Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Calculation of $\int_{0}^{\pi}\frac{1}{(5+4\cos x)^2}dx$ Calculation of $\displaystyle \int_{0}^{\pi}\frac{1}{(5+4\cos x)^2}dx$
$\bf{My\; Try}::$ Using $\displaystyle \cos x = \frac{1-\tan^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}$
Let $\displaystyle I = \int_{0}^{\pi}\frac{1}{\left(5+\frac{4-4\tan^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}\right)^2}dx = \int_{0}^{\pi}\frac{1+\tan^2 \frac{x}{2}}{\left(9+\tan^2 \frac{x}{2}\right)^2}dx$
$\displaystyle I = \int_{0}^{\pi}\frac{\sec^2 \frac{x}{2}}{\left(9+\tan^2 \frac{x}{2}\right)^2}dx$
Now Let $\tan \frac{x}{2} = t$ and $\sec^2 \frac{x}{2}dx = 2dt$
$\displaystyle I = 2\int_{0}^{\infty}\frac{1}{(9+t^2)^2dt}$
Now I did not understand how can i solve after that
Help Required
Thanks
| After setting
$$
t=\tan\frac{x}{2},
$$
we get
\begin{eqnarray}
I:&=&\int_0^\pi\frac{1}{(5+4\cos x)^2}\,dx=\int_0^\infty\frac{2}{1+t^2}\cdot\left(\frac{1+t^2}{9+t^2}\right)^2\,dt=2\int_0^\infty\frac{1+t^2}{(9+t^2)^2}\,dt\\
&=&2\int_0^\infty\frac{9+t^2-8}{(9+t^2)^2}\,dt=2\int_0^\infty\frac{1}{9+t^2}\,dt-16\int_0^\infty\frac{1}{(9+t^2)^2}\,dt\\
&=&\frac23\arctan\frac{t}{9}\Big|_0^\infty-16\int_0^\infty\frac{1}{(9+t^2)^2}\,dt=\frac\pi3-16\int_0^\infty\frac{1}{(9+t^2)^2}\,dt\\
&=&\frac\pi3-8\int_{-\infty}^\infty\frac{1}{(9+x^2)^2}\,dx=:\frac\pi3-8J.
\end{eqnarray}
Since the function
$$
f:\mathbb{C}\setminus\{3i\} \to \mathbb{C},\ z\mapsto \frac{1}{(9+z^2)^2}
$$
is holomorphic, we have thanks to the Residue Theorem:
$$
\int_{-r}^rf(x)\,dx+ir\int_0^\pi e^{it}f(re^{it})\,dt=2\pi i\text{Res}\left(f,3i\right)=2\pi i\cdot\frac{-i}{3\cdot36}=\frac{\pi}{54} \quad \forall r>3.
$$
Since
$$
\left|ir\int_0^\pi e^{it}f(re^{it})\,dt\right|\le \int_0^\pi\frac{r}{(r^2-9)^2}\,dt=\frac{\pi r}{(r^2-9)^2} \to 0 \text{ as } r \to \infty,
$$
we have
$$
J=\lim_{r\to \infty}\int_{-r}^rf(x)\,dx=\frac{\pi}{54}.
$$
Hence
$$
I=\frac\pi3-\frac{8\pi}{54}=\frac{5\pi}{27}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/574099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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} |
find limit of $a_n=\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+...+\frac{1}{(2n)^2}$ finding limit of $a_n=\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+...+\frac{1}{(2n)^2}$
I know that i have to use Stolz Cesaro theorem, but the problem is that i need second sequence.
| $$a_n=\frac{1}{n^2}\left( \frac{1}{(1+\frac1n)^2}+\frac{1}{(1+\frac2n)^2}+...+\frac{1}{(1+\frac{n}n)^2} \right)=\frac{1}{n}\left[\frac{1}{n}\left( \frac{1}{(1+\frac1n)^2}+\frac{1}{(1+\frac2n)^2}+...+\frac{1}{(1+\frac{n}n)^2} \right)\right]$$
Now, apply S-C to
$$a_n=\frac{ \left( \frac{1}{(1+\frac1n)^2}+\frac{1}{(1+\frac2n)^2}+...+\frac{1}{(1+\frac{n}n)^2} \right) }{n^2}$$
But this is an overkill, it is easy to see that the top is $\leq 1+1+1..+1=n$.
P.S.
$$\frac{1}{n}\left( \frac{1}{(1+\frac1n)^2}+\frac{1}{(1+\frac2n)^2}+...+\frac{1}{(1+\frac{n}n)^2} \right)$$
is a Riemann sum, thus convergent, and $\frac{1}{n}$ converges to $0$.
Alternately, if you don't know integrals,apply S-C to
$$a_n=\frac{ \left( \frac{1}{(1+\frac1n)^2}+\frac{1}{(1+\frac2n)^2}+...+\frac{1}{(1+\frac{n}n)^2} \right) }{n^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/576943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
How prove this $\tan{\frac{2\pi}{13}}+4\sin{\frac{6\pi}{13}}=\sqrt{13+2\sqrt{13}}$ Nice Question:
show that: The follow nice trigonometry
$$\tan{\dfrac{2\pi}{13}}+4\sin{\dfrac{6\pi}{13}}=\sqrt{13+2\sqrt{13}}$$
This problem I have ugly solution, maybe someone have nice methods? Thank you
My ugly solution:
let $$A=\tan{\dfrac{2\pi}{13}}+4\sin{\dfrac{6\pi}{13}},B=\tan{\dfrac{4\pi}{13}}+4\sin{\dfrac{\pi}{13}}$$
since
$$\tan{w}=2[\sin{(2w)}-\sin{(4w)}+\sin{(6w)}-\sin{(8w)}+\cdots\pm \sin{(n-1)w}]$$
where $n$ is odd,and $w=\dfrac{2k\pi}{n}$
so
$$\tan{\dfrac{2\pi}{13}}=2\left(\sin{\dfrac{4\pi}{13}}-\sin{\dfrac{5\pi}{13}}+\sin{\dfrac{\pi}{13}}+\sin{\dfrac{3\pi}{13}}-\sin{\dfrac{6\pi}{13}}+\sin{\dfrac{2\pi}{13}}\right)$$
$$\tan{\dfrac{4\pi}{13}}=2\left(\sin{\dfrac{5\pi}{13}}-\sin{\dfrac{3\pi}{13}}-\sin{\dfrac{2\pi}{13}}-\sin{\dfrac{6\pi}{13}}-\sin{\dfrac{\pi}{13}}+\sin{\dfrac{4\pi}{13}}\right)$$
then
$$A^2-B^2=(A+B)(A-B)=16\left(\sin{\dfrac{\pi}{13}}+\sin{\dfrac{3\pi}{13}}+\sin{\dfrac{4\pi}{13}}\right)\left(\sin{\dfrac{2\pi}{13}}-\sin{\dfrac{5\pi}{13}}+\sin{\dfrac{6\pi}{13}}\right)=\cdots=4\sqrt{13}$$
$$AB=\cdots=6\left(\cos{\dfrac{\pi}{13}}+\cos{\dfrac{2\pi}{13}}+\cos{\dfrac{3\pi}{13}}-\cos{\dfrac{4\pi}{13}}-\cos{\dfrac{5\pi}{13}}+\cos{\dfrac{6\pi}{13}}\right)=\cdots=3\sqrt{3}$$
so
$$A=\sqrt{13+2\sqrt{13}},B=\sqrt{13-2\sqrt{13}}$$
Have other nice metods?
and I know this is simlar 1982 AMM problem: How to prove that: $\tan(3\pi/11) + 4\sin(2\pi/11) = \sqrt{11}$
But My problem is hard then AMM problem。Thank you very much!
| Straightforward for WA, not so by hand: Let $x = \exp(i \theta)$ be a primitive $13$th root of unity and $y = \tan(\theta) + 4 \sin(3\theta)$. Then $(x, y)$ is a common zero of the polynomials $$i (x^2-1) x^3 + 2i (x^6-1)(x^2+1) + y (x^2+1)x^3 \textrm{ and}\\
x^{12}+x^{11}+x^{10}+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1.$$
The resultant of these polynomials in the variable $x$ is $(y^4-26y^2+117)^3$. The roots of this resultant are the possible values of $y$ for different choices of $x$. Since $\sin(6 \pi/13)$ is close to one the value of $y$ in this specific case must be $\sqrt{13+2\sqrt{13}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/578286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 2,
"answer_id": 1
} |
Proving that $(abc)^2\geq\left(\frac{4\Delta}{\sqrt{3}}\right)^3$, where $a$, $b$, $c$ are the sides, and $\Delta$ the area, of a triangle Let $a$, $b$, $c$ be the sides of $\triangle ABC$.
Prove $$(abc)^2\geq\frac{4\Delta}{\sqrt{3}}$$
where $\Delta$ is the area of the triangle.
(Editor's note: As observed in the answers, the target relation is in error. It should be $$(abc)^2\geq\left(\frac{4\Delta}{\sqrt{3}}\right)^3$$ I incorporated the correction into the title. —@Blue)
*
*Clearly, $a$, $b$, $c$ are the sides opposite to the angles $A$, $B$, $C$
*I considered the point $F$ inside the triangle as the Fermat point of the triangle and named $FA=x$, $FB=y$, and $FC=z$.
*Then $\angle AFB = \angle AFC = \angle BFC=120^\circ$, so we have $$a^2=y^2+z^2+yz$$ and similarly for $b^2$ and $c^2$.
*Observe that
$$xy+yz+zx=\frac{4\Delta}{\sqrt{3}}$$
*So, we have a big inequality to prove:
$$\left(x^2+xy+y^2\right)\left(y^2+yz+z^2\right)\left(z^2+zx+x^2\right)\geq xy+yz+zx$$ but I can't prove it.
Thanks for help
| The actual inequality is $(abc)^2\ge \left(\dfrac{4\Delta}{\sqrt{3}}\right)^3$.I will show this one.
*
*Lemma:
$\displaystyle \sin{\alpha}+\sin{\beta}+\sin{\gamma} \le \frac{3\sqrt{3}}{2}$
when $\alpha$,$\beta$,$\gamma$ are angles of a triangle.Now we see
that $\sin x$ is concave in $(0,\pi)$ so applying Jensen's inequality
we get $\dfrac{\sin{\alpha}+\sin{\beta}+\sin{\gamma}}{3}\le
\sin\left({\dfrac{\alpha+\beta+\gamma}{3}}\right)=\dfrac{\sqrt{3}}{2}$
So our lemma is proved.
Now to the actual problem.$a+b+c=2R(\sin{\alpha}+\sin{\beta}+\sin{\gamma})\le 3\sqrt{3}R$ (from the lemma).
Now $$\dfrac{4\Delta}{\sqrt{3}}=\dfrac{abc}{R\sqrt{3}}\le \dfrac{3abc}{a+b+c}\le (abc)^{2/3}$$ where in the last step we have used AM-GM.Now cubing both sides we get the desired inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/578876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\int\frac{\cot{x}}{1+\sin{x}+\cos{x}} \mathrm dx$ Find this integral:
$$\int\dfrac{\cot{x}}{1+\sin{x}+\cos{x}}\mathrm dx$$
My try: since
$$1+\sin{x}+\cos{x}=2\cos^2{\dfrac{x}{2}}+2\sin{\dfrac{x}{2}}\cos{\dfrac{x}{2}}$$
$$\cot{x}=\dfrac{1-\tan^2{\dfrac{x}{2}}}{2\tan{\dfrac{x}{2}}}$$
so
$$\dfrac{\cot{x}}{1+\sin{x}+\cos{x}}=\dfrac{1-\tan^2{\dfrac{x}{2}}}{2\tan{\dfrac{x}{2}}\left(2\cos^2{\dfrac{x}{2}}+2\sin{\dfrac{x}{2}}\cos{\dfrac{x}{2}}\right)}$$
then I fell very ugly.Thank you
| Let
$$I=\int \frac{\cot x}{1+\sin x+\cos x} \operatorname{d}x$$
Substitute $t = \tan\left(\frac{x}{2}\right)$ and $\operatorname{d}t = \frac{1}{2} \sec^2\left(\frac{x}{2}\right) \operatorname{d}x$, and transform the integrand using the substitutions $\sin x = \frac{2 t}{t^2+1}, \cos x = \frac{1-t^2}{t^2+1}$ and $\operatorname{d}x = \frac{2 }{t^2+1}\operatorname{d}t$:
$$I= \int -\frac{t^2-1}{t(t^2+1)\left(1+\frac{2 t}{t^2+1}+\frac{1-t^2}{t^2+1}\right)} \operatorname{d}t= \int \frac{1}{2}\left(\frac{1}{t}-1\right) \operatorname{d}t=\frac{1}{2}(\ln t+t)+C$$
Substitute back for $t = \tan\left(\tfrac{x}{2}\right)$ and finally
$$
I = \frac{1}{2}\left[\ln\left(\tan\left(\tfrac{x}{2}\right)\right)-\tan\left(\tfrac{x}{2}\right)\right]+C.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/580331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots$ Question is to Evaluate :
$$\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots$$
what all i could do is :
$$\frac{1}{3}+\frac{1}{4}\frac{1}{2!}+\frac{1}{5}\frac{1}{3!}+\dots=\sum_{n=1}^{\infty} \frac{1}{(n+2)n!}=\sum_{n=1}^{\infty} \frac{n+1}{(n+2)!}=\sum_{n=1}^{\infty} \frac{n}{(n+2)!}+\sum_{n=1}^{\infty} \frac{1}{(n+2)!}$$
I have $$\sum_{n=1}^{\infty} \frac{1}{(n+2)!}=\sum_{n=0}^{\infty} \frac{1}{n!}-1-\frac{1}{2}=e-\frac{3}{2}$$
Now, I am not able to see what $$\sum_{n=1}^{\infty} \frac{n}{(n+2)!}$$ would be.
I would be thankful if some one can help me to clear this.
Thank you.
| Since $\displaystyle e^x=\sum_{n=0}^\infty \frac {x^n}{n!}$, then $\displaystyle e^x-1-x=\sum_{n=0}^\infty\frac{x^{n+2}}{(n+2)!}$, and
$$ \frac{e^x-1-x}{x}=\sum_{n=0}^\infty\frac{x^{n+1}}{(n+2)!}, $$
so
$$ \left(\frac{e^x-1-x}{x}\right)'=\sum_{n=0}^\infty\frac{(n+1)x^n}{(n+2)!}. $$
Evaluate at $x=1$, and subtract $1/2$ (the term corresponding to $n=0$) to find the value you are after. You obtain
$$ \frac12=1-\frac12=\left(\frac{e^x(x-1)+1}{x^2}\right)_{x=1}-\frac12=\sum_{n=1}^\infty\frac{n+1}{(n+2)!}. $$
(By the way, you have a small typo in your analysis, since $\displaystyle e=\sum_{n=0}^\infty\frac1{n!}$; on line $6$ you wrote the series starting at $n=1$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/581603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
The indefinite integration of $\frac{1}{\sqrt{n^4-1}}$ I need the indefinite integral:
$$\int\frac{1}{\sqrt{n^4-1}}dn $$
I know it has a relation with the $\tanh^{-1}$ function, but can't find a proper substitution.
| Introduce variables $c, s$ and $\theta$ such that
$$\cos\theta = c = \frac{1}{n}\quad\text{ and }\quad \sin\theta = s = \sqrt{1-c^2} = \frac{\sqrt{n^2-1}}{n}.$$
We have
$$\begin{align}
\int \frac{dn}{\sqrt{n^4-1}}
&= - \int \frac{dc}{c^2\sqrt{\frac{1}{c^4}-1}}
= - \int \frac{dc}{\sqrt{1-c^4}}\\
&= - \frac{1}{\sqrt{2}} \int \frac{dc}{\sqrt{(1-c^2)(1 - \frac12(1-c^2))}}\\
&= \frac{1}{\sqrt{2}}\int \frac{ds}{\sqrt{(1-s^2)(1-\frac12 s^2)}}\\
&= \frac{1}{\sqrt{2}}\int \frac{d\theta}{\sqrt{1-\frac12 \sin^2\theta}}\\
\end{align}$$
The integrals in last two lines are in the Jacobi's form and regular form of the incomplete elliptic integral of the first kind for modulus $m = \frac12$:
$$
\begin{align}
F(x;m) &= \int_0^x \frac{dt}{\sqrt{(1-t^2)(1-mt^2)}}\\
F(\phi\mid m) &= \int_0^\phi \frac{d\theta}{\sqrt{1-m\sin^2\theta}} = F(\sin\phi;m)
\end{align}
$$
As a result
$$\int_1^x \frac{dn}{\sqrt{n^4-1}}
= \frac{1}{\sqrt{2}}F\left(\frac{\sqrt{x^2-1}}{x};\frac12\right)
= \frac{1}{\sqrt{2}}F\left(\cos^{-1}\frac{1}{x}\bigg|\frac12\right)
$$
On WolframAlpha, the regular form of the incomplete elliptic integral can be accessed with the function EllipticF. The integral can be evaluated using following expression (for real $x > 1$):
$$\bf 1/\text{Sqrt}[2]*\text{EllipticF}[\text{ArcCos}[1/x],1/2]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/583604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Please help me with this inequality $a,b,c > 0$ (no other conditions)
$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq\sqrt{3\left(a^2+b^2+c^2\right)}$
I tried this:
$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq\frac{\left (a+b+c\right)^2}{a+b+c}=a+b+c$
then
$a+b+c\geq\sqrt{3\left(a^2+b^2+c^2\right)}$ which is not correct.
| if $$a^2\to a,b^2\to b,c^2\to c$$
then $a+b+c=3$,we have
$$\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{c}}+\dfrac{c}{\sqrt{a}}\ge 3$$
so you can see this Inequality. $\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}+\frac{c}{\sqrt{a}}\geq3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/584881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
$\left(\frac1\alpha-\frac1\beta\right)^2$ for $p(x)=x^2+x-2$ If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $p(x)=x^2+x-2$,
then $\left(\frac1\alpha-\frac1\beta\right)^2 is:$
A) $\frac94$
B) $\frac{-9}4$
C) $\frac25$
D) $\frac{-2}5$
This is a homework question from our school's home assignment for class 10. I tried this:
We can write $p(x)=x^2+x-2$ as $$p(x)=\underbrace{(1)x^2}_{ax^2}
+\underbrace{(1)x}_{bx}+\underbrace{(2)}_c$$ Therefore, $$a=1\quad
b=1\quad c=2$$ We know that $$\alpha\beta=\frac ca=2$$ and
$$\alpha+\beta=\frac{-b}{a}=-1$$ Now $$\left(\frac1\alpha-\frac1
\beta\right)^2=\left(\frac{\alpha-\beta}{\alpha\beta}\right)^2=
\frac{(\alpha-\beta)^2}{(\alpha\beta)^2}=\frac{\alpha^2+\beta^2-2
\alpha\beta}4=\frac{\alpha^2+\beta^2-4}4$$
I got stuck here. What to do now?
| $(x+\frac{1}{2})^2-\frac{1}{4} -2 =0 \iff (x+\frac{1}{2})^2=\frac{9}{4} \iff x=1$ or $x=-2$
Since we know that $\alpha$ and $\beta$ are the roots of the polynomial, it must hold that $\alpha =1$ and $\beta=-2$ or $\alpha=-2$ and $\beta=1$
Lets first consider the case in which $\alpha =1$ and $\beta=-2. \Rightarrow (\frac{1}{\alpha}-\frac{1}{\beta})^2=(1+\frac{1}{2})^2=\frac{9}{4}$
In the case in which $\alpha=-2$ and $\beta=1 \Rightarrow (\frac{1}{\alpha}-\frac{1}{\beta})^2=(-\frac{1}{2}-1)^2=\frac{9}{4}$, so we arrive at the same result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/587915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Minimum value of $\left|z^2-z+1\right|+\left|z^2+z+1\right|$ for $z\in \mathbb{C}$ (1) If $\left|z\right| = 1$. Then find minimum value of $\left|z^2+z+4\right|$
(2) If $z\in \mathbb{C}.$ Then minimum value of $\left|z^2-z+1\right|+\left|z^2+z+1\right|.$
$\bf{My\; Try}::$ (1) Given $\left|z\right| = 1\Rightarrow z \bar{z} = 1$.
So $\left|z^2+z+4z\bar{z}\right| = |z|\cdot \left|z+4\bar{z}+1\right| = \left|z+4\bar{z}+1\right|$
Now Let $z = x+iy$. Then $\bar{z} = x-iy$ and $|z| =1\Rightarrow x^2+y^2 = 1$
So $\left|x+iy+4x-4iy+1\right| = \left|5x+1-3iy\right| = \sqrt{(5x+1)^2+9y^2}$
So Let $f(x) = \sqrt{(5x+1)^2+9(1-x^2)} = \sqrt{25x^2+1+10x+9-9x^2}$
So $\displaystyle f(x) = \sqrt{16x^2+10x+10}=4\sqrt{x^2+\frac{5}{8}x+\frac{5}{8}}$
So $\displaystyle f(x) = 4\sqrt{\left(x+\frac{5}{16}\right)^2+\left(\frac{5}{8}-\frac{25}{256}\right)}\geq 4\sqrt{\frac{27 \times 5}{256}} = 4\times \frac{3\sqrt{15}}{16} = \frac{3\sqrt{15}}{4}$
which is occur at $\displaystyle x = -\frac{5}{16}$
(2) Now I did not understand how can i solve (II) one
Help Required
Thanks
| Let $z=x+yi\ (x,y\in\mathbb R)$.
Let us consider the case $|z|=r$ where $r$ is a fixed non-negative real number.
We have
$$\small\begin{align}&\left|z^2-z+1\right|+\left|z^2+z+1\right|
\\\\&=|(x+yi)^2-(x+yi)+1|+|(x+yi)^2+x+yi+1|
\\\\&=\sqrt{(x^2-y^2-x+1)^2+(2xy-y)^2}+\sqrt{(x^2-y^2+x+1)^2+(2xy+y)^2}
\\\\& \stackrel{y^2=r^2-x^2}=\sqrt{4x^2+(-2r^2-2)x+r^4-r^2+1}+\sqrt{4x^2+2(r^2+1)x+r^4-r^2+1}
\\\\&=\sqrt{\bigg(2x-\frac{r^2+1}{2}\bigg)^2+\bigg(0-\frac{(r^2-1)\sqrt 3}{2}\bigg)^2}+\sqrt{\bigg(2x-\frac{-r^2-1}{2}\bigg)^2+\bigg(0-\frac{(r^2-1)\sqrt 3}{2}\bigg)^2}\end{align}$$
This can be seen as the sum of the distance between $A$ and $B$ and the distance between $A$ and $C$ where
$$A(2x,0),\quad B\bigg(\frac{r^2+1}{2},\frac{(r^2-1)\sqrt 3}{2}\bigg),\quad C\bigg(\frac{-r^2-1}{2},\frac{(r^2-1)\sqrt 3}{2}\bigg)$$
For any fixed $r$, the minimum value of the sum is attained when $A$ is on the line passing through $B$ and $$D\bigg(\frac{-r^2-1}{2},\color{red}{-}\frac{(r^2-1)\sqrt 3}{2}\bigg)$$
The equation of the line $BD$ is given by
$$y-\frac{(r^2-1)\sqrt 3}{2}=\frac{(r^2-1)\sqrt 3}{r^2+1}\bigg(x-\frac{r^2+1}{2}\bigg)$$
So, if $A$ is on this line, then we get
$$0-\frac{(r^2-1)\sqrt 3}{2}=\frac{(r^2-1)\sqrt 3}{r^2+1}\bigg(2x-\frac{r^2+1}{2}\bigg)$$
from which $x=0$ follows.
So, all we need is to consider the case $x=0$.
For $x=0$, we get
$$|z^2-z+1|+|z^2+z+1|=\sqrt{(2y^2-1)^2+3}\ge \sqrt 3$$
Therefore, the minimum value of $|z^2-z+1|+|z^2+z+1|$ is $\color{red}{\sqrt 3}$ which is attained when $z=\pm\frac{i}{\sqrt 2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that Q($\sqrt{2}$, $\sqrt{3}$) is a field Prove that $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \{a+b\sqrt{2} +c\sqrt{3} +d\sqrt{6}\ |\ a,b,c,d \in \mathbb{Q}\}$ is a field.
I am doing the subfield test, but having trouble in showing how to express the inverse in such a form. Anyone can help?
| HINT: $$(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})(a-b\sqrt{2}-c\sqrt{3}+d\sqrt{6})=(a+d\sqrt{6})^2-(b\sqrt{2}+c\sqrt{3})^2:=r+s\sqrt{6}$$ where $r,s\in\mathbb{Q}$. Then $$(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})(a-b\sqrt{2}-c\sqrt{3}+d\sqrt{6})(r-s\sqrt{6})=r^2-6s^2=\alpha$$ and $\alpha$ is rational. Hence you have got the inverse.
| {
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If $3^n+81$ is a perfect square, then positive integer value $n$ is If $3^n+81$ is a perfect square, Then calculation of a positive integer value of $n$.
$\bf{My\; Try}::$ When $n≤4,$ then easy to know that $3^n+81$ is not a perfect square.
Now let $n=k+4(k∈Z^{+}),$ then $3^{n}+81=81(3^{k}+1).$
So $3^{n}+81$ is a perfect square, and $81$ is square,
there must be a positive integer $x$, such that $3^{k}+1=x^2⇒3^k=(x−1)⋅(x+1)$
Means $(x+1)$ and $(x-1)$ must be a power of $3$ form
Now I did not understand how can i solve after that
Help Required
Thanks.
| Note that if $3^n+81 = x^2$, we have
$$(x+9)(x-9) = 3^n$$
Hence, we have
$x+9 = 3^m$ and $x-9 = 3^{n-m}$. Hence, we need two powers of $3$ that differ by $18$, i.e., we need $3^m - 3^{n-m} = 18$.
Now observe that $3^m-3^{m-1} \geq 18$, if $m\geq 3$. Hence, we have limited options for $m$ and $n-m$.
Hence, $m=3$ and $n-m=2$ and we are done.
| {
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How to find the minimum of $a+b+\sqrt{a^2+b^2}$ let $a,b>0$, and such
$$\dfrac{2}{a}+\dfrac{1}{b}=1$$
Find this minimum
$$a+b+\sqrt{a^2+b^2}$$
My try: since
$$2b+a=ab$$
so
$$a+b+\sqrt{a^2+b^2}=\sqrt{a^2+b^2+2ab}+\sqrt{a^2+b^2}=\sqrt{a^2+b^2+4b+2a}+\sqrt{a^2+b^2}$$
then I can't
maybe this problem can use AM-GM or Cauchy-Schwarz inequality solve it.Thank you very much
| since
$$\dfrac{1}{a}+\dfrac{2}{b}=1$$
then a straight line
$$\dfrac{x}{a}+\dfrac{y}{b}=1$$ cross $P(1,2)$
then
$$a+b+\sqrt{a^2+b^2}=|OA|+|OB|+|AB|$$
In the follow
we have
$$|OC|+|OD|+|CD|\ge|OE|+|OF|+|EF|=|OA|+|OB|+AB|=10$$
| {
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Triangle and Maxium value Given any triangle ABC with $a \ge b \ge c$ such that $\frac{a^3+b^3+c^3}{\sin^3(A)+\sin^3(B)+\sin^3(C)}=7$, what is the maximum value of $a$?
| Using the Sine Law, i.e.
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k $$
we find that the condition in the question is merely equivalent to $k^3 = 7$.
So $a = k \sin A \le k = \sqrt[3]7$.
| {
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How find this integral $\int\frac{1}{1+\sqrt{x}+\sqrt{x+1}}dx$ Question:
Find the integral
$$I=\int\dfrac{1}{1+\sqrt{x}+\sqrt{x+1}}dx$$
my solution:
let $\sqrt{x}+\sqrt{x+1}=t\tag{1}$
then
$$t(\sqrt{x+1}-\sqrt{x})=1$$
$$\Longrightarrow \sqrt{x+1}-\sqrt{x}=\dfrac{1}{t}\tag{2}$$
$(1)-(2)$ we have
$$2\sqrt{x}=t-\dfrac{1}{t}\Longrightarrow x=\dfrac{1}{4}(t-\dfrac{1}{t})^2$$
so
$$dx=\dfrac{1}{2}(t-\dfrac{1}{t})(1+\dfrac{1}{t^2})dt=\dfrac{t^4-1}{2t^3}dt$$
$$I=\int\dfrac{1}{1+t}\cdot\dfrac{t^4-1}{2t^3}dt=\dfrac{1}{2}\int\left(1+\dfrac{1}{t}+\dfrac{1}{t^2}+\dfrac{1}{t^3}\right)dt=\dfrac{1}{2}\left(t+\ln{t}-\dfrac{1}{t}-\dfrac{1}{2t^2}+C\right)$$
so
$$I=\dfrac{1}{2}\left(\sqrt{x}+\sqrt{x+1}+\ln{(\sqrt{x}+\sqrt{x+1})}-\dfrac{1}{\sqrt{x}+\sqrt{x+1}}-\dfrac{1}{2(\sqrt{x}+\sqrt{x+1})^2}+C\right)$$
My question: have other methods? Thank you very much
| Using $\sqrt{x}=u=\tan(\theta)$ and $v=\sin(\theta)$,
$$
\begin{align}
&\int\frac1{1+\sqrt{x}+\sqrt{x+1}}\,\mathrm{d}x\\
&=\int\frac{1+\sqrt{x}-\sqrt{x+1}}{2\sqrt{x}}\,\mathrm{d}x\\
&=\sqrt{x}+\frac x2-\int\frac{\sqrt{x+1}}{2\sqrt{x}}\,\mathrm{d}x\\
&=\sqrt{x}+\frac x2-\int\sqrt{u^2+1}\,\mathrm{d}u\\
&=\sqrt{x}+\frac x2-u\sqrt{u^2+1}+\int\frac{u^2}{\sqrt{u^2+1}}\,\mathrm{d}u\\
&=\sqrt{x}+\frac x2-\sqrt{x^2+x}+\int\tan^2(\theta)\sec(\theta)\,\mathrm{d}\theta\\
&=\sqrt{x}+\frac x2-\sqrt{x^2+x}+\int\frac{\sin^2(\theta)}{\cos^4(\theta)}\,\mathrm{d}\sin(\theta)\\
&=\sqrt{x}+\frac x2-\sqrt{x^2+x}+\int\frac{v^2}{1-2v^2+v^4}\,\mathrm{d}v\\
&=\sqrt{x}+\frac x2-\sqrt{x^2+x}+\frac14\int\left(\frac1{(1-v)^2}+\frac1{(1+v)^2}-\frac1{1-v}-\frac1{1+v}\right)\,\mathrm{d}v\\
&=\sqrt{x}+\frac x2-\sqrt{x^2+x}-\frac14\log\left(\frac{1+v}{1-v}\right)+\frac14\frac1{1-v}-\frac14\frac1{1+v}+C\\
&=\sqrt{x}+\frac x2-\sqrt{x^2+x}-\frac12\log\left(\frac{1+\sin(\theta)}{\cos(\theta)}\right)+\frac12\frac{\sin(\theta)}{\cos^2(\theta)}+C\\
&=\sqrt{x}+\frac x2-\frac12\sqrt{x^2+x}-\frac12\log(\sqrt{x}+\sqrt{x+1})+C
\end{align}
$$
since $\tan(\theta)=\sqrt{x}$ and $\sec(\theta)=\sqrt{x+1}$
| {
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Point on the graph of $y=\sqrt{4x+13}$ closest to $(5,0)$? Just did this question on an exam earlier today, I'm curious to see if I'm correct.
What point on the graph of $y=\sqrt{4x+13}$ is closest to $(5,0)$?
My answer: $(-1,3)$
| Your method described in the comments seems correct. You start by noting that the distance between $(x,y(x))$ and $(5,0)$ is given by
$$d(x) = \sqrt{(5 - x)^2 + (0 - y(x))^2} = \sqrt{(5 - x)^2 + 4x + 13} = \sqrt{x^2 - 6x + 38}. \tag{1}$$
You want to minimize $d(x)$, so indeed you want to look at the derivative:
$$d'(x) = \frac{1}{2 \sqrt{x^2 - 6x + 38}} \cdot (2x - 6) = \frac{x - 3}{\sqrt{x^2 - 6x + 38}}$$
Solving $d'(x) = 0$ for $x$ we find $x = 3$, and this is indeed a global minimum; this could be proved by showing that $d''(3) > 0$.
So in the end we find a minimum at $(3, \sqrt{4\cdot3+13}) = (3,5)$, and this point has distance $d(x) = \sqrt{3^2 - 6\cdot 3 + 38} = \sqrt{29}$ from $(5,0)$.
In this case you can take a shortcut from $(1)$ to the solution as noted by Arthur, by noting that
$$d(x)^2 = x^2 - 6x + 38 = (x - 3)^2 + 29 \geq 29$$
where the inequality follows from $(x-3)^2 \geq 0$, and equality holds only if $x - 3 = 0$, i.e. if $x = 3$.
| {
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Find value of integral: $I=\int_0^{2\pi}\frac{dx}{(2+\cos x)^2}$ Find value of integral: $$I_1=\int_0^{2\pi}\frac{dx}{(2+\cos x)^2}$$ and $$I_2=\int_0^{2\pi}\frac{dx}{(2+\sin x)^2}$$
I don't know how, i need a solution, please
| Let $\displaystyle I = \frac{\sin x}{(2+\cos x)}$
Now Diff. both side w.r. to $x$ , $\displaystyle \frac{dI}{dx} = \frac{d}{dx}\left(\frac{\sin x}{2+\cos x}\right) = \frac{(2+\cos x)\cdot \cos x-\sin x\cdot (-\sin x)}{(2+\cos x)^2}$
$\displaystyle \frac{dI}{dx} = \frac{2\cos x+1}{(2+\cos x)^2}\Rightarrow \frac{dI}{dx} = \frac{2\cdot \left(2+\cos x\right)-3}{(2+\cos x)^2} = \frac{2}{2+\cos x}-3\cdot \frac{1}{(2+\cos x)^2}$
Now Integrate both side w. r. to $x$
$\displaystyle \int \frac{dI}{dx}dx = 2\int\frac{1}{(2+\cos x)}dx - 3\int\frac{1}{(2+\cos x)^2}dx$
So $\displaystyle \int\frac{1}{(2+\cos x)^2}dx = \frac{2}{3}\int\frac{1}{(2+\cos x)}dx-\frac{1}{3}\cdot I$
Now Put $\displaystyle \cos x = \frac{1-\tan^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}$
$\displaystyle \int\frac{1}{(2+\cos x)^2}dx = \frac{2}{3}\int\frac{1+\tan^2 \frac{x}{2}}{2+2\tan^2 \frac{x}{2}+1-\tan^2 \frac{x}{2}}dx-\frac{1}{3}\cdot \frac{\sin x}{2+\cos x}+\mathbb{C}$
$\displaystyle = \frac{2}{3}\int\frac{\sec^2 \frac{x}{2}}{3+\tan^2 \frac{x}{2}}dx-\frac{1}{3}\cdot \frac{\sin x}{2+\cos x}+\mathbb{C}$
Let $\displaystyle \tan \frac{x}{2} = t$ and $\sec^2\frac{x}{2}dx = 2dt$
$\displaystyle = \frac{4}{3}\int \frac{1}{t^2+\left(\sqrt{3}\right)^2}dt-\frac{1}{3}\cdot \frac{\sin x}{2+\cos x}+\mathbb{C}$
$\displaystyle \int\frac{1}{(2+\cos x)^2}dx = \frac{4}{3}\cdot \frac{1}{\sqrt{3}}\cdot \tan^{-1}\left(\frac{\tan \frac{x}{2}}{\sqrt{3}}\right)-\frac{1}{3}\cdot \frac{\sin x}{2+\cos x}+\mathbb{C}$
Yes Mrnhan $\displaystyle \tan \frac{x}{2}$ is not defined at $\displaystyle x = \pi$
Thanks Alraxite.
| {
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All solutions of $a+b+c=abc$ in natural numbers I was observing some nice examples of equalities containing the numbers $1,2,3$ like $\tan^{-1}1+\tan^{-1}2+\tan^{-1}3=\pi$ and $\log 1+\log 2+ \log 3=\log (1+2+3)$. I found out this only happens because $1+2+3=1*2*3=6$. I wanted to find other examples in small numbers, but I failed. How can we find all of the solutions of $a+b+c=abc$ in natural numbers?The question seemed easy, but it seems difficult to find. I would prefer an elementary way to find them! What I did: We know if $a+b+c=abc$, $a|a+b+c$ so $a|b+c$. Similarly, $b|a+c$ and $c|a+b$. Other than that, if we multiply both sides by $b$, we get $b^2+1=(bc-1)(ab-1)$. If we also divide both sides by $abc$, we get $\frac{1}{bc}+\frac{1}{ac}+\frac{1}{ab}=1$. I don't know how to go further using any of these, but I think they are a good start. I would appreciate any help.
| If $a=0$ then you require $b+c=0$ and hence $b=c=0$.
Note that you can assume $a\leq b \leq c$. If $a, b, c \geq 2$ then $abc \geq 4c > c + b + a$. Hence at least one of $a,b,c$ is equal to $1$.
Wlog assume $a=1$, and look for solutions to $b+c+1 = bc$. If $b,c\geq 3$ then $bc \geq 3c > b + c + 1$, hence at least one of $b,c$ is less than $3$
Wlog assume $b=2$, and look for solutions to $c+3 = 2c$, which implies $c=3$.
So the only solutions are $(0,0,0)$ and $(1,2,3)$ and their permutations.
| {
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if $3^3 2^2 \ | a^2$ then $3^2 2 \ | a $ where a is integer if $3^3 2^2 \ | a^2$ then $3^2 2 \ |a $ where a is integer.
I just cannot see it. please explain this trivial remark.
| Here is a Bezout's Identity approach.
Suppose $9\nmid a$. Then $\gcd(a,9)\mid3$. Thus, there exist $x,y$ so that
$$
ax+9y=3\tag{1}
$$
Then
$$
a^2x^2+27\left(2y-3y^2\right)=9\tag{2}
$$
Thus, $\gcd\!\left(a^2,27\right)\mid9$. Then $27\nmid a^2$. By contraposition, we have
$$
27\mid a^2\implies9\mid a\tag{3}
$$
Suppose $2\nmid a$. Then $\gcd(a,2)=1$. Thus, there exists $x,y$ so that
$$
ax+2y=1\tag{4}
$$
Then
$$
a^2x^2+4\left(y-y^2\right)=1\tag{5}
$$
Thus, $\gcd\!\left(a^2,4\right)=1$. Then $4\nmid a^2$. By contraposition, we have
$$
4\mid a^2\implies2\mid a\tag{6}
$$
$$
\frac a{18}=\frac a2-4\frac a9\in\mathbb{Z}\tag{7}
$$
Therefore, $(3)$, $(6)$, and $(7)$ imply that if $2^2\cdot3^3\mid a^2$, then $2\cdot3^2\mid a$.
| {
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Real roots of the equation $1+\sum_{r=1}^{7}\frac{x^{r}}{r} = 0$ The number of real roots of the equation $\displaystyle 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}+\frac{x^6}{6}+\frac{x^7}{7} = 0$
$\bf{My\; Try}::$ Let $\displaystyle f(x) = 1+\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}+\frac{x^6}{6}+\frac{x^7}{7}$
Now $\displaystyle f^{'}(x) = 1+x+x^2+x^3+x^4+x^5+x^6$
and $\displaystyle f^{''}(x) = 1+2x+3x^2+4x^3+5x^4+6x^5$
$\displaystyle f^{'''}(x) = 0+2+6x+12x^2+20x^3+30x^4 = 2\left(1+3x+6x^2+10x^3+15x^4\right)$
Now I did not understand how can i solve it
Help Required
Thanks
| Note that
$$f'(x)=\sum_{k=0}^6 x^k=\frac{x^7-1}{x-1}.$$
Now, it is not too hard to see that $f'(x)>0$ for all $x\in\mathbb{R}$ (consider regions $x<-1$, $-1<x<0$, $0<x<1$ and $x>1$). So, there is a real root because the degree of the polynomial is odd, but there is only one because the function is monotonically increasing.
| {
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The speed of the top of a sliding ladder
A $5$m ladder is leaning against a wall. If the bottom of the ladder is pulled along the ground away from the wall at a constant rate of $0.4$m/s, how fast will the top of the ladder be moving down the wall when its bottom is $3$m away from the wall.
Is my solution wrong? How come I can't get the correct answer?
x=horizontal axis , y=vertical axis
Area=XY/2, when X=3, A=6m^2, dx/dt=0.4m/s
Y=2A/X
dy/dx=-2A/x^2.....A=6,x=3
=-12/9
dy/dt=dy/dx * dx/dt
=-12/9 * 0.4
=-0.533333
| First we use Pythagorean Theorem
$$5^2=x^2+y^2 \implies y = \sqrt{25-x^2}.$$
Next we use that we can write $x=0.4t=\frac{2}{5}t$:
$$y=\sqrt{25-\left(\frac{2}{5}t\right)^2} = \sqrt{25-\frac{4}{25}t^2}.$$
Now we calculate the derivative (using the chain rule)
$$ \frac{dy}{dt}= \frac{\frac{d}{dt}\left(25-\frac{4}{25}t^2\right)}{2\sqrt{25-\frac{4}{25}t^2}} = \frac{-\frac{8}{25}t}{2\sqrt{25-\frac{4}{25}t^2}}.$$
This can be simplified further, but it isn't necessary. Now for the last step: evaluating the derivative at $t=7.5$ (which comes from solving $0.4t=3$).
$$\frac{dy}{dt}|_{t=7.5} =\frac{-\frac{8}{25}\cdot 7.5}{2\sqrt{25-\frac{4}{25}\cdot7.5^2}}=-0.3. $$
So the conclusion is that the speed, when the bottom is $3\text{m}$ from the wall, is $-0.3 \text{ m}/\text{s}$.
| {
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Asymptotics of ${2^n \choose n}$? How can one compute the asymptotics of ${2^n \choose n}$? I know it is bounded below and above by $\left(\frac{2^{n}}{n}\right)^n$ and $\left(\frac{2^{n}e}{n}\right)^n$.
If I plug in Stirling's approximation I get
$$\frac{2^{2^n n+n/2-1/2}}{(2^n-n)^{2^n-n+1/2} n^{n+1/2}\sqrt{\pi}}.$$
I am hoping there is a simpler asymptotic formulation and in particular I would like to compare it to $2^{n^2}$.
| Writing a product as the exponential of the sum of the logarithms is often a fruitful method.
Here we can write
$$\begin{align}
\binom{2^n}{n} &= \prod_{m=1}^n \frac{2^n - (m-1)}{m}\\
&= \frac{2^{n^2}}{n!} \prod_{k=1}^{n-1} \left(1- \frac{k}{2^n}\right)\\
&= \frac{2^{n^2}}{n!} \exp \left(\sum_{k=1}^{n-1} \log \left(1-\frac{k}{2^n} \right)\right)\\
&= \frac{2^{n^2}}{n!} \exp \left(-\frac{n(n-1)}{2^{n+1}} + O\left(\frac{n^3}{2^{2n}}\right)\right)
\end{align}$$
to obtain an expression that allows good bounds. We have the first result
$$\binom{2^n}{n}\sim \frac{2^{n^2}}{n!}$$
by recognising that $\exp \left(- \frac{n^2}{2^n}\right)$ can be reasonably approximated by $1$, and using Stirling's approximation for the factorial, we can write that as
$$\binom{2^n}{n} \sim \frac{1}{\sqrt{2\pi n}}\left(\frac{2^ne}{n}\right)^n.$$
Using some terms of the approximation of the logarithms, more precise expressions can be obtained, e.g.
$$\binom{2^n}{n} \approx \frac{2^{n^2}}{n!} \left(1 - \frac{n(n-1)}{2^{n+1}}\right)$$
by approximating $\log (1-x) \approx -x$ for small $x$.
| {
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How find this $f^{(4)}(0)$ let
$$f(x)=\dfrac{e^x}{1-\sin{x}}$$
Find the value of
$$f^{(4)}(0)=?$$
My try: let
$$\dfrac{e^x}{1-\sin{x}}=a_{0}+a_{1}x+a_{2}x^2+a_{3}x^3+a_{4}x^4+\cdots$$
so
$$e^x=(1-\sin{x})(a_{0}+a_{1}x+a_{2}x^2+a_{3}x^3+a_{4}x^4+\cdots)$$
since
$$e^x=1+x+\dfrac{x^2}{2}+\dfrac{x^3}{3}+\dfrac{x^4}{4}+\cdots$$
$$1-\sin{x}=1-x+\dfrac{1}{3!}x^3-\dfrac{1}{5!}x^5+\cdots$$
Follow I fell very ugly,maybe this problem have nice methods?
Thank you
| First, note that your $e^x$'s expansion is wrong as Shuchang pointed out.
Why don't you compare the coefficients? You'll get
$$1=a_0$$
$$1=a_1-a_0$$
$$\frac{1}{2!}=a_2-a_1$$
$$\frac{1}{3!}=a_3-a_2+\frac{a_0}{3!}$$
$$\frac{1}{4!}=a_4-a_3+\frac{a_1}{3!}$$
Then, the answer is $4!\cdot a_4$.
| {
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$a,b$ are roots of $x^2-3cx-8d = 0$ and $c,d$ are roots of $x^2-3ax-8b = 0$. Then $a+b+c+d =$ (1) If $a,b$ are the roots of the equation $x^2-10cx-11d=0$ and $c,d$ are the roots of the equation
$x^2-10ax-11b=0$. Then the value of $\displaystyle \sqrt{\frac{a+b+c+d}{10}}=,$ where $a,b,c,d$ are distinct real numbers.
(2) If $a,b,c,d$ are distinct real no. such that $a,b$ are the roots of the equation $x^2-3cx-8d = 0$
and $c,d$ are the roots of the equation $x^2-3ax-8b = 0$. Then $a+b+c+d = $
$\bf{My\; Try}::$(1) Using vieta formula
$a+b=10c......................(1)$ and $ab=-11d......................(2)$
$c+d=10a......................(3)$ and $cd=-11b......................(4)$
Now $a+b+c+d=10(a+c)..........................................(5)$
and $abcd=121bd\Rightarrow bd(ab-121)=0\Rightarrow bd=0$ or $ab=121$
Now I did not understand how can i calculate $a$ and $c$
Help Required
Thanks
| The answer for (1) is $11$.
$$abcd=121bd\Rightarrow bd(ac-121)=0\Rightarrow bd=0\ \text{or}\ ac=121.$$
(Note that you have a mistake here too.)
1) The $bd=0$ case : If $b=0$, we have $x(x-10a)=0$. This leads that $c=0$ or $d=0$. This is a contradiction. The $d=0$ case also leads a contradiction.
2) The $ac=121$ case : We have $$c=\frac{121}{a}, b=\frac{1210}{a}-a, d=10a-\frac{121}{a}.$$ Hence, we have
$$1210-a^2-11\left(10a-\frac{121}{a}\right)=0$$
$$\Rightarrow a^3-110a^2-1210a+121\times 11=0$$
$$\Rightarrow a=11, \frac{11(11\pm 3\sqrt{13})}{2}.$$
If $a=11$, then $c=11$, which is a contradiction. Hence, we have
$$(a,c)=\left(\frac{11(11\pm 3\sqrt{13})}{2},\frac{11(11\mp 3\sqrt{13})}{2}\right).$$
Hence, we have
$$\sqrt{\frac{a+b+c+d}{10}}=\sqrt{a+c}=\sqrt{121}=11.$$
I think you can get an answer for (2) in the same way as above.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to reduce the radical? I was reading a pdf on Cardano's method of solving for the roots of a cubic polynomial, when I noticed an example.
$$x^3+6x-20=0$$
On solving, I got
$$x=\sqrt[3]{10+\sqrt{108}}+\sqrt[3]{10-\sqrt{108}}$$
I went through a calculator and it gave the answer $2$.
My question is how can we prove
$$\sqrt[3]{10+\sqrt{108}}+\sqrt[3]{10-\sqrt{108}}=2$$
I tried to simplify this radical, but arrive at the same cubic polynomial. No further advancement is taking place in my solving.
Another similar interesting cubic polynomial is
$$x^3-15x-4=0$$ It has the root
$$x=\sqrt[3]{2+\sqrt{-121}}+\sqrt[3]{2-\sqrt{-121}}$$
and read that
$$\sqrt[3]{2+\sqrt{-121}}+\sqrt[3]{2-\sqrt{-121}}=4$$ I hope someone will help me.
| As $\displaystyle10+\sqrt{108}=10+6\sqrt3,$
we can write $\displaystyle10+6\sqrt3=(a+b\sqrt3)^3$ where $a,b$ are real rationals
$\displaystyle\implies10+6\sqrt3=a^3+2a^2\cdot b\sqrt3+3a(b\sqrt3)^2+(b\sqrt3)^3$
$\displaystyle\implies10+6\sqrt3= a^3+9ab^2+3(a^2b+b^3)\sqrt3$
Comparing the rational & the irrational parts $\displaystyle 10=a^3+9ab^2\ \ \ \ (1)$ and $\displaystyle a^2b+b^3=2$
$\displaystyle\implies5(a^2b+b^3)=a^3+9ab^2\iff a^3-5a^2b+9ab^2-5b^3=0$
Clearly, $a=b$ is a solution
From $\displaystyle(1),10a^3=10\iff a^3=1\iff a=1$ as $a$ is real
So, from $(1)b^2=1$ and from $(2),b^3+b-2=0$ the common root being $1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/619662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Continuous $f$ satisfying $f(2x)=f(x-1/4)+f(x+1/4)$ on $(-1/2,1/2)$ What are the continuous functions $f\colon (-\frac{1}{2},\frac{1}{2}) \to \mathbb{C}$ that satisfy the following functional equation, and how are they derived?
$$f(2x)=f(x-\frac{1}{4})+f(x+\frac{1}{4})\,\,\,\,\,\,\,\,\text{for }x \in (-\frac{1}{4},\frac{1}{4})$$
I think this could be interesting because in addition to the obvious solutions $f(x)=zx$, this is also satisfied by $f(x)=\ln (2\cos(\pi x))$.
| If $f$ is as in the problem statement, then, from the functional equation and continuity of $f$ at $x=0$, we see that
$$\lim_{x \ \to -\frac{1}{4}} f(2x) - f(x-\frac{1}{4}) = f(0).$$
Now, suppose $f_0$ is any continuous function on $(-\frac{1}{2},0]$ such that
\begin{align*} \lim_{x \ \to -\frac{1}{4}} f_0(2x) - f_0(x-\frac{1}{4}) = f_0(0). && && (*) \end{align*}
A simple sufficient condition on $f_0$ implying the above condition is $\lim_{x \to - \frac{1}{2}} f_0(x)$ exists and $f_0(0) = 0$.
I claim that any $f_0$ as above extends uniquely to an $f$ as in the problem statement.
We partition $(-\frac{1}{2},\frac{1}{2})$ into subintervals so that $f$ can be defined on one interval at a time. Put
\begin{align*}
x_{-1}=-\frac{1}{2} && x_0 = 0 && x_1 = \frac{1}{4} && x_2 = \frac{3}{8} && x_3 = \frac{7}{16} && \ldots
\end{align*}
\begin{align*}
I_0 = (x_{-1},x_0] && I_1 = (x_0, x_1] && I_2 = (x_1,x_2] && I_3 = (x_2,x_3] && \ldots
\end{align*}
Generally:
\begin{align*} x_n = \frac{1}{2} - \frac{1}{2^{n+1}} && n \geq -1 \end{align*}
\begin{align*}I_n = (x_{n-1}, x_n ] && n \geq 0. \end{align*}
Now, whenever $x \in I_n$ for $n \geq 1$, a quick calculation shows that $2x -1/2 \in I_{n-1}$ and $x - 1/2 \in I_0$. So, we get a function $f_n$ on $I_n$ for all $n \geq 0$ by the recursive definition
\begin{align*}
f_n(x) = f_{n-1}(2x -\frac{1}{2}) - f_0(x-\frac{1}{2}) && x \in I_n && n \geq 1.
\end{align*}
By construction, the function $f$ on $(-\frac{1}{2},\frac{1}{2})$ obtained by glueing all the $f_n$ together satisfies the functional equation, this recursive definition having been derived directly from the functional equation. Since $f_0$ is continuous, it is clear that each $f_n$ is continuous. It remains to see the $f_n$ agree at the endpoints so that $f$ is also continuous. We check inductively that
\begin{align*} \lim_{x \to x_n} f_{n+1}(x) = f_n(x_n) && n \geq 0. && && (**)\end{align*}
When $n=0$, this says
$$ \lim_{x \to 0} f_0(2x - \frac{1}{2}) - f_0(x-\frac{1}{2}) = f_0(0)$$
which is just a restatement of $(*)$, so the induction starts. Suppose the claim holds when for some $n \geq 0$ look at
$$ f_{n+2}(x) = f_{n+1}(2x - \frac{1}{2}) - f_0(x - \frac{1}{2})$$
with $x \in I_{n+2}$ converging to $x_{n+1}$. Then $2x - \frac{1}{2}$ is converging to $x_n$ and $x - \frac{1}{2}$ is converging to $x_{n+1} - \frac{1}{2} \in I_0$. So, by induction and continuity of $f_0$, the whole expression above is converging to
$$f_n(x_n) - f_0(x_{n+1} - \frac{1}{2}) = f_n(2x_{n+1} - \frac{1}{2}) - f_0(x_{n+1} - \frac{1}{2}) = f_{n+1}(x_{n+1})$$
and the claim follows by induction.
We have obtained the continuous $f$ extending $f_0$ as claimed and by design it is the unique extension satisfing the functional equation.
| {
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"source": "stackexchange",
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Integration of $\int \frac{dx}{a+f^2(x)}$ I want to solve a integral of the form:
$$
\int \frac{dx}{a+f^2(x)}
$$
in my particular case I got
$$
\int \frac{dx}{5+\cos^2(x)}
$$
in my case I followed this process:
$$
\int \frac{dx}{5+\cos^2(x)} \\
let \ t = tg(\frac{x}{2}) =>
dx = \frac{2dt}{1+t^2}\\
\int \frac{dx}{5+\cos^2(x)} = \int \frac{\frac{2\,dt}{1+t^2}}{5+(\frac{1-t^2}{1+t^2})^2}\\
$$
Expanding the denominator
$$
5+(\frac{1-t^2}{1+t^2})^2 = \frac{5(1+t^2)^2+(1-t^2)^2}{(1+t^2)^2}
$$
Rewriting all:
$$
\int \frac{2\,dt}{1+t^2} \frac{(1+t^2)^2}{5(1+t^2)^2+(1-t^2)^2} = \\
2 \int \frac{(1+t^2)}{5(1+t^2)^2+(1-t^2)^2}dt
$$
Now I am stuck, I can try to expand the denominator again, but then I cannot divide it for the numerator, something like this:
$$
2 \int \frac{1+t^2}{5(1+2t^2+t^4)+(1-2t^2+t^4)} \,dt = \\
2 \int \frac{1+t^2}{5+10t^2+5t^4+1-2t^2+t^4} \,dt = \\
2 \int \frac{1+t^2}{6t^4+8t^2+6} \,dt = \\
\int \frac{1+t^2}{3t^4+4t^2+3}\, dt
$$
But now I really don't know how to move next...
| Let $f$ be the function such that for all real number $x$, $f(x) = \frac{1}{5+\cos^{2}(x)}$. The primitive you want to compute is
$$ \int f(x) \, dx $$
Since $f(x) \, dx$ is invariant under the change of variables $x \rightarrow \pi+x$, the change of variables $t=\tan(x)$ might lead to more simple computations. Let's work with this change of variables. We have : $dt = (1+t^{2})dx$. So,
$$
\begin{align*}
\int \frac{1}{5+\cos^{2}(x)} \, dx &= \int \frac{1}{5 + \frac{1}{t^{2}+1}} \frac{1}{1+t^{2}} \, dt \\
&= \int \frac{1}{5(t^{2}+1)+1} \, dt \\
&= \int \frac{1}{5t^{2}+6} \, dt \\
\end{align*}
$$
And (if I'm not mistaken) :
$$ \int \frac{1}{5t^2+6} \, dt = \frac{1}{\sqrt{30}} \arctan \left( \frac{\sqrt{5}}{\sqrt{6}} t \right)$$
So, the result is :
$$ \int \frac{1}{5+\cos^{2}(x)} \, dx = \frac{1}{\sqrt{30}} \arctan \left( \frac{\sqrt{5}}{\sqrt{6}} \tan(x) \right) $$
| {
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"source": "stackexchange",
"question_score": "2",
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Prove: $xy+yz+zx\leq \frac{16}{3}$ For $x,y,z\in R$ and $x^2+xy+y^2=1$; $y^2+yz+z^2=16$
Prove: $xy+yz+zx\leq \frac{16}{3}$
| $16=[(\frac x2+y)^2+\frac 34x^2][(\frac z2+y)^2+\frac 34z^2]$
Using the Cauchy-Schwarz inequality, we get
$16\ge \frac 34[(\frac x2+y)z+x(y+\frac z2)]^2$
$\Rightarrow xy+yz+zx\le \frac 8{\sqrt 3}<\frac {16}3$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How find this sum $I_n=\sum_{k=0}^{n}\frac{H_{k+1}H_{n-k+1}}{k+2}$ $$I_n=\sum_{k=0}^{n}\dfrac{H_{k+1}H_{n-k+1}}{k+2}$$
where $$H_{n}=1+\dfrac{1}{2}+\cdots+\dfrac{1}{n}$$
my try:since
$$I_n=\dfrac{1+\dfrac{1}{2}+\cdots+\dfrac{1}{n+1}}{2}+\dfrac{\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}+\cdots+\dfrac{1}{n}\right)}{3}+\cdots+\dfrac{1+\dfrac{1}{2}+\cdots+\dfrac{1}{n+1}}{n+2}$$
$$I_n=\dfrac{1+\dfrac{1}{2}+\cdots+\dfrac{1}{n+1}}{n+2}+\dfrac{\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}+\cdots+\dfrac{1}{n}\right)}{n+1}+\cdots+\dfrac{1+\dfrac{1}{2}+\cdots+\dfrac{1}{n+1}}{2}$$
so
$$2I_n=\left(1+\dfrac{1}{2}+\cdots+\dfrac{1}{n+1}\right)\left(\dfrac{1}{2}+\dfrac{1}{n+2}\right)+\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}+\cdots+\dfrac{1}{n}\right)\left(\dfrac{1}{3}+\dfrac{1}{n+1}\right)+\cdots+\left(1+\dfrac{1}{2}+\cdots+\dfrac{1}{n+1}\right)\left(\dfrac{1}{2}+\dfrac{1}{n+2}\right)$$
Maybe this try is not usefull, so I think use other methods to solve it .
Thank you very much!
| Maybe I have got the definitive trick. I recall my previous $(5)$:
$$I_n=\frac{3}{2}\sum_{t=3}^{n+3}\frac{H_{t-1}^2-H_{t-1}^{(2)}}{t}.\tag{5}$$
Partial summation gives (I set $H_{0}^{(j)}=0$ for consistency):
$$\sum_{n=1}^{m}\frac{H_{n-1}^{(2)}}{n}=H_m H_{m-1}^{(2)}-\sum_{n=1}^{m-1}\frac{H_n}{n^2}=H_m H_m^{(2)}-\sum_{n=1}^m\frac{H_n}{n^2}=H_m H_m^{(2)}-H_{m}^{(3)}-\sum_{n=1}^{m}\frac{H_{n-1}}{n^2}.\tag{7}$$
This leads to:
$$I_n=\frac{3}{2}\sum_{t=1}^{n+3}\left(\frac{H_{t-1}^2}{t}+\frac{H_{t-1}}{t^2}\right)+\frac{3}{2}H_{n+3}^{(3)}-\frac{3}{2}H_{n+3}H_{n+3}^{(2)}.\tag{8}$$
Now we have (since $a^3-b^3=(a-b)(a^2+ab+b^2)$):
$$H_{j+1}^3-H_{j}^3 = \frac{H_{j+1}^2+H_{j+1} H_j +H_{j}^2}{j+1}=3\frac{H_j^2}{j+1}+3\frac{H_j}{(j+1)^2}+\frac{1}{(j+1)^3}.\tag{9}$$
Summing both sides of $(9)$ with $j$ that runs from $0$ to $n+2$ we have:
$$H_{n+3}^3-H_{n+3}^{(3)}=3\sum_{t=1}^{n+3}\left(\frac{H_{t-1}^2}{t}+\frac{H_{t-1}}{t^2}\right).\tag{10}$$
(An alternative proof of $(10)$, always based on partial summation, is given below by Matt Groff)
If now we simply plug $(10)$ into $(8)$ we end with:
$$I_n = \frac{1}{2}H_{n+3}^3+H_{n+3}^{(3)}-\frac{3}{2}H_{n+3}H_{n+3}^{(2)},\tag{11}$$
that is way nicer than my previous $(6)$ and perfectly answers the question.
It is worth mentioning that, due to Greg Martin proof, this gives a closed expression for the sum
$$\sum_{i+j+k\leq n}\frac{1}{ijk}$$
and for the coefficients of the Taylor series of $\log^4(1-x)$ around zero.
Many many thanks to Greg Martin and Matt Groff for this brilliant piece of cooperative mathematics - should we ask to split the bounty in three?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
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$Z^2-YZ-Y^2+X^2+2XY$ is an irreducible polynomial How to show that $Z^2-YZ-Y^2+X^2+2XY$ is an irreducible polynomial in $\Bbb{C}[X,Y,Z]$?
| Suppose that $Z^2-YZ-Y^2+X^2+2XY$ is reducible. Since this is a degree two polynomial in $Z$ with coefficients in $\Bbb C[X,Y]$ it splits into a product of two polynomials of degree one in $Z$, that is, $Z^2-YZ-Y^2+X^2+2XY=(Z+f(X,Y))(Z+g(X,Y))$. We get $f(X,Y)+g(X,Y)=-Y$ and $f(X,Y)g(X,Y)=-Y^2+X^2+2XY$. Now plug in the second relation what you get from the first and find $f(X,Y)(f(X,Y)+Y)=Y^2-2XY-X^2$. This implies that $\deg_Yf(X,Y)=1$, so $f(X,Y)=a(X)+b(X)Y$. Now just calculate what you need and get $b(X)(b(X)+1)=1$, $2a(x)b(x)+a(X)=-2X$, and $a^2(X)=-X^2$. From $b(X)(b(X)+1)=1$ we deduce that $\deg b(X)=0$ and from now on write it $b$ instead of $b(X)$. We have $b(b+1)=1$. We also get $a(X)=\frac{-2}{2b+1}X$ and plugin this into the first relation we get $4=-(2b+1)^2$, so $4=-4(b^2+b)-1$ and from $b(b+1)=1$ we get $4=-4-1$, a contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the sum of $\sum_{n=1}^\infty \frac{x^{n-1}}{3^nn}$ - What is wrong with my solution? I have to find the sum of the following power series:
NOTE: please assume that x is in the convergence domain.
$$\sum_{n=1}^\infty \frac{x^{n-1}}{3^nn}$$
My solution:
*
*$$S(x) = \sum_{n=1}^\infty \frac{x^{n-1}}{3^nn} = \sum_{n=1}^\infty\frac{1}{3^nn}x^{n-1}$$
*$$xS(x) = \sum_{n=1}^\infty \frac{1}{3^nn}x^n $$
*$$(xS(x))' = \sum_{n=1}^\infty \frac{1}{3^nn}nx^{n-1} = \sum_{n=1}^\infty \frac{1}{3^n}x^{n-1}$$
*$$x \cdot (xS(x))' = \sum_{n=1}^\infty \frac{1}{3^n}x^n = \sum_{n=1}^\infty \frac{x^n}{3^n} = \sum_{n=1}^\infty \bigg(\frac{x}{3}\bigg)^n = \frac{\frac{x}{3}}{1 - \frac{x}{3}} = \frac{x}{3}(1 - \frac{3}{x}) = \frac{x}{3} - 1.$$
*$$(xS(x))' = \frac{1}{x}(\frac{x}{3}-1) = \frac{1}{3} - \frac{1}{x}.$$
*$$xS(x) = \int{\frac{1}{3} - \frac{1}{x}} = \int\frac{1}{3} - \int\frac{1}{x} = \frac{x}{3} -\ln|x|.$$
*$$S(x) = \frac{1}{x}(\frac{x}{3}-\ln|x|) = \frac{1}{3} - \frac{\ln|x|}{x}$$
So I got that $\sum_{n=1}^\infty \frac{x^{n-1}}{3^nn} = \frac{1}{3} - \frac{\ln|x|}{x}.$
My professor claims that $\sum_{n=1}^\infty \frac{x^{n-1}}{3^nn} = -\frac{1}{x} \ln|\frac{x}{3} - 1|.$
Where I made a mistake in my solution?
Thanks in advance!
| Your mistake is in step 4, where you claim
$$\frac{\frac{x}{3}}{1-\frac{x}{3}} = \frac{x}{3}\left(1 - \frac{3}{x}\right).$$
The correct value is
$$\frac{\frac{x}{3}}{1-\frac{x}{3}} = \frac{x}{3-x},$$
which then leads to
$$\begin{gather}
(x\cdot S(x))' = \frac{1}{3-x}\\
x\cdot S(x) = -\log (3-x) + C\\
S(x) = -\frac{\log (3-x)}{x} + \frac{C}{x}.
\end{gather}$$
Looking at the series, we can determine the integration constant $C$ and obtain
$$S(x) = -\frac{1}{x}\log \left(1-\frac{x}{3}\right).$$
By the series, we have $S(0) = \frac13$, a finite value, so we must choose $C$ so that
$$\lim_{x\to 0} \frac{C - \log (3-x)}{x}$$
exists, and that means the numerator must vanish if we insert $x = 0$, so $C - \log (3-0) = 0 \iff C = \log 3$. Now we can write
$$\log 3 - \log (3-x) = \log \frac{3}{3-x} = -\log \frac{3-x}{3} = -\log \left(1-\frac{x}{3}\right).$$
| {
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"source": "stackexchange",
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Finding the limit $\displaystyle\lim_{x\to 0+} \left(\frac{\sin x}x\right)^{1/{x^2}}$
Find the following limit: $$\displaystyle\lim_{x\to 0+} \left(\frac{\sin x}x\right)^{1/{x^2}}$$
Well I tried to do the $\exp\left(\frac{ \ln\frac{\sin x}{x}}{x^2}\right)$ then apply LHR but I seem to get to endless dervivations...
There's got to be a more simple approach.
| $\require{cancel}$
$$\displaystyle\lim_{x\to 0+} \left(\frac{\sin x}x\right)^{1/{x^2}}$$
$$f(x):=\left(\frac{\sin x}x\right)^{1/{x^2}}$$
$$\ln f(x)= \dfrac{1}{x^2}\ln \left( \dfrac{\sin x}{x}-1+1\right)$$
$$\ln f(x)=\dfrac{\ln \left( \dfrac{\sin x}{x}-1+1\right)}{\dfrac{\sin x}{x}-1}\times \dfrac{\dfrac{\sin x}{x}-1}{x^2}$$
$$\ln f(x)=\dfrac{\ln \left( \dfrac{\sin x}{x}-1+1\right)}{\dfrac{\sin x}{x}-1}\times \dfrac{\sin x-x}{x^3}$$
$$\lim_{x \to 0} \ln f(x)=\lim_{x\to 0}\cancelto{1}{\dfrac{\ln \left( \dfrac{\sin x}{x}-1+1\right)}{\dfrac{\sin x}{x}-1}}\times \cancelto{\dfrac{-1}{6}}{\dfrac{\sin x-x}{x^3}}=\dfrac{-1}{6}$$
$$\lim_{x \to 0} f(x)=\text{exp}(-\dfrac{1}{6})$$
Used from
Calculate the limit : $\lim_{x \to 0}\frac{x-\sin{x}}{x^3}$ WITHOUT using L'Hopital's rule
Limit of a function without using L'Hôpital Rule
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/632527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 5
} |
Taylor's series when x goes to infinity Let $f(x) = \frac {x^3}{(x+1)^2}$. Find constants a, b, c, so that $f(x) = ax + b + \frac cx + o(\frac 1x)$ as $x$ goes to $\pm \infty$. So i know that i can't take Taylor series as $x$ goes to infinity. So i am assuming i have to make some kind of substitution. I tried making $x = \frac 1u$ but i get nowhere.
$$f(x) = \frac {x^3}{(x+1)^2} = x^3(x+1)^{-2}=\frac{1}{u^3}(1 + \frac 1u)$$
$$\lim_{u\to 0}\frac{1}{u^3}(1 + \frac 1u) = \lim_{u\to 0}\frac{1}{u^3}(1 - \frac 2u + \frac{3}{u^2} + o({u^3})) =\lim_{u\to 0} u^3 - \frac {2}{u^4} + \frac{3}{u^5} + o(\frac 1u) = \lim_{u\to 0} \frac{1}{x^3} - 2x^4 + 3x^5 + o(x^5)$$
Thanks.
| I would start off here by using long division to simplify things a bit:
$$
\frac{x^3}{(x+1)^2}=\frac{x^3}{x^2+2x+1}=\cdots=x-2+\frac{3x+2}{x^2+2x+1}.
$$
Heuristically speaking, it is clear that as $x\rightarrow\infty$, that last bit is $\frac{3}{x}+o(\frac{1}{x})$. So, this would suggested that
$$
\frac{x^3}{(x+1)^2}=x-2+\frac{3}{x}+o\left(\frac{1}{x}\right)\text{ as }x\to\infty.
$$
Having used long division / heuristics to find the answer, now it is only a matter of justifying it; and, that shouldn't be too bad.
| {
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Integrate $\int \frac{(1+x^2)dx}{(1−x^2)\sqrt{1+x^4}}$ Integrate $\displaystyle \int \frac{(1+x^2)dx}{(1−x^2)\sqrt{1+x^4}}$
I don't know how to do this one. I need some suggestions. Thank you!
| These types of integrals are best evaluated using a substitution of the form $$u = x - x^{-1}, \quad du = 1 + x^{-2} \, dx.$$ Note that $u^2 = x^2 + x^{-2} - 2$, and we can write the integrand as $$\frac{1+x^2}{(1-x^2)\sqrt{x^4+1}} dx = \frac{x^2(1+x^{-2}) \, dx}{x^2(x^{-1}-x)\sqrt{x^2+x^{-2}}} = -\frac{du}{u\sqrt{u^2+2}} = -\frac{u \, du}{u^2 \sqrt{u^2+2}}.$$ Now let $v = \sqrt{u^2+2}$, $dv = \frac{u}{\sqrt{u^2+2}} \, du$, and the given integral becomes $$\begin{align*} \int \frac{dv}{2-v^2} &= \frac{\log(v+\sqrt{2}) - \log(v-\sqrt{2})}{2\sqrt{2}} + C \\ &= \frac{1}{2\sqrt{2}} \log \left| \frac{\sqrt{x^2+x^{-2}} + \sqrt{2}}{\sqrt{x^2+x^{-2}} - \sqrt{2}} \right| + C \\ &= \frac{1}{2\sqrt{2}} \log \left| \frac{\sqrt{x^4+1}+\sqrt{2}x}{\sqrt{x^4+1}-\sqrt{2}x} \right| + C. \end{align*}$$
| {
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Find the number of roots of equation $z^5 -12z^2+14=0$ that lie in the region {$z \in \Bbb C : 2 \leq |z|< \frac {5}{2}$}
Problem: Find the number of roots of equation $z^5 -12z^2+14=0$ that lie in the region
{$z \in \Bbb C : 2 \leq |z|< \frac {5}{2}$}
Solution :
In $f(z)=z^5 -12z^2+14=0$, there is $2$ change in sign
So it has two positive real roots
Now $f(2)f(2.5) <0$
So two positive real roots lie between $2$ and $2.5$
In $-z^5 -12z^2+14=0$, there is $1$ change in sign
So it has one negative real roots (It will not lie between $2$ and $2.5$)
I have no idea of other two conjugate complex roots
Please help me in finding the answer
| The idea is to use Rouché's theorem.
On the outer circle $\lvert z\rvert = \frac{5}{2}$, we have
$$\lvert z\rvert ^5 = \frac{3125}{32} = 97 + \frac{21}{32} > 89 = 75 + 14 = 12\lvert z\rvert^2+14 \geqslant \lvert -12z^2 + 14\rvert,$$
so by Rouché's theorem, $f(z) = z^5 - 12z^2 + 14$ has as many zeros in the disk $\lvert z\rvert < \frac{5}{2}$ as $z^5$ has, counting multiplicity. So all five zeros of $f$ lie in the larger disk.
On the inner circle, we have
$$\lvert z^5+14\rvert \leqslant \lvert z\rvert^5+14 = 32+14 = 46 < 48 = 12\lvert z\rvert^2 = \lvert -12z^2\rvert,$$
so by Rouché's theorem, in the smaller disk $\lvert z\rvert < 2$, $f$ has as many zeros (counting multiplicity) as $-12z^2$ has, namely $2$.
Thus in the annulus $2 <\lvert z\rvert < \frac{5}{2}$, there are three zeros of $f$.
So two positive real roots lie between $2$ and $2.5$
That is wrong, on the interval $[2,2.5]$ the function is strictly increasing ($5x^4 - 24x > 0$ there), so $f$ has only one real root between $2$ and $2.5$.
| {
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I'd like to get explain about complex roots If $x^6+1=0$ so $x^6=-1$, then we have to find the roots at $\mathbb{C}$.
I saw that the roots are $$\Large{e^{(\frac{\pi}{6}+\frac{2k\pi}{6})i}}\;\small{k=0,1,2,3,4,5}$$
this what I understand. maybe I wrong...
My question is why we are putting the $\frac{\pi}{6}$?
Thank you!
| As you said, $x^6 = -1$. We want to find all complex solutions. You may have been exposed to complex numbers as things of the form $a + bi$, where $a,b$ are real. However, there is another form you can write them in: $re^{i\theta}$, where $r,\theta$ are real, and $r \ge 0$.
The geometric interpretation of $a + bi$ is rectangular; you go $a$ units right and $b$ units up, and there you are. But just like coordinates, you can also express it in a polar fashion: $\theta$ is what direction you point, and $r$ is how far out you go (the diagram uses $\varphi$ instead of $\theta$, but the latter is more common, as far as I know).
Why does this work? One can show, by a variety of methods, that $e^{i\theta} = \cos \theta + i \sin \theta$ (Euler's formula). This corresponds to a point on the unit circle: $\cos \theta$ to the right and $\sin \theta$ up. Scaling by $r$ gives you a vector further away, but in the same direction.
So, back to the problem. Since $x$ is complex, we can represent it as $re^{i\theta}$. The problem is now $-1 = \left( r e^{i \theta} \right)^6 = r^6 e^{i \cdot 6\theta}$. Convince yourself, geometrically or algebraically*, that $r$ must be $1$. Thus, we are looking for all $\theta$ such that $e^{i\cdot 6\theta} = -1$.
We have to use Euler's formula here: $-1 = e^{i\cdot 6\theta} = \cos 6 \theta + i \sin 6 \theta$. Clearly, $\cos 6 \theta$ is $-1$ and $\sin 6 \theta$ is $0$. Well, it seems plausible that $6 \theta = \pi$, right? So $\theta = \frac{\pi}{6}$.
But wait! We know that going around the unit circle doesn't change sine or cosine. So what if $6 \theta = \pi + 2\pi$ (i.e., $3 \pi$)? Then $\theta = \frac{\pi}{2}$. But we can keep going! We can say $6 \theta = \pi + 2 k \pi$ for any $k \in \mathbb{Z}$, so $\theta = \frac{\pi}{6} + \frac{2 k \pi}{6}$.
There are only finitely many, thankfully. As we just noted, going around the unit circle doesn't change sine or cosine. Therefore, it doesn't change $e^{i\theta}$, or our $x$. So if any of our $\theta$ differ by $2\pi$, we can toss one out, because we already counted that solution.
*
*$k = 0, \ \theta = \frac{\pi}{6}$
*$k = 1, \ \theta = \frac{3\pi}{6}$
*$k = 2, \ \theta = \frac{5\pi}{6}$
*$k = 3, \ \theta = \frac{7\pi}{6}$
*$k = 4, \ \theta = \frac{9\pi}{6}$
*$k = 5, \ \theta = \frac{11\pi}{6}$
*$k = 6, \ \theta = \frac{13\pi}{6} = \frac{\pi}{6} + 2\pi$
*$k = 7, \ \theta = \frac{15\pi}{6} = \frac{3\pi}{6} + 2\pi$
*$k = 8, \ \theta = \frac{17\pi}{6} = \frac{5\pi}{6} + 2\pi$
*$\vdots$
So, at $k = 6$, they start repeating. You can check that $k = -1$ is the same as $k = 5$, $k = -2$ the same as $k = 4$, and so on. The only ones we care about are $k = 0,1,2,3,4,5$ (can you generalize this to $n$th roots?), and so the $6$ solutions are:
$$ x = {\Large e^{i \left(\frac{\pi}{6} + \frac{2 k \pi}{6} \right)}}, k = 0,1,2,3,4,5 $$
*The geometric approach is fast: $-1$ has distance $1$ from the origin, so it must have $r = 1$. The algebraic one is slower, but more rigorous, and shows why polar form works the way it does. Note that $|e^{i\alpha}| = |\cos \alpha + i \sin \alpha| = \sqrt{\cos^2 \alpha + \sin^2 \alpha} = 1$ for any $\alpha$. So, because $-1 = r^6 e^{i \cdot 6 \theta}$, we take norms on both sides: $$|-1| = |r^6 e^{i \cdot 6 \theta}| = |r^6| |e^{i \cdot 6 \theta}| = |r^6| \cdot 1$$
And since $|-1| = 1$, $|r^6| = 1$, and from there, it is quick to conclude that $r = 1$.
| {
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different results for the solution of bessel function with exponential I have this integral
$$
\int_0^\infty e^{-\alpha x}K_1(\beta \sqrt{x}) \, dx.
$$
for $\Re[\alpha] >0$, and $\Re[\beta]>0$
According to the (Table of Integrals, Series, and Products, Seventh Edition), equation 6.614.5 the solution of the above integral is:
$$
\frac{\sqrt{\pi}\beta}{8\alpha^{\frac{3}{2}}} \exp \left( \frac{\beta^2}{8\alpha}\right)\left[ K_1 \left(\frac{\beta^2}{8\alpha} \right)- K_0 \left( \frac{\beta^2}{8\alpha}\right)\right] \,,
$$
However, when using Mathematica to solve the above equation we get:
$$ \frac{\pi}{2\beta\sqrt{\alpha}} U\left(0.5,0,\frac{\beta^2}{4\alpha}\right).$$
Why this difference?
| There is a special relation between Tricomi's U and the Bessel K functions, see e.g. http://functions.wolfram.com/07.33.03.0006.01:
$$U(a, 2 a - 1, z) = \frac{e^{\frac{z}{2}} z^{\frac{3}{2} - a}}{2(a - 1) \sqrt{\pi}}\left(K_{a - \frac{1}{2}} \left(\frac{z}{2}\right) - K_{a - \frac{3}{2}}\left(\frac{z}{2}\right)\right)$$
With your $a=1/2$ this simplifies to
$$U\left(\frac{1}{2}, 0, z\right) = -\frac{e^{\frac{z}{2}} z}{ \sqrt{\pi}}\left(K_0 \left(\frac{z}{2}\right) - K_1\left(\frac{z}{2}\right)\right)\cdot$$
Now substitute $z=\frac{\beta^2}{4\alpha}$ and multiply by
$\frac{\pi}{2\beta\sqrt{\alpha}}$ and you get
$$\frac{\pi}{2\beta\sqrt{\alpha}} U\left(\frac{1}{2}, 0, \frac{\beta^2}{4\alpha}\right) = \frac{\sqrt{\pi}\beta}{8\alpha^{\frac{3}{2}}} \exp \left( \frac{\beta^2}{8\alpha}\right)\left( K_1 \left(\frac{\beta^2}{8\alpha} \right)- K_0\left( \frac{\beta^2}{8\alpha}\right)\right)\cdot$$
Therefore there is actually no difference in the results.
| {
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Calculating Bernoulli Numbers from $\sum\limits_{n=0}^\infty\frac{B_nx^n}{n!}=\frac x{e^x-1}$ How is the Bernoulli numbers? For example, found that in internet
$$\sum_{n=0}^\infty\frac{B_nx^n}{n!}=\frac x{e^x-1}$$
but if I want to find $B_2$ then
$$B_0+B_1x+\frac{B_2x^2}{2}+\sum_{n=3}^\infty\frac{B_nx^n}{n!}=\frac x{e^x-1}$$
and I believe this is not much help.
I want to learn how to calculate Bernoulli numbers to learn how to calculate $\zeta(2n)$.
| Note that
$$\frac{e^z-1}z=\frac1 z\sum_{n=1}^\infty\frac1{n!}z^n=\sum_{n=1}^\infty\frac1{n!}z^{n-1}=\sum_{n=0}^\infty\frac1{(n+1)!}z^n$$
and we can use Mertens’ multiplication theorem to get
$$1=\left(\sum_{n=0}^\infty\frac{B_n}{n!}z^n\right)\left(\sum_{n=0}^\infty\frac1{(n+1)!}z^n\right)=\sum_{n=0}^\infty\sum_{k=0}^n\left(\frac{B_k}{k!}\frac{1}{(n-k+1)!}\right)z^n$$
By the identity theorem, the $n=0$ term on the right must equal $1$ while all other
terms must vanish. The $n=0$ term on the right is just $B_0$, so $B_0 = 1$, and for
$n > 1$, we must have $\sum_{k=0}^n\frac{B_k}{k!}\frac{1}{(n-k+1)!}=0$. Multiplying this by $(n + 1)!$ we get
$$0=\sum_{k=0}^n\frac{B_k}{k!}\frac{1}{(n-k+1)!}=\sum_{k=0}^n\frac{(n+1)!}{k!(n-k+1)!}B_k=\sum_{k=0}^n\binom{n+1}kB_k$$
and adding $B_{n+1}=\binom{n+1}{n+1}B_{n+1}$ to both sides of this equation, we get
$$B_{n+1}=\sum_{k=0}^{n+1}\binom{n+1}kB_k$$
The right-hand side might look familiar from the binomial formula. Recall from
the binomial formula that for any complex number $a$, we have
$$(a+1)^{n+1}=\sum_{k=0}^{n+1}\binom{n+1}ka^k1^{n+1-k}=\sum_{k=0}^{n+1}\binom{n+1}ka^k$$
Notice that the right-hand side of this expression is exactly the right-hand side of
the previous equation if put $a = B$ and we make the superscript $k$ into a subscript
$k$. Thus, if we use the notation $\Doteq$ to mean equals after making superscripts into
subscripts, then we can write
$$\boxed{B^{n+1}\Doteq (B+1)^{n+1},n=1,2,3,...,B_0=1}$$
Use recent identity, one can in principle find all the Bernoulli numbers: When
$n = 1$, we see that
$$B^2\Doteq(B+1)^2=B^2+2B^1+1\Rightarrow0=2B_1+1\Rightarrow B_1=\frac{-1}2.$$
When $n = 2$, we see that
$$B^3\Doteq(B+1)^3=B^3+3B^2+3B^1+1\Rightarrow3B_2+3B_1+1=0\Rightarrow B_2=\frac1 6.$$
| {
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How to factorize $2x^2+5x+3$? I'm doing pre-calculus course at coursera.org and I'm in trouble with this solution $$2x^2 +5x +3 = (2x+3)(x+1)$$
By trial, using ac-method I got stuck:
$$
ac = (2)(3) = 6\\
6 + ? = 5 \Rightarrow~ ? = 5 - 6 = -1
$$
Then,
$$2x^2+6x-x+3 = 2x(x+3)-x+3$$
At this point I could not get the answer, any help?
| In general
$$ax^2+bx+c=a(x-x_1)(x-x_2)$$where $x_1,x_2$ are the roots of equation
$$ax^2+bx+c=0$$
in case $a=2,b=5,c=3$ and roots are
$$x_{1,2}=\frac{-5\pm\sqrt{5^2-4\cdot2\cdot3}}{2\cdot2}=\frac{-5\pm1}{4},x_1=-1,x_2=-\frac{3}{2}$$ and put these above we get
$$2x^2+5x+3=2(x-(-1))(x-(-\frac{3}{2}))=$$
$$=2(x+1)(x+\frac{3}{2})=(x+1)(2x+2\cdot\frac{3}{2})=(x+1)(2x+3)$$
| {
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Simplify the expression : $\tan(\theta) +2\tan(2\theta) +2^2\tan(2^2 \theta) +\cdots +2^{14} \tan(2^{14}\theta) +2^{15} \cot(2^{15} \theta)$ How to simplify the expression:
$\tan(\theta) +2\tan(2\theta) +2^2\tan(2^2 \theta) +\ldots +2^{14} \tan(2^{14}\theta) +2^{15} \cot(2^{15} \theta)$
I am not getting any clue how to proceed in such problem please suggest it will be of great help .. I got this problem from www.mathstudy.in
| HINT:
Use $$\cot A-\tan A=\frac{\cos^2A-\sin^2A}{\cos A\sin A}=2\cot2A$$ repeatedly
So, we have $$\cot\theta-\tan\theta=2\cot2 \theta$$
and $$2(\cot2\theta-\tan2\theta)=2(2\cot2^2\theta)$$
$$2^2(\cot2^2\theta-\tan2^2\theta)=2^2(2\cot2^3\theta)$$
and so on
Now add the relations.
Reference : Double-Angle Formulas
| {
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Solve $\frac3x - \frac4y = 1$ and $\frac7x + \frac2y = \frac{11}{12}$ How can we solve the following simultaneous equations:
$$\frac3x - \frac4y = 1$$
$$\frac7x + \frac2y = \frac{11}{12}$$
| $\frac{3}{x}-\frac{4}{y}=1$ ---(1)
$\frac{7}{x}+\frac{2}{y}=\frac{11}{12}$ ---(2)
2(2): $\frac{14}{x}+\frac{4}{y}=\frac{11}{6}$ ---(3)
(1)+(3): $\frac{17}{x}=\frac{17}{6}$
Hence $x=6$, and $y=-8$.
| {
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Show that the triangle which satisfy the inequality $\frac{\sin^2 A+\sin^2 B+\sin^2 C}{\cos^2 A+\cos^2 B+\cos^2 C}=2$ Show that the triangle which satisfy the inequality $\dfrac{\sin^2 A+\sin^2 B+\sin^2 C}{\cos^2 A+\cos^2 B+\cos^2 C}=2$ is right angled.
My work:
$\sin^2 A+\sin^2 B+\sin^2 C=2(\cos^2 A+\cos^2 B+\cos^2 C)$
$3(\sin^2 A+\sin^2 B+\sin^2 C)=6$
$(\sin^2 A+\sin^2 B+\sin^2 C)=2$
What to do now? Please help!
| The last equation gives: cos(2A) + cos(2B) + cos(2C) = - 1 => 2cos(A+B)cos(A-B) + 2(cos(C))^2 = 0 ==> -2cos(C)cos(A-B) + 2(cos(C))^2 = 0 ==> 2cos(C)(-cos(A-B) + cos(C)) = 0. So cos(C) = 0 gives C = pi/2 or cos(A-B) = cos(C). So A-B = C or -C and this means A = B + C or B = A + C so 2A = pi so A =pi/2 or 2B = pi so B = pi/2. So either way A or B or C = pi/2. ABC is a right triangle.
| {
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Are those rings fields? Let $f(x) = x^4+x^2+1 \in Z_{2}$ and $A = Z_{2}/f(x)$
*
*Is $A$ a field?
Let $f(x) = x^4-x^2+1 \in Z_{7}$ and $B = Z_{7}/f(x)$
*
*Is $B$ a field?
I know that a ring $A = Z_{n}/f(x)$ is a field $\Leftrightarrow$ $f(x)$ has no roots in $Z_{n}$ . But this happens only for 2nd-degree and 3rd-degree polynoms
What if you have a 4th-degree polynom?
| Good luck allowed this argument in answer to the second question.
In characteristic $7$, we get $(a+b)^6=a^6-a^5b+a^4b^2-a^3b^3+a^2b^4-ab^5+b^6$. In $\mathbb F_7[x]$, $(x^2+1)^6=x^{12}-x^{10}+x^8-x^6+x^4-x^2+1$, which is clearly congruent to $1$ modulo your polynomial $x^4-x^2+1$. Thus, in $B$, $x^2+1$ is a sixth root of unity. But all the six nonzero elements of $\mathbb F_7$ are sixth roots of unity, and we have just found a seventh. So $B$ is not a field.
| {
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How prove this $f(n)\le f(n+1)$ where $f(n)=\sum_{k=1}^{n}\frac{n}{n^2+k^2}$ let $$f(n)=\sum_{k=1}^{n}\dfrac{n}{n^2+k^2}$$
prove or disprove
$$f(n)\le f(n+1)$$
this inequality is found when I deal this follow limit:
$$\lim_{n\to\infty}f(n)=\lim_{n\to\infty}\dfrac{1}{n}\sum_{k=1}^{n}\dfrac{1}{1+(k/n)^2}=\int_{0}^{1}\dfrac{1}{1+x^2}dx=\dfrac{\pi}{4}$$
But I can't prove
$$f(n)\le f(n+1)$$
since
$$f(n+1)=\dfrac{n+1}{(n+1)^2+1}+\dfrac{n+1}{(n+1)^2+2^2}+\cdots+\dfrac{n+1}{(n+1)^2+n^2}+\dfrac{n+1}{(n+1)^2+(n+1)^2}$$
$$f(n)=\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2^2}+\cdots+\dfrac{n}{n^2+n^2}$$
so
$$f(n+1)-f(n)=\left(\dfrac{n+1}{(n+1)^2+1}-\dfrac{n}{n^2+1}\right)+\left(\dfrac{n+1}{(n+1)^2+2^2}-\dfrac{n}{n^2+2^2}\right)+\cdots+\left(\dfrac{n+1}{(n+1)^2+n^2}-\dfrac{n}{n^2+n^2}\right)+\dfrac{1}{2(n+1)}$$
so
$$f(n+1)-f(n)=\sum_{k=1}^{n}\dfrac{k^2-n^2-n}{(k^2+n^2)((n+1)^2+k^2)}+\dfrac{1}{2(n+1)}$$
This problem is my found it,can you help to solve this problem?
| This answer is essentially the same approach as vesszabo's; I include it just to give another perspective. Set $g(x)=1/(1+x^2)$. Consider the trapezoidal approximation
$$\int_0^1 g(x) dx \approx \frac{1}{n} \left( \frac{g(0)+g(1/n)}{2} + \frac{g(1/n)+g(2/n)}{n} + \cdots + \frac{g((n-1)/n)+ g(1)}{2} \right)$$
$$=\frac{1}{2n} + f(n) - \frac{1}{4n}.$$
It is well known that the error in the trapezoidal approximation is $-(1-0)^3 M/(12 n^2)$ where $M = g''(x)$ for some $x \in [0,1]$. NEXT PART IS UPDATED, thanks to zyx for pointing out the error. The most positive value of $g''$ is $1/2$, at $x=1$, and the most negative is $-2$ at $x=0$.
So
$$ \int_0^1 g(x) dx - \left( \frac{1}{2n} + f(n) - \frac{1}{4n} \right) \geq - \frac{1}{24 n^2}$$
or, after a little algebra,
$$f(n) \leq \frac{\pi}{4} - \frac{1}{4n} + \frac{1}{24 n^2}.$$
Similarly,
$$f(n+1) \geq \frac{\pi}{4} - \frac{1}{4(n+1)} - \frac{1}{6(n+1)^2}$$
So
$$f(n+1) - f(n) \geq \frac{n^2 + 4n -1}{24 n^2 (n+1)^2}$$
which is $>0$ for $n \geq 1$.
The basic point here is that, when comparing a sum to an integral, your first approach should be Riemann sums (tried by Did, but not good enough), your second should be the trapezoidal rule (used here), and the third should be the full power of Euler-Maclaurin summation (vesszabo's answer).
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On solving $\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+x}}}}=\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+x}}}}$ How do we show that there is only one solution to,$$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+x}}}}=\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+\sqrt[3]{6+x}}}}$$
I guess it is only $x=2$.
Please help.
| Let $\;f(x) = \sqrt{2+x}\;$ and $\;g(x) = \sqrt[3]{6+x}$, they are strictly increasing function in $x$ when $x \ge -2$.
Since $(x+2)^3 - (x+6)^2 = (x-2)(x^2 + 7x + 14)$ and $x^2 + 7x + 14 > 0$ for all $x$,
we have
$$\begin{cases} f(x) > g(x) > 2,& x > 2\\f(x) = g(x) = 2,& x = 2\\f(x) < g(x) < 2, & x <2\end{cases}$$
So for any $x > 2$, we have
$$\begin{align}
& f(x) > g(x) > 2\\
\implies & f(f(x)) > f(g(x)) > g(g(x)) > 2\\
\implies & f(f(f(x))) > f(g(g(x)) > g(g(g(x)) > 2\\
\implies & f(f(f(f(x))) > f(g(g(g(x))) > g(g(g(g(x)))) > 2\\
\implies & f(f(f(f(x))) \ne g(g(g(g(x))))
\end{align}$$
Please note that in above deductions, we are using following reasoning repeatedly.
$$\underbrace{g\circ\cdots\circ g(x)}_{k \text{ terms}} > 2
\implies
f(\underbrace{g\circ\cdots\circ g(x)}_{k \text{ terms}}) >
\underbrace{g\circ\cdots\circ g(x)}_{k+1 \text{ terms}} > 2.$$
Similar logic shows that $f(f(f(f(x)))) \ne g(g(g(g(x))))$ for $x < 2$. As a result,
$x = 2$ is the only solution for the equation $f(f(f(f(x)))) = g(g(g(g(x))))$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Taylor Series Expansion of $\frac{1}{1+x^2}$ about $x=a$ Let $$f(x)=\frac{1}{1+x^2}$$ Consider its Taylor series expansion about a point $a\in \mathbb{R}$. What is the radius of convergence of this series??
About $x=0$ we could expand it like
$$(1+x^2)^{-1}= 1-x^2+(-1)(-2) \frac{x^2}{2!}+(-1)(-2)(-3)\frac{x^3}{3!}+\dots$$ and get the radius as $1$.
But about $x=a$, there is a slight problem with the way as above...
$$\frac{1}{1+(x-a+a)^2}=\frac{1}{1+a^2(\frac{x-a}{a}+1)^2}.$$
Here is where I am getting stuck.
| Here is a pretty simple way to get the actual Taylor series centered at $a$:
Using partial fractions:
$\dfrac{1}{z^2+1} = \dfrac{i}{2}\left(\dfrac{1}{z+i} - \dfrac{1}{z-i}\right)$.
Now, we can expand each of the the two using the geometric series:
$\displaystyle\dfrac{1}{z+i} = \dfrac{1}{(z-a)+(a+i)} =\dfrac1{a+i}\sum_{n=0}^\infty(-1)^n \left(\dfrac{z-a}{a+i}\right)^{n}$, $|z-a| < |a+i| = \sqrt{1+a^2}$
$\displaystyle\dfrac{1}{z-i} = \dfrac{1}{(z-a)+(a-i)} =\dfrac1{a-i}\sum_{n=0}^\infty(-1)^n \left(\dfrac{z-a}{a-i}\right)^{n} $, $|z-a| < |a-i| = \sqrt{1+a^2}$
$\displaystyle\dfrac{1}{z^2+1} = \dfrac{i}{2}\sum_{n=0}^\infty(-1)^n \left(\left(\dfrac{1}{a+i}\right)^{n+1}-\left(\dfrac{1}{a-i}\right)^{n+1}\right)(z-a)^n$
Which maybe rewritten as:
$\displaystyle\dfrac{1}{z^2+1} = \sum_{n=0}^\infty(-1)^{n+1} Im\left(\left(\dfrac{1}{a+i}\right)^{n+1}\right)(z-a)^n $
EDIT: Radius of convergence from the series.
We can see that the the radius of convergence must be at least $\sqrt{1+a^2}$, since the two series we manipulated had this radius of convergence. We can also deduce this from the following inequality.
$\left|Im\left(\left(\dfrac{1}{a+i}\right)^{n+1}\right)\right| \le \left|\left(\dfrac{1}{a+i}\right)^{n+1}\right|= \left(\dfrac1{\sqrt{1+a^2}}\right)^{n+1}$.
On the other hand, the radius of convergence cannot be lager than $\sqrt{1+a^2}$ because of the singularity of $1/(1+z^2)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Formula for sides of a triangle where the Perimeter equals to the Area I was wondering if there is a formula that could generate the values of the sides of a triangle where his area equals to his perimeter. I only found that if the triangle is equilateral then
$$l=\frac{12}{√3}$$
where $l$ is the side of the triangle.
Thanks for support
Peterix
P.S. There is a similar problem here
| Given any triangle $T$ there is a triangle $T'$ similar to $T$ such that the area of $T'$ is the same as the perimeter of $T'$.
Suppose the perimeter of $T$ is $P$ and the area of $T$ is $A$. Then dilate $T$ by a factor of $\frac{P}{A}$ to produce the triangle $T'$. (That is, multiply all the lengths by $\frac{P}{A}$.)
The perimeter of $T'$ will be the perimeter $T$ multiplied by $\frac{P}{A}$ to give:
$$
\text{Perimeter of } T' = \frac{P}{A}\cdot P = \frac{P^2}{A}
$$
The ara of $T'$ will be the area of $T$ multiplied by $(\frac{P}{A})^2$ to give:
$$
\text{Area of } T' = \left(\frac{P}{A}\right)^2 A = \frac{P^2}{A}
$$
For example, an equilateral triangle with all edges equal to 1 has area $\sqrt3/4$ and perimeter 3. The equilateral triangle with perimeter equal to its area is dilated by $3 \div (\sqrt3/4) = 12/ \sqrt3$, so its side lengths are all $12/\sqrt3$.
Another example: a 3-4-5 triangle has perimeter 12 and area 6. The similar triangle with the area equal to its perimeter is dilated by $12/6 = 2$. Thus the new triangle is a 6-8-10 triangle and has area and perimeter 24.
A final example: a right-angled triangle with short sides $a$ and $b$ has perimeter $P = a+ b+ \sqrt{a^2+b^2}$ and area $A = \frac12 ab$. Dilate this by $P/A$ to give sides of:
$$
\begin{align}
a \cdot \frac{a + b + \sqrt{a^2 + b^2}}{\frac12 ab} &= 2\left( \frac{a}{b} + 1 + \sqrt{\left(\frac{a}{b}\right)^2 + 1}\right)\\
\text{and } \quad b \cdot \frac{a + b + \sqrt{a^2 + b^2}}{\frac12 ab} &= 2\left( \frac{b}{a} + 1 + \sqrt{\left(\frac{b}{a}\right)^2 + 1}\right)\\
\text{and }\quad\sqrt{a^2+b^2} \cdot \frac{a + b + \sqrt{a^2 + b^2}}{\frac12 ab} &= 2\left(\sqrt{\left(\frac{a}{b}\right)^2 + 1} + \sqrt{\left(\frac{b}{a}\right)^2 + 1} + \frac{a}{b} + \frac{b}{a}\right)
\end{align}
$$
| {
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Finding all positive integers $x,y,z$ that satisfy $3^x - 5^y = z^2$
Find all positive integers $x,y,z$ that satisfy:
$$3^x - 5^y = z^2.$$
I think that $(x,y,z)= (2,1,2)$ will be the only solution. But how to prove that?
| I will give a brief review to Yiyuan's answer, and then complete the proof. We have:
$$3^x-5^y=z^2$$
Working $\pmod4$, we have $(-1)^x-1\equiv 0,1 \pmod4$, so $x$ has to be even. Substituting $x=2k$ and $y=a+b$:
$$(3^k-z)(3^k+z)=5^y\implies 3^k-z=5^a, 3^k+z=5^b\implies 2\times3^k=5^a+5^b$$
If $a,b\ge1$, then the RHS is a multiple of $5$ but the LHS is not, so WLOG assume $b=0$. Then:
$$2\times3^k=1+5^a$$
There is no solutions with $k=0$ and one with $k=1(x=2,y=1,z=2)$. Assume henceforth that $k\ge 2$. We then have $-1\equiv 5^a \pmod9$, and since $a=3$ is the smallest solution, we can easily see that every solution is of the form $a=6m+3=3(2m+1)$. Finally
$$2\times3^k=1+5^{3(2m+1)}=(1+5^3)(\cdots)=126(\cdots)$$
But now the RHS is a multiple of $7$. Therefore the only solution is $(x=2,y=1,z=2)$.
| {
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Given $x+y$ and $x\cdot y$, what is $x^3+ y^3$ ? I have been looking at an assortment of high school number sense tests and I noticed a reoccurring problem that states what x+y is and what $x\cdot y$ is then asks for $x^3+ y^3$. I want to know how to work these problems. I have a couple of examples.
$x+y=5$ and $x\cdot y=1$, then $x^3+y^3=?$ [key says 110]
$x+y=-1$ and $x\cdot y=2$, then $x^3+y^3=?$ [key says 5]
$x-y=-1$ and $x\cdot y=2$, then $x^3 -y^3=?$ [key says -7]
$x+y=\frac{1}{3}$ and $x\cdot y=\frac{1}{9}$, then $x^3+y^3=?$ $\left[-\dfrac{2}{27}\right]$
| Its simple. Use the identity
${x^3 + y^3 =(x+y)(x^2 +y^2 -xy) }$
In the above identity;
make modifications and bring it to this form :
${x^3+y^3 = (x+y)((x+y)^2 -3xy) }$
Now putting the values of ${x+y}$ and ${xy}$ in the above equations, you should get the correct answer.
:)
| {
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"timestamp": "2023-03-29T00:00:00",
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Parabola $\sqrt {x}+\sqrt {y}=1 $ How do I prove that the equation $\sqrt {x}+\sqrt {y}=1 $ is part of parabola.
My attempt:rotation in 45 degrees brings the equation to $ -2a^2=1-2\sqrt {2}b $ when $ x= \frac {a-b} {\sqrt {2} } $ and $ y= \frac {a+b} {\sqrt {2} } $. It is a parabola, why is it only part of it? (also for $\sqrt {x}-\sqrt {y}=1 $)
| Rearranging, we get $\sqrt{y} = 1 - \sqrt{x}$, which becomes $y = 1 - 2\sqrt{x} + x$ when we square both sides. Rearranging again and squaring both sides, we get $(y-x-1)^2 = y^2 + x^2 + 1 - 2xy - 2y + 2x = 4x$.
Generally, when there is an equation of the form $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$, if $B^2 - 4AC = 0$, then the curve is one of the following:
*
*A parabola
*2 Parallel Lines (certainly this is not the case)
*1 Line (certainly this is not the case)
*No curve (certainly this is not the case)
Therefore, $\sqrt{x} + \sqrt{y} = 1$ is a part of a parabola.
(Credits partly go to WyzAnt)
| {
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Solving $\sqrt{3\cos^2 x - \sin 2x} = - \sin x$ Please, can you suggest something for solving this equation: I have to find the solutions included in interval $\left[3\pi/2, 2\pi\right]$:
$$\sqrt{3\cos^2 x - \sin 2x} = - \sin x$$
This is what I did:
$$\begin{array}{crcl}
\Longrightarrow & 3\cos^2 x - \sin 2x &=& \sin^2 x \\
\Longrightarrow &3\left(1-\sin^2 x\right)-\sin 2x &=& \sin^2 x \\
\Longrightarrow & 4\sin^2 x + \sin 2x - 3 &=& 0 \\
\Longrightarrow &2\left(1-\cos 2x\right)+\sin 2x - 3 &=& 0\\
\Longrightarrow &-2\cos 2x + \sin 2x &=& 1\end{array}$$
So, what's next?! Thank you in advance!
| from your last step
1−sin2x+2cos2x=0
=>(sin^2x+cos^2x)-2sinxcosx+2(cos^x-sin^2x)=0
=>3cos^2x-2sinxcosx-sin^2x=0
=>3cos^2x-3sinxcosx+sinxcosx-sin^2x=0
=>3cosx(cosx-sinx)+sinx(cosx-sinx)=0
=>(cosx-sinx)(3cosx+sinx)=0
=>cosx-sinx=0 or, 3cosx+sinx=0
=>tanx=1, or tanx=-3
=>x=π/4, or x=2π/3
| {
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"timestamp": "2023-03-29T00:00:00",
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Can I use this formula with pseudo determinants instead of usual determinants? Let $A$ be a matrix with $A^+$ Moore-Penrose inverse. Let also $Det()$ denote the pseudo-determinant of a matrix.
Does the formula (which assumes the existence of $A^{-1}$)
$$ det\left( \begin{array}{cc}
A & B \\
C & D \end{array} \right) = det(A)det(D-CA^{-1}B), $$
where $det()$ denotes the usual determinant, applies to the use of pseudo determinants and Moore-Penrose inverse? This is
$$ Det\left( \begin{array}{cc}
A & B \\
C & D \end{array} \right) = Det(A)Det(D-CA^{+}B)\,? $$
| The classical formula
${\rm Det}(R) = {\rm Det}( \left[ \begin{array}{cc} A & B \\ C & D \end{array} \right] )
= {\rm Det}(D) {\rm Det}(A-BD^+C)$ for block matrices
does not hold for pseudo determinants ${\rm Det}$ and Moore pseudo inverse $D^+$, as already
$$ \left[ \begin{array}{cc} A & B \\ C & D \end{array} \right]
\left[ \begin{array}{cc} I & 0 \\ -D^+ C & I \end{array} \right]
= \left[ \begin{array}{cc} A-B D^+ C & B \\ 0 & D \end{array} \right] $$
does not holds with the pseudo inverse $D^+$. But here is an example:
$$ A=\left[ \begin{array}{cc} -1 & 0 \\ -2 & -1 \\ \end{array} \right],
B=\left[ \begin{array}{cc} 2 & 1 \\ 0 & 2 \\ \end{array} \right],
C=\left[ \begin{array}{cc} 2 & -2 \\ 2 & -1 \\ \end{array} \right],
D=\left[ \begin{array}{cc} 1 & 1 \\ 0 & 0 \\ \end{array} \right] $$
so that the block matrix is
$$ R=\left[ \begin{array}{cccc} -1 & 0 & 2 & 1 \\ -2 & -1 & 0 & 2 \\ 2 & -2 & 1 & 1 \\
2 & -1 & 0 & 0 \\ \end{array} \right] $$
is even invertible with $\det(R)={\rm Det}(R)=2$. The pseudo inverse of
$D$ is $D^+ = \left[ \begin{array}{cc} \frac{1}{2} & 0 \\ \frac{1}{2} & 0 \\ \end{array} \right]$.
Also the matrix $A - B D^+ C= \left[ \begin{array}{cc} -4 & 3 \\ -4 & 1 \\ \end{array} \right]$
is invertible with determinant $8$. We have ${\rm Det}(D)=1$. The left hand side is $2$,
the right hand side is $8$. It is a nice question as we can answer also the question whether the block diagonal formula for diagonal block matrices generalizes to upper triangular block matrices.
There are a few surprises with pseudo determinants: see my
(paper) about it and the
(ArXiv pre-print).
| {
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Supremum calculation Calculate $\sup(\sum_{k=n+1}^{\infty}\frac{|x_{k}|^{2}}{4^{k} })$, where $x=(x_{1},x_{2},....)$ is a member of $l_{2}$ and the supremum is take over all $x$ with $||x||= 1$.
My intuition says the answer is $\frac{1}{4^{n+1}}$.
| Your intuition is correct. It follows from that
$\sum_{k=n+1}^\infty\frac{|x_k|^2}{4^k} \leq \frac{1}{4^{n+1}}\sum_{k=n+1}^\infty|x_k|^2\leq \frac{1}{4^{n+1}}\sum_{k=1}^\infty|x_k|^2=\frac{1}{4^{n+1}}\|x\|_{l_2}^2=\frac{1}{4^{n+1}}$
thus the supremum is at most $\frac{1}{4^{n+1}}$. We also have that this value is obtained for $x\in l_2$ such that $x_k=1$ if $k= n+1$ and zero for all other indices.
Therefore supremum of the sum must be equal to $\frac{1}{4^{n+1}}$.
| {
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parametric solution for the sum of three square Is there a parametric integer solution for $x,y,z,t$ when the sum of three square is equal to a square, i.e,
$$x^2+y^2+z^2=t^2$$?
| EDIT: I knew i had written up Jones and Pall 1939 before, but it was on a question with different intent: Every integer vector in $\mathbb R^n$ with integer length is part of an orthogonal basis of $\mathbb R^n$
J-P 1939: as long as $t$ is odd, find all solutions to
$$ a^2 + b^2 + c^2 + d^2 = t, $$
including permuting the letters and choosing various $\pm$ signs, while not requiring $a,b,c,d$ to be coprime. Then all solutions to your equation are given by
$$ \left( a^2 + b^2 - c^2 - d^2 \right)^2 + (-2ad+2bc)^2 +(2ac+2bd)^2 = t^2. $$
Edit, November 26, 2016. This is often called Lebesgue's Formula, after V. A. Lebesgue the number theorist. It seems to go back to Euler as well. The first acceptable proof that all solution arise this way was due to L. E. Dickson, about 1920.
This is Theorem 3 on page 176 of Regular and Semi-Regular Positive Ternary Quadratic Forms, Acta Mathematica, volume 70, (1939), pages 165-191. It is a theorem about quaternions with all coefficients integers.
Now, if $t$ is even, then $t^2$ is divisible by $4,$ which means you need to double all three numbers; if the sum of three squares is divisible by $4,$ then all the squares are even. So, given some even number $n = 2^k t$ with $t$ odd, solve as above, then multiply through all solutions found by the same $2^k.$
I finished the version with stereographic projection from the North Pole of the sphere, and I can see now why this is not typically taught: by the Gauss-Legendre Three Squares Theorem, either $t$ or $2t$ is the sum of three squares, possibly both.
The first case gives, when $\color{magenta}{t = p^2 + q^2 + r^2},$
$$ (2rp)^2 + (2rq)^2 + \left( p^2 + q^2 - r^2 \right)^2 = t^2. $$
Note that the recipe above cannot be used for $t=7,$ despite $2^2+3^2+6^2=7^2.$
If, instead,when $\color{magenta}{2t = p^2 + q^2 + r^2},$ then $p^2 + q^2 - r^2$ is even and
$$ (rp)^2 + (rq)^2 + \left( \frac{p^2 + q^2 - r^2}{2} \right)^2 = t^2. $$
This way, $2 \cdot 7 = 14 = 3^2 + 2^2 + 1^2,$ we do get $3^2 + 2^2 + (12/2)^2 = 49.$
Let's see, this is very different from Pythagorean triples. there are no restrictions on $t$ at all. Also, if you multiply $t$ by $4$ in either of the recipes above, all that happens is that you are forced to double all three of $p,q,r,$ so nothing important changes. that is, if the sum of three squares is divisible by 4, the squares are all even.
For those keeping score at home, it is quite easy to show that all rational points on the unit sphere $x^2 + y^2 + z^2 = 1$ other than the North Pole $(0,0,1)$ are given by
$$ \left( \; \frac{-2rp}{p^2 + q^2 + r^2} \; , \; \frac{-2rq}{p^2 + q^2 + r^2} \; , \; \frac{p^2 + q^2 - r^2}{p^2 + q^2 + r^2} \; \right) $$
for an integer triple $p,q,r.$
| {
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Find $\frac{a^3}{a^6 + 1}$ given $a$ is a root of a quadratic equation
If $a$ is a root of the equation $x^2 - 3x + 1 = 0$, then find the value of $\frac{a^3}{a^6 + 1}$.
So, I figured we can use the quadratic formula, and formed the following equation:
$$a=\frac{-(-3)+\sqrt{9-4}}{2(1)}\implies a=\frac{3+\sqrt5}2$$
But what I am thinking is, if I begin to find the required value, it will take me hours. And I believe, there must be some shortcut to this question. I had tried to solve this question with the manual process but it took me a lot of squares (one was $2012^2$!)
Can someone please help me.
Thank you.
| So
$$a^2 + 1 = 3a$$
and this gives:
$$\frac{a}{a^2 + 1} = \frac{1}{3},$$
and
$$(a^2 + 1)^2 = 9a^2 \implies a^4 + 1 = 7a^2.$$
So
$$\frac{a^2}{a^4 - a^2 + 1} = \frac{a^2}{6a^2} = \frac{1}{6}.$$ And finally
$$\frac{a^3}{a^6 + 1} = \frac{a}{a^2 +1}\cdot \frac{a^2}{a^4 - a^2 + 1} = \frac{1}{3}\cdot \frac{1}{6} = \frac{1}{18}.$$
| {
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Probability problem with binomial/multinomial distribution Mary knows the answers to $20$ of the $25$ multiple choice questions on the Psychology $101$ exam, but she has skipped several of the lectures, she must take random guesses for the other five. Assuming each question has four answers, what is the probability she will get exactly $3$ of the last $5$ questions right?
| We need to make some assumptions. We will assume the $5$ questions she does not know the answers to are equally likely to be any $5$ of the $25$ questions.
We want the probability she gets exactly $2$ of the last $5$ questions wrong. This can happen in various ways. The last $5$ questions may contain $2$ questions she has to guess on, and she gets both wrong. Or the last $5$ questions may contain $3$ questions she has to guess on, and she gets $1$ right. Or they may contain $4$ questions she has to guess on, and she gets $2$ right. Or they may contain $5$ questions she has to guess on, and she gets $3$ right.
$2$ questions: The probability there are $2$ questions she does not know the answer to is $\frac{\binom{5}{2}\binom{20}{3}}{\binom{25}{5}}$. Given this, the probability she gets them both wrong is $\left(\frac{3}{4}\right)^2$. This gives probability $\frac{\binom{5}{2}\binom{20}{3}}{\binom{25}{5}}\left(\frac{3}{4}\right)^2$.
$3$ questions: The probability there are $3$ questions she does not know the answer to is $\frac{\binom{5}{3}\binom{20}{2}}{\binom{25}{5}}$. Given this, the probability she gets $2$ of them wrong (so $1$ right) is $\binom{3}{1}\left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^2$. Now we can write down the associated probability by multiplying.
$4$ questions, $5$ questions: It's your turn.
Now add up.
| {
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The equation $3^n+4^m=5^k$ in positive integer numbers Please help me to prove that the equation $3^n + 4^m = 5^k$
where $n$, $m$, $k$ are positive integer numbers has only the solution $n=m=k=2$.
I know how to prove it for $n=m=k$.
If $3^x + 4^x = 5^x$ then $(3/4)^x + 1 = (5/4)^x$,
and this equation has at most one solution
since the function $(3/4)^x + 1$ decreases and the function $(5/4)^x$ increases
for all real $x$. So the solution $x = 2$ is unique.
Thank you very much in advance!
| Reducing the equation mod $3$ shows that $k$ must be even, and reducing mod $4$ shows the same for $n$, so let's rewrite the whole thing as
$$3^{2n}+2^{2m}=5^{2k}$$
and show that $(n,m,k)=(1,2,1)$ is the only solution in positive integers.
Now
$$2^{2m}=5^{2k}-3^{2n}=(5^k-3^n)(5^k+3^n)$$
implies $5^k-3^n=2^a$ and $5^k+3^n=2^b$ with $a\lt b=2m-a$. Subtracting these gives
$$2\cdot3^n=2^a(2^{b-a}-1)$$
which implies $a=1$ and $3^n=2^{b-a}-1$. It remains to show that the only positive power of $3$ of the form $2^u-1$ is $3^1=2^2-1$.
If $3$ divides $2^u-1$, then $u$ must be even, so let $u=2v$. In that case, $2^u-1=(2^v-1)(2^v+1)$. In order for $3$ to divide $2^v+1$, $v$ must be odd. But if $v$ is odd then $2^v-1$ is not divisible by $3$, so the only way it can be a power of $3$ is if it's equal to $1$, which implies $v=1$, which is exactly what we need.
Added later: One might alternatively rewrite the (modified) equation as
$$3^{2n}=5^{2k}-2^{2m}=(5^k-2^m)(5^k+2^m)$$
which implies $5^k-2^m=3^a$ and $5^k+2^m=3^b$ with $a\lt b=2n-a$. In this case subtracting gives
$$2\cdot2^m=3^a(3^{b-1}-1)$$
which implies $a=0$, so that $5^k-2^m=1$, or $5^k=2^m+1$. This leads to $3^{2n}=2^{m+1}+1$, or $2^{m+1}=(3^n-1)(3^n+1)$. As before, we must have $3^n-1=2^a$ and $3^n+1=2^b$ with $a\lt b$, but this time subtracting gives $2=2^a(2^{b-1}-1)$, which forces $a=1$, hence $n=1$.
Note, the only role of $5$ in either approach is to be congruent to $2$ mod $3$ and $1$ mod $4$ (so that we can replace $n$ and $k$ with $2n$ and $2k$), so the same proof works with anything congruent to $5$ mod $12$, e.g., $17$, $29$, $41$, etc. The only difference is that in these cases the final conclusion is that there are no solutions: Once you've got $n=1$ and $m=2$ on the left hand side, you can only have $5^2$ on the right.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/662121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How prove $\measuredangle CDE=2\measuredangle ABE$ In rectangular $ABCD$,and $E\in AC$,such
$$BE=\sqrt{2}\cdot AE$$
show that
$$\measuredangle CDE=2\measuredangle ABE$$
My try: let $$AB=a,AD=b,\dfrac{AE}{AC}=k,$$
then
$$AE=k\sqrt{a^2+b^2},BE=k\sqrt{2(a^2+b^2)}$$
I know have this nice relsut
$$AE^2+EC^2=BE^2+ED^2$$
then
$$ED^2=k^2\cdot AC^2+(1-k)^2\cdot AC^2-2k^2\cdot AC^2=(1-2k)AC^2$$
so
$$\cos{\measuredangle EBD}=\dfrac{AB^2+BE^2-AE^2}{2AB\cdot BE}=\dfrac{a^2+k^2(a^2+b^2)}{2a\cdot k\sqrt{2(a^2+b^2)}}$$
$$\cos{\measuredangle CDE}=\dfrac{DC^2+DE^2-EC^2}{2 DC\cdot DE}=\dfrac{a^2+(1-2k)(a^2+b^2)-(1-k)^2(a^2+b^2)}{2\sqrt{(1-2k)(a^2+b^2)}a}=\dfrac{a^2-k^2(a^2+b^2)}{2\sqrt{(1-2k)(a^2+b^2)}a}$$
we will prove
$$2\cos^2{\measuredangle ABE}-1=\cos{\measuredangle CDE}$$
$$\Longleftrightarrow 2\dfrac{((k^2+1)a^2+b^2)^2}{8a^2k^2(a^2+b^2)}-1=\dfrac{a^2-k^2(a^2+b^2)}{2\sqrt{(1-2k)(a^2+b^2)}a}$$
But I can't
Thank you for you
| I solve this problem in another way. Let us establish a coordinate system which $A$ is original point and $AB$ is y-axis and $AD$ is x-axis. And $AB=a$, $AD=b$.
In this frame, we have: $A(0,0)$, $B(0,a)$, $C(b,a)$, $D(b,0)$. For the sake of simplicity, we define $k:=\frac{b}{a}$ and equation of $AC$ can be written as $y=kx$. Also, assume $E(bx_0,ax_0)$ where $x_0$ is a parameter to be determined. And then the length of $AE$ and $BE$ can be written as :
\begin{equation}
AE=\sqrt{1+k^2}bx_0\\
BE=b\sqrt{x_0^2+k^2(1-x_0)^2}
\end{equation}
Note that the condition $BE=\sqrt{2}AE$ gives:
\begin{equation}
\frac{BE}{AE}=\sqrt{\frac{x_0^2+k^2(1-x_0^2)}{x_0^2+k^2x_0^2}}=\sqrt{2}\\
(1+k^2)x_0^2+2k^2x_0-k^2=0\\
x_0=\frac{k(-k+\sqrt{2k^2+1})}{k^2+1}\quad\mbox{(The negative root have been ignored since $E\in AC$)}
\end{equation}
Remark: I should point out the this result have been ignored in your deduction that the point $E$ can be determined by $a$ and $b$.
Let $\alpha$ and $\beta$ denote $\measuredangle ABE$ and $\measuredangle EDC$ respectively. We have:
\begin{equation}
\cos\alpha=\frac{AB^2+BE^2-AE^2}{2AB\cdot BE}=\frac{a^2+AE^2}{2\sqrt{2}a\cdot AE}\\
\cos\beta=\frac{ED^2+CD^2-EC^2}{2ED\cdot CD}=\frac{ED^2+a^2-EC^2}{2a\cdot ED}
\end{equation}
Now, the coordinate of $E$ is governed by $a$ and $b$. $AE$, $ED$ and $EC$ can be computed by coordinates of $A$, $B$, $C$, $D$, $E$ and all of them are functions of $k$. By a direct computation, we can check:
\begin{equation}
\cos\beta=2\cos^2\alpha-1
\end{equation}
Since the deduction is so long that I ignore them here. You can check it by using MAPLE if you feel it is hard.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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} |
If $|x + 1| + |x - 3| = 6$, solve for $x$ $|x + 1| + |x - 3| = 6$. Solve for X.
So I know when you have a problem like this: |x| = 6, you solve by doing x=6 and x=-6. That doesn't help us much in the above example.
You can also solve problems of this fashion by negative the variable portion of the equation. For ex:
$|x+3| = 6. $
You can solve it either by doing:
$x+3 = 6$ or
$x+3 = -6$
$x= 3,-9$
OR
$(x+3) = 6$ so $x=3$
or
$-(x+3) = 6$
$-x -3 =6$
$-x = 9$
$x = -9$
So back to the original problem of $|x + 1| + |x - 3| = 6$
if $X>3$, then $x+1 + x-3 = 6$
$2x=8$
$x=4$
or if $X<3$
$x+1 + x-3 = -6$
$2x=-4$
$x=-2$
Is that correct? Is that all I need to do? I also feel like I got the "critical points" of the equation wrong.
| $|x+1| = 6 - |x-3|$
$x + 1 = 6 - |x-3|$ or $x+1 = -6 + |x-3|$
For the first equation: $|x-3| = 5 - x$ so
$x-3 = 5 -x$ or $ x-3 = x-5$. Only the first part is valid, it gives you the solution $x=4$.
Second equation: $x+7 = |x-3|$. So
$x+7 = x-3$ or $-x - 7 = x-3$. Only the second equation is valid, that gives you $x=-2$. These are the two solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/665171",
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"source": "stackexchange",
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} |
$p \equiv 5 \mod8\Rightarrow p=(2x+y)^{2}+4y^{2}$ If $p \equiv 5 \mod8$ , then $p=(2x+y)^{2}+4y^{2}$,for some x and y integers.
Thanks
Here is my approach:
I know $p \equiv 5 \mod8\Rightarrow $
$p \equiv 1 \mod4\Rightarrow $
$n^{2}+m^{2}=p\equiv 5 \mod8 \Rightarrow$
from $t^{2}\equiv 0,1~or ~4 \mod8$ we get $n^{2} \equiv 1 \mod8$ and $m^{2} \equiv 4 \mod8$
$\Rightarrow n=4x-1$ and $m=2y$.
So I need $n=2x+y$.
| $p \equiv 5 \mod8\Rightarrow $
$p \equiv 1 \mod4\Rightarrow $
$n^{2}+m^{2}=p\equiv 5 \mod8 \Rightarrow$
from $t^{2}\equiv 0,1~or ~4 \mod8$ we get $n^{2} \equiv 1 \mod8$ and $m^{2} \equiv 4 \mod8 \Rightarrow$
n is odd and $ m=2y$ where y is odd $\Rightarrow$
n-y is even $\Rightarrow$
n-y=2x and m=2y $\Rightarrow$
$p=(2x+y)^{2}+4y^{2}\square$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How prove this $\sqrt{a^2(\cos{x}+\cos{y})^2+b^2(\sin{x}+\sin{y})^2}+\sqrt{a^2(\cos{x}-\cos{y})^2+b^2(\sin{x}-\sin{y})^2}\le 2\sqrt{a^2+b^2}$ let $a,b>0$ is give numbers,for any $x,y\in R$,show that
$$\sqrt{a^2(\cos{x}+\cos{y})^2+b^2(\sin{x}+\sin{y})^2}+\sqrt{a^2(\cos{x}-\cos{y})^2+b^2(\sin{x}-\sin{y})^2}\le 2\sqrt{a^2+b^2}$$
maybe can use Cauchy-Schwarz inequality to solve it.
This inequality background is from:How prove this $|ON|\le \sqrt{a^2+b^2}$
$$\Longleftrightarrow 2a^2(\cos^2{x}+\cos^2{y})+2b^2(\sin^2{x}+\sin^2{y})+2\sqrt{a^2(\cos{x}+\cos{y})^2+b^2(\sin{x}+\sin{y})^2}\sqrt{a^2(\cos{x}-\cos{y})^2+b^2(\sin{x}-\sin{y})^2}\le 4(a^2+b^2)$$
But I can't,Thank you for you help
| Let us make use of the trigonometric identities for the sum of angles i.e.
$\cos(a \pm b) = \cos a \cos b \mp \sin a \sin b$
$\sin(a \pm b) = \sin a \cos b \pm \cos a \sin b$
Thus
$\cos x+ \cos y = 2\cos (\frac{x+y}{2})\cos (\frac{x-y}{2})$
$\cos x- \cos y = -2\sin (\frac{x+y}{2})\sin (\frac{x-y}{2})$
$\sin x+ \sin y = 2\sin (\frac{x+y}{2})\cos (\frac{x-y}{2})$
$\sin x- \sin y = 2\cos (\frac{x+y}{2})\sin (\frac{x-y}{2})$
Going back to the LHS of the inequality, we have
$2\cos (\frac{x-y}{2})\sqrt{a^2\cos^2 (\frac{x+y}{2})+b^2\sin^2 (\frac{x+y}{2})}+2\sin (\frac{x-y}{2})\sqrt{a^2\sin^2 (\frac{x+y}{2})+b^2\cos^2 (\frac{x+y}{2})}$
We are in a position to use the Cauchy-Schwarz inequality, which we state below for reference
$\sum_{k=1}^nf_kg_k\leq (\sum_{k=1}^nf_k^2)^{1/2}(\sum_{k=1}^ng_k^2)^{1/2}$
If we let $f_1=2\cos (\frac{x-y}{2})$ and $f_2=2\sin (\frac{x-y}{2})$
and we let
$g_1=\sqrt{a^2\cos^2 (\frac{x+y}{2})+b^2\sin^2 (\frac{x+y}{2})}$ and
$g_2=\sqrt{a^2\sin^2 (\frac{x+y}{2})+b^2\cos^2 (\frac{x+y}{2})}$
The upper bound of the inequality is
$\sqrt{4\cos^2 (\frac{x-y}{2})+4\sin^2 (\frac{x-y}{2})}\sqrt{a^2\cos^2 (\frac{x+y}{2})+b^2\sin^2 (\frac{x+y}{2})+a^2\sin^2 (\frac{x+y}{2})+b^2\cos^2 (\frac{x+y}{2})}$
and recalling that $\sin^2 \theta + \cos^2 \theta = 1$, the upper bound becomes
$2\sqrt{a^2+b^2}$
which is what we set out to prove.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Find a formula for $\sum\limits_{k=1}^n \lfloor \sqrt{k} \rfloor$ I need to find a clear formula (without summation) for the following sum:
$$\sum\limits_{k=1}^n \lfloor \sqrt{k} \rfloor$$
Well, the first few elements look like this:
$1,1,1,2,2,2,2,2,3,3,3,...$
In general, we have $(2^2-1^2)$ $1$'s, $(3^2-2^2)$ $2$'s etc.
Still I have absolutely no idea how to generalize it for $n$ first terms...
| I am very bad in the area of discrete mathematics (as well in other) but I have been fascinated by the problem set in your post.
I am sure that Daniel Fisher's comment and Sami Ben Romdhane's answer are very useful; however, I have not been able to finish the work.
So, what I used is computer simulation and data regression in order to establish some relations. Later, RIES was used to identify the rational values of the obtained coefficients. As you see, this is a very empirical process but I hope it could help you.
I set the problem in the most general manner, loking for $$\sum\limits_{k=1}^n \lfloor k^{1/p} \rfloor$$.
What I found is that the sum starts with a first term which is $$(n+1) \left\lfloor \sqrt[p]{n}\right\rfloor$$ to which is added a polynomial (no constant term) of degree $(p+1)$ of a variable which is $$1+\left\lfloor \sqrt[p]{n}\right\rfloor$$ So, for the first successive values of $p$, I obtained after some simplifications (I am sure that more simplifications could be done)
$$
(n+1) \left\lfloor \sqrt{n}\right\rfloor +\frac{1}{3} \left(-\left(\left\lfloor
\sqrt{n}\right\rfloor +1\right)^3+\frac{3}{2} \left(\left\lfloor
\sqrt{n}\right\rfloor +1\right)^2-\frac{1}{2} \left(\left\lfloor
\sqrt{n}\right\rfloor +1\right)\right)
$$
$$
(n+1) \left\lfloor \sqrt[3]{n}\right\rfloor +\frac{1}{4} \left(-\left(\left\lfloor
\sqrt[3]{n}\right\rfloor +1\right)^4+2 \left(\left\lfloor \sqrt[3]{n}\right\rfloor
+1\right)^3-\left(\left\lfloor \sqrt[3]{n}\right\rfloor +1\right)^2\right)
$$
$$
(n+1) \left\lfloor \sqrt[4]{n}\right\rfloor +\frac{1}{5} \left(-\left(\left\lfloor
\sqrt[4]{n}\right\rfloor +1\right)^5+\frac{5}{2} \left(\left\lfloor
\sqrt[4]{n}\right\rfloor +1\right)^4-\frac{5}{3} \left(\left\lfloor
\sqrt[4]{n}\right\rfloor +1\right)^3+\frac{1}{6} \left(\left\lfloor
\sqrt[4]{n}\right\rfloor +1\right)\right)
$$
$$
(n+1) \left\lfloor \sqrt[5]{n}\right\rfloor +\frac{1}{6} \left(-\left(\left\lfloor
\sqrt[5]{n}\right\rfloor +1\right)^6+3 \left(\left\lfloor \sqrt[5]{n}\right\rfloor
+1\right)^5-\frac{5}{2} \left(\left\lfloor \sqrt[5]{n}\right\rfloor
+1\right)^4+\frac{1}{2} \left(\left\lfloor \sqrt[5]{n}\right\rfloor
+1\right)^2\right)
$$
I do not know how this will be of any use to you; however, I must confess that I had a great time with this problem.
| {
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"url": "https://math.stackexchange.com/questions/669460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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"answer_id": 1
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Determine value $\lim_{n\to +\infty} v_n$ Let $(u_n)$ sequence satisfy: $$\left\{\begin{matrix}u_1=3\\u_{n+1}=\frac{1}{5}\left(u_n^2+u_n+4\right),\: n=1, \: 2, ...\end{matrix}\right.$$
Set $v_n=\sum_{k=1}^n \frac{1}{u_k+3}$. Determine value $\lim_{n\to +\infty} v_n$
I think have to find a and b satisfy to $\frac{1}{u_k+3}=a\left(\frac{1}{u_k+b}-\frac{1}{u_{k+1}+b}\right)$
a and b=??
| Let us expand
$\large a(\frac{1}{u_k+b}-\frac{1}{u_{k+1}+b})$
Based on the given sequence recursion we obtain
$\large a(\frac{1}{u_k+b}-\frac{5}{u_k^2+u_k+4+5b})=a\frac{u_k^2+u_k+4+5b-5u_k-5b}{(u_k+b)(u_k^2+u_k+4+5b)}=a\frac{u_k^2-4u_k+4}{(u_k+b)(u_k^2+u_k+4+5b)}$
Simplifying further we have
$\large a\frac{(u_k-2)^2}{(u_k+b)(u_k^2+u_k+4+5b)}$
If you set $b$ to $-2$ the fraction becomes
$\large a\frac{(u_k-2)^2}{(u_k-2)(u_k^2+u_k-6)}=a\frac{(u_k-2)^2}{(u_k-2)^2(u_k+3)}=a\frac{1}{u_k+3}$
So $a$ is $1$.
Thus, $a=1$ and $b=-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/669628",
"timestamp": "2023-03-29T00:00:00",
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Scratch work for delta-epsilon proof for $\lim_{x \to 13} \sqrt{x-4} = 3$
Prove $\lim_{x \to 13} \sqrt{x-4} = 3$.
We need to show for all $E> 0$ there exists $D > 0$ such that
if $0 < |x - 13| < D$, then $|\sqrt{x-4} - 3| < E$. Let me write D for delta and E for epsilon please.
Scratch-work here: $\color{darkred}{\text{ I understand the formal proof. Ergo just asking about this. } }$
Note that $|\sqrt{x-4} - 3| = |\sqrt{x-4} - 3|\dfrac{|\sqrt{x-4} + 3|}{|\sqrt{x-4} + 3|} = \dfrac{|x - 13|}{|\sqrt{x-4} + 3|}$.
We can bound $|x - 13|$ for any choice of D, but we need a certain D to also bound $|\sqrt{x-4} + 3|$.
1. Ishfaaq's answer says 'the denominator will definitely cause trouble.' How?
Why won't bounding $|x - a|, |x - 13|$ bound the whole statements?
Beneath pp 83 of Spivak claims this too. How does '$|x + a|$ cause trouble'?
Assume that $D < 1.$
2. What sanctioned assuming $D < 1$? How to know if $1$ is too small, too big? $\\$ In pp 83 of Spivak beneath, $D$ looks randomly chosen?
4. Ishfaaq's last paragraph. $\color{darkred}{\left|{x^2 - a^2}\right| \lt \delta^2 + \left|{}2a\right|\delta. \text{ But we can hardly equate this to ϵ... }}$
Why not? $d^2 + \left|{}2a\right|d = e \iff d(d + |2a|) = e$?
Ishfaaq says $a = 1/2, e = d/2$ is a counterexample. But then $d^2 + 2|a|d < e \iff d^2 + d < d/2 \iff d(d - \frac{1}{2}) < 0 \iff 0 < d < 1/2.$ What founders?
If this works, then any smaller D will also work.
Then $|x - 13| < 1 \iff -1 < x - 13 < 1 \iff 12 < x < 14 \iff 8 < x - 4 < 10$
$\implies \sqrt(8) < \sqrt{x - 4} < \sqrt(10) \iff \color{blue}{\sqrt(8) + 3 < \sqrt{x - 4} + 3 < \sqrt(10) + 3} \\ \iff 0 < \color{blue}{\dfrac{1}{\sqrt(10) + 3} < \dfrac{1}{\sqrt{x - 4} + 3} < \dfrac{1}{\sqrt(8) + 3}} $
$\implies \color{green}{\frac{1}{|\sqrt{x - 4} + 3|} < \frac{1}{\sqrt(8) + 3}} $
So then $|x - 13|\color{green}{\dfrac{1}{|\sqrt{x - 4} + 3} < \dfrac{1}{\sqrt(8) + 3}}|x - 13|$ and need this $ < E.$
Thus, we need $|x - 13| < E * \color{green}{(\sqrt{8} + 3)}$. So this is our choice of D. But note that this only works when $D < 1$. Thus, we can take care of both conditions by choosing $D = \min\{1, E(\sqrt{8}+3)\}.$ █
$\color{darkred}{ \text{ 3. All this algebra fazed me. How to graph this to view all this algebra and $D$? Thanks. } }$
Also tried http://www.ocf.berkeley.edu/~yosenl/math/epsilon-delta.pdf.
| If you found the algebra a bit hard to decipher, maybe this will help.
Assume that $|x-13|<1$ just means that we assume that the distance between $x$ and 13 is less than one. $x$ is the input to your function, it's distance from the number 13 is less than 1. If you want to compute your limit, i.e. what happens to $f(x)$ as $x\rightarrow13$, the distance between $x$ and 13 is going to be less than 1 eventually anyway, so this is not an unreasonable thing to do.
$|x-13|<1\Longleftrightarrow-1<x-13<1$ These two statements mean the same thing. It's easier to work with the equation to the right.
You treat $-1<x-13<1$ like an ordinary equation except instead of doing the same thing to both sides of the equality sign you do the same thing to the left, in the middle and to the right.
$\begin{array}{ccccc}
-1 & < & x-13 & < & 1\\
-1+13 & < & x-13+13 & < & 1+13\\
12 & < & x & < & 14\\
12-4 & < & x-4 & < & 14-4\\
8 & < & x-4 & < & 10\\
\sqrt{8} & < & \sqrt{x-4} & < & \sqrt{10}\\
\sqrt{8}+3 & < & \sqrt{x-4}+3 & < & \sqrt{10}+3\\
& \text{flip} & & \text{flip}\\
\frac{1}{\sqrt{8}+3} & > & \frac{1}{\sqrt{x-4}+3} & > & \frac{1}{\sqrt{10}+3}\\
\end{array}$
All these moves are legal. What we have so far is that if the length between your input $x$ and 13 is less than one, i.e $|x-13|<1$, then $\frac{1}{|\sqrt{x-4}+3|} < \frac{1}{\sqrt{8}+3}$.
Now comes the epsilon-delta part. Remember that the limit you want to find is basically the same as asking "what happens to $f(x)$ when my input $x$ gets really close to 13", so let's consider the case where $x$'s distance from 13 is less than 1 (this has to happen eventually).
When $|x-13|<1$, we know from before that $|\sqrt{x-4}-3|=|x-13|\frac{1}{|\sqrt{x-4}+3|}$, so now we have $|\sqrt{x-4}-3|=\text{something less than 1}\cdot\text{something less than }\frac{1}{\sqrt{8}+3}<\frac{1}{\sqrt{8}+3}$
Let's consider when $|x-13|<(\sqrt{8}+3)\epsilon$:
We know from before that $|\sqrt{x-4}-3|=|x-13|\frac{1}{|\sqrt{x-4}+3|}$, so now we have $|\sqrt{x-4}-3|=\text{something less than }(\sqrt{8}+3)\epsilon\cdot\text{something less than }\frac{1}{\sqrt{8}+3}<(\sqrt{8}+3)\epsilon \frac{1}{\sqrt{8}+3}=\epsilon$
Now we choose $\delta=\min\{1,(\sqrt{8}+3)\epsilon\}$. Why? Because there is a flaw in the argument above! What if $\epsilon$ is greater than 1? Then we can't say that $\frac{1}{|\sqrt{x-4}+3|} < \frac{1}{\sqrt{8}+3}$ - but wait! if $\epsilon$ is greater than 1, we already have a $\delta$ that will satisfy our definition: 1.
Example:
If $\epsilon=2.2$, can you find a $\delta$ such that $|x-13|<\delta \Rightarrow |f(x)-3|<2.2?$ Yes, $|x-13|<1$, then $|\sqrt{x-4}-3|<\frac{1}{\sqrt{8}+3}\approx 0.17$
When $\epsilon$ is less than 1, then we know that $\frac{1}{\sqrt{x-4}+3} < \frac{1}{\sqrt{8}+3}$ and we use the $\delta$ we derived using the fact that $|x-13|<1$, namely $\delta=(\sqrt{8}+3)\epsilon$
Think about after $\epsilon=2.2$, when someone tries to pass me an $\epsilon$ like 0.001 so that $(\sqrt{8}+3)\epsilon<1$, then having chosen $|x-13|<\delta=\min\{1,(\sqrt{8}+3)\epsilon\}$, I can repeat my whole argument about how when $|x-13|<1$, then blah blah blah. I have a winning strategy against any player who tries to break me by passing me tiny or large epsilons.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "11",
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If $a+b+c = 7\;\;,a^2+b^2+c^2 = 23$, then $a^3+b^3+c^3=$ If $a,b,c\in \mathbb{R}$ and $a+b+c = 7\;\;,a^2+b^2+c^2 = 23$ and $\displaystyle \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1} = 31$. Then $a^3+b^3+c^3 = $
$\bf{My\; Trial\; Solution::}$ Given $a^2+b^2+c^2 = 23$ and
$a+b+c = 7\Rightarrow (a+b+c)^2 = 49\Rightarrow (a^2+b^2+c^2)+2(ab+bc+ca) = 49$
So $23+2(ab+bc+ca) = 49\Rightarrow (ab+bc+ca) = 13$
Now from $\displaystyle \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1} = 31\Rightarrow \frac{(a+1)\cdot (b+1)+(b+1)\cdot (c+1)+(c+a)\cdot (a+1)}{(a+1)(b+1)(c+1)} = 31$
So $\displaystyle \frac{(ab+bc+ca)+2(a+b+c)+3}{1+(a+b+c)+(ab+bc+ca)+abc} = 31\Rightarrow \frac{13+2\cdot 7+3}{1+7+13+abc} = 31$
So $\displaystyle \frac{30}{21+abc} = 31\Rightarrow 21\times 31+31(abc) = 30\Rightarrow (abc) = \frac{30-21\times 31}{31}=-\frac{621}{31}$
Now How can I calculate $a^3+b^3+c^3$
Is there is any better method by which we can calculate $abc$
Help me
Thanks
| Let $p(x) = (x-a)(x-b)(x-c)$. Then the following identity holds:
$$ a^{3} + b^{3} + c^{3} = (a+b+c)^{3} - 3p(a+b+c). $$
Since we know that $p(x) = x^{3}- 7 x^{2} + 13 x + \frac{621}{31}$, we now have the answer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to use $x + \frac{1}{x} = 7$ to compute $x^2 + \frac{1}{x^2}$. I am not sure how to approach this question:
You know that $x + \frac{1}{x} = 7$. Compute $x^2 + \frac{1}{x^2}$.
I have tried adding $x + \frac{1}{x}$ to get $\frac{x^2 +1}{x}$ but can't see if this was useful or not.
I need help in getting started.
| Notice $\left(x + \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} + 2$.
| {
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Evaluating a trigonometric limit What is the limit as $x$ approaches $0$ of: $$\frac{\sqrt{1+\tan x} - \sqrt{1+\sin x}}{1+x-\cos x}?$$
We cannot use L'Hôpital's rule or anything advanced like Taylor series. I reduced it to, by considering the numerator's conjugate:
$$\frac{1}{2}\lim_{x \to 0}\frac{\tan x - \sin x}{1+x-\cos x}$$
But I cannot go further. Please help.
EDIT: Thinking carefully, I think I can simplify further:
$$\frac{1}{2}\lim_{x \to 0}\frac{\frac{\tan x - \sin x}{x}}{\frac{1-\cos x}{x}+1}=\frac{1}{2}\lim_{x \to 0}\frac{\tan x - \sin x}{x}$$
And perhaps the solution:
$$\frac{1}{2}\lim_{x \to 0}\frac{\tan x - \sin x}{x}=\frac{1}{2}(\lim_{x \to 0}\frac{\tan x}{x} - 1)=\frac{1}{2}(\lim_{x \to 0}\frac{\sin x}{\cos x \cdot x}) - 1=\frac{1}{2}(1- 1)=0$$
Is this correct?
| The title of the question is a bit intimidating as this limit involves very basic trigonometric manipulation and standard limits like $\lim\limits_{x \to 0}\dfrac{\sin x}{x} = 1$. Here goes the simple solution $$\begin{aligned}L &= \lim_{x \to 0}\frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{1 + x - \cos x}\\
&= \lim_{x \to 0}\frac{\sqrt{1 + \tan x} - \sqrt{1 + \sin x}}{1 + x - \cos x}\cdot \frac{\sqrt{1 + \tan x} + \sqrt{1 + \sin x}}{\sqrt{1 + \tan x} + \sqrt{1 + \sin x}}\\
&= \lim_{x \to 0}\frac{\tan x - \sin x}{(1 + x - \cos x)(\sqrt{1 + \tan x} + \sqrt{1 + \sin x})}\\
&= \frac{1}{2}\lim_{x \to 0}\frac{\tan x - \sin x}{1 + x - \cos x}\\
&= \frac{1}{2}\lim_{x \to 0}\tan x\cdot \frac{1 - \cos x}{1 + x - \cos x}\\
&= \frac{1}{2}\lim_{x \to 0}\tan x\cdot \frac{2\sin^{2}(x/2)}{x + 2\sin^{2}(x/2)}\\
&= \lim_{x \to 0}\tan x\cdot \dfrac{\dfrac{\sin^{2}(x/2)}{(x/2)^{2}}\cdot(x/2)^{2}}{x + 2\dfrac{\sin^{2}(x/2)}{(x/2)^{2}}\cdot(x/2)^{2}}\\
&= \lim_{x \to 0}\tan x\cdot \frac{x}{4}\cdot\dfrac{\dfrac{\sin^{2}(x/2)}{(x/2)^{2}}}{1 + \dfrac{x}{2}\cdot\dfrac{\sin^{2}(x/2)}{(x/2)^{2}}}\\
&= 0\cdot 0\cdot\dfrac{1}{1 + 0\cdot 1} = 0\end{aligned}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sqrt{8}=1+\dfrac34+\dfrac{3\cdot5}{4\cdot8}+\dfrac{3\cdot5\cdot7}{4\cdot8\cdot12}+\ldots$ Prove that $\sqrt{8}=1+\dfrac34+\dfrac{3\cdot5}{4\cdot8}+\dfrac{3\cdot5\cdot7}{4\cdot8\cdot12}+\ldots$
My work:
$\sqrt8=\bigg(1-\dfrac12\bigg)^{-\frac32}$
Now, I suppose there is some "binomial expansion with rational co-efficients" or something similar for this, which I do not know. Please help.
N.B.: Any other solution is also acceptable, I do not have any restriction regarding the solution.
| The binomial series is "just" the Taylor series of $(1+x)^{\alpha}$ at $x=0$.
Start deriving $f(x)=(1+x)^{\alpha}$ by $x$ and you get $\alpha(1+x)^{\alpha-1}$, then $\alpha(\alpha-1)(1+x)^{\alpha-2}$ for the first derivative etc.
The $n$th derivative is
$$\frac{d^n}{dx^n} (1+x)^{\alpha} = \alpha(\alpha-1)\cdots(\alpha-n+1)(1+x)^{\alpha-n}$$
So the Taylor series is
$$\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} x^n= \sum_{n=0}^\infty \frac{\alpha(\alpha-1)(\alpha-2)\cdots(\alpha-n+1)}{n!} x^n$$
If $\alpha$ is a non-negative integer this series becomes a finite sum and the $n$th coefficient is simply ${\alpha\choose n}$, so it makes sense to call it a "generalized" binomial coefficient and denote it by the same symbol, also if $\alpha$ is not a non-negative integer.
Now notice that the Taylor series has convergence radius $1$ and that $(1+x)^{\alpha}$ is real-analytic, therefore
$$(1+x)^{\alpha} = \sum_{n=0}^\infty {\alpha\choose n} x^n$$
for all $|x|<1$, in particular $x=-\frac{1}{2}$ and $\alpha=-3/2$. Now notice that $${-3/2\choose n}=\frac{(-1)^n}{n!2^n} 3\cdot 5\cdots (2n+1)$$
Then,
$$\begin{align}
\sqrt{8} &=(1-1/2)^{-3/2} = \sum_{n=0}^\infty \frac{(-1)^n}{n!2^n} 3\cdot 5\cdots (2n+1) \frac{1}{(-2)^n} = \sum_{n=0}^\infty \frac{3\cdot 5\cdots (2n+1)}{n! 4^n}\\
&= 1 + \frac{3}{4} + \frac{3\cdot 5}{4\cdot 8} + \frac{3\cdot 5\cdot 7}{4\cdot 8\cdot 12} + \cdots
\end{align}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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An equation, where the solution does not exist, but on solving the equation we got a solution. why this is happening? The solution of the equation
$\sqrt{(x+1)} -\sqrt{(x-1)}= \sqrt{(4x-1)}$
is $\frac{5}{4}$,but when we put $x=\frac{5}{4}$ in the given equation, then it does not satisfy the equation.
Actually, if we take $f(x)=\sqrt{(x+1)} -\sqrt{(x-1)} -\sqrt{(4x-1)}$ then we can see that
$f(x)$ is defined when $x \geq 1$ and $f(1) \geq 0\mbox{ and }f'(x) \geq 0$ so, the function is monotone increasing and it will never appear zero.
so, my question is , In this type of equation where the solution actually does not exist, then why should we get this type of solution?
my solution procedure is,
$$
\begin{align}
\sqrt{(x+1)} -\sqrt{(x-1)}&= \sqrt{(4x-1)}\\
\implies 2x-2\sqrt{x^2-1}&=4x-1\\
\implies {-2}\sqrt{x^2-1}&= 2x-1\\
\implies 4(x^2-1)&=4x^2+1-4x\\ \implies x&=5/4
\end{align}$$
| Write the equation as
$$
\sqrt{x+1}=\sqrt{x-1}+\sqrt{4x-1}
$$
Then you must have
\begin{cases}
x+1\ge0\\
x-1\ge0\\
4x-1\ge0
\end{cases}
which boils down to $x\ge1$. Now square, you're sure not to add spurious solutions, because both sides represent non negative numbers:
$$
x+1=x-1+4x-1+2\sqrt{(x-1)(4x-1)}
$$
or
$$
-4x+3=2\sqrt{(x-1)(4x-1)}
$$
Now the right hand side is non negative, so also the left hand side must be, which means
$$
-4x+3\ge0
$$
or $x\le 3/4$. With the previous limitation, this has the consequence that no solution can exist.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the equation of a circle containing three given points Problem:
Determine the equation of the circle that passes through three points, $J(-3, 2)$, $K(4, 1)$, and $L(6, 5)$.
I thought of using systems like so:
$$\left\{ \begin{array}{rcl} (x+3)^2 + (y-2)^2 = r^2 \\ (x-4)^2 + (y-1)^2 = r^2 \\ (x-6)^2 + (y-5)^2 = r^2 \\ \end{array} \right.$$
After equating the expressions on the left hand side of each equation, expanding and simplifying, I found out that $y = 7x -2$. I decided to substitute the $y$ into two of the expressions to solve for $x$. But it only gives me $50x^2 - 50x + 25 = 50x^2 - 50x + 25$, which does not help. Can someone please help me out? Thanks!
| Use the equation $x^2+y^2+2gx+2fy+c=0$. Substitute those values, you will get three equations involving g,f,c. Substituting the values back to the equation will be your equation. The answer will be $x^2+y^2-2x-10y+1=0$ or $(x-1)^2+(y-5)^2=25$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\iint_D 6x-2x^2y^2+6y\,dx$$dy$
Evaluate $\iint_D 6x-2x^2y^2+6y\,dx dy$ where $D$ is the rectangle given by $-2 \leq x \leq 3$ and $-2 \leq y \leq 1$.
I've done this problem two ways. The first time I got $-630$ and then the second time I got $\frac{-1030}{9}$. Lon capa said that they were both wrong . This makes sense because volume cannot be negative.
| $$ \int_{-2}^1 \int_{-2}^3 6x-2x^2y^2+6y\ dxdy= \int_{-2}^1 \left(6\int_{-2}^3 x\ dx-2y^2\int_{-2}^3 x^2\ dx+6y\int_{-2}^3 dx\right)dy $$
$$ = \int_{-2}^1 \left(6\frac{x^2}{2}\Bigg|_{-2}^3 -2y^2\frac{x^3}{3}\Bigg|_{-2}^3 +6yx\Bigg|_{-2}^3 \right)dy = \int_{-2}^1 \left(3(9-4) -\frac{2}{3}y^2(27+8) +6y(3+2) \right)dy $$
$$ =\int_{-2}^1 \left(15 -\frac{70}{3}y^2 +30y \right)dy= 15\int_{-2}^1 dy -\frac{70}{3} \int_{-2}^1 y^2\ dy + 30\int_{-2}^1 y\ dy $$
$$ = 15y\Bigg|_{-2}^1 -\frac{70}{9}y^3 \Bigg|_{-2}^1 + 15y^2\Bigg|_{-2}^1 =15(1+2)-\frac{70}{9}(1+8)+15(1-4) $$
$$ = 15(3)-70-15(3)=-70 $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Maclaurin series of $\frac{1}{1+\sin x}$ Find the terms through degree four of the Maclaurin series of $f(x)$.
$$f(x) = \frac{1}{1+\sin x}$$
My work:
The Maclaurin series for $\sin x$ up to degree $4$ is $x - \frac{x^3}{6} + \frac{x^5}{120}$
The Maclaurin series for $\frac{1}{1+x}$ up to degree $4$ is $1 - x + x^2 - x^3 + x^4$
I substituted $x - \frac{x^3}{6} + \frac{x^5}{120}$ for $x$ in $1 - x + x^2 - x^3 + x^4$
Did I do this right?
Plugging this into WolframAlpha, I get this: http://goo.gl/SKddyh
Which doesn't seem like the answer in the text: $1-x+x^2-\frac{5x^3}{6}+\frac{2x^4}{3}$
| You did it right and nothing has to be added to the answers you received.
I do not know if you were obliged to use these steps since there is a direct way of doing this expansion applying the basic rules, that is to say that the Taylor series of $f(x)$ built at $x=0$ just write (up to the fourth degree)
$$f(0)+x f'(0)+\frac{1}{2} x^2 f''(0)+\frac{1}{6} f^{(3)}(0) x^3+\frac{1}{24}
f^{(4)}(0) x^4+O\left(x^5\right)$$ The value of the function and its first four derivatives are $1,-1 ,2,-5,16$ (they are easy to evaluate bcause $sin(0)=0$ and $cos(0)=1$).
For sure, you arrive to your result.
You also have another way : replace in the denominator $sin(x)$ by its Taylor expansion and perform the long division of numerator by denominator.
You have done a good job ! Congratulations.
| {
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Prove, that two equations are equivalent EDIT: Missed something very important! Sorry! We have $x^4+1=2(2x-1)^{1/4}$ not $x^4+1=2\sqrt{2x-1}$. One friend of mine told me that the equation $x^4+1=2(2x-1)^{1/4}$, where $x\geq \frac{1}{2}$
is equivalent to $$x^4+1=2x$$. How did he obtain this? The equation $x^4+1=2(2x-1)^{1/4}$ is equivalent to $\frac{x^4+1}{2}=\sqrt{2x-1}$ and we can consider two functions: One is $f:[\frac{1}{2},\infty)\rightarrow[0,\infty)$, $f(x)=(2x-1)^{1/4}$ and $g:[0,\infty)\rightarrow [\frac{1}{2},\infty), g(x)=\frac{x^4+1}{2}$. $ g $is the inverse of$ f$ so they can only intersect on the line $y=x$. Then we can consider the equation $x^4+1=2x$
Is that enough?
| This is false. Suppose the equivalence as given: $x^4 + 1 = 2\sqrt{2x-1}$ exactly when $x^4 + 1 = 2x$ for $x \ge 1/2$. In particular, this means that if we assume $x^4 + 1 = 2x$, then
$$x^4 + 1 = 2\sqrt{2x-1} = 2\sqrt{(x^4 +1)-1} = 2x^2,$$
so $x^4 - 2x^2 + 1 = (x^2 -1)^2 = 0$ and $x = 1$.
But noting the polynomial $p(x) = x^4 - 2x + 1$ has $p(1) = 0$ and dividing by the factor $x - 1$ gives a quotient $q(x) = x^3 + x^2 + x - 1$. Then $q(1/2) < 0$ and $q(1) > 0$ means $q$ has a root between 1/2 and 1, and so therefore does $p$. In other words, $x^4 + 1 = 2x$ and $x \ge 1/2$ does not imply $x = 1$, as we found by taking your friend's assumption.
| {
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Partial fractions Im working on some partial fraction calculus 2 homework problems and I am having some issues with one problem. The question reads:
Evaluate the integral $\displaystyle \int\dfrac{2-(5x^2+x)\;dx}{(x-1)(x+1)^2}$
I approached it by splitting it into the form
$$\dfrac{2-(5x^2+x)}{(x-1)(x+1)^2} = \dfrac A{x-1}+\dfrac B{x+1}+\dfrac C{(x+1)^2}.$$
Then I solved for $A,B$ and $C$ to get $A=-1, B=-2$ and $C=1$. Then I plugged those values back into the form I set up and started to solve for
$$\displaystyle \int\dfrac{2-(5x^2+x)\;dx}{(x-1)(x+1)^2} = - \ln|x-1|-2 \ln|x+1|-\dfrac 1{x+1}+C.$$
When I entered that answer to my online homework it said that it was incorrect and I don't understand why. Hopefully someone can point out where I when wrong and possibly explain how to answer this problem correctly. Thank you!
| Your answer is very close except the constant multiplied by $\ln|x + 1|$. First, look at
$$\dfrac{2 - x - 5x^2}{(x - 1)(x + 1)^2} = \dfrac{A}{x - 1} + \dfrac{B}{x + 1} + \dfrac{C}{(x + 1)^2}$$
Multiply both sides by $(x - 1)(x + 1)^2$ to get
$$2 - x - 5x^2 = A(x + 1)^2 + B(x - 1)(x + 1) + C(x - 1)$$
If $x = 1$, then $A = -1$. If $x = -1$, then $C = 1$. Finally, $B = -4$ because
$\begin{aligned}
A(x + 1)^2 + B(x - 1)(x + 1) + C(x - 1) &= Ax^2 + 2Ax + A + Bx^2 - B + Cx - C\\
&= x^2(A + B) + x(2A + C) + (A - B - C)
\end{aligned}$
which implies that if $A + B = -5$, then $B = -4$ (also works for $A - B - C = 2$). So we get
$$\int \left(-\dfrac{1}{x - 1} - \dfrac{4}{x + 1} + \dfrac{1}{(x + 1)^2} \right)\,dx$$
Thus, the answer is
$$-\ln|x - 1| - 4\ln|x + 1| - \dfrac{1}{x + 1} + \mbox{C}$$
| {
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Different solutions of $x+y+z=10$ where $x$, $y$, $z$ are all positive integers and $x, y, z \leq 10$ The number of solutions to the equation $x+y+z=10$ where $x,y,z$ are positive integers, is given by ${k−1 \choose n−1}$, where in this case $k=10,n=3$, giving us ${9 \choose 2} = 36$
Now we have
$x + y + z = 10$ with $x, y, z \leq 10$ (where $x,y,z$ are positive integers and can be the same)
What are the different methods by which we can solve this?
| Following up on Trismegistos' idea of using generating functions: note that limiting the values to 10 really makes no difference, each variable can at most take the value 8 (two at 1, other one is 8). But if there is no limit, we can just write:
\begin{align}
[z^{10}] (z + z^2 + \cdots)^3
&= [z^{10}] z^3 (1 + z + z^2 + \cdots)^3 \\
&= [z^7] (1 - z)^{-3} \\
&= (-1)^3 \binom{-3}{7} \\
&= 36
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Number theory: Odd number and powers of 2 Is is true that for any odd natural number $x > 2$, there exists a positive natural number $y$, such that $x^y = 2^n+1$ or $x^y=2^n-1$ where $n$ is a also natural $> 0$. This cannot be solved by simple group theory methods, since we demand that $x^y$ be exactly $2^n+1$ or $2^n-1$ and not only modulu $2^n$.
Thanks.
| This is to show without the Catalan conjecture/theorem that $11^y\pm 1$ cannot be a power of $2$. Begin with the polynomial $p=x^2-x-1$ whose value at $x=4$ is $11.$ Note that the last two terms in the expansion of $p^y$ are $yx+1$ if $y$ is even, and $-yx-1$ if $y$ is odd. This means that if $y$ is even then $p^y+1$ is $2$ mod $4$ [so not a power of $2$] and also that if $y$ is odd then $p^y-1$ is also $2$ mod $4$. So we may assume either $y$ is even and we're looking at $p^y-1$ or else $y$ is odd and we look at $p^y+1,$ in both cases wanting to rule out getting a power of $2$.
Now if $y$ is even, we have for $y=2$ that $p^2-1=(x-2)(x-1)x(x+1)$, and since for larger even $y$ we have $p^y-1$ divisible by $p^2-1$ we have in all these cases the four divisors $2,3,4,5$ (recalling here $x=4$), and so $p^y-1$ is not a power of $2$.
On the other hand if $y$ is odd we have for $y=1$ that $p^1+1=x(x-1)$, and for larger odd $y$ we have $p^y-1$ divisible by $p-1$, so in these cases $p^y+1$ is divisible by $3,4$ so cannot be a power of $2$.
So there is an elementary way to show that $11^y\pm 1$ is not a power of $2$, without the need of a deep theorem like Catalan's.
Edit: I realized after entering this that it can be done without the polynomial. For $y$ even, $11^y+1$ is 2 mod 4 (because $11$ is odd its even powers are $1$ mod 4) Then since $11^2-1$ is divisible by $2,3,5$ we have for larger even integers that $11^y-1$ is divisible by $11^2-1$, so also by $2,3,5.$ For $y$ odd we have that $11^y-1$ is 2 mod 4 while $11^1+1$ is divisible by $2,3$ and then for larger $y$ we have $11^y+1$ is divisible by $11^1+1$ and so also by $2,3.$
| {
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Functional equation $xf(y)+yf(x)=f(x+y)^2-f\left(x^2\right)-f\left(y^2\right)$ Here is a nice problem:
Let $f:\mathbb R\to\mathbb R$ be a function, $\mathbb R$ is the set of real numbers, satisfying the following properties: $f(1)$ is an integer and
$$xf(y)+yf(x)=f(x+y)^2-f\left(x^2\right)-f\left(y^2\right)\text,$$
for all real numbers $ x , y $.
$f(x)=0$ is a solution, another is $ f(x)=x $. These are all solutions?
Better asking: determine all functions that satisfy the above conditions.
I would like to see a complete solution! Thank you!
| You can show that the only functions $ f : \mathbb R \to \mathbb R $ satisfying
$$ x f ( y ) + y f ( x ) = f ( x + y ) ^ 2 - f \left( x ^ 2 \right) - f \left( y ^ 2 \right) \tag 0 \label 0 $$
are the constant zero function and the identity function. It's easy to check that those in fact are solutions. We try to prove the converse.
Defining $ a = f ( 0 ) $ and plugging $ y = 0 $ in \eqref{0} we get
$$ f \left( x ^ 2 \right) = f ( x ) ^ 2 - a ( x + 1 ) \text . \tag 1 \label 1 $$
This, in particular shows $ a \in \{ 0 , 2 \} $, as is seen by putting $ x = 0 $. Defining $ b = f ( - 1 ) $ and putting $ x = - 1 $ in \eqref{1}, we get $ f ( 1 ) = b ^ 2 $, which then letting $ x = 1 $ in \eqref{1} gives $ b ^ 4 - b ^ 2 - 2 a = 0 $. Using \eqref{1}, we can rewrite \eqref{0} as
$$ x f ( y ) + y f ( x ) = f ( x + y ) ^ 2 - f ( x ) ^ 2 - f ( y ) ^ 2 + a ( x + y + 2 ) \text . \tag 2 \label 2 $$
Now, substituting $ x - 1 $ for $ x $ and $ 1 $ for $ y $ in \eqref{2} we get
$$ b ^ 2 ( x - 1 ) + f ( x - 1 ) = f ( x ) ^ 2 - f ( x - 1 ) ^ 2 - b ^ 4 + a ( x + 2 ) \text , $$
while letting $ y = - 1 $ in \eqref{2} we have
$$ b x - f ( x ) = f ( x - 1 ) ^ 2 - f ( x ) ^ 2 - b ^ 2 + a ( x + 1 ) \text . \tag 3 \label 3 $$
Combining the last two equations we get
$$ f( x - 1 ) - f ( x ) + \left( b ^ 2 + b - 2 a \right) x = - b ^ 4 + 3 a \text . \tag 4 \label 4 $$
In particular, for $ x = 0 $ this shows that $ b ^ 4 + b - 4 a = 0 $, which together with $ b ^ 4 - b ^ 2 - 2 a = 0 $ gives $ b ( b + 1 ) \left( b ^ 2 - b - 1 \right) = 0 $ by eliminating $ a $ from the equations. Using these equations and $ a \in \{ 0 , 2 \} $, it's straightforward to check that $ b ^ 2 - b - 1 $ can't be equal to $ 0 $, and thus we must have $ b \in \{ 0 , - 1 \} $ and $ a = 0 $. If $ b = 0 $, then by \eqref{4} we will have $ f ( x - 1 ) = f ( x ) $, which together with \eqref{3} shows that $ f $ if the constant zero function. If $ b = - 1 $, then $ f ( 1 ) = b ^ 2 = 1 $, and \eqref{4} will give $ f ( x + 1 ) = f ( x ) + 1 $. Thus letting $ y = 1 $ in \eqref{2} we'll get
$$ x + f ( x ) = \left( f ( x ) ^ 2 + 2 f ( x ) + 1 \right) - f ( x ) ^ 2 - 1 \text , $$
which shows that $ f ( x ) = x $.
| {
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Is this fraction even possible to put into partial fractions? I'm to integrate $\int\frac{x}{x^2+6x+13}dx$
But I'm finding it impossible to do anything with it.
Since $x^2+6x+13$ is an irreducible quadratic factor, and the only factor it means that the partial fraction should be in the form...
$\frac{Ax+B}{x^2+6x+13}$
Hence $x = (Ax+B)(x^2+6x+13)$
$=>x= Ax^3+(6A+B)x^2+(13A+6B)x+13B$
Equating coefficients mean that A =0, 6A+B = 0, 13A+6B = 1 and 13B = 0.
Obviously this doesn't hold. Since 13A+6B apparently = 1 despite A and B both coming out as zero.
The other way I did it is
Let x = 0, hence 0 = 13B, therefore B = 0.
Let x = 1, hence 1 = A + 6A + 13A = 20A, therefore A = $\frac{1}{20}$
But that means $\int\frac{x}{x^2+6x+13}dx$ = $\int\frac{x}{20(x^2+6x+13)}dx$, which isn't possible unless x is always 0. And it doesn't help me integrate it.
Quite clearly I've done something wrong something along the lines, but I have no idea what?
Any pointers?
| Actually, it is a simple partial fraction and your mistake is
$$x = (Ax+B)\color{red}{(x^2+6x+13)}$$ the red part should be deleted and so
$$x = (Ax+B)\implies A=1,B=0$$
For the integration, you have
$$\frac{1}{2}\int\frac{(2x+6)-6}{x^2+6x+13}dx\\\frac{1}{2}\int\frac{(2x+6)}{x^2+6x+13}-\frac{6}{(x+3)^2+4}dx$$
| {
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Prove that $\frac{x^2+1}{x^2(1-x)}>8$ for $x\in(0,1)$
Show that $\displaystyle \frac{x^2+1}{x^2(1-x)}>8$ for $x\in(0,1)$
I thought about derivative but I think it's too complicated, do you have any ideas?
Progress
*
*instead show $x^2+1 > 8x^2(1-x) = 8x^2 - 8x^3$. -- k.stm
*I have $(2x+1)(4x^2-2x+1)>7x^2$ am I on good way? because I don't know what next
| We have, $\displaystyle \frac{x^2+1}{x^2(1-x)} = \displaystyle \frac{x+\frac{1}{x}}{x(1-x)}= \displaystyle \frac{2 + (\sqrt x - \frac{1}{\sqrt x})^2}{\frac{1}{4} - (x-\frac{1}{2})^2} > 8$.
(inequality is strict since $x\in(0,1)$).
| {
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Question from Spivak's Calculus. Hint makes no sense. I'm currently working through Spivak. This question has me a little bit tied up in knots. I was able to answer it, but I can't make heads or tails of the hint.
Here's the question:
Find out when $(x+y)^5=x^5+y^5$. Hint: from the assumption $(x+y)^5=x^5+y^5$ you should be able to derive the equation $x^3 + 2x^2y + 2xy^2+y^3= 0$ if $xy\neq 0$. (I got this part, the next part is what confuses me.) This implies that $(x+y)^3=x^2y + xy^2=xy(x+y)$.
The last part of the above is what has me confused. What he's stating to be equal to $(x+y)^3$looks to me to be the difference of the binomial expansion of $(x+y)^3$ and the other equation given in the hint. Am I missing something???
| $$(x+y)^3 = x^3+3x^2y+3xy^2+y^3.$$
Suppose $x^3 + 2x^2y+2xy^2+y^3 = 0$. Add to both sides $x^2y+xy^2$:
$$x^3+2x^2y+2xy^2+y^3 + \color{red}{x^2y+xy^2} = \color{red}{x^2y+xy^2} \\
x^3+3x^2y+3xy^2+y^3 = x^2y+xy^2 \\
(x+y)^3 = xy(x+y).$$
| {
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Proving that $\frac{\sec\theta\cdot\sin\theta}{\tan\theta+\cot\theta}=\sin^2\theta$ The question is:
Prove that: $$\dfrac{\sec\theta\cdot\sin\theta}{\tan\theta+\cot\theta}=\sin^2\theta$$
My proof is shown below. If anyone has an alternate proof please, please post it. Thanks!
| Proof:
$$\dfrac{\sec\theta\cdot\sin\theta}{\tan\theta+\cot\theta}=\dfrac{\left(\dfrac{1}{\cos\theta}\cdot\sin\theta\right)}{\tan\theta+\cot\theta}$$
$$=\dfrac{\left(\dfrac{\sin\theta}{\cos\theta}\right)}{\left(\dfrac{\sin\theta}{\cos\theta}+\dfrac{\cos\theta}{\sin\theta}\right)}$$
$$=\dfrac{\left(\dfrac{\sin\theta}{\cos\theta}\right)}{\left(\dfrac{\sin^2\theta}{\sin\theta\cdot\cos\theta}+\dfrac{\cos^2\theta}{\sin\theta\cdot\cos\theta}\right)}$$
$$=\dfrac{\left(\dfrac{\sin\theta}{\cos\theta}\right)}{\left(\dfrac{\sin^2\theta+\cos^2\theta}{\sin\theta\cdot\cos\theta}\right)}$$
$$=\dfrac{\left(\dfrac{\sin\theta}{\cos\theta}\right)}{\left(\dfrac{1}{\sin\theta\cdot\cos\theta}\right)}$$
$$=\dfrac{\sin\theta}{\cos\theta}\cdot\left(\sin\theta\cdot\cos\theta\right)$$
$$=\dfrac{\sin\theta}{\cos\theta}\cdot\sin\theta\cdot\cos\theta$$
$$=\sin^2\theta$$
$$\displaystyle \boxed{\therefore \dfrac{\sec\theta\cdot\sin\theta}{\tan\theta+\cot\theta}=\sin^2\theta}$$
| {
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Asymptotic behavior of $\sum_{k=1}^n \binom{n}{k} \left(\frac{ck}{n}\right)^k$ I am looking to show that
$$\lim_{n \rightarrow \infty}\frac{1}{e^n}\sum_{k=1}^n \binom{n}{k} \left(\frac{ck}{n}\right)^k = 0. $$
In my application, $c = (e+1)/2 \approx 1.85914\ldots$. I have been looking all over the place, but I can't seem to find a closed form expression or good upper bound for the sum.
The obvious estimation
$$\sum_{k=1}^n \binom{n}{k} \left(c\frac{k}{n}\right)^k \leq \sum_{k=0}^n \binom{n}{k} \left(c\right)^k \leq \left(1+c\right)^n $$
won't do the trick, since $1+c=1+(1+e)/2 > e$.
Any ideas?
| To estimate the sum , we first consider even $n$ and note the following:
*
*$c = \frac{(1+e)}{2} \approx 1.85914\dots < e$, and therefore, the last term of the sum for $k=n$ can be ignored as $\binom{n}{n}\left(\frac{c}{e}\right)^n \rightarrow 0$ for $n \rightarrow \infty$.
*$\binom{n}{k} = \binom{n}{n-k}$, and therefore, both terms $\binom{n}{k}\left(c \frac{k}{n}\right)^k$ and $\binom{n}{k}\left(c \frac{n-k}{n}\right)^{n-k}$ have the same binomial coefficient in the sum.
*The idea is to estimate the sum of both terms by an exponential function of the form $c \mapsto e^{m k + b}$, where $m$ and $b$ depend only on $n$ and $c$.
*Indeed, the log of the sum, the function $k \mapsto \log\left( \left(c\frac{k}{n}\right)^k + \left(c\frac{n-k}{n}\right)^{n-k} \right)$, is strictly convex and on the interval $\left[1,\frac{n}{2}\right]$ it is upper-bounded by the linear function
$$ f(k) = \left(\frac{\log(4)}{n}-\log(2c)\right) k + n \log(c).$$
*Thus, $$\binom{n}{k}\left(c\frac{k}{n}\right)^k + \binom{n}{n-k}\left(c\frac{n-k}{n}\right)^{n-k} \leq \binom{n}{k} e^{f(k)}.$$
We can bound the sum by
\begin{eqnarray}
\frac{1}{e^n} \sum_{k=1}^{ \frac{n}{2} } \binom{n}{k} e^{f(k)} & = & \frac{1}{e^n} \sum_{k=1}^{\frac{n}{2}} \binom{n}{k} 4^{\frac{k}{n}} c^n \left(\frac{1}{2c}\right)^k \\
& \leq & 4\frac{c^n \left(1+ \frac{1}{2c}\right)^n}{e^n} \\
& = & 4\left(\frac{\frac{1}{2} + c}{e}\right)^n \approx 4\left(\frac{2.35914}{e}\right)^n.
\end{eqnarray}
Since $2.35914 < e$, we get exponential convergence to $0$.
For odd $n$ the argument is essentially the same, except that we need to also consider the central term (for $k = \frac{n}{2}+1$) separately.
\begin{eqnarray}
\binom{n}{\frac{n}{2}+1} \left(c \frac{\frac{n}{2}+1}{2}\right)^{\frac{n}{2}+1} & \leq & 2^n \left( \sqrt{c \left(\frac{1}{2} + \frac{1}{n}\right)}\right)^n \cdot c \left(\frac{1}{2} + \frac{1}{n}\right) \\
& = & \left( \sqrt{c \left(2 + \frac{4}{n}\right)}\right)^n \cdot c \left(\frac{1}{2} + \frac{1}{n}\right).
\end{eqnarray}
With $\sqrt{c \left(2 + \frac{4}{n}\right)} \approx 1.92828 < e$, the result follows for odd $n$ as well.
Note on generality:
So basically, if you want to estimate the asymptotics of
$$ \frac{1}{x^n} \sum_{k=1}^n \binom{n}{k} \left(c\frac{k}{n}\right)^k, $$
the limit is guaranteed to be $0$ as long as $c \in [0,x-0.5)$.
| {
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Linear Algebra question on diagonlization Please check my work My first question is that is a basis for each eigenspace the same thing as a corresponding eigenvector for an eigenspace?
Could someone tell me if im doing this correctly?
I have the matrix $A=\begin{bmatrix}
4 & 3 & 3\\
-9 & -8 & -9\\
3 & 3 & 4\\
\end{bmatrix}$
I found out that the characteristic polynomial is $-(\lambda-1)^2(\lambda+2)$
So the eigenvalues are -2 and 1.
I solved for the the eigenvector for $\lambda=-2$ and I got the eigenvector to be $\begin{bmatrix}
1 \\
-3 \\
1\\
\end{bmatrix}$
When I solve for $\lambda=1$, I get $\begin{bmatrix}
-1 \\
1 \\
0\\
\end{bmatrix}$ and $\begin{bmatrix}
-1 \\
0 \\
1\\
\end{bmatrix}$ as my basis.
So from this is my $P=\begin{bmatrix}
1 & -1 & -1\\
-3 & 1 & 0\\
1 & 0 & 1\\
\end{bmatrix}$ and my $D=\begin{bmatrix}
-2 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
\end{bmatrix}$?
And lastly,how does the negative sign in my characteristic polynomial affect my answer?
Could you please check my work, I feel like I am doing something wrong.
| The negative sign as no bearing on your answer because you're interested in the roots of the polynomial and $-(\lambda-1)^2(\lambda+2)=0\iff (\lambda-1)^2(\lambda+2)=0$.
You can check what you did yourself by computing $P^{-1}AP$.
In fact
$$\begin{align}P^{-1}AP &=\begin{bmatrix} -1& -1 &-1\\ -3 & -2 & -3\\ 1 & 1 & 2\end{bmatrix}\begin{bmatrix} 4 & 3 & 3\\ -9 & -8 & -9\\ 3 & 3 & 4\end{bmatrix}P\\
&=\begin{bmatrix} 2& 2 & 2\\ -3 & -2 & -3\\ 1 & 1 & 2\end{bmatrix}\begin{bmatrix} 1 & -1 & -1\\ -3 & 1 & 0\\ 1 & 0 & 1\end{bmatrix}=D.\end{align}$$
| {
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How prove this inequality $\sum_{k=2}^{49}\frac{1}{k^2}\ge\frac{9}{10}\ln{2}$? show that
$$\sum_{k=2}^{49}\dfrac{1}{k^2}\ge\dfrac{9}{10}\ln{2}$$
my idea: since
$$\dfrac{1}{k^2}\ge\dfrac{1}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}$$
so
$$\sum_{k=2}^{49}\dfrac{1}{k^2}\ge\sum_{k=2}^{49}\left(\dfrac{1}{k}-\dfrac{1}{k+1}\right)=\dfrac{1}{2}-\dfrac{1}{50}=\dfrac{12}{25}$$
But $$\dfrac{9}{10}\ln{2}>\dfrac{12}{25}$$
other idea:
$$\dfrac{1}{k^2}\ge\dfrac{3}{(2k-1)(2k+1)}$$
so How prove it? Thank you
and I found the constant $\dfrac{9}{10}\ln{2}$ is best value.can see
http://www.wolframalpha.com/input/?i=%28%5Csum_%7Bk%3D2%7D%5E%7B49%7D%5Cdfrac%7B1%7D%7Bk%5E2%7D%29-%289%2F10%29*ln2&dataset=
I Looking forward to answer this problem by hand! Thank you
| $$\sum_{k=2}^{49} \dfrac{1}{k^2} = \sum_{k=1}^\infty \dfrac{1}{k^2} - 1 - \sum_{k=50}^\infty \dfrac{1}{k^2} = \dfrac{\pi^2}{6} - 1 - \sum_{k=50}^\infty \dfrac{1}{k^2}$$.
The function $f(x) = 1/x^2$ is strictly decreasing on $[49, +\infty)$ so the right hand Riemann sum $\sum_{k=50}^\infty \dfrac{1}{k^2}$ is less than $\int_{49}^\infty \dfrac{1}{x^2}\, dx$. The latter is equal to $1/49$. Therefore,
$$\dfrac{\pi^2}{6} - 1 - \sum_{k=50}^\infty \dfrac{1}{k^2} > \dfrac{\pi^2}{6} - 1 - \dfrac{1}{49}$$
which one can check is numerically bigger than $\dfrac{9}{10} \log(2)$ by about $0.00069344$. This is not a direct estimate but perhaps of some help.
| {
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Prove $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}-\frac{1}{7}+\cdots$ converges. Consider the series:
$1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}-\frac{1}{7}+\cdots$
Is it convergent?
I believe I need to find a way to split the terms into additive and subtraction terms, however I'm not sure how to do this and as a result prove the outcome.
| \begin{align}
\dots &= \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots \right) - \left(\frac{1}{2} - \frac{1}{4} + \frac{1}{6} - \dots \right) \\
&= \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots \right)- \frac{1}{2} \, \left(1 - \frac{1}{2} + \frac{1}{3} - \dots \right) \\
&= \tan^{-1}(1) - \frac{1}{2} \, \log\left(1+1\right)= \frac{\pi}{4} - \frac{1}{2} \, \log(2)
\end{align}
| {
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There is 10 people that pick-up random number between 1 to 20 There is 10 people that pick-up random number between $1$ to $20$. More then one person can pick up same number (i.e. the pick-ups are independent).
What is the probability that the minimum number of all the people is grater then $8$?
My answer is $\frac{12^{10}}{20^{10}}$, I'm right?
And where I confused is about "What is the probability that the minimum number will be exactly $8$? I don't know which is right:
$$\frac{10\cdot 13^9}{20^{10}}\;\text{or}\;\left(\frac{13}{20}\right)^{10}-\left(\frac{12}{20}\right)^{10}$$
Here is the conflict:
$\frac{10\cdot 13^9}{20^{10}}$ - One person must choose $8$ so we have 10 options. then, all the others can choose a number that is $8$ or bigger: $13^9$ [from 8 to 20] and we divide it by $\Omega$.
$\left(\frac{13}{20}\right)^{10}-\left(\frac{12}{20}\right)^{10}$ - is the number of people that can pick-up a number that is equal or greater then $8$ minus the number of people the can pick up a number that is grater then $8$.
Please help to understand which is the right one...
Thank you!!
| Your answer $\dfrac{12^{10}}{20^{10}}$ to the first question is right: all $10$ people must pick numbers from among the $12$ numbers greater than $8$, the probability of which is $\left(\dfrac{12}{20}\right)^{10}$.
For the second question (minimum exactly $8$), your second answer $\left(\dfrac{13}{20}\right)^{10} - \left(\dfrac{12}{20}\right)^{10}$ is right, for the reason you stated: it's the probability that the minimum is greater than $7$, minus the probability that the minimum is greater than $8$.
In your other approach, when there are multiple people with the same choice $8$ (and the minimum is still exactly $8$), you count the same possibility multiple times for each choice of which of them you focus on as the "special" person with the minimum. You can take that into account as follows: suppose the number of people who chose exactly $8$ is $k$, which means that the remaining $10-k$ people chose numbers greater than $8$. The probability of this is
$$\binom{10}{k} \left(\frac{1}{20}\right)^{k} \left(\frac{12}{20}\right)^{10-k}$$
As all $k$ such that $1 \le k \le 10$ are possible, the probability is got by adding them up, which gives, via the binomial theorem
$$(x+y)^{n} = \sum_{k=0}^{n} \binom{n}{k} x^k y^{n-k},$$
the same answer
$$\left(\frac{1}{20} + \frac{12}{20}\right)^{10} - \binom{10}{0}\left(\frac{1}{20}\right)^{0} \left(\frac{12}{20}\right)^{10}$$
$$ = \left(\dfrac{13}{20}\right)^{10} - \left(\dfrac{12}{20}\right)^{10}$$
as before.
| {
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How would you determine the transformation matrix? Suppose there exists a linear transformation $T$ where $T: \mathbb{R^3} \to \mathbb{R^5}$ and $T(\textbf{x}) = \text{A} \textbf{x}$. Given
$$ \text{A} \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ 1\\ 2\\ 1 \\3 \end{pmatrix} ,
\quad
\text{A} \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} \hspace{6pt}1 \\ \hspace{6pt}2\\ \hspace{6pt}2\\ \hspace{6pt}2 \\-1 \end{pmatrix} ,
\quad
\text{A} \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 2 \\ 3\\ 4\\ 3 \\2 \end{pmatrix} ,
$$
how can we then determine the transformation matrix $\text{A}$? What are some "standard methods" to do it? Isn't this the same as finding the change-of-coordinates matrix when making a change of basis?
| Let $v_1=(1,1,1)^T, v_2=(1,2,3)^T$ and $v_3=(0,1,1)^T$ then we verify easily that $B=(v_1,v_2,v_3)$ is a basis of $\Bbb R^3$. Let $B_c$ the canonical basis and $P$ the change matrix from $B$ to $B_c$. Now if $\textbf{x}$ is a vector of $\Bbb R^3$ such that $X=(x,y,z)^T$ is its coordinates in $B_c$ then $PX=(\alpha,\beta,\gamma)^T$ is its coordinates in $B$ and then
$$T(\textbf{x})=A(PX)=\alpha A v_1+\beta A v_2+\gamma A v_3$$
| {
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Find constants $a$ and $b$ such that $ \lim_{x \to 0} \frac{1}{bx-\sin x} \int^x_0 \frac{t^2dt}{ \sqrt{a+t}}=1$
Find constants $a$ and $b$ such that $$ \lim_{x \to 0} \frac{1}{bx-\sin x} \int^x_0 \frac{t^2dt}{ \sqrt{a+t}}=1$$
First,$a$ should be positive to make sure the limit is meaningful as $x \to 0^-$ .
Then I check the limit of the numerator,say $- \frac{a}{2}<x<0$. For t in the interval(x,0), there's $\frac{1}{\sqrt{a+t}}<M$, where $M>0$,so $$| \int^x_0 \frac{t^2dt}{ \sqrt{a+t}} | <M |\int^x_0t^2dt| =M \frac{-x^3}{3},$$ by using the sandwich theorem I get $\lim_{x \to 0^-} \int^x_0 \frac{t^2dt}{ \sqrt{a+t}} =0$,thing should be the same when $x>0$ .
So the limit is the form $0/0$ . Apply L'Hopital's rule twice I get
$$\begin{align} \lim_{x \to 0} \frac{1}{bx-\sin x} \int^x_0 \frac{t^2dt}{ \sqrt{a+t}} &= \lim_{x \to 0} \frac{x^2}{ \sqrt{a+x}(bx-\sin x)} \\ &= \lim_{x \to 0} \frac{2x} { \frac{bx-\sin x}{2 \sqrt{a+x}}+ \sqrt{a+x} (b- \cos x)} \\ &= \lim_{x \to 0} \frac{4x \sqrt{a+x}}{3bx+2ab-2(a+x)\cos x -\sin x} \end{align}$$Now the limit becomes $ \frac{0}{2ab-2a}$,so $2a(b-1)=0$ and $b=1$ cause $a$ is positive.
Apply L'Hopital's rule again $$\begin{align} \lim_{x \to 0} \frac{4x \sqrt{a+x}}{3x+2a-2(a+x)\cos x -\sin x} &= \lim_{x \to 0} \frac{4( \sqrt{a+x} + \frac{x}{2 \sqrt{a+x}})}{3+2a \sin x- 2 \cos x + 2 x \sin x - \cos x} \\ & = \lim_{x \to 0} \frac{2(2a+3x)}{\sqrt{a+x} (3-3 \cos x + 2a \sin x + 2x \sin x )} \\ &= \frac{4a}{\sqrt{a} 0} \quad ?!\end{align}$$
and I fail to solve it.
Thanks in advance.
| Using the L'Hôpital's rule we have
$$ \lim_{x \to 0} \frac{1}{bx-\sin x} \int^x_0 \frac{t^2dt}{ \sqrt{a+t}}=\lim_{x \to 0}\frac{x^2}{(b-\cos x) \sqrt{a+x}}=\lim_{x \to 0} \frac{1}{\frac{(b-\cos x)}{x^2} \sqrt{a+x}}=1\\\iff (b=1)\land\left(\frac{\sqrt a}{2}=1\right)\iff (b=1)\land(a=4)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/715835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Safe prime mod 24 Given a safe-prime $p = 2q + 1$ where $q$ is also a prime and $p \gt 7$, I've read in a crypto.se answer that either $p \equiv 11 \pmod {24}$ or $p \equiv 23 \pmod {24}$.
I understand the proofs of why $p^2 \equiv 1 \pmod {24}$, and $p \equiv 1 \pmod 6$ or $p \equiv 5 \pmod 6$ for any prime $p$, and I can see that $p \equiv 11 \pmod {24}$ and $p \equiv 23 \pmod {24}$ are consistent with that, but can anyone explain why the other possible congruences for a prime $p$ (such as $p \equiv 1 \pmod {24}$) are excluded by $p$ being a safe prime?
My reasoning so far is:
For a prime $p \equiv 1 \pmod 6$ and $p \equiv (1,7,13,19) \pmod {24}$, or $p \equiv 5 \pmod 6$ and $p \equiv (5,11,17,23) \pmod {24}$.
For a safe prime $p = 2q + 1$ it cannot be true that $p \equiv 1 \pmod 6$, otherwise $2q$ would be divisible by 6 and $q$ would not be prime. This eliminates 1, 7, 13 and 19.
Likewise $p = 2q+1 \equiv 5 \pmod {24}$ and $p = 2q+1 \equiv 17 \pmod {24}$ cannot hold, otherwise $q$ would have to be even: $q \equiv 2 \pmod {24}$ or $q\equiv 8\pmod {24}$ respectively.
This leaves $p \equiv 11 \pmod {24}$ or $p \equiv 23 \pmod {24}$ as possible congruences.
Is this correct and sufficient, and/or is there a better way of demonstrating that either $p \equiv 11 \pmod {24}$ or $p \equiv 23 \pmod {24}$ can and must hold?
| A prime must be $\{1,5\} \text{ mod } 6$, so $\{1,5,7,11\} \text{ mod } 12$, so $\{1,5,7,11,13,17,19,23\} \text{ mod } 24$. This is true for $q$. Double each of these and add 1:
$1 \rightarrow 3 \not \in \{1,5,7,11,13,17,19,23\}$
$5 \rightarrow 11 \in \{1,5,7,11,13,17,19,23\}$
$7 \rightarrow 15 \not \in \{1,5,7,11,13,17,19,23\}$
$11 \rightarrow 23 \in \{1,5,7,11,13,17,19,23\}$
$13 \rightarrow 3 \not \in \{1,5,7,11,13,17,19,23\}$
$17 \rightarrow 11 \in \{1,5,7,11,13,17,19,23\}$
$19 \rightarrow 15 \not \in \{1,5,7,11,13,17,19,23\}$
$23 \rightarrow 23 \in \{1,5,7,11,13,17,19,23\}$
So the only options for $p$ are $11$ and $23 \text{ mod } 24$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate the limit $\lim\limits_{x\to0+}\left(\frac{3^x+5^x}{2}\right)^{\frac1x}$ Evaluate
$$
\displaystyle\lim_{x\to0+}\left(\frac{3^x+5^x}{2}\right)^{\displaystyle\frac{1}{x}}
$$
And actually I have my answer and just need someone to verify this for me since I haven't done something like this for a long time.
First, to deal with the pesky $1/x$, I take the natural log inside the limit:
\begin{align}
\lim_{x\to0+}\ln\left(\frac{3^x+5^x}{2}\right)^{\displaystyle\frac{1}{x}}
&= \lim_{x\to0+}\frac{1}{x}\ln\left(\frac{3^x+5^x}{2}\right)\\
&= \lim_{x\to0+}\frac{\ln(3^x+5^x)-\ln2}{x}\\
&= \lim_{x\to0+}\frac{3^x\ln3+5^x\ln5}{3^x+5^x}......L'Hopital's \;Rule\\
&=\frac{\ln3+\ln5}{2}\\
&=\frac{1}{2}\ln3+\frac{1}{2}\ln5
\end{align}
And since what we calculated was the limit the of the natural log, the final answer would be $\displaystyle e^{\frac{1}{2}ln3+\frac{1}{2}ln5}=e^{\sqrt{3}+\sqrt{5}}$. Please tell me if I did this correctly, thanks.
| If the limit $L=\lim_{n\rightarrow a} f(x)^{g(x)}$,
is of the form $1^{\infty}$ maning $\lim_{x\rightarrow a}f(x)=1, \lim_{x\rightarrow a} g(x) = \infty.$ Then $L=\exp[\lim_{x\rightarrow a}(f(x)-1)g(x)]$. So in this case
$$L= \exp \left [\lim_{x\rightarrow 0} \frac{1}{x} \left( \frac{3^x + 5^x}{2}-1 \right)\right]=
\exp \left [ \frac{1}{2} \left (\lim_{x\rightarrow 0} \frac {3^x-1}{x} + \lim_{x\rightarrow 0} \frac{5^x-1}{x} \right ) \right]=\exp[\frac{1}{2}(\ln 3 + \ln 5)] =\sqrt{15}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/720185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the determinant without row expansion Show that the determinant of the matrix
\begin{bmatrix} 1& a& a^3\\
1& b& b^3\\
1& c& c^3\end{bmatrix}
is $(a-b)(b-c)(c-a)(a+b+c)$ without expanding.
I was able to get out $(a-b)(b-c)(c-a)$ but couldn't complete.
| \begin{bmatrix} 1& a& a^2&a^3\\
1& b& b^2 &b^3\\
1& c& c^2 &c^3\\
1& X& X^2 &X^3
\end{bmatrix}is the well known Vandermonde determinant. When expend with respect to the last line, this is a polynomial
whose $X^2$ coefficient is the opposite of the result.
Now, expending
$$
(c-a)(b-a)(c-b)(X-c)(X-b)(X-a) \\= (c-a)(b-a)(c-b)(X^3 - (a+b+c)X+\cdots)
$$gives the result $$
(a+b+c)(c-a)(b-a)(c-b)
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Does there exist bivariate polynomials $p$ and $q$ such that $p(x,y)^2 = q(x, y)^2 ( x^2 + y^2)$? Does there exist bivariate polynomials $p$ and $q$ such that $p(x,y)^2 = q(x, y)^2 ( x^2 + y^2)$ for all real $x$ and $y$?
| The answer is no.
If $p(x,y)^2 = q(x, y)^2 ( x^2 + y^2)$ for all real $x$ and $y$, then $p(X,Y)^2 = q(X,Y)^2 (X^2 + Y^2)$ in $\mathbb R[X,Y]$.
Since $X^2+Y^2$ is irreducible (hence prime) in $\mathbb R[X,Y]$ we get $p(X,Y)=(X^2+Y^2)p_1(X,Y)$, and therefore $(X^2+Y^2)p_1(X,Y)^2=q(X,Y)^2$. Using the same argument we have $q(X,Y)=(X^2+Y^2)q_1(X,Y)$, and thus we get $p_1(X,Y)^2 = q_1(X,Y)^2 (X^2 + Y^2)$. Obviously, $\deg_X p_1<\deg_Xp$ and $\deg_X q_1<\deg_Xq$. Continuing the procedure we obtain a relation of one of the above forms, say $p(X,Y)^2 = q(X,Y)^2 (X^2 + Y^2)$ with $\deg_Xp=1$ and $\deg_Xq=0$. Can you continue from here?
Remark. A slight generalization: in a UFD is not possible to have a relation of the form $a^2=b^2p$ with $a,b$ non-zero and $p$ prime.
| {
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Given six numbers $x,y,z,a,b,c$ which satisfy the following relations. Express $x+y+z$ in terms of $a,b,c$ Given six numbers $x,y,z,a,b,c$ which satisfy the following relations
$y^2+yz+z^2=a^2$
$z^2+zx+x^2=b^2$
$x^2+xy+y^2=c^2$
Express $x+y+z$ in terms of $a,b,c$
My attempt:
$\dfrac{(y^3-z^3)}{(y-z)}=a^2,\dfrac{(z^3-x^3)}{(z-x)}=b^2,\dfrac{(x^3-y^3)}{(x-y)}=c^2$
But this step didn't seem to help.
So I tried adding and expressing $k(a+b+c)^2$ but some other unmanagable elements came along. I could not think of any other way to approach the problem. Please help.
| I'm using $p$, $q$, $r$ for $x$, $y$, $z$, because GeoGebra doesn't like the latter for labels. Also, I'm assuming $p$, $q$, $r$ non-negative. (I believe that minor tweaks to the argument would allow for negative values.)
Arranging edges of length $p$, $q$, $r$ on symmetric rays about point $O$ gives $\triangle ABC$.
The problem's relations for $a^2$, $b^2$, $c^2$ are in fact the Law of Cosines applied to the sub-triangles (whose angles at $O$ have measure $120^\circ$). For instance,
$$a^2 = q^2 + r^2 - 2 q r \cos 120^\circ = q^2 + r^2 - 2 q r \cdot \left(-\frac{1}{2}\right) = q^2 + r^2 + q r$$
A little more trig gives
$$\begin{align}
|\triangle ABC| &= |\triangle OBC| + |\triangle OCA| + |\triangle OAB| \\[6pt]
&= \frac{1}{2}qr\sin 120^\circ + \frac{1}{2}rp\sin 120^\circ + \frac{1}{2}pq\sin 120^\circ \\[6pt]
&= \frac{1}{2}\frac{\sqrt{3}}{2}\left(\;pq+qr+rp\;\right) \\[6pt]
&= \frac{\sqrt{3}}{4} \left(\;pq+qr+rp\;\right)
\end{align}$$
so that
$$pq+qr+rp = \frac{4}{\sqrt{3}} |\triangle ABC| = \frac{4\sqrt{3}}{3}|\triangle ABC|$$
Summing-up the three original relations gives
$$\begin{align}
a^2 + b^2 + c^2 &= 2 p^2 + 2 q^2 + 2 r^2 + p q + q r + r p \\
&= 2 ( p + q + r )^2 - 3\left( pq + qr + rp \right) \\
&= 2 ( p + q + r )^2 - 4\sqrt{3} |\triangle ABC|
\end{align}$$
so that
$$(p+q+r)^2 = \frac{1}{2}\left(\; a^2 + b^2 + c^2 + 4 \sqrt{3} |\triangle ABC|\;\right)$$
Now, we need only recall Heron's formula for the area of a triangle ...
$$|\triangle ABC|^2 = \frac{1}{16}(a+b+c)(-a+b+c)(a-b+c)(a+b-c)$$
and then we can write
$$(p+q+r)^2 = \frac{1}{2}\left(\; a^2 + b^2 + c^2 + \sqrt{3(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}\;\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluate $\int\frac{\sqrt {25 - x^2}}{ x^4}$ I'm pretty sure the method used is trig substitution. But I'm having trouble setting up and solving the problem.
| Subsititute $\theta=\sin^{-1} \dfrac{x}5\implies \dfrac{d\theta}{dx}=\dfrac{1}{\sqrt{25-x^2}}$, so we get,
\begin{align}
\\\\\\&\int\frac{\sqrt{25-x^2}}{x^4}dx
\\=&\int\frac{25-x^2}{x^4}\dfrac{1}{\sqrt{25-x^2}}d\theta
\\=&\int \dfrac{25-25\sin^2 \theta}{625\sin^4\theta}d\theta
\\=&\dfrac{1}{25}\int\dfrac{\cos^2 \theta}{\sin^4\theta}d\theta
\\=&\dfrac{1}{25}\int\cot^2\theta \csc^2\theta d\theta
\end{align}
Now subsitute $u=\cot \theta$ to finish.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Prove $\frac{\sin(a)}{\sin(b)}<\frac{a}{b}<\frac{\tan(a)}{\tan(b)}$ for $0not sure how to approach the following $\frac{\sin(a)}{\sin(b)}<\frac{a}{b}<\frac{\tan(a)}{\tan(b)}$ for $0<b<a<\pi/2$. Hints would be appreciated!
| Note that the function $\frac{\sin x}x$ is decreasing:
$$\left(\frac{\sin x}x\right)'=\frac{x\cos x-\sin x}{x^2}=\frac{\cos x}{x^2}(x-\tan x)<0,$$
because $\cos x>0$ and $\tan x>x$ on $(0,\frac\pi2)$.
So for $\frac\pi2>a>b>0$ we have $\frac{\sin a}a<\frac{\sin b}b\Longrightarrow\frac{\sin a}{\sin b}<\frac ab$.
Additionally $\frac{\tan x}x$ is increasing:
$$\left(\frac{\tan x}x\right)'=\frac{\frac{x}{\cos^2 x}-\tan x}{x^2}=\frac{x-\sin x\cos x}{x^2\cos^2 x}=\frac{2x-\sin(2x)}{2x^2\cos^2x}>0,$$
because $\sin x<x$ on $(0,\infty)$.
So for $\frac\pi2>a>b>0$ we have $\frac{\tan a}a>\frac{\tan b}b\Longrightarrow\frac ab<\frac{\tan a}{\tan b}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find a value $c$ such that $\left\|\begin{pmatrix} x^2 - y^2\\2xy \end{pmatrix}\right\| \leq |c|\left\|\begin{pmatrix} x\\y \end{pmatrix}\right\|$ This is part of a larger problem where I am trying to find the derivative of a vector valued function. I feel like I'm missing something simple. NOTE: $c$ can be a function of $x$ and $y$.
| We compute
$$(x^2 - y^2)^2 + (2xy)^2 = x^{4} - 2x^{2}y^{2}+ y^{4} + 4x^{2}y^{2} = x^{4} + 2x^2 y^2 + y^4 = (x^2 + y^2)^2$$
Hence
$$\sqrt{(x^2 - y^2)^2 + (2xy)^2} = \sqrt{(x^2 + y^2)^2} = x^2 + y^2$$
We are trying to find $c$ for which
$$ (x^2 + y^2) \leq c \sqrt{x^2 + y^2}$$
If we let $z = x^2 + y^2$ we obtain
$$ z \leq c \sqrt{z}$$
Does this seem possible?
| {
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"timestamp": "2023-03-29T00:00:00",
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Factorising a complex polynomial over $\mathbb{C}$ I'm given $f(z)=z^6-1$ to factorise over $\mathbb{C}$.
My working is as follows up to the point I don't understand:
$f(-1)=0$ and $f(1)=0$
So $(z+1)$ and $(z-1)$ are factors
$(z+1)(z-1)=z^2-1$
$(z^2-1)(z^4+pz^3+qz^2+rz+s)=z^6-1$
$z^6+pz^5+qz^4-z^4+rz^3-pz^3+sz^2-qz^2-rz-s=z^6-1$
$s=1$ because $-s=-1$
$q=1$ because $qz^4-z^4=0$
$p=0$ and $r=0$ because $pz^5-rz=0$
$(z^2-1)(z^4+z^2+1)=z^6-1$
If $z^4+z^2+1=0$
Let $y=z^2$
$y^2+y+1=0$
$y=$$-1\pm\sqrt3i\over2$
$z^2=$$-1\pm\sqrt3i\over2$
Now here's where I get stuck. In the answers section of my textbook the remaining 4 factors are listed as:
$z+{\frac{1}{2}\pm\frac{\sqrt3}{2}i},$
$z-{\frac{1}{2}\pm\frac{\sqrt3}{2}i}$
but I have:
$z=\pm\sqrt{-\frac12+\frac{\sqrt3}{2}i},$
$z=\pm\sqrt{-\frac12-\frac{\sqrt3}{2}i}$
and I don't know if I've made a mistake somewhere, or if I just don't know how to link what I have so far to the answers given by the textbook.
| You're answer is right; what you're missing is that you can actually simplify those square roots by actually computing their real and imaginary parts.
| {
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Find the number of series Find the number of series $(a_1,..., a_{2n})$ that have terms from ${\{0,...9\}}$ so that:
$$
11|\sum_{i=1}^{n}a_i-\sum_{i=n+1}^{2n}a_i
$$
(this is not a homework)
There is a similar problem (Problem 6) in the IMO 1995.
| For a fixed $n$, let $ A_r $ be the number of sequences of $n$ terms, whose sum is $r \pmod{11}$. Then, we want the value of $ \sum A_r ^2 $.
As pointed out by ABC, the generating function $f(x) = \left ( 1 + x + x^2 + \ldots + x^9 \right) ^n $ gives us the number of sequences of $n$ terms, with a total sum of $r$. We find the sum of those coefficients with the power mod 11, we use the typical counting trick. Let $\omega$ be a primitive 11th root of unity. We know that $ 1 + \omega + \omega^2 + \ldots + \omega^{10} = 0 $. For $ i \not \equiv 0 \pmod{11}$, we have $ f( \omega^i) = ( -\omega^{10i})^n$
We have $$ 11 A_r = \sum_{i=0}^{10} \omega^{-ri} f( \omega^i) = 10^n + \sum_{i=1}^{10} \omega^{-ri} (- \omega^{10i})^n = 10^n + (-1)^n \sum_{i=1}^{10} \omega^{(10n-r)i}. $$
In particular, this shows that for $r \neq 10n \pmod{11}$, we have $A_r = \frac{10^n-(-1)^{n}}{11}$. Otherwise, for $R \equiv 10n \equiv -n \pmod{11}$, we have $A_R = \frac{10^n+10 \times (-1)^n}{11}$.
Hence, we can now find the sum as
$$ \sum_{r=0}^{10} A_r^2 = 10 \left (\frac{10^n-(-1)^n}{11} \right)^2 + \left( \frac{10^n+10\times (-1)^n}{11} \right)^2$$
Note: This solution is nice 9 because 2 less than 11, which makes $f(\omega^i)$ easy to evaluate. For a much more general case, the equations would not be pretty.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the area of the surface obtained by rotating the curve of parametric equations Rotate about the $x$ axis
$x = 2t-2/3t^3$
$y = 2t^2$
$0 \leq t \leq 1$
I did the integral of $\sqrt{(2-2t^2)^2+(4t)^2}$ and got $(2x(x^2+3))/3$ and then I did the integral of $2\pi 2t^2 ((2x(x^2+3))/3)$ and I get $22\pi/9$ but apparently that's not the correct answer
| The surface area integral for a figure revolved about the $ \ x-$ axis is
$$ S \ = \ 2 \pi \ \int \ y \ \ ds \ , $$
which, for a parametric curve, will be
$$ S \ = \ 2 \pi \ \int \ y(t) \ \sqrt{ \ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dx}{dt} \right)^2} \ \ dt \ \ = \ \ 2 \pi \ \int \ y(t) \ \sqrt{ \ \left( 2 - 2t^2 \right)^2 + \left( 4t \right)^2} \ \ dt , $$
$$ = \ 2 \pi \ \int \ y(t) \ \sqrt{ \ \left( 2 + 2t^2 \right)^2} \ \ dt \ \ = \ 2 \pi \ \int \ y(t) \ \cdot \ \underbrace{(2 + 2t^2) \ \ dt}_{ds} \ \ , $$
as you have set up. However, while your arclength has been determined correctly, you have to multiply the factors of the integrand first before you set about performing the integration itself. So you want
$$ S \ = \ 2 \pi \ \int_0^1 \ ( \ 2t^2 \ ) \ \cdot \ (2 + 2t^2) \ \ dt \ \ = \ \ \ 2 \pi \ \int_0^1 \ 4t^2 \ + \ 4t^4 \ \ dt $$
$$ = \ \ 2 \pi \ \left .\left( \ \frac{4}{3}t^3 \ + \ \frac{4}{5}t^5 \right) \ \right \vert_0^1\ \ \ = \ \frac{64 \pi}{15} . $$
[corrected from what I did late last night]
| {
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Group-Isomorphism problem I want to find an group-isomorphism $$ \psi : (\mathbb{Z}/8\mathbb{Z},+) \longrightarrow \mathbb{F}_9^\times $$
which should be used to multiply elements in $\mathbb{F}_9$ or to find the inverse element in an easy way.
| As Seth and Pedro indicated the existence of such an isomorphism follows from (and is equivalent to) the cyclicity of the multiplicative group $\Bbb{F}_9^*$. To exhibit an explicit isomorphism you need to specify a construction of $\Bbb{F}_9$ and find a generator of the multiplicative group (aka a primitive element).
Elements of order $8$ are zeros of the cyclotomic polynomial $\phi_8(x)=x^4+1$. As we are working modulo $3$, we can factor it as follows
$$
\begin{aligned}
x^4+1&=x^4+4=(x^4+4x^2+4)-4x^2\\
&=(x^2+2)^2-(2x)^2=(x^2+2-2x)(x^2+2+2x)\\
&=(x^2-x-1)(x^2+x-1).
\end{aligned}
$$
So, if we identify $\Bbb{F}_9$ with $\Bbb{Z}_3[x]/\langle x^2-x-1\rangle$, then the coset
$\alpha=x+\langle x^2-x-1\rangle$ will be a generator.
The desired isomorphism $\psi:(\Bbb{Z}_8,+)\to (\Bbb{F}_9^*,\cdot)$ is given by
$$
\psi(\overline{n})=\alpha^n
$$
for all $n=0,1,\ldots,7$. This comes to the following:
$$
\begin{array}{c|ccc|c}
n&\psi(\overline{n})&&n&\psi(\overline{n})\\
\hline
0&1&&1&\alpha\\
2&\alpha^2=\alpha+1&&3&\alpha^3=\alpha^2+\alpha=2\alpha+1\\
4&\alpha^4=2&&5&\alpha^5=2\alpha\\
6&\alpha^6=2\alpha^2=2\alpha+2&&7&\alpha^7=\alpha+2
\end{array}
$$
Here I used the equation $\alpha^2=\alpha+1$ repeatedly. Note also that $\alpha^4=2=-1$, so
$\alpha^{4+j}=-\alpha^j$ for all $j$. As a final check we do that
$$
\alpha^8=\alpha\cdot\alpha^7=\alpha(\alpha+2)=\alpha^2+2\alpha=3\alpha+1=1
$$
as it should.
The way to use this in finding products and inverses is to treat the inverse of $\psi$ as a logarithm, i.e. a mapping that transforms multiplication in $\Bbb{F}_9$ into addition in $\Bbb{Z}_8$. For example (read the above table from right to left)
$$
(\alpha+2)(\alpha+1)=\alpha^7\cdot\alpha^2=\alpha^9=\alpha^{8+1}=\alpha,
$$
and
$$
(\alpha+1)^{-1}=(\alpha^2)^{-1}=\alpha^{-2}=\alpha^{8-2}=\alpha^6=2\alpha+2.
$$
You mentioned that you had found $x+2=x-1$ and $2x+1$ to be generators of $\Bbb{F}_9^*$.
That may be true, but to make that meaningful you need to specify the minimal polynomial of $x$. Above I used the minimal polynomial $x^2-x-1$, and $\alpha$ was one of its zeros ($\alpha^3=2\alpha+1$ being the other. If a cyclic group of order $8$ is generated by an element $c$, then $c^m$ is also a generator iff $\gcd(m,8)=1$. So here any of $\alpha$, $\alpha^3$, $\alpha^5$ and $\alpha^7$ is a generator. Not surprisingly those are the zeros of $x^4+1$ in this copy of $\Bbb{F}_9$.
Changing the irreducible polynomial always changes the scene. Your copy of $\Bbb{F}_9$ is $\Bbb{Z}_3[x]/\langle x^2+1\rangle$. Because $x^2+1\mid x^4-1$ in this case $x+\langle x^2+1\rangle$ will not work as a generator. If we let $\beta=x+\langle x^2+1\rangle$, then
for $\alpha=\beta+2$ (that you might call $x+2$ even though that abuses notation slightly)
we get
$$
0+\langle x^2+1\rangle =(x^2+1)+\langle x^2+1\rangle=(\alpha+1)^2+1=\alpha^2+2\alpha+2=\alpha^2-\alpha-1.
$$
Thus $x^2-x-1$ is a minimal polynomial of $\alpha$. In terms of $\beta$ the above the homomorphism looks like
$$
\begin{array}{c|ccc|c}
n&\psi(\overline{n})&&n&\psi(\overline{n})\\
\hline
0&1&&1&\beta+1\\
2&(\beta+1)^2=2\beta&&3&(\beta+1)^3=2\beta^2+2\beta=2\beta+1\\
4&(\beta+1)^4=2&&5&(\beta+1)^5=2\beta+2\\
6&(\beta+1)^6=\beta&&7&(\beta+1)^7=\beta+2
\end{array}
$$
All the elements on the right column ($\beta+1,2\beta+1,2\beta+2,\beta+2$) can take the role of the generator. Thus you get four different such isomorphisms $\psi$. Each will work equally well as the inverse of a discrete logarithm.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/730809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
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