Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Proof using formal definition: Infinite limit I was wondering how get the proof of this limit:
$$\lim\limits_{x\to -\infty}\dfrac{{x^2} - x + 1}{x + 4} = -\infty$$
The problem is that I don't know what to do for find the appropriated values to make valid the implication of the formal definition (epsilon-delta).
I would appreciate if somebody can help me.
| You want to find $N(M)<0$ (a function in terms of $M<0$) such that
$$x<N(M)\implies \frac{x^2-x+1}{x+4}<M$$
Let $N(M)\le -4$. Then $x<-4$ and $$\frac{x^2-x+1}{x+4}<M\iff x^2-x+1>M(x+4)$$
$$\iff x^2-x(M+1)+(1-4M)>0$$
If $M\le -9-2\sqrt{21}$, then $\Delta=M^2+18M-3\ge 0$ and let $$N(M)=\min\left\{-4,\frac{M+1-\sqrt{M^2+18M-3}}{2}\right\}$$
If $M\in(-9-2\sqrt{21},0)$, then let $$N(M)=\min\left\{-4,\frac{k+1-\sqrt{k^2+18k-3}}{2}\right\}$$
for any $k\le-9-2\sqrt{21}$, e.g. you can let $k=-19$:
$$N(M)=\min\left\{-4,-11\right\}=-11$$
Answer: you can let $$N(M)=\begin{cases}\min\left\{-4,\frac{M+1-\sqrt{M^2+18M-3}}{2}\right\}, && M\le -9-2\sqrt{21}\\-11, && M\in(-9-2\sqrt{21},0)\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1572866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Can you add a tile to a $7\times 7$ checkerboard so that it is tile-able by $5$ ominoes ($5\times1$ rectangles)? Can you add a tile to a $7\times7$ checkerboard so that it is tile-able by $5$ ominoes ($5\times1$ rectangles)? If so, where must you place the extra tile? If not, explain why.
you can add a tile because you would need the checkerboard to be a multiple of $5$. and a $7\times7$ checkboard means there are $49$ squares which is not a multiple of $5$ so by adding $1$, we get 50 which is a multiple of $50$.
Now im having difficult of where to place the extra tile
| Hint:
$$\begin{array}{ccccc} 1 & 2 &3&4&5&1&2\\2&3&4&5&1&2&3\\
3&4&5&1&2&3&4\\ 4&5&1&2&3&4&5\\
5&1&2&3&4&5&1\\ 1&2&3&4&5&1&2\\ 2&3&4&5&1&2&3 \end{array}$$
Each $1\times 5$ tile will cover a $1$, a $2$, a $3$, a $4$, and a $5$. Is there a tile you can add, within this numbering scheme, and get everything to work out?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1573059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve for $x$ where $x=(1-2)(1+2+4)+(2-3)(4+6+9)+(3-4)(9+12+16)+....+(49-50)(2401+2450+2500)$ Consider the following equality:
$$x=(1-2)(1+2+4)+(2-3)(4+6+9)+(3-4)(9+12+16)+....+(49-50)(2401+2450+2500)$$
Solve for $x$.
The only thing I noticed is the first part like $(1-2)$,$(3-4)$ gives us $-1$ but then I just don't see what the trick behind this problem is.
| Each term is of the form
$$(n-(n+1))(n^2 + (n^2+n) + (n+1)^2) = -(3n^2+3n+1)$$
Hence,
$$x = - \sum_{n=1}^{m} (3n^2+3n+1) = - m(m^2+3m+3)$$
where $m=49$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1573141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Limit problem with fifth root $$\lim_{x\to 0}\frac{x^2}{\sqrt[5]{1+5x}-1-x}$$
How to solve this limit without L'Hopital rule and Taylor series?
| We have
$$\dfrac1{a-b} = \dfrac{a^4+a^3b+a^2b^2+ab^3+b^4}{a^5-b^5}$$
Taking $a=\sqrt[5]{1+5x}$ and $b=1+x$, we obtain that
\begin{align}
\dfrac1{\sqrt[5]{1+5x}-1-x} = \dfrac{(1+5x)^{4/5} + (1+5x)^{3/5}(1+x) + (1+5x)^{2/5}(1+x)^{2} + (1+5x)^{1/5}(1+x)^3 + (1+x)^4}{(1+5x)-(1+x)^5}
\end{align}
Hence,
\begin{align}
\dfrac{x^2}{\sqrt[5]{1+5x}-1-x} & = \dfrac{(1+5x)^{4/5} + (1+5x)^{3/5}(1+x) + (1+5x)^{2/5}(1+x)^{2} + (1+5x)^{1/5}(1+x)^3 + (1+x)^4}{-10-15x-5x^2-x^3}
\end{align}
Now taking the limit as $x \to 0$, we obtain the limit to be
$$L = \dfrac{1+1+1+1+1}{-10} = -\dfrac12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1573469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How would you solve for x in this case using the trig identities?
Solve for all values of $x$:
$$\cos 2x = 2\sin x$$
$$1 - 2\sin^2x - 2\sin x = 0$$
$$-2\sin^2x - 2\sin x + 1 = 0$$
$$\sin^2x - 2\sin x - 2 = 0$$
How would you factor this above to solve for $x$?
| $$\cos(2x)=2\sin(x)\Longleftrightarrow$$
$$\cos(2x)-2\sin(x)=0\Longleftrightarrow$$
$$1-2\sin(x)-2\sin^2(x)=0\Longleftrightarrow$$
$$-\frac{1}{2}+\sin(x)+\sin^2(x)=0\Longleftrightarrow$$
$$\sin(x)+\sin^2(x)=\frac{1}{2}\Longleftrightarrow$$
$$\frac{1}{4}+\sin(x)+\sin^2(x)=\frac{3}{4}\Longleftrightarrow$$
$$\left(\frac{1}{2}+\sin(x)\right)^2=\frac{3}{4}\Longleftrightarrow$$
$$\frac{1}{2}+\sin(x)=\pm\frac{\sqrt{3}}{2}\Longleftrightarrow$$
$$\sin(x)=\pm\frac{\sqrt{3}}{2}-\frac{1}{2}\Longleftrightarrow$$
$$x=\begin{cases}
\ \pi+\arcsin\left(\frac{1}{2}+\frac{\sqrt{3}}{2}\right)+2\pi n_1 \\
\ \pi+\arcsin\left(\frac{1}{2}-\frac{\sqrt{3}}{2}\right)+2\pi n_2 \\
\ 2\pi n_3-\arcsin\left(\frac{1}{2}+\frac{\sqrt{3}}{2}\right) \\
\ 2\pi n_4-\arcsin\left(\frac{1}{2}-\frac{\sqrt{3}}{2}\right)
\end{cases}$$
With $n_1,n_2,n_,n_4\in\mathbb{Z}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1574011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
$3^x-2^y=1$, $x \in \mathbb{N}$ and $y \in \mathbb{N}$ $3^x-2^y=1$
or
$y=\log_2{\left(3^x-1\right)}$
$x$ and $y$ must be natural numbers.
I know this two solutions:
$x=1$ and $y=1$
$x=2$ and $y=3$
Are there more solutions?
How can I find them?
| We look for $y\ge 2$. Then we must have $3^x\equiv 1\pmod{4}$, so $x$ must be even, say $x=2a$. Now we can rewrite our equation as $3^{2a}-1=2^y$, or equivalently as $(3^a-1)(3^a+1)=2^y$.
It follows that $3^a-1$ and $3^a+1$ must each be powers of $2$. They differ by $2$, and the only pair of powers of $2$ that differ by $2$ are $2$ and $4$. That forces $a=1$. So there are no solutions in positive integers other than the ones you listed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1577298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\lim_{R\to\infty}\left(\int_0^R\left|\frac{\sin x}{x}\right|dx-\frac{2}{\pi}\log R\right)$ Is there a closed form of
$$\lim_{R\to\infty}\left(\int_0^R\left|\frac{\sin x}{x}\right|dx-\frac{2}{\pi}\log R\right)$$
I am pretty interested whether we can find out a closed form of this limit.
We can show that for $R=n\pi,n\in\mathbb{N}$, we have
$$\begin{aligned}
\int_0^R\left|\frac{\sin x}{x}\right|dx&=\sum_{k=0}^{n-1}\int_{k\pi}^{(k+1)\pi}\frac{|\sin x|}{x}dx\\
&=\sum_{k=0}^{n-1}\int_{0}^{\pi}\frac{\sin x}{(k+1)\pi-x}dx\\
&\leq \int_0^\pi\frac{\sin x}{\pi-x}dx+\sum_{k=1}^{n-1}\int_{0}^{\pi}\frac{\sin x}{k\pi}dx\\
&=\int_0^\pi\frac{|\sin(\pi-x)|}{x}dx+\sum_{k=1}^{n-1}\frac{2}{k\pi}dx\\
&=\int_0^\pi\frac{\sin x}{x}dx+\frac{2}{\pi}\sum_{k=1}^{n-1}\frac{1}{k}
\end{aligned}$$
On the other hand we have
$$\begin{aligned}
\int_0^R\left|\frac{\sin x}{x}\right|dx&\geq \sum_{k=0}^{n-1}\int_{0}^{\pi}\frac{\sin x}{(k+1)\pi}dx\\
&=\sum_{k=1}^n\frac{1}{k\pi}\int_0^\pi\sin xdx\\
&=\frac{2}{\pi}\sum_{k=1}^n\frac{1}{k}
\end{aligned}$$
Then I tried to apply the squeeze rule, but this does not lead to anything appetizing. Anybody know any tricks for this problem?
| We have
\begin{align*}
\int_0^{\pi N} {\left| {\frac{{\sin x}}{x}} \right|{\rm d}x} & = \frac{2}{\pi }\sum\limits_{k = 1}^N {\frac{1}{k}} + \sum\limits_{k = 1}^N {\int_0^\pi {\left[ {\frac{{\sin x}}{{\pi k - x}} - \frac{{\sin x}}{{\pi k}}} \right]{\rm d}x} }
\\ & = \frac{2}{\pi }\sum\limits_{k = 1}^N {\frac{1}{k}} + \sum\limits_{k = 1}^N {\int_0^\pi {\frac{{x\sin x}}{{\pi k(\pi k - x)}}{\rm d}x} } .
\end{align*}
It is not difficult to show that the second sum converges as $N\to+\infty$ and the error committed by stopping at the $N$th term is $\mathcal{O}(N^{-1})$. Consequently, by using the standard approximation for the harmonic numbers,
\begin{align*}
\int_0^{\pi N} {\left| {\frac{{\sin x}}{x}} \right|{\rm d}x} & = \frac{2}{\pi }\log N + \frac{2}{\pi }\gamma + \sum\limits_{k = 1}^\infty {\int_0^\pi {\frac{{x\sin x}}{{\pi k(\pi k - x)}}{\rm d}x} } + \mathcal{O}\!\left( {\frac{1}{N}} \right)
\\ & = \frac{2}{\pi }\log N + \frac{2}{\pi }\gamma + \sum\limits_{k = 1}^\infty {\int_0^1 {\frac{{x\sin (\pi x)}}{{k(k - x)}}{\rm d}x} } + \mathcal{O}\!\left( {\frac{1}{N}} \right)
\end{align*}
as $N\to +\infty$ ($\gamma$ being the Euler–Mascheroni constant). It is easy to see that
$$
\int_0^R {\left| {\frac{{\sin x}}{x}} \right|{\rm d}x} - \int_0^{\pi \left\lfloor {R/\pi } \right\rfloor } {\left| {\frac{{\sin x}}{x}} \right|{\rm d}x} = \mathcal{O}\!\left( {\frac{1}{R}} \right),
$$
whence
$$\boxed{
\int_0^R {\left| {\frac{{\sin x}}{x}} \right|dx} = \frac{2}{\pi }\log R + C + \mathcal{O}\!\left( {\frac{1}{R}} \right)}
$$
as $R\to +\infty$, with
$$\boxed{
C = \frac{2}{\pi }(\gamma - \log \pi ) + \sum\limits_{k = 1}^\infty {\int_0^1 {\frac{{x\sin (\pi x)}}{{k(k - x)}}{\rm d}x} } .}
$$
We may obtain a some alternative expressions for $C$ as follows. Note that
\begin{align*}
&\sum\limits_{k = 1}^\infty {\int_0^1 {\frac{{x\sin (\pi x)}}{{k(k - x)}}{\rm d}x} } = \int_0^1 {\left( {\sum\limits_{k = 1}^\infty {\frac{x}{{k(k - x)}}} } \right)\sin (\pi x){\rm d}x}\\ & = - \int_0^1 {(\gamma + \psi (1 - x))\sin (\pi x){\rm d}x}
= - \frac{{2\gamma }}{\pi } - \int_0^1 {\psi (1 - x)\sin (\pi x){\rm d}x} \\ & = - \frac{{2\gamma }}{\pi } - \int_0^1 {\psi (t)\sin (\pi t){\rm d}t} = - \frac{{2\gamma }}{\pi } + \pi \int_0^1 {\log \Gamma (t)\cos (\pi t){\rm d}t} .
\end{align*}
Here $\psi$ and $\Gamma$ are the digamma function and the gamma function, respectively. Consequently,
$$\boxed{
C = - \frac{{2\log \pi }}{\pi } -\int_0^1 {\psi (t)\sin (\pi t){\rm d}t}= - \frac{{2\log \pi }}{\pi } + \pi \int_0^1 {\log \Gamma (t)\cos (\pi t){\rm d}t} .}
$$
Using the Fourier series of $\log\Gamma(t)$, we find
\begin{align*}
\int_0^1 {\log \Gamma (t)\cos (\pi t){\rm d}t} & = \frac{{2(\gamma + \log (2\pi ))}}{{\pi ^2 }} + \frac{4}{{\pi ^2 }}\sum\limits_{n = 1}^\infty {\frac{{\log n}}{{4n^2 - 1}}}
\\ & = \frac{{2(\gamma + \log (2\pi ))}}{{\pi ^2 }} + \frac{4}{{\pi ^2 }}\sum\limits_{n = 1}^\infty {\frac{{\log n}}{{(2n - 1)(2n + 1)}}}
\\ & = \frac{{2(\gamma + \log (2\pi ))}}{{\pi ^2 }} - \frac{2}{{\pi ^2 }}\sum\limits_{n = 2}^\infty {\frac{{\log (1 - 1/n)}}{{2n - 1}}} .
\end{align*}
Hence,
$$\boxed{
C =\frac{2}{\pi }\left[ {\gamma + \log 2 +2 \sum\limits_{n = 1}^\infty {\frac{{\log n}}{{4n^2 - 1}}}} \right]= \frac{2}{\pi }\left[ {\gamma + \log 2 - \sum\limits_{n = 2}^\infty {\frac{{\log (1 - 1/n)}}{{2n - 1}}} } \right].}
$$
Numerically, $C= 1.1129238102\ldots$. Note that
$$
\int_0^1 {\psi (t)\sin (\pi t){\rm d}t} = \int_0^1 {\psi (t + 1)\sin (\pi t){\rm d}t} - \int_0^\pi {\frac{{\sin t}}{t}{\rm d}t} ,
$$
and the latter is easier to compute numerically.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1579025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
How to prove $\sin(9^\circ) \sin(27^\circ) \sin(63^\circ) \sin(81^\circ)=\frac{1}{16}$ How can one prove the following equation?
$$\sin(9^\circ) \sin(27^\circ) \sin(63^\circ) \sin(81^\circ)=\frac{1}{16}$$
| Let $\sin5y=\sin45^\circ$
$\implies5y=180^\circ n+(-1)^n45^\circ$ where $n$ is any integer
$y=72^\circ m+9^\circ$ where $-2\le m\le2$
But $\sin5y=16\sin^5y − 20\sin^3y + 5\sin y$
So, the roots of $$16\sin^5y − 20\sin^3y + 5\sin y-\sin45^\circ=0$$
are $y=72^\circ m+9^\circ$ where $-2\le m\le2$
$\implies\prod_{m=-2}^2\sin(72^\circ m+9^\circ)=\dfrac{\sin45^\circ}{16}$
Now use $\sin(-A)=-\sin A,\sin(180^\circ-B)=\sin B$
See also :this
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1579633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solve the equation: $ \sin x - 3\sin 2x + \sin 3x = \cos x - 3\cos 2x + \cos3x$ How do I solve this equation:
$$ \sin x - 3\sin 2x + \sin 3x = \cos x - 3\cos 2x + \cos3x$$
| First, we may use some identities that you can find here. That's how we proceed
$$\begin{align}
\sin x - 3\sin 2x + \sin 3x &= (\sin x + \sin 3x) - 3\sin 2x \\
&=2\sin2x\cos x-3\sin2x \\
&=\sin2x(2\cos x - 3)
\end{align}$$
$$\begin{align}
\cos x - 3\cos 2x + \cos 3x &= (\cos x + \cos 3x) - 3\cos 2x \\
&=2\cos2x\cos x-3\cos2x \\
&= \cos2x(2\cos x - 3)
\end{align}$$
and then we can get
$$
\sin2x(2\cos x - 3) = \cos2x(2\cos x - 3) \\
(\sin2x-\cos2x)(2\cos x - 3) =0
$$
and hence
$$\begin{cases}
\sin2x-\cos2x=0 & \to \sqrt{2}\sin(2x-\frac{\pi}{4})=0\\
\text{or} \\
\cos x=\frac{3}{2} & \to \text{impossible}
\end{cases}$$
Where in the first implication, I used the identity mentioned here. This finally leads to
$$2x - \frac{\pi}{4} = n\pi, \qquad n=0,\pm1,\pm2,...$$
Or equivalently
$$x=\frac{\pi}{8} + n\frac{\pi}{2}, \qquad n=0,\pm1,\pm2,...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1579901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Integral $\int_0^1\arctan(x)\arctan\left(x\sqrt3\right)\ln(x)dx$ I need to evaluate this integral:
$$\int_0^1\arctan(x)\arctan\left(x\sqrt3\right)\ln(x)dx$$
Apparently, Maple and Mathematica cannot do anything with it, but I saw similar integrals to be evaluated in terms of polylogarithms (unfortunately, I have not yet mastered them enough to do it myself). Could anybody please help me with it?
| Following the method outlined in another answer, and simplifying the resulting expression, we get the following closed form:
$$\frac{5 G}{6 \sqrt{3}}-\frac{\Im\operatorname{Li}_3(1+i)}{\sqrt{3}}+\Im\operatorname{Li}_3\left(i
\sqrt{3}\right)-\frac{\Im\operatorname{Li}_3\left(i \sqrt{3}\right)}{4 \sqrt{3}}-\frac{1}{2}
\Im\operatorname{Li}_3\left(1+i \sqrt{3}\right)\\
-3 \Im\operatorname{Li}_3\left(\left(-\frac{1}{2}+\frac{i}{2}\right)
\left(-1+\sqrt{3}\right)\right)+\sqrt{3} \Im\operatorname{Li}_3\left(\left(-\frac{1}{2}+\frac{i}{2}\right)
\left(-1+\sqrt{3}\right)\right)\\
+\frac{1}{\sqrt{3}}\Im\operatorname{Li}_3\left(\tfrac{(1+i) \sqrt{3}}{1+\sqrt{3}}\right)-3
\Im\operatorname{Li}_3\left(\left(\frac{1}{2}+\frac{i}{2}\right) \left(1+\sqrt{3}\right)\right)+\frac{2}{\sqrt{3}}\Im\operatorname{Li}_3\left(\left(\frac{1}{2}+\frac{i}{2}\right) \left(1+\sqrt{3}\right)\right)\\
-\frac{1}{288} \pi
\left[-2 \left\{\vphantom{\Large|}3 \left(4+\sqrt{3}\right) \cdot \operatorname{Li}_2\left(\tfrac{1}{3}\right)+6 \ln ^23-6 \left(7 \sqrt{3}-24\right) \cdot \ln
^2\left(1+\sqrt{3}\right)\\+24 \ln 3
-4 \left(9+4 \sqrt{3}\right) \cdot \ln \left(1+\sqrt{3}\right)\right\}+3 \left(5 \sqrt{3}-36\right)
\cdot \ln ^22\\
-4 \left\{\vphantom{\Large|}9+7 \sqrt{3}-6 \ln 3+3 \left(7 \sqrt{3}-24\right) \cdot \ln
\left(1+\sqrt{3}\right)\right\}\cdot\ln 2\right]\\
-\frac{1}{216} \left(18+5 \sqrt{3}\right) \pi ^2+\left(\frac{5}{36}-\frac{31}{384
\sqrt{3}}\right) \pi ^3+\frac{5 \psi ^{(1)}\left(\frac{1}{3}\right)}{48 \sqrt{3}},$$
that might be possible to simplify further.
Mathematica expression is here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1581669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Determine point of maxima and minima of the function Determine point of maxima and minima of the function
$f(x)=\frac 1 8 \log x -bx +x^2$, $x> 0$, where $b \geq0$ and is a constant.
I found out $f'(x)$ and equated it to $0$ and got it as
$$8 f'(x)= 1/x - 8b+16x=0.$$
Upon solving this I got $16x^2-8bx+1=0$
Solving for $x$ I got $x=(b±(b^2-1)^{1/2})/4$
I don't know how to eliminate the term $b$ and got stuck here.
I also tried $f''$ and got it as
$8 f''= -1/x^2+16$ and tried using the inequalties for $f''>0$ and $f''<0$ and got $x ∈ (-∞,-1/4)∪(1/4,∞)$ for minima condition and $x ∈ (-1/4,1/4)$ for maxima condition but i was not able to make use of it.
| You given that $b>0$.
You got that the function has two extreme values at
$$x_{\pm}=\frac{b\pm \sqrt{b^2-1}}4$$
For $0<b<1$, $x_\pm$ is not real, so restrict the discussion to $b\ge1$.
For $b=1$, $x_\pm=\frac14$ and $f''(\tfrac14)=0$, so the second derivative test is inconclusive (use extremum test to decide that this is a saddle point, since $f'''(\tfrac14)\ne0$).
Now for $b>1$, $$f''(x_+)=16-\frac{1}{\left(\frac{1}{4} \left(b+\sqrt{b^2-1}\right)\right)^2}=\frac{16 \left(\left(b+\sqrt{b^2-1}\right)^2-1\right)}{\left(b+\sqrt{b^2-1}\right)^2}>0,\quad b>1,$$
since $\left(b+\sqrt{b^2-1}\right)^2-1 =2 b^2+2 b\sqrt{b^2-1} -2>2 b \sqrt{ b^2-1}>0$. Thus, $x_+$ is a minimum for any $b>1$.
Next
$$f''(x_-)=16-\frac{1}{\left(\frac{1}{4} \left(b-\sqrt{b^2-1}\right)\right)^2}=\frac{16 \left(\left(b-\sqrt{b^2-1}\right)^2-1\right)}{\left(b-\sqrt{b^2-1}\right)^2}$$
So we need to check when $\left(b-\sqrt{b^2-1}\right)^2-1<0$ or equivalently
$$b-\sqrt{b^2-1}<1$$
$$b-1<\sqrt{b^2-1}$$
$$b^2-2b + 1 =(b-1)^2<b^2-1$$
$$2-2b <0$$
Finally, $x_-$ is a maximum for all $b>1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1583050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$x^4-(a+b+c+d)x^3-(a+b+c)x^2-(a+b)x-a=0$,then the number of integer roots in the equation is?
Let a,b,c,d be natural numbers.Now consider an equation
$x^4-(a+b+c+d)x^3-(a+b+c)x^2-(a+b)x-a=0$,then the number of integer
roots in the equation is ?
Me: Plugged in $0,1,-1$.No use.At 0 the sign is negative and at infinity the sign is positive. At least one real root.Not sure if that will be integer though :/.Help please!
| Let $P(x) = x^4-(a+b+c+d)x^3-(a+b+c)x^2-(a+b)x-a$.
For $x \le -1$, we have $-x \le x^2$, and thus, $P'(x) = 4x^3-3(a+b+c+d)x^2-2(a+b+c)x-(a+b)$ $\le 4x^3-(a+b+c+3d)x^2-(a+b) < 0$.
Then, since $P(-1) = 1+b+d > 0$, we have that $P(x) > 0$ for all $x \le -1$, and thus, $P(x)$ has no negative integer roots.
Trivially, $P(0) = -a < 0$, so $0$ is not a root of $P(x)$.
By Descartes Rule of signs, $P(x)$ has at exactly one positive root.
However, $P(a+b+c+d) = -(a+b+c)(a+b+c+d)^2-(a+b)(a+b+c+d)-a < 0$, and since $P(x) \ge x^4-(a+b+c+d)(x^3+x^2+x+1) = x^4-(a+b+c+d)\dfrac{x^4-1}{x-1}$ for $x \ge 0$, we have $P(a+b+c+d+1) \ge 1$.
Thus, the one positive root of $P(x)$ is not an integer.
Therefore, $P(x)$ has no integer roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1584056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How is $9^{40}\equiv\ 1 \pmod {100}$? By Euler's theorem, $a^{\varphi(100)} \equiv\ 1 \pmod {100}$.
We know that the last two digits of $9^{40}$ are non-zero. So they can't even be $01$.
Since $1\equiv\ 1 \pmod {100}$, how come $9^{40}\equiv\ 1 \pmod {100}$?
I have looked at:
Find the last two digits of $9^{{9}^{9}}$,
Find the last two digits of the number $9^{9^9}$ ,
Find the last two digits of $9^{9^{9}}$
| It is straightforward
$9^{40} = 81^{20} = 6561^{10} \equiv 61^{10} \pmod{100} = 3721^5 \pmod{100} \equiv\ 21^5 \pmod{100}$
$= 441 \times 441 \times 21 \pmod{100}\equiv\ 41 \times 41 \times 21 \pmod{100}$
$= 35301 \pmod{100} \equiv\ 1 \ \pmod{100}$
and this is just one of the many many routes you could take.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1585795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Calculate $\int_{-\infty}^{\infty}\;\left( \frac{x^2}{1+4x+3x^2-4x^3-2x^4+2x^5+x^6}\right) \;dx$
Calculate $$\displaystyle \int_{-\infty}^{\infty}\;\left(
\frac{x^{2}}{1+4x+3x^{2}-4x^{3}-2x^{4}+2x^{5}+x^{6}}\right) \;dx$$
The answer given is $\pi$. How does one calculate this?
| There is an inner structure that enable this integral to be evaluated into such nice form.
Let $$f(x) = 1+4x+3x^2-4x^3-2x^4+2x^5+x^6$$ The first miracle is: $f(x)$ factorizes nicely in $\mathbb{Q}[i]$:
$$f(x) = \underbrace{\left(x^3+(1-i) x^2-(2+i) x-1\right)}_{g(x)} \underbrace{\left(x^3+(1+i) x^2-(2-i) x-1\right)}_{h(x)}$$
The second miracle is: the root of $g(x)$ all lie in the same half plane. In this case, all roots of $g$ are in the upper plane. Denote them by $\alpha_1, \alpha_2, \alpha_3$, by contour integration $$I:=\int_{-\infty}^\infty \frac{x^2}{f(x)}dx = 2\pi i\left[ {\frac{{{\alpha _1}^2}}{{g'({\alpha _1})h({\alpha _1})}} + \frac{{{\alpha _2}^2}}{{g'({\alpha _2})h({\alpha _2})}} + \frac{{{\alpha _3}^2}}{{g'({\alpha _3})h({\alpha _3})}}} \right]$$
Now the right hand side is symmetric in $\alpha_i$, which are roots of $g$. Since $g,h\in \mathbb{Q}[i][x]$, we have
$$\frac{I}{\pi} \in \mathbb{Q}$$
This explain the nice result of the integral. Note that the numerator $x^2$ can be replaced by any polynomial in $\mathbb{Q}[x]$, $I/\pi$ is still rational.
Using similar construction, we obtain the analogous integrals:
Let $$f(x) = 4 + 8x - 11{x^2} - 18{x^3} + 13{x^4} + 8{x^5} + {x^6}$$ then $f$ satisfies the above two "mircales" so we have
$$\int_{ - \infty }^\infty {\frac{1}{{f(x)}}dx} = \frac{{5\pi }}{6} \qquad \int_{ - \infty }^\infty {\frac{x}{{f(x)}}dx} = - \frac{\pi }{3} \qquad \int_{ - \infty }^\infty {\frac{{{x^2}}}{{f(x)}}dx} = \frac{\pi }{3}$$
Another example with
$$f(x) = 4 + 12x - 6{x^2} - 26{x^3} + 11{x^4} + 8{x^5} + {x^6}$$
$$\int_{ - \infty }^\infty {\frac{1}{{f(x)}}dx} = \frac{{3\pi }}{4} \qquad \int_{ - \infty }^\infty {\frac{x}{{f(x)}}dx} = - \frac{\pi }{4} \qquad \int_{ - \infty }^\infty {\frac{{{x^2}}}{{f(x)}}dx} = \frac{\pi }{4}$$
An octic example:
$$f(x) = 13 + 12 x + 7 x^4 + 2 x^5 - 3 x^6 + x^8$$
$$\int_{ - \infty }^\infty {\frac{1}{{f(x)}}dx} = \frac{{487\pi }}{4148} \qquad \int_{ - \infty }^\infty {\frac{x}{{f(x)}}dx} = - \frac{325\pi }{4148} \qquad \int_{ - \infty }^\infty {\frac{{{x^2}}}{{f(x)}}dx} = \frac{515\pi }{4148}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1587694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 2,
"answer_id": 1
} |
If $a(\cos x-1)+b^2=\cos(ax+b^2)-1$ hold true for all $x\in \mathbb{R}$, Then $(a,b)$
Number of positive integer ordered pair $(a,b)$ for which the equation
$a(\cos x-1)+b^2=\cos(ax+b^2)-1$ hold true for all $x\in \mathbb{R}$
$\bf{My\; Try::}$ We can write above equation as $a(\cos x-1)+b^2+1 = \cos(ax+b^2)$
Now Using $\cos x\leq 1\Rightarrow a(\cos x-1)+b^2+1 \leq b^2+1$
and $-1 \leq \cos (ax+b^2)\leq 1$
Now How can I solve after that, Help me
Thanks
| $x=0$ gives $\cos(b^2)=b^2+1$. However $b^2+1\geq 1 \geq \cos(x)$, with equality only if $b=0$.
Since your question requires positive integers, we're technically done.
However, I shall proceed to find all integers $a$ for which it holds.
Now we get $a(\cos(x)-1)=\cos(ax)-1$. $x=\frac12 \pi$ gives $-a=\cos(a\frac12 \pi)-1$.
But $a$ is an integer, so $\cos(a\frac12 \pi)$ is -1,0 or 1. so $-a=-1$, $-a=-2$ or $-a=0$, so $a=0$, $a=1$ or $a=2$. $a=0$ and $a=1$ indeed work.
For $a=2$ let $x=\pi$ and we get $2\cdot(-2)=\cos(2\pi)-1=0$, which is not true.
One can also prove that this are in fact all real solutions.
Conclusion: For positive integers $(a,b)$ there exists no such pair.
In the real numbers, the only pairs are $(0,0)$ and $(1,0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Very strong inequality Let $a$, $b$ and $c$ be non-negative numbers. Prove that:
$$a^3+b^3+c^3+3abc\geq(a+b-c)\sqrt{ab(a+c)(b+c)}+(a+c-b)\sqrt{ac(a+b)(b+c)}+(b+c-a)\sqrt{bc(a+b)(a+c)}$$
I have a proof, but my proof is long. Maybe there is something simple?
Thanks!
| Here is a brutal force method and I am not sure it is shorter than op:
it is easy to verify if $abc=0$, the inequality holds.
then suppose $a,b,c$ are sides of a triangle, then we can take the advantage of Thunderstruck's proof.
if $a,b,c$ are not sides of a triangle,WLOG, $a \ge b+c$, now we prove :
LHS $>(a+b-c)\sqrt{ab(a+c)(b+c)}+(a+c-b)\sqrt{ac(a+b)(b+c)}$
we have:
$((a+b-c)^2+(a+c-b)^2)(ab(a+c)(b+c)+ac(a+b)(b+c))=2a(a^2+(b-c)^2)(b+c)(ab+ac+2bc) \ge ((a+b-c)\sqrt{ab(a+c)(b+c)}+(a+c-b)\sqrt{ac(a+b)(b+c)})^2$
$a=b+c+x \implies x\ge0 $
$(a^3+b^3+c^3+3abc)^2-2a(a^2+(b-c)^2)(b+c)(ab+ac+2bc)=x^6+6cx^5+6bx^5+13c^2x^4+32bcx^4+13b^2x^4+14c^3x^3+56bc^2x^3+56b^2cx^3+14b^3x^3+7c^4x^2+42bc^3x^2+79b^2c^2x^2+42b^3cx^2+7b^4x^2+12bc^4x+52b^2c^3x+52b^3c^2x+12b^4cx+16b^2c^4+32b^3c^3+16b^4c^2>0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
integrate $\int_0^\pi\frac {1-\alpha^2}{1-2\alpha\cos x +\alpha^2} dx$ I am stuck at this example from Wikipedia on differentiation under the integral sign.
$$\int_0^\pi\frac {1-\alpha^2}{1-2\alpha\cos x + \alpha^2} dx$$
Any help?
Edits:
*
*The last term in the denominator is changed to $\alpha^2$.
*This integral appears in example 3 of the Wikipedia article https://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign.
| Let us write the integrand as
\begin{equation}
g(x) = \frac{1 - \alpha^2}{1 - 2\alpha\cos x + \alpha^2}
\end{equation}
We first write $\cos x = \cos^2(x/2) - \sin^2(x/2)$, $1 = \cos^2(x/2) + \sin^2(x/2)$ and $\alpha^2 = \alpha^2(\cos^2(x/2) + \sin^2(x/2))$ to get
\begin{equation}
g(x) = \frac{(1 - \alpha^2)\sec^2(x/2)}{(1 - \alpha)^2 + (1 + \alpha)^2\tan^2(x/2)}
\end{equation}
which is same as
\begin{equation}
g(x) = 2\frac{1 + \alpha}{2}\frac{(1 - \alpha)\sec^2(x/2)}{(1 - \alpha)^2 + (1 + \alpha)^2\tan^2(x/2)}
\end{equation}
or,
\begin{equation}
g(x) = 2\frac{1}{1 + \frac{(1 + \alpha)^2}{(1 - \alpha)^2}\tan^2(x/2)}\frac{1 + \alpha}{1 - \alpha}\sec^2(x/2)\frac{1}{2}
\end{equation}
or
\begin{equation}
g(x) = 2 \frac{1}{1 + \left(\frac{1 + \alpha}{1 - \alpha}\right)^2\tan^2\left(\frac{x}{2}\right)}\frac{d}{dx}\left[\frac{1 + \alpha}{1 - \alpha}\tan\left(\frac{x}{2}\right)\right]
\end{equation}
Therefore,
\begin{equation}
\int g(x)dx = 2\tan^{-1}\left(\frac{1 + \alpha}{1 - \alpha}\tan\left(\frac{x}{2}\right)\right) + c,
\end{equation}
where $c$ is a constant of integration. Therefore,
\begin{equation}
\int_0^\pi \frac{1 - \alpha^2}{1 - 2\alpha\cos x + \alpha^2}
dx = \begin{cases}
\pi & |\alpha| < 1 \\
-\pi & |\alpha| > 1
\end{cases}
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
How to prove $\left(1+\frac{1}{\sin a}\right)\left(1+\frac{1}{\cos a}\right)\ge 3+2\sqrt{2}$? Prove the inequality:
$$\left(1+\dfrac{1}{\sin a}\right)\left(1+\dfrac{1}{\cos a}\right)\ge 3+2\sqrt{2}; \text{ for } a\in\left]0,\frac{\pi}{2}\right[$$
| By the AM-GM inequality, $$\dfrac{1}{\cos a}+\dfrac{1}{\sin a} \geq 2 \sqrt{\dfrac{1}{\sin a\cos a}}$$
Since $\sin a \cos a = \frac12 \sin(2a) \leq \frac12$, we have $\dfrac{1}{\sin a\cos a}\geq 2$.
Hence $$\left(1+\dfrac{1}{\sin a}\right)\left(1+\dfrac{1}{\cos a}\right) = 1+\dfrac{1}{\cos a}+\dfrac{1}{\sin a}+\dfrac{1}{\sin a\cos a} \\\geq 1+2\sqrt{\dfrac{1}{\sin a\cos a}}+\dfrac{1}{\sin a\cos a}\geq 1+2\sqrt{2} + 2 = 3+ 2\sqrt{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Prove that $\left (\frac{1}{a}+1 \right)\left (\frac{1}{b}+1 \right)\left (\frac{1}{c}+1 \right) \geq 64.$
Let $a,b,$ and $c$ be positive numbers with $a+b+c = 1$. Prove that $$\left (\dfrac{1}{a}+1 \right)\left (\dfrac{1}{b}+1 \right)\left (\dfrac{1}{c}+1 \right) \geq 64.$$
Attempt
Expanding the LHS we obtain $\left (\dfrac{1+a}{a} \right)\left (\dfrac{1+b}{b} \right)\left (\dfrac{1+c}{c} \right)$. We are given that $a+b+c = 1$, so substituting that in we get $\left (\dfrac{b+c+2a}{a} \right)\left (\dfrac{a+c+2b}{b} \right)\left (\dfrac{a+b+2c}{c} \right)$. Then do I say $\left (\dfrac{b+c+2a}{a} \right)\left (\dfrac{a+c+2b}{b} \right)\left (\dfrac{a+b+2c}{c} \right) \geq 64$ and see if I can get a true statement from this?
| Here is a "simple" proof utilizing GM-HM inequality.
Use GM-HM inequality on the set $\{\frac{a+1}{a},\frac{b+1}{b},\frac{c+1}{c}\} $ to get:
$$\left(\,\left(\frac{a+1}{a}\right)\left(\frac{b+1}{b}\right)\left(\frac{c+1}{c}\right)\,\right)^\frac{1}{3} \geq \frac {3} {\left(\frac{a}{a+1}\right) +\left(\frac{b}{b+1}\right) + \left(\frac{c}{c+1}\right)} $$
$$\implies \,\left(\frac{a+1}{a}\right)\left(\frac{b+1}{b}\right)\left(\frac{c+1}{c}\right) \geq \frac {27} {\left( \,\, \left(\frac{a}{a+1}\right) +\left(\frac{b}{b+1}\right) + \left(\frac{c}{c+1}\right)\,\,\right)^{3}} \,\,\,(♦)$$
Now, maximizing $\left(\frac{a}{a+1}\right) +\left(\frac{b}{b+1}\right) + \left(\frac{c}{c+1}\right)$ according to the constraint $a+b+c=1$ , its maximum value comes out to be $\frac{3}{4}$.
$$\therefore \left(\frac{a}{a+1}\right) +\left(\frac{b}{b+1}\right) + \left(\frac{c}{c+1}\right) \leq \frac {3}{4}\,\,\,(♣)$$
Plug $(♣)$ in $(♦)$ to get :
$$\left(\frac {a+1}{a}\right) \left( \frac{b+1}{b}\right) \left(\frac{c+1}{c}\right) \geq \frac {27} {(3/4)^3} =\frac {27\times 64}{27} = 64$$
$$OR$$
$$\left(\frac{1}{a}+1\right)\left(\frac{1}{b}+1\right)\left(\frac{1}{c}+1\right) \geq 64$$
as desired.
$$HENCE \,\,\,\, PROVED$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 7,
"answer_id": 2
} |
simple proof that $\sqrt{1+\frac{1}{x+1/2}}(1+1/x)^x\le e$ It is well known that for $x>0$ that $\left(1+\frac{1}{x}\right)^x\le e\le\left(1+\frac{1}{x}\right)^{x+1}$ (see wikipedia). However, one can obtain the stronger inequality
$$
\sqrt{1+\frac{1}{x+\frac{1}{2}}}\left(1+\frac{1}{x}\right)^x\le e\le\sqrt{1+\frac{1}{x}}\left(1+\frac{1}{x}\right)^{x}
$$
The second inequality can be found in Proposition B.3 of "Randomized Algorithms", by Raghaven and Motwani (which itself refers to the book "Analytic Inequalities" by Mitrinović) , and can be proven straight-forwardly by calculus (showing a first derivative is non-negative and such).
While I can also prove the first inequality using familiar calculus methods, it is a bit messy (ultimately requiring that $\frac{1}{y+2}+\frac{1}{3y+2}\ge \frac{1}{y+1}$ for $y\ge 0$).
Does anyone know a "simple" proof of $\sqrt{1+\frac{1}{x+\frac{1}{2}}}\left(1+\frac{1}{x}\right)^x\le e$?
| In response to @WimC, I'm posting my own proof of this inequality. The first observation is to see that $\sqrt{1+\frac{1}{x+1/2}}(1+1/x)^x\le e$ for $x\ge 0$ is equivalent to (under $y:=1/x$), $$\left(1+\frac{2y}{y+2}\right)^{y/2}(1+y)\le e^y$$ for $y>0$. One can then hope to establish this inequality for all $y \ge 0$.
Taking logarithms, this reduces to showing that $$f(y):=y-\ln(y+1)-\frac{y}{2}\ln(3y+2)+\frac{y}{2}\ln(y+2)$$ has $f(y)\ge 0$ for $y\ge 0$. As $f(0)=0$, one needs only show the derivative $f'(y)\ge 0$ for $y\ge 0$. Taking the derivative, one finds that
\begin{align*}
f'(y)&=1-\frac{1}{y+1}-\frac{1}{2}\ln\left(1+\frac{2y}{y+2}\right)+\frac{1}{2}\cdot\frac{y}{y+2}-\frac{1}{2}\cdot \frac{3y}{3y+2}\\
&\ge 1-\frac{1}{y+1}-\frac{1}{2}\cdot \frac{2y}{y+2}+\frac{1}{2}\cdot\frac{y}{y+2}-\frac{1}{2}\cdot \frac{3y}{3y+2}\\
&=\frac{y^2}{(y+1)(y+2)(3y+2)}
\end{align*}
where we have used $\ln(1+z)\le z$, then simplified. This last expression is clearly non-negative, as desired.
My issue with the above proof is that the algebra is messy, and I don't gain any intuition on why the inequality is true.
In particular, I would ideally like to be able to get the best $\alpha$ so that
$$\sqrt{1+\frac{1}{x+1/2}}(1+1/x)^x\le e$$
is true for all $x>0$. Using the above $y:=1/x$, this is equivalent to
$$\sqrt{1+\frac{1}{1/y+\alpha}}(1+y)^{1/y}\le e$$
If we take $y\rightarrow \infty$ then as $1+y\sim e^{\ln y}$ we get $(1+y)^{1/y}\rightarrow 1$, so that we must have that for large $y$ that
$$\sqrt{1+\frac{1}{0+\alpha}}\le e$$
Rearranging, this says that $\alpha$ must obey $\alpha\ge\frac{1}{e^2-1}\approx\frac{1}{6.39}$ for large $y$. To me it seems natural to expect that any such $\alpha$ works for all $y$, but that might be naive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1592345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
How find the sum of the last two digits of $(x^{2})^{2013} + \frac{1}{(x^{2})^{2013}}$ for $x + \frac{1}{x} = 3$? Let x be a real number so that $x + \frac{1}{x} = 3$. How find the sum of the last two digits of $(x^{2})^{2013} + \frac{1}{(x^{2})^{2013}}$?
| I haven't solved this yet but my approach so far is along the lines of:
$$\begin{align}
x+\frac{1}{x}&=3\\
\therefore \left(x+\frac{1}{x}\right)^2&=3^2\\
\therefore x^2+\frac{1}{x^2}+2&=9\implies x^2+\frac{1}{x^2}=7
\end{align}$$
Now we make use of this result and notice the following pattern:
$$\begin{align}
x^2+\frac{1}{x^2}&=7\\
x^4+\frac{1}{x^4}&=\left(x^2+\frac{1}{x^2}\right)\left(x^2+\frac{1}{x^2}\right)-2&&=7\left(x^2+\frac{1}{x^2}\right)-2\\
x^6+\frac{1}{x^6}&=\left(x^2+\frac{1}{x^2}\right)\left(x^4+\frac{1}{x^4}\right)-\left(x^2+\frac{1}{x^2}\right)&&=7\left(x^4+\frac{1}{x^4}\right)-\left(x^2+\frac{1}{x^2}\right)\\
x^8+\frac{1}{x^8}&=\left(x^2+\frac{1}{x^2}\right)\left(x^6+\frac{1}{x^6}\right)-\left(x^4+\frac{1}{x^4}\right)&&=7\left(x^6+\frac{1}{x^6}\right)-\left(x^4+\frac{1}{x^4}\right)\\
&\cdots\\
x^{2n}+\frac{1}{x^{2n}}&=\left(x^2+\frac{1}{x^2}\right)\left(x^{2(n-1)}+\frac{1}{x^{2(n-1)}}\right)-\left(x^{2(n-2)}+\frac{1}{x^{2(n-2)}}\right)&&=7\left(x^{2(n-1)}+\frac{1}{x^{2(n-1)}}\right)-\left(x^{2(n-2)}+\frac{1}{x^{2(n-2)}}\right)\\
\end{align}$$
If we now let:$$a_n=x^{2n}+\frac{1}{x^{2n}}$$then this can be written as the following recurrence relation:$$a_n=7a_{n-1}-a_{n-2}$$which can be solved to yield:$$a_n=x^{2n}+\frac{1}{x^{2n}}=\left(\frac{7+3\sqrt{5}}{2}\right)^{n}+\left(\frac{7-3\sqrt{5}}{2}\right)^{n}, n=0,1,2,3,\cdots$$
We now need to compute the sum of last two digits for $n=2013$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1592663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Simple question on trigonometry identities of sec and tan Please, I want to know different methods to prove following identity
$$\frac{\tan \theta + \sec\theta - 1}{\tan\theta-\sec\theta + 1}=\frac{1+\sin\theta}{\cos\theta}$$
| Multiplying by $\cos\theta$ the num and the denom
$$\frac{\tan \theta+\sec \theta-1}{\tan \theta-\sec \theta+1} = \frac{\sin \theta+1-\cos \theta}{\sin \theta-1+\cos \theta}$$
Now setting the equality and deleting the fractions
\begin{align}
\frac{\sin \theta+1-\cos \theta}{\sin \theta-1+\cos \theta}&=\frac{1+\sin \theta}{\cos \theta}\\
\cos \theta + \cos \theta - \cos^2 \theta &= \sin \theta + \sin^2 \theta-\sin \theta-1+ \cos \theta+\sin \theta \cos \theta\\
-\cos^2 \theta &=\sin^2 \theta-1.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
How to solve $\lim\limits_{x \to -\infty} \left(x\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)\right)$? I have a problem with this limit, i have no idea how to compute it.
Can you explain the method and the steps used?
$$\lim\limits_{x \to -\infty} \left(x\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)\right)$$
| Given
$$\lim_{x\to-\infty}(x(\sqrt{x^2-x}-\sqrt{x^2-1}))$$
As
$$x(\sqrt{x^2-x}-\sqrt{x^2-1}) = \frac{x(1-x)}{\sqrt{x^2-x}+\sqrt{x^2-1}}$$
$$=\frac{x-x^2}{\sqrt{x^2-x}+\sqrt{x^2-1}}$$
$$=\frac{\frac{1}{x}-1}{\sqrt{\frac{1}{x^2}-\frac{1}{x^3}}+\sqrt{\frac{1}{x^2}-\frac{1}{x^4}}}$$
$$\lim_{x\to-\infty}\frac{\frac{1}{x}-1}{\sqrt{\frac{1}{x^2}-\frac{1}{x^3}}+\sqrt{\frac{1}{x^2}-\frac{1}{x^4}}}=-\infty$$
As $x\to-\infty$ then $\frac{1}{x},\frac{1}{x^2},\frac{1}{x^3},\frac{1}{x^4}\to0$ And $\frac{-1}{0}=-\infty$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 7
} |
Prove that $\frac{a}{(1+b)(1+c)}+\frac{b}{(1+a)(1+c)}+\frac{c}{(1+a)(1+b)} \geq \frac{3}{4}.$
Let $a,b,c$ be positive numbers that satisfy $abc = 1$, prove that $$\dfrac{a}{(1+b)(1+c)}+\dfrac{b}{(1+a)(1+c)}+\dfrac{c}{(1+a)(1+b)} \geq \dfrac{3}{4}.$$
Attempt
I tried doing $$\dfrac{a}{(1+b)(1+c)}+\dfrac{b}{(1+a)(1+c)}+\dfrac{c}{(1+a)(1+b)} = \dfrac{1}{bc(1+b)(1+c)}+\dfrac{1}{ac(1+a)(1+c)}+\dfrac{c}{bc(1+a)(1+b)}$$ then using the fact that $a^2+b^2+c^2 \geq ab+bc+ca$ but that only seemed to get me $$\dfrac{a}{(1+b)(1+c)}+\dfrac{b}{(1+a)(1+c)}+\dfrac{c}{(1+a)(1+b)} \geq \dfrac{1}{bc+bc^2+c+1}+\dfrac{1}{ac+a^2c+c+1}+\dfrac{1}{ab+ab^2+b+1}.$$
| The inequality is equivalent to
$$4(a(1+a)+b(1+b)+c(1+c))\ge 3(1+a)(1+b)(1+c).\tag{1}$$
Since
$$(1+a)(1+b)(1+c) = 1+ (a+b+c)+(ab+bc+ca)+abc$$
and $abc=1$, Inequality (1) is equivalent to
$$4(a^2+b^2+c^2)+(a+b+c) \ge 6+3(ab+bc+ca).$$
The last inequality follows from $a^2+b^2+c^2\ge ab+bc+ca$ and AM-GM.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Evalute $\int\frac{3x-2}{x^2-4x+5} \, dx$ Evalute $$\int\frac{3x-2}{x^2-4x+5} \, dx$$
So, I tried using $x^2-4x+5 = t$, which unfortunately didn't help much.
I also tried making the denominator $(x-2)^2 +1$ and then using $x-2 = t$, which made everything messy. Any suggestions? It looks so simple, and I bet my first approach is right.
| Hint. You may write the integrand as
$$
\frac{3x-2}{x^2-4x+5}=\frac32\frac{2x-4}{x^2-4x+5}+\frac{4}{x^2-4x+5}
$$ giving
$$
\frac{3x-2}{x^2-4x+5}=\frac32\frac{(x^2-4x+5)'}{x^2-4x+5}+\frac{4}{(x-2)^2 +1}
$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Solve $ \int{\frac{7x^2 + 1}{(x+1)(x-1)(x+3)}}\,dx $ I don’t know how to solve this integral:
$$\int{\frac{7x^2 + 1}{(x+1)(x-1)(x+3)}}\,dx$$
I know this is a rational integral but I don’t know how to write it in a different way
| Hint:
Partial Fractions yields:
$$\frac{(7x^2 + 1)}{(x+1)(x-1)(x+3)}=\frac{-2}{x+1}+\frac{1}{x-1}+\frac{8}{x+3}$$
Therefore,
$$\int \frac{(7x^2 + 1)}{(x+1)(x-1)(x+3)}=\int \left(\frac{-2}{x+1}+\frac{1}{x-1}+\frac{8}{x+3}\right)\,dx$$
$$=-2\log|x+1|+\log|x-1|+8\log|x+3|+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
how to find the roots of the following floor-equation: How to find the roots of $$\lfloor x\rfloor+\lfloor 2x\rfloor+\lfloor 3x\rfloor=6$$
| As my previous solution was not complete i would like to write a new one.
Let assume $\left\lfloor x \right\rfloor=n$ so that we know $n\le x<n+1$ and let $\delta$ the fractional part of x, so that $\delta \epsilon \left( 0,1 \right) $ then
$$\left\lfloor x \right\rfloor +\left\lfloor 2x \right\rfloor +\left\lfloor 3x \right\rfloor =6\quad \Longrightarrow \quad \left\lfloor n+\delta \right\rfloor +\left\lfloor 2\left( n+\delta \right) \right\rfloor +\left\lfloor 3\left( n+\delta \right) \right\rfloor =6$$
obviosly,it is equal to $$\left\lfloor n+\delta \right\rfloor +\left\lfloor 2n+2\delta \right\rfloor +\left\lfloor 3n+3\delta \right\rfloor =6$$
Due to the fact that $n$,$2n$ and $3n$ is an integer we can write it as follows:
$$n+\left\lfloor \delta \right\rfloor +2n+\left\lfloor 2\delta \right\rfloor +3n+\left\lfloor 3\delta \right\rfloor =6$$
Now,there are three sub-cases,i.e. we will divide $\left( 0,1 \right) $ in to three parts by $\delta $:
$1$.If $0\le \delta <\frac { 1 }{ 3 } $ then
$$
\left\lfloor n+\delta \right\rfloor +\left\lfloor 2n+2\delta \right\rfloor +\left\lfloor 3n+3\delta \right\rfloor =6
\Rightarrow n+2n+3n=6 \Rightarrow n=1 $$
so
$$1\le x<\frac { 4 }{ 3 } $$
$2.$ If $\frac { 1 }{ 3 } \le \delta <\frac { 2 }{ 3 } $ then
$$
n+\left\lfloor \delta \right\rfloor +2n+\left\lfloor 2\delta \right\rfloor +3n+\left\lfloor 3\delta \right\rfloor =6 \Rightarrow
n+2n+3n+1=6 \Rightarrow n=\frac { 5 }{ 6 }
$$
but n should be integer
$3.$ If $\frac { 2 }{ 3 } \le \delta <1$ then
$$
n+\left\lfloor \delta \right\rfloor +2n+\left\lfloor 2\delta \right\rfloor +3n+\left\lfloor 3\delta \right\rfloor =6 \Rightarrow
n+2n+1+3n+2=6 \Rightarrow n=\frac { 1 }{ 2 }
$$ in that case n is not integer either.
So our final answer is
$$1\le x<\frac { 4 }{ 3 } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to factor $x^6+x^5-3\,x^4+2\,x^3+3\,x^2+x-1$ by hand? I know that
$x^6+x^5-3\,x^4+2\,x^3+3\,x^2+x-1 = (x^4-x^3+x+1)(x^2+2x-1)$
but I would not know how to do that factoring without a software.
Some idea? Thank you!
| A possibility is to write a generic product
$x^6+x^5-3x^4+2x^3+3x^2+x-1=(x^4+a_3x^3+a_2x^2+a_1x+a_0)(x^2+b_1x+b_0)$
then expand the right-hand side and compare the two polynomials, obtaining
\begin{cases}
&a_0 b_0=-1,\\
&a_0 b_1+a_1 b_0=1,\\
&a_0+a_1 b_1+a_2 b_0=3,\\
&a_1+a_2 b_1+a_3 b_0=2,\\
&a_2+a_3 b_1+b_0=-3,\\
&a_3+b_1=1
\end{cases}
that can be solved rather easily for integer solutions.
The same could be tried for a product of two third degree polynomials, without any (integer) solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 1
} |
How to compute $\lim\limits_{x \to +\infty} \left(\frac{x^4+x^5\sin\left(\frac{1}{x}\right)}{x^4\ln\left(\frac{x}{2x+1}\right)-x} \right)$? I have a problem with this limit, I don't know what method to use.
Can you show a method for the resolution ?
$$\lim\limits_{x \to +\infty} \left(\frac{x^4+x^5\sin\left(\frac{1}{x}\right)}{x^4\ln\left(\frac{x}{2x+1}\right)-x} \right)$$ Thanks
| Note that we can write the relationships
$$\sin\left(\frac1x\right)=\frac1x+O\left(\frac1{x^3}\right) \tag 1$$
and
$$\begin{align}
\log\left(\frac{x}{2x+1}\right)&=-\log 2+\log\left(1-\frac{1}{2x+1}\right)\\\\
&=-\log 2-O\left(\frac{1}{2x+1}\right) \tag 2
\end{align}$$
Using $(1)$ and $(2)$ reveals
$$\begin{align}
\frac{x^4+x^5\sin\left(\frac1x\right)}{x^4\log\left(\frac{x}{2x+1}\right)-x}&=\frac{2x^4+O\left(x^2\right)}{-\log(2)x^4+O\left(x^3\right)}\\\\
&=-\frac{2}{\log(2)}+O\left(\frac1x\right)\\\\
&\to -\frac{2}{\log(2)}\,\,\text{( as}\,\,x\to \infty\,\text{)}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1598600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Evaluating the integral $\int_{-\infty}^\infty x^2\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2} \ dx$
How does one evaluate $$\int_{-\infty}^\infty x^2\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2} \ dx ?$$
The result is $1$ and it corresponds to $E[X^2]$, where $X$ is a random variable with $X\sim\mathcal{N}(0,1)$. I have tried to do some substituions and I've tried integration by parts but didn't succeed to integrate it. With the integration by parts I ended up with a harder integral in both cases and I couldn't find a good substitution.
| Here are two solutions that exploit the same two tools in different orders:
*
*It's a standard trick using elementary multivariable calculus and converting to polar coordinates (see, e.g., this answer) to show that the Gaussian integral has value
$$\require{cancel}\int_{-\infty}^{\infty} e^{-t^2} dt = \sqrt{\pi} .$$
Now, applying integration by parts with $u = e^{-t^2}, dv = dt$ to the above integral gives $du = -2 t e^{-t^2} dt, v = t$ and hence
$$\sqrt{\pi} = \cancelto{0}{\left.\left(e^{-t^2}\right)(t)\right\vert_{-\infty}^{\infty}} - \int_{-\infty}^{\infty} (t)(-2t e^{-t^2}) dt = 2 \int_{-\infty}^{\infty} t^2 e^{-t^2} dt,$$
and so
$$\int_{-\infty}^{\infty} t^2 e^{-t^2} dt = \frac{\sqrt{\pi}}{2} .$$
Now, substituting $t = \frac{x}{\sqrt{2}}, dt = \frac{dx}{\sqrt{2}}$ in the integral gives
$$\frac{1}{2 \sqrt{2}} \int_{-\infty}^{\infty} x^2 e^{-\frac{x^2}{2}} dx = \frac{\sqrt{\pi}}{2},$$
and multiplying by $\frac{2}{\sqrt{\pi}}$ gives the desired result:
$$\color{#bf0000}{\boxed{\int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} x^2 e^{-\frac{x^2}{2}} dx = 1}} .$$
*We can apply the "multivariable polar" trick directly to the integral at hand rather than to the Gaussian integral: If one denotes $$I := \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} x^2 e^{-\frac{x^2}{2}} dx ,$$ we have
$$
I^2
= \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} x^2 e^{-\frac{x^2}{2}} dx \cdot \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} y^2 e^{-\frac{y^2}{2}} dy
= \frac{1}{2 \pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} x^2 y^2 e^{-\frac{x^2 + y^2}{2}} dx \, dy .
$$
Now, converting to polar coordinates gives
\begin{align}
I^2
&= \frac{1}{2 \pi} \int_0^{2 \pi} \int_0^{\infty} (r \cos \theta)^2 (r \sin \theta)^2 e^{-\frac{r^2}{2}} \cdot r \,dr \,d\theta \\
&= \frac{1}{2 \pi} \int_0^{2 \pi} \int_0^{\infty} r^5 e^{-\frac{r^2}{2}} \sin^2 \theta \cos^2 \theta \,dr \,d\theta \\
&= \frac{1}{2 \pi} \int_0^{2 \pi} \sin^2 \theta \cos^2 \theta \,d\theta \int_0^{\infty} r^5 e^{-\frac{r^2}{2}} dr .
\end{align}
Now, the first integral is a standard exercise that can be handled with double-angle identities and has value $\frac{\pi}{4}$. The second integral can be evaluated readily with the substitution $u = -\frac{r^2}{2}, du = -r \,dr$ and has value $8$, so
$$I^2 = \frac{1}{2 \pi} \left(\frac{\pi}{4}\right) (8) = 1 .$$
On the other hand, the integrand in $I$ is everywhere nonnegative, and so again $I$ has value
$$\color{#bf0000}{\boxed{\int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} x^2 e^{-\frac{x^2}{2}} dx = 1}}$$
as claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
How to calculate $\sum_{n \in P}\frac{1}{n^2}, P=\{n \in \mathbb{N}: \exists (a,b) \in\ \mathbb{N^+} \times \mathbb{N^+} \mbox{ with } a^2+b^2=n^2\}$ How I can evaluate
$$\sum_{n \in P}\frac{1}{n^2} \quad \quad P=\{n \in \mathbb{N^+}: \exists (a,b) \in\ \mathbb{N^+} \times \mathbb{N^+} \mbox{ with } a^2+b^2=n^2\}$$
It's clearly convergent. I thought about seeing the sum as a sum of complex numbers using $(a+ib)(a-ib)=a^2+b^2$.
| We just have to understand which numbers $n\in\mathbb{N}^+$ are such that $n^2$ cannot be written as $a^2+b^2$ with $a,b\in\mathbb{N}^+$. If there is some prime $p\equiv 1\pmod{4}$ that divides $n$, such prime splits in $\mathbb{Z}[i]$ (the ring of gaussian integers), hence $n^2$ can be represented for sure as $a^2+b^2$ with $a,b\neq 0$. If $n$ is even then $n^2\equiv 0\pmod{4}$, hence $n^2=a^2+b^2$ gives that both $a$ and $b$ are even and the problem boils down to representing $(n/2)^2$. At last we have that the only numbers whose square cannot be represented as $a^2+b^2$ with $a,b\neq 0$ are the ones whose prime factors lie in the set made by $2$ and the primes $\equiv 3\pmod{4}$. That gives:
$$ \sum_{n\in P}\frac{1}{n^2} = \frac{\pi^2}{6}-\!\!\!\!\prod_{p\in\{2,3,7,\ldots\}}\left(1-\frac{1}{p^2}\right)^{-1}=\frac{\pi^2}{6}-\frac{4}{3}\cdot\!\!\!\prod_{p\equiv 3\!\!\pmod{4}}\left(1-\frac{1}{p^2}\right)^{-1}.\tag{1}$$
Now, let $\chi$ be the non-principal Dirichlet character $\!\!\pmod{4}$ and $L(s,\chi)$ the associated Dirichlet L-function. We have:
$$ L(2,\chi) = \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^2} = \prod_{p}\left(1-\frac{\chi(p)}{p^2}\right)^{-1},\tag{2} $$
$$ L(4,\chi) = \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^4} = \prod_{p}\left(1-\frac{\chi(p)}{p^4}\right)^{-1},\tag{3} $$
$$ \frac{L(4,\chi)}{L(2,\chi)}=\prod_{p}\left(1+\frac{\chi(p)}{p^2}\right)^{-1}=\prod_{p\equiv 3\!\!\pmod{4}}\left(1-\frac{1}{p^2}\right)^{-1}\prod_{p\equiv 1\!\!\pmod{4}}\left(1+\frac{1}{p^2}\right)^{-1}\tag{4}$$
hence:
$$ S= \sum_{n\in P}\frac{1}{n^2}=\frac{\pi^2}{6}-\frac{4\, L(4,\chi)}{3\, L(2,\chi)}\cdot\!\!\prod_{p\equiv 1\!\!\pmod{\!\! 4}}\!\!\left(1+\frac{1}{p^2}\right).\tag{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How to compute $\int_0^{\frac{\pi}{2}} \frac{\ln(\sin(x))}{\cot(x)}\ \text{d}x$ I am trying to compute this integral.
$$\int_0^{\frac{\pi}{2}} \frac{\ln(\sin(x))}{\cot(x)}\ \text{d}x$$
Any thoughts will help. Thanks.
| Note that
\begin{eqnarray}
&&\int_0^{\frac{\pi}{2}} \frac{\ln(\sin(x))}{\cot(x)}\ \text{d}x\\
&=&\int_0^{\frac{\pi}{2}}(\cos x)^{-1}\sin x\ln(\sin x)\ \text{d}x=\lim_{a\to0,b\to2}\frac{d}{db}\int_0^{\frac{\pi}{2}}(\cos x)^{a-1}(\sin x)^{b-1}\ \text{d}x\\
&=&\lim_{a\to0,b\to2}\frac{d}{db}\frac{1}{2}B\left(\frac{a}{2},\frac{b}{2}\right)\\
&=&\lim_{a\to0,b\to2}\frac{1}{4}B\left(\frac{a}{2},\frac{b}{2}\right)\left(\psi\left(\frac{b}{2}\right)-\psi\left(\frac{a+b}{2}\right)\right)\\
&=&\lim_{a\to0}\frac{1}{4}B\left(\frac{a}{2},1\right)\left(\psi\left(\frac{b}{2}\right)-\psi\left(\frac{a+2}{2}\right)\right)\\
&=&\lim_{a\to0}\frac{1}{4}B\left(\frac{a}{2},1\right)\left(\psi(1)-\psi\left(\frac{a}{2}+1\right)\right)\\
&=&-\lim_{a\to0}\frac{1}{4}B\left(\frac{a}{2},1\right)\left(\gamma+\psi\left(\frac{a}{2}+1\right)\right)\\
&=&-\lim_{a\to0}\frac{1}{4}\frac{\Gamma(\frac{a}{2})\Gamma(1)}{\Gamma(\frac{a+2}{2})}\left(\gamma+\psi\left(\frac{a}{2}+1\right)\right)\\
&=&-\lim_{a\to0}\frac{1}{4}\left(\frac{2}{a}+O(a^2)\right)\left(\frac{\pi^2}{12}a+O(a^2)\right)\\
&=&-\frac{\pi^2}{24}.
\end{eqnarray}
Here we use the following fact
$$ \Gamma(a)\approx\frac{1}{a}, \gamma+\psi(\frac{a}{2}+1)\approx\frac{\pi^2}{12}a.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Prove that $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}} \leq \frac{3}{2}.$
Let $a,b,$ and $c$ be positive real numbers such that $a+b+c = abc$. Prove that $$\dfrac{1}{\sqrt{1+a^2}}+\dfrac{1}{\sqrt{1+b^2}}+\dfrac{1}{\sqrt{1+c^2}} \leq \dfrac{3}{2}.$$
This question seems tricky because how do we incorporate the $a+b+c = abc$ condition to solve this? I was thinking of solving for the variables and substituting like $a = \dfrac{b+c}{1-bc}$ etc., but I am not sure how that will help.
| Change the notation using $x=\frac{1}{a}$ , $y=\frac{1}{b}$ and $z=\frac{1}{c}$
Now the condition is that $xy+yz+zx=1$.
And the inequality to prove is :
$$\frac{x}{\sqrt{x^2+1}}+\frac{y}{\sqrt{y^2+1}}+\frac{z}{\sqrt{z^2+1}} \leq \frac{3}{2}$$
Now use the condition and notice that :
$$x^2+1=x^2+xy+yz+zx=(x+y)(x+z)$$
So now using the AM-GM inequality :
$$\frac{x}{\sqrt{x^2+1}}=\frac{x}{\sqrt{(x+y)(x+z)}} =\sqrt{\frac{x^2}{(x+y)(x+z)}} \leq \frac{1}{2} \left (\frac{x}{x+y}+\frac{x}{x+z} \right )$$
Do the same thing for the other terms and add them to get the conclusion .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Dilemma about the value of $\frac{4- 4}{4 - 4}$ I can't find where the mistake is here. Can someone explain how it is possible?
| The expression $\frac{4-4}{4-4}$ has no well-defined value because it is just $\frac{0}{0}$.
I like the trick of factorising the numerator and denominator to give $\frac{(2-2)(2+2)}{2(2-2)}$.
However, when you try to cancel $(2-2)$ from both the numerator and denominator, you must ask yourself what it means to "cancel". When you "cancel the twos" to give $\frac{2}{6} = \frac{1}{3}$, what have you done? You have divided both the numerator and denominator by 2.
When you try to cancel the $(2-2)$ in your fraction, you are dividing both the numerator and denominator by $(2-2)$. The problem is $2-2=0$, and so you have divided both the numerator and denominator by zero. It's bad enough dividing by zero once, but you did it twice!
This kind of cancellation can work when you are dealing with limits. For example, what happens to $\frac{x^2}{x}$ as $x$ gets very small. It makes sense for all $ x \neq 0$. For example, when $x=3$, $\frac{x^2}{x}=\frac{9}{3}=3$. When $x=2$, $\frac{x^2}{x}=\frac{4}{2}=2$. When $x=1$, $\frac{x^2}{x}=\frac{1}{1}=1$. What about $x=0$? When $x=0$, $\frac{x^2}{x}$ is not well-defined and takes no value. However, we can first pretend that $x \neq 0$ and get $\frac{x^2}{x} \equiv x$. Then as $x \to 0$, we say that
$$\lim_{x \to 0} \frac{x^2}{x} = \lim_{x \to 0} x = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1606159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the value of the infinite product $\sqrt\frac12\cdot\sqrt{\frac12+\sqrt\frac12}\cdot\sqrt{\frac12+\sqrt{\frac12+\sqrt\frac12}}\cdots$ Find $\sqrt{\frac{1}{2}}.\sqrt{\frac{1}{2}+\sqrt\frac{1}{2}}.\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt\frac{1}{2}}}....\infty$
Let $x=\sqrt{\frac{1}{2}}.\sqrt{\frac{1}{2}+\sqrt\frac{1}{2}}.\sqrt{\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt\frac{1}{2}}}....\infty$
$\log x=\frac{1}{2}\log(\frac{1}{2})+\frac{1}{2}\log(\frac{1}{2}+\sqrt\frac{1}{2})+\frac{1}{2}\log(\frac{1}{2}+\sqrt{\frac{1}{2}+\sqrt\frac{1}{2}})+....$
I do not know how to solve it further.
| Denote the $n$-th factor by $x_n$. The sequence $(x_n)$ is increasing, and
$$
x_2 = \sqrt{\frac{1}{2}+\sqrt\frac{1}{2}} > \sqrt{\frac{1}{2}+\frac{1}{2}} = 1
$$
Therefore
$$
x_1 x_2 \cdots x_n \ge x_1 x_2^{n-1} \to \infty \, .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1606695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Relation between areas of regular polygons wrt constant height. If we have a regular $n$-gon with height 1 (midpoint to furthest vertex for odd-gons/midpoints to midpoints for even-gons), how does the area of different regular $n$-gons compare to each other from triangle (3-gon) to circle ($\infty$-gon)?
Is it true that for even-gons, the area decreases with increasing $n$, and that for odd gons, area increases with increasing $n$?
Can you put the areas in an ascending order?
| Lines from the center of a regular polygon, with n sides, divide it into n isosceles triangles with central angle $\frac{2\pi}{n}$. If we take the length of a side to be s and the length of the two equal sides of the triangle to be l, then, by the cosine law, the $s^2= l^2+ l^2- 2l^3 cos\left(\frac{2\pi}{n}\right)= 2l^2(1- cos\left(\frac{2\pi}{n}\right)$. Further, taking the length of a line from the center of the polygon to the midpoint of a side to be h, by the Pythagorean theorem, $l^2= h^2+ \frac{s^2}{4}$ so that $l^2= h^2+ \frac{s^2}{4}= h^2+ \frac{1}{2}l^2\left(1- cos\left(\frac{2\pi}{n}\right)\right)$ so that $h^2= \frac{1}{2}l^2\left(1+ cos\left(\frac{2\pi}{n}\right)\right)$ and, finally, $h= l\sqrt{\frac{1}{2}\left(1+ cos\left(\frac{2\pi}{n}\right)\right)}$.
Now in an odd sided polygon, with your "height" equal to 1, $h= \frac{1}{2}$ and $l= \frac{4}{1+ cos(\left(\frac{2\pi}{n}\right)}$ so that $s= 2\sqrt{l^2- h^2}= 2l\sqrt{\frac{1}{2}+ \frac{1}{2}cos\left(\frac{2\pi}{n}\right)}$. The area of each of those n triangles is $\frac{1}{2}sl= \sqrt{\frac{1}{2}+ \frac{1}{2}cos\left(\frac{2\pi}{n}\right)}$ and the area of the entire polynomial is $n\sqrt{\frac{1}{2}+ \frac{1}{2}cos\left(\frac{2\pi}{n}\right)}$.
In an even sided polynomial, $l+ h= l\left(1+ \sqrt{1+ cos\left(\frac{2\pi}{n}\right)}\right)= 1$ so $l= \frac{1}{\left(1+ \sqrt{1+ cos\left(\frac{2\pi}{n}\right)}\right)}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1608790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
P is a natural number. 2P has 28 divisors and 3P has 30 divisors. How many divisors of 6P will be there? While answering Aptitude questions in a book I faced this question but not able to find the solution So I searched Google and got two answers but didn't get an idea how the answer came.
Question:
P is a natural number. 2P has 28 divisors and 3P has 30 divisors. How many divisors of 6P will be there?
Solution 1:
2P is having 28(4*7) divisors but 3P is not having a total divisors which is divisible by 7, so the first part of the number P will be 2^5.
Similarly, 3P is having 30 (3*10) divisors but 2P does not have a total divisors which is divisible by 3. So 2nd part of the number P will be 3^3. So, P = 2^5*3^3 and the solution is 35.
Solution 2:
2P has 28 divisors =4x7,
3P has 30 divisors
Hence P=2^5 3^3
6p =2^6 3^4
Hence 35 divisors
I have been trying to understand the steps but not able to get.
| First, we want to know how to easily calculate the number of divisors of any number. If we have $n=p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$ where all $p_i$ are distinct, then to construct a divisor we have $e_1+1$ choices for the number of factors $p_1$ in our divisor, $e_2+1$ choices for the number of factors $p_2$, etc., making the number of divisors equal to $(e_1+1)(e_2+1)\cdots (e_k+1)$. So for example, $12=2^2\cdot 3$ so $12$ has $(2+1)(1+1)=6$ divisors.
Let's look at the first hint now, $2P$ having 28 divisors. Let's write $2P=p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$. The number of divisors is now $(e_1+1)(e_2+1)\cdots (e_k+1)=28$. So, we can say that one power, say $e_1$, must be either $6,13$ or $27$ (since 28 has a factor 7, and we assumed it was contained in $e_1+1$. Our options are now:
$$2P=p_1^6\cdot p_2\cdot p_3$$
$$2P=p_1^6\cdot p_2^3$$
$$2P=p_1^{13}\cdot p_2$$
$$2P=p_1^{27}$$
The second hint says that 3P has 30 divisors. Since this does not contain a factor 7, we know that the $2$ in $2P$ must be responsible for this (notice that follows $e_1$ is the power of 2 in $2P$). Thus we know that $p_1=2$. Now we can, by our previously stated options, calculate P.
$$P=2^5\cdot p_2\cdot p_3$$
$$P=2^5\cdot p_2^3$$
$$P=2^{12}\cdot p_2$$
$$P=2^{26}$$
In the third option, the number of divisors of $3P$ must be divisible by $12+1=13$, and in the last case, the number of divisors of $3P$ must be divisible by $27$. We conclude the last two are not possible, since 30 is not divisible by either 13 or 27. In the first case, the number of divisors of $3P$ is divisible by $6$ because of the amount of factors 2, and we also know that $3P$ there has (at least) one prime factor that it only contains one time, so we have another factor 2 in the number of divisors of $3P$. Now the number of divisors of $3P$ is divisible by $12$, which is also impossible ($12$ does not divide $30$). We conclude that $P$ must be of the form $p_1^5\cdot p_2^3$. We also know that if $p_2\neq 3$, then the number of divisors of $3P$ is divisible by $6$ because of the factors 2 and by $4$ because of the factors $p_2$. This is again impossible.
Finally, we must have $p_2=3$ so $P=2^5\cdot 3^3$. Now we can easily calculate the number of divisors of $6P$; it must be $(6+1)(4+1)=35$.
Hope this helped!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1609248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
integrate $\int \frac{x-3}{\sqrt{1-x^2}}$
$$\int \frac{x-3}{\sqrt{1-x^2}} \mathrm dx$$
I know that $\int \frac{1}{1-x^2}\mathrm dx=\arcsin(\frac{x}{1})$ but how can I continue from here?
| Notice, $$\int \frac{x-3}{\sqrt {1-x^2}}\ dx$$$$=\int \frac{x}{\sqrt {1-x^2}}\ dx-\int \frac{3}{\sqrt {1-x^2}}\ dx$$
$$=-\frac{1}{2}\int \frac{(-2x)}{\sqrt {1-x^2}}\ dx-3\int \frac{1}{\sqrt {1-x^2}}\ dx$$
$$=-\frac{1}{2}\int (1-x^2)^{-1/2}\ d(1-x^2)-3\int \frac{1}{\sqrt {1-x^2}}\ dx$$
$$=-\frac{1}{2}\frac{(1-x^2)^{1/2}}{1/2}-3\sin^{-1}(x)+C$$
$$=-\sqrt{1-x^2}-3\sin^{-1}(x)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1610700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
$(x+y)(x+1)(y+1) = 3$ and $x^3 + y^3 = \frac{45}{8}$ I've came across this problem :
If $x$ and $y$ are real numbers such that
$(x+y)(x+1)(y+1) = 3$ and
$x^3 + y^3 = \frac{45}{8}$,
find $xy$.
This is what I've tried so far:
$$x^3 + y^3 = (x+y)(x^2-xy+y^2) =\frac{45}{8}$$
So $$\frac{45}{8(x^2 - xy + y^2)} = \frac{3}{(x+1)(y+1)}$$
$$24x^2 - 24xy + 24y^2 = 45xy + 45(x+y+1)$$
$$24x^2 + 24y^2 - 45(x+y+1) = 69xy$$
But this doesn't seem to help. I tend to think it is just a matter of factorization and/or substitution, but I can't get it right.
A piece of advice would be apreciated.
Thanks!
| Expand the first equation to get
$$
x^2y+xy^2+x^2+2xy+y^2+x+y = 3
$$
Note that this can be rewritten
$$
x^2y+xy^2+(x+y)^2+(x+y) = 3
$$
Multiply this by $3$ and add to the second equation to get
$$
(x+y)^3+3(x+y)^2+3(x+y) = \frac{117}{8}
$$
Add $1$ to both sides to obtain
$$
(x+y+1)^3 = \frac{125}{8} = \left(\frac{5}{2}\right)^3
$$
whereupon
$$
x+y = \frac{3}{2}
$$
This means that
$$
xy+x+y+1 = (x+1)(y+1) = 2
$$
which finally yields
$$
xy = -\frac{1}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1611093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find $\int x\ln(x^2e^{x^2})\,dx.$ How can I find the following integral $$\int x\ln(x^2e^{x^2})\,dx.$$ Which substitution may I use to solve the integral?
| Note that by applying logarithms addition rule you get
$$\begin{align}
\int x\ln(x^2e^{x^2})dx &= \int x\left(\ln(x^2) + \ln(e^{x^2})\right)dx\\
&=\int x\left(2\ln x + x^2\right)dx\\
&=\int 2x\ln x + x^3dx\\
&= 2\int x\ln x + \int x^3dx\\
&= 2\left( \frac{1}{2} x^2 \log (x)-\frac{x^2}{4} \right) + \frac{x^4}{4} + C\\
&= x^2 \log x - \frac{x^2}{2} + \frac{x^4}{4}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1612057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Hundredth Derivative From Taylor's polynomial for $\frac{x^2}{1+x^4}$ I'm trying to solve this problem:
Find the hundredth derivative (at $x=0$) from Taylor's polynomial for $\dfrac{x^2}{1+x^4}$.
I keep getting the wrong answer; can someone help?
I have tried to solve this by extracting the coefficient of $x^{100}$ in the sum
$$x^2\cdot\sum_{i=0}^{\infty}(-x^4)^i.$$
I see that the coefficient is $1$ and since, by Taylor's formula, this should be $1=\frac{f^{(100)}(0)}{100!}$ I get that the hundredth derivative is $100!$, but this doesn't appear to be correct.
| Notice that $x^4+1=\Phi_8(x)$, i.e. the roots of $x^4+1$ are the primitive eighth roots of unity $\frac{\pm 1\pm i}{\sqrt{2}}$.
The residue theorem or equivalent techniques give:
$$\begin{eqnarray*} f(x)&=&\frac{x^2}{x^4+1}\\
&=& \frac{1-i}{4\sqrt{2}}\cdot\frac{1}{x-\frac{1+i}{\sqrt{2}}}+\frac{1+i}{4\sqrt{2}}\cdot\frac{1}{x-\frac{1-i}{\sqrt{2}}}-\frac{1+i}{4\sqrt{2}}\cdot\frac{1}{x-\frac{-1+i}{\sqrt{2}}}-\frac{1-i}{4\sqrt{2}}\cdot\frac{1}{x-\frac{-1-i}{\sqrt{2}}} \end{eqnarray*}$$
hence $f^{(100)}(x)$ equals:
$$-100!\cdot\left[\frac{1-i}{4\sqrt{2}}\cdot\frac{1}{\left(x-\frac{1+i}{\sqrt{2}}\right)^{101}}+\frac{1+i}{4\sqrt{2}}\cdot\frac{1}{\left(x-\frac{1-i}{\sqrt{2}}\right)^{101}}-\frac{1+i}{4\sqrt{2}}\cdot\frac{1}{\left(x-\frac{-1+i}{\sqrt{2}}\right)^{101}}-\frac{1-i}{4\sqrt{2}}\cdot\frac{1}{\left(x-\frac{-1-i}{\sqrt{2}}\right)^{101}}\right].$$
Since $f(x)$ is an analytic function and the radius of convergence of its Taylor series at $x=0$ is one,
$$ f(x) = \frac{x^2}{1+x^4} = x^2-x^6+x^{10}-x^{14}+\ldots = \sum_{n\geq 0}(-1)^{n} x^{4n+2} $$
for any $x$ such that $|x|<1$. That also gives:
$$ f^{(100)}(x) = 100!\cdot\sum_{n\geq 25}(-1)^n\binom{4n+2}{100} x^{4n+2-100} $$
for any $|x|<1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1612272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find $6^{1000} \mod 23$
Find $6^{1000} \mod 23 $
Having just studied Fermat's theorem I've applied $6^{22}\equiv 1 \mod 23 $, but now I am quite clueless on the best way to proceed.
This is what I've tried:
Raising everything to the $4th$ power I have $$6^{88} \equiv 1 \mod 23 $$ $$6^{100} \equiv 6^{12} \mod 23 $$ $$6^{1000}\equiv 6^{120} \mod 23 $$ $$6^{1000} \equiv 6^{10} \mod 23$$ How do I simplify now the right hand side of the congruence ?
| Since, by Fermat, $6^{22}\equiv1\pmod{23}$, we can try and see whether the order is $11$. Write $11=1+2+8$; then
$$
6^{11}\equiv 6\cdot 6^2\cdot ((6^2)^2)^2\pmod{23}
$$
Compute $6^2\equiv 13$, $13^2\equiv 8$ and $8^2\equiv18$, so
$$
6^{11}\equiv 6\cdot 13\cdot 18\equiv1
$$
Therefore, since $1000\equiv 10\pmod{11}$, we just observe that, since $6\cdot 4=24\equiv1\pmod{23}$, we have $6^{10}\equiv 4\pmod{23}$ (because the order is $11$, as shown before).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
A circulant matrix and its transpose It is well-known that a circulant matrix $A$ of size $n \times n$ is isomorphic to a polynomial $$p(x) \bmod x^n - 1.$$ If we consider the transpose $A^T$, what is the corresponding polynomial called? Is there something such as transpose of a polynomial?
| Using the equipment provided by @CarlLöndahl, another way of presenting that polynomial's transpose is
$a(x)^T = a(x^{-1}) \mod (x^n - 1)$
Furthermore, the product
$a(x)^T a(x) \sim A^TA$
is also an $n \times n$ circulant, one which is of some interest in autocorrelation modelling, and even musical Pitch Class Theory.
It allows you to evaluate the difference distribution of a set. For example where $a(x) = 1 + x + x^3 + x^4$ represents a set of integers, $\{0, 1, 3, 4 \}$ from $\mathbb{Z}_6$.
The multiset of differences is
$\{ 1-0, 3-0, 4-0, 3-1, 4-1, 4-3 \} \cup \{ 0-1, 0-3, 0-4, 1-3, 1-4, 3-4 \}$
$= \{ 1, 3, 4, 2, 3, 1 \} \cup \{ -1, -3, -4, -2, -3, -1 \}$
$\equiv_6 \{ 1, 3, 4, 2, 3, 1 \} \cup \{ 5, 3, 2, 4, 3, 5 \}$
$= \{ 1, 1, 2, 2, 3, 3, 3, 3, 4, 4, 5, 5 \}$
From which we may account 2 of 1, 2 of 2, 4 of 3, 2 of 4 and 2 of 5. Present also is 4 of 0 if you include differences between identical elements, to take the total to the expected 16.
But instead of counting them 'by hand' we can multiply $a(x)$ by its transpose, $a(x^5)$ or, more conveniently in this case by $a(x^{-1})$ - it saves a little time with the exponents:
$(1 + x + x^3 + x^4)(1 + x^{-1} + x^{-3} + x^{-4})$
$= (1 + x^{-1} + x^{-3} + x^{-4}) + (x + 1 + x^{-2} + x^{-3}) + (x^3 + x^2 + 1 + x^{-4}) + (x^4 + x^3 + x + 1)$
$= 4 + 2x + x^2 + x^{-4} + 2x^3 + 2x^{-3} + x^4 + x^{-2} + 2x^{-1}$
$\equiv_6 4 + 2x + 2x^2 + 4x^3 + 2x^4 + 2x^5$
whereupon we recover the frequency of differences of size $d$ directly by reading off the coefficients of $x^d$ in $a^T(x)a(x)$.
One may of course recover the same result by multiplication of the two $6 \times 6$ circulant matrices (in either order - it's commutative) derived from row shifts of (1, 1, 0, 1, 1, 0) with its transpose - the circulant matrix derived similarly from (1, 0, 1, 1, 1, 0). The result will be the (perforce symmetric) circulant matrix (4, 2, 2, 4, 2, 2).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding roots of a Complex Polynomial
Question
Given that x = 3 is a solution to $$ x^3 - (7+3i)x^2 + (16+15i)x - 6(2+3i) = 0 $$ find the other two solutions.
What I have attempted;
If $x = 3$ is a root then $(x-3)$ is a factor
so $$ (x-3)(x^2 + Ax + B) = 0 $$
$$ x^3 + (A-3)x^2 + (B-3A)x - 3B = 0 $$
Equation Coefficients
$$ x^3 - (7+3i)x^2 + (16+15i)x - 6(2+3i) = 0 $$
$$ x^2 :−3+A=−7−3i $$
$$ ⇒ A=−4−3i $$
$$\text{Constant: } -3B = -6(2+3i) $$
$$ ⇒ B=4+6i $$
$$ (x-3)(x^2 -(4+3i)x + 4+6i) = 0 $$
$$ x = {-b\pm\sqrt{b^2-4ac} \over 2a} $$
$$ x = {4+3i\pm\sqrt{(4+3i)^2-4(4+6i)} \over 2} $$
$$ x = {4+3i\pm\sqrt{-9} \over 2} $$
$$ x = {4+3i±3i\over 2} $$
$$ x_1 = 3 $$
$$ x_2 = 2+3i $$
$$ x_3 = 2 $$
Hopefully this is correct , and if so I was wondering if there is an alternative method in solving this.
I was thinking because we are given x = 3 is a root and (x-3) is a factor we can do polynomial division with complex numbers? Is it possible to do polynomial division with complex numbers? If so can someone show how to get the same answer by using that method..
|
Is it possible to do polynomial division with complex numbers?
Yes, polynomial division works with complex numbers as well.
If so can someone show how to get the same answer by using that method..
A efficient method for polynomial evaluation and division is Horner's method.
Applied to your polynomial
$$
f(x) = x^3 - (7+3i)x^2 + (16+15i)x - 6(2+3i)
$$
and $x_1 = 3$ it gives:
1 -7-3i 16+15i -12-18i
3 -12-9i 12+18i
-------------------------------
1 -4-3i 4+6i 0
From the Wikipedia article:
The entries in the third row are the sum of those in the first two. Each entry in the second row is the product of the x-value (3 in this example) with the third-row entry immediately to the left. The entries in the first row are the coefficients of the polynomial to be evaluated.
...
As a consequence of the polynomial remainder theorem, the entries in the third row are the coefficients of the second-degree polynomial, the quotient of f(x) on division by x-3 .
It follows that $f(3) = 0$ and
$$
f(x) = (x-3)(x^2 -(4+3i)x + (4+6i))
$$
which is exactly what you got by equating coefficients.
If you know (or guess) another root then you can repeat the process.
Otherwise your method of solving the quadratic equation is fine.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1614984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
mysterious sum of two sequences Let
$$S_1 = \sum_{n=1}^\infty \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \cdots$$
$$S_2 = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} - \frac{1}{8} + \cdots$$
So
\begin{align}
S_1 - S_2 = {} & 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \cdots \\
& {} - \left[ 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} + \frac{1}{8} - \cdots \right] \\
= {} & 0 + 1 + 0 + \frac{1}{2} + 0 + \frac{1}{3} + 0 + \frac{1}{4} + \cdots \\
= {} & S_1
\end{align}
That makes $S_2$ zero, right?
The reason I am asking is because I was under the impression that $S_2 = \ln 2$.
What did I miss?
Is it that $S_1$ is divergent and $S_2$ is convergent and it's not allowed?
| You can't rearrange the terms of a sequence that doesn't converge absolutely without changing it's value, and you can't group series like that. For example:
$$
\sum_{i=0}^\infty (-1)^i = 1 -1 + 1 - 1 +\cdots \ne (1-1) + (1-1) + \cdots
$$
which would make it look like the sum converges to zero, but the above sum clearly diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1616279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Solve $\sin(3x)=\cos(2x)$
Question: Solve $\sin(3x)=\cos(2x)$ for $0≤x≤2\pi$.
My knowledge on the subject; I know the general identities, compound angle formulas and double angle formulas so I can only apply those.
With that in mind
\begin{align}
\cos(2x)=&~ \sin(3x)\\
\cos(2x)=&~ \sin(2x+x) \\
\cos(2x)=&~ \sin(2x)\cos(x) + \cos(2x)\sin(x)\\
\cos(2x)=&~ 2\sin(x)\cos(x)\cos(x) + \big(1-2\sin^2(x)\big)\sin(x)\\
\cos(2x)=&~ 2\sin(x)\cos^2(x) + \sin(x) - 2\sin^2(x)\\
\cos(2x)=&~ 2\sin(x)\big(1-\sin^2(x)\big)+\sin(x)-2\sin^2(x)\\
\cos(2x)=&~ 2\sin(x) - 2\sin^3(x) + \sin(x)- 2 \sin^2(x)\\
\end{align}
edit
\begin{gather}
2\sin(x) - 2\sin^3(x) + \sin(x)- 2 \sin^2(x) = 1-2\sin^2(x) \\
2\sin^3(x) - 3\sin(x) + 1 = 0
\end{gather}
This is a cubic right?
So $u = \sin(x)$,
\begin{gather} 2u^3 - 3u + 1 = 0 \\
(2u^2 + 2u - 1)(u-1) = 0
\end{gather}
Am I on the right track?
This is where I am stuck what should I do now?
| Use $\sin 3x=3 \sin x - 4 \sin^3x$ and $\cos 2x=1-2\sin^2x$. To get
$$3 \sin x - 4 \sin^3x=1-2\sin^2x.$$
Now call $\sin x=t$. Thus we have
$$4t^3-2t^2-3t+1=0.$$
Observe that $t=1$ is definitely a solution, so we have
$$(t-1)(4t^2+2t-1)=0.$$
The quadratic factor will be zero for
$$t=\frac{-1\pm \sqrt{5}}{4}$$
I hope you can solve from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
To evaluate the sum $\frac{1}{5}-\frac{1 \cdot 4}{5 \cdot 10}+\frac{1 \cdot 4 \cdot 7}{5 \cdot 10 \cdot 15}-\ldots$ Right now I am working through archived papers of a math aptitude quiz. For some reason I seem to be haveing a hard time with these series problems. I have managed to write the above series in a compact form but thats as far as I got.
$$\frac{1}{5}-\frac{1 \cdot 4}{5 \cdot 10}+\frac{1 \cdot 4 \cdot 7}{5 \cdot 10 \cdot 15}-\ldots = \sum\limits_{i=0}^\infty (-1)^i\frac{\prod\limits_{j=0}^i (3j+1)}{5^{i+1}(i+1)!}$$
Help please!
| Since:
$$ \prod_{j=0}^{i}\frac{3j+1}{5j+5} = \frac{1}{\left(\frac{5}{3}\right)^{i+1}\Gamma\left(\frac{1}{3}\right)\Gamma\left(\frac{2}{3}\right)}\cdot B\left(\frac{4}{3}+i,\frac{2}{3}\right) \tag{1}$$
we have:
$$ S=\sum_{i\geq 0}(-1)^i \prod_{j=0}^{i}\frac{3j+1}{5j+5} = \frac{\sqrt{3}}{2\pi}\int_{0}^{1}\sum_{i\geq 0}(-1)^i \left(\frac{3}{5}\right)^{i+1}x^{1/3+i}(1-x)^{-1/3}\,dx \tag{2}$$
hence:
$$ S = \frac{\sqrt{3}}{2\pi}\int_{0}^{1}\frac{3x^{1/3}}{(1-x)^{1/3}(5+3x)}\,dx =\frac{\sqrt{3}}{2\pi}\int_{0}^{+\infty}\frac{3z^{1/3}}{(1+z)(5+8z)}\,dz\tag{3}$$
where in the last step we set $x=\frac{z}{1+z}$. That leads to:
$$ S = \frac{\sqrt{3}}{2\pi}\int_{0}^{+\infty}\frac{9t^3}{(1+t^3)(5+8t^3)}\,dt=\frac{\sqrt{3}}{2\pi}\left(\int_{0}^{+\infty}\frac{3\,dt}{1+t^3}-\int_{0}^{+\infty}\frac{15\,dt}{5+8t^3}\right) \tag{4}$$
and the last integrals can be easily computed through the residue theorem or other techniques, leading to:
$$ S = \color{red}{1-\frac{1}{2}\sqrt[3]{5}}.\tag{5} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1618071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve in positive integers $x^2+3y^2=z^2$ and $x^2+y^2=5z^2$ Solve the following equations in postive integers:
*
*$x^2+3y^2=z^2$
*$x^2+y^2=5z^2$
I have solved them using this as a reference, but I am interested in other solutions, more "elegant" ones.
The equations are to be solved separate!!!
| People are not careful with these. With $x^2 + 3 y^2 = z^2;$ with odd $z,$ we indeed get
$$ (r^2 - 3 s^2)^2 + 3(2rs)^2 = (r^2 + 3 s^2)^2. $$ This does not give primitive solutions with even $z,$ which come from
$$ \left( \frac{r^2 - 3 s^2}{2} \right)^2 + 3 (rs)^2 = \left( \frac{r^2 + 3 s^2}{2} \right)^2 $$ with both $r,s$ odd
For $x^2 + y^2 = 5 z^2$ primitive means $z$ odd, we take $x$ to be the even one.
$$ (2u^2 - 2 u v - 2 v^2)^2 + (u^2 + 4 uv - v^2)^2 = 5 (u^2 + v^2)^2 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1620395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Sum of elements of a recursive sequence Say we have $$a_n=\frac{n-1+b}{n-b}a_{n-1}$$ with $a_0>0$ and $-1<b<0$. Then it holds that $$\sum_{n=1}^{\infty}a_n=-\frac12 a_0.$$
I could establish this using simple manupulations with $\Gamma$ and $Beta$ funtions. I was however wondering if one could do it differently. I appreciate if you share your fun on this problem with me :-)
| The result depends
on two different ways
of writing the recurrence.
The first way is
$a_n
=(1-\frac{1-2b}{n-b})a_{n-1}
$.
From this
we show that
$a_m
<a_0m^{-(1-2b)}
$
or
$ma_m
<a_0m^{-2b}
$.
The second way is
$na_n-(n-1)a_{n-1}
=b(a_n+a_{n-1})
$.
Summing this,
we get
$\sum_{n=1}^{m-1}a_{n}
=\frac{(m-b)a_m}{2b}-\frac{a_0}{2}
$.
Letting $m \to \infty$,
since
$ma_m \to 0$
as $m \to \infty$
(from the first result,
since $b < 0$),
we get the desired result.
Now, the details.
$\begin{array}\\
a_n
&=\frac{n-1+b}{n-b}a_{n-1}\\
&=\frac{n-b-1+2b}{n-b}a_{n-1}\\
&=(1+\frac{2b-1}{n-b})a_{n-1}\\
&=(1-\frac{1-2b}{n-b})a_{n-1}\\
\text{so}\\
\frac{a_n}{a_{n-1}}
&=1-\frac{1-2b}{n-b}\\
\text{or}\\
\ln(a_n)-\ln(a_{n-1})
&=\ln(1-\frac{1-2b}{n-b})\\
&\lt -\frac{1-2b}{n-b}
\qquad\text{since }
\ln(1-x) < -x.\\
\text{Summing,}\\
\ln(a_m)-\ln(a_0)
&\lt -\sum_{n=1}^m\frac{1-2b}{n-b}\\
&<-(1-2b)\ln(m)\\
\text{or}\\
a_m
&<a_0m^{-(1-2b)}\\
\end{array}
$.
We have
$(n-b)a_n
=(n-1+b)a_{n-1}
$
or
$na_n-(n-1)a_{n-1}
=b(a_n+a_{n-1})
$.
Summing,
$\begin{array}\\
\sum_{n=1}^m(na_n-(n-1)a_{n-1})
&=\sum_{n=1}^m(b(a_n+a_{n-1}))\\
&=b\sum_{n=1}^ma_{n-1}+b\sum_{n=1}^ma_{n}\\
&=b\sum_{n=0}^{m-1}a_{n}+b\sum_{n=1}^ma_{n}\\
&=b(a_0+a_m+2\sum_{n=1}^{m-1}a_{n})\\
\text{or}\\
ma_m
&=b(a_0+a_m+2\sum_{n=1}^{m}a_{n})\\
\text{or}\\
\sum_{n=1}^{m-1}a_{n}
&=\frac{ma_m}{2b}-\frac{a_m}{2}-\frac{a_0}{2}\\
&=\frac{(m-b)a_m}{2b}-\frac{a_0}{2}\\
\end{array}
$
We have shown that
$ma_m
<a_0m^{2b}
\to 0
$
since $b < 0$.
Therefore,
letting $m \to \infty$,
we are done!!!!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1621308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\int \frac{2-3x}{2+3x}\sqrt\frac{1+x}{1-x}dx$ Evaluate $$\int \frac{2-3x}{2+3x}\sqrt\frac{1+x}{1-x}d x$$
What substitution should I use ? $\sqrt\frac{1+x}{1-x}$ suggests $x=cos2\theta $ but its not useful in $\frac{2-3x}{2+3x}$
| The substitution $t = \sqrt{\frac{1 + x}{1 - x}}$, that is, $$x = \frac{t^2 - 1}{t^2 + 1}, \qquad dx = \frac{4 t \,dt}{(t^2 + 1)^2},$$
rationalizes the integrand.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1622622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Quartic with $4 $ equidistant roots Today I got the problem $(x^2 -1)(x^2 -4)=k$, and I have no idea how to prove this algebraically. $K$ is a real, non-zero number that makes the equation have $4$ distinct real equidistant roots. Solve for $k$ algebraically.
What I have attempted:
I have tried solving for x and then in turn attempting to solve for k
$$(x^2 -1)(x^2 -4)=k$$
$$ (x^4 -4x^2-1x^2+4) = k$$
$$ x^4 - 5x^2 + 4 - k = 0 $$
$$ x^2 = {-b\pm\sqrt{b^2-4ac} \over 2a} $$
$$ x^2 = {5\pm\sqrt{25-4(4-k)} \over 2} $$
$$ x^2 = {5\pm\sqrt{9+4k} \over 2} $$
$$ x^2 = {5+\sqrt{9+4k} \over 2} $$ and $$ x^2 = {5-\sqrt{9+4k} \over 2} $$
Therefore
$$ x_1 = \sqrt {5+\sqrt{9+4k} \over 2} $$
$$ x_2 = -\sqrt {5+\sqrt{9+4k} \over 2} $$
$$ x_3 = \sqrt {5-\sqrt{9+4k} \over 2} $$
$$ x_4 = -\sqrt {5-\sqrt{9+4k} \over 2} $$
Now I am stuck trying to solve for k.. The answer is $k$ = 7/4 but having trouble solving for it algebraically..
| You have found $x^2=A+B$ and $x^2=A-B$ with $$A={5\over2},\qquad B={1\over2}\sqrt{9+4k}\ .\tag{1}$$ It follows that the four roots are (in increasing order):
$$x_1=-\sqrt{A+B},\quad x_2=-\sqrt{A-B},\quad x_3=\sqrt{A-B},\quad x_4=\sqrt{A+B}\ .$$
The condition $x_3-x_2=x_4-x_3$ leads to
$$\sqrt{A+B}=3\sqrt{A-B}\ ,$$
and this in turn implies $10B=8A$. Plugging in $(1)$ and solving for $k$ leads to $k={7\over4}$, so that $A+B={9\over2}$, $A-B={1\over2}$. This then leads to
$$x_1=-{3\over2}\sqrt{2},\quad x_2=-{1\over2}\sqrt{2},\quad x_3={1\over2}\sqrt{2},\quad x_4={3\over2}\sqrt{2}\ .$$
One easily checks that these findings fulfill all requirements.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1623320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding equation of a circle I am supposed to find an equation of a circle passing though points M(3,0,0),N(0,3,0) and P(0,0,3) I have tried to find an intersection of bisectors of lines MN and NP but ended up with nothing.
| It's not as hard as you may believe, in fact, you would use the following equation:
$$(x-a)^2+(y-b)^2+(z-c)^2=r^2$$
Just an altered version of the circle equation where $C(a,b,c)$ is the center of our circle.
So all you do is plug in you numbers to get 3 equations:
$$(3-a)^2+b^2+c^2=r^2\tag{1}$$
$$a^2+(3-b)^2+c^2=r^2\tag{2}$$
$$a^2+b^2+(3-c)^2=r^2\tag{3}$$
Subtract $(1)$ and $(2)$ to get:
$$(3-a)^2-a^2+b^2-(3-b)^2=0$$
$$9-6a-9+6b=0$$
$$6b=6a\to a=b$$
Subtract $(2)$ and $(3)$:
$$(3-b)^2-b^2+c^2-(3-c)^2=0$$
Following from above steps, we get:
$$a=b=c$$
From this, I want to assume that $a=b=c=0$ because I think it is obvious with the three points you chose and I want to have $r=3$, such that we have:
$$x^2+y^2+z^2=3^2=9$$
Plugging in your points say that this works and we have the following:
$$3=\pm\sqrt{x^2+y^2+z^2}=\pm\sqrt{x^2+(\sqrt{y^2+z^2})^2}$$
Basically, this is applying Pythagorean Theorem twice so that the points $x_1,y_1,z_1$ are all equidistant to the origin.
Since ask specifically for the circle, here you go:
$$(x-1)^2+(y-1)^2+(z-1)^2=6$$
Just remember to use the following equation: $x+y+z=3$ to maintain the fact that it is a circle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1623466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding $\sum_{k=1}^{\infty}k^2 \frac{2^{k-1}}{3^k}$ Finding $$\sum_{k=1}^{\infty}k^2 \frac{2^{k-1}}{3^k}$$
I think if $$\int_{1}^{\infty}x^2 \frac{2^{x-1}}{3^x}dx$$
exists that this sequence is convergent, but I doubt that this integral is equal to the number to which the sequence converges to. I do not know a way to solve this sequence, thought about doing geometric progression, but can't because of $k^2$
| This is the same as
$$ \frac{1}{2} \sum_{k \geq 1} k^2 \left( \frac{2}{3} \right)^k,$$
which looks almost like the geometric series
$$ f(x) = \frac{1}{2} \sum_{k \geq 1} x^k = \frac{1}{2} \frac{x}{1 - x}$$
at $x = \frac{2}{3}$. Notice that
$$ \frac{d}{dx} f(x) = \frac{1}{2} \sum_{k \geq 1} k x^{k - 1},$$
which is closer to what we want. We want another factor of $k$, so we multiply by $x$ and differentiate again,
$$ \frac{d}{dx} x \frac{d}{dx} f(x) = \frac{1}{2} \sum_{k \geq 1} k^2 x^{k-1}.$$
We are only missing a single factor of $x$, so we multiply by $x$ again to see that
$$ \left( x \frac{d}{dx} \right)^2 f(x) = \frac{1}{2} \sum_{k \geq 1} k^2 x^k.$$
To evaluate, you need only to compute
$$ \left( x \frac{d}{dx} \right)^2 \frac{1}{2} \frac{x}{1 - x}$$
and evaluate it at $x = \frac{2}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1623851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Parametrize an intersection of a plane and an elliptic paraboloid I'm supposed to parametrize the intersection of the plane that has the equation $z = 5x + 3y$ and the 'elliptic paraboloid' with the equation $z = 3x^2+2xy+3y^2$
These two equations can also be written in terms of $u$ and $v$, the plane then has equation $z = 4u + v$ and the paraboloid has $z= 2u^2 + v^2$
How can I go about parametrising these in either domain?
Thanks in advance
| The intersection satisfies
$$ 4u + v = 2u^2 + v^2 \implies 2u^2 - 4u + v^2 - v = 0. $$
Completing the square, we have
$$ 2u^2 - 4u + v^2 - v = 2(u - 1)^2 - 2 + \left( v - \frac{1}{2} \right)^2 - \frac{1}{4} = 0 $$
which implies that
$$ 2(u-1)^2 + \left( v - \frac{1}{2} \right)^2 = \frac{9}{4}. $$
This is the equation of an ellipse in the $u$-$v$ plane. Manipulating the equality algebraically, we obtain
$$ \left( \frac{\sqrt{2}}{3}(u-1) \right)^2 + \left( \frac{2v - 1}{3} \right)^2 = 1 $$
and so this can be parametrized by letting
$$ \frac{\sqrt{2}}{3}(u - 1) = \cos(\theta), \,\,\, \frac{2v - 1}{3} = \sin(\theta) $$
or, more explicitly,
$$ u = 1 + \frac{3}{\sqrt{2}}\cos(\theta), \,\,\, v = \frac{3 \sin(\theta) + 1}{2} $$
for $\theta \in [0,2\pi]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1626694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Elementary symmetric polynomial task with three variables Can anyone help me to wite this as sum or product of elementary symmetric polynomial.
$$\frac xy+\frac yx +\frac xz + \frac zx +\frac yz + \frac zy =7$$
I tried to set under one fraction, but I didn't go much far.
| First, we get rid of the fractions. We get $$x^2 y+x^2 z+x y^2+x z^2+y^2 z+y z^2= 7xyz$$
Now we need to express the left hand side in terms of elementary symmetric polynomials. Since it is third-degree, we try computing $(x+y+z)^3$ and we get $$x^3+y^3+z^3 +3x^2 y+3 x^2 z+3 x y^2+6 x y z+3 x z^2+3 y^2 z+3 y z^2$$ We see that, if we get rid of the terms $x^3+y^3+z^3$ and $6xyz$, we have what we want. So, computing $$\tfrac13\left((x+y+z)^3-6xyz-(x^3+y^3+z^3)\right)=x^2 y+x^2 z+x y^2+x z^2+y^2 z+y z^2$$ which is what we wanted. So the expression becomes $$\tfrac13\left((x+y+z)^3-6xyz-(x^3+y^3+z^3)\right)=7xyz$$
or, simplified:
$$(x+y+z)^3-(x^3+y^3+z^3)=27xyz$$
Hope this helped!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1630313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Sum of $1-\frac{2^2}{5}+\frac{3^2}{5^2}-\frac{4^2}{5^3}+....$ $1-\frac{2^2}{5}+\frac{3^2}{5^2}-\frac{4^2}{5^3}+....$
How can we find sum of above series upto infinite terms?
I don't know how to start and just need some hint.
| Say sum to $n$ terms of the series is $S_n$.
Now we have that $$S_n=1+\sum_{r=1}^n (-1)^n\cdot\frac{(n+1)^2}{5^{n}}=1+R_n$$
where $R_n=\sum_\limits{r=1}^n (-1)^n\cdot\frac{(n+1)^2}{5^{n}}=-\frac{2^2}{5}+\sum_\limits{r=1}^n (-1)^{n+1}\cdot\frac{(n+2)^2}{5^{n+1}}$
Now $$-\frac{1}{5}\cdot R_n=\sum_{r=1}^n (-1)^{n+1}\cdot\frac{(n+1)^2}{5^{n+1}}$$
Subtracting, we get $$\frac{6}{5}\cdot R_n=-\frac{2^2}{5}+\sum_{r=1}^n (-1)^{n+1}\cdot\frac{2n+3}{5^{n+1}}=-\frac{2^2}{5}+Q_n$$
Similarly $$Q_n=\frac{1}{5}+\sum_{r=1}^n (-1)^{n+2}\cdot\frac{2n+5}{5^{n+2}}$$
and $$-\frac{1}{5}\cdot Q_n=\sum_{r=1}^n (-1)^{n+2}\cdot\frac{2n+3}{5^{n+2}}$$
Again we get that $$\frac{6}{5}Q_n=\frac{1}{5}+\sum_{r=1}^n (-1)^{n+2}\cdot\frac{2}{5^{n+2}}$$
When $n\to\infty$, we have that $$Q_\infty=\frac{5}{6}(\frac{1}{5}-\frac{1}{75})=\frac{7}{45}$$
Hence we can calculate $R_\infty$ and $S_\infty$ which comes out to be $\frac{25}{54}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1630497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
Summation of more than one series How can I find $1+3x+6x^2+10x^3+.... $ for $x=6/7$. I have seen that this can be written as a sum of three series $\sum_0^\infty [(n^2/2)x^n+(3/2)nx^n+x^n]$. But I do not know how to proceed further.
| Note that you have sum of terms of type $\frac{(n+1)(n+2)}{2}x^n$, so your sum would be:
$$1+(1+2)x+(1+2+3)x^2+(1+2+3+4)x^3+...=\\=(1+x+x^2+x^3+...)+2(x+x^2+x^3+...)+3(x^2+x^3+x^4+...)+...=\\=(1+x+x^2+x^3+...)+2x(1+x+x^2+x^3...)+3x^2(1+x+x^2+x^3+...)+...$$ Since $x=6/7$ we have $1+x+x^2+x^3+...=7$, so the above expression is:
$$=7(1+2x+3x^2+4x^3+...)$$ Now, calculate $1+2x+3x^2+4x^3+...$ by similar method, and you will obtain result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1631298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How to find pythagorean triples and n-tuples A pythagorean triple is any set of three positive integers $(a,b,c)$ where $a^2 + b^2 = c^2$
I'm wondering, is there a formula to find all pythagorean triples, and can it be generalized to $n$-tuples? (i.e $a_1^2+a_2^2+a_3^2+a_4^2+....+a_n^2=x^2$)
| Let $t$ be a scaling factor. Then the complete rational solutions are,
$$((a^2-b^2)t)^2+(2abt)^2 = ((a^2+b^2)t)^2$$
$$((a^2-b^2-c^2)t)^2+(2abt)^2+(2act)^2 = ((a^2+b^2+c^2)t)^2$$
$$((a^2-b^2-c^2-d^2)t)^2+(2abt)^2+(2act)^2+(2adt)^2 = ((a^2+b^2+c^2+d^2)t)^2$$
and so on.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1632281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Complex number ( prove ) Let
$$
{x-yi\over{x+yi}}=a+bi\;\;.
$$
Prove $a^2+b^2=1$
I don't know how to start prove it, can anyone help me?
| Another method:
Note that $a^2 + b^2 = (a + bi)(a - bi)$, as indeed for any complex number $w = a + bi$,
$$ww^* = |w|^2 = a^2 + b^2$$
where $^*$ is the complex conjugate.
In this case $$(a + bi)^* = \left( \frac{x - yi}{x + yi} \right)^* = \frac{(x - yi)^*}{(x + yi)^*} = \frac{x + yi}{x - yi}$$
Hence
$$a^2 + b^2 = (a + bi)(a + bi)^* = \frac{x - yi}{x + yi}\frac{x + yi}{x - yi} = 1$$
And another:
$$a^2 + b^2 = |a + bi|^2 = \left| \frac{x - yi}{x + yi} \right|^2 = \frac{|x-yi|^2}{|x + yi|^2} = \frac{x^2 + y^2}{x^2 + y^2} = 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1633926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
solving rectangle Diagonal of a rectangle is $13$ cm. If we extend the length of the rectangle for $4$ cm and width for $7$ cm, then diagonal will be longer for $7$ cm as well. Find sides (length and width) of the rectangle.
So $d^2=a^2+b^2 \implies b^2=d^2-a^2 \implies b^2=13^2-a^2 \implies b^2=169-a^2$ and $a^2+b^2=169$
\begin{align*}
d'^2 & =a'^2+b'^2\\
(d+7)^2 & =(a+4)^2+(b+7)^2\\
20^2 & = a^2+8a+16+b^2+14b+49\\
400 & = 8a+16+14b+49+169\\
\frac{166}{2} & = \frac{8a+14b}{2}\\
83 & =4a+7b
\end{align*}
...
That's what I've done. What should I do next?
| You have
$$
a^2+b^2=169
$$
$$
4a+7b=83 \Longleftrightarrow a=\frac{83-7b}{4}
$$
After substitution:
$$
\frac{(83-7b)^2}{16}+b^2=169
$$
$$
\frac{49b^2}{16}-\frac{1162b}{16}+\frac{6889}{16}-169=0
$$
Finally, use 'abc-formula' with a = 49/16, b = 1162/16 and c = 6889/16-169.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1635123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the remainder when ${{5^5}^5}^5$ is divided by $24$
Find the remainder when ${{5^5}^5}^5$ is divided by $24$
I tried using congruence modulo. $$5^2\equiv1\mod{24}$$
$$5^5=125\mod{24}$$
But this does not give the correct answer.
| To know $a^{b^{c^d}}\bmod 24$ you only need to know $b^{c^d}\bmod \varphi(24)$.
To know $b^{c^d}\bmod \varphi(24)$ you only need to know $c^d\bmod \varphi(\varphi(24))$
Lets actually do it:
First note $\varphi(24)=8$, $\varphi(\varphi(24))=4$.
Since $5^5\equiv 1\bmod 4$ we conclude $5^{5^5}\equiv 5^1\bmod 8$.
From here $5^{5^5}\equiv5^1\equiv 5\bmod 8$.
Finally we get $5^{5^{5^5}}\equiv 5^5\bmod 24$
Finally use logarithmic exponentiation:
$5^2\equiv 1,5^4\equiv 1\bmod 24$
therefore $5^5=5\times5^4\equiv 5\times 1\equiv 5 \bmod 24$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1635506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
} |
Fibonacci sequence in the factorization of certain polynomials having a root at the Golden Ratio I was playing around with the Golden Ratio $\Phi = \frac{1 + \sqrt 5}{2}$ on Wolfram Alpha and I noticed that if $F_n$ denotes the $n{th}$ Fibonacci number, then the polynomial $P_n(x) = x^n - F_n x - F_{n-1}$ seems always have a root at $\Phi$ (and hence is divisible by the minimal polynomial $f(x) = x^2 - x - 1$ of $\Phi$ )
After testing several $P_n$ I noticed that factoring out the $f(x)$ gave products of the form:
$$
P_n(x) = (x^2 - x - 1)(F_1 x^{n-2} + F_2 x^{n-3} + ... + F_{n-2} x + F_{n-1})
$$
where the $F_k$ are the $k^{th}$ Fibonacci numbers.
I've been stumped as to why the coefficients of the second factor should list the elements of the Fibonacci sequence like that, and so I'm hoping someone here can provide insight as to why it all seems to work out so nicely.
| Distributing the product in the r.h.s. of the factorization formula
$$
P_n(x) = (x^2 - x - 1)(F_1 x^{n-2} + F_2 x^{n-3} + ... + F_{n-2} x + F_{n-1})
$$
gives
\begin{align*}(x^2 - x - 1)\left(\sum_{k = 1}^{n - 1} F_k x^{n - 1 - k}\right)
&= x^2 \sum_{k = 1}^{n - 1} F_k x^{n - 1 - k} - x \sum_{k = 1}^{n - 1} F_k x^{n - 1 - k} - \sum_{k = 1}^{n - 1} F_k x^{n - 1 - k} \\
&= \sum_{k = 1}^{n - 1} F_k x^{n + 1 - k} - \sum_{k = 1}^{n - 1} F_k x^{n - k} - \sum_{k = 1}^{n - 1} F_k x^{n - 1 - k}
\end{align*}
Judiciously reindexing to match exponents in $x$ gives that this is
$$\sum_{k = -1}^{n - 3} F_{k + 2} x^{n - 1 - k} - \sum_{k = 0}^{n - 2} F_{k + 1} x^{n - 1 - k} - \sum_{k = 1}^{n - 1} F_k x^{n - 1 - k} .$$
If we peel off the terms whose index does not appear in every sum, we can combine the resulting summations and collect like terms:
\begin{multline*}
\left(F_1 x^n + F_2 x^{n - 1} + \sum_{k = 1}^{n - 3} F_{k + 2} x^{n - 3}\right)
- \left(F_1 x^{n - 1} + \sum_{k = 1}^{n - 3} F_{k + 1} x^{n - 3} + F_{n - 1} x \right) \\
\hspace{9cm}- \left(\sum_{k = 1}^{n - 3} F_k x^{n - 3} + F_{n - 2} x + F_{n - 1} \right) \\
= F_1 x^n + (F_2 - F_1) x^{n - 1} + \sum_{k = 1}^{n - 3} (F_{k + 2} - F_{k + 1} - F_k) x^{n - 3} - (F_{n - 1} + F_{n - 2}) x - F_{n - 1} .\end{multline*}
We have $F_1 = F_2 = 1$, so the leading term is $x^n$ and the $x^{n - 1}$ term vanishes. The coefficient of the $k$th term of the summation is $F_{k + 2} - F_{k + 1} - F_k$, but this vanishes by the definition of the Fibonacci sequence. The coefficient $F_{n - 1} + F_{n - 2}$ of the $x$ term, again by definition, is equal to $F_n$.
So, as desired, the product is
$$\color{#bf0000}{\boxed{(x^2 - x - 1)(F_1 x^{n-2} + F_2 x^{n-3} + ... + F_{n-2} x + F_{n-1}) = x^n - F_n x - F_{n - 1}}} .$$
NB we can view this as a variation of the familiar factorization
$$(x - 1)(x^{n - 1} + x^{n - 1} + \cdots + x + 1) = x^n - 1 .$$
More precisely, we can rederive this latter identity by characterizing the sequence $(1, 1, 1, \ldots)$ via the recurrence relation $G_1 = 1$, $G_{k + 1} = G_k$ ($k \geq 1$) and proceeding as above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1638705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\lim_{x\to1}{\frac{\sqrt{x^2+3}-2}{\sqrt{x^2+8}-3}}$ without L'Hospital's Theorem I've been trying to evaluate$$\lim_{x\to1}{\frac{\sqrt{x^2+3}-2}{\sqrt{x^2+8}-3}}$$
I tried:
(a) Rationizing the numerator -> Error
(b) Rationizing the denominator -> Error
(c) Factoring out $x$ -> Error
Finally, I used L'Hospital's Rule and got the answer $3/2$, but that is not what I am supposed to do, for not a single word about this Theorem was mentioned during my lectures.
Is there any other way to solve this limit without this Theorem?
| Notice, $$\lim_{x\to 1}\frac{\sqrt{x^2+3}-2}{\sqrt{x^2+8}-3}$$
$$=\lim_{x\to 1}\frac{(\sqrt{x^2+3}-2)(\sqrt{x^2+3}+2)}{(\sqrt{x^2+8}-3)(\sqrt{x^2+8}+3)}\cdot \frac{(\sqrt{x^2+8}+3)}{(\sqrt{x^2+3}+2)}$$
$$=\lim_{x\to 1}\frac{x^2+3-4}{x^2+8-9}\cdot \frac{(\sqrt{x^2+8}+3)}{(\sqrt{x^2+3}+2)}$$
$$=\lim_{x\to 1}\frac{(x^2-1)}{(x^2-1)}\cdot \frac{(\sqrt{x^2+8}+3)}{(\sqrt{x^2+3}+2)}$$
$$=\lim_{x\to 1}\frac{\sqrt{x^2+8}+3}{\sqrt{x^2+3}+2}$$
$$=\frac{3+3}{2+2}=\color{red}{\frac{3}{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1638863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 0
} |
How can I solve $\int \frac{3x+2}{x^2+x+1}dx$ I want to compute this primitive $$I=\int \frac{3x+2}{x^2+x+1}dx.$$
I split this integral into two part:
$$\int \frac{3x+2}{x^2+x+1}dx=\int \frac{2x+1}{x^2+x+1}dx+\int \frac{x+1}{x^2+x+1}dx,$$
For the first part:
$$\int \frac{2x+1}{x^2+x+1}dx= \ln(|x^2+x+1|)+c_1$$
For the second part: due a change of variable $u=x+1$, I find
$$\int \frac{x+1}{x^2+x+1}dx= \int \frac{x+1}{x(x+1)+1}dx=\int \frac{u}{u(u-1)+1}dx=\dots$$
| Let $3x+2=A\dfrac{d(x^2+x+1)}{dx}+B$
$\iff3x+2=A(2x+1)+B=2Ax+A+B$
$\implies2A=3,A+B=2$
$$\implies\int\dfrac{3x+2}{x^2+x+1}dx=A\cdot\dfrac{d(x^2+x+1)}{x^2+x+1}+B\dfrac{dx}{x^2+x+1}$$
$$\int\dfrac{dx}{x^2+x+1}=\int\dfrac4{(2x+1)^2+(\sqrt3)^2}dx$$
Set $2x+1=\sqrt3\tan y$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1638942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
I want to solve $\int \frac{2}{x^2(x^2+1)^2}dx$ I want to solve this primitive $$I=\int \frac{2}{x^2(x^2+1)^2}dx.$$
I substitute $u=x^2$ then,
$$I=\int \frac{2}{x^2(x^2+1)^2}dx=\int \frac{du}{u^{3/2}(u+1)^2}=\cdots$$
How do I use partial fraction decomposition? Is there another idea to compute this primitive?
| Since both factors are repeated, the partial fractions decomposition of the integrand has the form
$$\frac{1}{x^2 (x^2 + 1)^2} = \frac{A}{x^2} + \frac{B}{x} + \frac{C x + D}{(x^2 + 1)^2} + \frac{E x + F}{x^2 + 1} .$$
The resulting linear system is somewhat complicated (six equations in six variables), but we can streamline solving it by noticing that the integrand is an even function, and hence so is its partial fractions decomposition, which forces $B = C = E = 0$. So, the above simplifies to the more tractable $3 \times 3$ system:
$$\frac{1}{x^2 (x^2 + 1)^2} = \frac{A}{x^2} + \frac{D}{(x^2 + 1)^2} + \frac{F}{x^2 + 1} .$$
Cross-multiplying gives the (even) polynomial equation
$$(A + F) x^4 + (2 A + D + F) x^2 + A = 1 .$$
Alternatively, one can immediately use trigonometric substitution here (the form $x^2 + 1$ suggests $x = \tan \theta$, $dx = \sec^2 \theta\,d\theta$), but I'm not sure that turns out to be any faster in this case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1639036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Show $\cos(x^2)/(1+ x^2)$ is uniformly continuous on $\Bbb R$. now here's how I did proceed.
By definition a function $f: E →\Bbb R$ is uniformly continuous iff for every $ε > 0$, there is a $δ > 0$ such that $|x-a| < δ$ and $x,a$ are elements of $E$ implies $|f(x) - f(a)| < ε.$
Then suppose $x, a$ are elements of $\Bbb R. $
Now
\begin{align}
|f(x) - f(a)|
&= \left|\frac{\cos(x^2)}{1 + x^2} - \frac{\cos(a^2)}{1 + a^2}\right|
\\&= \left| \frac{\cos(x^2)(1+a^2 )- \cos(a^2)(1+x^2)}{(1 + x^2)(1 + a^2)}\right|
\\&≤ \left| \frac{a^2 - x^2}{(1 + x^2)(1 + a^2)}\right|
\end{align}
can it be written..? the last step?
If not, please help me solve the sum.
| Hint: use triangular inequality and $2|x| < 1 + x^2$ to get $|f'(c)| < \dfrac{2}{1+c^2} < 2 \Rightarrow |f(x) - f(a)| < 2|x-a|$ , and conclude.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Showing that this sequence is eventually decreasing I'm trying to show that this sequence $$a_n = \frac{3^n-7}{4^n+5}$$ is decreasing for all $n$ greater than some $N\in \Bbb N$.
All I can see to do is something like
$$a_{n+1} = \frac{3^{n+1}-7}{4^{n+1}+5} = \frac{3\cdot3^n-7}{4\cdot 4^n+5}\le \frac{3\cdot3^n-7}{4^n+5}$$
But that last expression is not less than $a_n$ for large $n$. Is there some better way to do this?
| We have
$$\frac{a_{n}}{a_{n+1}}=\frac{3^n-7}{3^{n+1}-7}\cdot \frac{4^{n+1}+5}{4^n+5}.$$
The ratio $\dfrac{a_n}{a_{n+1}}$ has limit $4/3$ as $n\to\infty$. (Divide top and bottom of the first term on the right by $3^n$, and top and bottom of the second term by $4^n$.)
So by the definition of limit there is an $N$ such that if $n\gt N$ then
$$\frac{a_n}{a_{n+1}}\gt \frac{4}{3}-0.1\gt 1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1643212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove $\frac{1}{3}(x+y+z)^2 \geq xy + yz + xz.$
Prove that for nonnegative $x,y,z$ that $\frac{1}{3}(x+y+z)^2 \geq xy + yz + xz.$
I saw this result in a problem but didn't know how to prove it. I tried expanding and collecting to get the trivial inequality but it didn't work.
The reason I am asking this is because it comes from the solution below where it claims it is true.
We note that $S_a=ad_a/2$, $S_b=bd_b/2$, and $S_c=cd_c/2$ are the areas of the triangles $MBC$, $MCA$, and $MAB$ respectively. The desired inequality now follows from $$S_aS_b+S_bS_c+S_cS_a \le \frac{1}{3}(S_a+S_b+S_c)^2=\frac{S^2}{3}.$$ Equality holds if and only if $S_a=S_b=S_c$, which is equivalent to $M$ being the center of the triangle.
| Here's my attempt to expand and collect terms:
\begin{align}
\frac 13(x + y + z)^2 & \stackrel{?}{\ge} xy + xz + yz\\
x^2 + y^2 + z^2 + 2xy + 2xz + 2yz & \stackrel{?}{\ge} 3xy + 3xz + 3yz \\
x^2 + y^2 + z^2 - xy - xz - yz & \stackrel{?}{\ge} 0 \\
\frac 12(x - y)^2 + \frac 12(x - z)^2 + \frac 12(y - z)^2 & \stackrel{?}{\ge} 0.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1644039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Finding the minimum of $x^2+y^2$ for $(x^2y-xy^2)(x^3-y^3)=x^3+y^3$ and $xy>0$. If $x,y \in \mathbb {R}$, find the minimum of $x^2+y^2$ when $(x^2y-xy^2)(x^3-y^3)=x^3+y^3$ and $xy>0$.
This problem was inspired by a problem which asked if $x,y \in \mathbb {R}$ and $xy \neq 0$, find the minimum of $x^2+y^2$ when $xy(x^2-y^2)=x^2+y^2$.
By setting $x=a\sin\theta$, $y=a\cos\theta$, the equation can be simplified to $a^2=\frac{1}{\sin\theta\cos\theta(\sin\theta^2-\cos\theta^2)}$
However, notice that $\sin\theta\cos\theta=\frac{\sin2\theta}{2}$, $\sin\theta^2-\cos\theta^2=-\cos2\theta$.
This implies that $a^2=\frac{2}{-\sin2\theta\cos2\theta}$, thus that $a^2=\frac{4}{-\sin4\theta}\ge 4$.
Thus the minimum of $x^2+y^2$ is $4$, with the equality holding when $x=\sqrt{2-\sqrt{2}}$, $y=\sqrt{2+\sqrt{2}}$.
However, since there are no formulas I know of where $\sin^3 x+\cos^3 x$, I did not know how to find the minimum of $x^2+y^2$ when $(x^2y-xy^2)(x^3-y^3)=x^3+y^3$.
Graphing it seems to imply that such a minimum exists, but I am not aware of how to find it( and when such a minimum exists).
Any help would be appreciated.
| Using Lagrange multiplier got a link condition:
$$ \dfrac{x}{y} = \dfrac{x^4 y -2 x y^4 + y^5 - 3 x^2}{-2 x^4 y +5 x y^4 + x^5 -3 y^2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1645369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Solve the differential equation : $0.5 \frac{dy}{dx}=4.9-0.1y^2$
The question is to solve the differential equation :
$$0.5 \frac{dy}{dx}=4.9-0.1y^2$$
What I have attempted:
$$0.5 \frac{dy}{dx}=4.9-0.1y^2$$
$$ \frac{dy}{dx} = \frac{4.9-0.1y^2}{0.5} $$
$$ \frac{dy}{dx} = 9.8 - 0.2y^2 $$
$$ \int \frac{1}{9.8 - 0.2y^2} dy = \int 1 dx $$
$$ \int \frac{10}{98 - 2y^2} dy = \int 1 dx $$
$$ \int \frac{5}{49 - y^2} dy = \int 1 dx $$
$$ \int \frac{1}{49 - y^2} dy = \int \frac{1}{5} dx $$
How should I continue?
EDIT Thanks @andrenicolas for the hint
By using partial fractions
$$ \int \frac{1}{(7-y)(7+y)} dy = \int \frac{1}{5} dx $$
$$ \frac{1}{14}\int \frac{1}{7-y} + \frac {1}{7+y} dy = \int \frac{1}{5} dx $$
$$ \frac{1}{14} [ ln |y+7| - ln|y-7| ] = \frac{x}{5} + c$$
$$ [ ln |y+7| - ln|y-7| ] = \frac{14x}{5} + 14c $$
$$ ln\frac{y+7}{y-7} = \frac{14x}{5} + 14c $$
$$ \frac{y+7}{y-7} = Ae^{\frac{14x}{5}} $$ (Letting $A = 14c$)
$$ y + 7 = yAe^{\frac{14x}{5}} - 7Ae^{\frac{14x}{5}} $$
$$ 7Ae^{\frac{14x}{5}} + 7 = yAe^{\frac{14x}{5}} - y $$
$$ 7Ae^{\frac{14x}{5}} + 7 = y(Ae^{\frac{14x}{5}} - 1) $$
$$ y = \frac{7Ae^{\frac{14x}{5}}}{(Ae^{\frac{14x}{5}} - 1)} $$
| Continuing your calculations
$$ \int \frac{1}{49 - y^2} dy = \int \frac{1}{5} dx $$
$$ \int \frac{1}{7^2 - y^2} dy = \int \frac{1}{5} dx $$
$$ \int \frac{1}{7 - y} + \frac{1}{7 + y} dy = \int \frac{14}{5} dx $$
$$ \log(\frac{7+y}{7-y}) = \frac{14}{5} x + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1645621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to expand $x_1^3 + x_2^3$ with the parameters of quadratic equation Given:
$X_1$ and $X_2$ are the roots of the equation $ax^2+bx+c = 0$ $a\neq 0$
expand $X_1^3 + X_2^3$ using the parameters a,b and c
Here's what I tried to do:
$X_1^3 + X_2^3 = $ $(X_1\cdot X_2)\cdot(X_1^2-X_1\cdot X_2+X_2^2)$
From Viet's formulas:
$(X_1\cdot X_2)(X_1^2-X_1\cdot X_2+X_2^2)$ = $\frac{-b}{a}\cdot(X_1^2+X_2^2- \frac{c}{a})=$
$\frac{c\cdot b}{a^2}\cdot(X_1^2+X_2^2)$
And this i all I can do the answer in my textbook is $\frac{3abc-b^3}{a^3}$
| HINT:
Use
$$p^3+q^3=(p+q)^3-3pq(p+q)$$
Alternatively,
$x^3=y$ and cubing $$(ax^2+bx)^3=(-c)^3$$
$$-c^3=a^3(x^3)^2+b^3(x^3)+3ab(x^3)(ax^2+bx)=a^3y^2+y(b^3-3abc)$$
$$a^3y^2+y(b^3-3abc)+c^3=0$$ whose roots are $x_1^3,x_2^3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1646209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
integrate $\int \frac{dx}{4+3\sin2x}$
$$\int \frac{dx}{4+3\sin (2x)}$$
$u=2x$
$du=2dx$
$$\frac{1}{2}\int \frac{du}{4+3\sin(u)}$$
$v=\tan(\frac{u}{2})$
$du=\frac{2dv}{1+v^2}$
\begin{align*}
\frac{1}{2}\int \frac{du}{4+3\sin(u)}={}&\frac{1}{2}\int \frac{\frac{2dv}{1+v^2}}{4+\frac{6v}{1+v^2}}=\frac{1}{2}\int \frac{\frac{2dv}{1+v^2}}{\frac{4v^2+6v+4}{1+v^2}}={} \\
{}={}&\frac{1}{2}\int \frac{2dv}{4v^2+6v+4}=\frac{1}{2}\int \frac{2dv}{4v^2+6v+4}={} \\
{}={}&\int \frac{2dv}{4v^2+6v+4}=\int \frac{dv}{4(v+\frac{3}{4})^2-1}
\end{align*}
$m=v+\frac{3}{4}$
$dm=dv$
\begin{align*}
\int \frac{dm}{4m^2-1}={}&4\int \frac{dm}{m^2-\frac{1}{4}}=4\int \frac{dm}{m^2-(\frac{1}{2})^2}={} \\
{}={}&\frac{1}{2}\ln\left(\frac{m-\frac{1}{2}}{m+\frac{1}{2}}\right)+C=\frac{1}{2}\ln\left(\frac{v+\frac{3}{4}-\frac{1}{2}}{v+\frac{3}{4}+\frac{1}{2}}\right)+C={} \\
{}={}&\frac{1}{2}\ln\left(\frac{v-\frac{5}{4}}{v+\frac{11}{4}}\right)+C\frac{1}{2}\ln\left(\frac{v-\frac{5}{4}}{v+\frac{11}{4}}\right)+C={} \\
{}={}&\frac{1}{2}\ln\left(\frac{\tan(x)+\frac{5}{4}}{\tan(x)+\frac{11}{4}}\right)+C
\end{align*}
I think I got it wrong, but I cannot find the mistake.
| $\frac{1}{2}\int \frac{\frac{2dv}{1+v^2}}{4+\frac{6v}{1+v^2}}=\frac{1}{2}\int \frac{\frac{2dv}{1+v^2}}{\frac{4v^2+6v+4}{1+v^2}}=\frac{1}{2}\int \frac{2dv}{4v^2+6v+4}=\frac{1}{2}\int \frac{2dv}{4v^2+6v+4}=\int \frac{2dv}{4v^2+6v+4}=\int \frac{dv}{4(v+\frac{3}{4})^2+\frac{7}{4}}$
Edit - You completed the square wrong. It should be $4(v+\frac{3}{4})^2+\frac{7}{4}$ not $4(v+\frac{3}{4})^2-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1649118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation $7t+[2t] =52 $ ,where $[x]$ denotes the floor function for $x$.
Solve the equation $7t+\left\lfloor 2t\right\rfloor =52 $.
My effort
Using the fact that for any number $x$ we have that $x=\left\lfloor x\right\rfloor+\{x\}$ (where $\{x\}$ is the fractional part of $x$) for $7t$ ,I have that:
\begin{array}{c}
7t+\left\lfloor 2t\right\rfloor &=52 \\
\left\lfloor 7t\right\rfloor+\{7t\} +\left\lfloor 2t\right\rfloor &=52
\end{array}
where $\{7t\}=0$ ,since we have no fractional part, and from this it also follows that $\left\lfloor 7t\right\rfloor=7t$
So the equation breaks down to
\begin{array}{c}
7t+\left\lfloor 2t\right\rfloor=52 \\
\left\lfloor 2t\right\rfloor =52-7t \\
\end{array}
Now, applying the definition of the floor function, I have that
\begin{array}{c}
52-7t \le 2t <53-7t \\
52\le 9t <53 \\
52/9 \le t < 53/9 \\
\end{array}
Question
Is my effort correct? Are there other ways to approach the problem?
| $$7t + \lfloor 2t \rfloor = 52$$
Note that $f(t) = 7t + \lfloor 2t \rfloor$ is an increasing function. Hence there can be no more than one answer.
We note that $f(5) = 45$ and $f(6) = 54$. So the answer is somewhere between $t=5$ and $t=6$.
So lets say $t=5+\delta$ where $0 < \delta < 1$.
If $0 \lt \delta < \dfrac 12$, then
\begin{align}
f(t) &= 52 \\
35+7\delta + \lfloor 10 + 2\delta\rfloor &= 52 \\
45 + 7\delta &= 52 \\
7\delta &=7 \\
\delta &= 1
\end{align}
So there is no solution.
If $\dfrac 12 \lt \delta < 1$, then
\begin{align}
f(t) &= 52 \\
35+7\delta + \lfloor 10 + 2\delta\rfloor &= 52 \\
46 + 7\delta &= 52 \\
7\delta &= 6 \\
\delta &= \dfrac 67
\end{align}
So $t = 5\dfrac 67$.
CHECK:
\begin{align}
7\left(5\dfrac 67 \right) + \left\lfloor 2 \left(5\dfrac 67 \right) \right\rfloor
&=(35 + 6) + \left\lfloor 10 + \dfrac{12}{7} \right\rfloor \\
&= 41 + 11 \\
&= 52
\end{align}
As to your solution.
$$\color{red}{\begin{array}{c}
7t+\left\lfloor 2t\right\rfloor=52 \\
\left\lfloor 2t\right\rfloor =52-7t \\
\end{array}}$$
$$\color{red}{\begin{array}{c}
52-7t \le 2t <53-7t \\
52\le 9t <53 \\
52/9 \le t < 53/9 \\
\end{array}}$$
From $\left\lfloor 2t\right\rfloor =52-7t$, we gather that $7t$ must be an integer.
So we start by making a denominator that has a factor of $7$ in it
$$\dfrac{52\cdot 7}{9\cdot 7} \le t < \dfrac{53\cdot 7}{9\cdot 7}$$
$$\dfrac{364}{9\cdot 7} \le t < \dfrac{378}{9\cdot 7}$$
Next we look for a numerator between $364$ and $377$ that is a multiple of $9$. (Then the nines will cancel and we will have a fraction with a denominator of $7$.) That number is $369$
$$\dfrac{364}{9\cdot 7} \le \dfrac{369}{9\cdot 7} < \dfrac{378}{9\cdot 7}$$
So $t = \dfrac{369}{63} = \dfrac{41}{7} = 5 \dfrac 67$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1649233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Proving that $\cos(\frac{\arctan(\frac{11}{2})}{3}) = \frac{2}{\sqrt{5}}$ I am trying to solve the cubic equation $x^3-15x-4=0$ using Cardano's formula. I already know that the solutions are $x=4$, $x= \sqrt{3}-2$ and $x= -\sqrt{3}-2$ and that using the formula in this problem requires finding the cube roots of $2+11i$ and $2-11i$, which are $2+i$ and $2-i$. But when I try to use the formula on my calculator, a TI-89 Titanium, I get $2\sqrt 5 \sin \left( \frac{\arctan(\frac{2}{11})}{3}+\pi/3 \right)$ instead of $4$. For some reason, the fact that $(2+i)^3 = 2 +11i$ and $x = 4$ is a zero of $x^3-15x-4$ feels like a byproduct of something else. So I have tried for more than a month to prove that $\cos(\frac{\arctan(\frac{11}{2})}{3}) = \frac{2}{\sqrt{5}}$ without using either of these results.
| If $\cos(\theta / {3} )=x;$ then $\cos(\theta) = 4x^3 - 3x$
If $\tan(\theta) = $$11\over 2$; $\cos(\theta) = $$2\over 5\sqrt{5}$
Now, solving$ 4x^3 - 3x =$$2\over 5\sqrt{5}$ , You get x= $2\over \sqrt{5}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1650087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 3
} |
How to form a recurrence for a $n$-digit sequence using digits $0,1,2,3$ so that we have even no of $0$'s? If we assume $T(n)$ to be the function representing the case where we have even number of $0$'s then $T(1)=3$ precisely strings $1, 2$ and $3$.
$T(2)= 10$ ($00,11,22,33,12,21,13,31,23,32$). Likewise I got $T(3)=36$ and $T(4)=124$.
Now how to proceed from here?
|
Here is a variation of the theme based upon regular expressions. We can describe all valid strings as
\begin{align*}
(0(1+2+3)^*0+1+2+3)^*
\end{align*}
Comment:
*
*Valid strings consist of zero or more occurrences of $1$ or $2$ or $3$ or
*when we start with a $0$ it has to be followed by another $0$ possibly with zero or more occurrences of $1,2$ or $3$ in between.
This was the hard part, namely finding an appropriate regular expression. Everything what follows is a matter of technique.
Using the notation from P. Flajolet's and R. Sedgewicks Analytic Combinatorics with the sequence $SEQ(x)$ defined as
\begin{align*}
SEQ(x)=1+x+x^2+\cdots=\frac{1}{1-x}
\end{align*}
we can translate the regular expression into the generating function
\begin{align*}
SEQ&(xSEQ(3x)x+3x)\\
&=\frac{1}{1-\left(\frac{x^2}{1-3x}+3x\right)}\\
&=\frac{1-3x}{1-6x+8x^2}\\
&=1+3x+10x^2+36x^3+136x^4+\cdots
\end{align*}
We use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series.
We obtain the wanted sequence numbers $T(n)$ by partial fraction decomposition of the generating series
\begin{align*}
T(n)&=[x^n]\frac{1-3x}{1-6x+8x^2}\\
&=[x^n]\left(\frac{1}{2(1-4x)}+\frac{1}{2(1-2x)}\right)\\
&=[x^n]\frac{1}{2}\sum_{k\geq 0}4^kx^k+[x^n]\frac{1}{2}\sum_{k\geq 0}2^kx^k\\
&=\frac{1}{2}\left(4^n+2^n\right)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1650356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Showing Trigonometric Identity Prove that:
$$\cos^2\theta\sin^4\theta=\frac{1}{32}(\cos6\theta-\cos2\theta+2-2\cos4\theta)$$
Attempt:
\begin{align*}
L.H.S & = \cos^2\theta\sin^4\theta\\
& = \cos^2\theta\sin^2\theta\sin^2\theta\\
& = \frac{1+\cos2\theta}{2}.\frac{1-\cos2\theta}{2}.\frac{1-\cos2\theta}{2}\\
& = \frac{1}{8} (1-\cos^22\theta)(1-\cos2\theta)
\end{align*}
Now, what should I do?
| Use the linearisation and the duplication formulae:
\begin{align*}
\cos^2\theta\sin^4\theta
& = \cos^2\theta\sin^2\theta\sin^2\theta = \frac14 \sin^2 2\theta\sin^2\theta\\
&=\frac14\frac{1-\cos4\theta}{2}\frac{1-\cos2\theta}{2}= \frac{1}{32} (2\cos4\theta\cos2\theta-2\cos4\theta-2\cos2\theta+2)\\
&=\frac1{32}(\cos 6\theta+\cos2\theta-2\cos4\theta-2\cos2\theta+2)\\
&=\frac1{32}(\cos 6\theta-2\cos4\theta- \cos2\theta+2)\
\end{align*}
If you are allowed to use complex numbers, this is much easier:
set $u=\mathrm e^{\mathrm i\theta}$. Then we have, by Euler's formulae:
$$\bar u=\mathrm e^{-\mathrm i\theta},\quad \cos\theta=\frac{u+\bar u}2,\quad \sin\theta =\frac{u-\bar u}{2\mathrm i},$$
whence (note $u\bar u=1$)
\begin{align*}
\cos^2\theta\sin^4\theta
& = \frac{(u+\bar u)^2}4 \frac{(u-\bar u)^4}{16}=\frac1{64}(u^2-\bar u^2)^2(u-\bar u)^2\\
&=\frac1{64}(u^4-2+\bar u^4)(u^2-2+\bar u^2)\\
&=\frac1{64}(u^6-2u^4+u^2-2u^2+4-2\bar u^2+\bar u^2-2\bar u^4+\bar u^6)\\
&=\frac1{64}(u^6+\bar u^6-2(u^4+\bar u^4) -(u^2+\bar u^2)+4)\\
&=\frac1{32}(\cos6\theta-2\cos4\theta-\cos2\theta+2).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1650766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Pythagorean triplets of the form $a^2+(a+1)^2=c^2$ and the space between them I was searching for pythagorean triples where $b=a+1$, and I found using a python program I made the first 10 integer solutions:
*
*$0^2+1^2=1^2$
*$3^2+4^2=5^2$
*$20^2+21^2=29^2$
*$119^2+120^2=169^2$
*$696^2+697^2=985^2$
*$4059^2+4060^2=5741^2$
*$23660^2+23661^2=33461^2$
*$137903^2+137904^2=195025^2$
*$803760^2+803761^2=1136689^2$
*$4684659^2+4684660^2=6625109^2$
Now what's so interesting? I discovered that any $c$, divided by the previous (for example $5/1$ or $29/5$) limits to $5.828427...=\left(\frac{1}{\sqrt2-1}\right)^2=\sqrt8+3$. My question: why?
| Actually, you should write a general formula, and these issues are constantly emerging.
Solutions of the equation: $$x^2+(x\pm{a})^2=z^2$$
Can be written using the solutions of Pell's equation: $p^2-2s^2=\pm{a}$
And have the form:
$$x=2s(p+s)$$
$$z=p^2+2ps+2s^2$$
If we introduce a replacement:
$$q=p^2+2ps+2s^2$$
$$t=2ps$$
Solutions have the form:
$$x=t^2-q^2$$
$$z=q^2+t^2$$
Solutions have the form:
$$x=2qt$$
$$z=q^2+t^2$$
It must be remembered that the number of $p,s$ can be of any sign.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1651227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 4,
"answer_id": 3
} |
Finding the shortest distance between two Parabolas Recently, a problem asked me to find the minimum distance between the parabolas $y=x^2$ and $y=-x^2-16x-65$.
I proceeded with the problem as thus.
Let $P(a,a^2), Q(b, -b^2-16b-65), a-b=x$.
$\therefore PQ^2=x^2+(2a^2+2ax+16a+x^2+16x+65)^2$.
$PQ^2=x^2+(2(a+\frac{x+8}{2})^2+\frac{(x+8)^2+2}{2})^2 \ge (x^2+(\frac{(x+8)^2+2}{2})^2)(1+\frac{1}{4}) \times \frac{4}{5}$
Applying Cauchy gives us that
$PQ^2 \ge (\frac{1}{4}x^2+3x+\frac{33}{2})^2 \times \frac{4}{5} \ge (\frac{15}{2})^2 \times \frac{4}{5}=75$
This implies that the answer is $\sqrt{75}$.
However, it took me a long time to find the values for Cauchy, and the calculations proved tedious.
What are other approaches to this problem?
EDIT: $(\frac{15}{2})^2 \times \frac{4}{5} \neq 75$, it`s $45$ actually!
| You could define a new function as the distance between a point on one parabola and a point on the other.
We know a generic point for the first parabola is given by $(x,x^2)$ and a generic point for the second parabola is given by $(y,-y^2-16y-65)$
We want to minimize
$$min{\sqrt{(y-x)^2+(-y^2-16y-65-x^2)^2}}$$
Note that this is the same as minimizing
$$min{(y-x)^2+(-y^2-16y-65-x^2)^2}$$
This is another approach.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1651369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
What number comes next in the sequence $7, 16, 8, 27, 9,...$? What number comes next in this series?
$$7, 16, 8, 27, 9,...$$
I thought it was $38$, but I'm wrong.
It is a multiple choice, and options are $27, 10, 40, 37$.
Don't worry - I'm not cheating on anything, but helping my daughter with homework.
And I can't help her since I can't figure it out myself!
| Another view, also giving $40$ as the next term: $2n+2;\frac{n}{2};3n+3;\frac{n}{3};4n+4;\frac{n}{4}\cdots$
\begin{align}\text{initial value}&=7\\
7*2+2&=16\\
\frac{16}{2}&=8\\
8*3+3&=27\\
\frac{27}{3}&=9\\
9*4+4&=40\\
\frac{40}{4}&=10
\end{align}
Or: $2(n+1);\frac{n}{2};3(n+1);\frac{n}{3};4(n+1);\frac{n}{4}\cdots$
\begin{align}\text{initial value}&=7\\
2(7+1)&=16\\
\frac{16}{2}&=8\\
3(8+1)&=27\\
\frac{27}{3}&=9\\
4(9+1)&=40\\
\frac{40}{4}&=10
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1652363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Principal roots in the derivation of the quadratic formula The following is an abbreviation of a common derivation of the quadratic formula:
$$
ax^2+bx+c=0\\
\vdots\\
\sqrt{(x+\frac{b}{2a})^2}=\sqrt{\frac{b^2-4ac}{4a^2}}\\
x+\frac{b}{2a}=\frac{\pm\sqrt{b^2-4ac}}{2a}\\
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\
$$
Following the square root extraction, the equation should have three plus-minus signs:
$$
\pm(x+\frac{b}{2a})=\frac{\pm\sqrt{b^2-4ac}}{\pm2a}\\
$$
But instead there's only one. Is it because the signs cancel each other out, or are moved from one expression to the other (through multiplication in $-1$ or $\frac{-1}{-1}$) in six of the eight permutations that the three plus-minus signs allow; or is it a matter of choosing a principal root for some arcane reason beyond the scope of elementary algebra (eg. branch points)?
| Cancellation and "moving" signs (as you put it) are one explanation for why only one $\pm$ sign is required. We know that
\begin{gather}
\frac{-p}{-q} = \frac{p}{q}, \\[.5ex]
-\left(\frac{p}{q}\right) = \frac{-p}{q} = \frac{p}{-q}, \\[.5ex]
p = -q \quad\text{if and only if}\quad -p = q.
\end{gather}
Using these facts we can show that the next four equations
are all equivalent, and hence by stating any one of them we also imply
that each of the others is true:
\begin{align}
x+\frac{b}{2a} &= \frac{\sqrt{b^2-4ac}}{2a}, &
x+\frac{b}{2a} &= \frac{-\sqrt{b^2-4ac}}{-2a}, \\
-\left(x+\frac{b}{2a}\right) &= \frac{-\sqrt{b^2-4ac}}{2a}, &
-\left(x+\frac{b}{2a}\right) &= \frac{\sqrt{b^2-4ac}}{-2a}.
\end{align}
The following four equations also are all equivalent to each other:
\begin{align}
x+\frac{b}{2a} &= \frac{-\sqrt{b^2-4ac}}{2a}, &
x+\frac{b}{2a} &= \frac{\sqrt{b^2-4ac}}{-2a}, \\
-\left(x+\frac{b}{2a}\right) &= \frac{\sqrt{b^2-4ac}}{2a}, &
-\left(x+\frac{b}{2a}\right) &= \frac{-\sqrt{b^2-4ac}}{-2a}.
\end{align}
By asserting the equation
$$\pm\left(x+\frac{b}{2a}\right)=\frac{\pm\sqrt{b^2-4ac}}{\pm2a},$$
then, assuming we can choose each of the $+$ or $-$ signs independently,
we assert that at least one of the previous eight equations is true.
So either one of the first four of those equations is true, in which case
$x+\frac{b}{2a} = \frac{\sqrt{b^2-4ac}}{2a}$
(as implied by any of those four equations),
or one of the other four of those equations is true, in which case
$x+\frac{b}{2a} = \frac{-\sqrt{b^2-4ac}}{2a}$.
That is, the equation with three $\pm$ signs is simply asserting that
$$
x+\frac{b}{2a} = \frac{\sqrt{b^2-4ac}}{2a} \quad\text{or}\quad
x+\frac{b}{2a} = \frac{-\sqrt{b^2-4ac}}{2a},
$$
which is just what we mean when we write
$$x+\frac{b}{2a} = \frac{\pm\sqrt{b^2-4ac}}{2a}.$$
It is simpler, however, if we apply the rules of sign cancellation, etc., earlier in the procedure, as detailed in the previous comments and answer, so that we never end up with three $\pm$ signs in the first place.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1654541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Solutions of $y'=3x(y-1)^{\frac{1}{3}}$ where $y(0)=1$ So i am looking for the solutions of this differential equation and i am having problems. I need to find $9$ solutions but i can only find $3$. I separated the variables and got this equation $(y-1)^{\frac{2}{3}}=x^2$ well from that we get only two functions $y=1-x^3$ and $y=1+x^3$ and also $y=1$. Where are the $6$ solutions, could someone please explain? Thanks
| What about of
\begin{align}
y_4(x)&=\begin{cases}1&x<0\\x^3+1&x\ge0\end{cases},&y_5(x)&=\begin{cases}1&x<0\\-x^3+1&x\ge 0\end{cases},&y_6(x)&=\begin{cases}x^3+1&x<0\\1&x\ge 0\end{cases}\\[5pt]
y_7(x)&=\begin{cases}-x^3+1&x<0\\1&x\ge0\end{cases},&y_8(x)&=\begin{cases}x^3+1&x<0\\-x^3+1&x\ge 0\end{cases},&y_9(x)&=\begin{cases}x^3+1&x<0\\-x^3+1&x\ge 0\end{cases}?
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1657870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proof of $n(n^2+5)$ is divisible by 6 for all integer $n \ge 1$ by mathematical induction
Prove the following statement by mathematical induction:
$n(n^2+5)$ is divisible by 6 for all integer $n \ge 1$
My attempt:
Let the given statement be p(n).
(1) $1(1^2+5)$=6 Hence, p(1) is true.
(2) Suppose for all integer $k \ge 1$, p(k) is true.
That is, $k(k^2+5)$ is divisible by 6
We must show that p(k+1) is true.
$(k+1)((k+1)^2+5)$=$k^3+3k^2+3k+1+5(k+1)$
=$k^3+3k^2+8k+6$
=$k(k^2+5)+3k^2+3k+6$
I'm stuck on this step. I feel I have to show $3k^2+3k+6$ is divisible by 6. But, how can I show $3k^2+3k+6$ is divisible by 6?
| If $f(n)=n(n^2+5)$
$f(k+1)-f(k)$
$=(k+1)\{(k+1)^2+5\}-k(k^2+5)=3k^2+3k+1+5=6\cdot\dfrac{k(k+1)}2+6$ which is divisible by $6$ as $k(k+1)$ is even
$\implies6\mid f(k)\iff6\mid f(k+1)$
If induction is not mandatory,
$$n(n^2+5)=\underbrace{(n-1)n(n+1)}_{\text{Product of Three consecutive integers }}+6n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1658054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
integrate $\int \frac{x^4+4x+6}{(2-x)^2(4+x^2)}$
$$\int \frac{x^4+4x+6}{(2-x)^2(4+x^2)}$$
$(2-x)^2(4+x^2)=x^4-4x^3+8x^2-16x+16$
$$\int \frac{x^4+4x+6}{x^4-4x^3+8x^2-16x+16}$$
Is there is anything to do else than long division?
| $$\int \frac{x^4+4x+6}{(2-x)^2(4+x^2)}$$
$$(2-x)^2(4+x^2)=x^4-4x^3+20x^2-16x+16=(x^4+4x+6)-(4x^3-20x^2-20x+22)$$
$$\int \frac{x^4+4x+6}{(2-x)^2(4+x^2)}=x-\int \frac{4x^3-20x^2-20x+22}{(2-x)^2(4+x^2)}$$
$$-4(2-x)(4+x^2)=4x^3-8x^2+16x-32=(4x^3-20x^2-20x+22)-(-12x^2-36x+54)$$
$$\int \frac{x^4+4x+6}{(2-x)^2(4+x^2)}=x-\int{\frac{1}{4(2-x)}}+\int \frac{-12x^2-36x+54}{4(2-x)^2(4+x^2)}$$
And so on...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1658648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\int_{0}^{\infty} \sin(t^2)\,dt$ equals $\sqrt{\pi/8}$ Why is $$\int_{0}^{\infty} \sin(t^2)\,dt=\sqrt{\frac{\pi}{8}}$$ I am kinda new to this kind of calculus and I only know that $$\int_{-1}^{1}\sqrt{1-x^2}dx=\frac{\pi}{2}$$
The problem in which I have to use it is:
$$\lim_{x\to\infty}\frac{\int_{0}^{x} \sin(t^2)\,dt}{x^3}$$
| Hint. By setting, for any real number $t$ such that $t>0$,
$$
I_t:=\int_0^\infty e^{-tx^2}\cos (x^2)\:dx,\quad J_t:=\int_0^\infty e^{-tx^2}\sin (x^2)\:dx. \tag1
$$ One may prove that
$$
\begin{align}
\left(I_t\right)^2-\left(J_t\right)^2 &=\left(\int_0^\infty e^{-tx^2}\cos (x^2)\:dx\right)^2-\left(\int_0^\infty e^{-tx^2}\sin (x^2)\:dx\right)^2
\\\\&=\int_0^\infty\!\int_0^\infty e^{-t(x^2+y^2)}\left(\cos (x^2)\cos (y^2)-\sin (x^2)\sin (y^2)\right)dxdy
\\\\&=\int_0^\infty\!\int_0^\infty e^{-t(x^2+y^2)}\cos (x^2+y^2)\:dxdy
\\\\&=\int_0^{\pi/2}d\theta\int_0^\infty r\:e^{-tr^2}\cos(r^2)\:dr
\\\\&=\frac{\pi}4\:\frac t{1+t^2}. \tag2
\end{align}
$$ Similarly
$$
\begin{align}
2\:I_tJ_t &=\int_0^\infty\!\int_0^\infty e^{-t(x^2+y^2)}\sin (x^2+y^2)\:dxdy
\\\\&=\int_0^{\pi/2}d\theta\int_0^\infty r\:e^{-tr^2}\sin (r^2)\:dr
\\\\&=\frac{\pi}4\:\frac1{1+t^2}. \tag3
\end{align}
$$ From $(2)$ and $(3)$, one easily gets
$$
I_t=\sqrt{\frac{\pi}8}\:\sqrt{\frac{\sqrt{1+t^2}+t}{1+t^2}},\quad J_t=\sqrt{\frac{\pi}8}\:\sqrt{\frac{\sqrt{1+t^2}-t}{1+t^2}} \tag4
$$ As $t\to 0$, this leads to
$$
\int_0^\infty \cos (x^2)\:dx=\int_0^\infty \sin (x^2)\:dx=\sqrt{\frac{\pi}8}. \tag5
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1658722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
For all $x$ which are real numbers, prove that $\lfloor 2x\rfloor = \lfloor x\rfloor + \lfloor x+0.5\rfloor.$
For all $x$ which are real numbers, prove that
$$\lfloor 2x\rfloor = \lfloor x\rfloor + \lfloor x+0.5\rfloor.$$
I know that
Let $\lfloor x\rfloor = n$
$n \leq x < n+1$
| $n\leq x< n+1$ , then $f(x)=n$ for all $x$
I) If $n<x< \frac{2n+1}{2}$
$f(x)+f(x+0.5)= n+n=2n=f(2x)$
since $x+0.5<\frac{2n+1}{2} + \frac{1}{2}=n+1$
II) If $n+1>x\geq \frac{2n+1}{2}$
$f(x)+f(x+0.5)= n+n+1=2n+1=f(2x)$
Since $x+0.5\geq\frac{2n+1}{2} + \frac{1}{2}=n+1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1659107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Sum of Arithmetic-Geometric Series So I am having trouble getting the sum of the series:
$1 + 2\left(\frac32\right) + 3\left(\frac{3^2}{2^2}\right) + ... + k\left(\frac{3^{k-1}}{2^{k-1}}\right)$
I cant figure out for the life of me how to figure out the sum of this.
| $$\sum\limits_{i=1}^{k}{i \cdot \left(\frac{3}{2}\right)^{(i-1)}}=\sum\limits_{i=0}^{k-1}{\sum\limits_{j=i}^{k-1}{{\left(\frac{3}{2}\right)}^j}}$$
$$=\sum\limits_{i=0}^{k-1}{\frac{ \left(\frac{3}{2}\right)^i -\left(\frac{3}{2}\right)^k}{1-\frac{3}{2}}}$$
$$=\frac{\frac{1- \left(\frac{3}{2}\right)^k }{ 1-\frac{3}{2}}-k\left(\frac{3}{2}\right) ^k}{-\frac{1}{2}}$$
$$=(2k-4)\left(\frac{3}{2}\right)^k+4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1659545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find the limits for this function Find the limits
$ \lim_{x\to 0} \frac {x^2 sin^2 (x)} {x^2 - sin^2 (x)}$
| $\lim\limits_{x\to 0}\dfrac{x^2 \sin^2 x}{x^2-\sin^2 x} = \lim\limits_{x\to 0}\dfrac{x^2}{x^2\csc^2 x-1} = \lim\limits_{x\to 0}\dfrac{x^2}{(x\csc x-1)(x\csc x +1)} = \dfrac 1 2\lim\limits_{x\to 0} \dfrac{x^2}{x\csc x-1}$
Then apply l'Hospital's Rule ($\times 2$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1660738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Riemann zeta-function functional equation form In Titchmarch's book the functional equation is given as
$$\zeta(s)=2^s\pi^{s-1}\sin \left( \frac{s\pi}2\right) \Gamma(1-s)\zeta(1-s).$$
However, in the third proof, he derives a following equation
$$\pi^{-\frac{s}2}\Gamma\left( \frac{s}2\right) \zeta(s)=\pi^{-\frac12+\frac{s}2}\Gamma\left( \frac12-\frac{s}2\right) \zeta(1-s)$$
and simply states that our initial equation follows from our derived equation. Can somebody explain to me how exactly?
| We obviously have
$$ \zeta(s) = \pi^{- \frac{1}{2} + s} \frac{\Gamma(\frac{1}{2} - \frac{s}{2})}{\Gamma(\frac{s}{2})} \zeta(1 - s).$$
We now use the duplication formula in the form $\Gamma(z + \frac{1}{2}) = 2^{1 - 2z} \sqrt{\pi} \frac{\Gamma(2z)}{\Gamma(z)}$, where $z = -\frac{s}{2}$. This yields
$$\pi^{- \frac{1}{2} + s} \frac{\Gamma(\frac{1}{2} - \frac{s}{2})}{\Gamma(\frac{s}{2})} = 2^{1 + s} \pi^s \frac{\Gamma(-s)}{\Gamma(-\frac{s}{2})\Gamma(\frac{s}{2})}.$$
Now using Euler's reflection formula $\Gamma(z) = \frac{\pi}{\sin(\pi z) \Gamma(1 - z)}$ for $z = -\frac{s}{2}$ gives us
\begin{align*}2^{1 + s} \pi^s \frac{\Gamma(-s)}{\Gamma(-\frac{s}{2})\Gamma(\frac{s}{2})}
&= 2^{1 + s} \pi^{s-1} \frac{\Gamma(-s)\Gamma(1 + \frac{s}{2})}{\Gamma(\frac{s}{2})} \sin(-\frac{\pi s}{2}) \\
&= 2^{1 + s} \pi^{s-1} \frac{\frac{s}{2} \Gamma(-s)\Gamma(\frac{s}{2})}{\Gamma(\frac{s}{2})} \sin(-\frac{\pi s}{2})\\
&= -2^{s} \pi^{s-1} \Gamma(1-s)\sin(-\frac{\pi s}{2}) \\
&= 2^{s} \pi^{s-1} \Gamma(1-s)\sin(\frac{\pi s}{2}).\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to find the indefinite integral for $\int \frac{x^2}{\sqrt{r^2 - x^2}}dx$? I'm trying to find the indefinite integral of
$$ \int \frac{x^2}{\sqrt{r^2 - x^2}}dx$$
where r is a positive number.
EDIT: I'm wondering how to find the integral, not just what it is.
| Let $x=r\sin t$, then $dx=r\cos t \, dt$
\begin{align*}
\int \frac{x^{2}}{\sqrt{r^{2}-x^{2}}} \, dx &=
\int \frac{r^{2}\sin^{2} t}{r\cos t} \times r\cos t \, dt \\ &=
r^{2} \int \sin^{2} t \, dt \\ &=
\frac{r^{2}}{2} \int (1-\cos 2t) \, dt \\ &=
\frac{r^{2}}{2} \left( t-\frac{1}{2} \sin 2t \right) \\ &=
\frac{r^{2}}{2} ( t-\sin t \cos t ) \\ &=
\frac{1}{2} \left( r^{2}\sin^{-1} x- x\sqrt{r^{2}-x^{2}} \, \right)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1663700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Evaluate $\int\frac{da}{a\sqrt{a+1}}$
$$\int\dfrac{da}{a\sqrt{a+1}}$$
I don't know how to solve this integral. The fact that $\dfrac1a$ is the derivative of $\ln(a)$ and $\dfrac{1}{\sqrt{a+1}}$ is the derivative of $\cos^{-1}a$ suggested Integration by Parts.
$$\int\dfrac{da}{a\sqrt{a+1}} = \dfrac{\ln(a)}{\sqrt{a+1}}-\int $$
However I got stuck after this.
Any help with the Integral would be greatly appreciated. Many thanks!
| Trig sub approach:
$$\int\frac{1}{a\sqrt{a+1}}\space\text{d}a=$$
Substitute $a=\tan{\theta}$ and $da=\frac{2\tan{\theta}}{\cos^2{\theta}}du $:
$$\int\frac{1}{\tan^2{\theta}\sqrt{\tan^2{\theta}+1}}\frac{2\tan{\theta}}{\cos^2{\theta}} d\theta =$$
$$=\int\frac{1}{\tan^2{\theta}\frac{1}{\cos{\theta}}}\frac{2\tan{\theta}}{\cos^2{\theta}} d\theta=$$
$$=2\int \frac{1}{\tan{\theta}\cos{\theta}}d\theta=$$
$$= 2\int \frac{1}{\sin{\theta}}d\theta$$
Now with these integral, we do:
$$\int \frac{1}{\sin{\theta}}d\theta = \int \frac{\sin{\theta}}{\sin^2{\theta}}d\theta = \int \frac{{\sin{\theta}}}{1-\cos^2{\theta}}d\theta$$
Using substitution $t = \cos{\theta}$ and $dt$ = $-\sin{\theta} d\theta$:
$$-\int \frac{1}{1-t^2}dt = \int \frac{-1}{(1-t)(1+t)}dt = \int \frac{1}{2(t-1)} - \frac{1}{2(1+t)} dt = \frac{1}{2} \ln{(t-1)} - \frac{1}{2} \ln{(t+1)} - C$$(-C because i'm special)
So the original integral is:
$$\int \frac{1}{a\sqrt{a+1}} = \int \ln{(t-1)} - \ln{(t-1)}dt = \ln{\Bigl(\frac{\arctan{(\cos{a})} -1}{\arctan{(\cos{a})}+1} \Bigr)} - C $$
Not really a nice solution, but still.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1665419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Find integers $x,y$ such that $p=mx+ny$ Given a positive value $p$ and positive integers $m, n$ less than or equal to $p$, find integers (possibly negative) $x,y$ such that $p=mx+ny$.
What method can be used to find $x$ and $y$?
| Are you familiar with the Euclidean algorithm? https://en.wikipedia.org/wiki/Euclidean_algorithm.
The Euclidean algorithm gives a way of finding $\gcd(m,n)$ and integers $x,y$ such that $mx + ny = \gcd(m,n)$. You can then scale to find a solution to $mx + ny = p$, assuming $\gcd(m,n) \mid p$. If not, then there is no solution.
Here is an example of how it works: Let $m = 6, n = 64$. We have
\begin{equation*}
\begin{aligned}
64 &= 6 \cdot 10 + 4 \\
6 &= 4 \cdot 1 + 2 \\
4 &= 2 \cdot 2 + 0.
\end{aligned}
\end{equation*}
Since $2$ is the last nonzero remainder, we have $2 = \gcd(6,64)$.
Now observe that
\begin{equation*}
\begin{aligned}
2 &= 6 - 4\cdot 1 \\
&= 6 - (64 - 6 \cdot 10) \cdot 1 \\
&= 6 - 1 \cdot 64 + 6 \cdot 10 \\
&= 11 \cdot 6 - 1 \cdot 64.
\end{aligned}
\end{equation*}
So $x = 11, y = -1$ is a solution to the equation $6m + 64y = \gcd(6,64)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1666190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Computing Limits of Multivariable Functions I have seen this problem on an exam I was looking at:
Let $f(x,y) = \frac{y}{1+xy} - \frac{1 - y\sin(\frac{\pi x}{y})}{\arctan(x)}$
Compute the limit:
$g(x) = \lim_{y\to\infty} f(x,y)$
I am confused on how to solve this limit however. Wouldn't the value of the limit also depend on $x$ ? (this problem lets $x$ be independent?)
| Let $f(x,y) = \frac{y}{1+xy} - \frac{1-ysin(\frac{\pi x}{y})}{\arctan(x)}$, let's modify our function
$$\frac{y}{1+xy} - \frac{1-ysin(\frac{\pi x}{y})}{\arctan(x)} = \frac{1}{x}\frac{1}{\frac{1}{xy}+1} - \frac{1-ysin(\frac{\pi x}{y})}{\arctan(x)},$$
now let $z = \frac{1}{y}$, hence
$$\lim_{y\rightarrow \infty} f(x,y) = \lim_{z\rightarrow 0} f(x,\frac{1}{z})=\lim_{z\rightarrow 0}\frac{1}{x}\frac{1}{\frac{z}{x}+1} - \frac{1-\frac{1}{z}sin(z\pi x)}{\arctan(x)},$$
clearly $\frac{1}{x}\frac{1}{\frac{z}{x}+1} \rightarrow \frac{1}{x},$
let's have a closer look at $\frac{1-\frac{1}{z}sin(z\pi x)}{\arctan(x)}$.
Using Taylor series we can write
$$\frac{1}{z}sin(z\pi x) = \frac{1}{z}\Big(z\pi x + \mathcal{O}(z^3)\Big)=\pi x + \mathcal{O}(z^2),$$
Can you go from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1668941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Compare five ways of solving cubic equation by iterations (nested expressions) Say we have a depressed cubic equation in the general form:
$$x^3-bx-c=0$$
There are basically five ways of solving it by iterations. Let's consider them in no particular order (the names are my own).
1) The continued fraction / nested radical method:
$$x=\sqrt{b+\frac{c}{x}}=\sqrt{b+\cfrac{c}{\sqrt{b+\cfrac{c}{x}}}}=\sqrt{b+\cfrac{c}{\sqrt{b+\cfrac{c}{\sqrt{b+\cfrac{c}{x}}}}}}$$
Take some $x_0$ and solve the recurrence:
$$x_n=\sqrt{b+\frac{c}{x_{n-1}}}$$
2) The continued fraction / nested square method:
$$x=-\cfrac{c}{b-x^2}=-\cfrac{c}{b-\cfrac{c^2}{(b-x^2)^2}}=-\cfrac{c}{b-\cfrac{c^2}{\left( b-\cfrac{c^2}{(b-x^2)^2} \right)^2}}$$
3) The branching continued fraction method:
$$x=\cfrac{b}{x}+\cfrac{c}{x^2}=\cfrac{b}{\cfrac{b}{x}+\cfrac{c}{x^2}}+\cfrac{c}{\left( \cfrac{b}{x}+\cfrac{c}{x^2} \right)^2}$$
It can also be expressed in several different ways, like this one:
$$x=\cfrac{c+bx}{x^2}=\cfrac{c+b\cfrac{c+bx}{x^2}}{\cfrac{(c+bx)^2}{x^4}}$$
But anyway, for some $x_0$:
$$x_n=\cfrac{b}{x_{n-1}}+\cfrac{c}{x_{n-1}^2}$$
4) Nested cubic roots method:
$$x=\sqrt[3]{c+bx}=\sqrt[3]{c+b\sqrt[3]{c+bx}}=\sqrt[3]{c+b\sqrt[3]{c+b\sqrt[3]{c+bx}}}$$
5) And nested cubes method:
$$x=\frac{1}{b} \left(-c+x^3 \right)=\frac{1}{b} \left(-c+\frac{1}{b^3} \left(-c+x^3 \right)^3 \right)$$
They all work in different ways, so I wonder, what are general rules of using one method or the other? How to get all three roots by using them?
For example we have the equation with three real roots $(-3.65617,0.42297,3.23319)$:
$$x^3-12x+5=0$$
Let's take $x_0=1$. Then methods 1, 3 and 4 will converge to the third root, while methods 2 and 5 will converge to the second root.
Changing $x_0$ to other values will not lead to the first root, but some methods will not converge (mainly, method 3).
| All your iterations are of the form $x_{i+1}=f(x_i)$ for some $f$ chosen so that at the root $x=f(x)$ If $|f'(x)| \lt 1$ at the root, the iteration will converge to the root for starting values that are close enough. The smaller you can make $|f'(x)|$, the faster the method will converge. Your first iteration has $f(x)= \sqrt{12-\frac 5x}$, which has $f'(x)$ about $0.074$ at the third root. That small value is why it converges so quickly. In contrast, your third method has $f'(x) \approx -0.85$ at the third root. The minus sign is what makes the oscillation in your plot-if $x_i$ is below the root $x_{i+1}$ is above. The value close to $1$ in absolute value is why the convergence is slow. You are waiting for $(-0.85)^n$ to get small.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1669163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Number of solutions of $8\sin(x)=\frac{\sqrt{3}}{\cos(x)}+\frac{1}{\sin(x)}$ Number of solution in $[0,2\pi]$ satisfying the equation
$$8\sin(x)=\frac{\sqrt{3}}{\cos(x)}+\frac{1}{\sin(x)}$$
Options are $5$ or $6$ or $7$ or $8$.
Doing some manipulations I reached $4\sin^2(x)\cos(x)=\sin(x+\pi/6)$ but I think this direction will not lead towards result. How to proceed?
| Multiply through by $\sin x\cos x$. We get
$$8\cos x(1-\cos^2 x)=\sqrt{3}\sin x+\cos x.$$
Recall that $\cos 3x=4\cos^3 x-3\cos x$. So the left-hand side becomes $8\cos x-2(\cos 3x+3\cos x)$, that is, $2\cos x-2\cos 3x$. Thus our equation can be rewritten as
$$\cos3x=\frac{1}{2}\cos x-\frac{\sqrt{3}}{2}\sin x.$$
The right-hand side is equal to $\cos(x+\pi/3)$, so our equation simplifies to
$$\cos 3x=\cos(x+\pi/3).$$
To finish, recall that $\cos s=\cos t$ if and only if $s=\pm t+2k\pi$ for some integer $k$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1672613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Max and min of $5 \sin\left(\frac{\pi}{6} x\right)+10 \cos\left(\frac{\pi}{6}x\right) +11.2$ without graphing
$$P(x)= 5 \sin\left(\frac{\pi}{6} x\right)+10 \cos\left(\frac{\pi}{6} x\right) +11.2$$
How would you mathematically find the the max and min points of $P(x)$ without graphing?
I know that individually if say $5 \sin\left(\frac{\pi}{6} x\right)+11.2$ is of $c(x)$ then max and min would be when $\sin\left(\frac{\pi}{6} x\right)= 1$ or $-1$ ( max = 16.2 min = 6.2)
thanks in advance
| Assume that $$ \cos\ t=\frac{5}{\sqrt{5^2+10^2}},\ \sin\ t = \frac{10}{\sqrt{5^2+10^2}} $$
Hene $$ P(t)=\sqrt{5^2+10^2} \{ \cos\ t\sin\ \frac{\pi x}{6} + \sin\ t\cos\ \frac{\pi x}{6} \} + 11.2 $$ $$=5\sqrt{5} \sin\ (\frac{\pi x}{6} + t) + 11.2 $$
That is max$ =5\sqrt{5} +11.2$ and min$=-5\sqrt{5}+11.2 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1673928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Differentiaion Calculus: Trig Inverse Function. Well, yesterday at a Mathematics exam, i had to find $\frac{dy}{dx}$ of a cotangent inverse function
$$ y=\text{arccot}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-sin x} }\right]$$
My Approach:
$$ \text{Multiply and Divide by}\frac{1}{\sqrt{1+\sin x}}$$
$$ \text{but}\space \frac{\sqrt{1-sinx}}{\sqrt{1+\sin x}}=\sec x-\tan x$$
$$\text{so} \space \space y=\text{arccot}\left[\frac{1+\sec x-\tan x}{1-\sec x+\tan x}\right]$$
$\text{Then no idea, Anwers is 1/2}$
Can anyone Give me Full answer with explanation
| Let $x=\dfrac\pi2-2y\implies\sin x=\cos2y$ and $\dfrac{dx}{dy}=-2\ \ \ \ (1)$
WLOG $0\le2y\le\pi\implies0\le y\le\dfrac\pi2\iff\dfrac\pi4\le\dfrac\pi4+y\le\dfrac{3\pi}4$
$$\implies\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} =\dfrac{\sqrt{1+\cos2y}+\sqrt{1-\cos2y}}{\sqrt{1+\cos2y}-\sqrt{1-\cos2y}}$$
$$=\dfrac{\cos y+\sin y}{\cos y-\sin y} =\dfrac{1+\tan y}{1-\tan y}=\tan\left(\dfrac\pi4+y\right)$$
$$\implies\text{arccot}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x} }\right]=\dfrac\pi2-\arctan\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x} }\right]$$
$$=\dfrac\pi2-\arctan\left[\tan\left(\dfrac\pi4+y\right)\right]$$
$$u=\arctan\left[\tan\left(\dfrac\pi4+y\right)\right]=\begin{cases}\dfrac\pi4+y &\mbox{if } -\dfrac\pi2\le\dfrac\pi4+y\le\dfrac\pi2 \\
\dfrac\pi4+y-\pi & \mbox{if } \dfrac\pi2<\dfrac\pi4+y<\pi \end{cases}$$
$$\implies\dfrac{du}{dy}=-1$$
Now use $\dfrac{du}{dx}=\dfrac{du}{dy}\cdot\dfrac{dy}{dx}=(-1)\cdot\left(-\dfrac12\right)=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1675333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that for any integers $a, b, c, d$ number $(a^2 + b^2)(c^2 + d^2)$ is a sum of two squares of integer. Prove that for any integers $a, b, c, d$ number $(a^2 + b^2)(c^2 + d^2)$ is a sum of two squares of integer.
In fact I have no idea how to do this and I'll appreciate any tips or the solution.
| \begin{align}
(a^2+b^2)(c^2+d^2)& =a^2c^2+a^2d^2+b^2c^2+b^2d^2 \\[10pt]
& =a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2 \\[10pt]
&=(ac+bd)^2+(ad-bc)^2
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1677192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the point $(x_0, y_0)$ on the line $ax + by = c$ that is closest to the origin. Find the point $(x_0, y_0)$ on the line $ax + by = c$ that is closest to the origin.
According to this source, I thought that $\left( -\frac{ac}{a^{2}+b^{2}}, -\frac{bc}{a^{2}+b^{2}} \right)$ was the point but it doesn't seem to be correct. Thanks for any help.
| $$(h-0)^2+\left(\dfrac{c-ah}b\right)^2=\dfrac{(a^2+b^2)h^2-2cah+c^2}{b^2}$$
Now $(a^2+b^2)h^2-2cah=\left(h\sqrt{a^2+b^2}-\dfrac{ca}{\sqrt{a^2+b^2}}\right)^2-\dfrac{(ca)^2}{a^2+b^2}\ge-\dfrac{(ca)^2}{a^2+b^2}$
the minimum value occurs iff
the equality occurs if $$h\sqrt{a^2+b^2}-\dfrac{ca}{\sqrt{a^2+b^2}}=0\iff h=?$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1678687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
If $x=\omega-\omega^2-2$, then the value of $x^4+3x^3+2x^2-11x-6$ is?
If $x=\omega-\omega^2-2$, then the value of $x^4+3x^3+2x^2-11x-6$ is? ($\omega$ represents the cube roots of unity not equal to $1$).
Directly substituting the given value will work. But there is no fun in that. The calculations will also get lengthy.
$$\omega-\omega^2-2=2\omega-1$$
I tried simplifying the given equation.
$$x^4+3x^3+2x^2-11x-6=x^3(x+3)+2\left(x+\frac{1}{2}\right)(x-6)$$
I got stuck here.
| As @David said, $x^2=4ω^2-4ω+1=-8ω-3=-4x-7$
Note that $$x+1=2ω \Leftrightarrow x^4+3x^3+3x^2+x=8x \Leftrightarrow x^4+3x^3=7x-3x^2$$
This implies that $$x^4+3x^3+2x^2-11x-6=-x^2-4x-6=1$$
However, another way to appraoch it using @David's method is by noting that $$x^4+3x^3+2x^2-11x-7+1=(x^2+4x+7)(x^2-x-1)+1=1$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1679489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove or disprove $d\mid (a^2-1)\Longrightarrow d\mid (a^4-1)$ I need to prove or to give a counter-example:
$$d\mid (a^2-1)\Longrightarrow d\mid (a^4-1)$$
My attempt:
Yes, this is correct,
First: $(a^2-1)=(a-1)(a+1)\\
(a^4-1)=(a-1)(a+1)(a^2+1)$
If $d\mid (a-1)(a+1)$ so $\exists k_1 : d\cdot k_1=(a-1)(a+1)$
Therfore $d$ must divide also $(a-1)(a+1)(a^2+1)$
Is my attempt correct?
| The factorization $a^2-1=(a-1)(a+1)$ is not helpful in proving or disproving this claim. Essential is the question if $a^2-1$ divides $a^4-1$. So, the factorization of
\begin{align*}
a^4-1
\end{align*}
is relevant. In case the following Lemma is already known to you, you could state:
Since the following Lemma holds
\begin{align*}
d|m\Rightarrow d|mn\quad\qquad \text{for all } d,m,n\in\mathbb{Z}
\end{align*}
and
\begin{align*}
a^4-1=(a^2-1)(a^2+1)
\end{align*}
the claim is valid by taking
\begin{align*}
&m=a^2-1\\
&n=a^2+1\\
&\qquad\qquad\qquad\qquad\Box
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1681578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.