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Mistake with Integration with Beta, Gamma, Digamma Fuctions Problem: Evaluate:
$$I=\int_0^{\pi/2} \ln(\sin(x))\tan(x)dx$$
I tried to attempt it by using the Beta, Gamma and Digamma Functions. My approach was as follows:
$$$$
Consider $$I(a,b)=\int_0^{\pi/2} \sin^a(x)\sin^b(x)\cos^{-b}(x)dx$$
$$$$
$$=\dfrac{1}{2}\beta\bigg (\dfrac{a+b+1}{2},\dfrac{1-b}{2}\bigg )= \dfrac{1}{2} \dfrac{\Gamma(\frac{a+b+1}{2})\Gamma(\frac{1-b}{2})}{\Gamma(\frac{a+2}{2})}$$ Now,
$$\dfrac{1}{2}\dfrac{\partial}{\partial a} \beta\bigg (\dfrac{a+b+1}{2},\dfrac{1-b}{2}\bigg )\bigg |_{a=0,b=1} = \int_0^{\pi/2}\ln(\sin(x))\sin^a(x)\tan^b(x)dx \bigg |_{a=0,b=1} = I$$
Now,
$$\dfrac{\partial}{\partial a} \beta\bigg (\dfrac{a+b+1}{2},\dfrac{1-b}{2}\bigg )$$
$$$$
$$=\dfrac{1}{2} \dfrac{\Gamma(\frac{1-b}{2})}{(\Gamma(\frac{a+2}{2}))^2}\bigg (\Gamma '\bigg (\frac{a+b+1}{2}\bigg )\Gamma \bigg ( \frac{a+2}{2}\bigg ) - \Gamma '\bigg ( \frac{a+2}{2}\bigg )\Gamma \bigg (\frac{a+b+1}{2}\bigg )\bigg ) $$
$$$$
$$= \dfrac{1}{2}\dfrac{\Gamma(\frac{1-b}{2})}{\Gamma(\frac{a+2}{2})}\bigg (\psi\bigg ( \frac{a+b+1}{2}\bigg )\Gamma \bigg (\frac{a+b+1}{2}\bigg ) - \psi\bigg ( \frac{a+2}{2}\bigg )\Gamma \bigg (\frac{a+b+1}{2}\bigg )\bigg )$$
$$$$
$$=\dfrac{1}{2} \dfrac{\Gamma(\frac{a+b+1}{2})\Gamma(\frac{1-b}{2})}{\Gamma(\frac{a+2}{2})}\bigg (\psi\bigg ( \frac{a+b+1}{2}\bigg )-\psi\bigg ( \frac{a+2}{2}\bigg )\bigg )$$
$$$$
$$\Rightarrow\dfrac{\partial}{\partial a} \beta\bigg (\dfrac{a+b+1}{2},\dfrac{1-b}{2}\bigg )=\dfrac{1}{2} \dfrac{\Gamma(\frac{a+b+1}{2})\Gamma(\frac{1-b}{2})}{\Gamma(\frac{a+2}{2})}\bigg (\psi\bigg ( \frac{a+b+1}{2}\bigg )-\psi\bigg ( \frac{a+2}{2}\bigg )\bigg )$$
$$$$
$$\Longrightarrow \dfrac{1}{2} \dfrac{\partial}{\partial a} \beta \bigg ( \dfrac{a+b+1}{2} ,\dfrac{1-b}{2} \bigg ) \bigg |_{a=0,b=1} = I $$
$$$$
$$=\dfrac{1}{2}\times \dfrac{1}{2} \dfrac{\Gamma(\frac{a+b+1}{2})\Gamma(\frac{1-b}{2})}{\Gamma(\frac{a+2}{2})}\bigg (\psi\bigg ( \frac{a+b+1}{2}\bigg )-\psi\bigg ( \frac{a+2}{2}\bigg )\bigg ) \bigg |_{a=0,b=1}$$
$$$$
$$ =\dfrac{1}{4}\dfrac{\Gamma(0)\Gamma(1)}{\Gamma(1)}\bigg (\psi\bigg ( 1 \bigg )-\psi\bigg (1\bigg )\bigg )$$
$$$$
Could somebody please be so kind as to tell me where I have gone wrong? I would be truly grateful for your assistance. Thanks very, very much in advance!
| $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
I & = \int_{0}^{\pi/2}\ln\pars{\sin\pars{x}}\tan\pars{x}\,\dd x
\,\,\,\stackrel{x\ =\ \arcsin\pars{t}}{=}\,\,\,
\int_{0}^{1}\ln\pars{t}{t \over \root{1 - t^{2}}}
\,{\dd t \over \root{1 - t^{2}}}
\\[5mm] & =
\int_{0}^{1}\ln\pars{t}\,{t \over 1 - t^{2}}\,\dd t =
\int_{0}^{1}\ln\pars{t}\pars{{1/2 \over 1 - t} - {1/2 \over 1 + t}}\,\dd t
\\[5mm] & =
-\,{1 \over 2}\int_{0}^{1}\bracks{-\,{{\ln\pars{1 - t} \over t}}}\,\dd t +
{1 \over 2}\int_{0}^{-1}{\ln\pars{-t} \over 1 - t}\,\dd t
\\[5mm] & =
-\,{1 \over 2}\int_{0}^{1}\bracks{-\,{{\ln\pars{1 - t} \over t}}}\,\dd t -
{1 \over 2}\int_{0}^{-1}\bracks{-\,{\ln\pars{1 - t} \over t}}\,\dd t
\\[5mm] & =
-\,{1 \over 2}\int_{0}^{1}\mrm{Li}_{2}'\pars{t}\,\dd t -
{1 \over 2}\int_{0}^{-1}\mrm{Li}_{2}'\pars{t}\,\dd t =
-\,{1 \over 2}\bracks{\mrm{Li}_{2}\pars{1} + \mrm{Li}_{2}\pars{-1}}
\\[5mm] & =
-\,{1 \over 2}\sum_{n = 1}^{\infty}
\bracks{{1 \over n^{2}} + {\pars{-1}^{n} \over n^{2}}} =
-\,{1 \over 2}\sum_{n = 1}^{\infty}
\bracks{{1 \over \pars{2n}^{2}} + {1 \over \pars{2n}^{2}}} =
-\,{1 \over 4}\ \underbrace{\sum_{n = 1}^{\infty}{1 \over n^{2}}}
_{\ds{\pi^{2} \over 6}} = \bbx{-\,{\pi^{2} \over 24}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1324719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Questionnaire probability When sent a questionnaire, the probability is .5 that any particular individual to whom it is sent will respond immediately to that questionnaire. For an individual who did not respond immediately, there is a probability of .4 that the individual will respond when sent a follow-up letter. If the questionnaire is sent to 4 persons and a follow-up letter is sent to any of the 4 who do not respond immediately, what is the probability that at least 3 never respond?
The probability that at least 3 never respond is
P[3 never respond]+P[4 never respond]
P[4 never respond] is just $(.3)^4$. To find P[3 never respond], it's P[1 doesn't respond from first round] + P[1 doesn't respond from the second round].
So my steps:
P[3 never respond]+P[4 never respond]
[1 doesn't respond from first round] + P[1 doesn't respond from the second round] +P[4 never respond]
$(.5)(.5)^3(.6)^3$+$(.5^4)(.6)^3(0.4)$+$(.3)^4$
=$(.5)^4(.6)^3$+$(.5^4)(.6)^3(0.4)$+$(.3)^4$
However, according to the book, the answer is
4[$(.5)^4(.6)^3$]+4[$(.5^4)(.6)^3(0.4)$]+$(.3)^4$
=$4(.3)^3(.7)$+$(.3)^4$
Where did I go wrong? Where did the 4's come from? How do you get from the second to last step to the last one?
| Let $X$ be the number of those who respond immediately and Let $Y$ be the number of those who never respond. The probability that at least $3$ will never respond is
$$P(Y\ge 3)=P(Y=3)+P(Y=4).$$
So far, so good. Then
$$P(Y=3)=P(Y=3\mid X=0)P(X=0)+P(Y=3 \mid X=1)P(X=1).$$
and
$$P(Y=4)=P(Y=4\mid X=0)P(X=0).$$
Now
*
*$P(X=0)=\frac{1}{2^4}=\frac{1}{16},$
*$P(X=1)={4 \choose 1}\frac{1}{2^4}=\frac{1}{4},$
and
*
*$P(Y=3\mid X=0)={4 \choose 3}\left(\frac{6}{10}\right)^3\frac{4}{10}=4\left(\frac{6}{10}\right)^3\frac{4}{10},$
*$P(Y=3\mid X=1)=\left(\frac{6}{10}\right)^3,$
*$P(Y=4\mid X=0)=\left(\frac{6}{10}\right)^4.$
So,
$$P(Y\ge 3)=4\left(\frac{6}{10}\right)^3\frac{4}{10}\frac{1}{16}+\left(\frac{6}{10}\right)^3\frac{1}{4}+\left(\frac{6}{10}\right)^4\frac{1}{16}=0.0837.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to find out the greater number from $15^{1/20}$ and $20^{1/15}$? I have two numbers $15^{\frac{1}{20}}$ & $20^{\frac{1}{15}}$.
How to find out the greater number out of above two?
I am in 12th grade. Thanks for help!
| This is just another way of looking at alkabary's proof.
\begin{align}
\left[ \dfrac{20^{\frac{1}{15}}}
{15^{\frac{1}{20}}} \right]^{20}
&=\dfrac{20^{\frac 43}}{15} \\
&= \dfrac 43 20^{\frac 13} \\
&> \dfrac 43 \cdot 2 \\
&> 1
\end{align}
$\left[ \dfrac{20^{\frac{1}{15}}}
{15^{\frac{1}{20}}} \right]^{20} > 1
\implies \dfrac{20^{\frac{1}{15}}} {15^{\frac{1}{20}}} > 1
\implies 20^{\frac{1}{15}} > 15^{\frac{1}{20}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1327118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
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Solution to Fibonacci Recursion Equations Let the sequence $(a_n)_{n\geq0}$ of the fibonacci numbers: $a_0 = a_1 = 1, a_{n+2} = a_{n+1} + a_n, n \geq 0$
Show that:
i) $$a^2_n - a_{n+1}a_{n-1} = (-1)^n \text{ for }n\geq1$$
I try to show this with induction:
$n=1:$
\begin{align*}
a^2_1 - a_{2}a_{0}&=1 - a_2a_0\\
&= 1- (a_1 + a_0 )a_0\\
&= 1-(1+1)1 = -1 = (-1)^1\\
\end{align*}
Assume that: $a^2_n - a_{n+1}a_{n-1} = (-1)^n ,\forall n\geq1$
Inductive step:
\begin{align*}
a^2_{n+1} - a_{n+2}a_{n}
&=(a_n + a_{n-1}) ^2 - (a_{n+1}+a_{n})(a_{n-1}-a_{n-2})\\
&=a^2_{n}+2a_na_{n-1}+a^2_{n-1}-a_{n+1}a_{n-1}+a_{n+1}a_{n-2}-a_{n}a_{n-1}+a_{n}a_{n-2}\\
&=(a^2_{n}-a_{n+1}a_{n-1})+(a^2_{n-1}-a_{n}a_{n-1})+2a_na_{n-1}+a_{n+1}a_{n-2}+a_{n}a_{n-2}\\
&=\underbrace{(-1)^n + (-1)^{n-1}}_{=0}+2a_na_{n-1}+a_{n+1}a_{n-2}+a_{n}a_{n-2}\\
&=2a_na_{n-1}+a_{n+1}a_{n-2}+a_{n}a_{n-2}\\
&=??\\
\end{align*}
ii) $$\sum \limits_{i=0}^na_i=a_{n+2}-1 , n\geq0$$
iii) $$a^2_{n-1}+a^2_n=a_{2n} \text{ and } a_{n-1}a_n+a_na_{n+1}=a_{2n+1}, n\geq1$$
Thank you for any help
| Another flavor on the inductive step argument:
$$
\begin{split}
a_n^2-a_{n+1}a_{n-1}
&= a_n^2-(a_n + a_{n-1})a_{n-1} \\
&= a_n^2 - a_n a_{n-1} - a_{n-1}^2 \\
&= a_n(a_n-a_{n-1}) - a_{n-1}^2 \\
&= a_n a_{n-2} - a_{n-1}^2 \\
&= -\left( a_{n-1}^2 - a_n a_{n-2}\right),
\end{split}
$$
reducing to the inductive hypothesis...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1328131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Solve the inequation: $5+3\sqrt{1-x^2}\geq x+4\sqrt{1-x}+3\sqrt{1+x}$ Solve the inequation: $5+3\sqrt{1-x^2}\geq x+4\sqrt{1-x}+3\sqrt{1+x}$
I tried to substitute $x=\cos t$ but don't get any result. Who can help me?
| First of all, we need to have $1-x^2\ge 0$, $1-x\ge 0$ and $1+x\ge 0$, i.e. $-1\le x\le 1$.
Now since the both sides of
$$5-x+3\sqrt{1-x^2}\ge 4\sqrt{1-x}+3\sqrt{1+x}\tag 1$$
are non-negative, this is equivalent to
$$\left(5-x+3\sqrt{1-x^2}\right)^2\ge \left(4\sqrt{1-x}+3\sqrt{1+x}\right)^2,$$
i.e.
$$6(1-x)\sqrt{1-x^2}\ge 8x^2+3x-9\tag 2$$
Here, the solutions of $8x^2+3x-9=0$ are $x=\frac{-3\pm 3\sqrt{33}}{16}$ where $$\frac{-3-3\sqrt{33}}{16}\lt -1\lt\frac{-3+3\sqrt{33}}{16}\lt 1.$$
So, for $-1\le x\le\frac{-3+3\sqrt{33}}{16}$, since the LHS of $(2)$ is non-negative and the RHS of $(2)$ is non-positive, the given inequality $(1)$ holds.
For $\frac{-3+3\sqrt{33}}{16}\lt x\le 1$, since the both sides of $(2)$ are non-negative, $(2)$ is equivalent to
$$\left(6(1-x)\sqrt{1-x^2}\right)^2\ge \left(8x^2+3x-9\right)^2,$$
i.e.
$$(4x^2-3)(25x^2-6x-15)\le 0,$$
i.e.
$$-\frac{\sqrt 3}{2}\le x\le \frac{3-8\sqrt 6}{25}\ \ \ \text{or}\ \ \ \frac{\sqrt 3}{2}\le x\le \frac{3+8\sqrt 6}{25}.$$
Hence, in this case, we have $\frac{-3+3\sqrt{33}}{16}\le x\le \frac{3+8\sqrt 6}{25}$.
Hence, the answer is
$$-1\le x\le\frac{3+8\sqrt 6}{25}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1329947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the cubic equation of $x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$ Find the cubic equation which has a root $$x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$$
My attempt is
$$x^3=2-\sqrt{3}+3\left(\sqrt[3]{(2-\sqrt{3})^2}\right)\left(\sqrt[3]{(2+\sqrt{3})}\right)+3\left(\sqrt[3]{(2-\sqrt{3})}\right)\left(\sqrt[3]{(2+\sqrt{3})^2}\right)+2-\sqrt{3}$$
$$x^3=4+\left(\sqrt[3]{(2-\sqrt{3})^2}\right)\left(\sqrt[3]{(2+\sqrt{3})}\right)+3\left(\sqrt[3]{(2-\sqrt{3})}\right)\left(\sqrt[3]{(2+\sqrt{3})^2}\right)$$
then what I will do??
| $$x^3=4+3\left ( \sqrt[3]{2-\sqrt{3}} \right )\left ( \sqrt[3]{2+\sqrt{3}} \right )\left ( \sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}} \right )$$
$$=4+3\sqrt[3]{2^2-3}(x)=4+3x$$
$$x^3=4+3x$$
$$x^3-3x-4=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 2
} |
Substituting the value $x=2+\sqrt{3}$ into $x^2 + 1/x^2$ My teacher gave me a question which I am not able to solve:
If $x=2+\sqrt{3}$ then find the value of $x^2 + 1/x^2$
I tried to substitute the value of x in the expression, but that comes out to be very big.
| Here is a slightly weird way of doing it. $x$ looks like the quadratic formula, so if we can cook up a quadratic equation that it satisfies, we won't actually have to square it. The solutions to
$$ y^2+2by+c = 0 $$
are
$$ y = -b \pm \sqrt{b^2-c} $$
(because $a=1$ and the $b$ has a $2$ multiplying it that cancels the $2$ normally present). In this case, the number not in the square root is $2$, so $b=-2$. Then $b^2=4$, so to agree with the number inside the square root, we have $ 4-c=3 $, so $c=1$. Therefore $x$ satisfies the quadratic equation
$$ x^2 -4x+1 = 0 \tag{1} $$
From here, we can read off that
$$ x^2 = 4x-1, $$
and if we divide (1) by $x$, we also find
$$ \frac{1}{x} = 4-x, \tag{2} $$
and dividing (1) by $x^2$ gives
$$ \frac{1}{x^2} = \frac{4}{x}-1 = 4(4-x)-1 = 15-4x, $$
applying (2) to the $1/x$ term. Therefore,
$$ x^2 + \frac{1}{x^2} = 4x-1 + 15-4x = 14. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 1
} |
Improper integral and residues
Evaluate $\int_0^\infty \frac{dx}{x^4+1}$
By the residue theorem $$\int_{-R}^Rf(x)dx+\int_{C_R}dz=2\pi i\sum Res(f,z_i)$$
but I have problems to evaluate it because
$$z^4+1=0\Rightarrow z^4=-1=e^{i\pi},e^{i3\pi},e^{i5\pi},e^{i7\pi}$$
$$z=e^{\frac{i\pi}{4}},e^{\frac{i3\pi}{4}},e^{\frac{i5\pi}{4}},e^{\frac{i7\pi}{4}}$$
are the four roots, and just $z_0=e^{\frac{i\pi}{4}}$ and $z_1=e^{\frac{i3\pi}{4}}$ are in the upper half plane, I know that $z_0$ and $z_1$ are simple poles, but I don't know how to evaluate the resiudes of them.
| Well, I'm learning the subject now, too, we might as well teach each other, no? :)
Note that: $$\int_0^{+\infty} \frac{1}{x^4+1}\,{\rm d}x = \frac{1}{2}\int_{-\infty}^{+\infty}\frac{1}{x^4+1}\,{\rm d}x,$$so we'll compute that second integral. Let $R > 1$, $\gamma_1$ be the line segment joining $-R$ to $R$, and $\gamma_2$ be the semicircle centered at the origin with radius $R$, in the superior semi-plane. Let $\gamma = \gamma_1 \ast \gamma_2$ be the whole thing. Consider the function $1/(z^4+1)$. Granted, we want to find its singularities. Instead, look at the eighth roots of $1$ and give "larger steps":
Our singularities are the gray points. Write $\omega = \sqrt{2}(1+i)/2$. The only singularities that'll come into play are $\omega$ and $-\overline{\omega}$. Do notice that $\omega^4 = (-\overline{\omega})^4=-1$, we'll use that soon. What you need to know is:
If $a$ is a simple pole of $1/f(z)$, then ${\rm Res}(1/f(z),a) = 1/f'(a)$.
With this in hands: $${\rm Res}\left(\frac{1}{z^4+1},\omega\right) = \frac{1}{4\omega^3} = -\frac{\omega}{4}$$and: $${\rm Res}\left(\frac{1}{z^4+1},-\overline{\omega}\right) = \frac{1}{4(-\overline{\omega})^3} = -\frac{1}{4\overline{\omega}^3} = \frac{\overline{\omega}}{4}.$$Now, to the integrals. $$\begin{align} \oint_\gamma \frac{{\rm d}z}{z^4+1} &= \int_{\gamma_1} \frac{{\rm d}z}{z^4+1} + \int_{\gamma_2}\frac{{\rm d}z}{z^4+1} \\ 2\pi i \left(\frac{\overline{\omega}-\omega}{4}\right) &= \int_{-R}^R \frac{{\rm d}x}{x^4+1} + \int_{\gamma_2}\frac{{\rm d}z}{z^4+1} \\ \frac{\pi i}{2}(-2i\,{\rm Im}(\omega))&= \int_{-R}^R \frac{{\rm d}x}{x^4+1} + \int_{\gamma_2}\frac{{\rm d}z}{z^4+1} \\ \frac{\pi\sqrt{2}}{2}&= \int_{-R}^R \frac{{\rm d}x}{x^4+1} + \int_{\gamma_2}\frac{{\rm d}z}{z^4+1}\end{align}$$ If $|z| = R>1$, we have that $|z^4+1| \geq R^4-1 > 0$, and so $1/|z^4+1| \leq 1/(R^4-1)$. Since the lenght of $\gamma_2$ is $\pi R$, we have: $$\left|\int_{\gamma_2}\frac{{\rm d}z}{z^4+1}\right|\leq \frac{\pi R}{R^4-1} \stackrel{R\to+\infty}{\longrightarrow} 0.$$ So make $R \to +\infty$. We get: $$\int_{-\infty}^{+\infty} \frac{{\rm d}x}{x^4+1} = \frac{\pi\sqrt{2}}{2} \implies \int_0^{+\infty}\frac{{\rm d}x}{x^4+1} = \frac{\pi\sqrt{2}}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1333305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the infinite sum of a sequence Define a sequence $a_n$ such that $$a_{n+1}=3a_n+1$$ and $a_1=3$ for $n=1,2,\ldots$. Find the sum $$\sum_{n=1} ^\infty \frac{a_n}{5^n}$$ I am unable to find a general expression for $a_n$. Thanks.
| Let $x=\sum_{n=1}^{\infty}\dfrac{a_{n}}{5^n}$,since
$$\dfrac{a_{n+1}}{5^{n+1}}=\dfrac{3}{5}\dfrac{a_{n}}{5^n}+\dfrac{1}{5^{n+1}}$$
so
$$\sum_{n=1}^{\infty}\dfrac{a_{n+1}}{5^{n+1}}=\dfrac{3}{5}\sum_{n=1}^{\infty}\dfrac{a_{n}}{5^n}+\sum_{n=1}^{\infty}\dfrac{1}{5^{n+1}}$$
then we have
$$x-\dfrac{a_{1}}{5}=\dfrac{3}{5}x+\dfrac{\frac{1}{25}}{1-\frac{1}{5}}\Longrightarrow x=\dfrac{3}{8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1336101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Combining sums and/or differences of squares I'd like to combine a sum of as many squares as possible into a sum of as few squares as possible. The signs of the squares doesn't matter.
For example, the Brahmagupta-Fibonacci Identity combines a sum of four squares into a sum of two squares. Thus the compression ratio is 2:1. Using this identity, we can, with careful calculations, apply it recursively to get an even better compression ratio. In other words, take two sums of four squares and combine them together to get one sum of four squares, and then combine this to get one sum of two squares. Thus we have a ratio of 4:1.
THE QUESTION'
Are there are any other formulas that take a sum of as many squares as possible and combine them into as few squares as possible?
| The "compression ratio" can be made as high as you want.
I. Euler-Aida Ammei
$$(-x_0^2+x_1^2+x_2^2+\dots+x_n^2)^2+(2x_0x_1)^2+(2x_0x_2)^2+\dots+(2x_0x_n)^2 = (x_0^2+x_1^2+x_2^2+\dots+x_n^2)^2$$
Thus, there can be arbitrarily many squares on the LHS.
II. Fauquembergue $n$-Squares Identity
Using the Brahmagupta-Fibonacci,
$$(a^2+b^2)(c^2+d^2) = u_1^2+u_2^2 = (ac+bd)^2+(ad-bc)^2$$
then,
$$(a^2+b^2-c^2-d^2+x)^2 + (2u_1)^2 + (2u_2)^2 + 4x(c^2+d^2) = (a^2+b^2+c^2+d^2+x)^2$$
where $x$ is arbitrary and can be chosen as any sum of $n$ squares.
III. Using the Euler-Four Square
From,
$$(a^2+b^2+c^2+d^2)(e^2+f^2+g^2+h^2) = v_1^2+v_2^2+v_3^2+v_4^2$$
then,
$$(a^2+b^2+c^2+d^2-e^2-f^2-g^2-h^2+x)^2 + (2v_1)^2 + (2v_2)^2 + (2v_3)^2 + (2v_4)^2 + 4x(e^2+f^2+g^2+h^2) = (a^2+b^2+c^2+d^2+e^2+f^2+g^2+h^2+x)^2$$
where $x$ can again be any sum of squares. (I trust you know the expressions for the $v_i$.) Of course, a similar $n$-square identity can be made using the Degen-Graves Eight-Square.
For more on sums of $n$ squares, more details at my site.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1337824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
3 variable symmetric inequality Show that for positive reals $a,b,c$,
$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\geq \frac{3a^3+3b^3+3c^3}{2a^2+2b^2+2c^2}$
What I did was WLOG $a+b+c=1$ (since the inequality is homogenous)
Then I substituted into the LHS to get $\sum_{\text{cyc}} \frac{a^2}{1-a}\geq \frac{3a^3+3b^3+3c^3}{2a^2+2b^2+2c^2}$. Now I'm not sure if I should clear the denominators and expand and try to use Muirhead+Schur? (Clearing the denominators seems quite tedious)?
Any ideas are appreciated.
| your inequalitiy is equivalent to
$$\left(a^3-2 a b c+b^3+c^3\right) \left(2 a^3-a^2 b-a^2 c-a b^2-a c^2+2 b^3-b^2
c-b c^2+2 c^3\right)\geq 0$$
since $$a^3+b^3+c^3\geq 3abc>2abc$$ is the first factor positive,
and
$$2(a^3+b^3+c^3)\geq ab(a+b)+ac(a+c)+bc(b+c)$$
since $a^3+b^3=(a+b)(a^2+b^2-ab)\geq ab(a+b)$ etc is the second factor also positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1337913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integral of $\int \frac{x+1}{(x^2+x+1)^2}dx$ I'm currently learning Calculus II and I have the following integral:
Integal of $\large{\int \frac{x+1}{(x^2+x+1)^2}dx}$
I've tried with partial fractions but it led nowhere, I've tried with substitution, but I failed again.
If it helps, I know the answer, but I don't know how to get to it.
Answer: $\large{\frac{1}{3}\frac{x-1}{x^2+x+1}} + \frac{1}{3}2\sqrt{3} \arctan{2x+1\sqrt{3}} + C$
| Hint:
\begin{align*}
\int\frac{x+1}{(x^2+x+1)^2}dx&=\frac{1}{2}\int{\frac{2x+1}{(x^2+x+1)^2}}dx+\frac{1}{2}\int\frac{1}{(x^2+x+1)^2}dx\\
&=\frac{1}{2}\int\frac{d(x^2+x+1)}{(x^2+x+1)^2}+\frac{1}{2}\int{\frac{1}{\left[\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right]^2}}dx
\end{align*}
The last of these integrals can be found using the trigonometric sustitution $x+\frac{1}{2}=\frac{\sqrt{3}}{2}\tan t$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1337986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve for $x$ from an equation containing inverse trigonometric functions
How to solve the following for $x$?
$$
\sin^{-1}\left(\frac{2a}{1+a^{2}}\right)+
\sin^{-1}\left(\frac{2b}{1+b^{2}}\right)=
2 \tan^{-1}(x )
$$
What conditions apply?
| Hint: Let $\theta = \sin^{-1}\left(\dfrac{2a}{1+a^2}\right) \Rightarrow \sin \theta = \dfrac{2a}{1+a^2}$, and similarly $\sin \beta = \dfrac{2b}{1+b^2}$. Thus:
$\tan^{-1}x = \dfrac{\theta}{2}+\dfrac{\beta}{2} \Rightarrow x = \tan\left(\dfrac{\theta}{2}+\dfrac{\beta}{2}\right)$. You can compute $\tan\left(\dfrac{\theta}{2}\right)$, and $\tan\left(\dfrac{\beta}{2}\right)$ using the above equations, then find $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1339791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Proving that $\frac{1}{a^2}+\frac{1}{b^2} \geq \frac{8}{(a+b)^2}$ for $a,b>0$ I found something that I'm not quite sure about when trying to prove this inequality.
I've proven that
$$\dfrac{1}{a}+\dfrac{1}{b}\geq \dfrac{4}{a+b}$$
already. My idea now is to replace $a$ with $a^2$ and $b^2$, so we now have
$$\dfrac{1}{a^2}+\dfrac{1}{b^2}\geq \dfrac{4}{a^2+b^2}$$
So to prove the required result, we just need to show that
$$\dfrac{4}{a^2+b^2} \geq \dfrac{8}{(a+b)^2}$$
or equivalently
$$(a+b)^2 \geq 2(a^2+b^2)$$
But this inequality cannot be true since the pair $(a,b)=(1,2)$ doesn't work. If anything, the reverse is always true!
What have I done wrong here?
| For $a,b>0$, let
\begin{equation}
A(a,b)=\frac{a+b}{2}, \quad G(a,b)=\sqrt{ab}\,, \quad H(a,b)=\frac{2}{\dfrac1{a}+\dfrac1{b}}
\end{equation}
are respectively the arithmetic, geometric, and harmonic means of two positive numbers $a,b$.
It is commom knowledge that
\begin{equation}
A(a,b)\ge G(a,b) \quad\text{and}\quad H(a,b)=\frac{G^2(a,b)}{A(a,b)}.
\end{equation}
Therefore, we derive
\begin{equation}
A\bigl(a^2,b^2\bigr)\ge G\bigl(a^2,b^2\bigr)=G^2(a,b)
\Longleftrightarrow 1\ge\frac{G^2(a,b)}{A\bigl(a^2,b^2\bigr)}
\end{equation}
and
\begin{equation}
A^2(a,b)\ge G^2(a,b).
\end{equation}
Consequently, we deduce
\begin{equation}
A^2(a,b)\ge G^2(a,b)\frac{G^2(a,b)}{A\bigl(a^2,b^2\bigr)}
=\frac{G^2\bigl(a^2,b^2\bigr)}{A\bigl(a^2,b^2\bigr)}
=H\bigl(a^2,b^2\bigr)
\end{equation}
which can be reformulated as
\begin{equation}
\biggl(\frac{a+b}{2}\biggr)^2\ge\frac{2}{\dfrac1{a^2}+\dfrac1{b^2}}
\Longleftrightarrow
\dfrac1{a^2}+\dfrac1{b^2}\ge\frac{8}{(a+b)^2}.
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1340034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Compute definite integral Question: Compute
$$\int_0^1 \frac{\sqrt{x-x^2}}{x+2}dx.$$
Attempt: I've tried various substitutions with no success. Looked for a possible contour integration by converting this into a rational function of $\sin\theta$ and $\cos \theta$.
| Let $x=\frac{1}{2}\sin t+\frac{1}{2}$, then we have
\begin{align*}
\int_0^1\frac{\sqrt{x-x^2}}{x+2}dx&=\int_0^1\frac{\sqrt{\left(\frac{1}{2}\right)^2-\left(x-\frac{1}{2}\right)^2}}{x+2}dx\\
&=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\sqrt{\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\sin t\right)^2}}{\frac{1}{2}\sin t+\frac{1}{2}+2}\frac{1}{2}\cos t\,dt\\
&=\frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos t}{\sin t+5}\cos t\,dt\\
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1340612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 4
} |
Minimization on compact region I need to solve the minimization problem
$$\begin{matrix} \min & x^2 + 2y^2 + 3z^2 \\ subject\;to & x^2 + y^2 + z^2 =1\\ \; & x+y+z=0
\end{matrix}$$
I was trying to verify the first order conditions using lagrange multipliers
$$\left\{\begin{matrix}
2x = 2\lambda_1x + \lambda_2 \\
4y = 2 \lambda_1 y + \lambda_2 \\
6z = 2\lambda_1 z + \lambda_2
\end{matrix}\right.$$
Summing the 3 equations we get
$$ 2y + 4z = 3 \lambda_2$$
Multiplying the first one for x, the second for $y$ and third for $z$ and summing:
$$ 2 + 2y^2 + 4z^2 = 2 \lambda_1$$
I just got stuck there and I have no idea how to proceed. We can check the answer here but I couldn't find none of the variables. Any idea?
Thanks in advance!
| since you have
$$x=\dfrac{\lambda_{2}}{2(1-\lambda_{1})},y=\dfrac{\lambda_{2}}{4-2\lambda_{1}},z=\dfrac{\lambda_{2}}{6-2\lambda_{1}}$$
and $x+y+z=0$ so we have
$$\dfrac{\lambda_{2}}{2}\left(\dfrac{1}{1-\lambda_{1}}+\dfrac{1}{2-\lambda_{1}}+\dfrac{1}{3-\lambda_{1}}\right)=0$$
so we have$\lambda_{2}=0$ or
$$\dfrac{(2-\lambda_{1})(3-\lambda_{1})+(1-\lambda_{1})(3-\lambda_{1})+(1-\lambda_{1})(2-\lambda_{1})}{(1-\lambda_{1})(2-\lambda_{2})(3-\lambda_{3})}=0$$
then we have
$$3\lambda^2_{1}-12\lambda_{1}+11=0\Longrightarrow \lambda_{1}=2\pm\dfrac{1}{\sqrt{3}}$$
then you can easy to find this respect $\lambda_{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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This general inequality maybe is true? $\sum_{i=1}^{n}\frac{i}{1+a_{1}+\cdots+a_{i}}<\frac{n}{2}\sqrt{\sum_{i=1}^{n}\frac{1}{a_{i}}}$
Let $a_{1},a_{2},\ldots,a_{n}>0$ and prove or disprove
$$\dfrac{1}{1+a_{1}}+\dfrac{2}{1+a_{1}+a_{2}}+\cdots+\dfrac{n}{1+a_{1}+a_{2}+\cdots+a_{n}}\le\dfrac{n}{2}\sqrt{\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}}\tag{1}$$
This problem from when I solve this two variable inequality
since $n=1$ it is clear
$$\dfrac{1}{1+a_{1}}\le\dfrac{1}{2}\sqrt{\dfrac{1}{a_{1}}}$$
because $1+a_{1}\ge 2\sqrt{a_{1}}$
$n=2$ case,can see this links my answer.
For general
simaler this two variable inequality methods, then I use Cauchy-Schwarz inequality we have
$$\left(\sum_{i=1}^{n}\dfrac{1}{1+a_{1}+\cdots+a_{i}}\right)^2\le\left(\sum_{i=1}^{n}\dfrac{1}{a_{i}}\right)\cdot\left(\sum_{i=1}^{n}\dfrac{i^2a_{i}}{(1+a_{1}+a_{2}+\cdots+a_{i})^2}\right)$$
it suffices to show that
$$\sum_{i=1}^{n}\dfrac{i^2a_{i}}{(1+a_{1}+\cdots+a_{i})^2}\le\dfrac{n^2}{4}\tag{2}$$
it seem hard.
because I tried following also fail;
$$\sum_{i=1}^{n}\dfrac{i^2a_{i}}{(1+a_{1}+\cdots+a_{i})^2}<\sum_{i=1}^{n}i^2\left(\dfrac{1}{1+a_{1}+\cdots+a_{i-1}}-\dfrac{1}{1+a_{1}+a_{2}+\cdots+a_{i}}\right)$$
and use Abel transformation.not can to prove $(2)$,
Note $(1)$ Left side hand was simaler Hardy's inequality when $p=-1$,But there are different problem.
EDIT:Numerical tests $(2)$ is not right.so my idea can't works
| suppose $n=k, \\ \dfrac{1}{1+a_{1}}+\dfrac{2}{1+a_{1}+a_{2}}+\cdots+\dfrac{k}{1+a_{1}+a_{2}+\cdots+a_{k}} \\ \le\dfrac{k}{2}\sqrt{\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{k}}}$
when $n=k+1 $
LHS=$\dfrac{1}{1+a_{1}}+\dfrac{2}{1+a_{1}+a_{2}}+\cdots+\dfrac{k}{1+a_{1}+a_{2}+\cdots+a_{k}}+\dfrac{k+1}{1+a_{1}+a_{2}+\cdots+a_{k+1}} \\<\dfrac{k}{2}\sqrt{\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{k}}}+\dfrac{k+1}{1+a_{1}+a_{2}+\cdots+a_{k+1}} \\ <\dfrac{k}{2}\sqrt{\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{k}}+\dfrac{1}{a_{k+1}}}+\dfrac{k+1}{1+a_{1}+a_{2}+\cdots+a_{k+1}}$
RHS$=\dfrac{k}{2}\sqrt{\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{k}}+\dfrac{1}{a_{k+1}}}+\dfrac{1}{2}\sqrt{\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{k}}+\dfrac{1}{a_{k+1}}}$
so is remains:
$ \dfrac{k+1}{1+a_{1}+a_{2}+\cdots+a_{k+1}} \\ \le \dfrac{1}{2}\sqrt{\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{k}}+\dfrac{1}{a_{k+1}}}$
$1+a_{1}+a_{2}+\cdots+a_{k+1} \ge 2\sqrt{a_{1}+a_{2}+\cdots+a_{k+1}} \\ \implies \dfrac{k+1}{1+a_{1}+a_{2}+\cdots+a_{k+1}} \le \dfrac{k+1}{2\sqrt{a_{1}+a_{2}+\cdots+a_{k+1}}} \le \dfrac{1}{2}\sqrt{\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{k}}+\dfrac{1}{a_{k+1}}} \\ \iff \dfrac{(k+1)^2}{\sum_{i=1}^{k+1} a_i} \le \sum_{i=1}^{k+1} \dfrac{1}{a_i}$
the last one is true ,$HM\le AM$
QED
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 0
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Convert the equation to rectangular form $r = \frac {6}{1-\sinθ}$ Convert the equation to rectangular form $r = \frac {6}{1-\sinθ}$
The answer should be: $y = \frac{1}{12} x^2 -3$
But how to arrive at the answer?
I tried replacing r with $\sqrt{x^2 + y^2}$, then $\sinθ$ as $\frac{y}{r}$ but to no avail.
I also ended up with
$r = \frac{6}{\frac{r-y}{r}} $ ->
$ 1 = \frac{6}{\sqrt{x^2+y^2}-y}$ -> x = +- 6
but that's not the answer...
| $$r = \frac {6}{1-\sinθ}$$
$${r-r\sinθ}=6$$
$$\sqrt{x^2 + y^2}-y=6$$
$$\sqrt{x^2 + y^2}=y+6$$ now squaring both side and simplyfying we get
$$y = \frac{1}{12} x^2 -3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Solving a system of non-linear equations Let
$$(\star)\begin{cases}
\begin{vmatrix}
x&y\\
z&x\\
\end{vmatrix}=1, \\
\begin{vmatrix}
y&z\\
x&y\\
\end{vmatrix}=2, \\
\begin{vmatrix}
z&x\\
y&z\\
\end{vmatrix}=3.
\end{cases}$$
Solving the above system of three non-linear equations with three unknowns.
I have a try.
Let$$A=\begin{bmatrix}
1& 1/2& -1/2\\
1/2& 1& -1/2\\
-1/2& -1/2& -1
\end{bmatrix}$$
We have $$(x,y,z)A\begin{pmatrix}
x\\
y\\
z
\end{pmatrix}=0.$$
There must be a orthogonal matrix $T$,such that $T^{-1}A T=diag \begin{Bmatrix}
\frac{1}{2},\frac{\sqrt{33}+1}{4},-\frac{\sqrt{33}-1}{4}
\end{Bmatrix}.$
$$\begin{pmatrix}
x\\
y\\
z
\end{pmatrix}=T\begin{pmatrix}
x^{'}\\
y^{'}\\
z^{'}
\end{pmatrix}\Longrightarrow\frac{1}{2} {x'}^{2}+\frac{\sqrt{33}+1}{4} {y'}^{2}-\frac{\sqrt{33}-1}{4}{z'}^{2}=0.$$
But even if we find a $\begin{pmatrix}
x_0^{'}\\
y_0^{'}\\
z_0^{'}
\end{pmatrix} $ satisfying $\frac{1}{2} {x_0'}^{2}+\frac{\sqrt{33}+1}{4} {y_0'}^{2}-\frac{\sqrt{33}-1}{4}{z_0'}^{2}=0,\begin{pmatrix}
x_0\\
y_0\\
z_0
\end{pmatrix}=T\begin{pmatrix}
x_0^{'}\\
y_0^{'}\\
z_0^{'}
\end{pmatrix}$ may not be the solution of $(\star)$
If you have some good ideas,please give me some hints. Any help would be appreciated!
| Let's rewrite the system as $$\mathbf r \times \mathbf P\mathbf{r} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\x&y&z\\z&x&y\end{vmatrix} = 2\mathbf{i} + 3\mathbf{j} + 1\mathbf{k} \equiv \mathbf f$$
where $\mathbf{r} = (x,y,z)^\top$ and
$$
\mathbf P = \begin{pmatrix}
0 & 0 & 1\\
1 & 0 & 0\\
0 & 1 & 0
\end{pmatrix}
$$
So we want to solve
$$
\mathbf{r} \times \mathbf{Pr} = \mathbf{f}.
$$
Let's dot multiply both sides with $\mathbf{r}$
$$
0 = \mathbf{r}^\top \mathbf{f}
$$
also when multiplying by $\mathbf{Pr}$:
$$
0 = \mathbf{r}^\top \mathbf P^\top \mathbf{f}
$$
That's a pair of linear equations in $x,y,z$. Since the equations are homogeneous, the solution will be of the form $\mathbf r = \alpha \mathbf{h}, \alpha \in \mathbb{R}$.
$$
2x + 3y + z = 0\\
2z + 3x + y = 0
$$
Row reducing the system we get
$$
\mathbf h = \begin{pmatrix}
-5\\1\\7
\end{pmatrix}.
$$
Since $\mathbf h \times \mathbf {Ph} = 18 \mathbf f$ we deduce that
$$
\alpha^2 = \frac{1}{18}\qquad\alpha = \pm\frac{1}{3\sqrt{2}}.
$$
Finally
$$
x = \mp \frac{5}{3\sqrt{2}}\\
y = \pm \frac{1}{3\sqrt{2}}\\
z = \pm \frac{7}{3\sqrt{2}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1347813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 1
} |
value of an $\sum_3^\infty\frac{3n-4}{(n-2)(n-1)n}$ I ran into this sum $$\sum_{n=3}^{\infty} \frac{3n-4}{n(n-1)(n-2)}$$
I tried to derive it from a standard sequence using integration and derivatives, but couldn't find a proper function to describe it.
Any ideas?
| Using the Heaviside Method of Partial Fractions to solve
$$
\frac{3n-4}{n(n-1)(n-2)}=\frac{A}{n}+\frac{B}{n-1}+\frac{C}{n-2}
$$
we get
multiply by $n$ and set $n=0\implies\frac{3\cdot\color{#C00000}{0}-4}{(\color{#C00000}{0}-1)(\color{#C00000}{0}-2)}=A\implies A=-2$
multiply by $n-1$ and set $n=1\implies\frac{3\cdot\color{#C00000}{1}-4}{\color{#C00000}{1}(\color{#C00000}{1}-2)}=B\implies B=1$
multiply by $n-2$ and set $n=2\implies\frac{3\cdot\color{#C00000}{2}-4}{\color{#C00000}{2}(\color{#C00000}{2}-1)}=C\implies C=1$
Thus,
$$
\frac{3n-4}{n(n-1)(n-2)}=-\frac2n+\frac1{n-1}+\frac1{n-2}
$$
Then applying Telescoping Series to get
$$
\begin{align}
\sum_{n=3}^N\frac{3n-4}{n(n-1)(n-2)}
&=-2\sum_{n=3}^N\frac1n+\sum_{n=3}^N\frac1{n-1}+\sum_{n=3}^N\frac1{n-2}\\
&=\color{#C00000}{-2\sum_{n=3}^N\frac1n}\color{#00A000}{+\sum_{n=2}^{N-1}\frac1n}\color{#0000F0}{+\sum_{n=1}^{N-2}\frac1n}\\
&=\color{#0000F0}{1+\frac12}\color{#00A000}{+\frac12+\frac1{N-1}}\color{#C00000}{-\frac2{N-1}-\frac2N}
\end{align}
$$
Now take the proper limit.
Note that we can't sum from $n=0$ since the denominators will be $0$, so I assumed we were to start at $n=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1348046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the area of triangle, given an angle and the length of the segments cut by the projection of the incenter on the opposite side.
In a triangle $ABC$, one of the angles (say $\widehat{C}$) equals $60^\circ$.
Given that the incircle touches the opposite side ($AB$) in a point that splits it in two segments having length $a$ and $b$, what is the area of $ABC$?
|
Area $S_{ABC}$ in terms of $|AB|$ and altitude $h_C$ at the point $C$:
\begin{align}
h_C&=(b+r\sqrt{3})\sin(60)
=\frac{\sqrt{3}}{2}(b+r\sqrt{3})
\\
S&=
\frac{\sqrt{3}}{4}
(a+r\sqrt{3})
(b+r\sqrt{3})
\\
&=
\frac{3}{4}\left(
\sqrt{3}r^2+(a+b)r
\right)
+\frac{\sqrt{3}}{4}ab
\quad(1)
\end{align}
Also, $S_{ABC}=S_{OAB}+S_{OBC}+S_{OCA}$:
\begin{align}
S_{ABC}&=
\frac{r}{2}
(2a+2b+2r\sqrt3)
\\
&=
\sqrt{3}r^2 + (a+b)r
\quad(2)
\end{align}
$\frac{4}{3} (1)-(2):$
\begin{align}
\frac{1}{3}S_{ABC}
&=
\frac{\sqrt3}{3}ab,
\end{align}
thus, $S_{ABC}=ab\sqrt{3}$.
Edit:
This approach also provides a generalization for
$\angle BAC =\alpha$. By replacing $r\sqrt{3}$
with $r/\tan(\alpha/2)$, similar steps result in
\begin{align}
S_{ABC}(\alpha)&=\frac{ab}{\tan{\alpha/2}}
=
\frac{\sin{\alpha}}{1-\cos{\alpha}}ab.
\end{align}
Particularly, $S(90°)=ab$.
| {
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"question_score": "1",
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According to Stewart Calculus Early Transcendentals 5th Edition on page 140, in example 5, how does he simplify this problem? In Stewart's Calculus: Early Transcendentals 5th Edition on page 140, in example 5, how does
$$\lim\limits_{x \to \infty} \frac{\dfrac{1}{x}}{\dfrac{\sqrt{x^2 + 1} + x}{x}}$$
simplify to
$$\lim\limits_{x \to \infty} \frac{\dfrac{1}{x}}{\sqrt{1 + \dfrac{1}{x^2}} + 1}$$
I understand how he simplifies $\frac{x}{x}$ to 1 but how does he simplify $\frac{\sqrt{x^2 + 1}}{x}$?
| $$
\frac{\sqrt{x^2 + 1}}{x} = \frac{\sqrt{x^2 + 1}}{\sqrt{x^2}} = \sqrt{\frac{x^2+1}{x^2}} = \sqrt{\frac{x^2}{x^2} + \frac 1 {x^2}} = \cdots
$$
$\frac{\left(\dfrac{1}{x}\right)}{\left(\dfrac{\sqrt{x^2 + 1} + x}{x}\right)}$ is exactly the same thing as $\dfrac 1 {\sqrt{x^2+1}+x}$. To get that just multiply the numerator and the denominator by $x$. Then:
$$
0 < \frac 1 {\sqrt{x^2+1}+x}<\frac 1 {2x} \to 0 \text{ as } x\to\infty.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Show that $\frac{n}{n^2-3}$ converges Hi I need help with this epsilon delta proof. The subtraction in the denominator as well as being left with $n$ in multiple places is causing problems.
| @Ian has provided a very efficient way to tackle a problem such as this. I thought it might be instructive to see how tight a bound we can achieve. To that end, let's write for $n\ge2$
$$\begin{align}
\left|\frac{n}{n^2-3}\right|<\epsilon \implies n&>\frac{1}{2\epsilon}+\frac{1}{2\epsilon}(1+12\epsilon^2)^{1/2} \tag 1\\\\
&=\frac{1}{2\epsilon}+\frac{1}{2\epsilon}\left(1+6\epsilon^2+O\left(\epsilon^4\right)\right)\\\\
&=\frac{1}{\epsilon}+3\epsilon+O\left(\epsilon^3\right)
\end{align}$$
where $(1)$ provides the tightest bound for $N$ as $N=\left\lceil \left(\frac{1}{2\epsilon}+\frac{1}{2\epsilon}(1+12\epsilon^2)^{1/2} \right)\right \rceil$.
Now, we could also proceed as follows. We note that $n^2-3>\frac{n^2}{1+\alpha}$, for any $\alpha>0$, provided that $n\ge \left \lceil \sqrt{3\frac{1+\alpha}{\alpha}}\right \rceil$. For $\alpha =1$, $n^2-3>\frac12 n^2$ for $n\ge \left \lceil \sqrt{6} \right \rceil =3$, which is the inequality imposed by @Ian.
So, given $\epsilon>0$, we could actually take $N=\max\left(\frac{1+\alpha}{\epsilon},\left \lceil \sqrt{3\frac{1+\alpha}{\alpha}}\right \rceil\right)$, for any $\alpha >0$ and
$$\left|\frac{n}{n^2-3}\right|<\epsilon$$
whenever $n>N$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the number of solutions of the trigonometric equation in $(0,\pi)$ Find the number of solutions of the equation $$\sec x+\csc x=\sqrt {15}$$ in $(0,\pi)$. The question is easy. But when you solve, you get would get $4$ as the answer. I am sure the method gives $4$ as answer, but the correct answer is $3$. I don't think there is an extraneous root of the equation. So, which of the $4$ solutions is not satisfying the equation? Thanks.
| If you set $X=\cos x$ and $Y=\sin X$, the equation is
$$
\frac{1}{Y}+\frac{1}{X}=\sqrt{15}
$$
to be considered together with $X^2+Y^2=1$.
Note that we must have $Y>0$, since $x\in(0,\pi)$.
The equation becomes $X+Y=\sqrt{15}\,XY$ and it's convenient to make another substitution, namely $S=X+Y$, $P=XY$, so we get
$$
\begin{cases}
S=P\sqrt{15}\\
S^2-2P=1
\end{cases}
$$
Thus we have $15P-2P-1=0$, that means
$$
P=\frac{1\pm4}{15}
$$
so $P=1/3$ or $P=-1/5$. If $P=1/3$, we have $S=\sqrt{15}/3$ and
$$
\begin{cases}
X+Y=\sqrt{15}/3\\
XY=1/3
\end{cases}
$$
This is solved by considering the roots of $3t^2-\sqrt{15}\,t+1=0$, giving
$$
X=\frac{\sqrt{15}-\sqrt{3}}{6},\qquad Y=\frac{\sqrt{15}+\sqrt{3}}{6}\\
\text{or}\\
X=\frac{\sqrt{15}+\sqrt{3}}{6},\qquad Y=\frac{\sqrt{15}-\sqrt{3}}{6}
$$
that correspond to two solutions in $(0,\pi)$.
For $P=-1/5$, we get $S=-\sqrt{15}/5$ and so
$$
\begin{cases}
X+Y=-\sqrt{15}/5\\
XY=-1/5
\end{cases}
$$
which is solved by considering $5t^2-\sqrt{15}\,t-1=0$, giving
$$
X=\frac{\sqrt{15}-\sqrt{20}}{10},\qquad Y=\frac{\sqrt{15}+\sqrt{20}}{10} \\
\text{or}\\
X=\frac{\sqrt{15}+\sqrt{20}}{10},\qquad Y=\frac{\sqrt{15}-\sqrt{20}}{10}
$$
and only the first one corresponds to an angle in $(0,\pi)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Closed form expression for products How can I find a closed form expression for products of the following form $$\prod_{k=1}^n (ak^2+bk+c)\space \text{?}$$
| As commented by r9m, the key idea is to factorize the polynomial. When this is done, you can either use Pochhammer functions which would give $$\prod_{k=1}^n (ak^2+bk+c)=a^n \left(\frac{2 a+b-\sqrt{b^2-4 a c}}{2 a}\right){}_n \left(\frac{2 a+b+\sqrt{b^2-4
a c}}{2 a}\right){}_n$$ or transform to Gamma functions $$\prod_{k=1}^n (ak^2+bk+c)=a^n\frac{ \Gamma \left(n+\frac{b}{2 a}-\frac{\sqrt{b^2-4 a c}}{2 a}+1\right) \Gamma
\left(n+\frac{b}{2 a}+\frac{\sqrt{b^2-4 a c}}{2 a}+1\right)}{\Gamma
\left(\frac{b}{2 a}-\frac{\sqrt{b^2-4 a c}}{2 a}+1\right) \Gamma \left(\frac{b}{2
a}+\frac{\sqrt{b^2-4 a c}}{2 a}+1\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1355975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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How can I get the exact value of this infinite series? I want to compute the exact value of this infinite series
$$\sum_{n=2}^\infty\arcsin{\left(\dfrac{2}{\sqrt{n(n+1)}(\sqrt{n}+\sqrt{n-1})}\right)}$$
By comparison test, we can get the series is convengence.
I tried to find some hints from the exact value of $\displaystyle\sum_{n=2}^\infty\arcsin{\left(\dfrac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n^2-1}}\right)}$
,but the split phase method maybe difficult to solve this question.
I am not sure whether it has a closed form.But if not so,how can I evaluate the sum ?
| The sum to 1000 terms,
according to Wolfy,
is
1.55713
or $0.481892\pi$.
Would be amusing if the sum were
$\pi/2$.
$\sum_{n=2}^\infty\arcsin{\left(\dfrac{2}{\sqrt{n(n+1)}(\sqrt{n}+\sqrt{n-1})}\right)}
$
More seriously,
I will try
$\arcsin(x)
=\arctan(\frac{x}{\sqrt{1-x^2}})
$
to see if
$\arctan(u)+\arctan(v)
=\arctan(\frac{u+v}{1-uv})
$
can be used.
If
$x
=\dfrac{2}{\sqrt{n(n+1)}(\sqrt{n}+\sqrt{n-1})}
=\dfrac{2}{f(n)}
$,
$\begin{array}\\
\frac{x}{\sqrt{1-x^2}}
&=\frac{\frac{2}{f(n)}}{\sqrt{1-(\frac{2}{f(n)})^2}}\\
&=\frac{2}{\sqrt{f^2(n)-2}}\\
&=\frac{2}{\sqrt{n(n+1)(2n-1+2\sqrt{n(n-1)}-2)}}\\
&=\frac{2}{\sqrt{n(n+1)(2n-3+2\sqrt{n(n-1)})}}\\
\end{array}
$
For $n=2$,
this is
$\frac{2}{\sqrt{6(1+2\sqrt{2})}}
$.
For $n=3$,
this is
$\frac{2}{\sqrt{12(3+2\sqrt{6})}}
$.
Combining these,
to get the sum up to 3,
we get
$\begin{array}\\
\dfrac{\frac{2}{\sqrt{6(1+2\sqrt{2})}}+\frac{2}{\sqrt{12(3+2\sqrt{6})}}}{1-\frac{2}{\sqrt{6(1+2\sqrt{2})}}\frac{2}{\sqrt{12(3+2\sqrt{6})}}}
&=2\dfrac{\sqrt{6(1+2\sqrt{2})}+\sqrt{12(3+2\sqrt{6})}}{\sqrt{6(1+2\sqrt{2})}\sqrt{12(3+2\sqrt{6})}-4}\\
&=2\dfrac{\sqrt{6(1+2\sqrt{2})}+\sqrt{12(3+2\sqrt{6})}}{6\sqrt{2(1+2\sqrt{2}) (3+2\sqrt{6})}-4}\\
&=2\dfrac{\sqrt{6(1+2\sqrt{2})}+\sqrt{12(3+2\sqrt{6})}}{6\sqrt{2(3+6\sqrt{2}+2\sqrt{6}+4\sqrt{12})}-4}\\
&=\dfrac{\sqrt{6(1+2\sqrt{2})}+\sqrt{12(3+2\sqrt{6})}}{3\sqrt{2(3+6\sqrt{2}+2\sqrt{6}+8\sqrt{3})}-2}\\
\end{array}
$
This doesn't look like
anything I would like
to meet in a dark alley.
So I'll give it up here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1357068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Probability that team $A$ has more points than team $B$
Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and the loser gets 0 points. The total points are accumulated to decide the ranks of the teams. In the first game of the tournament, team $A$ beats team $B.$ The probability that team $A$ finishes with more points than team $B$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$
I got that since, Team $B$ already has one loss, it doesnt matter how many games team $B$ wins.
We must find the probability that team $A$ wins the rest $5$ games.
Since it says: "outcome of games is independent, I am confused."
My first approach was:
$$P(\text{A Wins 5}) = \frac{1}{32} \implies m + n = 33$$
This was wrong.
Second approach.
Suppose $A$ has a match with Team $C$.
$$P(\text{Team A wins, Team C loses}) = \frac{1}{2}\frac{1}{2} = \frac{1}{4}$$
But then overall: $\frac{1}{4^5} > 1000$ too big of an answer ($m + n < 1000$ requirement).
HINTS ONLY PLEASE!!
EDIT:
I did some casework and the work is very messy and I don't think I got the right answer anyway. I have to find:
$$\binom{5}{1} = 5, \binom{5}{2} = 10, \binom{5}{3} = 10, \binom{5}{4} = 5, \binom{5}{5} = 1.$$
Let $A = x$
$$P(B=1, x \ge 1) + P(B = 2, x\ge 2) + P(B=3, x\ge 3) + ... + P(B=5, x = 5)$$
$$P(B=1, x \ge 1) = \binom{5}{1}(0.5)^{5} \cdot \bigg(\binom{5}{1} (0.5)^5 + \binom{5}{2} (0.5)^5 + ... + (0.5)^5\binom{5}{1} \bigg) = \frac{5}{1024} \cdot \bigg(31\bigg) = \frac{155}{1024} $$
$$P_2 = \frac{10}{1024} \bigg(\binom{5}{2} + ... + \binom{5}{5}\bigg) = \frac{260}{1024}$$
$$P_3 = \frac{10}{1024} \bigg(\binom{5}{3} + ... + \binom{5}{5} \bigg) = \frac{160}{1024}$$
$$P_4 = \frac{5}{1024} \bigg( \binom{5}{4} + \binom{5}{5}\bigg) = \frac{60}{1024}$$
$$P_5 = \frac{1}{1024} \bigg( 1\bigg) = \frac{1}{1024}$$
$$P(\text{Total}) = \frac{636}{1024} = \frac{318}{512} = \text{wrong}$$
What is wrong with this method?
| You can actually get away with computing only one binomial coefficient; I'll leave out one detail so this qualifies as a hint.
As @GrahamKemp observes, both $X$ and $Y$ are distributed as $\mathcal{Bin}(5,1/2)$; we want the probability that $X\ge Y$, i.e. that $Z:=X-Y\ge 0$. But $Z+5$ is distributed as $\mathcal{Bin}(10,1/2)$ (why?) From this you can compute $p=\textrm{Pr}(X=Y)=\textrm{Pr}(Z=0)$. Finally, by symmetry, we have $\textrm{Pr}(X<Y)=\textrm{Pr}(X>Y)$, so both of these are equal to $(1-p)/2$. It follows that
$$\textrm{Pr}(X\ge Y)=(1+p)/2.$$
Incidentally, I got $m+n=831.$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Find all postive integers $n$ such $(2n+7)\mid (n!-1)$ Find all postive integers $n$ such that
$$(2n+7)\mid(n!-1).$$
I have $n=1,5$, but can not find any other and can not prove whether there is any other solution or not.
| Building on nayrb's observation that $2n+7$ must be prime (for $n\ge 4$), let $p = 2n+7$. Then by assumption, $n! \equiv 1 \pmod p$. Symmetrically, we also have $$(2n+6)(2n+5)\cdots(n+7) \equiv (-1)(-2)\cdots(-n) \equiv (-1)^n n! \equiv \pm 1 \pmod p.$$
Now combine this with Wilson's theorem that $(2n+6)! \equiv -1 \pmod p$, and we have
$$(n+1)(n+2)(n+3)(n+4)(n+5)(n+6) \equiv \pm 1 \pmod p.$$
Multiply both sides by $2^6$ and this becomes
$$(2n+2)(2n+4)(2n+6)(2n+8)(2n+10)(2n+12) = \pm 64 \pmod p,$$
which simplifies to
$$(p-5)(p-3)(p-1)(p+1)(p+3)(p+5) \equiv -225 \equiv \pm 64 \pmod p,$$
so necessarily $p \mid 225\pm 64$, that is either $p \mid 289 = 17^2$ or $p \mid 161 = 7\cdot 23$.
This proves that the number of solutions is finite. I leave it to you to identify the actual solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1359021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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This inequality $a+b^2+c^3+d^4\ge \frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}+\frac{1}{d^4}$
let $0<a\le b\le c\le d$, and such $abcd=1$,show that
$$a+b^2+c^3+d^4\ge \dfrac{1}{a}+\dfrac{1}{b^2}+\dfrac{1}{c^3}+\dfrac{1}{d^4}$$
it seems harder than This inequality $a+b^2+c^3\ge \frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}$
| it is trivial $d \ge 1$
there are two cases: $c\ge 1$ , or $c\le 1$
$c\ge 1$
let $x=\dfrac{1}{a},y=\dfrac{1}{b} \implies x\ge y\ge 1 ,xy=cd$
let $f(x)=x-\dfrac{1}{x}$, it is trivial $f(x)$ is mono increasing
function
now we need to prove :
$f(c^3)+f(d^4) \ge f(x)+f(y^2)$
1)if $ c^3< x \cap c^3<y^2 \implies yc^3 <xy=cd \iff d> c^2y>y \cap x^2c^3 <x^2y^2=c^2d^2 \iff d^2>cx^2 \iff d>x $
$f(d^4)=f(d^2)(d^2+\dfrac{1}{d^2}) \ge 2 f(d^2) > f(x^2)+f(y^2)>f(x) +f(y^2)$
$f(c^3) \ge 0$ and this part is true.
2)if $x>y^2 \cap y^2 <c^3<x \implies c^3d^3=x^3y^3 <xd^3 \iff d^3>y^3x^2 \iff d^4>x^2 $
$f(c^3)+f(d^4) >f(y^2)+f(x^2)>f(x)+f(y^2)$
3)if $x<y^2 \cap x<c^3<y^2 \implies x^3y^3=c^3d^3 <y^2d^3\iff d^3>x^2y \ge y^3 \iff d>y$
$f(c^3)>f(x), f(d^4)>f(y^4)>f(y^2)$ it is true also.
4)it is trivial true when $c^3>x,c^3>y^2$
when $c \le 1 ,z=\dfrac{1}{c} \implies x \ge y \ge z \ge 1, d=xyz \ge 1$
we need to prove : $ f(d^4) \ge f(x)+f(y^2)+f(z^3)$
$f(z^4) \ge f(z^3),\iff f((xyz)^4) \ge f(x)+f(y^2)+f(z^4) \\ \iff (xy)^4((xy)^4-1)v^2+x^2y^2(x^2+xy^2-x^3y^2-x^2y^4)v+(xy)^4-1 \ge 0 ,v=z^4$
$(xy)^4((xy)^4-1)>0,$ if $\Delta \le 0 $ then it is true.
$\Delta \le 0 \\ \iff (x^2+xy^2-x^3y^2-x^2y^4)^2 \le 4((xy)^4-1)^2 \\ \iff x^2y^4+x^3y^2-x^2-xy^2 \le 2((xy)^4-1) \\ \iff (2x^4-x^2)y^4-(x^3-x)y^2+x^2-2 \ge 0 $
$g(t=y^2)=(2x^4-x^2)t^2-(x^3-x)t+x^2-2$
note $\dfrac{(x^3-x)}{2(2x^4-x^2)}<1 \iff 4x^3 -x^2 -2x+1 \ge 0 \iff 2x^2(2x-1)+(x-1)^2 \ge 0$ is true,
$g_{min}=g(1)=2x^4-x^3+x-2=(x^2-1)(2x^2-x+2) \ge 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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is there a general formula for cases like $\sqrt{2}$ = $\frac {2}{\sqrt{2}}$? I just noticed that $\sqrt{2}$ is equal to $\frac {2}{\sqrt{2}}$:
$\sqrt{2} = 1.414213562$
$\frac {2}{\sqrt{2}} = 1.414213562$
It is confirmed by a hand-calculator.
I tried to proof this as follows:
$\sqrt{2}$ = $\frac {2}{\sqrt{2}}$
$2^{\frac 12} = \frac {2}{2^{\frac 12}}$
$2 = 2^{\frac 12} \cdot 2^{\frac 12}$
$2 = 2^{\frac 12 + \frac 12} $
$ 2 = 2^1 $
$ 2 = 2$
It is also true for:
$-\frac {2}{\sqrt5} = \frac {-2}{5}\sqrt 5$
I didn't know this relationship beforehand and it was new for me, my question is: Is there a general rule for this?
| The steps are: $$\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}.$$You do the same thing for any positive number $a>0$: $$\frac{1}{\sqrt{a}} = \frac{1}{\sqrt{a}}\frac{\sqrt{a}}{\sqrt{a}} = \frac{\sqrt{a}}{a}.$$The trick is "multiplying by $1$ in a smart way".
P.s.: Congratulations on finding this out on your own! This is how we should always do math. Good studies.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1362633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
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Evaluating the infinite product $ \prod\limits_{n=1}^{\infty} \cos(\frac{y}{2^n}) $ How does one evaluate
$ \displaystyle\prod\limits_{n=1}^{\infty} \cos(\frac{y}{2^n}) $?
Seems fairly straightforward, as I just plugged in some initial values $n = 1, 2, 3,\dotsc$
$n = 1$
$ \sin(y)= 2\sin(\frac{y}{2})\cos(\frac{y}{2})$
$ \cos(\frac{y}{2}) = \sin(y)/2\sin(\frac{y}{2}) $
$n = 2$
$\sin(\frac{y}{2})= 2\sin(\frac{y}{4})\cos(\frac{y}{4}) $
$\cos(\frac{y}{4}) = \sin(\frac{y}{2})/2\sin(\frac{y}{4}) $
Hence,
$ \prod\limits_{n=1}^{\infty} \cos(\frac{y}{2^n}) = \sin(y)/2\sin(\frac{y}{2}) \times \sin(\frac{y}{2})/2\sin(\frac{y}{4}) ... \times \sin(\frac{y}{2^{n-1}})/2\sin(\frac{y}{2^n}) $
Then some denominators/numerators cancel, reducing the expression into
$ \prod\limits_{n=1}^{\infty} \cos(\frac{y}{2^n}) = \sin(y)/2^n\sin(\frac{y}{2^n})$
Is this expression no longer reducible?
Because apparently, somehow, the denominator seems to reduce to $y$
How does that work?
| Let's take a look at the finite product first:
$$\prod_{i=0}^k \cos\frac{x}{2^i} = \frac1{2^k}\frac{\sin 2 x}{\sin \frac{x}{2^k}}$$
which is the standard trick of multiplying the product by $\sin \frac{x}{2^k}$.
Now take limit as $k \rightarrow +\infty$ and you get:
$$\prod_{i=0}^\infty \cos\frac{x}{2^i} = \frac{\sin{2 x}}{x}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find lim$_{x \to 0}\left(\frac{1}{x} - \frac{\cos x}{\sin x}\right).$ $$\lim_{x \to 0} \left(\frac{1}{x} - \frac{\cos x}{\sin x}\right) = \frac{\sin x - x\cos x}{x\sin x}= \frac{0}{0}.$$
L'Hopital's: $$\lim_{x \to 0} \frac{f'(x)}{g'(x)} = -\frac{1}{x^2} + \frac{1}{\sin ^2x} = \frac{0}{0}.$$
Once again, using L'Hopital's: $$\lim_{x \to 0} \frac{f''(x)}{g''(x)} = \frac{2}{x^3}- \frac{2\cos x}{\sin ^3x} = \frac{0}{0}\,\ldots$$ The terms are getting endless here. Any help? Thanks.
| Learn to love asymptotics;
$$\frac {\tan x - x}{x \tan x} \sim \frac{x + x^3/3 - x}{x^2} \sim x/3 \to 0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove by induction that $4$ divides $n^3+(n+1)^3+(n+2)^3+(n+3)^3$ Just looking for someone to check my work and for feedback, thanks!
Base case: $n=0$
$0+1+8+27 = 36$
$4$ divides $36.$
Inductive step: Assume $4$ divides $k^3+(k+1)^3+(k+2)^3+(k+3)^3$ for some number where $k$ is a natural number including zero. So $k^3+(k+1)^3+(k+2)^3+(k+3)^3 = 4b$ where $b$ is some integer. We need to show $4$ divides $(k+1)^3+(k+2)^3+(k+3)^3+(k+4)^3$.
\begin{align}
(k+1)^3 & +(k+2)^3+(k+3)^3+(k+4)^3\\
&= 4k^3+30k^2+90k+100\\
&=(k^3+(k+1)^3+(k+2)^3+(k+3)^3)+12k^2+48k+36\\
&=4b+12k^2+48k+36 \qquad \text{(by inductive hypothesis)}\\
&=4(b+3k^2+12k+9)
\end{align}
Since $b$ is an element of any integer this holds true for $(k+1)$. Hence proven.
| Your answer seems right and looks like a short route, I would use modules to avoid the algebra but it wouldn't be induction.
For $K \equiv 0 \pmod 4$ the modules of $4$ for the elements of the equation would be $0^3 + 1^3 + 2^3 + (-1)^3 \equiv 0 + 1 + 0 + (-1) = 0 \pmod 4$.
And because there are always $4$ consecutive elements, the four module values are going to be the same for any $K$.
| {
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"answer_id": 0
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Repeated roots of biquadratic equation What is condition for repeated roots of the fourth order polynomial
$$ x^4 + a x^3 + b x^2 + c x + d = 0 ?$$
| A polynomial has roots with multiplicity greater than one iff its discriminant is zero.
The discriminant of your generic fourth-degree polynomial is:
$$a^2 b^2 c^2-4 b^3 c^2-4 a^3 c^3+18 a b c^3-27 c^4-4 a^2 b^3 d+16 b^4 d+18 a^3 b c d-80 a b^2 c d-6 a^2 c^2 d+144 b c^2 d-27 a^4 d^2+144 a^2 b d^2-128 b^2 d^2-192 a c d^2+256 d^3.$$
| {
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"url": "https://math.stackexchange.com/questions/1364618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the values of $\cos(\alpha+\beta) $ if the roots of an equation are given in terms of tan It is given that $ \tan\frac{\alpha}{2} $ and $ \tan\frac{\beta}{2} $ are the zeroes of the equation $ 8x^2-26x+15=0$ then find the value of $\cos(\alpha+\beta$).
I attempted to solve this but I don't know if my solution is right.Can someone verify this.
sum of roots = $ \frac{-b}{a}$
so $ \tan\frac{\alpha}{2} + \tan\frac{\beta}{2} $ = $ \frac{26}{8}$
product of roots = $ \frac{c}{a}$
so $ \tan\frac{\alpha}{2} $ . $ \tan\frac{\beta}{2} $ = $ \frac{15}{8}$
$$ \tan(\frac{\alpha+\beta}{2} ) = \frac{\tan\frac{\alpha}{2}+\tan\frac{\beta}{2}}{1-\tan\frac{\alpha}{2}.\tan\frac{\beta}{2}}$$
Puting the values in :-
$$ \tan(\frac{\alpha+\beta}{2} ) = -\frac{26}{7}$$
$$ \sec^2(\frac{\alpha+\beta}{2}) = \tan^2(\frac{\alpha+\beta}{2} ) + 1 $$
$$ \sec^2(\frac{\alpha+\beta}{2}) = \frac{(-26)^2 + 7^2}{7^2} $$
$$ \cos^2{\frac{\alpha+\beta}{2}} = \frac{7^2}{25^2}$$
$$ \frac{\cos(\alpha+\beta) + 1}{2} = \frac{49}{725}$$
$$ \cos(\alpha+\beta) = \frac{-627}{725}$$
| Since, $\tan\frac{\alpha}{2}$ & $\tan\frac{\beta}{2}$ are roots of the equation $8x^2-26x+15=0$ Hence, we have $$\tan\frac{\alpha}{2}+\tan\frac{\beta}{2}=\frac{-(-26)}{8}=\frac{13}{4}$$
$$\tan\frac{\alpha}{2}\tan\frac{\beta}{2}=\frac{15}{8}$$
$$\implies \tan \left(\frac{\alpha}{2}+\frac{\beta}{2}\right)=\frac{\tan\frac{\alpha}{2}+\tan\frac{\beta}{2}}{1-\tan\frac{\alpha}{2}\tan\frac{\beta}{2}}$$
$$\implies \tan \left(\frac{\alpha}{2}+\frac{\beta}{2}\right)=\frac{\frac{13}{4}}{1-\frac{15}{8}}$$ $$\implies \color{red}{\tan \left(\frac{\alpha}{2}+\frac{\beta}{2}\right)=-\frac{26}{7}}$$
Now, we have
$$\cos (\alpha+\beta)=\cos2\left(\frac{\alpha}{2}+\frac{\beta}{2}\right)$$
$$=\frac{1-\tan^2\left(\frac{\alpha}{2}+\frac{\beta}{2}\right)}{1+\tan^2\left(\frac{\alpha}{2}+\frac{\beta}{2}\right)}$$
$$=\frac{1-\left(\frac{-26}{7}\right)^2}{1+\left(\frac{-26}{7}\right)^2}$$ $$\implies \color{blue}{\cos(\alpha+\beta)=-\frac{627}{725}}$$
The answer above is same as you obtained. Your answer is correct.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Span of similar matrices Given $D=\begin{pmatrix} 3 & 0 \\ 0 & 2\\ \end{pmatrix}$
How can i find the span of all matrices A that are similar to D.
thus $\exists P$ $A=P^{-1}DP$
| Here are four matrices similar to $D$ which span all of $\mathbb{R}^{2\times 2}$:
$$ \begin{pmatrix} 3 & 0 \\ 0 & 2 \end{pmatrix},
\begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix},
\begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix},
\begin{pmatrix} 2 & 0 \\ 1 & 3 \end{pmatrix} $$
So the span of matrices $A$ similar to $D$ is all of (real) $2\times 2$ matrices.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solution verification for $y'$ when $y=\sin^{-1}(\frac{2x}{1+x^2})$ I was required to find $y'$ when $y=\sin^{-1}(\frac{2x}{1+x^2})$
This is my solution.
Above when I put $\sqrt{x^4-2x+1}=\sqrt{(1-x^2)^2}$ then I get the correct answer but when I put $\sqrt{x^4-2x+1}=\sqrt{(x^2-1)^2}$ then I get something else.
Now my question is that which one is correct and why? How you will come to know that we should put $\sqrt{x^4-2x+1}=\sqrt{(1-x^2)^2}$ and not $\sqrt{x^4-2x+1}=\sqrt{(x^2-1)^2}$
Kindly help me.
| Hint:
Given
$$
y = \sin^{-1}\left( \frac{2x}{1+x^2}\right),
$$
then
$$
\sin(y) = \frac{2x}{1+x^2},
$$
so
$$
\cos(y) y' = \left( \frac{2x}{1+x^2} \right)'.
$$
Now
$$
\cos(y) = \sqrt{ 1 - \sin^2(y) } =
\sqrt{ 1 - \left( \frac{2x}{1+x^2} \right)^2 },
$$
so you get
$$
y' = \frac{ \displaystyle \left( \frac{2x}{1+x^2} \right)' }
{ \displaystyle \sqrt{ 1 - \left( \frac{2x}{1+x^2} \right)^2 }}.
$$
And
$$
\left( \frac{2x}{1+x^2} \right)' = 2 \frac{1 - x^2}{(1+x^2)^2}.
$$
So final is
$$
y' = 2 \frac{ 1 - x^2 }{ \displaystyle (1+x^2) \sqrt{ (1+x^2)^2 - 4x^2 }}
= 2 \frac{ 1 - x^2 }{ \displaystyle (1+x^2) \sqrt{ (1-x^2)^2 }}.
$$
Thus
$$
y' =
\left\{
\begin{array}{rcl}
|x| < 1 : \displaystyle \frac{2}{1+x^2}\\\\
|x| > 1 : \displaystyle \frac{-2}{1+x^2}
\end{array}
\right.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove the inequality $\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$ with the constraint $abc=1$ If $a,b,c$ are positive reals such that $abc=1$, then prove that $$\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$$ I tried substituting $x/y,y/z,z/x$, but it didn't help(I got the reverse inequality). Need some stronger inequality. Thanks.
| Replace $(a,b,c) \to \left(\frac{a^2}{bc},\frac{b^2}{ca},\frac{c^2}{ab}\right)$ the inequality become
$$ \sum \frac{a}{\sqrt{a^2+8bc}} \geqslant 1.$$
We have
$$\left[5(a^2+b^2+c^2)+4(ab+bc+ca)\right]^2-(a^2+8bc)(5a+2b+2c)^2$$
$$=[5a(b+c)+17bc](2a-b-c)^2+[82a^2+25(b^2+c^2)+35a(b+c)+41bc](b-c)^2 \geqslant 0.$$
Therefore
$${\frac {a}{\sqrt {a^{2} + 8bc}}}\geq {\frac {a(5a + 2b + 2c)}{5\left(a^2 + b^2 + c^2\right) + 4(bc + ca + ab)}}.$$
Note
$$\sum{\frac {a(5a + 2b + 2c)}{5\left(a^2 + b^2 + c^2\right) + 4(ab+bc+ca)}} = 1,$$
The proof is completed.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the Jordan Form and basis $$A= \begin{pmatrix}
2&1&2\\ -1&0&2 \\ 0&0&1
\end{pmatrix}$$
I found that $$f_A(x)=m_A(x) = (x-1)^3.$$
So the Jordan form must be:
$$J= \begin{pmatrix}
1&0&0\\ 1&1&0 \\ 0&1&1
\end{pmatrix}$$
Now,
$$A-I= \begin{pmatrix}
1&1&2\\ -1&-1&2 \\ 0&0&0
\end{pmatrix} $$
$$(A-I)^2 = \begin{pmatrix}
0&0&4\\ 0&0&-4 \\ 0&0&0
\end{pmatrix}$$
$$(A-I)^3 = 0$$
We choose $x_3$ such that $(A-I)^3 x_3= 0$ but $(A-I)^2 x_3\ne 0$. We may choose for example $(0,0,1)^T$. Then we choose $x_2 = (A-I)x_3 = (2,2,0)^T$ and $x_1 = (A-I)x_2 = (4,-4,0)$.
We denote $$Q = \begin{pmatrix}
4&2&0\\ -4&2&0 \\ 0&0&1
\end{pmatrix}$$
What I got is: $$Q^{-1}AQ = \begin{pmatrix}
1&1&0\\ 0&1&1 \\ 0&0&1
\end{pmatrix} \ne J$$
Questions:
*
*I want the $1$'s to be below the main-diagonal. Why did I get them above? (And how do I change it?)
*Is this algorithm "bullet-proof"? Could I always find the Jordan form by observing the characteristic polynomial and minimal polynomial?
| *
*If you choose a basis such that $Ab_2 = b_1 + \lambda b_2$ the ones will be above the diagonal. If you choose a basis such that $A b_1 = b_2 + \lambda b_1$ they will be below. It's a matter of whether the chains go up or down your indexing.
*No. These two matrices have different Jordan forms, but the same characteristic and minimal polynomial
$$A = \left(\begin{array}{cccc} 0&1&0&0\\0&0&0&0\\0&0&0&1\\0&0&0&0 \end{array}\right)$$
$$B = \left(\begin{array}{cccc} 0&1&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0 \end{array}\right)$$
The characteristic polynomial is $t^4$ and minimal polynomial $t^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1369850",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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$U_n=\int_{n^2+n+1}^{n^2+1}\frac{\tan^{-1}x}{(x)^{0.5}}dx$ . $U_n= \int_{n^2+n+1}^{n^2+1}\frac{\tan^{-1}x}{(x)^{0.5}}dx$ where
Find $\lim_{n\to \infty} U_n$ without finding the integration
I don't know how to start
| HINT:
For $n^2+1<x<n^2+n+1$, we hav
$$\arctan(n^2+1)<\arctan(x)<\arctan(n^2+n+1)$$
and
$$\frac{1}{n+1}<\frac{1}{\sqrt{n^2+n+1}}<\frac{1}{\sqrt{x}}<\frac{1}{\sqrt{n^2+1}}<\frac1n$$
SPOLIER ALERT
SCROLL OVER SHADED AREA
We have $$I(n)=\int_{n^2+1}^{n^2+n+1}\frac{\arctan(x)}{x^{1/2}}dx$$Then, since the arctangent is increasing and $x^{-1/2}$ is decreasing, we have $$\frac{\arctan(n^2+1)}{n+1}<\frac{\arctan(x)}{\sqrt{x}}<\frac{\arctan(n^2+n+1)}{n}$$Finally, we can write$$\arctan(n^2+1)\frac{n}{n+1}<I(n)<\arctan(n^2+n+1)$$which by the squeeze theorem implies $$I(n)\to \pi/2$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Polynomial of $11^{th}$ degree Let $f(x)$ be a polynomial of degree $11$ such that $f(x)=\frac{1}{x+1}$,for $x=0,1,2,3.......,11$.Then what is the value of $f(12)?$
My attempt at this is:
Let $f(x)=a_0+a_1x+a_2x^2+a_3x^3+......+a_{11}x^{11}$
$f(0)=\frac{1}{0+1}=1=a_0$
$f(1)=\frac{1}{1+1}=\frac{1}{2}=a_0+a_1+a_2+a_3+......+a_{11} $
$f(2)=\frac{1}{2+1}=\frac{1}{3}=a_0+2a_1+4a_2+8a_3+......+2^{11}a_{11} $
.
.
.
$f(11)=\frac{1}{11+1}=\frac{1}{12}=a_0+11a_1+11^2a_2+11^3a_3+......+11^{11}a_{11} $
for calculating $f(12)$, I need to calculate $a_0,a_1,a_2,....,a_11$ but I could solve further.Is my approach right,how can I solve further or there is another right way to solve it.
$(A)\frac{1}{13}$
$(B)\frac{1}{12}$
$(C)0 $
$(D)\frac{1}{7}$
which one is correct answer?
| For convenience, let us shift the variable: $g(y)=\dfrac1y$ for $y=1,2,\cdots12$.
Then $$\frac{A(y-1)(y-2)\cdots(y-12)+1}y$$ is a polynomial of degree $11$ and coincides with $\dfrac1y$ at the given points, provided that the numerator has no independent term, i.e. $12!A+1=0$.
From this,
$$f(12)=g(13)=\frac{A(13-1)(13-2)\cdots(13-12)+1}{13}=\frac{12!A+1}{13}=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Find a polynomial from an equality Find all polynomials for which
What I have done so far:
for $x=8$ we get $p(8)=0$
for $x=1$ we get $p(2)=0$
So there exists a polynomial $p(x) = (x-2)(x-8)q(x)$
This is where I get stuck. How do I continue?
UPDATE
After substituting and simplifying I get
$(x-4)(2ax+b)=4(x-2)(ax+b)$
For $x = 2,8$ I get
$x= 2 \to -8a+b=0$
$x= 8 \to 32a+5b=0$
which gives $a$ and $b$ equal to zero.
| The route you take is fruitful.
$p\left(x\right)=\left(x-2\right)\left(x-8\right)q\left(x\right)$
leads to:
$$\left(x-4\right)q\left(2x\right)=2\left(x-2\right)q\left(x\right)$$
Then $4$ must be a root of $q$, so $q\left(x\right)=\left(x-4\right)r\left(x\right)$ leading to:
$$r\left(2x\right)=r\left(x\right)$$
Then $r\left(x\right)$ must be a constant polynomial and we end up with: $$p\left(x\right)=c\left(x-2\right)\left(x-4\right)\left(x-8\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find $x$ in the triangle
the triangle without point F is drawn on scale, while I made the point F is explained below
So, I have used $\sin, \cos, \tan$ to calculate it
Let $\angle ACB = \theta$, $\angle DFC = \angle BAC = 90^\circ$, and $DF$ is perpendicular to $BC$ (the reason for it is to have same $\sin, \cos, \tan$ answer)
$$\sin \angle ACB = \frac {DF}{CD} = \frac{AB}{BC}$$
$$\cos \angle ACB = \frac {CF}{CD} = \frac{AC}{BC}$$
$$\tan \angle ACB = \frac {DF}{CF} = \frac{AB}{AC}$$
putting known data into it
\begin{align}
\frac {DF}{CD} &= \frac {AB}{12} \quad(1) \\
\frac {EF+3}{CD} &= \frac {2CD}{12} \\
\frac {EF+3}{CD} &= \frac {CD}{6} \quad(2)\\
\frac {DF}{EF+3} &= \frac {AB}{2CD} \quad(3)
\end{align}
I've stuck at here, how do I find their length?
| Let $AD=y$ so that $\cos C=\frac{2y}{12}$
Then using the cosine rule, $$x^2=y^2+3^2-6y\cos C$$
So $x=3$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Not the toughest integral, not the easiest one Perhaps it's not amongst the toughest integrals, but it's interesting to try to find an elegant
approach for the integral
$$I_1=\int_0^1 \frac{\log (x)}{\sqrt{x (x+1)}} \, dx$$
$$=4 \text{Li}_2\left(-\sqrt{2}\right)-4 \text{Li}_2\left(-1-\sqrt{2}\right)+2 \log ^2\left(1+\sqrt{2}\right)-4 \log \left(2+\sqrt{2}\right) \log \left(1+\sqrt{2}\right)-\frac{\pi ^2}{3}$$
Do you see any such a way? Then I wonder if we can think of some elegant ways for the evaluation of the quadratic and cubic versions, that is
$$I_2=\int_0^1 \frac{\log^2 (x)}{\sqrt{x (x+1)}} \, dx$$
$$I_3=\int_0^1 \frac{\log^3 (x)}{\sqrt{x (x+1)}} \, dx.$$
How far can we possibly go with the generalization such that we can get integrals in closed form?
| By making the Euler substitution $\sqrt{x^{2}+x} = x+t$, we find
$$ \begin{align} \int_{0}^{1} \frac{\log(x)}{\sqrt{x^{2}+x}} \, dx &= 2 \int_{0}^{\sqrt{2}-1} \frac{\log \left(\frac{t^{2}}{1-2t}\right)}{1-2t} \, dt \\ &= 4 \int_{0}^{\sqrt{2}-1} \frac{\log t}{1-2t} \, dt - 2\int_{0}^{\sqrt{2}-1} \frac{\log(1-2t)}{1-2t} \, dt. \end{align}$$
The first integral can be evaluated by integrating by parts.
$$\begin{align} \int_{0}^{\sqrt{2}-1} \frac{\log t}{1-2t} \, dt &= -\frac{1}{2}\log(t) \log(1-2t) \Bigg|^{\sqrt{2}-1}_{0} + \frac{1}{2}\int_{0}^{\sqrt{2}-1} \frac{\log(1-2t)}{t} \, dt \\ &= -\frac{1}{2}\log(\sqrt{2}-1) \log(3 - 2 \sqrt{2}) - \frac{1}{2}\text{Li}_{2} \big(2(\sqrt{2}-1)\big) \end{align}$$
Therefore,
$$\begin{align}\int_{0}^{1} \frac{\log(x)}{\sqrt{x^{2}+x}} \, dx &= -2 \log(\sqrt{2}-1) \log(3 - 2 \sqrt{2}) - 2 \text{Li}_{2} \big( 2(\sqrt{2}-1) \big) + \frac{1}{2} \log^{2}(3- 2 \sqrt{2}) \\ &\approx -3.8208072259. \end{align}$$
EDIT:
David H used an Euler substitution to evaluate a similar but much more difficult integral here.
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate the limit $\lim_{x\to 0} \left(\frac 1{x^2}-\cot^2x\right)$ The answer of the given limit is $2/3$, but I cannot reach it. I have tried to use the L'Hospital rule, but I couldn't drive it to the end. Please give a detailed solution!
$$\lim_{x\to 0} \left(\dfrac 1{x^2}-\cot^2x\right)$$
|
Solution by using L-Hospital's rule (no series expansion)
$$\lim_{x\to 0} \left(\frac{1}{x^2}-\cot^2 x\right)$$ $$=\lim_{x\to 0} \left(\frac{1}{x^2}-\frac{\cos^2 x}{\sin^2x}\right)$$ $$=\lim_{x\to 0} \left(\frac{\sin^2x-x^2\cos^2 x}{x^2\sin^2x}\right)$$ $$=\lim_{x\to 0} \left(\frac{\sin^2x-x^2(1-\sin^2 x)}{x^2\sin^2x}\right)$$ $$=\lim_{x\to 0} \left(\frac{\sin^2x-x^2+x^2\sin^2 x}{x^2\sin^2x}\right)$$ $$=\lim_{x\to 0} \left(\frac{\sin^2x-x^2}{x^2\sin^2x}+1\right)$$ $$=1+\lim_{x\to 0} \left(\frac{\sin^2x-x^2}{x^2\sin^2x}\right)$$ Now, using L-Hospital's rule as follows $$1+\lim_{x\to 0} \left(\frac{\frac{d}{dx}(\sin^2x-x^2)}{\frac{d}{dx}(x^2\sin^2x)}\right)$$ $$=1+\lim_{x\to 0} \left(\frac{\sin 2x-2x}{x^2\sin 2x+2x\sin^2x}\right)$$ Apply L-H $$=1+\lim_{x\to 0} \left(\frac{2\cos 2x-2}{2x^2\cos 2x+2x\sin 2x+2x\sin 2x+2\sin^2x}\right)$$ $$=1+\lim_{x\to 0} \left(\frac{2\cos 2x-2}{2x^2\cos 2x+4x\sin 2x+2\sin^2x}\right)$$ Apply L-H $$=1+\lim_{x\to 0} \left(\frac{-4\sin 2x}{-4x^2\sin 2x+4x\cos 2x+8x\cos 2x+4\sin 2x+2\sin 2x}\right)$$ $$=1-4\lim_{x\to 0} \left(\frac{\sin 2x}{-4x^2\sin 2x+12x\cos 2x+6\sin 2x}\right)$$ Apply L-H $$=1-4\lim_{x\to 0} \left(\frac{2\cos 2x}{-8x^2\cos 2x-8x\sin 2x-24x\sin 2x+12\cos 2x+12\cos 2x}\right)$$ $$=1-8\lim_{x\to 0} \left(\frac{\cos 2x}{-8x^2\cos 2x-32x\sin 2x+24\cos 2x}\right)$$ $$=1-8\left(\frac{1}{0-0+24}\right)=1-\frac{8}{24}=1-\frac{1}{3}=\color{blue}{\frac{2}{3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
} |
how can i prove this trigonometry equation I need help on proving the following:
$$\frac{\cos {7x} - \cos {x} + \sin {3x}}{ \sin {7x} + \sin {x} - \cos {3x} }= -\tan {3x}$$
So far I've only gotten to this step:
$$\frac{-2 \sin {4x} \sin {3x} + \sin {x}}{ 2 \sin {4x} \cos {3x} - \cos {x}}$$
Any help would be appreciated as trigonometry is not my forte.
| Notice, the following formula $$\color{blue}{\cos A-\cos B=2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{B-A}{2}\right)}$$ & $$\color{blue}{\sin A+\sin B=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)}$$ Now, we have
$$\frac{\cos 7x-\cos x+{\sin 3x}}{\sin 7x+\sin x-\cos 3x}=-\tan 3x$$
$$\implies LHS=\frac{(\cos 7x-\cos x)+{\sin 3x}}{(\sin 7x+\sin x)-\cos 3x}=\frac{2\sin 4x\sin (-3x) +\sin 3x}{2\sin 4x\cos 3x-\cos 3x}$$
$$= \frac{-\sin 3x(2\sin 4x-1)}{\cos 3x(2\sin 4x-1)}=-\frac{\sin 3x}{\cos 3x}=-\tan 3x=RHS$$
| {
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Prove using mathematical induction that $x^{2n} - y^{2n}$ is divisible by $x+y$ Prove using mathematical induction that
$(x^{2n} - y^{2n})$ is divisible by $(x+y)$.
Step 1: Proving that the equation is true for $n=1 $
$(x^{2\cdot 1} - y^{2\cdot 1})$ is divisible by $(x+y)$
Step 2: Taking $n=k$
$(x^{2k} - y^{2k})$ is divisible by $(x+y)$
Step 3: proving that the above equation is also true for $(k+1)$
$(x^{2k+2} - y^{2k+2})$ is divisible by $(x+y)$.
Can anyone assist me what would be the next step? Thank You in advance!
| For $n=k$, assume $P(k)$ is true, we have
$$x^{2k}-y^{2k}=A(x-y)$$, where A is a polynomial.
For $n=k+1$,
\begin{align}
x^{2k+2}-y^{k+2}&=x^2[A(x-y)+y^{2k}]-y^{2k+2}\\&=A(x-y)x^2+x^2y^{2k}-y^{2k+2}\\&=A(x-y)x^2+y^{2k}(x^2-y^2)\\&=A(x-y)x^2+y^{2k}(x-y)(x+y)\\&=(x-y)[Ax^2+y^{2k}(x+y)]\\&=B(x-y)\text{, where } B \text{ is a polynomial}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
} |
How to integrate ${x^3}/(x^2+1)^{3/2}$? How to integrate $$\frac{x^3}{(x^2+1)^{3/2}}\ \text{?}$$
I tried substituting $x^2+1$ as t, but it's not working
| $$\int \frac{x^3}{(x^2+1)^{3/2}}dx=\int \frac{x(x^2+1)-x}{(x^2+1)^{3/2}}dx=\int \frac{x}{(x^2+1)^{1/2}}dx-\int \frac{x}{(x^2+1)^{3/2}}dx\\
=\sqrt{x^2+1}+\frac1{\sqrt{x^2+1}}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is $\left(45+29\sqrt{2}\right)^{1/3} + \left(45-29\sqrt{2}\right)^{1/3}$ an integer? The problem is the following:
Prove that this number
$$x = \left(45+29\sqrt{2}\right)^{1/3} + \left(45-29\sqrt{2}\right)^{1/3}$$
is an integer. Show which integer it is.
I thought that it has some relations with something like complex numbers such as the set $N(\sqrt{2})$ or the some kind of integer polynomial which has root $x$ and therefore show that it has integer solutions.
| Let $a,b$ be the first and second term of the sum then $x = a+b$. Observe that: $ab = 7 \to x^3 = (a+b)^3 = a^3+b^3 + 3ab(a+b)=90+3\times 7\times x \to x^3-21x-90 = 0\to (x-6)(x^2+6x+15)=0\to (x-6)((x+3)^2+6)=0\to x = 6.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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The value of the definite integral The value of the definite integral $\displaystyle\int\limits_0^\infty \frac{\ln x}{x^2+4} \, dx$ is
(A) $\dfrac{\pi \ln3}{2}$ (B) $\dfrac{\pi \ln2}{3}$ (C) $\dfrac{\pi \ln2}{4}$ (D) $\dfrac{\pi \ln4}{3}$
I tried using integration by parts,
\begin{align}
& \int_0^\infty \frac{\ln x}{x^2+4}dx = \ln x\int_0^\infty \frac{1}{x^2+4}dx-\int_0^\infty \left(\frac{d}{dx}\ln x\int_0^\infty \frac{1}{x^2+4}\right) \, dx \\[10pt]
= {} & \left[\ln x \frac{1}{2}\tan^{-1}\frac{x}{2}\right]-\int_0^\infty \frac{1}{x}\frac{1}{2}\tan^{-1}\frac{x}{2} \, dx
\end{align}
I could not move ahead. Can someone help me to get final answer?
| In General $$\int_0^{\infty}\frac{\ln(x)}{x^2+b^2}dx=\frac{\pi\ln(b)}{2b}$$ for $b>0, b\neq 1$
$\bf{Solution::}$ Let we introduce a parameter, say $a > 0$ as follows:
Set $$F(a) = \int_{0}^{\infty} \frac{\ln ax}{x^2 + b^2} dx$$
So that we get $$\frac{dF}{da} = \frac{1}{a}\int_{0}^{\infty} \frac{1}{x^2 + b^2} dx = \frac{1}{a} \frac{\pi}{2b}$$.
On integrating with respect to $a$,
We get $$\displaystyle F(a) = \frac{\pi}{2b} \ln a + C$$ and plugging in $a = 1$ gives $C = 0$
So we get $$F(a) = \int_{0}^{\infty} \frac{\ln ax}{x^2 + b^2} dx=\frac{\pi}{2b} \ln a$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to factor intricate polynomial $ ab^3 - a^3b + a^3c -ac^3 +bc^3 - b^3c $ I would like to know how to factor the following polynomial.
$$ ab^3 - a^3b + a^3c -ac^3 +bc^3 - b^3c $$
What is the method I should use to factor it? If anyone could help.. Thanks in advance.
| \begin{align*}
ab^3 - a^3b + a^3c -ac^3 +bc^3 - b^3c&=-a^3(b-c)+b^3(a-c)+c^3(b-a)\\
&=-a^3(b-c)+b^3(a-c)+c^3[(b-c)+(c-a)]\\
&=(c^3-a^3)(b-c)+(b^3-c^3)(a-c)\\
&=(c-a)(c^2+ac+a^2)(b-c)+(b-c)(b^2+bc+c^2)(a-c)\\
&=(a-c)(b-c)(-c^2-ac-a^2+b^2+bc+c^2)\\
&=(a-c)(b-c)(b^2+bc-ac-a^2)\\
&=(a-c)(b-c)[(b+a)(b-a)+c(b-a)]\\
&=(a-c)(b-c)(b-a)(a+b+c)\\
&=(a-b)(b-c)(c-a)(a+b+c)\\
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1379045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integration by Partial Fractions $\int\frac{1}{(x+1)^3(x+2)}dx$ I'm trying to do a problem regarding partial fractions and I'm not sure if I have gone about this right as my answer here doesn't compare to the answer provided by wolfram alpha. Is it that I can't Seperate things that are raised to powers on the denominator?
http://www.wolframalpha.com/input/?i=integrate+1%2F%28%28x%2B1%29%5E3%28x%2B2%29%29
My Work:
$$\int\frac{1}{(x+1)^3(x+2)}dx$$
$$\frac{1}{(x+1)^3(x+2)}=\frac{A}{(x+1)^3}+\frac{B}{(x+2)}$$
$$1=A(x+2)+B(x+1)^3$$
Plugging $x=-1$
$$1=A(1)+B(0)^3$$
$$A=1$$
Plugging $x=-2$
$$1=A(0)+B(-1)^3$$
$$B=-1$$
$$\int\frac{1}{(x+1)^3(x+2)}dx=\int\left(\frac{1}{(x+1)^3}-\frac{1}{(x+2)}\right)$$
$$\int\left(\frac{1}{(x+1)^3}-\frac{1}{(x+2)}\right)dx=\frac{-1}{2(x+1)^2}-ln|x+2|+C$$
| Let $$\frac{1}{(x+1)^3(x+2)}=\frac{A_1}{(x+1)^3}+f_1(x)$$
$$A_1=\frac{1}{(x+2)}|_{x=-1}=1,f_1(x)=-\frac{1}{(x+1)^2(x+2)}$$
Let $$f_1(x)=-\frac{1}{(x+1)^2(x+2)}=\frac{A_2}{(x+1)^2}+f_2(x)$$
$$A_2=-\frac{1}{(x+2)}|_{x=-1}=-1,f_2(x)=\frac{1}{(x+1)(x+2)}$$
$$f_2(x)=\frac{1}{(x+1)}-\frac{1}{(x+2)}$$
thus:
$$\frac{1}{(x+1)^3(x+2)}=\frac{1}{(x+1)^3}-\frac{1}{(x+1)^2}+\frac{1}{(x+1)}-\frac{1}{(x+2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1379702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Minimize the area of a wire divided into a circle and square.
A wire is divided into two parts. One part is shaped into a square, and the other part is shaped into a circle. Let r be the ratio of the circumference of the circle to the perimeter of the square when the sum of the areas of the square and circle is minimized. Find r.
My attempt so far
The total length of the wire is $y$ the length used for the square is $x$ therefore the perimeter of the square is $x$ the area of the square is then $\dfrac{x^2}{16}$, the length for the circle is $y-x$ and the circumference is therefore $y-x$
$$R=\dfrac{y-x}{2\pi}$$
Area of the circle is then $$\dfrac{(y-x)^2}{4\pi}$$
I need to minimize $$\dfrac{x^2}{16} + \dfrac{(y-x)^2}{4\pi}$$
In order to solve for $$r=\dfrac{y-x}{x}$$
Now I'm a little stuck. Thanks for any help.
| For Minimization of $$\displaystyle \frac{x^2}{16}+\frac{(y-x)^2}{4\pi}\;,$$ We Also use $\bf{Cauchy-Schwartz}$ Modified Inequality.
Which is $$\displaystyle \frac{A^2}{X}+\frac{B^2}{Y}\geq\frac{(A+B)^2}{X+Y}$$ and equality hold When $\displaystyle \frac{A}{X} = \frac{B}{Y}$.
So $$\displaystyle \frac{x^2}{16}+\frac{(y-x)^2}{4\pi}\geq \frac{(x+y-x)^2}{16+4\pi}=\frac{y^2}{16+4\pi}.$$ and equality hold when $\displaystyle \frac{x}{16}=\frac{y-x}{4\pi}$
So we get $$\displaystyle \displaystyle y-x = \frac{4\pi \cdot x}{16}$$ . So we get $$\displaystyle r = \frac{y-x}{x} = \frac{4\pi}{16} = \frac{\pi}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Two circles touching a line and the axes If the circle $C_1$ touches x-axis and the line $y=x \tan\theta$,$\theta \in (0,\frac{\pi}{2})$ in the first quadrant and circle $C_2$ touches the line $y=x \tan\theta$,y-axis and circle $C_1$ in such a way that ratio of radius of $C_1$ to the radius of $C_2$ is 2:1,then value of $\tan\frac{\theta}{2}=\frac{\sqrt a-b}{c}$,where $a,b,c$ are relatively prime natural numbers.Then find $a+b+c$.
I tried the condition of external touching but center of circles not given,so failed.Can someone guide me to solve this question?
| Let the radius of circle $C_2$ be $R_2$ & center $O_2$ & the radius of the circle $C_1$ will be $R_1$ & center $O_2$ $\forall\ \frac{R_1}{R_2}=2$
Now, let the circles touch at any point say $P(d, d)$ on the line $y=x$ then in right $\triangle OPO_1$ $$\tan\frac{\theta}{2}=\frac{R_1}{d}$$ $$d=R_1\cot\frac{\theta}{2}$$ Similarly, in right $\triangle OPO_2$ $$\tan\left(\frac{90^\circ-\theta}{2}\right)=\frac{R_2}{d}$$
$$\tan\left(45^\circ-\frac{\theta}{2}\right)=\frac{R_2}{R_1\cot\frac{\theta}{2}}$$ $$\frac{\tan 45^\circ-\tan\frac{\theta}{2}}{1+\tan 45^\circ\tan\frac{\theta}{2}}=\frac{1}{2}\tan\frac{\theta}{2}$$
$$\frac{1-\tan\frac{\theta}{2}}{1+\tan\frac{\theta}{2}}=\frac{1}{2}\tan\frac{\theta}{2}$$ $$2-2\tan\frac{\theta}{2}=\tan\frac{\theta}{2}+\tan^2\frac{\theta}{2}$$ $$\tan^2\frac{\theta}{2}+3\tan\frac{\theta}{2}-2=0$$ Now, solving the above quadratic equation for $\tan \frac{\theta}{2}$ we get $$\tan \frac{\theta}{2}=\frac{-3\pm\sqrt{3^2-4(1)(-2)}}{2(1)}$$ $$=\frac{-3\pm\sqrt{17}}{2}$$ But, we have $0<\theta<\frac{\pi}{2}\implies \tan \frac{\theta}{2}>0$, hence, we get $$\tan \frac{\theta}{2}=\frac{-3+\sqrt{17}}{2}=\frac{\sqrt{17}-3}{2}$$ Now, compare above with $\tan\frac{\theta}{2}=\frac{\sqrt a-b}{c}$, we get $\color{red}{a=17, b=3, c=2}$ Hence $$\bbox[5px, border: 2px solid #C0A000]{\color{blue}{a+b+c=17+3+2=\color{red}{22}}}$$
| {
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"url": "https://math.stackexchange.com/questions/1381778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Showing that $\frac{1}{2^n +1} + \frac{1}{2^n +2} + \cdots + \frac{1}{2^{n+1}}\geq \frac{1}{2}$ for all $n\geq 1$
Show that $$\frac{1}{2^n +1} + \frac{1}{2^n +2} + \cdots + \frac{1}{2^{n+1}}\geq \frac{1}{2}$$ for all $n\geq 1$
I need this in order to complete my proof that $1 + \frac{n}{2} \leq H_{2^n}$, but I don't have any ideas.
I know it is true at least for the first $20$ cases, but I can't prove it. Any suggestions?
| $$\sum_{k=1}^{2^n} \frac{1}{2^n+k} \geq \sum_{k=1}^{2^n} \frac{1}{2^n+2^n} =\sum_{k=1}^{2^n} \frac{1}{2^{n+1}}$$
the rightmost sum equals $\frac{1}{2}$, in fact we have a strict inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Big an open ball inside small open ball in metric space When i was reading the book "Elements of the Theory of Functions and Functional Analysis" (A. N. Kolmogorov, S. V. Fomin), I encountered a very interesting (for me) problem.
Problem:
Create a metric space with two open balls $B(x,\rho_1), B(y,\rho_2)$ such that $\rho_1>\rho_2$ and $B(x,\rho_1)\subset B(y,\rho_2)$.
I will post my example, and i would like to know more examples.
Thanks for attention!
| One example:
metric space $(X,d)$ such that $X(0,1)$ with metrica $d\left( x,y \right) =\begin{cases} 2-\left| x-y \right| \quad ,x\neq y \\ 0\quad \quad \quad \quad \quad \quad ,x=y \end{cases}$
$$d\left( \frac { 1 }{ 2 } ,\frac { 1 }{ 2 } +\frac { 1 }{ 5 } \right) =2-\frac { 1 }{ 5 } =\frac { 9 }{ 5 } >\frac { 7 }{ 4 } =2-\frac { 1 }{ 4 } =d\left( \frac { 1 }{ 2 } ,\frac { 1 }{ 2 } +\frac { 1 }{ 4 } \right) \\$$
however $$\left( \frac { 1 }{ 2 } -\frac { 1 }{ 5 } ,\frac { 1 }{ 2 } +\frac { 1 }{ 5 } \right) \subset \left( \frac { 1 }{ 2 } -\frac { 1 }{ 4 } ,\frac { 1 }{ 2 } +\frac { 1 }{ 4 } \right) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1383535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why does $x^2+47y^2 = z^5$ involve solvable quintics? This is related to the post on $x^2+ny^2=z^k$. In response to my answer on,
$$x^2+47y^2 = z^3\tag1$$
where $z$ is not of form $p^2+nq^2$, Will Jagy provided one for,
$$x^2+47y^2 = z^5\tag2$$
$$ (14p^5 + 405p^4q + 3780p^3q^2 + 13410p^2q^3 + 11550pq^4 - 14647q^5)^2 + 47 ( p^5 - 270p^3q^2 - 2520p^2q^3 - 8115pq^4 - 8344q^5)^2 = (3p^2 + 28pq + 81q^2)^5\tag3 $$
As noticed by Elaqqad, the cubic polynomials I used for $(1)$ involve the discriminant $d=-47$ and I assumed it would be same with the (irreducible) quintic polynomials used by Jagy for $(3)$. Then I wondered if they were solvable in radicals as well. (I knew Ramanujan played with a solvable quintic with $d=-47$.) It turns out they are.
This Magma calculator computes the Galois group and the command is:
Z := Integers(); P < x > := PolynomialRing(Z); f := 14*x^5 + 405*x^4 + 3780*x^3 + 13410*x^2 + 11550*x - 14647; G, R := GaloisGroup(f); G;
Testing both polynomials, it shows the group has order 20 and hence is solvable.
Q: Given $x^2+dy^2=z^k$ where $z\neq p^2+dq^2$, is it true that if
$$\big(P_1(x)\big)^2+d\big(P_2(x)\big)^2=\big(P_3(x)\big)^k$$
then the equations $P_1(x) = P_2(x) = P_3(x) = 0$ are solvable in radicals?
P.S. Or is this $5$th parameterization special only because the class number $h(-47) = 5$? One way to check would be to solve $x^2+47y^2 = z^\color{red}7$ analogous to $(3)$ (Will, care to oblige?) and see if it involves solvable septics.
| (A partial answer.) Thanks to Jagy's two parameterizations for degs $5$ and $7$, a general identity has been found. The clue was to transform $(3)$ using $p=u-14v,\,q=3v$ to the form,
$$(14 u^5 + 235 u^4 v - 6580 u^3 v^2 - 22090 u^2 v^3 + 154630 u v^4 + 47^3 v^5)^2 + 47(u^5 - 70 u^4 v - 470 u^3 v^2 + 6580 u^2 v^3 + 11045 u v^4 - 14\cdot47^2 v^5)^2\\=3^5(u^2+47v^2)^5$$
(and similarly for the deg $7$). Inspecting the coefficients, patterns were found. Thus,
If $a^2+db^2=c^5$, then,
$$(\color{blue}+a u^5 \color{blue}+ 5 b d u^4 v \color{red}- 10 a d u^3 v^2 \color{red}- 10 b d^2 u^2 v^3 \color{blue}+ 5 a d^2 u v^4 \color{blue}+ b d^3 v^5)^2
+ d (\color{blue}+ b u^5 \color{red}- 5 a u^4 v \color{red}- 10 b d u^3 v^2 \color{blue}+ 10 a d u^2 v^3 \color{blue}+ 5 b d^2 u v^4 \color{red}- a d^2 v^5)^2 \\= c^5 (u^2 + d v^2)^5$$
where the above used $14^2+47\times1^2=3^5$.
(Edited later.) In general, if,
If $a^2+db^2=c^k$, then,
$$\Big(a\,\phi_1-b\sqrt{-d}\,\phi_2\Big)^2+d \Big(b\,\phi_1-\frac{a}{\sqrt{-d}}\,\phi_2\Big)^2 =c^k(u^2+dv^2)^k$$
where,
$$\phi_1 = \frac{(u+\sqrt{-d}\,v)^k+(u-\sqrt{-d}\,v)^k}{2}$$
$$\phi_2 = \frac{(u+\sqrt{-d}\,v)^k-(u-\sqrt{-d}\,v)^k}{2}$$
Since an initial solution to $a^2+db^2=c^k$ is easily found for any $d$ (such as using the obvious $a,b,c = 1,0,1$), then the class number $h(-d)$ need not matter, answering part of my original question. For example, using $a^2+47b^2=c^7$ where $a,b,c = 866458,55861,51$, then one can find a $k=7$ parameterization similar to the one found by Jagy for $d=71$, even though $h(-47) = 5$.
The discriminants $D$ of the first polynomials $P_1(u,v)$ for $k=5,7$ are,
$$D_5 = 2^{12}\cdot5^5c^{20}d^{10}$$
$$D_7 = 2^{30}\cdot7^7c^{42}d^{21}$$
Their form and discriminants highly suggest that, equated to zero, then $P_i(u,v)=0$ is solvable in radicals for any $a,b,d$. But I do not (yet) have a rigorous proof that this is the case.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Solving simple mod equations Solve $3x^2 + 2x + 1 \equiv 0 \mod 11$
Additionally, I have an example problem, but a step in the middle has confused me:
$3x^2 + 5x - 7 \equiv 0 \mod 17$. Rearrange to get $3x^2 + 5x \equiv 7 \mod 17$.
$\implies 6\cdot 3x^2 + 6\cdot 5x \equiv 6*7 \mod 17$.
The next line reads $x^2 + 30x \equiv 8 \mod 17$, which I am confused about;
I understand how $8$ came to be ($42 - 34$, reduced to $8$) but what happened to the coefficient of $3x^2$? This example is from my notes in NT class, so perhaps I copied something incorrectly, or did I?
For the first problem, you don't need to finish the whole problem. Once you get to a point where it looks something like $(x+15)^2 \equiv 10 \mod 17$ (result of the example problem), I know how to proceed. It's the beginning parts that have muddled me a bit...
Help would be greatly appreciated.
| $3x^2 + 2x + 1 \equiv 0 \pmod{11}$
Since $3 \cdot 4 \equiv 12 \equiv 1 \pmod{11}, \quad \frac 13 \equiv 4 \pmod{11}$
So, multiplying both sides by $4$, we get
$x^2 + 8x + 4 \equiv 0 \pmod{11}$
Half of $8$ is $4$ and $4^2 = 16 \equiv 5 \pmod{11}$. So
$x^2 + 8x + 4
\equiv x^2 + 8x + 5 - 1
\equiv (x+4)^2 - 1 \pmod{11}$
So you get
$(x+4)^2 - 1 \equiv 0 \pmod{11}$
$(x+4)^2 \equiv 1 \pmod{11}$
$x+4 \equiv 1 \pmod{11} \quad \text{or} \quad x+4 \equiv -1 \pmod{11}$
$x \equiv 8 \pmod{11} \quad \text{or} \quad x \equiv 6 \pmod{11}$
Note that this implies
$3x^2+2x+1 \equiv 3(x-6)(x-8) \pmod{11}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove the series has positive integer coefficients How can I show that the Maclaurin series for
$$
\mu(x) = (x^4+12x^3+14x^2-12x+1)^{-1/4}
\\
= 1+3\,x+19\,{x}^{2}+147\,{x}^{3}+1251\,{x}^{4}+11193\,{x}^{5}+103279\,
{x}^{6}+973167\,{x}^{7}+9311071\,{x}^{8}+\cdots
$$
has positive integer coefficients? (I have others to do, too, but this one will be a start.)
possibilities
(a) The coefficients $Q(n)$ satisfy the recurrence
$$
(n+1)Q(n)+(12n+21)Q(n+1)+(14n+35)Q(n+2)+(-12n-39)Q(n+3)+(n+4)Q(n+4) = 0
$$
(b) $\mu(x)$ satisfies the differential equation
$$
(x^3+9x^2+7x-3)\mu(x)+(x^4+12x^3+14x^2-12x+1)\mu'(x)=0
$$
(c) factorization of $x^4+12x^3+14x^2-12x+1$ is
$$
\left( x-\sqrt {5}+3+\sqrt {15-6\,\sqrt {5}} \right) \left( x-\sqrt
{5}+3-\sqrt {15-6\,\sqrt {5}} \right) \left( x+\sqrt {5}+3-\sqrt {15+
6\,\sqrt {5}} \right) \left( x+\sqrt {5}+3+\sqrt {15+6\,\sqrt {5}}
\right)
$$
(d) $(1-8X)^{-1/4}$ has positive integer coefficients, But if $X$ is defined by $1-8X=x^4+12x^3+14x^2-12x+1$, then $X$ does not have integer coefficients.
(e) Can we compute the series for $\log\mu(x)$
$$3\,x+{\frac {29}{2}}{x}^{2}+99\,{x}^{3}+{\frac {3121}{4}}{x}^{4}+{
\frac {32943}{5}}{x}^{5}+{\frac {348029}{6}}{x}^{6}+\dots
$$
and then recognize that its exponential has integer coefficients?
| Here is a proof for positivitiy. Use the recurrence
$$
(n+1)Q(n)+(12n+21)Q(n+1)+(14n+35)Q(n+2)+(-12n-39)Q(n+3)+(n+4)Q(n+4) = 0
$$
$Q(0)=1,Q(1)=3,Q(2)=19,Q(3)=147$. Prove by induction that
$Q(n) \ge 3 Q(n-1)$ for $n \ge 1$. This is true for the first few terms.
Assume true up to $Q(n+3)$, then prove it for $Q(n+4)$ as follows:
$$
(n+4)Q(n+4) = (12n+39)Q(n+3)-(14n+35)Q(n+2)-(12n+21)Q(n+1)-(n+1)Q(n)
\\
\ge \left[(12n+39)-\frac{14n+35}{3}-\frac{12n+21}{9}-\frac{n+1}{27}\right]Q(n+3)
\\
=\left[\frac{161}{27} n + \frac{692}{27}\right] Q(n+3)
\gt (3n+12)Q(n+3)=(n+4)3Q(n+3)
$$
and therefore $Q(n+4) > 3 Q(n+3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1386819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
} |
Find the value below
Let $a,b,c$ be the roots of the equation $$8x^{3}-4x^{2}-4x+1=0$$
Find $$\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}$$
It's just for sharing a new ideas, thanks:)
| It has been observed by others that the reciprocals $x_i$ $(1\leq i\leq 3)$ of $a$, $b$, $c$ are the solutions of the equation
$$x^3-4x^2-4x+8=0\ .\tag{1}$$
The power sum $$p_3:=x_1^3+x_2^3+x_3^3={1\over a^3}+{1\over b^3}+{1\over c^3}$$ is a symmetric function of the $x_i$. Therefore it can be expressed as a polynomial in terms of the elementary functions $\sigma_j$ $(1\leq j\leq 3)$ of the $x_i\,$: One easily verifies that
$$x_1^3+x_2^3+x_3^3=(x_1+x_2+x_3)^3-3(x_1+x_2+x_3)(x_1x_2+x_2x_3+x_3x_1)+3x_1x_2x_3\ ,$$
or $$p_3=\sigma_1^3-3\sigma_1\sigma_2+3\sigma_3\ .$$
Now $(1)$ and Vieta's theorem tell us that
$$\sigma_1=4,\quad \sigma_2=-4,\qquad \sigma_3=-8\ .$$
It follows that
$$p_3=64-3\cdot4\cdot(-4)+3\cdot(-8)=88\ .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1387198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Proving that $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}$ I have written the left side of the equation as $$\left(1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n-1}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n}\right).$$ I don't know how to find the sums of these sequences. I know the sums for odd and even integers, but I can't figure this out.
| $$\begin{align}
S&=\sum_{r=1}^n\frac 1{2r-1}-\sum_{r=1}^n\frac 1{2r}\color{orange}{+\sum_{r=1}^n\frac 1{2r}-\sum_{r=1}^n\frac 1{2r}}\\
&=\underbrace{\sum_{r=1}^n\frac 1{2r-1}+\sum_{r=1}^n\frac 1{2r}}_{}-\underbrace{2\sum_{r=1}^n\frac 1{2r}}\\
&=\qquad\quad\sum_{r=1}^{2n}\frac 1r\qquad \qquad-\sum_{r=1}^n\frac 1r\\
&=\color{blue}{\sum_{r=n+1}^{2n}\frac 1r}\qquad\blacksquare
\end{align}$$
The proof above is illustrated visually in the expansion below.
If $n$ is odd,
$$\require{cancel}\begin{align}
S&=1-\frac12+\frac13-\frac14+\frac15-\frac16+\cdots+\frac1n\color{green}{-\frac 1{n+1}+\cdots-\frac 1{2n}}\\\\
&=1+\frac12+\frac13+\frac14+\frac15+\frac16+\cdots+\frac1n\color{blue}{+\frac1{n+1}+\cdots+\frac 1{2n}}\\
&\;- 2 \left(\frac12\qquad+\frac14\qquad+\frac16+\cdots\qquad\;+\frac 1{n+1}+\;\cdots+\frac 1{2n}\right)\\\\\;
&=\cancel{1+\frac12+\frac13+\frac14+\frac15+\frac16+\cdots+\frac1n}\color{blue}{+\frac 1{n+1}+\cdots+\frac 1{2n}}\\
&\quad\;\;- \cancel{\left(1\qquad+\frac12\qquad+\frac13+\cdots\qquad+\frac 1{\frac12(n+1)}+\cdots+\frac 1{n}\right)}\\\\
&=\color{blue}{\frac1{n+1}+\frac1{n+2}\cdots+\frac 1{2n}}\qquad\blacksquare
\end{align}$$
If $n$ is even,
$$\require{cancel}\begin{align}
S&=1-\frac12+\frac13-\frac14+\frac15-\frac16+\cdots-\frac1n\color{green}{+\frac 1{n+1}+\cdots-\frac 1{2n}}\\\\
&=1+\frac12+\frac13+\frac14+\frac15+\frac16+\cdots+\frac1n\color{blue}{+\frac1{n+1}+\cdots+\frac 1{2n}}\\
&\;- 2 \left(\frac12\qquad+\frac14\qquad+\frac16+\cdots+\frac 1n+\cdots\qquad\qquad+\frac 1{2n}\right)\\\\
&=\cancel{1+\frac12+\frac13+\frac14+\frac15+\frac16+\cdots+\frac1n}\color{blue}{+\frac1{n+1}+\cdots+\frac 1{2n}}\\
&\quad\;\;- \cancel{\left(1\qquad+\frac12\qquad+\frac13+\cdots+\frac 1{\frac12n}+\cdots\quad\qquad+\frac 1n\right)}\\\\
&=\color{blue}{\frac1{n+1}+\frac1{n+2}\cdots+\frac 1{2n}}\qquad\blacksquare
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1387744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
How do I find the remainder of $4^0+4^1+4^2+4^3+ \cdots + 4^{40}$ divided by 17? Recently I came across a question,
Find the remainder of $4^0+4^1+4^2+4^3+ \cdots + 4^{40}$ divided by 17?
At first I applied sum of G.P. formula but ended up with the expression $1\cdot \dfrac{4^{41}-1}{4-1}$. I couldn't figure out how to proceed further. Secondly I thought of using the fact $(a+b+\cdots) \pmod {17} = (r_a+r_b\dots) \pmod {17}$ but it is getting more messier.
Please explain in detail. And also mention the formula being used.
| If you calculate $4^{k}\ (\operatorname{mod}\ 17)$ for some $k$'s, you might quickly notice that the values are periodically $1,4,-1,-4$. This is not coincidence, since $4^4 \equiv 1 \ (\operatorname{mod}\ 17)$, and thus $$4^{4k+l}\equiv 4^{4k}\cdot 4^l \equiv (4^4)^k\cdot 4^l \equiv 4^l \ (\operatorname{mod}\ 17)$$ So, we conclude $$4^{4k+1}+4^{4k+2}+4^{4k+3}+4^{4k+4}\equiv 4 + (-1) + (-4) + 1 \equiv 0 \ (\operatorname{mod}\ 17)$$ and can calculate $$4^0+4^1+4^3+\cdots 4^{40} \equiv (4^0+4^1+4^3+4^4)+\sum_{k=1}^9(4^{4k+1}+4^{4k+2}+4^{4k+3}+4^{4k+4}) \equiv 4^0+4^1+4^3+4^4 \equiv 1 + 4 + (-4) + 1 \equiv 2\ (\operatorname{mod}\ 17)$$ Of course, if $4^2$ is accidentally missing in the question, sum just as easily evaluates to $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Find the sum of binomial coefficients
Calculate the value of the sum
$$
\sum_{i = 1}^{100} i\binom{100}{i} = 1\binom{100}{1} +
2\binom{100}{2} +
3\binom{100}{3} +
\dotsb +
100\binom{100}{100}
$$
What I have tried:
$$\begin{align}
S &= 0\binom{100}{0}+1\binom{100}{1}+ \dotsb +99\binom{100}{99}+100\binom{100}{100} \\ \\
&=100\binom{100}{100}+99\binom{100}{99}+ \dotsb +1\binom{100}{1}+0\binom{100}{0}
\end{align}$$
and I'm stuck here, I don't know if it's true or not, any help will be appreciated.
| $\bf{My\; Solution::}$ Using $$\displaystyle (1+x)^n = \binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+..........+\binom{n}{n}x^n$$
Now Diff. both side w.r to $x\;,$ We get
$$\displaystyle n(1+x)^{n-1} = \binom{n}{1}x+\binom{n}{2}\cdot 2x+\binom{n}{3}\cdot 3x^2+.........+\binom{n}{n}\cdot nx^{n-1}$$
Now Put $x=1$ and $n=100\;,$ We get
$$\displaystyle 100\cdot 2^{99} = \binom{100}{1}\cdot 1+\binom{100}{2}\cdot 2+\binom{100}{3}\cdot 3+..........+\binom{100}{100}\cdot 100$$
Can anyone have any idea about combinatorial prove, If Yes Then plz explain here,
Thanks
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find the general solution of $\cos(x)-\cos(2x)=\sin(3x)$ Problem:
Find the general solution of $$\cos(x)-\cos(2x)=\sin(3x)$$
I tried attempting this by using the formula$$\cos C-\cos D=-2\sin(\dfrac{C+D}{2})\sin(\dfrac{C-D}{2})$$
Thus, $$-2\sin\left(\dfrac{x}{2}\right)\sin\left(\dfrac{3x}{2}\right)=\sin 3x$$
$$\Rightarrow -2\sin\left(\dfrac{x}{2}\right)\sin\left(\dfrac{3x}{2}\right)-\sin 3x=0$$
Unfortunately, I couldn't get further with this problem. Any help with this would be truly appreciated. Many thanks in advance!
| \begin{align}
−2 \sin\left(\frac{x}{2}\right) \, \sin\left(\frac{3x}{2}\right)−\sin 3x &=0 \\
−2 \sin\left(\frac{x}{2}\right) \, \sin\left(\frac{3x}{2}\right)-2 \sin\left(\frac{3x}{2}\right) \, \cos\left(\frac{3x}{2}\right) &=0 \\
\sin\left(\frac{3x}{2}\right) =0 \quad \text{or} \quad \sin\left(\frac{x}{2}\right)-\cos\left(\frac{3x}{2}\right) &=0 \\
\sin\left(\frac{3x}{2}\right) = \sin(0) \quad \text{or} \quad \cos\left(\frac{3x}{2}\right) &= \cos\left(\frac{π-x}{2}\right)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
If $(y^2-5y+3)\cdot (x^2+x+1)<2x$ for all $x\in \mathbb{R}\;,$ Then Range of $y$ is
If $(y^2-5y+3)\cdot (x^2+x+1)<2x$ for all $x\in \mathbb{R}\;,$ Then Range of $y$ is
$\bf{My\; Try::}$ We can write the given inequality as $\displaystyle y^2-5y+2<\frac{2x}{x^2+x+1}.$
Now we will calculate range of $$\displaystyle z = \frac{2x}{x^2+x+1}\Rightarrow zx^2+zx+z = 2x$$
So $$zx^2+x(z-2)+z=0\;,$$ Now if given equation has real roots, Then $\bf{Discriminant \geq 0}$
So $$(z-2)^2-(2z)^2\geq 0 \Rightarrow (2z)^2-(z-2)^2\leq 0$$
So $$\displaystyle (z+2)\cdot (3z-2)\leq 0 \Rightarrow (z+2)\cdot \left(z-\frac{2}{3}\right)\leq 0 \Rightarrow -2 \leq z\leq \frac{2}{3}$$
Here above we have given $$\displaystyle y^2-5y+3<\frac{2x}{x^2+x+1} \leq -2$$
So $\displaystyle y^2-5y+3< -2\Rightarrow y^2-5y+5 < 0$
So we get $$\displaystyle \left(\frac{5+\sqrt{5}}{2}\right)< y < \left(\frac{5+\sqrt{5}}{2}\right)$$
Is my solution is Right, If not then plz explain me , Thanks
| If $x<0$, then by AM-GM, $|x|+\frac{1}{|x|}\geq 2$, whence
$$\frac{2x}{x^2+x+1}=\frac{2}{1-\left(|x|+\frac{1}{|x|}\right)}\geq \frac{2}{1-2}=-2\,.$$
The equality holds iff $x=-1$. If $x\geq 0$, then $\frac{2x}{x^2+x+1}\geq 0>-2$. That is, $y^2-5y+3<-2$, meaning that all possible values of $y$ are $y\in\mathbb{R}$ such that $$\frac{5-\sqrt{5}}{2}<y<\frac{5+\sqrt{5}}{2}\,.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What substitution should I do to solve the following integral? Evaluate $$ \int{\frac{(1+x^2)}{(1-x^2)\sqrt{1+x^2+x^4}}}dx $$
I substituted $x- \frac{1}{x} $ with u so $$-\int{\frac{du}{u\sqrt{u^2+3}}} $$
Now I put $u=\frac1t$
so $$\int{\frac{dt}{\sqrt{1+3t^2}}} $$
| HINT:
$$\text{Set }\int\left(1+\dfrac1{x^2}\right)dx=u\text{ so that }\dfrac{du}{dx}=\cdots$$
$$\text{in }\frac{(1+x^2)}{(1-x^2)\sqrt{1+x^2+x^4}}=\dfrac{1+\dfrac1{x^2}}{\left(x-\dfrac1x\right)\sqrt{\dfrac1{x^2}+1+x^2}}$$
and use
$$\dfrac1{x^2}+x^2=\left(x-\dfrac1x\right)^2+2$$
For $\int\dfrac{du}{u\sqrt{u^2+3}}$
either set $u=\sqrt3\tan y$
or for $\int\dfrac{du}{u\sqrt{u^2+3}}=\int\dfrac{u\ du}{u^2\sqrt{u^2+3}}$
set $\sqrt{u^2+3}=v$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Counting integral solutions Suppose $a + b + c = 15$
Using stars and bars method, number of non-negative integral solutions for the above equation can be found out as $15+3-1\choose15$ $ =$ $17\choose15$
How to extend this principle for finding number of positive integral solutions of
$a + b + 3c = 15$?
I tried to do it by substituting $3c$ with another variable $d$. But could not succeed.
| One way to solve this is using generating functions. If you multiply $A(z) = \sum_{n \ge 0} a_n z^n$ by $B(z) = \sum_{n \ge 0} b_n z^n$, you get:
$$
A(z) \cdot B(z)
= \sum_{n \ge 0} \left(\sum_{0 \le k \le n} a_k b_{n - k}\right) z^n
$$
If $a_n$ is the number of ways of picking $a$ to have value $n$, and similarly for $b$, the coefficient of $z^n$ is the number of combining a value of $a$ and one of $b$ to make up $n$.
For your original example, there is $1$ way to get $a$ (and $b$, $c$) of any particular value, so that:
$$
A(z) = B(z) = C(z)
= \sum_{n \ge 0} z^n
= \frac{1}{1 - z}
$$
so that the number of ways to get $n$ is ($[z^n]$ is just shorthand for "get the coefficient of $z^n$"):
$$
[z^n] A(z) \cdot B(z) \cdot C(z)
= [z^n] \frac{1}{(1 - z)^3}
= (-1)^n \binom{-3}{n}
= \binom{n + 3 - 1}{3 - 1}
= \binom{n + 2}{2}
$$
For the second case you have:
$$
C(z) = 1 + z^3 + z^6 + \dotsb
= \sum_{n \ge 0} z^{3 n}
= \frac{1}{1 - z^3}
$$
so here you need:
$\begin{align}
A(z) \cdot B(z) \cdot C(z)
&= \frac{1}{(1 - z)^2 (1 - z^3)} \\
&= \frac{1 + 2 z}{9 (1 + z + z^2)}
+ \frac{2}{9 (1 - z)}
+ \frac{1}{3 (1 - z)^2}
+ \frac{1}{3 (1 - z)^3}
\end{align}$
This is just the partial fractions mostly used when integrating.
The first term is troublesome as written. Using complex numbers you can write it as:
$$
\frac{1 + 2 z}{1 + z + z^2}
= - \frac{\omega}{1 - \omega z}
- \frac{\omega^2}{1 - \omega^2 z}
$$
Here $\omega = - 1/2 + \mathrm{i} \sqrt{3} /2$ is a cube root of $1$, and $\omega^2 = \overline{\omega}$ is just it's conjugate. As the sum of a complex number and it's conjugate is just twice their real part, this term contributes:
$$
- \omega \cdot \omega^n - \overline{\omega} \cdot \overline{\omega}^n
= - \omega^{n + 1} - \overline{\omega}^{n + 1}
= - 2 \Re \left( \omega^{n + 1} \right)
$$
We can write:
$$
\omega = \exp \left( \frac{2 \pi \mathrm{i}}{3} \right)
$$
so that:
$$
\omega^{n + 1}
= \exp \left( \frac{2 \pi (n + 1) \mathrm{i}}{3} \right)
= \cos \left( \frac{2 \pi (n + 1)}{3} \right)
+ \mathrm{i} \sin \left( \frac{2 \pi (n + 1)}{3} \right)
$$
and finally the contribution is:
$$
- 2 \Re \left( \omega^{n + 1} \right)
= - 2 \cos \left( \frac{2 \pi (n + 1)}{3} \right)
$$
Adding the contributions of the other terms gives:
$$
\frac{2}{9}
+ \frac{1}{3} \binom{n + 2 - 1}{2 - 1}
+ \frac{1}{3} \binom{n + 3 - 1}{3 - 1}
- \frac{2}{9} \cos \left( \frac{2 \pi (n + 1)}{3} \right)
= \\
\frac{3 n^2 + 15 n + 16}{18}
- \frac{2}{9} \cos \left( \frac{2 \pi (n + 1)}{3} \right)
$$
The last expression looks quite weird, but you'll notice that it repeats with period $3$, so you could write this out as three cases depending on the remainder when $n$ is divided by $3$.
OK, back to your specific case now. You asked for a sum of $15$. The number of ways of getting that one is:
$$
\frac{3 \cdot 15^2 + 15 \cdot 15 + 16}{18}
- \frac{2}{9} \cos \left( \frac{2 \pi (15 + 1)}{3} \right)
= 51
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Find Sum of n terms of the Series: $ \frac{1}{1\times2\times3} + \frac{3}{2\times3\times4} + \frac{5}{3\times4\times5} + \cdots $ I want to find the sum to n terms of the following series.
$$ \frac{1}{1\times2\times3} + \frac{3}{2\times3\times4} + \frac{5}{3\times4\times5} + \frac{7}{4\times5\times6} + \cdots $$
I have found that the general term of the sum is
\begin{equation}
\sum\limits_{i=1}^n \frac{2i - 1}{i(i+1)(i+2)}
\end{equation}
so the nth term must be
\begin{equation}
\frac{2n - 1}{n(n+1)(n+2)}
\end{equation}
Now.. how should i evaluate the sum of the n terms? What is the method i must use in order to do so? Thank you in advance.
| $$\displaystyle \frac{2n-1}{n(n+1)(n+2)} = \frac{2}{(n+1)(n+2)}-\frac{1}{n(n+1)(n+2)} = A-B$$
Now $$\displaystyle A = \frac{2}{(n+1)(n+2)} = 2\left[\frac{(n+2)-(n+1)}{(n+1)(n+2)}\right] = 2\left[\frac{1}{n+1}-\frac{1}{n+2}\right]$$
And $$\displaystyle B = \frac{1}{n(n+1)(n+2)} = \frac{1}{2}\left[\frac{(n+2)-(n)}{n(n+1)(n+2)}\right] = \frac{1}{2}\left[\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right]$$
$$\displaystyle = \frac{1}{2}\left\{\left[\frac{1}{n}-\frac{1}{n+1}\right]-\left[\frac{1}{n+1}-\frac{1}{n+2}\right]\right\}$$
Now $$\displaystyle \sum_{r=1}^{n}S_{n} = 2\sum_{r=1}^{n}\left[\frac{1}{r+1}-\frac{1}{r+2}\right]-\frac{1}{2}\sum_{r=1}^{n}\left[\frac{1}{r}-\frac{1}{r+1}\right]+\frac{1}{2}\sum_{r=1}^{n}\left[\frac{1}{r+1}-\frac{1}{r+2}\right]$$
Now Use Telescopic Sum
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\int e^x \sin^2 x \mathrm{d}x$ Is the following evaluation of correct?
\begin{align*} \int e^x \sin^2 x \mathrm{d}x &= e^x \sin^2 x -2\int e^x \sin x \cos x \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int e^x (\cos^2 x - \sin^2x) \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int e^x (1 - 2\sin^2x) \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int -2 e^x \sin^2x \mathrm{d}x + 2 e^x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 e^x -4 \int e^x \sin^2x \mathrm{d}x \end{align*}
First two steps use integration by parts. In the first step we differentiate $\sin^2 x$. In the second step we differentiate $\sin x \cos x$.
Using this, we reach $$5\int e^x \sin^2 x \mathrm{d}x = e^x \sin^2 x -2e^x \sin x \cos x + 2 e^x$$
$$\int e^x \sin^2 x \mathrm{d}x = \frac{e^x \sin^2 x -2e^x \sin x \cos x + 2 e^x}{5}+C$$
I can't reach the form that most integral calculators give, which has terms $\cos(2x)$ and $\sin(2x)$ by just using trig identities, so I wonder whether the result is correct. I would also be interested in a method that immediately gives the form $$-\frac{e^x[2 \sin(2x)+ \cos(2x)-5]}{10}+C$$
| Let $$\displaystyle I = \frac{1}{2}\int e^x 2\sin^2 xdx = \frac{1}{2}\int e^x-\frac{1}{2}\int e^x \cos 2x dx$$
Now Using $$\displaystyle \cos \phi +i\sin \phi = e^{i\phi}$$ and $$\cos \phi-i\sin \phi = e^{-i\phi}.$$
So $$\displaystyle \cos \phi = \frac{e^{i\phi}+e^{-i\phi}}{2}$$
So we get $$\displaystyle \cos 2x = \left(\frac{e^{i2x}+e^{-i2x}}{2}\right)$$
so we get $$\displaystyle J = \bf{Re}\left[\int e^{x}\cdot \left(\frac{e^{i2x}+e^{-i2x}}{2}\right)dx\right] = \frac{1}{2}\bf{Re}\int \left[e^{(1+2i)x}+e^{(1-2i)x}]\right]dx$$
So we get $$\displaystyle J = \frac{1}{2}\bf{Re}\left[\frac{e^{(1+2i)x}}{(1+2i)}+\frac{e^{1-2i}}{(1-2i)}\right]$$
$$\displaystyle = \frac{e^x}{10}\bf{Re}\left[\left(\cos 2x+i\sin 2x\right)\cdot (1-2i)+\left(\cos 2x-i\sin 2x\right)\cdot (1+2i)\right]$$
So $$\displaystyle J = \frac{e^x}{10}\left[2\cos 2x+4\sin 2x\right] = \frac{e^x}{5}\left[\cos 2x+\sin 2x\right]$$
So $$\displaystyle I = \frac{e^x}{2}-\frac{e^x}{10}\left[\cos 2x+2\sin 2x\right]+\mathcal{C} = -\frac{e^x}{10}\left[\cos 2x+2\sin 2x-5\right]+\mathcal{C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1398965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
How do I evaluate this improper integral $\int_{-1}^{1}\frac{dx}{(2-x)\sqrt{1-x^{2}}}$ Given integral is $$\int_{-1}^{1}\frac{dx}{(2-x)\sqrt{1-x^{2}}}.$$ I tried to split it up at $0$, but I donot know what to do ahead.
Thanks.
| Let $\displaystyle I =\int_{-1}^{1}\frac{1}{(2-x)\sqrt{1-x^2}}dx....(1)$
Noe Let $x = -t\;,$ Then $dx = -dt$ and changing Limit, we get
$\displaystyle I = \int_{1}^{-1}\frac{1}{(2+t)\sqrt{1-t^2}}\times -dt = \int_{-1}^{1}\frac{1}{(2+x)\sqrt{1-x^2}}......(2)$
Now Add $(1)$ and $(2)\;,$ we get $\displaystyle 2I = \int_{-1}^{1}\left[\frac{1}{2-x}+\frac{1}{2+x}\right]\cdot \frac{1}{\sqrt{1-x^2}}dx = 4\int_{-1}^{1}\frac{1}{(4-x^2)\sqrt{1-x^2}}dx$
So $\displaystyle 2I = 8\int_{0}^{1}\frac{1}{(4-x^2)\sqrt{1-x^2}}dx$
Now Put $x=\sin \phi$ and $dx = \cos \phi\phi$ and Changing Limit, we get
$\displaystyle 2I = 8\int_{0}^{\frac{\pi}{2}}\frac{1}{4-\sin^2 \phi}d\phi. = \int_{0}^{\frac{\pi}{2}}\frac{\sec^2 \phi}{4(1+\tan^2 \phi)-1}d\phi = 8\int_{0}^{\frac{\pi}{2}}\frac{\sec^2 \phi}{3+4\tan^2 \phi}d\phi$
Now Put $\tan \phi = z\;,$ Then $\sec^2 \phi d\phi = dx$ and changing limit, we get
$\displaystyle 2I = 8\int_{0}^{\infty}\frac{1}{3+4z^2}dz = 2\int_{0}^{\infty}\frac{1}{z^2+\left(\frac{\sqrt{3}}{2}\right)^2}dz = \frac{4}{\sqrt{3}}\cdot \frac{\pi}{2}\Rightarrow I = \frac{\pi}{\sqrt{3}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1399250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to get to this equality $\prod_{m=1}^{\infty} \frac{m+1}{m}\times\frac{m+x}{m+x+1}=x+1$?
How to get to this equality $$\prod_{m=1}^{\infty} \frac{m+1}{m}\times\frac{m+x}{m+x+1}=x+1?$$
I was studying the Euler Gamma function as it gave at the beginning of its history, and need to solve the following product operator, tried a few things, but I could not ...
$$\prod_{m=1}^{\infty} \frac{m+1}{m}\times\frac{m+x}{m+x+1}=\prod_{m=1}^{\infty} \frac{m^2+mx+m+x}{m^2+mx+m} =\prod_{m=1}^{\infty} \frac{m(m+x+1)+x}{m(m+x+1)}$$
WolframAlpha checked in and the result really is correct.
http://www.wolframalpha.com/input/?i=simplify+prod+m%3D1+to+inf+%5B%28m%2B1%29%2Fm+*+%28m%2Bx%29%2F%28m%2B1%2Bx%29%5D
| If you write out the first few terms of the product, you'll see that most of the factors cancel out:
$$
\require{cancel}
\begin{align}
\prod_{m=1}^{\infty} \frac{m+1}{m}\cdot\frac{m+x}{m+x+1} &= \left(\frac{\cancel2}{1}\cdot\frac{x+1}{\cancel{x+2}}\right)\left(\frac{\cancel3}{\cancel2}\cdot\frac{\cancel{x+2}}{\cancel{x+3}}\right)\left(\frac{\cancel4}{\cancel3}\cdot\frac{\cancel{x+3}}{\cancel{x+4}}\right)\cdots\\
&=x+1
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove area of a quadrilateral is $\frac14[4m^2n^2-(b^2+d^2-a^2-c^2)^2]^{\frac12}$ Someone asked me this question which I am really stuck at, any help is appreciated.
If $a,b,c,d$ are the sides of a quadrilateral and $m,n$ are diagonals of the quadrilateral, then prove that area of the quadrilateral is $$\frac14[4m^2n^2-(b^2+d^2-a^2-c^2)^2]^{\frac12}$$
I do not recall anything like this from my school days. I need some hints to get started on this, I do not remember high school geometry formulas, may be one of them could help here.
It works fine for squares and recatngles, so I assume it is a valid formula.
| For convex $\square ABCD$, we can proceed as follows:
Let the diagonals meet a $P$, which subdivides the diagonals into segments of length $m_1$, $m_2$, $n_2$, $n_2$; and let the diagonals make an angle of $\theta$ at $P$ as shown. Then we can write:
$$\begin{align}
|\square ABCD| &= |\triangle APB| + |\triangle BPC| + |\triangle CPD| + |\triangle DPA| \\[4pt]
&= \frac12 \sin\theta \;\left( m_1 n_1 + m_2 n_1 + m_2 n_2 + m_1 n_2 \right) = \frac12 (m_1 + m_2)(n_1+n_2) \sin\theta \\[4pt]
&= \frac12 m n \sin\theta \\[4pt]
\therefore 4\;|\square ABCD| &= 2 m n \sin\theta \tag{$\star$}
\end{align}$$
Applying the Law of Cosines in each triangle gives:
$$\begin{align}
a^2 &= m_1^2 + n_1^2 - 2 m_1 n_1 \cos\theta \qquad\qquad
b^2 = m_2^2 + n_1^2 + 2 m_2 n_1 \cos\theta \\
c^2 &= m_2^2 + n_2^2 - 2 m_2 n_2 \cos\theta \qquad\qquad
d^2 = m_1^2 + n_2^2 + 2 m_1 n_2 \cos\theta
\end{align}$$
so that
$$a^2 - b^2 + c^2 - d^2 = -2 m n \cos \theta$$
Finally, recalling $(\star)$,
$$\begin{align}
4 m^2 n^2 = 4 m^2 n^2\left( \sin^2\theta + \cos^2 \theta \right) = 16\;|\square ABCD|^2 + \left( a^2 - b^2 + c^2 - d^2 \right)^2
\end{align}$$
which gives the desired formula. $\square$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Power series of z/sin(z)? So I need to compute the coefficient of the $z^4$ in the power series of $\frac{z}{\sin z}$.
I tried differentiating the function and obtaining coefficients like in Taylor's expansions but had a really hard time. In general I'm finding it extremely hard to obtain the power series of various complex functions, especially with something like $\sin z$ in the denominator. Any tips / tricks?
| Here is a simple method that works well to find the first few terms in the power-series.
We first expand the denominator in a power-series around $z=0$: $$\frac{z}{\sin(z)} = \frac{1}{1 - \frac{z^2}{3!} + \frac{z^4}{5!} - \ldots}$$
This has the form of a geometrical series $\frac{1}{1-x} = 1+x + x^2 + \ldots$ for $x = \frac{z^2}{3!} - \frac{z^4}{5!} + \mathcal{O}(z^6)$. The expansion above holds as long as $|x|<1$ which is the case for all $z$ sufficiently close to $z=0$ (which is all we need).
Expanding the different terms $x^k$ is a power-series in $z$ we find
$$\begin{align}x^2 &= \left(\frac{z^2}{3!}\right)^2 - 2\frac{z^2}{3!}\frac{z^4}{5!} + \mathcal{O}(z^8)\\x^3 &= \left(\frac{z^2}{3!}\right)^3 + \mathcal{O}(z^8)\end{align}$$
The term $x^n$ will only give rise to terms of order $z^{2n}$ and larger so to get the series to order $z^6$ we can stop at $x^3$. Adding up all the terms above gives
$$\frac{z}{\sin(z)} = 1 + \frac{z^2}{6} + \frac{7z^4}{360} + \frac{31z^6}{15120} + \mathcal{O}(z^8)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
limit involving $e$, ending up without $e$. Compute the limit
$$ \lim_{n \rightarrow \infty} \sqrt n \cdot \left[\left(1+\dfrac 1 {n+1}\right)^{n+1}-\left(1+\dfrac 1 {n}\right)^{n}\right]$$
we have a bit complicated solution using Mean value theorem. Looking for others
| Taylor expansion always works:
\begin{align}
& \left(1 + \frac{1}{n + 1}\right)^{n + 1} - \left(1 + \frac{1}{n}\right)^{1/n}\\
= & \exp\left[(n + 1)\log\left(1 + \frac{1}{n + 1}\right)\right] - \exp\left[n\log\left(1 + \frac{1}{n}\right)\right] \\
= & \exp\left[(n + 1)\left(\frac{1}{n + 1} - \frac{1}{2(n + 1)^2} + o\left(\frac{1}{(n + 1)^2}\right)\right)\right] - \exp\left[n \left(\frac{1}{n} - \frac{1}{2n^2} + o\left(\frac{1}{n^2}\right)\right)\right] \\
= & \exp\left[1 - \frac{1}{2(n + 1)} + o\left(\frac{1}{n + 1}\right)\right]
-\exp\left[1 - \frac{1}{2n} + o\left(\frac{1}{n}\right)\right] \\
= & \left\{e + e\left[1 - \frac{1}{2(n + 1)} + o\left(\frac{1}{n + 1}\right) - 1\right] + O\left(\frac{1}{(n + 1)^2}\right)\right\} \\
- & \left\{e + e\left[1 - \frac{1}{2n} + o\left(\frac{1}{n}\right) - 1\right] + O\left(\frac{1}{n^2}\right)\right\} \\
= & o\left(\frac{1}{n}\right)
\end{align}
Therefore the original expression is $o\left(\frac{1}{\sqrt{n}}\right)$, hence the limit is $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Determine the equation of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ such that it has the least area but contains the circle $(x-1)^2+y^2=1$ Determine the equation of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ such that it has the least area but contains the circle $(x-1)^2+y^2=1$
Since the area of ellipse is $A=\pi ab\Rightarrow A^2=\pi^2a^2b^2$ and the circle and the ellipse touch each other internally. This much I can visualise, but how to find out $a$ and $b$ from this?
| Firstly, define $f(x):=b\sqrt{1-\frac{x^2}{a^2}}$ and $g(x):=\sqrt{1-(x-1)^2}$ such that $f$ and $g$ represent the ellipse and the circle respectively in the upper half of the coordinate system.
In order to guarantee that the ellipse contains the circle, we need to have:
$$
f(x)≥g(x)\iff b\sqrt{1-\frac{x^2}{a^2}}≥\sqrt{1-(x-1)^2} \iff b^2-\frac{b^2}{a^2}x^2≥2x-x^2\iff \\
\left(1-\frac{b^2}{a^2}\right)x^2-2x+b^2≥0
$$
In the last inequality, we have a quadratic function, which needs to be nonnegative everywhere, thus the discriminant $D$ has to be nonpositive:
$$
D=4-4\left(1-\frac{b^2}{a^2}\right)b^2≤0\iff 1≤b^2-\frac{b^4}{a^2} \iff \frac{b^4}{a^2}≤b^2-1
$$
We can see that $0<b^2-1\iff 1<b$, therefore we can divide by $b^2-1$ without any changes:
$$
a^2≥\frac{b^4}{b^2-1}\iff a≥\frac{b^2}{\sqrt{b^2-1}}
$$
This last inequality is sufficient and necessary for the given condition.
Therefore, we have:
$$
A=\pi ab≥\frac{\pi b^3}{\sqrt{b^2-1}}
$$
Thus, the minimum of $A$ has to be greater than or equal to the minimum of $\frac{\pi b^3}{\sqrt{b^2-1}}$. This minimum can be found as follows:
Define $h(b):=\frac{\pi b^3}{\sqrt{b^2-1}}$.
$$
h'(b)=\frac{\pi b^2\left(2b^2-3\right)}{\left(b^2-1\right)^{\frac{3}{2}}} \implies \left(h'(b)=0\iff b=\sqrt{\frac{3}{2}}\right)
$$
$b=0$ is impossible because $b>1$. Therefore, the minimum has to be at $b=\sqrt{\frac{3}{2}}$ so we have $\min A≥h\left(\sqrt{\frac{3}{2}}\right)=\frac{\pi 3\sqrt{3}}{2}$. This minimum can indeed be achieved for $a=\frac{3}{\sqrt 2}$ and $b=\sqrt{\frac{3}{2}}$, for which we also have:
$$
a=\frac{3}{\sqrt 2}≥\frac{\sqrt{\frac{3}{2}}^2}{\sqrt{\sqrt{\frac{3}{2}}^2-1}}=\frac{b^2}{\sqrt{b^2-1}}
$$
Thus, the equation of the ellipse has to be:
$$
\frac{2x^2}{9}+\frac{2y^2}{3}=1
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1401256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A circle of radius $r$ is dropped into the parabola $y=x^{2}$. Find the largest $r$ so the circle will touch the vertex. If $r$ is too large, the circle will not fall to the bottom, if $r$ is sufficiently small, the circle will touch the parabola at its vertex $(0, 0)$. Find the largest value of $r$ s.t. the circle will touch the vertex of the parabola.
| Let the center of the circle $(0, r)$ & radius $r$ hence the equation of the circle $$(x-0)^2+(y-r)^2=r^2$$ $$x^2+(y-r)^2=r^2$$ Now, solving the equations of parabola $y=x^2$ & equation of the circle we get $$x^2+(x^2-r)^2=r^2$$ $$x^2+x^4+r^2-2rx^2=r^2$$ $$x^4-(2r-1)x^2=0$$ $$x^2(x^2+1-2r)=0$$
$$x^2=0\ \ \vee\ \ x^2+1-2r=0$$ $$x=0\ \ \vee\ \ x^2=2r-1$$
$$x=0\ \ \vee\ \ x^=\pm\sqrt{2r-1}$$ Corresponding values of $y$ are calculated as follows
$$(x=0, y=0), (x=\sqrt{2r-1}, y=2r-1), (x=-\sqrt{2r-1}, y=2r-1)$$
Thus, we find the three points of intersection of the circle with the parabola $(0, 0)$, $(\sqrt{2r-1}, 2r-1)$ & $(-\sqrt{2r-1}, 2r-1)$
But the circle is touching the parabola at the vertex $(0, 0)$ only hence, it is possible when other two points $(\sqrt{2r-1}, 2r-1)$ & $(-\sqrt{2r-1}, 2r-1)$ are coinciding with the vertex $(0, 0)$ hence, we get $$\pm \sqrt{2r-1}=0\iff r=\frac{1}{2}$$
& $$2r-1=0\iff r=\frac{1}{2}$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{Maximum radius of circle, }r_{\text{max}}=\color{blue}{\frac{1}{2}}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1402486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to find the MacLaurin series of $\frac{1}{1+e^x}$ Mathcad software gives me the answer as: $$ \frac{1}{1+e^x} = \frac{1}{2} -\frac{x}{4} +\frac{x^3}{48} -\frac{x^5}{480} +\cdots$$
I have no idea how it found that and i don't understand. What i did is expand $(1+e^x)^{-1} $ as in the binomial MacLaurin expansion. I want until the $x^4$ term. So what i found is the following:
$$ (1+e^x)^{-1} = 1 - e^x + e^{2x} + e^{3x} + e^{4x}+\cdots\tag1 $$
Then by expanding each $e^{nx}$ term individually:
$$ e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24}+\cdots\tag2 $$
$$ e^{2x} = 1 + 2x + 2x^2 + \frac{3x^3}{2} + \frac{2x^4}{3}+\cdots\tag3 $$
$$ e^{3x} = 1 + 3x + \frac{9x^2}{2} + \frac{9x^3}{2} + \frac{27x^4}{8}+\cdots\tag4 $$
$$ e^{4x} = 1 + 4x + 8x^2 + \frac{32x^3}{3} + \frac{32x^4}{3}+\cdots\tag5 $$
So substituting $(2),(3),(4),(5)$ into $(1)$ i get:
$$ (1+e^x)^{-1} = 1 +2x+5x^2+ \frac{22x^3}{3} + \frac{95x^4}{12} +\cdots$$
Which isn't the correct result. Am i not allowed to expand this series binomially? I've seen on this site that this is an asymptotic expansion. However, i don't know about these and i haven't been able to find much information on this matter to solve this. If someone could help me understand how to solve it and why my approach isn't correct, i would be VERY grateful. Thanks in advance.
| Let $\dfrac1{1+e^x}=a_0+a_1x+a_2x^2+\cdots$
$$\implies(1+e^x)\left(a_0+a_1x+a_2x^2+\cdots\right)=1$$
As $e^x=\sum_{r=0}^\infty\dfrac{x^r}{r!}$
$$\implies\left(2+\dfrac x1+\dfrac{x^2}2+\dfrac{x^3}{3!}+\cdots\right)\left(a_0+a_1x+a_2x^2+\cdots\right)=1$$
Comparing the coefficients of different powers of $x,$
$\implies1=2a_0\iff a_0=?$
$0=2a_1+a_0\iff a_1=?$
$0=2a_2+a_1+\dfrac{a_0}2$
and so on
| {
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"url": "https://math.stackexchange.com/questions/1404886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Integrating $\frac{1}{(x^4 -1)^2}$ How to solve the the following integral? $$\int{\frac{1}{(x^4 -1)^2}}\, dx$$
| In general, the following method will always work. Solve for the roots of the denominator in the complex plane:
$$
\begin{split}
z^4 &= 1 \Longrightarrow\\
z&=z_{n}\equiv\exp\left(\frac{i\pi n}{2}\right)
\end{split}
$$
In this case things are simple, you have $z_0=1$, $z_1=i$, $z_2=-1$, $z_3 = -i$.
Then the trick is to consider the function $\frac{1}{x^4-1}$ instead of its square first. A partial fraction expansion of a rational function that tends to zero at infinity can be obtained by writing down the expansion around all the singularities, keeping only the singular terms and adding them up. This works because the difference between the sum of the terms you obtain and the rational function then only has removable singularities left, it is therefore some polynomial, but it still tends to zero at infinity (because both the rational function and the singular terms from the expansions tend to zero at infinity). But the only polynomial that tends to zero at infinity is identical to zero.
Now, each expansion of $\frac{1}{\left(z^4-1\right)^2}$around each singularity $z_n$, is more easily computed by expanding $\frac{1}{z^4-1}$ instead and the squaring both sides. The expansion around $z_n$ only has one singular term, but because we want to square both sides we need to also compute the constant term. If we put $z = z_n + t$ and expand in powers of $t$ we get:
$$\frac{1}{z^4-1} = \frac{1}{4 z_n^3 t + 6 z_n^2 t^2 +\mathcal{O}(t^3)}=\frac{1}{4 z_n^3 t}-\frac{3}{8} +\mathcal{O}(t)$$
Squaring both sides and keeping only the singular terms yields:
$$\frac{1}{\left(z^4-1\right)^2} = \frac{1}{64 z_n^6 t^2}-\frac{3}{16 z_n^3 t} +\mathcal{O}(1)$$
So, the partial fraction expansion is:
$$\frac{1}{\left(x^4-1\right)^2} =\sum_{n=0}^3 \frac{1}{64 z_n^6}\frac{1}{\left(x-z_n\right)^2} - \sum_{n=0}^3 \frac{3}{16 z_n^3}\frac{1}{x-z_n} $$
We thus have:
$$\int \frac{dx}{\left(x^4-1\right)^2} = -\sum_{n=0}^3 \frac{1}{64 z_n^6}\frac{1}{x-z_n} - \sum_{n=0}^3 \frac{3}{16 z_n^3}\log\left(x-z_n\right) $$
Then the fractions with the imaginary terms are easily simplified so that they become manifestly real. You can deal with the logarithms of complex arguments as follows. The sum of the logarithmic terms with the complex arguments in this case is:
$$\frac{3 i}{16}\log\left(\frac{x+i}{x-i}\right)$$
If we write:
$$x + i = r\exp(i\alpha)$$
then since $x-i$ is the complex conjugate, we have:
$$x - i = r\exp(-i\alpha)$$
Therefore:
$$\frac{x+i}{x-i} = \exp(2 i\alpha)$$
Since $\alpha = \arctan\left(\frac{1}{x}\right)$, the sum of the two terms is:
$$-\frac{3}{8}\arctan\left(\frac{1}{x}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1406034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why is $\binom{2n}{n} \asymp \Theta \big(\frac{2^{2n}}{\sqrt{n}}\big)$? I saw this statement : $$\binom{2n}{n} \asymp \Theta \bigg(\frac{2^{2n}}{\sqrt{n}}\bigg) \asymp \Theta\bigg(\frac{4^n}{\sqrt{n}}\bigg)$$
How did we go from the first statement to the second? I tried Stirling's aproximation, but that didn't get me anywhere.
| Stirling's approximation gives
$$\binom{2n}{n} = \frac{(2n)!}{n!^2} \sim \frac{\sqrt{2\pi \cdot 2n} \left( \frac{2n}{e} \right)^{2n}}{\left[\sqrt{2\pi n} \left( \frac{n}{e} \right)^n \right]^2} = \frac{2\sqrt{\pi n} \cdot 2^{2n} \cdot \left( \frac{n}{e} \right)^{2n}}{2 \pi n \cdot \left( \frac{n}{e} \right)^{2n}} = \frac{1}{\sqrt{\pi n}} 2^{2n}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Local extremes of $f(x) = (x-2)^{\frac{1}{5}}(x-7)^{\frac{1}{9}}$ The task is to find local extremes of $f: \mathbb R \to \mathbb R$, $f(x) = (x-2)^{\frac{1}{5}}(x-7)^{\frac{1}{9}}$
There is theorem that if $x_{0}$ is local extreme of $f(x)$ then $f'(x_0) = 0$
So I start from calculating derivative of $f(x)$
$$f'(x) = ((x-2)^{\frac{1}{5}}(x-7)^{\frac{1}{9}})' = \frac19(x-7)^{-\frac{8}{9}}(x-2)^\frac15 + \frac15(x-2)^{-\frac45}(x-7)^\frac19 = \\ = (x-2)^\frac15(x-7)^\frac19\cdot[\frac19(x-7)^{-1}+\frac15(x-2)^{-1}] = \\ = (x-2)^\frac15(x-7)^\frac19\cdot[\frac{14x-73}{(9x-63)(5x-10)}]= (*)$$
So $(*) = 0 <=> x = 2 \lor x = 7 \lor x = \frac{73}{14}$
How can I check if any of them is local extreme?
(according to wolframalpha $\frac{73}{14}$ is local minimum and there is no local maximum)
| for $$f'(x)$$ i have got $$f'(x)=1/45\,{\frac {14\,x-73}{ \left( x-2 \right) ^{4/5}} \left( x-7
\right) ^{-{\frac {8}{9}}}}
$$ and for the second derivative we have
$$f''(x)=-{\frac {434\,{x}^{2}-4526\,x+15416}{2025\, \left( x-2 \right) ^{9/5}}
\left( x-7 \right) ^{-{\frac {17}{9}}}}
$$ here you must plugg in $$x_0=\frac{73}{14}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $ a + b + c \mid a^2 + b^2 + c^2$ then $ a + b + c \mid a^n + b^n + c^n$ for infinitely many $n$
Let $ a,b,c$ positive integer such that $ a + b + c \mid a^2 + b^2 + c^2$.
Show that $ a + b + c \mid a^n + b^n + c^n$ for infinitely many positive integer $ n$.
(problem composed by Laurentiu Panaitopol)
So far no idea.
| For any positive integer x this is true: $x \leqslant x^2$ (From $1 \leqslant x$ for any positive ineger x ). So $a + b + c \leqslant a^2 + b^2 + c^2$. But for 2 positive integers $x$, $y$ $x$ is divisible by $y$ only if $x \geqslant y$. So $a + b + c \geqslant a^2 + b^2 + c^2$ if $a + b + c \mid a^2 + b^2 + c^2$.
From these 2 inequalities:
$a + b + c = a^2 + b^2 + c^2$
So $a + b - a^2 - b^2 = c^2 - c$
Evaluation for the right part $c^2 - c \geqslant 0$ (because $x \leqslant x^2$ for any positive integer x).
Evaluation for the left part $a + b - a^2 - b^2 \leqslant 0$ (adding 2 inequalities $a - a^2 \leqslant 0$ and $b - b^2 \leqslant 0$)
So the left part is $\leqslant 0$ and the right part $\geqslant 0$. But they are equal so $a + b - a^2 - b^2 = c^2 - c = 0$ and $c^2 = c$. c is positive so we can divide both part of the last equation by c and get $c = 1$.
Similarly $b = 1$ and $a = 1$. So $a + b + c = 3$ and $a^n + b^n + c^n = 3$ for any positive $n$. 3 is divisible by 3 so $a + b + c \mid a^n + b^n + c^n$ for infinitely many positive integer n.
| {
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"url": "https://math.stackexchange.com/questions/1408323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Evaluating the ratio $ {{a_{n+1}}\over{a_n}}$ in calculating the radius of convergence for a power series In calculating the radius of convergence for the power series
$$ \sum_{n=1}^\infty {{(2n)!}\over(n!)^2}\ x^n $$
By the ratio test, we let
$$ a_n = \lvert {{(2n)!}\over(n!)^2}\ x^n \rvert \quad\quad a_{n+1} = \lvert{{(2n+2)!}\over(n+1)!^2}\ x^{n+1} \rvert$$
$$ {{a_{n+1}}\over{a_n}} = \lvert {(2n+2)!\over(2n)!}{(n!)^2\over(n+1)!^2}\ x \rvert \longrightarrow \boxed{4\lvert x\rvert < 1} \longrightarrow \boxed{\lvert x\rvert<\frac 14}$$
My method for deriving 4 as the limit of convergence was obtained crudely simply by inputting large values for n and calculating manually; I am unsure as to how to derive 4 formally by simplifying the ratio.
Similarly
$$ \sum_{n=1}^\infty {(n!)^3\over(3n)!}\ x^n$$
By the ratio test
$$ a_n=\lvert {(n!)^3\over(3n)!}\ x^n \rvert \quad\quad a_{n+1} = \lvert {(n+1)!^3\over(3n+3)!}\ x^{n+1} \rvert$$
$$ {{a_{n+1}}\over{a_n}} = \lvert {(n+1)!^3\over(n!)^3}{(3n)!\over(3n+3)!}\ x \rvert \longrightarrow \boxed{{1\over27}\lvert x\rvert<1} \longrightarrow \boxed{\lvert x\rvert<27}$$
Again, a formal method for deriving ${1\over27}$ would be much appreciated.
| Hint
\begin{equation}
\frac{(2n + 2)!}{(2n)!} \frac{(n!)^2}{(n + 1)!} = \frac{(2n + 2)(2n + 1)}{(n + 1)(n + 1)} = \frac{4n^2 + 6n + 2}{n^2 + 2n + 1} = \frac{4 + \frac{6}{n} + \frac{2}{n^2}}{1 + \frac{2}{n} + \frac{1}{n^2}}
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1408817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find min of $M=\frac{1}{2+\cos2A}+\frac{1}{2+\cos2B}+\frac{1}{2-\cos2C}$
Find min of $$M=\frac{1}{2+\cos2A}+\frac{1}{2+\cos2B}+\frac{1}{2-\cos2C}$$, where $A, B, C$ are three angle of triangle $ABC$
Using Cauchy-Schwarz, we obtain: \begin{align*}
M &= \frac{1}{2+\cos2A}+\frac{1}{2+\cos2B}+\frac{1}{2-\cos2C}\\
&\ge \frac{9}{6+2\cos(A+B)\cos(A-B)+1-2\cos^2C}\\
& \ge \frac{9}{7-2\cos C-2\cos^2C}\ge \frac{9}{\frac{15}{2}}=\frac{6}{5}
\end{align*}
The equality holds if and only if $\left\{\begin{matrix}A=B\\\cos C=-\frac{1}{2} \end{matrix}\right.\iff \left\{\begin{matrix}A=B=30^{\circ}\\ C=120^{\circ}\end{matrix}\right.$
But the change $\cos(A+B)\cos(A-B)\le-\cos C $ may be not correct. :(
| The change might not be correct, but before that the numerator is
$$
f(A,B)=6+\cos 2A+\cos 2B-\cos 2C=6+\cos 2A+\cos 2B-\cos 2(A+B)
$$
$$
\frac{\partial f}{\partial A}=2 \sin (2 A+2 B)-2 \sin (2 A)
$$
$$
\frac{\partial f}{\partial B}=2 \sin (2 A+2 B)-2 \sin (2 B)
$$
We can only have extrema at the border of the set, namely $A=0$, $B=0$, $A+B=\pi$ or where both partials are $0$, meaning $\sin 2A=\sin 2B$, equivalently $A=B$ or $A+B=k\frac{\pi}{2}$.
From checking these we get that the minimum is at $\frac{\pi}{6},\frac{\pi}{6}$ and is in fact $\frac{6}{5}$
| {
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"url": "https://math.stackexchange.com/questions/1411021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Prove that $\cos \arctan 1/2 = 2/\sqrt{5}$ How can we prove the following? $$\cos \left( \arctan \left( \frac{1}{2}\right) \right) =\frac{2}{\sqrt{5}}$$
| Let $\displaystyle \tan^{-1}\left(\frac{1}{2}\right) = \theta\;,$ Then $\displaystyle \frac{1}{2} = \tan \theta$
So $\displaystyle \cos\left[\tan^{-1}\left(\frac{1}{2}\right)\right] = \cos \theta = \frac{1}{\sec \theta} = \frac{1}{\sqrt{1+\tan^2 \theta}}= \frac{2}{\sqrt{5}}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If complex no. ($z$) satisfying $\frac{1}{2}\leq |z|\leq 4\;,$ Then Max. and Min. of $\left|z+\frac{1}{z}\right|$
Let $z$ be a complex no. satisfying $\displaystyle \frac{1}{2}\leq |z|\leq 4\;,$ then the Sum of greatest and
least value of $\displaystyle \left|z+\frac{1}{z}\right|$ is
$\bf{My\; Try::}$ Let $z=re^{i\theta} = r\left(\cos \theta+i\sin \theta\right)\;,$ Where $-\pi<\theta \leq \pi.$ and $\displaystyle \frac{1}{2}\leq r \leq 4$
Then $$\displaystyle \left|z+\frac{1}{z}\right| = \left|re^{i\alpha}+\frac{e^{-i\alpha}}{r}\right| = \left|\frac{r^2\cos \alpha+r^2\cdot i\sin \alpha+\cos \alpha-i\sin \alpha}{r}\right|$$
so we get $$\displaystyle \left|\frac{(r^2+1)\cos \alpha+i(r^2-1)\sin \alpha}{r}\right| = \frac{\sqrt{(r^2+1)^2\cos^2 \alpha+(r^2-1)^2\sin^2 \alpha}}{r}$$
Now Let $$\displaystyle f(r,\alpha) = \frac{\sqrt{r^4+1+2r^2\cos 2\alpha}}{r}$$
Now i did not understand how can i solve after that, Help me , Thanks
| $g(r,\alpha)= r^2+\dfrac{1}{r^2}+2cos2\alpha$
for min:$r^2+\dfrac{1}{r^2} \ge 2,cos2\alpha \ge -1$
for max: $r^2+\dfrac{1}{r^2}$ will get max at two ends, $cos2\alpha \le 1$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $s(n-1)s(n)s(n+1)$ is always an even number Let $n$ be a natural number, and let $s(n)$ denote the sum of all positive divisors of $n$. Show that for any $n>1$ the product $s(n-1)s(n)s(n+1)$ is always an even number.
I calculated the sum of the divisors(in the form a geometric Progression) and then tried proving. But, I was not able to comment on the divisors of $n-1$ and $n+1$. Please give some hints. Thanks.
| Express the numbers $n-1,n,n+1$ in their prime factorisations.
Let $x$ be any one of $n-1,n,n+1$ and express $$x=2^a\cdot3^b\cdot5^c\cdot\ldots \tag{1}$$
Then $$s(x)=(1+2+\ldots 2^a)\times(1+3+\ldots 3^b)\times(1+5+\ldots 5^c)\times\ldots \tag{2}$$
* Assume now that all of $s(n-1),s(n),s(n+1)$ are odd *
The first term is always odd, while the remaining terms are only odd if the exponents $b,c,\ldots$ are even, so $x$ must be a multiple of a power of 2 and an odd perfect square. As Erick Wong has commented this means that $x$ must be an odd perfect square if it is odd.
Now if $n-1,n+1$ are odd, they must both be odd perfect squares. But this is impossible as no two positive perfect squares differ by 2. Therefore, $n$ is odd, and hence an odd perfect square.
Express $n=\lambda^2$ where $\lambda=2k+1,\quad k\in\mathbb{Z^+}$. Then we have:
$$\begin{align}
n-1 = \lambda^2-1 = (2k+1)^2 - 1 = 4k(k+1) \tag{3} \\
n = \lambda^2 = (2k+1)^2 = 4k^2+4k+1 \tag{4}\\
n+1 = \lambda^2+1 = (2k+1)^2 = 2(2k^2+2k+1) \tag{5}
\end{align}$$
In (5), $2k^2+2k+1$ is odd, so must be an odd perfect square.
So $n+1$ is twice an odd perfect square.
In (3), one of $k,k+1$ must be odd. Also, $\gcd(k,k+1)=1$. So any odd prime factors in $k$ are not in $k+1$, and vice-versa.
Let $y$ be the even number in $\{k,k+1\}$ and $z$ be the other. Note that $y$ must be of the form $2^q\cdot r^2$ where $r$ is an odd number, and $z$ must be an odd perfect square, of the form $s^2$ where $s$ is odd. If $q$ is even, then $y$ is a perfect square, which is impossible because $y$ and $z$ would both then be positive perfect squares differing by 1. Therefore, we have shown that $q$ is odd, and so
$$4k(k+1)=4yz=4\cdot2^q\cdot r^2\cdot s^2=2\cdot2^{q+1}\cdot r^2\cdot s^2=2(2^{(q+1)/2}rs)^2$$
Since $q$ is odd, $(q+1)/2$ is an integer, and hence we can conclude that $n-1=4k(k+1)$ is twice a square number.
To recap, we now have that $n-1$ is twice a square number, and $n+1$ is twice an odd perfect square, with $n$ being odd. Therefore, $\frac{n-1}{2}$ and $\frac{n+1}{2}$ are both positive perfect squares. However, this is impossible because they differ by one. We have arrived at a fatal contradiction of the original assumption, so $s(n-1),s(n),s(n+1)$ cannot all be odd. $\blacksquare$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve this system of nonlinear equations? How to solve these equations for $a$, $b$, $c$ and $x$?
I have the following:
\begin{align}
1 &= 2a+b+c\\
a &= (a+b)x + 0.25(a+c)\\
a&=(a+c)(1-x)\\
b&=a(1-x)+c(x-0.25)\\
c&=b(1-x)+a(x-0.25)
\end{align}
I tried, but ended circular at the point I started. Can someone help me?
| Using Grobner basis in maxima gives:
grobner_basis([2*a+b+c-1,(a+b)x+(a+c)/4-a,(a+c)(1-x)-a,a*(1-x)+c*(x-1/4)-b,b*(1-x)+a*(x-1/4)-c]);
(- 44) x + (- 144) a^2 + 213 a + (- 36),
(- 11) c + (- 48) a^2 + 60 a + (- 12),
(- 11) b + 48 a^2 + (- 82) a + 23,
(- 144) a^3 + 381 a^2 + (- 312) a + 64
The last equation is a cubic in "a" alone.
The second last solves for b given a.
The third last solves for c in a.
The top equation solves for x in a.
Grobner Basis and Buchberger's algorithm explain how this works.
regards arthur
lower case x
grobner_basis([2*A+B+C-1,(A+B)x+(A+C)/4-A,(A+C)(1-x)-A,A*(1-x)+C*(x-1/4)-B,B*(1-x)+A*(x-1/4)-C]);
(- 3) C + 4 x^2 ,
(- 3) B + 4 x^2 + (- 8) x + 3,
(- 3) A + (- 4) x^2 + 4 x,
(- 8) x^3 + 13 x^2 + (- 12) x + 3
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1415612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"Trig Substitutions", I tried half- angle and trig indentity in this one, but doesn't work I´m really lost in this one.
$\int \sin^3 (2x) \cos^2 (2x) dx$
I know that the answer is:
$\frac{1}{10}cos^5(2x)-\frac{1}{6}cos^3(2x) + c$
Please help
| HINT:
Write $\sin^3(2x)\cos^2(2x)=(1-\cos^2(2x))\cos^2(2x)\sin(2x)$
SPOLIER ALERT: SCROLL OVER THE SHADED AREA TO REVEAL SOLUTION
We have $$\begin{align}\sin^3(2x)\cos^2(2x)&=\left(1-\cos^2(2x)\right)\cos^2(2x)\sin(2x)\\\\&=\left(\cos^2(2x)-\cos^4(2x)\right)\sin(2x)\end{align}$$Then, $$\int \sin^3(2x)\cos^2(2x)\,dx=\int \left(\cos^2(2x)-\cos^4(2x)\right)\sin(2x)\,dx$$Now, enforce the substitution $u=\cos (2x)$ so that $du=-2\sin(2x)\,dx$. Then, the integral becomes $$\begin{align}\int \left(u^2-u^4\right)\,\left(-\frac12 du\right)&=\frac12 \left(\frac15 u^5-\frac13 u^3\right)+C\\\\&=\frac{1}{10}\cos^5 (2x)-\frac16 \cos^3(2x)+C\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1416646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find tangent to trigonometric function I want to find the tangent to the curve:
$x\sin{y} + y\sin{x} = \frac{\pi}{4}(1+\sqrt{2})$
through the point $(\frac{\pi}{2}, \frac{\pi}{4})$
Now I know I can fill certain information into this formula
$y-y_0 = a(x-x_0)$ to get a tangent, but I dont really know how to find
$a$ in this situation. How can I find the derivate here when I cant isolate $y$ on one side (that im aware of) ?
I tried to make an expression on the form $\sin(x+\theta) = \text{something}$ but without success
Thanks in advance for any help or hints
| We will use implicit differentiation in this problem to extract $\frac{dy}{dx}$:
$$
x\sin{y}+y\sin{x}=\frac{\pi}{4}(1+\sqrt{2})\\
\cos{y}+x\frac{dy}{dx}\cos{y}+\frac{dy}{dx}\sin{x}+y\cos{x}=0\\
\frac{dy}{dx}(x\cos{y}+\sin{x})=-(\sin{y}+y\cos{x})\\
\frac{dy}{dx}=-\frac{\sin{y}+y\cos{x}}{x\cos{y}+\sin{x}}
$$
And for $x=\frac{\pi}{2}\quad y=\frac{\pi}{4}$ you get:
$$
a={\frac{dy}{dx}}_{(x,y)=(\frac{\pi}{2},\frac{\pi}{4})}=-\frac{2\sqrt{2}}{4+\pi\sqrt{2}}
$$
So the equation of the tangent becomes:
$$
y=-\frac{2\sqrt{2}}{4+\pi\sqrt{2}}x+C
$$
Since this is a tangent line it will pass from $(\frac{\pi}{2},\frac{\pi}{4})$:
$$
\frac{\pi}{4}=-\frac{2\sqrt{2}}{4+\pi\sqrt{2}}\frac{\pi}{2}+C\\
C=\frac{\pi}{4}+\frac{\pi\sqrt{2}}{4+\pi\sqrt{2}}
$$
And we finally get:
$$
y=-\frac{2\sqrt{2}}{4+\pi\sqrt{2}}x+\frac{\pi}{4}+\frac{\sqrt{2}\pi}{4+\sqrt{2} \pi}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to evaluate $45^\frac {1-a-b}{2-2a}$ where $90^a=2$ and $90^b=5$ without using logarithm? Let $90^a=2$ and $90^b=5$, Evaluate
$45^\frac {1-a-b}{2-2a}$
I know that the answer is 3 when I used logarithm, but I need to show to a student how to evaluate this without involving logarithm. Also, no calculators.
| Let me try.
$$10 = 90^{a+b} \Rightarrow 3^2 = 90^{1-a-b} \Rightarrow 3 = 90^{\frac{1-a-b}{2}}.$$
Then, $$45 = 90^{(1-a-b)+b} = 90^{1-a}.$$
So, $$45^{\frac{1}{1-a}} = 90 \Rightarrow 45^{\frac{1-a-b}{2(1-a)}} = 90^{\frac{1-a-b}{2}} = 3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1421740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $z$ when $z^4=-i$? Consider $z^4=-i$, find $z$.
I'd recall the fact that $z^n=r^n(\cos(n\theta)+(i\sin(n\theta))$
$\implies z^4=|z^4|(\cos(4\theta)+(i\sin(4\theta))$
$|z^4|=\sqrt{(-1)^2}=1$
$\implies z^4=(\cos(4\theta)+(i\sin(4\theta))$
$\cos(4\theta)=Re(z^4)=0 \iff \arccos(0)=4\theta =\frac{\pi}{2} \iff \theta=\frac{\pi}{8}$
Since $z^n=r^n\cdot e^{in\theta}$, $z^4$ can now be rewritten as $z^4=e^{i\cdot4\cdot\frac{\pi}{8}} \iff z=e^{i\frac{\pi}{8}}$ However, my answer file says this is wrong. Can anyone give me a hint on how to find $z$?
| the last step, you need $$-i = \cos(3\pi/2) + \sin (3\pi/2) = \cos (4 \theta) + i \sin(4 \theta) $$ which will give you $$4\theta = 3\pi/2, 3\pi/2 + 2\pi, 3\pi/2 + 4\pi, 3\pi/2 + 6\pi$$ now you can solve for $\theta.$
| {
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"url": "https://math.stackexchange.com/questions/1421839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve using: complete quadratic equation? $$x^2 - x - 20 > 0$$
$$x^2 - x > 20$$
Step 3:
So to complete it, let's use: $1/2$ so it ends up with $x$ as the middle term.
$$x^2 - x + 1/2 > 20 + 1/2$$
Step 4:
But then:
It would be transform to:
$$(x - 1/2)^2 > 20 + 1/2$$
Step 5:
But resolving it, the initial inequation of "Step3" is different:
$$x^2 - 2(x)(1/2) + (-1/2)^2 > 20 + 1/2$$
Giving us:
$$x^2 - x + 1/4 > 20 + 1/2$$
Which is different than:
$$x^2 - x + 1/2 > 20 + 1/2$$
Any hint?
| Given $$\displaystyle x^2-x-20=0\Rightarrow \underbrace{x^2-x+\left(\frac{1}{2}\right)^2}-\underbrace{20-\left(\frac{1}{2}\right)^2} =0$$
So we get $$\displaystyle \left(x-\frac{1}{2}\right)^2-\left(\frac{9}{2}\right)^2>0$$
So we get $$\displaystyle \left[x-\frac{1}{2}-\frac{9}{2}\right]\cdot \left[x-\frac{1}{2}+\frac{9}{2}\right]>0$$
So we get $$(x-5)(x+4)>0\Rightarrow x<-4\cup x>5$$
So our Solution is $$x\in \left(-\infty,-4\right)\cup (5,\infty)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1422226",
"timestamp": "2023-03-29T00:00:00",
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Conjecture $\Re\,\operatorname{Li}_2\left(\frac12+\frac i6\right)=\frac{7\pi^2}{48}-\frac13\arctan^22-\frac16\arctan^23-\frac18\ln^2(\tfrac{18}5)$ I numerically discovered the following conjecture:
$$\Re\,\operatorname{Li}_2\left(\frac12+\frac i6\right)\stackrel{\color{gray}?}=\frac{7\pi^2}{48}-\frac{\arctan^22}3-\frac{\arctan^23}6-\frac18\ln^2\!\left(\frac{18}5\right).$$
It holds numerically with a precision of more than $30000$ decimal digits.
Could you suggest any ideas how to prove it?
Can we find a closed form for $\Im\,\operatorname{Li}_2\left(\frac12+\frac i6\right)$?
Is there a general method to find closed forms of expressions of the form $\Re\,\operatorname{Li}_2(p+iq)$, $\Im\,\operatorname{Li}_2(p+iq)$ for $p,q\in\mathbb Q$?
| I don't know how you got your conjecture but I checked and on further simplifications I found it to be correct.
We know that :
$$
{Li}_{2}(\bar{z})=\bar{{Li}_{2}(z)}
$$
So :
$$
\Re{{Li}_{2}(z)}=\frac{\bar{{Li}_{2}(z)}+{Li}_{2}(z)}{2}
$$
So:
$$
\Re{{Li}_{2}(\frac{1}{2}+\frac{i}{6})}=\frac{{Li}_{2}(\frac{1}{2}+\frac{i}{6})+{Li}_{2}(\frac{1}{2}-\frac{i}{6})}{2}\\
=\frac{{Li}_{2}(\frac{1}{2}+\frac{i}{6})+{Li}_{2}(1-(\frac{1}{2}+\frac{i}{6}))}{2}
$$
Now let's use:
$$
{Li}_{2}(z)+{Li}_{2}(1-z)=\frac{{\pi}^{2}}{6}-\ln{z}\ln{(1-z)}
$$
So we get:
$$
\Re{{Li}_{2}(\frac{1}{2}+\frac{i}{6})}=\frac{\frac{{\pi}^{2}}{6}-\ln{(\frac{1}{2}+\frac{i}{6})}\ln{(1-(\frac{1}{2}+\frac{i}{6}))}}{2}\\
=\frac{\frac{{\pi}^{2}}{6}-\ln{(\frac{1}{2}+\frac{i}{6})}\ln{(\frac{1}{2}-\frac{i}{6})}}{2}
$$
Let's compute those logarithms:
$$
\ln(\frac{1}{2}\pm \frac{i}{6})=\ln{(\sqrt{\frac{1}{2^2}+\frac{1}{6^2}}{e}^{\pm i\arctan{\frac{1}{3}}})}\\
=\ln{(\sqrt{\frac{5}{18}}{e}^{\pm i\arctan{\frac{1}{3}}})}\\
=\frac{1}{2}\ln{(\frac{5}{18})}\pm i\arctan{\frac{1}{3}}
$$
Taking their product:
$$
\ln{(\frac{1}{2}+\frac{i}{6})}\ln{(\frac{1}{2}-\frac{i}{6})}=\frac{1}{4}{\ln{(\frac{5}{18}})}^{2}+{(\arctan{\frac{1}{3}})}^{2}
$$
Finally:
$$
\Re{{Li}_{2}(\frac{1}{2}+\frac{i}{6})}=\frac{{\pi}^{2}}{12}-\frac{1}{8}{\ln{(\frac{18}{5}})}^{2}-\frac{1}{2}{(\arctan{1/3})}^{2}\\
=\frac{7(\pi)^2}{48}-\frac{{(\arctan{2})}^{2}}{3}-\frac{{(\arctan{3})}^{2}}{6}-\frac{1}{8} {\ln{\frac{18}{5}}}^{2}
$$
There's a general formula using the same method. But the imaginary part doesn't have a known closed form:
$$
\Re{{Li}_{2}(\frac{1}{2}+iq)}=\frac{{\pi}^{2}}{12}-\frac{1}{8}{\ln{(\frac{1+4q^2}{4})}}^{2}-\frac{{\arctan{(2q)}}^{2}}{2}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
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What is the integral for this expression? (I tried complete the square) But complete the square, doesn´t lead me to something coherent.
I got this:
$$\int\frac{x \, dx}{\sqrt{3-2x-x^2}}$$
| $$
\int\frac{x \, dx}{\sqrt{3-2x-x^2}}
$$
First let $u=3-2x-x^2$ so that $du = -2(1+x)\,dx$ and $\dfrac{du}{-2} = (1+x)\,dx$. Then we can say
\begin{align}
\int\frac{x \, dx}{\sqrt{3-2x-x^2}} = \int \frac{(1+x) \, dx}{\sqrt{3-2x-x^2}} - \int \frac{dx}{\sqrt{3-2x-x^2}}.
\end{align}
For the first integral, use the substitution.
Then we must deal with the second integral. Complete the square:
$$
3-2x-x^2 = -(x^2 + 2x + 1) + 3+1 = 4 - (x+1)^2.
$$
So we have
$$
\int \frac{dx}{\sqrt{4-(x+1)^2}} = \int \frac{dx/2}{\sqrt{1 - \left( \frac{x+1}2 \right)^2}} = \int \frac{\cos\theta\,d\theta}{\sqrt{1-\sin^2\theta}} = \int 1\,d\theta = \text{etc.}
$$
After that we need to change $\theta$ to $\arcsin\dfrac{x+1}{2}$ and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1427629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to get the RHS from the LHS $\sum_{k=0}^n2^kx^{k+2}=\frac{x^2-2^{n+1}x^{n+3}}{1-2x}\tag{1}$ How to get this required RHS from the given LHS.
$$\sum_{k=0}^n2^kx^{k+2}=\frac{x^2-2^{n+1}x^{n+3}}{1-2x}\tag{1}$$
This was used in a solution to the following question I asked. I couldn't understand the step and hence to understand in detail I put it as a question.
Find the sum of the n terms of the series $2\cdot2^0+3\cdot2^1+4\cdot2^2+\dots$
| $$
\sum_{k=0}^{n}{2^k{x}^{k+2}} = x^2\sum_{k=0}^{n}{2^k{x}^{k}}\\
\sum_{k=0}^{n}{2^k{x}^{k+2}} = x^2\sum_{k=0}^{n}{{(2x)}^{k}}\\
\sum_{k=0}^{n}{2^k{x}^{k+2}} = x^2\frac{1-{(2x)}^{n+1}}{1-2x}\\
\sum_{k=0}^{n}{2^k{x}^{k+2}} = \frac{x^2-x^2{2}^{(n+1)}{x}^{n+1}}{1-2x}\\
\sum_{k=0}^{n}{2^k{x}^{k+2}} = \frac{x^2-{2}^{n+1}{x}^{n+3}}{1-2x}
$$
I used the formula for summing geometric series:
$$
\sum_{k=0}^{n}{t^k} = \frac{1-{t}^{n+1}}{1-t}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1428052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Calculating the $n^\text{th}$ derivative How do we calculate the $n^{\text{th}}$ derivative for
$$
\frac{x^3}{(x-a)(x-b)(x-c)}?
$$
How can I obtain the partial fraction for the given term?
| If we assume that $a,b,c$ are distinct numbers, from:
$$ \text{Res}\left(\frac{z^3}{(z-a)(z-b)(z-c)},z=a\right)=\frac{a^3}{(a-b)(a-c)}\tag{1}$$
it follows that:
$$\begin{eqnarray*} f(z)&=&\frac{z^3}{(z-a)(z-b)(z-c)}\tag{2}\\&=&1+\frac{a^3}{(a-b)(a-c)(z-a)}+\frac{b^3}{(b-a)(b-c)(z-b)}+\frac{c^3}{(c-a)(c-b)(z-c)}\end{eqnarray*}$$
and we are able to compute the derivatives of $f(z)$ from the derivatives of $\frac{1}{(z-a)}$, that are straightforward to compute.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1429800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Convergence/divergence of: $\sum_{n = 2}^\infty \ln\left( 1-\frac{1}{n^2}\right)$? How to prove the convergence/divergence of:
$$\sum_{n = 2}^{+\infty} \ln\left( 1-\frac{1}{n^2}\right)$$ ?
I've tried:
$$\sum_{n = 2}^{+\infty} \ln\left( 1-\frac{1}{n^2}\right) = \sum_{n = 2}^{+\infty} \ln\left(\frac{n^2-1}{n^2}\right) $$
$$= \sum_{n = 2}^{+\infty} [\ln(n^2-1) + \ln(n^2)] $$
$$= [\ln(3)-\ln(4)]+[\ln(8)-\ln(9)]+[\ln(15)-\ln(16)]+\cdots$$
But I can't see a pattern.
Thought about finding an upper limit as well, but since $$\sum_{n = 2}^{+\infty} \ln\left( 1-\frac{1}{n^2}\right)$$ isn't monotonic, then (monotonic + bounded => converges) or Squeeze theorem doesn't help.
| \begin{array} \\
\displaystyle\sum_{n=2}^N \ln\left(1-\frac{1}{n^2}\right) &=& \displaystyle\sum_{n=2}^N \ln\left(\frac{n^2-1}{n^2}\right) \\
&=& \displaystyle\sum_{n=2}^N \ln\left(n^2-1\right)-\ln\left(n^2\right) \\
&=& \displaystyle\sum_{n=2}^N \ln\left((n-1)(n+1)\right)-2\ln n \\
&=& \displaystyle\sum_{n=2}^N \ln(n-1) + \ln(n+1) - 2\ln n \\
&=& \displaystyle [\ln(1)+\ln(3)-2\ln(2)] + [\ln(2)+\ln(4)-2\ln(3)] + [\ln(3)+\ln(5)-2\ln(4)] + \cdots \\
&&+ [\ln(N-2) + \ln(N) - 2 \ln(N-1)] + [\ln(N-1) + \ln(N+1) - 2 \ln(N)]
\end{array}
You can check that most of the terms cancel (or prove it by induction), and you have:
$$\displaystyle\sum_{n=2}^N \ln\left(1-\frac{1}{n^2}\right) = -\ln(2) + \ln(N+1) - \ln(N) = -\ln(2) + \ln\left(1 + \frac{1}{N}\right)$$
And thus:
$$\displaystyle\sum_{n=2}^\infty \ln\left(1-\frac{1}{n^2}\right) = \lim_{N\rightarrow\infty} \sum_{n=2}^N \ln\left(1-\frac{1}{n^2}\right) = \lim_{N\rightarrow\infty} \left(-\ln(2) + \ln\left(1 + \frac{1}{N}\right)\right) = -\ln(2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1431013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Simple limit of a sequence
Need to solve this very simple limit $$ \lim _{x\to \infty
\:}\left(\sqrt[3]{3x^2+4x+1}-\sqrt[3]{3x^2+9x+2}\right) $$
I know how to solve these limits: by using
$a−b= \frac{a^3−b^3}{a^2+ab+b^2}$. The problem is that the standard way (not by using L'Hospital's rule) to solve this limit - very tedious, boring and tiring. I hope there is some artful and elegant solution. Thank you!
| $$\lim _{ x\to \infty \: } \left( \sqrt [ 3 ]{ 3x^{ 2 }+4x+1 } -\sqrt [ 3 ]{ 3x^{ 2 }+9x+2 } \right) =$$
$\lim _{ x\to \infty \: } \frac { \left( \sqrt [ 3 ]{ 3x^{ 2 }+4x+1 } -\sqrt [ 3 ]{ 3x^{ 2 }+9x+2 } \right) \left( \sqrt [ 3 ]{ { \left( 3x^{ 2 }+4x+1 \right) }^{ 2 } } +\sqrt [ 3 ]{ \left( 3x^{ 2 }+4x+1 \right) \left( 3x^{ 2 }+9x+2 \right) } +\sqrt [ 3 ]{ { \left( 3x^{ 2 }+9x+2 \right) }^{ 2 } } \right) }{ \left( \sqrt [ 3 ]{ { \left( 3x^{ 2 }+4x+1 \right) }^{ 2 } } +\sqrt [ 3 ]{ \left( 3x^{ 2 }+4x+1 \right) \left( 3x^{ 2 }+9x+2 \right) } +\sqrt [ 3 ]{ { \left( 3x^{ 2 }+9x+2 \right) }^{ 2 } } \right) } =$$$\ =\lim _{ x\to \infty : } \frac { 3x^{ 2 }+4x+1-3x^{ 2 }-9x-2 }{ \left( \sqrt [ 3 ]{ { \left( 3x^{ 2 }+4x+1 \right) }^{ 2 } } +\sqrt [ 3 ]{ \left( 3x^{ 2 }+4x+1 \right) \left( 3x^{ 2 }+9x+2 \right) } +\sqrt [ 3 ]{ { \left( 3x^{ 2 }+9x+2 \right) }^{ 2 } } \right) } =\lim _{ x\rightarrow \infty }{ \frac { -5x-1 }{ \left( \sqrt [ 3 ]{ { \left( 3x^{ 2 }+4x+1 \right) }^{ 2 } } +\sqrt [ 3 ]{ \left( 3x^{ 2 }+4x+1 \right) \left( 3x^{ 2 }+9x+2 \right) } +\sqrt [ 3 ]{ { \left( 3x^{ 2 }+9x+2 \right) }^{ 2 } } \right) } } $$
now ,the power of denominator of polynomial is higher than the numerator so the limit is equal to $0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1433216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
Show that $e^{\sqrt 2}$ is irrational I'm trying to prove that $e^{\sqrt 2}$ is irrational. My approach:
$$
e^{\sqrt 2}+e^{-\sqrt 2}=2\sum_{k=0}^{\infty}\frac{2^k}{(2k)!}=:2s
$$
Define $s_n:=\sum_{k=0}^{n}\frac{2^k}{(2k)!}$, then:
$$
s-s_n=\sum_{k=n+1}^{\infty}\frac{2^k}{(2k)!}=\frac{2^{n+1}}{(2n+2)!}\sum_{k=0}^{\infty}\frac{2^k}{\prod_{k=1}^{2k}(2n+2+k)}\\<\frac{2^{n+1}}{(2n+2)!}\sum_{k=0}^{\infty}\frac{2^k}{(2n+3)^{2k}}=\frac{2^{n+1}}{(2n+2)!}\frac{(2n+3)^2}{(2n+3)^2-2}
$$
Now assume $s=\frac{p}{q}$ for $p,q\in\mathbb{N}$. This implies:
$$
0<\frac{p}{q}-s_n<\frac{2^{n+1}}{(2n+2)!}\frac{(2n+3)^2}{(2n+3)^2-2}\iff\\
0<p\frac{(2n)!}{2^n}-qs_n\frac{(2n)!}{2^n}<\frac{2}{(2n+1)(2n+2)}\frac{(2n+3)^2}{(2n+3)^2-2}
$$
But $\left(p\frac{(2n)!}{2^n}-qs_n\frac{(2n)!}{2^n}\right)\in\mathbb{N}$ which is a contradiction for large $n$. Thus $s$ is irrational. Can we somehow use this to prove $e^\sqrt{2}$ is irrational?
| Since the sum of two rational numbers is rational, one or both of $e^{\sqrt{2}}$
and $e^{-\sqrt{2}}$ is irrational. But, $e^{-\sqrt{2}}=1/e^{\sqrt{2}}$,
and hence both are irrational.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1441292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "44",
"answer_count": 3,
"answer_id": 0
} |
Minimum value of $\frac{4}{x}+2x+10+\frac{3+x}{4x^2+1}$ If the minimum value of $\frac{4}{x}+2x+10+\frac{3+x}{4x^2+1}$ when $x>0$ is $\frac{p}{q}$ where $p,q\in N$ then find the least value of $(p+q).$
I can find the minimum value of $\frac{4}{x}+2x$ using $AM- GM$ inequality but that too comes irrational and when i tried to find the minimum value of $\frac{3+x}{4x^2+1}$ using first derivative test,i am facing difficulty.Please help me in solving this problem.
| The derivative of $f(x) = \frac{4}{x}+2x+10+\frac{3+x}{4x^2+1}$ is
$$
\begin{align}
f'(x) &= \frac{-4}{x^2}+2+\frac{(4x^2+1)(1) - (3+x)(8x)}{(4x^2+1)^2}\\
&= \frac{-4}{x^2}+2+\frac{-4x^2-24x+1}{(4x^2+1)^2}\\
\end{align}
$$
So when $f'(x) = 0$, we have
$$
\begin{align}
-4(4x^2+1)^2 + 2x^2(4x^2+1)^2 + (-4x^2-24x+1)x^2 &= 0\\
32x^6-52x^4-24x^3-29x^2-4&=0
\end{align}
$$
There is a solution at $x \approx 1.577$, which can be found numerically.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1442089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
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