alpha x + \gamma x^3 = F \cos \Omega t,
$$
where $\zeta, \alpha, \gamma, F$ and $\Omega$ are parameters. We assume "the steady-state harmonic solution" as
$$
x = Y\cos{(\Omega t - \theta)}.
$$
My goal is to obtain the following equation for $Y$:
$$
Y^2 = \frac{F^2}{4\zeta^2\Omega^2 + \left( \Omega^2 - \alpha -
%
\frac{3}{4}\gamma Y^2 \right)^2}.
$$
This is done by 1) substituting the second equation into the first, and 2) equating the coefficients of $\sin\Omega t$ and $\cos\Omega t$ from both sides.
The substitution part seems straightforward, but I can't quite wrap my head around the last step. Especially the "$\cos(\Omega t-\theta)$" on the right hand side is causing headaches.
Suggestions are welcome!
| This problem is simple differentiation and substitution...except that the cubic term makes things a little hairy. First thing is to expand the cosine term as
$$\cos{(\Omega t-\theta)} = \cos{\Omega t} \cos{\theta} + \sin{\Omega t} \sin{\theta}$$
Now differentiate and plug in. The linear terms involving the derivative are simple and I'll just list them here:
$$\ddot{x} + 2 \zeta \dot{x} + \alpha x = Y \left [ \left ( -\Omega^2 \cos{\theta} - 2 \zeta \Omega \sin{\theta} + \alpha \cos{\theta} \right ) \cos{\Omega t} \\+ \left (-\Omega^2 \sin{\theta} + 2 \zeta \Omega \cos{\theta} + \alpha \sin{\theta} \right ) \sin{\Omega t} \right ] $$
The cubic piece is expanded as usual:
$$\gamma x^3 = \gamma Y^3 (\cos^3{\Omega t} \cos^3{\theta} + 3 \cos^2{\Omega t} \sin{\Omega t} \cos^2{\theta} \sin{\theta} + 3 \sin^2{\Omega t} \cos{\Omega t} \sin^2{\theta} \cos{\theta}+ \sin^3{\Omega t} \sin^3{\theta}$$
Now we do not want powers of the sines and cosines, but rather linear terms. Use the fact that
$$\cos{3 y} = 4 \cos^3{y} - 3 \cos{y}$$
$$\sin{3 y} = 3 \sin{y} - 4 \sin^3{y}$$
and, doing some rearranging, get that
$$\gamma x^3 = \gamma Y^3 \left [\frac{3}{4} \cos{\theta} \cos{\Omega t} + \frac{3}{4} \sin{\theta} \sin{\Omega t} \\+ \frac{1}{4} \cos{\theta}\left (\cos^2{\theta} - 3 \sin^2{\theta} \right ) \cos{3 \Omega t} + \frac{1}{4} \sin{\theta}\left (3 \cos^2{\theta}-\sin^2{\theta} \right ) \sin{3 \Omega t} \right ]$$
Note that the $3 \Omega$ terms represent the frequency multiplication induced by nonlinear terms. We will ignore these terms because...(e.g., $\gamma$ is small, etc. etc.). What we are left with is
$$Y \left [ \left ( \left (\alpha-\Omega^2 + \frac{3}{4} \gamma Y^2\right ) \cos{\theta} - 2 \zeta \Omega \sin{\theta} \right ) \cos{\Omega t} \\+ \left ( \left (\alpha-\Omega^2 + \frac{3}{4} \gamma Y^2\right ) \sin{\theta} + 2 \zeta \Omega \cos{\theta} \right ) \sin{\Omega t} \right ] = F \cos{\Omega t}$$
Now we simply eliminate $\theta$ by setting the coefficient of $\sin{\Omega t}$ to zero:
$$\left (\alpha-\Omega^2 + \frac{3}{4} \gamma Y^2\right ) \sin{\theta} + 2 \zeta \Omega \cos{\theta} =0$$
or
$$\tan{\theta} = -\frac{ 2 \zeta \Omega}{\alpha-\Omega^2 + \frac{3}{4} \gamma Y^2}$$
which means that, to within a sign,
$$\cos{\theta} = \frac{\alpha-\Omega^2 + \frac{3}{4} \gamma Y^2}{\sqrt{4 \zeta^2 \Omega^2 + \left (\alpha-\Omega^2 + \frac{3}{4} \gamma Y^2 \right )^2}} $$
$$\sin{\theta} = -\frac{2 \zeta \Omega}{\sqrt{4 \zeta^2 \Omega^2 + \left (\alpha-\Omega^2 + \frac{3}{4} \gamma Y^2 \right )^2}} $$
and therefore, plugging these values into the equation for the coefficient of $\cos{\Omega t}$, we get that, within a sign:
$$Y = \frac{F}{\sqrt{4 \zeta^2 \Omega^2 + \left (\Omega^2 -\alpha - \frac{3}{4} \gamma Y^2 \right )^2}} $$
The posted equation for $Y$ follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/529857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Are there any integers $a,b$ s.t. ${ a }^{ 2 }-{ b }^{ 2 }=8$? For $a$ and $b$ are integers greater than $1$, ${ a }^{ 2 }-{ b }^{ 2 }=8$ holds?
| Hint: We have $8= a^2 - b^2 = (a+b)(a-b)$. As $a,b \ge 1$, we have $a+b > a-b$, that is either $a+b=8$ and $a-b = 1$, or $a+b=4$ and $a-b=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/530775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Why do the conditions $x_1+x_2=b$ and $x_1\cdot x_2=ac$ hold for any quadractic equation? Consider the equation $$ax^2+bx+c=0.$$ The factorization of the left hand side is of the form $(x+x_1)(x+x_2)$, where the solutions $x_1$ and $x_2$ must satisfy $$(1)\quad x_1+x_2=-\tfrac bc\quad\mbox{and}\quad(2)\quad x_1\cdot x_2=\tfrac ca.$$
I solve the equation by putting $$x_1=\tfrac12 b+k\quad\mbox{and}\quad x_2=\tfrac12 b-k.$$ This means (1) holds and by substitution into $(2)$, I get
$$(\tfrac12 b+k)(\tfrac12 b-k)=-ac\iff k^2=ac+\tfrac14 b^2.$$ This yields a solution for $k$, and then I can find $x_1$ and $x_2$.
I have been using this method with success and I like it a lot, since I always forget the quadratic formula. but I have no clue what I am doing. More specifically, why do the conditions (1) and (2) hold for any quadratic equation?
| First of all there are lot of mistakes in your notation the quadratic equation should be factorized as:
$$ax^2 + bx + c = a(x-x_1)(x-x_2)$$
And we want to prove that the following holds:
$$x_1 + x_2 = -\frac ba \tag{1}$$
and
$$x_1x_2 = \frac ca \tag{2}$$
From the formula for quadratic equations we have:
$$x_{1/2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
So WLOG let's have:
$$x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \quad \quad \text {and} \quad \quad x_1 = \frac{-b - \sqrt{b^2 - 4ac}}{2a}$$
For $(1)$ just add this together:
$$x_1 + x_2 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} + \frac{-b - \sqrt{b^2 - 4ac}}{2a} = - \frac{2b}{2a} - \frac ba \tag{1}$$
For $(2)$ just multiply them together:
$$x_1x_2 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \times \frac{-b - \sqrt{b^2 - 4ac}}{2a}$$
$$x_1x_2 = \frac{b^2 - (\sqrt{b^2 - 4ac})^2}{4a^2}$$
$$x_1x_2 = \frac{b^2 - b^2 - 4ac}{4a^2} = \frac{c}{a} \tag{2}$$
This relations are know as Vieta's Formulas and back at school, they taught us that this is how Vieta find out this relations.
Actually the Vieta's Formulas don't just hold for polynomials of second degree, they hold for polynomials of any degree. Consider the following polynomial:
$$P(x) = a_nx^n + a_{n-1}x^{n-1} +...a_1x + a_0x^0$$
The Vieta stated that the the $(n − k)^{th}$ coefficient $a_{n−k}$ is related to a signed sum of all possible subproducts of roots, taken $k-at-a-time$. Or using mathematical notation written as:
$$\sum_{1\le i_1\le i_2 \le i_3 ... \le i_k \le n} x_{i1}x_{i2}x_{i3}...x_{ik} = (-1)^k \frac{a_{n-k}}{a_n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/531315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
finding the inverse of function. the domain is given and the question requires to find the inverse. Find the inverse of the function.
$$f(x)=x^2 + 4x$$
Domain: $x \geq -2$
I have done:
*
*$X = y^2 + 4y$
*$X - 4y = y^2$
*$\sqrt{X - 4y} = y$
My answer: $f^{-1}(x) = \sqrt{X - 4x} $
Real answer: $f^{-1}(x) = -2 \pm \sqrt{x + 4}$
| $y=x^2+4x \Rightarrow x^2+4x-y=0 \Rightarrow x=\frac{-4\pm\sqrt{16+4y}}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/532083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding derivative by using L'Hospital's rule $$ f(x)= \begin{cases} \frac{1}{x\ln 2} - \frac{1}{2^x -1} \text { , if } x \neq 0 \\ \\ \frac{1}{2} \text{ , if }x=0 \end{cases}$$
I'm supposed to find derivative at point 0. I thought I should start by finding limits of both directions, showing that they equal by applying L'Hospital as many times as needed.
| Apply the approximation
$$
2^{x}-1\sim x\ln2+x^{2}\frac{\left(\ln2\right)^{2}}{2}+x^{3}\frac{\left(\ln2\right)^{3}}{6}+\cdots
$$
then
\begin{eqnarray*}
& & \lim_{x\rightarrow0}\frac{f\left(x\right)-f\left(0\right)}{x}\\
& = & \lim_{x\rightarrow0}\frac{\frac{1}{x\ln2}-\frac{1}{2^{x}-1}-\frac{1}{2}}{x}\\
& = & \lim_{x\rightarrow0}\frac{2\left(2^{x}-1\right)-2\ln2\cdot x-\ln2\cdot x\left(2^{x}-1\right)}{2\ln2\cdot x^{2}\left(2^{x}-1\right)}\\
& = & \lim_{x\rightarrow0}\frac{2\left(x\ln2+x^{2}\frac{\left(\ln2\right)^{2}}{2}+x^{3}\frac{\left(\ln2\right)^{3}}{6}\right)-2\ln2\cdot x-\ln2\cdot x\left(x\ln2+x^{2}\frac{\left(\ln2\right)^{2}}{2}\right)}{2\ln2\cdot x^{2}\left(x\ln2\right)}\\
& = & \lim_{x\rightarrow0}\frac{\frac{1}{3}\left(\ln2\right)^{3}x^{3}-\frac{1}{2}\left(\ln2\right)^{3}x^{3}}{2\left(\ln2\right)^{2}x^{3}}\\
& = & -\frac{1}{12}\ln2
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/532311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
The roots of $z^4+z^3+1=0$ are in $\frac{3}{4}<|z|< \frac{3}{2}$. How can we show that all the roots of the complex polynomial
$$p(z)=z^4+z^3+1$$
lie in $\frac{3}{4}<|z|< \frac{3}{2}$? This is an exercise in complex analysis.
| Use Rouché's theorem.
Note that for $|z|=\frac{3}{2}$, we have $|(z^4+z^3+1)-z^4| = |z^3+1| \le 1+ |z|^3 = 1+ (\frac{3}{2})^3 < (\frac{3}{2})^4 = |z|^4 $, hence $z \mapsto z^4+z^3+1$ and $z \mapsto z^4$ have the same number of zeros in $B(0,\frac{3}{2})$ (that is, all four zeros)
Let $\epsilon>0$.
Then for $|z| = \frac{3}{4}+\epsilon$, we have $|(z^4+z^3+1)-1| = |z^4+z^3| \le |z|^4+|z|^3 =
(\frac{3}{4}+\epsilon)^3+ (\frac{3}{4}+\epsilon)^4$. Since $f(\epsilon ) = (\frac{3}{4}+\epsilon)^3+ (\frac{3}{4}+\epsilon)^4$ is continuous, and $f(0) <1$, it follows that for some $\epsilon>0$, we have $f(\epsilon)<1$.
Hence for $|z| = \frac{3}{4}+\epsilon$ we have $|(z^4+z^3+1)-1| < 1 = |1|$, hence $z \mapsto z^4+z^3+1$ and $z \mapsto 1$ have the same number of zeros in $B(0,\frac{3}{4}+\epsilon)$ (that is, there are no zeros in $B(0,\frac{3}{4}+\epsilon)$).
It follows that all zeros are in $B(0,\frac{3}{2}) \setminus \overline{B}(0,\frac{3}{4})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/536212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the sum of cubes Evaluate $1^3 + 2^3 + 3^3 + . . . + n^3.$
Can I get a hint? I'm really stuck and don't know how to break this problem down.
| We'll prove the following using induction:
$$1^3 + 2^3 + 3^3 + 4^3 + 5^3+...+n^3 = [S(n)]^2$$
Where $S(n)$ is the sum of the consecutive numbers from $1$ to $n$. We know that $S(n) = \frac{n(n+1)}{2}$. So we have:
$$1^3 + 2^3 + 3^3 + 4^3 + 5^3+...+n^3 = \left(\frac{n(n+1)}{2}\right)^2$$
Now let's prove it:
Basis $n=1$
$$1^3 = \left(\frac{1(1+1)}{2}\right)^2$$
$$1 = 1$$
Which is true:
Hypothesis $n=k$
Assume that the follwoing holds for some number $k$:
$$1^3 + 2^3 + 3^3 + 4^3 + 5^3+...+k^3 = \left(\frac{k(k+1)}{2}\right)^2$$
Unductive step $n=k+1$
We'll prove that it holds for every number $k+1$ so we have:
$$\left(\frac{(k+1)(k+2)}{2}\right)^2 = \left(\frac{(k+1)k+2(k+1)}{2}\right)^2 = \left(\frac{k(k+1)}{2}\right)^2 + \frac{4k(k+1)(k+1)}{4} + \left(\frac{2(k+1)}{2}\right)^2$$
Substitute for the first term from the hypothesis and expand the other two. and we have:
$$\left(\frac{(k+1)(k+2)}{2}\right)^2 = 1^3 + 2^3 + 3^3 + 4^3 + 5^3+...+k^3 + k(k+1)(k+1) + (k+1)^2$$
$$\left(\frac{(k+1)(k+2)}{2}\right)^2 = 1^3 + 2^3 + 3^3 + 4^3 + 5^3+...+k^3 + (k+1)^2(k + 1) = 1^3 + 2^3 + 3^3 + 4^3 + 5^3+...+k^3 + (k+1)^3$$
Q.E.D.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/538425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
find area of the region $x=a\cos^3\theta$ $y=a\sin^3\theta$ Find the area of the region enclosed by $x=a\cos^3\theta$ and $y=a\sin^3\theta$
What steps should I take in order to find the area?
| To get the area:
$$
\begin{align}
A
&=\int_{}^{}x\,\mathrm{d}y\\
&=\int_0^{2\pi}a\cos^3(\theta)\,3a\sin^2(\theta)\cos(\theta)\,\mathrm{d}\theta\\
&=3a^2\int_0^{2\pi}\sin^2(\theta)\cos^4(\theta)\,\mathrm{d}\theta\tag{1}
\end{align}
$$
We can compute $A$ without using any trigonometric integrals.
Substituting $\theta\mapsto\theta+\pi/2$ in $(1)$, we get
$$
A=3a^2\int_0^{2\pi}\cos^2(\theta)\sin^4(\theta)\,\mathrm{d}\theta\tag{2}
$$
Adding $(1)$ and $(2)$, remembering that $\sin^2(\theta)+\cos^2(\theta)=1$, yields
$$
2A=3a^2\int_0^{2\pi}\cos^2(\theta)\sin^2(\theta)\,\mathrm{d}\theta\tag{3}
$$
Multiplying $(3)$ by $4$ gives
$$
8A=3a^2\int_0^{2\pi}\sin^2(2\theta)\,\mathrm{d}\theta\tag{4}
$$
Substituting $\theta\mapsto\theta+\pi/4$ in $(4)$, we get
$$
8A=3a^2\int_0^{2\pi}\cos^2(2\theta)\,\mathrm{d}\theta\tag{5}
$$
Adding $(4)$ and $(5)$ yields
$$
\begin{align}
16A
&=3a^2\int_0^{2\pi}1\,\mathrm{d}\theta\\[4pt]
&=6\pi a^2\\[8pt]
A&=\frac{3\pi a^2}{8}\tag{6}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/538554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Problem based on Algebraic identities Algebraic identities:
$$(a+b)^2 = a^2 + b^2 + 2ab$$
$$(a-b)^2 = a^2 + b^2 - 2ab$$
Other identities can also come to solve this question?
If $x + 1/x = 5$ and $x^2 + 1/x^3 = 8$, then what would be the value of $x^3 + 1/x^2$?
Possible answers:
a- 215
b- 125
c- 256
d- 525
| see ($x^2$+$x^-3$)($x^3$+$x^-2$)
*
*=$x^5$+$x^-5$+2
*NOW x+$\frac{1}{x}$=5. FROM THIS WE GET $x^2$+$x^-2$=23 and $x^3$+$x^-3$=110
*NOW $x^5$+$x^-5$=($x^2$+$x^-2$)($x^3$+$x^-3$)$-$(x+$x^-1$)=23.110$-$5=2525
*SO($x^2$+$x^-3$)($x^3+$$x^-2$)=2525+2=2527
*so $x^3$+$x^-2$=$\frac{2527}{8}${as $x^2$+$x^-3$=8}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/539012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Prove $\sqrt{2} + \sqrt{5}$ is irrational How do you prove that $\sqrt{2} + \sqrt{5}$ is irrational?
I tried to prove it by contradiction and got this equation: $a^2/b^2 = \sqrt{40}$.
| If $\sqrt{2}+\sqrt{5}\in\mathbb{Q}$, i.e., $ \sqrt{2}+\sqrt{5}=\frac{b}{a}$ for $a,b\in\mathbb{Z} \mbox{ and } a\neq 0$ then $ 2-5=\frac{b}{a}(\sqrt{2}-\sqrt{5})$ and
$$
\sqrt{2}-\sqrt{5} =\frac{-3\cdot a}{b}
$$
Here $b$ is necessarily greater than zero by cause $\sqrt{2}+\sqrt{5}>0$.Finally we get,
$$
\sqrt{2}=\frac{1}{2}\left( (\sqrt{2}+\sqrt{5})+(\sqrt{2}-\sqrt{5}) \right)= \frac{1}{2}\left( \frac{b}{a}+\frac{-3\cdot a}{b} \right)\in\mathbb{Q}.
$$
But this is a contradiction since $\sqrt{2}$ is not rational.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/539266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Finit Sum $\sum\limits_{i=1}^{100}i^8-2 i^2$ Can anyone help me?
How can I find
$$\sum_{i=1}^{100}i^8-2i^2 $$
| Hint: Expand following formula for $k=2,5,8$
$$\sum_{i=1}^n i^k=\frac{(n+1+B)^{k+1}-B^{k+1}}{k+1}$$
$$B^0=1,B^1=\frac{-1}{2},B^2=\frac{1}{6},B^3=0,B^4=\frac{-1}{30},B^5=0,B^6=\frac{1}{42},B^7=0,B^8=\frac{-1}{30},B^9=0,....$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/539474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
How to prove $\cos ^6x+3\cos ^2x\space \sin ^2x+\sin ^6x=1$ Prove the following equation.
\begin{eqnarray}
\\\cos ^6x+3\cos ^2x\space \sin ^2x+\sin ^6x=1\\
\end{eqnarray}
I can't prove it by many methods I use.
Please give me some hints.
Thank you for your attention
| It seems one sign in your equality is wrong. One actually has
\begin{align}
1=\left(\cos^2x+\sin^2x\right)^3&=\cos^6x+3\cos^4x\sin^2x+3\cos^2x\sin^4x+\sin^6x=\\
&=\cos^6x+3\sin^2x\cos^2x\underbrace{\left(\cos^2x+\sin^2x\right)}_{=1}+\sin^6x=\\
&=\cos^6x+3\sin^2x\cos^2x+\sin^6x.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/539953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Error in finding sum of $1\cdot 2+3\cdot 4+ \cdots \text{to}\space n\space \text{terms}$ To find sum of the series $1\cdot 2+3\cdot 4+ \cdots \text{to}\space n\space \text{terms}$
My approach,
Let S=$1^2+2^2+3^2 + \cdots +n^2$
If $n$ is even
S=$(1-2)^2+(3-4)^2+ \cdots +[(n-1)-n]^2+2(1 \cdot 2+3 \cdot 4+5 \cdot 6+ \cdots \text{to}\space \dfrac{n}{2}\space \text{terms})$
=$\frac{n}2+2(1 \cdot 2+3 \cdot 4+5 \cdot 6+ \cdots \text{to}\space \dfrac{n}{2}\space \text{terms})$
And we know,
$$S=\frac{n(n+1)(2n+1)}6$$
Therefore,$$2(1 \cdot 2+3 \cdot 4+5 \cdot 6+ \cdots \text{to}\space \frac{n}2\space \text{terms}) =\frac{n(n+1)(2n+1)}6-\frac{n}2$$
or $$2(1 \cdot 2+3 \cdot 4+5 \cdot 6+ \cdots \text{to}\space \frac{n}2\space \text{terms})=\frac{n}2.\frac{2n^2+3n-2}3$$
or $$(1 \cdot 2+3 \cdot 4+5 \cdot 6+ \cdots \text{to}\space \frac{n}2\space \text{terms})=\frac{n(n+2)(2n-1)}{24}$$
But, this is the sum of $\dfrac{n}{2}$ terms. To get sum of $n$ terms, we replace $n$ by $2n$
Thus,
$$(1 \cdot 2+3 \cdot 4+5 \cdot 6+ \cdots \text{to}\space n\space \text{terms})=\frac{n(n+1)(4n-1)}6$$
So, if input of $n$ is even result should be correct,
if $n=2$, sum (from formula) = $7$
However, actual sum is $14$ (=$1 \cdot 2+3 \cdot 4$).
What is the error in the above approach?
EDIT: I rechecked my calculations and also have written the steps but still the answer isn't coming right. Please tell in which step the problem lies.
Thanks for your time and patience.
| Interesting approach and it will work for all values of n if you do it right.
The sum of first n terms in the sequence 1.2 + 3.4 + ... is calculated by the taking first sum of 2n terms of the sum of squares sequence (S in your case), subtracting n from it and taking the half.
say n = 2 sum of first 4 squares is 30 subtract 2 and take half 14.
for n = 3 sum of first 6 squares is 91 subtract 2 and take half 44.
and so on.
Your method works, however when you write "Combining the above results, we get," there you have done a calculation mistake.
There are simpler ways of doing the problem, however I like your approach.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/540593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Show that $(\sqrt{y^2-x}-x)(\sqrt{x^2+y}-y)=y \iff x+y=0$
Let $x,y$ be real numbers such that
$$\left(\sqrt{y^{2} - x\,\,}\, - x\right)\left(\sqrt{x^{2} + y\,\,}\, - y\right)=y$$
Show that $x+y=0$.
My try:
Let
$$\sqrt{y^2-x}-x=a,\sqrt{x^2+y}-y=b\Longrightarrow ab=y$$
and then
$$\begin{cases}
y^2=a^2+(2a+1)x+x^2\cdots\cdots (1)\\
x^2=b^2+(2b-1)y+y^2\cdots\cdots
\end{cases}$$
$(1)+(2)$
then
$$x=-\dfrac{a^2+b^2+(2b-1)ab}{2a+1}\cdots\cdots (3)$$
so
$$x+y=ab-\dfrac{a^2+b^2+(2b-1)ab}{2a+1}=\dfrac{(a-b)(2ab-a+b)}{2a+1}$$
we take $(3)$ in $(2)$,we have
$$b^2+(2b-1)y+y^2-x^2=\dfrac{(2ab-a+b)(2a^3b+a^3+3a^2b-2ab^3+ab^2+4ab-b^3+b)}{(2a+1)^2}=0$$
so
$$(2ab-a+b)=0$$
or
$$2a^3b+a^3+3a^2b-2ab^3+ab^2+4ab-b^3+b=0$$
if
$$2ab-a+b=0\Longrightarrow x+y=\dfrac{(a-b)(2ab-a+b)}{2a+1}=0$$
and if
$$2a^3b+a^3+3a^2b-2ab^3+ab^2+4ab-b^3+b=0$$
I don't prove
$$x+y=\dfrac{(a-b)(2ab-a+b)}{2a+1}=0?$$
| Assuming continuity in the area interval $ (0 < x < 1 )$ and $ (0 > y > -1 ) $ would create problems as $x$ and $y$ are not always real in these areas.
Nay, union of inside parabola areas of $ y_1 = - x^2 $ and $ y_2 = \sqrt{x} $ would violate $ x + y = 0, $ which is only the common chord of intersection of $ y_1,y_2$. So the shown line joining $(0,0)$ to $(1,-1)$ does not exist as real.
E.g., $(\frac12, -\frac12)$ does not lie on the
common line.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/543607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 7,
"answer_id": 3
} |
How prove this inequality $\left(1-\frac{1}{3}\right)\left(1-\frac{1}{3^2}\right)\cdots\left(1-\frac{1}{3^n}\right)\ge\frac{14}{25}$
show that
$$\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{3^2}\right)\cdots\left(1-\dfrac{1}{3^n}\right)\ge\dfrac{14}{25}\tag{1}$$
My try:
I only prove following not strong inequality:
$$\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{3^2}\right)\cdots\left(1-\dfrac{1}{3^n}\right)\ge\dfrac{1}{2}$$
proof: use Bernoulli inequality
$$(1+x_{1})(1+x_{2})\cdots (1+x_{n})\ge 1+x_{1}+x_{2}+\cdots +x_{n},$$where $x_{i}\ge -1$
so
$$\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{3^2}\right)\cdots\left(1-\dfrac{1}{3^n}\right)\ge1-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\cdots+\dfrac{1}{3^n}\right)$$
so
$$LHS\ge1-\sum_{n=1}^{\infty}\dfrac{1}{3^n}=1-\dfrac{\dfrac{1}{3}}{1-\dfrac{1}{3}}=\dfrac{1}{2}$$
But for $(1)$,I can't prove it,Thank you
| Here's another, somewhat complicated way to prove things. We can rewrite the inequality (with $n$ taken to infinity) as
$${25\over14}\ge\prod_{k=1}^\infty{1\over1-(1/3)^k}=\sum_{k=0}^\infty p(k)(1/3)^k$$
where $p(k)$ is the partition function $1, 1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56,77,101,\ldots$. Let's accept for the moment the inequality
$$p(k)\le{1\over50}2^k\text{ for } k\gt11$$
Then we have
$$\sum_{k=0}^\infty p(k)(1/3)^k \le 1+{1\over3}+{2\over3^2}+{3\over3^3}+{5\over3^4}+{7\over3^5}+{11\over3^6}+{15\over3^7}+{22\over3^8}+{30\over3^9}+{42\over3^{10}}+{56\over3^{11}}+{3\over50}\left({2\over3}\right)^{12}$$
where the final term comes from
$$p(12)(1/3)^{12}+\cdots \le {1\over50}\left({2\over3}\right)^{12}\left(1+{2\over3}+\left({2\over3}\right)^2+\cdots\right)={1\over50}\left({2\over3}\right)^{12}{1\over1-(2/3)}={3\over50}\left({2\over3}\right)^{12}$$
The sum of all those fractions is not a lot of fun to work out, but it turns out to give
$$\sum_{k=0}^\infty p(k)(1/3)^k \le {316217\over177147}+{3\over50}\left({2\over3}\right)^{12}\approx1.7855\lt1.7857\approx{25\over14}$$
The inequality on partition numbers that made this work can be proven by induction from Euler's pentagonal number theorem:
$$\begin{align}
p(k)&=p(k-1)+p(k-2)-p(k-5)-p(k-7)+p(k-12)+p(k-15)-\cdots\cr
&=p(k-1)+p(k-2)-\left(p(k-5)-p(k-12)\right)-\left(p(k-7)-p(k-15)\right)-\cdots\cr
&\le p(k-1)+p(k-2)\cr
\end{align}$$
Clearly one can conclude from this an inequality of the form $p(k)\le cr^k$ with any ratio $r$ greater than or equal to the golden ratio $\phi=(1+\sqrt5)/2$, but with a constant $c$ and/or starting value for $k$ that need to be checked against the initial terms of the sequence. To the extent that I played around with different possibilities, I always found I had to go out to the twelfth term to make things less than $25/14$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/545084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 1
} |
How find this sum $\sum_{n=1}^{\infty}\frac{1}{n^2-n+a}$ Today Question
if $a>\dfrac{1}{4}$, show that
$$\sum_{n=1}^{\infty}\dfrac{1}{n^2-n+a}=\dfrac{\pi}{\sqrt{4a-1}}\cdot\dfrac{e^{\pi\sqrt{4a-1}}-1}{e^{\pi\sqrt{4a-1}}+1}\tag{1}$$
I have konw that solution:
$$\sum_{n=1}^{\infty}\dfrac{1}{n^2+1}=\dfrac{\pi}{2}\dfrac{e^{2\pi}+1}{e^{2\pi}-1}-\dfrac{1}{2}$$
solution:
note
$$e^{ax}=\dfrac{\pi^{2\pi a}-1}{\pi}\left(\dfrac{1}{2a}+\sum_{n=1}^{\infty}\dfrac{a\cos{nx}-n\sin{x}}{n^2+a^2}\right)$$
let $a=1,x=0$,then we have
$$\sum_{n=1}^{\infty}\dfrac{1}{n^2+1}=\dfrac{\pi}{2}\dfrac{e^{2\pi}+1}{e^{2\pi}-1}-\dfrac{1}{2}$$
But for $(1)$ I have
$$\sum_{n=1}^{\infty}\dfrac{1}{(n-\frac{1}{2})^2+a-\frac{1}{4}}$$
Then I can't ,Thank you for your help.
| Complete the square in the summand to get
$$\sum_{n=1}^{\infty} \frac{1}{\left ( n-\frac12\right)^2+\left (a-\frac14\right)}$$
Note that the sum is symmetric and equal to
$$\frac12 \sum_{n=-\infty}^{\infty} \frac{1}{\left ( n-\frac12\right)^2+\left (a-\frac14\right)}$$
so that we may use the residue theorem to evaluate the sum. I will skip the steps involved in deriving the general result, but will just state it here:
$$\sum_{n=-\infty}^{\infty} f(n) = -\sum_{k} \operatorname*{Res}_{z=z_k} \pi \, \cot{(\pi z)} \, f(z)$$
where the $z_k$ are poles of $f$, which obeys certain integrability conditions. Details may be found elsewhere within this site.
In this case,
$$f(z) = \frac{1}{\left ( z-\frac12\right)^2+\left (a-\frac14\right)}$$
which has poles $z_{\pm} = \frac12 \pm i \sqrt{a-\frac14}$. The sum is therefore equal to
$$-\frac{\pi}{2} 2 \Re{\left [\frac{\cot{\pi \left ( \frac12 + i \sqrt{a-\frac14}\right )}}{i 2 \sqrt{a-\frac14}} \right ]} = \frac{\pi}{\sqrt{4 a-1}} \tanh{\left ( \pi \sqrt{4 a-1}\right)}$$
which matches the result to be shown.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/545836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to find recurrence relation of a given solution. Determine values of the constants $A$ and $B$ such that $a_n=An+B$ is a solution of the recurrence relation $a_n=2a_{n-1}+n+5$.
I know that the characteristic equation is $r-2 = 0$ which has the root $r = 2$.
Usually I find constants $A$ and $B$ by $a_n=Ar^n+Br^n$. It
looks like I cannot apply this here because $a_n=An+B$ is given.
The solution is $A = -1$ and $B = -7$. I don't know how to get this answer.
Can anyone help me please.
| You have $\forall n ~:~ a_n = 2~a_{n-1} + n + 5$ and $\forall n ~:~ a_n = A~n+B$. So construct the equations
$$\begin{array} {c|cc}
n & a_n = 2~a_{n-1} + n + 5 & a_n = A~n+B \\ \hline
1 & a_1 = 2~a_{0} + 6 & a_1 = A+B \\
2 & a_2 = 2~a_{1} + 7 & a_2 = 2~A+B \\
3 & a_3 = 2~a_{2} + 8 & a_3 = 3~A+B \\
\end{array}$$
Simple system of 6 linear equations, 6 variables. Solve for $A$ and $B$.
Or you can be a bit more indirect: Plug $\forall n ~:~ a_n = A~n+B$ into $\forall n ~:~ a_n = 2~a_{n-1} + n + 5$
$$\forall n ~:~ A~n + B = 2~A~(n-1) + 2~B + n + 5$$
Plug $n=0$ in to get
$$B = 2~B - 2A + 5$$
Plug $n=1$ in to get
$$A+B = 2B + 1 + 5$$
So 2 equations, 2 variables.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/547184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Finding the singular value(s) of a given matrix without SVD. I am struggling on a problem that asks to find the singular value(s) that are unequal to 0 in the following matrix:
$M = \begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 \end{bmatrix}$
I am not supposed to do a singular value decomposition on the matrix.
The given answer shows how you should rewrite the matrix as
$
\sqrt{20}
\begin{bmatrix} \frac{1}{2}\\ \frac{1}{2}\\ \frac{1}{2}\\ \frac{1}{2} \end{bmatrix}
\begin{bmatrix}
\frac{1}{\sqrt{5}} & \frac{1}{\sqrt{5}} & \frac{1}{\sqrt{5}} & \frac{1}{\sqrt{5}} & \frac{1}{\sqrt{5}}
\end{bmatrix}
$, and that therefore the singular value must be $\sqrt{20}$.
I see how the two vectors are orthonormal but other than that I don't really see why $\sqrt{20}$ must be a singular value here. It is also unclear to me why this is the only singular value unequal to zero.
Any help would be much appreciated.
| For any $n$, let $e=(1,...,1)^T$, and let $x_2,...,x_n$ be an orthonormal basis of $\operatorname{sp}\{e\}^\bot$. Let $x_1 = \frac{1}{\|e\|} e$. Then $x_1,...,x_n$
is an orthonormal basis for $\mathbb{R}^n$. Let $U_n = \begin{bmatrix} x_1 & \cdots & x_n \end{bmatrix}$.
Note that $U_4^* M U_5 = \begin{bmatrix} \begin{matrix} \sqrt{4} & \cdots & \sqrt{4} \end{matrix} \\ 0 \end{bmatrix} U_5 = \Sigma = \begin{bmatrix} \sqrt{20} & 0 & \cdots & 0 \\
0 & 0 & \cdots & 0 \\
\vdots & \vdots & & \vdots\\
0 & 0 & \cdots & 0
\end{bmatrix} $
and so $M = U_4 \Sigma U_5^*$.
$U_4,U_5$ are unitary, $\Sigma$ is diagonal with non-negative entries, hence this is a singular value decomposition.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/548736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Waring's problem The first comment on OEIS A002379 states:
It is an important unsolved problem related to Waring's problem to show that a(n) = floor((3^n-1)/(2^n-1)) holds for all n >= 1. This has been checked for 10000 terms and is true for all sufficiently large n, by a theorem of Mahler. [Lichiardopol]
Edit -- a new equality
$$
\frac{2^n( 3^n \mod 2^n)}{4^n-2^n} -\frac{3^n \mod 2^n}{4^n-2^n} -\frac{2^n( (-2+3^n) \mod (-1+2^n))}{4^n-2^n}
-\left(\frac{3}{2}\right)^n +\frac{3^n}{2^{n-1}} -\frac{1}{2^{n-1}} +\frac{2^{2 n}}{4^n-2^n} -\frac{2^{n+1}}{4^n-2^n} = 1
$$
The items with $4^n-2^n$ in the denominators are distances and the items with $2$ or $2^{n-1}$ are locations on the number line.
An alternate form is:
$$
\left\lfloor\left(\frac{3}{2}\right)^n\right\rfloor=\left\lfloor\left(\frac{3^n}{2^n-1}-\frac{1}{2^{n-1}}\right)\right\rfloor
$$
| $a_n$ is defined as the floor of $3^n/2^n$. The unsolved problem is whether that is always equal to the floor of ${3^n-1\over2^n-1}$.
EDIT: where "always" seems to mean "for all $n\ge2$".
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/548912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solve the equation $\tan \theta = 2\sin \theta$.
Solve the equation $\tan \theta = 2\sin \theta$.
What I did was rewrite it to the form $$\sin \theta = 2 \sin \theta \cos \theta$$ You'll get $$\sin \theta = \sin\ 2 \theta.$$
How am I supposed to solve this when I have $\sin$ on both sides? My main problem with these types of 'solve' equations are that I don't know which forms I should rewrite them too. Usually just to the form $\sin \theta = n$, but I wonder if having a $\sin$ on both sides can result in an answer too.
| We have the equality $$\sin x - \sin y = 2\sin \frac{x - y}{2} \cos\frac{x + y}{2}.$$ Hence, $\sin x = \sin y$ if and only if either $\sin \frac{x-y}{2} = 0$ or $\cos \frac{x+y}{2} = 0$.
In particular, for the solutions of $\sin\theta = \sin(2\theta)$, substitute $x=\theta$, $y=2\theta$. We need $\sin(\theta/2) = 0$ or $\cos(3\theta/2) = 0$. Now, $$\sin(\theta/2) = 0 \iff \exists n\in\mathbb{Z},\,\theta/2 = \pi n \iff \exists n,\,\theta = 2\pi n,$$ and $$\cos(3\theta/2) = 0 \iff \exists n,\,3\theta/2 = \frac{\pi}{2}+\pi n \iff \exists n,\,\theta = \frac{\pi}{3}+\frac{2\pi n}{3}.$$ The solution set is thus
$$\left\{2\pi n:n\in\mathbb{Z}\right\}\cup\left\{\frac{\pi}{3}+\frac{2\pi n}{3}:n\in\mathbb{Z}\right\}.$$
Finally, we verify that $\frac{\pi}{2}+\mathbb{Z}\pi$ is not a subset of our solution set, hence the $\tan$ is well-defined.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/552167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
What is the maximum value of $4(\sin x)^2 + 3(\cos x)^2$ The question is: What is the maximum value of: $4\sin^2\theta + 3\cos^2\theta$
This is the way I did it:
$4\sin^2\theta + 3\cos^2\theta = \sin^2\theta + 3\sin^2\theta + 3\cos^2\theta = \sin^2\theta + 3$
The max value of $\sin^2\theta$ is $1$, so the answer must be $4$. However my book says that the answer is $5$. Where did I go wrong?
| Your argument is OK. You can check it taking the derivative of the function:
$$f(x)=4\sin(x)^2+3\cos(x)^2$$
$$y'(x)=2\sin(x)\cos(x)$$
which is null for $x=0,x=\frac{\pi}{2}$
The first value gives $y(x)=3$, with the second one you get $y(x)=4$. So, $4$ is the maximum value of $f(x)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/552838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Determinant of matrix $A^3 + 2A^2 - A - 5I$ Given the eigenvalues of A So A is a 3 by 3 matrix with eigenvalues -1, 1, 2. And I have to find the determinant of $$A^3 + 2A^2 - A - 5I$$
Let $u$ be the eigenvector for the eigenvalue -1. Let
$S = A^3 + 2A^2 - A - 5I$ then
$Su = \lambda u$.
$=(A^3 + 2A^2 - A - 5I)u\\
=A^3u + 2A^2u - Au - 5u$
Where do I go from here?
| As $A$ has eigen values $-1,1,2$ you can actually see $A$ as $\begin{pmatrix}-1&0&0\\ 0&1&0\\ 0&0&2\end{pmatrix}$
If this is the case, what would be $A^3$?? what would be $A^2$??
$A^3=\begin{pmatrix}-1&0&0\\ 0&1&0\\ 0&0&8\end{pmatrix}$ and $A^2=\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&4\end{pmatrix}$
So what would be $A^3+2A^2-A-5I$??
what would be determinant of $A^3+2A^2-A-5I$??
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/553775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How to find a matrix $X$ such that $X+X^2+X^3 = \begin{bmatrix} 1&2005\\ 2006&1 \end{bmatrix}$?
Find a matrix $X \in M_{2}(\mathbb Z)$ such that $$X+X^2+X^3=\begin{bmatrix} 1&2005\\ 2006&1\end{bmatrix}$$
My try:
Let
$$X=\begin{bmatrix}
a&b\\
c&d
\end{bmatrix}$$
where $a,b,c,d\in Z$
then
$$X^2=\begin{bmatrix}
a^2+bc&ab+bd\\
ac+cd&bc+d^2
\end{bmatrix}$$
then $$X^3=\begin{bmatrix}
a^3+abc+abc+bcd&a^2b+b^2c+abd+bd^2\\
a^2c+acd+bc^2+cd^2&abc+bcd+bcd+d^3
\end{bmatrix}$$
so
$$X+X^2+X^3=\begin{bmatrix}
a^3+2abc+bcd+a^2+bc+a&a^2b+b^2c+abd+bd^2+ab+bd+b\\
a^2c+acd+bc^2+cd^2+ac+cd+c&abc+2bcd+d^3+bc+d^2+d
\end{bmatrix}$$
then we have
$$\begin{cases}
a^3+2abc+bcd+a^2+bc+a=1\\
a^2b+b^2c+abd+bd^2+ab+bd+b=2005\\
a^2c+acd+bc^2+cd^2+ac+cd+c=2006\\
abc+2bcd+d^3+bc+d^2+d=1
\end{cases}$$
Note $2005=5\cdot 401 $ is prime number,so
$$b(a^2+bc+ad+d^2+a+d+1)=2005$$
we can $b=\pm 1$ ,or $b=\pm 2005$ ,or $b=\pm 5$ or $b=\pm 401$
and note $2006=2\times 1003$
then $c=\pm 2$ ,or $c=\pm 1003$or $c=\pm 1$,or $c=\pm 2006$
so following is very ugly. But I can't.Thank you ,maybe this problem have other nice methods,Thank you
| There's no such $X$, even with rational entries.
If there were, then it would have an eigenvalue that's either
rational or a quadratic irrationality.
But if $\lambda$ is an eigenvalue of $X$ then
$\lambda + \lambda^2 + \lambda^3$ is an eigenvalue of
$\left[\begin{array}{cc}1&2005\cr2006&1\end{array}\right]$.
But those eigenvalues are the roots $x = 1 \pm \sqrt{2005\cdot 2006}$ of
$(x-1)^2 = 2005 \cdot 2006$, and the polynomial
$(\lambda^3+\lambda^2+\lambda-1)^2 - 2005 \cdot 2006$
turns out to be irreducible, so none of its roots can be
the eigenvalue of a $2 \times 2$ matrix with rational entries, QED.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/555195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Proving $\int_a^b \frac {x dx}{\sqrt{(x^2-a^2)(b^2-x^2)}} = \frac {\pi}{2} $ Can anybody here help me to prove that
$$
\int_a^b \frac {x\, \mathrm dx}{\sqrt{(x^2-a^2)(b^2-x^2)}} = \frac {\pi}{2}
$$
Thanks for your help.
| After substituting $y=x^2$, this integral may be solved by a type-III Euler substitution, of the form:
$$\sqrt{\left(y-a^2\right)\left(b^2-y\right)}=\left(y-a^2\right)t.\tag{1}$$
In my opinion, Euler substitutions offer an elegant alternative to trigonometric substitutions, and they are unfortunately under-taught in basic calculus courses. So when applicable, I like to post solutions using the Euler method for the sake of raising awareness, even for problems such as this one that have already been adequately solved by other methods. Using the substitution relation $(1)$ above, we find,
$$\begin{cases}
t=\frac{\sqrt{\left(y-a^2\right)\left(b^2-y\right)}}{y-a^2},\\
y=\frac{a^2t^2+b^2}{t^2+1},\\
\mathrm{d}y=\frac{2t\left(a^2-b^2\right)}{\left(t^2+1\right)^2}\,\mathrm{d}t,\\
\end{cases}$$
and so the integral evaluates to:
$$\begin{align}
I
&=\int_{a}^{b}\frac{x\,\mathrm{d}x}{\sqrt{\left(x^2-a^2\right)\left(b^2-x^2\right)}}\\
&=\frac12\int_{a^2}^{b^2}\frac{\mathrm{d}y}{\sqrt{\left(y-a^2\right)\left(b^2-y\right)}}\\
&=\frac12\int_{\infty}^{0}\frac{t^2+1}{|b^2-a^2||t|}\cdot\frac{2t\left(a^2-b^2\right)}{\left(t^2+1\right)^2}\,\mathrm{d}t\\
&=\frac{b^2-a^2}{|b^2-a^2|}\int_{0}^{\infty}\frac{\mathrm{d}t}{t^2+1}\\
&=\operatorname{sgn}{\left(b^2-a^2\right)}\cdot\frac{\pi}{2}.\\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/555708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Expected number of points on circle to form an acute angled triangle This problem was asked to me in an interview.
We keep on adding points on a circle uniformly until there exist three points on the circle which form an acute angled triangle. What is the expected number of points on the circle when the process stops?
| Well, let's consider the probability of any 3 points forming an acute triangle. First, due to the symmetry of a circle, the position of the first point is irrelevant. Then, the next point is somewhere in $[0,\pi]$ radians from the first point. Finally, a triangle in the circle is acute iff it contains the center of the circle, so the angle $\theta$ between the first two points, reflected over the origin, is the region the third point must occupy to form an acute triangle. Denoting $P(n)$ as the probability of forming an acute triangle after $n$ points, and $p(n)$ being the probability that the $n^{\mathrm{th}}$ point forms the first such triangle,
$$p(3)=P(3)=\frac{\int_0^\pi \frac{\theta}{2\pi} \mathrm d\theta}{\pi}=\frac{\frac{1}{2}\theta^2|_0^\pi}{2\pi^2}=\frac{1}{4}$$
Now, the next point, if it doesn't form an equilateral triangle, can occupy $2\pi-\theta$ rad, $\pi$ of which would extend the maximum angle between points.
For generality, let's call $\theta_n$ the largest angular separation between any two points after $n$ points have been placed, given that no three points form an acute triangle. We can obtain a recurrence relation for $\overline{\theta_n}$, as placing a new point increases $\theta$ linearly when it is between $\theta$ and $\pi$ from one of the most extreme points. When it is between these points, $\theta$ increases by $0$. Thus, $\overline{\theta_n}=\theta_{n-1}+\frac{2\int_0^{\pi-\theta_{n-1}}x\mathrm dx+\int_0^{\theta_{n-1}}0\mathrm dx}{2\pi-\theta_{n-1}}=\theta_{n-1}+\frac{(\pi-\theta_{n-1})^2}{2\pi-\theta_{n-1}}$. The original average $\theta_2$ was $\pi/2$, so $\overline{\theta_3}=2\pi/3$, and so chances of forming an acute angle with the fourth point is $1/3$. By the fourth point, the chance of having formed an acute angled triangle is now
$$ \frac{1}{4}+\frac{3}{4}\cdot\frac{1}{3}=\frac{1}{2} $$
So you can expect a $50/50$ chance of an acute triangle after 4 points.
Now let's compute expected value:
In fact, by substituting $\overline{\theta_n}$ for $\pi\left(1-\frac{1}{a_n}\right)$, we obtain
$$\begin{align} \overline{\theta_n}&=\pi\left(1-\frac{1}{a_{n-1}}\right)+\frac{\left(\pi-\pi\left(1-\frac{1}{a_{n-1}}\right)\right)^2}{2\pi-\pi\left(1-\frac{1}{a_{n-1}}\right)} \\
&=\pi\left(1-\frac{1}{a_{n-1}}+\frac{1}{a_{n-1}(1+a_{n-1})}\right) \\
&=\pi\left(1-\frac{1}{a_{n-1}+1}\right) \end{align}$$
And therefore $a_n=a_{n-1}+1$. Since $a_2=2$, $\overline{\theta_n}=\frac{(n-1)\pi}{n}$:
$$\begin{align}
P(n+2)&=\sum_{k=1}^n \left(\prod_{i=1}^k (1-p(i+1))\cdot\frac{p(k+1)}{1-p(k+1)}\right) \\
&=\sum_{k=1}^n \left(\left(\prod_{i=1}^k \left(1-\frac{i}{2(i+1)}\right)\right)\frac{k}{k+2} \right) \\
&=\sum_{k=1}^n 2^{-k} \left(\prod_{i=1}^k \frac{i+2}{i+1}\right) \frac{k}{k+2} \\
&=\sum_{k=1}^n \frac{k}{2^{k+1}} \\
&=1-\frac{n+2}{2^{n+1}}
\end{align}$$
The product is $1-\frac{\overline{\theta_{i+1}}}{2\pi}$, or the probability of not getting an acute triangle for $k+2$ points, multiplied by $k/(k+2)$ so that the last term is the probability of getting an acute triangle rather than not. Telescoping series allows the simplification of the product. The sum can be proven by induction on $n$.
We then have that $P(n)=1-2^{1-n}\cdot n$. We can also determine the expected value of the number of points, $E(X)$:
$$E(X)=\sum_{n=3}^\infty n\cdot p(n)=\sum_{n=3}^\infty \frac{n(n-2)}{2^{n-1}}=\sum_{n=1}^\infty \frac{n(n-1)+3n}{2^{n+1}}$$
Since $\sum_{n=0}^\infty {n\choose r}2^{-n}=2$, $\sum_{n=0}^\infty n\cdot 2^{-n}=\sum_{n=0}^\infty \frac{1}{2}n(n-1)2^{-n}=2$.
$$ \Rightarrow E(X)=\sum_{n=0}^\infty \left({n\choose 2}2^{-n}+\frac{3}{2}{n\choose 1}2^{-n} \right)=2+3=5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/555772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find the sum of the series $\sum \frac{1}{n(n+1)(n+2)}$ I got this question in my maths paper
Test the condition for convergence of $$\sum_{n=1}^\infty \frac{1}{n(n+1)(n+2)}$$
and find the sum if it exists.
I managed to show that the series converges but I was unable to find the sum. Any help/hint will go a long way. Thank you.
| There is an alternate method and is as follows.
Notice that
$$ \frac{1}{n(n+1)(n+2)} = \frac{(n-1)!}{(n+2)!} = \frac{1}{2!} \, B(n,3) $$
where $B(x,y)$ is the Beta function. Using an integral form of the Beta function the summation becomes
\begin{align}
S &= \sum_{n=1}^{\infty} \frac{1}{n \, (n+1) \, (n+2)} \\
&= \frac{1}{2} \, \int_{0}^{1} \left( \sum_{n=1}^{\infty} x^{n-1} \right) \, (1-x)^{2} \, dx \\
&= \frac{1}{2} \, \int_{0}^{1} \frac{(1-x)^{2}}{1-x} \, dx = \frac{1}{2} \, \int_{0}^{1} (1-x) \, dx \\
&= \frac{1}{4}
\end{align}
This leads to the known result
\begin{align}
\sum_{n=1}^{\infty} \frac{1}{n \, (n+1) \, (n+2)} = \frac{1}{4}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/560816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 10,
"answer_id": 1
} |
Sum of a cosine series I have a short question:
It turns out that the following holds:
$\sum_{i=1}^{k-1} \cos(\frac{2\pi i}{k}) = - 1$.
Why is that?
Thank you!
| Clearly, the Series vanishes if $k=1$
If $k>1, \sin\frac{\pi}k\ne0$
As $2\cos A\sin B=\sin(A+B)-\sin(A-B),$ using this, $$2\cos\frac{2i\pi}k\sin\frac{\pi}k=\sin\frac{(2i+1)\pi}k-\sin\frac{(2i-1)\pi}k $$
Putting $i=1,2,\cdots,k-1$ and summing we get $$2\sin\frac{\pi}k\sum_{1\le i\le k-1}\cos\frac{2i\pi}k=\sin\frac{(2k-1)\pi}k-\sin\frac{\pi}k=2\sin\frac{(k-1)\pi}k\cos\frac{k\pi}k$$
Using $\displaystyle\sin C-\sin D=2\sin\frac{C-d}2\cos\frac{C+D}2,$
$\displaystyle\sin\frac{(2k-1)\pi}k-\sin\frac{\pi}k=2\sin\frac{(k-1)\pi}k\cos\frac{k\pi}k$
Now, $\displaystyle \cos\frac{k\pi}k=\cos\pi=-1$ and $\displaystyle \sin\frac{(k-1)\pi}k=\sin\left(\pi-\frac\pi k\right)=\sin\frac\pi k\ne0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/561163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integral $\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right) \mathrm dx$ I need help with this integral:
$$I=\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right)\ \mathrm dx.$$
The integrand graph looks like this:
$\hspace{1in}$
The approximate numeric value of the integral:
$$I\approx8.372211626601275661625747121...$$
Neither Mathematica nor Maple could find a closed form for this integral, and lookups of the approximate numeric value in WolframAlpha and ISC+ did not return plausible closed form candidates either. But I still hope there might be a closed form for it.
I am also interested in cases when only numerator or only denominator is present under the logarithm.
| For the purposes of alternative methods, it may be of interest to note that the integrand
$$f(x)=\frac{1}{x}\sqrt{\frac{1+x}{1-x}}\log\left(\frac{2x^2+2x+1}{2x^2-2x+1}\right)$$ may be rewritten in terms of hyperbolic trigonometric functions. Using
$$\tanh^{-1}(z) = \frac{1}{2}\log\left(\frac{1+z}{1-z}\right),$$
and we obtain
$$f(x)=\frac{1}{x}e^{\tanh^{-1}x}\log\left(\frac{1+\frac{2x}{1+2x^2}}{1-\frac{2x}{1+2x^2}}\right) = e^{\tanh^{-1} x}\left(\frac{2\tanh^{-1}\left(\frac{2x}{1+2x^2}\right)}{x}\right).$$
The rational function in the bracket, which we will denote $s(x)$, is symmetric about $x=0$.
The desired integral is
$$I=\int_{-1}^1 f(x)dx = \int_{-1}^1e^{\tanh^{-1}x}s(x)dx,$$
which, by adding the indicated useful definite integral to both side, gives
$$I + \int_{-1}^1 e^{-\tanh^{-1}x}s(x)dx = 2\int_{-1}^1 \frac{s(x)dx}{\sqrt{1-x^2}}.$$
Now using the change of variable $x=-y$ we have $$\int_{-1}^1 e^{-\tanh^{-1} x}s(x)dx = -\int_1^{-1} e^{\tanh y}s(-y)dy = \int_{-1}^1 e^{\tanh y}s(y)dy = I,$$ by the symmetry of $s(x)$. Hence, we finally obtain
$$I = \int_{-1}^1\frac{s(x)dx}{\sqrt{1-x^2}} = 2\int_{-1}^1\frac{1}{x\sqrt{1-x^2}}\tanh^{-1}\left(\frac{2x}{1+2x^2}\right)dx.$$
This integral is symmetric about $x=0$, so we have
$$I=4\int_0^1\frac{1}{x\sqrt{1-x^2}}\tanh^{-1}\left(\frac{2x}{1+2x^2}\right)dx,$$ which can be rewritten $$I=-4\int_0^1\left(\frac{d}{dx}\text{sech}^{-1}x\right)\tanh^{-1}\left(\frac{2x}{1+2x^2}\right)dx.$$
Using integration by parts this results in
$$I=8\int_0^1\frac{\text{sech}^{-1}(x)(1-2x^2)}{1+4x^4}dx.$$
We could also make the change of variable $y=\text{sech}^{-1}x$ to obtain
$$I=8\int_0^\infty\frac{y(\cosh^2(y)-2)\sinh y}{\cosh^4(y)+4}dy= 8\int_0^\infty\frac{y\sinh^3 y}{\cosh^4y+4}dy-8\int_0^\infty\frac{y\sinh y}{\cosh^4 y+4}dy.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/562694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "516",
"answer_count": 6,
"answer_id": 3
} |
Procedure for showing a polynomial is irreducible in $\mathbb{Q[x]}$ Let $f(x) = x^5 − 6x^2 + 21x + 13$
What is the procedure for showing $f(x)$ is irreducible in $\mathbb{Q}[x]$?
| Suppose $f(x) = p(x) q(x)$ over $\mathbb{Z}$, where we can take $p$ and $q$ monic and non-constant. Use subscript $l$ to indicate taking coefficients mod $l$. Note that $\deg p_l(x) = \deg p(x)$, etc., and $f_l(x) = p_l(x) q_l(x)$.
One computes
$$ f_2(x) = x^5+x+1 = (x^2+x+1)(x^3+x^2+1) $$
$$ f_3(x) = x^5 + 1 = (x+1)(x^4-x^3+x^2-x+1) $$
The factors of $f_2(x)$ have no roots, so are irreducible. The quartic factor of $f_3(x)$ also has no root, so if it factors, it's the product of monic irreducible quadratics. The quadratics must have discriminant $-1$ since otherwise they would factor with the quadratic equation, hence we get factors of the form $x^2+bx+c$ where $b^2-c = -1$. The possible quadratics are $x^2+1$, $x^2+x-1$, $x^2-x-1$, and no two of these give the quartic as product, so the quartic is irreducible
Hence the degrees of $p$ and $q$ are ($2$ and $3$ from $l=2$) and ($1$ and $4$ from $l=3$), a contradiction, so $f$ is irreducible over $\mathbb{Z}$, hence over $\mathbb{Q}$.
Maybe you can find a less computational way to show the quartic is irreducible....
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/566670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How do you prove the inequality $x^2+y^2+z^2-y(x+z)>0$? This actually comes out of a matrix multiplication of $x^TAx$, but when you multiply it out, you get the following.
$x^2+y^2+z^2-y(x+z)>0$
I just can't figure out how to actually prove that that inequality holds for all $x,y,z\neq0$. Also, note that this is for a linear algebra class, so I don't think you're allowed to use derivatives. Also, in case it is important, here are the original matrices used to get that inequality.
$A=$ $
\left( \begin{array}{ccc}
2 & -1 & 0 \\
-1 & 2 & -1 \\
0 & -1 & 2 \end{array} \right)$
and $x = \left(\begin{array}{c}
x\\
y\\
z\ \end{array}\right)$
| The inequality is
$$\frac12\left(x^2 + (x-y)^2 + (y-z)^2 + z^2 \right) > 0.$$
It is easy to see that that holds for all real $(x,y,z) \neq (0,0,0)$.
A more linear-algebra-y way is to consider the minors of $A$; we have $2 > 0$,
$$\begin{align}
\det \begin{pmatrix}2 & -1\\-1 & 2 \end{pmatrix} &= 2^2 - (-1)^2 = 3 > 0\\
\det \begin{pmatrix}2 & -1 & 0\\ -1 & 2 & -1\\ 0 & -1 & 2 \end{pmatrix} &= 8 + 0 + 0 - 0 - 2 - 2 = 4 > 0,
\end{align}$$
so $A$ is positive definite.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/567337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Finding a closed form for $\int \frac{dx}{(1+x)(1+x^a)}$ if one exists. Find a closed form for this integral
$$\int \frac{dx}{(1+x)(1+x^a)}$$
This integral has the possibility of not having a closed form in which case can it be proven?
Feeble attempt so far:
$$\int \frac{dx}{(1+x)(1+x^a)}=\frac{\log(x+1)}{1+x^a}-\int\frac{ax^{a-1}\log(x+1)}{(x^a+1)^3}dx$$It is starting to feel analytical.
WA is not happy with it. No elementary function representation found it says. WA
| It is clearly known that this integral should have closed form when $a$ is a rational number.
Let $a=\dfrac{p}{q}$ , where $p\in\mathbb{Z}$ , $q\in\mathbb{Z}^+$ and $\text{gcd}(p,q)=1$ ,
Then $\int\dfrac{dx}{(1+x)(1+x^a)}=\int\dfrac{dx}{(1+x)(1+x^\frac{p}{q})}$
Let $u=x^\frac{1}{q}$ ,
Then $x=u^q$
$dx=qu^{q-1}~du$
$\therefore\int\dfrac{dx}{(1+x)(1+x^\frac{p}{q})}=\int\dfrac{qu^{q-1}}{(1+u^q)(1+u^p)}du$ , which is an integral of rational function and it should have closed form.
When $a$ is an irrational number, it is afraid that you can only solve this integral by these approaches:
When $|x|<1$ and $a>0$ ,
Then $\int\dfrac{dx}{(1+x)(1+x^a)}$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^n}{1+x^a}dx$
$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty(-1)^{n+k}x^{n+ak}~dx$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-1)^{n+k}x^{n+ak+1}}{n+ak+1}+C$
When $|x|>1$ and $a>0$ ,
Then $\int\dfrac{dx}{(1+x)(1+x^a)}$
$=\int\dfrac{dx}{x^{a+1}(1+x^{-1})(1+x^{-a})}$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{-n}}{x^{a+1}(1+x^{-a})}dx$
$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-1)^{n+k}x^{-n-ak}}{x^{a+1}}dx$
$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty(-1)^{n+k}x^{-n-a(k+1)-1}~dx$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-1)^{n+k}x^{-n-a(k+1)}}{-n-a(k+1)}+C$
$=-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-1)^{n+k}}{(n+a(k+1))x^{n+a(k+1)}}+C$
When $|x|<1$ and $a<0$ ,
Then $\int\dfrac{dx}{(1+x)(1+x^a)}$
$=\int\dfrac{dx}{x^a(1+x)(1+x^{-a})}$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^n}{x^a(1+x^{-a})}dx$
$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-1)^{n+k}x^{n-ak}}{x^a}dx$
$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty(-1)^{n+k}x^{n-a(k+1)}~dx$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-1)^{n+k}x^{n-a(k+1)+1}}{n-a(k+1)+1}+C$
When $|x|>1$ and $a<0$ ,
Then $\int\dfrac{dx}{(1+x)(1+x^a)}$
$=\int\dfrac{dx}{x(1+x^{-1})(1+x^a)}$
$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{-n}}{x(1+x^a)}dx$
$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-1)^{n+k}x^{-n+ak}}{x}dx$
$=\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty(-1)^{n+k}x^{-n+ak-1}~dx$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-1)^{n+k}x^{-n+ak}}{-n+ak}+C$
$=-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-1)^{n+k}}{(n-ak)x^{n-ak}}+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/570971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How find this sum $\sum_{k=1}^{n}\sqrt[3]{a_{k}}$ let sequence $\{a_{n}\}$ such
$$(6k^2+2-a_{k})^3=27(4k^6+3k^4+3k^2-1)a_{k}$$
for all $k\ge 1$, then compute
$$\sum_{k=1}^{n}\sqrt[3]{a_{k}}$$
My try: I have note
$$(4k^6+3k^4+3k^2-1)=(2k-1)(2k+1)(k^2-k+1)(k^2+k+1)$$
then
$$6k^2+2-a_{k}=3\sqrt[3]{(2k-1)(2k+1)(k^2-k+1)(k^2+k+1)a_{k}}$$
Then I can't.Thank you very much
| An interesting observation is if one define $u_k=k^3+(k+1)^3$, then the defining equation of $a_k$ can be simplified as
$$(u_k - u_{k-1} - a_k)^3 = 27 u_k u_{k-1} a_k\tag{*1}$$
Compare this with the algebraic identity
$$\left(x^3 - y^3 - (x-y)^3\right)^3 = 27 x^3 y^3 (x-y)^3$$
One will suspect
$$\sqrt[3]{a_k} = \sqrt[3]{u_k} - \sqrt[3]{u_{k-1}}$$
is a solution of $(*1)$. For $k \ge 1$ where $u_k > u_{k-1} > 0$, we can
verify this is indeed the case by direct substitution. As a result,
$$\sum_{k=1}^n\sqrt[3]{a_k} = \sqrt[3]{u_n} - \sqrt[3]{u_0} = \sqrt[3]{n^3 + (n+1)^3} - 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/571561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How find this limit $\lim_{x\to 0}\frac{1}{x^4}\left(\frac{1}{x}\left(\frac{1}{\tanh{x}}-\frac{1}{\tan{x}}\right)-\frac{2}{3}\right)=？$ Find this following limit
$$\displaystyle \lim_{x\to 0}\dfrac{1}{x^4}\left(\dfrac{1}{x}\left(\dfrac{1}{\tanh{x}}-\dfrac{1}{\tan{x}}\right)-\dfrac{2}{3}\right)=？$$
My try: since $$\tanh{x}=\dfrac{e^{x}-e^{-x}}{e^x+e^{-x}}$$
then
$$\dfrac{1}{\tanh{x}}-\dfrac{1}{\tan{x}}=\dfrac{e^{x}-e^{-x}}{e^x+e^{-x}}-\dfrac{1}{\tan{x}}$$
then I can't,Thank you
| $$
\coth x=\frac{1}{x}+\frac{x}{3}-\frac{x^3}{45}+\frac{2x^5}{945}-\frac{x^7}{4725}+... \\
\cot x =\frac{1}{x}-\frac{x}{3}-\frac{x^3}{45}-\frac{2x^5}{945}-\frac{x^7}{4725}-... \\[0.9cm]
\text{Given limit is}\\ \lim_{x \to 0} \frac{1}{x^4}\left(\frac{1}{x}\left(\frac{2x}{3}+\frac{4x^5}{945} + ...\right)-\frac{2}{3}\right) \\
= \lim_{x \to 0} \frac{1}{x^4}\left(\frac{4x^4}{945} + ...\right) \\
=\frac{4}{945}
$$
Don't worry if you can't remember these power series. You can derive them in a jiffy using Bernoulli numbers. But that's for another day.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/572180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluating the surface integral $\iint_\Sigma \mathbf{f} \cdot d \mathbf{a}$ where $\mathbf{f}(x,y,z)=(x^2,xy,z)$
Evaluate the surface integral $\iint_\Sigma \mathbf{f} \cdot d \mathbf{a}$ where $\mathbf{f}(x,y,z)=(x^2,xy,z)$ and $\Sigma$ is the part of the plane $6x+3y+2z=6$ with $x,y,z\geq 0$.
I changed the function to parametric form, $\mathbf{r} (u,v)=(u^2,uv,3-3u-3v/2)$.
Then $\mathbf{r}_u\times \mathbf{r}_v=(-3v/2+3u,3u,2u^2)$.
So the integral is $\int^2_0\int^1_0(-3u^2v/2+6u^2-3u^3)\,dudv=3/2$.
But the answer turn out to be $15/4$, what's wrong with my solution?
| You can choose a parametrization of the plane as
$$\textbf{r}(x,y) = \left( x, y, \frac{6-6x-3y}{2} \right).$$
We find the expression of the field in the surface by finding the composite, yielding
$$\textbf{f}(\textbf{r}(x,y)) = \left( x^2, xy, \frac{6-6x-3y}{2} \right).$$
To find the projected area in the $xy$ plane we make $z=0$, finding
$$\frac{6-6x-3y}{2} = 0 \iff 6x+3y=6 \iff 2x+y=2.$$
The normal vector is computed as $\textbf{r}_x \times \textbf{r}_y = (3, 3/2, 1)$.
We have that the surface integral is
$$\iint\limits_{\Sigma} \textbf{f} \cdot d \textbf{a} = \int_0^1 \int_0^{2-2x} 3x^2 + \frac{3xy}{2} + \frac{6-6x-3y}{2} \, dy \, dx.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/574575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Three consecutive integers with power of 5 mod 11 Let $(n - 1)$, $n$ and $(n + 1)$ be three consecutive integers, and $(n - 1)^5 \equiv n^5 \equiv (n + 1)^5 \equiv a \pmod{11}$, what are the possible values of $a$?
I know the facts that $3^5 \equiv 4^5 \equiv 5^5 \equiv 1 \pmod{11}$ and $6^5 \equiv 7^5 \equiv 8^5 \equiv -1 \pmod{11}$ are two solutions, however, is there a way to actually solve this equation?
| There is sufficiently to check the numbers $-5,-4,\ldots,4,5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/575056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the solution to $\frac1{a^2 +2} + \frac1{b^2 +2} + \frac1{c^2 +2} \le \frac{\sqrt2}{2}\frac{\sqrt a+\sqrt b+\sqrt c}{\sqrt{abc}}$ one of my friends asked me if I could solve him a mathematics problem.
It looks like this:
$$\frac1{a^2 +2} + \frac1{b^2 +2} + \frac1{c^2 +2} \le \frac{\sqrt2}{2}\frac{\sqrt a+\sqrt b+\sqrt c}{\sqrt{abc}}$$
As I think it looks like an inequality between means, hope it helps.
And sorry for my bad english by the way. :)
| @michael Rozenberg: Let's bury this one for good by presenting a point with all positive $a,b,c$ such that the inequality is violated.
Take
$$
a = 25\\
b=25\\
c= \frac{59}{81} \\
\frac{1}{a^2+2}+\frac{1}{b^2+2}+\frac{1}{c^2+2} = \frac{4146953}{10410081}
> 0.3983 \\
\frac{\sqrt{2}}{2}\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{abc}}=
\frac{59+90\sqrt{59}}{2950}\sqrt{2} < 0.3597
\\\mbox{LHS } > 0.39 > 0.36 > \mbox{ RHS}
$$
which violates the inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/575216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Missing step in rearrangement Can someone explain the missing step in the following rearrangement ?
From $ \displaystyle a^2 + b^2 + \frac {a^2y}{x} + \frac {b^2x}{y} \geq (a+b)^2 $ to $ \displaystyle \frac{(a+b)^2}{x+y} \leq \frac{a^2}{x} + \frac{b^2}{y} $ ?
| We have
$$ \left(\frac{a^2}{x} + \frac{b^2}{y}\right)(x+y)
=\frac{a^2}{x}x +\frac{a^2}{x}y+ \frac{b^2}{y}x +\frac{b^2}{y}y
=a^2 +\frac{a^2}{x}y+ \frac{b^2}{y}x +b^2y $$
and therefore
$$ \left(\frac{a^2}{x} + \frac{b^2}{y}\right)(x+y) = a^2 + b^2 + \frac {a^2y}{x} + \frac {b^2x}{y} \geq (a+b)^2. $$
And this gives, assuming $x+y>0$
$$ \frac{(a+b)^2}{x+y} \leq \frac{a^2}{x} + \frac{b^2}{y}.$$
If $x+y<0$, the inequality would be reversed!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/575800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Indefinite intergral of $\int { \sqrt{ x^2-a^2} \over x } \mathrm dx $ I need to integrate $$\int { \sqrt{x^2-a^2} \over x } \mathrm dx $$ using substition, and show it equals $$ \sqrt{x^2 - a^2} - a(\operatorname{arcsec} ({x\over a })) +c $$
I've tried $x=a\sin t$ but I couldn't finish it out.
Thank in advance for any help.
| Use the substitution:
$$x = a \sec(\theta)\implies dx = a \sec\theta\tan\theta\,d\theta$$
And use the identity
$$\sec^2(\theta)-1 = \tan^2(\theta).$$
$\require{cancel}$
Then your integral becomes: $$\begin{align}\int { \sqrt{x^2-a^2} \over x } dx & = \int \dfrac{\sqrt{a^2\sec^2 \theta - a^2}(\cancel{a \sec\theta}\tan\theta \,d\theta) }{\cancel{a\sec\theta}}\, \\ \\ & = a\int \sqrt{\tan^2\theta}\, \tan\theta\,d\theta \\ \\ &= a\int \tan^2 \theta \,d\theta \\ \\ & = a\int (\sec^2 \theta - 1)\,d\theta \\ \\ & = a \int \sec^2 \theta \,d\theta - a\int d\theta\\ \\ & = a \tan\theta - a\theta + c\end{align}$$
Now, back substitute to get $$\sqrt{x^2 - a^2} - a\left(\operatorname{arcsec} \left({x\over a }\right)\right) +c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/575946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Show that $\frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} \geq x+y+z $ by considering homogeneity Well, I'm preparing for an undergrad competition that is held in April and because of that I've been trying to solve the inequalities I find on the internet. I found this problem:
$$\displaystyle \frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} \geq x+y+z \text{ for all } x,y,z \in \mathbb{R^+}$$
It's easy to show that this inequality holds by applying the AM-GM inequality three times like the following:
$$ \frac{xy}{z} + \frac{xz}{y} \geq 2 \sqrt{\frac{xy}{z} \cdot\frac{xz}{y}} = 2x$$
$$ \frac{xy}{z} + \frac{yz}{x} \geq 2 \sqrt{\frac{xy}{z} \cdot\frac{yz}{x}} = 2y$$
$$ \frac{zy}{x} + \frac{xz}{y} \geq 2 \sqrt{\frac{zy}{x} \cdot\frac{xz}{y}} = 2z$$
Then summing these inequalities and canceling a factor of $2$ at the end will give us the desired inequality.
But if you look at the problem from another point of view, if we substitute $x'=\lambda x$, $y'=\lambda y$ and $z' = \lambda z$ the inequality stays the same. So, I was wondering if there could be another way of solving this inequality and the inequalities that are homogenous like this one in a more systematic way that could be applied to a broader range of problems.
I decided to add a constraint $x^2+y^2+z^2=1$ to the problem because the inequality in the problem could be thought of as a function in the three variables $x,y,z$ in $\mathbb{R}^3$. So, it's sufficient to study this function on the unit sphere because homogeneity allows us to define it for other points of $\mathbb{R}^3$, but I have no idea how this could lead me anywhere.
| Well, I found another proof of the famous Nesbitt's inequality by considering its homogeneity. So, I thought it might be worth sharing it on here for future reference and also to show how homogeneity can be employed to prove a famous inequality by what I've learned today from @Macavity and @SteveKass. I'm going to post it as an answer because it is too long for an edit to my question.
Theorem: For $a,b,c \in \mathbb{R}^+$ we have:
$$ \frac{c}{a+b} + \frac{a}{b+c} + \frac{b}{c+a} \geq \frac{3}{2} $$
Proof:
The LHS stays the same if we substitute $a'=\lambda a$, $b' = \lambda b$ and $c' = \lambda c$.
Suppose that $a'+b'+c'=1$. This implies $\lambda a + \lambda b + \lambda c = \lambda (a+b+c) = 1$, which gives us $\displaystyle \lambda = \frac{1}{a+b+c}$.
In other words, that means for any $a,b,c \in \mathbb{R}^+$ we can restrict our attention only to the case $a+b+c=1$. Because for any other triple $(x,y,z)$ we can always scale it by $\displaystyle \lambda= \frac{1}{x+y+z}$ to get to that case.
So, from now on, let's assume, Without Loss Of Generality, that $a+b+c=1$.
Now let's prove the inequality. We start by manipulating the inequality from the LHS:
$$ \frac{c}{a+b} + \frac{a}{b+c} + \frac{b}{c+a} \geq \frac{3}{2} \iff \frac{c}{a+b} +1+ \frac{a}{b+c}+1 + \frac{b}{c+a}+1 \geq \frac{3}{2}+3$$
$$ \frac{a+b+c}{a+b} + \frac{a+b+c}{b+c} + \frac{a+b+c}{c+a} \geq \frac{9}{2} \iff \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} \geq \frac{9}{2} $$
The last statement is correct because we have assumed $a+b+c=1$.
It's trivial that $2=2(a+b+c)=(a+b)+(b+c)+(c+a)$. So, multiplying both sides of the inequality by $2$ we get:
$$((a+b)+(b+c)+(c+a))(\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a}) \geq 9$$
The last inequality can be checked easily by applying AM-GM inequality or Cauchy-Schwarz inequality. Q.E.D.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/576592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 3
} |
Limit of a nested radical containing fractions How can I give a proof of the following inequality?
$$\displaystyle L=1+\sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+\sqrt[4]{\frac{1}{4}}+...}}$$
$$L\le\sqrt[3]{4\pi}$$
Numerical computations suggest it's true, but is it possible to give an analytical evidence?
| First note that the nested root $\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}$ converges to $\frac{1}{2}(1+\sqrt{5})$ (see (*) below).
To get the proposed inequality, we crudely estimate:
$$
\begin{align}
L &= 1+\sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+\sqrt[4]{\frac{1}{4}+\cdots}}}\le1+\sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+\sqrt[4]{1+\sqrt[5]{1+\cdots}}}}\\
&\le1+\sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+\sqrt{1+\sqrt{1+\cdots}}}}=1+\sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+\frac{1+\sqrt{5}}{2}}}<2.323<\sqrt[3]{4\pi}.
\end{align}
$$
Better estimates can be obtained by introducing the $1$s a later position. I don't know where the expression $\sqrt[3]{4\pi}\approx 2.3249$ comes from but if we can produce better estimates as simple as that, it seems to me like that is not important anyway.
Note:
(*) The value can be determined using the recurrence relation $x_{n+1}=\sqrt{1+x_n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/577317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How prove this inequality $\sum\limits_{cyc}\frac{x^2}{(2x+y)(2x+z)}\le\frac{1}{3}$ let $x,y,z>0$,show that
$$\dfrac{x^2}{(2x+y)(2x+z)}+\dfrac{y^2}{(2y+z)(2y+x)}+\dfrac{z^2}{(2z+x)(2z+y)}\le\dfrac{1}{3}$$
My try: $$\Longleftrightarrow\sum_{cyc}\dfrac{4x^2}{4x^2+2x(y+z)+yz}\le\dfrac{4}{3}$$
$$\Longleftrightarrow\sum_{cyc}\left(1-\dfrac{2x(y+z)+yz}{4x^2+2x(y+z)+yz}\right)\le\dfrac{4}{3} $$
$$\Longleftrightarrow \sum_{cyc}\dfrac{2x(y+z)+yz}{4x^2+2x(y+z)+yz}\ge\dfrac{5}{3}$$
then I can't.Thank you
| I have consider nice methos: use Cauchy-Schwarz inequality
\begin{align*}
\dfrac{a^2}{(2a+b)(2a+c)}&=\dfrac{a^2}{4a^2+2ab+2ac+bc}=\dfrac{a^2}{(2a^2+bc)+2a(a+b+c)}\\
&\le\dfrac{a^2}{9}\left(\dfrac{1}{2a^2+bc}+\dfrac{2}{a(a+b+c)}\right)\\
&=\dfrac{1}{9}\left(\dfrac{2a}{a+b+c}+\dfrac{a^2}{2a^2+bc}\right)
\end{align*}
so
$$\sum_{cyc}\dfrac{a^2}{(2a+b)(2a+c)}\le\dfrac{1}{9}\left(2+\sum\dfrac{a^2}{2a^2+bc}\right)$$
it suffices to prove that
$$\dfrac{a^2}{2a^2+bc}+\dfrac{b^2}{2b^2+ca}+\dfrac{c^2}{2c^2+ab}\le 1$$
which is equivalent to
$$\dfrac{bc}{bc+2a^2}+\dfrac{ca}{ca+2b^2}+\dfrac{ab}{ab+2c^2}\ge 1$$
Using the AM-GM inequality,we hve
$$\dfrac{bc}{bc+2a^2}=\dfrac{b^2c^2}{b^2c^2+2a^2bc}\ge\dfrac{b^2c^2}{b^2c^2+a^2(b^2+c^2)}=\dfrac{b^2c^2}{b^2c^2+a^2c^2+a^2b^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/577798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Proof of an inequality in arithmetic, geometric and harmonic means Prove that $$\left[\frac{x^2+y^2+z^2}{x+y+z}\right]^{x+y+z}\gt x^xy^yz^z\gt \left[\frac{x+y+z}{3}\right]^{x+y+z}$$
I could prove the inequality between the first two terms using the fact that $A.M\gt G.M$ in the following way. $$\left[\frac{x(\frac{x^2}{x})+y(\frac{y^2}{y})+z(\frac{z^2}{z})}{x+y+z}\right]\gt (x^xy^yz^z)^{\frac{1}{x+y+z}}$$
But I am unable to prove the inequality between the next two terms.
| Just to spell out what Ivan says:
Consider the sequence $a_1,a_2,a_3 = x, y, z$ and weights $w_1,w_2,w_3 = x/s, y/s, z/s$, where $s = x+y+z$. The inequality between harmonic and geometric (weighted) mean says that:
$$ \prod_{i} a_i^{w_i} \geq \frac{1}{\sum_i w_i \frac{1}{a_i}}.$$
Thus:
$$ x^{x/s}y^{y/s}z^{z/s} \geq \frac{1}{\frac{x}{s} \frac{1}{x} + \frac{y}{s} \frac{1}{y}+ \frac{z}{s} \frac{1}{z}} = \frac{1}{3 \cdot{1}{s} }= \frac{s}{3}.$$
Rising both sides to power $ s $ gives the claim:
$$ x^x y^y z^z \geq \left( \frac{x+y+z}{3} \right)^{x+y+z}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/578212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is the right solution for derivative of implicit function? There's need to find y' of:
$$\arctan(y/x)=\ln\sqrt{x^2 + y^2}$$
Tried:
$\dfrac{1}{(1+(y/x)^2)}*(\dfrac{y}{x})'=(x^2+y^2)^\dfrac{-1}{2}*(\dfrac{1}{2})*(x^2+y^22)^\dfrac{-1}{2} * (2x+2y'*y')$
| $$\arctan(\frac{y}{x})=\ln\sqrt{x^2+y^2}$$
$$\frac{1}{1+(y/x)^2}\frac{y'x-y}{x^2}=\frac{x+yy'}{\sqrt{x^2+y^2}}$$
solve for $y'$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/578282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
What is the $n$ times iterate of $f(x)=\frac{x}{\sqrt{1+x^2}}$? We were asked to determine the composition $f \circ f \circ f \circ \cdots \circ f $, $n$ times, where $f(x)=\dfrac{x}{\sqrt{1+x^2}}$.
Has anyone an idea?
| Let's denote $f^n = f \circ ... \circ f$ with $n$ times the function $f$, and suppose (which is obviously true for $n=1$), that:
$$f^n(x) = \frac{x}{\sqrt{1+n x^2}}$$
Then
$$f^{n+1}(x)=f \circ f^n(x)= \frac{\frac{x}{\sqrt{1+n x^2}}}{\sqrt{1+\left(\frac{x}{\sqrt{1+n x^2}}\right)^2}}$$
$$=\frac{\frac{x}{\sqrt{1+n x^2}}}{\sqrt{\frac{1+(n+1)x^2}{1+nx^2}}}=\frac{x}{\sqrt{1+(n+1)x^2}}$$
And you are done, by induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/581037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.