Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find the area of the region described by $|5x|+|6y| \le 30 $
Find the area of the region described by $|5x|+|6y| \le 30 $ (where
$|z|$ denotes the absolute value of $z$).
My effort
Imagining a number line and interpreting the problem as the request to have the sum of the distances of $5x,6y$ from $0$ to be less or equal to $30$ ,we have that :
i) for $|5x| \le 30-h $, where $0 \le h \le 15 $, then $ |6y| \le h$ (and also the other way around : $ |6y| \le 30-h $ and $|5x| \le h$).
Therefore $15\le 5x \le 30 $ and $-30 \ge 5x \ge -15 $ which implies $x \in [-6,-3]\cap [3,6] $ and so we also we find $y \in [-15/6,15/6] $.
Inverting the cases we find $x \in [3,-3]$ and $y\in [-5,-15/6]\cap [15/6,5]$.
By plotting several points for $ 0 \le x \le 6 $ and using the fact that we have symmetry about the origin $(0,0)$ we find that the requested area is a square with side $\sqrt{61}$ (since one side will have coordinates $(-6,0)$ and $(0,-5)$)
Question
Is my effort correct ?How could I have simplified it (I think I have
made the problem harder than it needed to be in some part) ? Are there
other (easier ?) ways to tackle the problem ?
|
$$|5x| + |6y| \le 30$$
There are a couple ways to approach this
Approach 1:
Consider only the $x \ge 0$ and $y \ge 0$ quadrant. The shape enclosed is a triangle, with $x$-intersept at $5\cdot 0 + 6y = 30$, that is $y=5$. Similarly the $y$-intersept is at $x = 6$. So the area of the triangle is $\frac{1}{2}5\times 6 = 15$, and the area across all 4 quadrants is then $4\times 15 = 60$.
Approach 2:
$|x| + |y| \le 1$ is a diamond with edgelength $\sqrt 2$. Thus the area of the diamond is $2$. $|x| + |y| \le 30$ increases the radius of the diamond $30$ times so the area is $2\times 30 \times 30 = 1800$. $|5x| + |y| \le 30$ decreases all $x$ values by a factor of $5$ so the area is $1800 / 5 = 360$. $|5x| + |6y| \le 30$ decreases all $y$ values by a factor of $6$ so the area is $360 / 6 = 60$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1648182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Proof of limit of $x^2\left(\arcsin\left(\frac{1}{x}\right)-\arcsin\left(\frac{1}{x+1}\right)\right)$ correct? I want to show that $$\lim_{x\to\infty} x^2\left(\arcsin\left(\frac{1}{x}\right)-\arcsin\left(\frac{1}{x+1}\right)\right)= 1 $$
But I'm not sure if it's correct:
Because $\arcsin:[-1;1]\to \mathbb{R}$ is continuous and differentiable on $(-1,1)$, the mean value theorem says that there is a $c$ in $\left(\frac{1}{x+1} , \frac{1}{x}\right)$ for $f(t)=\arcsin(t)$ such that
$$x^2\left(\arcsin\left(\frac{1}{x}\right)-\arcsin\left(\frac{1}{x+1}\right)\right)= f'(c_x) {\frac{x}{1+x}} $$
then
$f'(c_x)=\frac{1}{(\sqrt{1-c_x})}(\frac{x}{x+1}))\leqslant \frac{1}{(\sqrt{1-c_x})}\times\frac{x}{x} = \frac{1}{(\sqrt{1-c_x})} $
with Sandwichlemma for $x\to\infty, c_x\to0$
so the limit of $(x^2(\arcsin{\frac{1}{x}} -\arcsin{\frac{x}{1+x}} )= 1$
| Noting that $\frac{d}{dx}\arcsin\left(\frac{1}{x}\right)=\frac{-1}{x\sqrt{x^2-1}}$, we can write:
$$x^2\left(\arcsin\left(\frac{1}{x}\right)-\arcsin\left(\frac{1}{x+1}\right)\right)=x^2\int_x^{x+1}\frac{dt}{t\sqrt{t^2-1}}$$
The integrand is nestled between $\frac{1}{(x+1)\sqrt{x^2+2x}}$ and $\frac{1}{x\sqrt{x^2-1}}$, both of which are readily asymptotic to $\frac{1}{x^2}$. That the limit is $1$ follows directly from this by considering the squeeze theorem
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1650854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Application of z In which lines of the $w$ plane is the line $|z| = \frac{1}{2} $ transformed with the function $w= \frac{1}{z} $?
$$| z | = \frac{1}{2}$$
$$\sqrt{x^2 + y ^2} = \frac{1}{2}$$
$$x = \sqrt{\frac{1}{4} - y^2}$$
$$y = \sqrt{\frac{1}{4} - x^2}$$
$$w = u+iv = \frac{1}{z}=\frac{1}{u+iv}=\frac{x-iy}{x^2+y^2}$$
$$u=\frac{x}{x^2+y^2}$$
$$v=-\frac{y}{x^2+y^2}$$
such that
$$w = \frac{\sqrt{\frac{1}{4}-y^2}}{\frac{1}{2}-x^2-y^2} -i \frac{\sqrt{\frac{1}{4}-x^2}}{\frac{1}{2}-x^2-y^2}$$
However, I don't see how $w = u^2 + v^2 = 4$, which is a circle and the answer according to the problem set.
| $$z=\frac 1w\Rightarrow \left|\frac 1w\right|=\frac 12\Rightarrow |w|=2\Rightarrow u^2+v^2=4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1651152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
integrate $\int \frac{dx}{x\sqrt{1-x}}$
$$\int \frac{dx}{x\sqrt{1-x}}$$
$$\int \frac{dx}{x\sqrt{1-x}}$$
$u=1-x$
$du=-dx$
$$-\int \frac{du}{(1-u)\sqrt{u}}$$
$a(1-u)+b\sqrt{u}=1\Rightarrow a-au+b\sqrt{u}=1$
$a=1\Rightarrow b\sqrt{u}-u=0\Rightarrow b=\sqrt{u}$
$$\int \frac{\sqrt{u}}{1-u}du-\int \frac{du}{\sqrt{u}}=\int \frac{\sqrt{u}}{1-u}du-2{\sqrt{u}}$$
$$\int \frac{\sqrt{u}}{1-u}du=\int \frac{1+\sqrt{u}-1}{(1+\sqrt{u})(1-\sqrt{u})}du=\int \frac{1+\sqrt{u}}{(1+\sqrt{u})(1-\sqrt{u})}du-\int \frac{du}{(1+\sqrt{u})(1-\sqrt{u})}du=\int \frac{1}{(1-\sqrt{u})}du-\int \frac{du}{(1+\sqrt{u})(1-\sqrt{u})}du$$
$$\int \frac{1}{(1-\sqrt{u})}du$$
How do I continue from here? it seems that I have made it harder
| Hint 1:- Let, $\sqrt{u}=t\implies\dfrac{du}{\sqrt{u}}=2dt$. So, $$\int\dfrac{du}{(1-u)\sqrt{u}}=\int\dfrac{du}{(1-\sqrt{u})(1+\sqrt{u})\sqrt{u}}=2\int\dfrac{dt}{(1-t)(1+t)}$$
Hint 2:- $$I=\int\dfrac{du}{(1-u)\sqrt{u}}=\int\dfrac{du}{(1-\sqrt{u})(1+\sqrt{u})\sqrt{u}}=\int\dfrac{2(1-\sqrt{u})+(1+\sqrt{u})+\sqrt{u}}{(1-\sqrt{u})(1+\sqrt{u})\sqrt{u}}$$Hence we get, $$I=2\color{red}{\int\dfrac{du}{(1+\sqrt{u})\sqrt{u}}}+\color{blue}{\int\dfrac{du}{(1-\sqrt{u})\sqrt{u}}}+\color{green}{\int\dfrac{du}{(1-\sqrt{u})(1+\sqrt{u})}}=2\color{red}{I_1}+\color{blue}{I_2}+\color{green}{I_3}$$Further, $$\color{red}{I_1=\int\dfrac{du}{(1+\sqrt{u})\sqrt{u}}=\int\dfrac{(1+\sqrt{u})-\sqrt{u}}{(1+\sqrt{u})\sqrt{u}}du=\int\dfrac{du}{\sqrt{u}}-\int\dfrac{du}{1+\sqrt{u}}}$$$$\color{blue}{I_2=\int\dfrac{du}{(1-\sqrt{u})\sqrt{u}}=\int\dfrac{(1-\sqrt{u})+\sqrt{u}}{(1-\sqrt{u})\sqrt{u}}du=\int\dfrac{du}{\sqrt{u}}+\int\dfrac{du}{1-\sqrt{u}}}$$$$\color{green}{I_3=\int\dfrac{du}{(1+\sqrt{u})(1-\sqrt{u})}=\dfrac{1}{2}\int\dfrac{(1+\sqrt{u})+(1-\sqrt{u})}{(1+\sqrt{u})(1-\sqrt{u})}du=\int\dfrac{du}{1+\sqrt{u}}+\int\dfrac{du}{1-\sqrt{u}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1651932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Calculate $\int_0^1 \ \int_0^1 \ x \sin \lvert x^2-y^2 \lvert \; dx \; dy$ $$\int_0^1 \ \int_0^1 \ x \ \sin \lvert x^2-y^2 \lvert dx \ dy $$
$$\int_0^1 \frac{1}{2} \Big[ \sin \lvert x^2-y^2 \lvert \Big]_0^1 \ dy=
\int_0^1 \frac{1}{2} \Big( \sin \lvert 1-y^2 \lvert - \sin \ (y^2) \Big) \ dy $$
How can I continue the integration?
Thanks!
| The integral is equal to
$$\begin{align}I &=\int_0^1 dx \, x \, \int_0^x dy \, \sin{(x^2-y^2)} - \int_0^1 dx \, x \, \int_x^1 dy \, \sin{(x^2-y^2)}\\ &= \int_0^1 dx \, x \, \int_0^x dy \, \sin{(x^2-y^2)} - \int_0^1 dy \, \int_0^y dx \, x \, \sin{(x^2-y^2)} \\ &= \int_0^1 dx \, x \, \int_0^x dy \, \sin{(x^2-y^2)} + \int_0^1 dx \, \int_0^x dy \, y \, \sin{(x^2-y^2)} \\ &= \int_0^1 dx \, \int_0^x dy \, (x+y) \sin{(x^2-y^2)} \\ &= \frac12 \int_0^1 dv \, \int_0^{2-v} du \, u \sin{(u v)}\\ &= \frac12 \int_0^1 dv \left [\frac{\sin{[v (2-v)]}}{v^2} - \frac{\cos{[v (2-v)]}}{v} \right ] \end{align}$$
which finally, is expressible in terms of the Fresnel integrals
$$C(z) + i S(z) = \int_0^z dt \, e^{i \pi t^2/2}$$
as
$$I = 2 - \sin{1} - \sqrt{2 \pi} \left [C \left (\sqrt{\frac{2}{\pi}} \right ) \cos{1}+ S \left (\sqrt{\frac{2}{\pi}} \right ) \sin{1} \right ] $$
Here is a summary the steps I took:
1) Split up integration region according to whether $x^2-y^2$ is pos or neg
2) Reverse order of integration in second double integral
3) Switch $y$ and $x$
4) Combine the double integrals into a single double integral
5) Change coordinates to $u=x+y$, $v=x-y$. factor of $1/2$ in front is the Jacobian of the coordinate transformation.
6) Integrate wrt $u$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove that $2^{n(n+1)}>(n+1)^{n+1}\left(\frac{n}{1}\right)^n\left(\frac{n-1}{2}\right)^{n-1}\cdots \left(\frac{2}{n-1}\right)^{2}\frac{1}{n}$ If $n$ be a positive integer $>1$, prove that
$$2^{n(n+1)}>(n+1)^{n+1}\left(\frac{n}{1}\right)^n\left(\frac{n-1}{2}\right)^{n-1}\left(\frac{n-2}{3}\right)^{n-2}\cdots \left(\frac{2}{n-1}\right)^{2}\frac{1}{n}$$
Please help me to prove the above. I have to use laws of inequality like AM-GM. But how to use it for this particular problem.
Edit:
Only use laws of inequality.
Edit 2
I want to solve this by using laws of inequality like weighted AM-GM. My attempt is the following
Consider positive numbers $\left(\frac{n}{1}\right), \left(\frac{n-1}{2}\right), \left(\frac{n-2}{3}\right), \cdots \left(\frac{2}{n-1}\right), \frac{1}{n}$ with corresponding weights $n, n-1, n-2, \cdots 2,1$, respectively and applying weighted AM>GM, we get,
$$\frac{\left(\frac{n^2}{1}\right)+\left(\frac{(n-1)^2}{2}\right)+\left(\frac{(n-2)^3}{3}\right)+\cdots \left(\frac{2^2}{n-1}\right)\frac{1^1}{n}}{n+(n-1)+\cdots +2+1}>\left[\left(\frac{n}{1}\right)^n\left(\frac{n-1}{2}\right)^{n-1}\left(\frac{n-2}{3}\right)^{n-2}\cdots \left(\frac{2}{n-1}\right)^{2}\frac{1}{n}\right]^{\frac{n(n+1)}{2}}$$
I am unable to get the result because I am unable to get the sum $\left(\frac{n^2}{1}\right)+\left(\frac{(n-1)^2}{2}\right)+\left(\frac{(n-2)^3}{3}\right)+\cdots \left(\frac{2^2}{n-1}\right)\frac{1^1}{n}$
Please suggest me some possible approach.
| Here is a proof by induction
that starts with my formula
for the right side.
The right side is
$s(n)
=(n+1)^{n+1}\prod_{k=1}^{n} \left(\frac{n-k+1}{k} \right)^{n-k+1}
$.
$\begin{array}\\
s(n)
&=(n+1)^{n+1}\prod_{k=1}^{n} \left(\frac{n-k+1}{k} \right)^{n-k+1}\\
&=(n+1)^{n+1}\frac{\prod_{k=1}^{n}(n-k+1)^{n-k+1}}{\prod_{k=1}^{n} k^{n-k+1}}\\
&=(n+1)^{n+1}\frac{\prod_{k=1}^{n}k^k}{\prod_{k=1}^{n} k^{n-k+1}}\\
&=(n+1)^{n+1}\frac{\prod_{k=1}^{n}k^{2k}}{\prod_{k=1}^{n} k^{n+1}}\\
&=(n+1)^{n+1}\frac{\prod_{k=1}^{n}k^{2k}}{(n!)^{n+1}}\\
\end{array}
$
Therefore
$\begin{array}\\
\frac{s(n+1)}{s(n)}
&=\frac{(n+2)^{n+2}\frac{\prod_{k=1}^{n+1}k^{2k}}{((n+1)!)^{n+2}}}{(n+1)^{n+1}\frac{\prod_{k=1}^{n}k^{2k}}{(n!)^{n+1}}}\\
&=\frac{(n!)^{n+1}(n+2)^{n+2}(n+1)^{2(n+1)}}{((n+1)!)^{n+2}(n+1)^{n+1}}\\
&=\frac{(n!)^{n+1}(n+2)^{n+2}(n+1)^{n+1}}{((n+1)!)^{n+1}(n+1)!}\\
&=\frac{(n+2)^{n+2}(n+1)^{n+1}}{(n+1)^{n+1}(n+1)!}\\
&=\frac{(n+2)^{n+2}}{(n+1)!}\\
\end{array}
$
Since,
if
$t(n)
=2^{n(n+1)}
$,
$\frac{t(n+1)}{t(n)}
=2^{(n+2)(n+3)-(n+1)(n+2)}
=2^{2(n+2)}
$,
if we can show that
$\frac{(n+2)^{n+2}}{(n+1)!}
< 2^{2(n+2)}
$,
we are done.
This is
$n+2
< 4((n+1)!)^{1/(n+2)}
$
or,
replacing $n+1$ by $n$,
$n+1
< 4(n!)^{1/(n+1)}
$.
Since
$n!
\gt (n/e)^n
$,
this is implied by
$\begin{array}\\
n+1
< 4(n/e)^{n/(n+1)}\\
or\\
(n+1)^{n+1}
< 4^{n+1}(n/e)^n\\
or\\
(n+1)e^n(1+1/n)^n
<4^{n+1}\\
or\\
(n+1)e^{n+1}
<4^{n+1}\\
or\\
n+1
< (4/e)^{n+1}\\
or,
\text{again replacing }n+1
\text{ by }n,\\
n
< (4/e)^n\\
or\\
n^{1/n}
< 4/e\\
\end{array}
$
Since
$n^{1/n}
\le e^{1/e}
< 1.45
$
and
$4/e
> 1.47
$,
this is true for all $n$.
Therefore
$\frac{s(n+1)}{s(n)}
< \frac{t(n+1)}{t(n)}
$
or
$s(n+1)
< t(n+1)\frac{s(n)}{t(n)}
$.
Therefore,
if
$s(n) < t(n)$,
$s(n+1) < t(n+1)$.
Since
$s(2)
=3^32^2\frac12
=54
$
and
$t(2)
=2^{6}
=64
\gt s(2)
$,
$t(n) > s(n)$
for
$n \ge 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1654070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 1
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Is there a closed-form of $\frac{1}{1}+\frac{1}{1+2^2}+\frac{1}{1+2^2+3^2}+.....$ How can I find the closed-form of?
$$\frac{1}{1}+\frac{1}{1+2^2}+\frac{1}{1+2^2+3^2}+.....$$
Any help thanks
| hypergeometric has a good idea, but we cannot work with divergent series this way. Similar rearrangements can give wrong answers. But taking hypergeometric's ideas, a valid proof looks like this:
$$
\begin{align}
\log 2 &= \sum_{n=1}^\infty\left(\frac{1}{2n-1}- \frac{1}{2n}\right)
\\
\sum_{n=1}^N \frac{1}{\sum_{r=1}^n r^2} &=
\sum_{n=1}^N \frac{6}{n(n+1)(2n+1)}
\\ &=
12 \sum_{n=1}^N \left(\frac{1}{2n} + \frac{1}{2n+2} - \frac{2}{2n+1}\right)
\\ &=
12 \sum_{n=1}^N \frac{1}{2n} + 12\left(-\frac{1}{2}+\frac{1}{2N+2}+\sum_{n=1}^N \frac{1}{2n}\right)
- 24\left(-1+\frac{1}{2N+1}+\sum_{n=1}^N\frac{1}{2n-1}\right)
\\ &=
12\left(-\frac{1}{2}+\frac{2}{2N+2}+2+\frac{1}{2N+1}\right)
-24\sum_{n=1}^N\left(\frac{1}{2n-1}-\frac{1}{2n}\right)
\\
\lim_{N \to \infty}\sum_{n=1}^N \frac{1}{\sum_{r=1}^n r^2} &=
18-24\log 2
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1657477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Prove that $\int\frac{3x^2+1}{(x^2-1)^3}dx=\frac{-x}{(x^2-1)^2}+c$ Prove that $\int\frac{3x^2+1}{(x^2-1)^3}dx=\frac{-x}{(x^2-1)^2}+c$
My Try:
$\int\frac{3x^2+1}{(x^2-1)^3}dx$
Put $x=\sec\theta$
$\int\frac{3\sec^2\theta+1}{\tan^6\theta}\sec\theta\tan\theta d\theta$
Converting to $\sin$ and $\cos$ we get
$=\int\frac{(3+\cos^2\theta)\cos^2\theta}{\sin^5\theta}d\theta$
I could not solve it further.
| Let $$I = \int\frac{3x^2+1}{(x^2-1)^3}dx = \int\frac{3x^2+1}{x^{\frac{3}{2}}\left(x^{\frac{3}{2}}-x^{-\frac{1}{2}}\right)^3}dx$$
So $$I = \int\frac{3x^{\frac{1}{2}}+x^{-\frac{3}{2}}}{\left(x^{\frac{3}{2}}-x^{-\frac{1}{2}}\right)^3}dx\;,$$ Now Put $\left(x^{\frac{3}{2}}-x^{-\frac{1}{2}}\right)=t\;,$ Then $\left(3x^{\frac{1}{2}}+x^{-\frac{3}{2}}\right)dx = 2dt$
So we get $$I = 2\int t^{-3}dt = -\frac{1}{t^2}+\mathcal{C} = -\frac{x}{(x^2-1)^2}+\mathcal{C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1658367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $a^2 + b^2 \geq ab$ and $a^2 + b^2 \geq -ab$ I know that $a^2 + b^2 \geq 2ab$, and I have been trying to use that to prove $a^2 + b^2 \geq ab$, but I haven't made any progress on using that. Can I have a hint?
|
To prove $a^{2} + b^{2} \geq ab$ for any real numbers $a$ and $b$, regardless of if they are positive or negative:
Note that when you square a number, you get a nonnegative number, so this implies $(a - b)^{2} \geq 0$.
But $(a - b)^{2} = a^{2} -2ab + b^{2}$, so we get $a^{2} + b^{2} \geq 2ab$. But:
$$a^{2} + b^{2} \geq 2ab \implies \frac{1}{2}(a^{2} + b^{2}) \geq ab $$
But $a^{2} + b^{2} \geq \frac{1}{2}(a^{2} + b^{2})$ since half of a positive number is smaller than that positive number (and $a^{2} + b^{2}$ is positive), so this gives $$a^{2} + b^{2}\geq ab. $$
Also, to show $a^{2} + b^{2} \geq -ab$:
Since a number squared is always nonnegative, $(a+b)^{2} \geq 0$.
But $(a +b)^{2} = a^{2} + 2ab + b^{2}$, so $a^{2} + 2ab + b^{2} \geq 0$, which implies $a^{2} + b^{2} \geq -2ab$. Multiplying both sides by $\frac{1}{2}$ gives: $$\frac{1}{2}(a^{2} + b^{2}) \geq -ab $$ and again, as above, since $a^{2} + b^{2} \geq \frac{1}{2}(a^{2} + b^{2})$, we get the desired result: $$ a^{2} + b^{2} \geq -ab.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1658817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove the following trigonometric identity without a calculator involved I have to prove the following statement.
$$1+\cos{2\pi\over5}+\cos{4\pi\over5}+\cos{6\pi\over5}+\cos{8\pi\over5}=0$$
I have tried to use the sum of angles formula for cosine, but didn't get to a point where I'd be able to show that it is equal to $0$.
| We may use formulas, mainly double angle formula and Chebyshev Polynomials, to simplify this problem:
$$\cos(2x)=2\cos^2(x)-1$$
$$\cos(4x)=2\cos^2(2x)-1=8\cos^4(x)-8\cos^2(x)+1$$
$$\cos(6x)=32\cos^6(x)-48\cos^4(x)+18\cos^2(x)-1$$
$$\cos(8x)=64\cos^8(x)-128\cos^6(x)+22\cos^4(x)-16\cos^2(x)+1$$
Use $x=\frac{\pi}5$.
$$1+\cos{2\pi\over5}+\cos{4\pi\over5}+\cos{6\pi\over5}+\cos{8\pi\over5}$$
$$=1+2\cos^2(x)-1+8\cos^4(x)-8\cos^2(x)+1+32\cos^6(x)-48\cos^4(x)+18\cos^2(x)-1+64\cos^8(x)-128\cos^6(x)+22\cos^4(x)-16\cos^2(x)+1$$
$$=1-4\cos^2(x)-18\cos^4(x)-96\cos^6(x)+64\cos^8(x)$$
We know that $\cos(\frac{\pi}5)=\frac{1+\sqrt{5}}4$. Plug this in:
$$=1-4\left(\frac{1+\sqrt{5}}4\right)^2-18\left(\frac{1+\sqrt{5}}4\right)^4-96\left(\frac{1+\sqrt{5}}4\right)^6+64\left(\frac{1+\sqrt{5}}4\right)^8$$
Avoiding the messy stuff, I'll leave it to the reader to determine if the above is equal to zero.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "25",
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If $a^2+b^2=5c^2$ where $a,b,c$ are the sides of a triangle, prove that the area of triangle is $c^2\tan C$
If $a^2+b^2=5c^2$ where $a,b,c$ are the sides of a triangle, prove that the area of triangle is $c^2tan C$
Let median through $C$ be $CF$. $AF=FB=\frac{c}{2}$
$$CF=\frac{1}{2}\sqrt{2(a^2+b^2)-c^2}=\frac{3c}{2}$$
$CG=c$ where $G$ is the centroid and $GF=\frac{c}{2}$
$$\frac{3}{4}(a^2+b^2+c^2)=M_a^2+M_b^2+M_c^2$$
$$\frac{9c^2}{2}=M_a^2+M_b^2+\frac{9c^2}{4}$$
$$c^2=(\frac{2}{3}M_a^2)+(\frac{2}{3}M_b^2)$$
$$BC^2=AG^2+BG^2$$
So medians through $A$ and $B$ are perpendicular.
I don't know if this information is useful for finding area.
Is $CF$ perpendicular to $AB$?
| The law of cosines says:
$a^2 + b^2 - 2ab\cos(C) = c^2$
One also has the following area formula for a triangle:
$\text{Area}(ABC) = \frac{1}{2}ab\sin(C)$
Can you figure it out from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Different eigenvectors obtained with Mathematica Suppose that we have a matrix
$$
\begin{pmatrix}
a&c\\c&b
\end{pmatrix}.
$$
The eigenvalues of this matrix are given by $\lambda_1=(a+b-D)/2$ and $\lambda_1=(a+b+D)/2$, where $D=\sqrt{(a-b)^2+4c^2}$. The eigenvectors are given by
$$
\tag{1}
\begin{pmatrix}1\\(\lambda_1-a)/c\end{pmatrix}\quad\text{and}\quad\begin{pmatrix}1\\(\lambda_2-a)/c\end{pmatrix}
$$
provided that $c\ne0$. However, when I use Mathematica to calculate the eigenvectors of this matrix, I get
$$
\tag{2}
\begin{pmatrix}-(\lambda_2-a)/c\\1\end{pmatrix}\quad\text{and}\quad\begin{pmatrix}-(\lambda_1-a)/c\\1\end{pmatrix}.
$$
As far as I know, the eigenvectors are defined up to a multiplication by a non-zero constant. But multiplication by a constant does not transform eigenvectors $(1)$ into eigenvectors $(2)$. Or am I wrong?
Could anyone explain why I get different eigenvectors when I use Mathematica? Or are these eigenvectors the same?
Any help is much appreciated!
| The characteristic polynomial of your matrix $A$ is
$$ p_A(x) = x^2 - (a + b)x + (ab - c^2) = (x - \lambda_1)(x - \lambda_2) = x^2 - (\lambda_1 + \lambda_2)x + \lambda_1 \lambda_2. $$
Assume that $c \neq 0$. If you multiply your first eigenvector by $-\frac{\lambda_2 - a}{c}$, you get
$$ -\frac{\lambda_2 - a}{c} \begin{pmatrix} 1 \\ \frac{\lambda_1 - a}{c} \end{pmatrix} = \begin{pmatrix} -\frac{\lambda_2 - a}{c} \\ - \frac{(\lambda_1 - a)(\lambda_2 - a)}{c^2} \end{pmatrix}. $$
Now,
$$ (\lambda_1 - a)(\lambda_2 - a) = \lambda_1 \lambda_2 - a (\lambda_1 + \lambda_2) + a^2 = (ab - c^2) - a(a + b) + a^2 = -c^2$$
so indeed we get
$$ -\frac{\lambda_2 - a}{c} \begin{pmatrix} 1 \\ \frac{\lambda_1 - a}{c} \end{pmatrix} = \begin{pmatrix} -\frac{\lambda_2 - a}{c} \\ - \frac{(\lambda_1 - a)(\lambda_2 - a)}{c^2} \end{pmatrix} = \begin{pmatrix} -\frac{\lambda_2 - a}{c} \\ 1 \end{pmatrix} $$
so your eigenvector and Mathematica's eigenvector are equal up to a multiplication by a non-zero constant.
| {
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"timestamp": "2023-03-29T00:00:00",
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Family of circles with $AB$ as diameter The circle $S_1 :x^2+y^2-4=0$ cuts the circle $S_2 :x^2+y^2+2x+3y-5=0$ in A and B. Then find the equation of circle with $AB$ as diameter.
Answer is $13(x^2+y^2)-4x-6y-50=0$
Equation of AB will $S_2-S_1$ and then I used the fact that centre of one of the circles in the family of circles $S_1+kS_2=0$ must lie on $AB$ but I am not getting the correct answer. Any suggestions?
| Your reasoning is correct, so I suspect you just made an algebra error.
As you say, the line $AB$ has equation $S_1 - S_2=0$, which is
$$
2x +3y - 1 = 0
$$
Any circle $C_k$ through $A$ and $B$ has equation $S_1 + kS_2=0$ for some $k$. Doing the algebra tells us that $C_k$ has equation
$$
(1+k)x^2 + (1+k)y^2 +2x +3y -5 - 4k = 0
$$
The center of $C_k$ is at the point
$$
P_k = - \left( \frac{2}{2(1+k)}, \frac{3}{2(1+k)} \right)
$$
If $C_k$ has diameter $AB$, then $P_k$ lies on $AB$, so
$$
-\frac{4}{2(1+k)} - \frac{9}{2(1+k)} = 1
$$
This gives $1+k = -13/2$ and $k = -15/2$. Substituting back in the equation of $C_k$ gives
$$
13(x^2+y^2)-4x-6y-50=0
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Genetics and probability (“other than” case) (Moved from Bio SE due to the mathematical nature of the problem.)
If a man and woman, both carriers of a autosomal recessive
disorder (i.e. having genotype $Aa$), produce three children, what is the probability of one or
more children having the disorder?
The answer key suggests finding the probability of the children all being normal and then subtracting that from 1:
$$Aa \times Aa$$
$$P_{AA\:or\:Aa}=\frac{3}{4}$$
$$\frac{3}{4} \times \frac{3}{4} \times \frac{3}{4} = \frac{27}{64}\tag{3 normal}$$
$$1-\frac{27}{64}=\frac{37}{64}\tag{≥1 abnormal}$$
I understand the logic behind that: If they aren’t all normal, then at least one of them has to be abnormal.
However, what if I didn’t want to use the $1-x$ method? What if I wanted to do it the hard way?
I tried breaking it into three cases: 1 abnormal, 2 abnormal, and all abnormal.
$$P_{aa}=\frac{1}{4}$$
$$\frac{1}{4} \times \frac{3}{4} \times \frac{3}{4} = \frac{9}{64}\tag{1 abnormal, 2 normal}$$
$$\frac{1}{4} \times \frac{1}{4} \times \frac{3}{4} = \frac{3}{64}\tag{2 abnormal, 1 normal}$$
$$\frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{64}\tag{3 abnormal, 0 normal}$$
$$\sum P=\frac{13}{64}$$
which obviously isn’t the correct answer. What’s wrong with the way I’m doing it?
| $$P_{aa}=\frac{1}{4}$$
$$\frac{1}{4} \times \frac{3}{4} \times \frac{3}{4} = \frac{9}{64}\tag{1 abnormal, 2 normal}$$
$$\frac{1}{4} \times \frac{1}{4} \times \frac{3}{4} = \frac{3}{64}\tag{2 abnormal, 1 normal}$$
$$\frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{64}\tag{3 abnormal, 0 normal}$$
The error in your argument lies in the first and second statements. Suppose the three children are $A,B,C$.
In the first case, where 1 is abnormal and 2 are normal, we can have,
Abnormal - $A$, Normal - $B,C$
Abnormal - $B$, Normal - $C,A$
Abnormal - $C$, Normal - $A,B$
Thus, this condition is satisfied in three outcomes, while only one has been counted. The probability for this case then becomes,
$$3\times\frac{1}{4} \times \frac{3}{4} \times \frac{3}{4} = \frac{27}{64}$$
Similarly, in the second case,
$$3\times\frac{1}{4} \times \frac{1}{4} \times \frac{3}{4} = \frac{9}{64}$$
which gives the desired result.
In general, to get the number of outcomes for the particular case, we need to find out under how many combinations that case can occur. In the first case here, 1 child is abnormal, and can be selected from 3 children in $\binom{3}{1}=3$ ways. Thus, we have three such outcomes possible.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Showing $f(x) = \frac{1}{1 + x^2}$ is continuous I need to show with a $\epsilon$-$\delta$ proof that the function $$f : \mathbb{R} \rightarrow \mathbb{R} : x \mapsto \frac{1}{1 + x^2} $$ is continuous.
Here is what I did. Let $a \in \mathbb{R}$ be arbitrary. Then for every $x \in \mathbb{R}$ we have $$ | f(x) - f(a) | = \bigg| \frac{1}{1 + x^2} - \frac{1}{1 + a^2} \bigg| = \frac{| a^2 - x^2 |}{(1+x^2)(1+a^2)} \leq \frac{| x - a| (|a| + |x|)}{(1+x^2)(1+a^2)}. $$ Now I think I need to use the trick that $$ \frac{|t|}{1 + t^2} \leq \frac{1}{2} $$ for all $t \in \mathbb{R}$. If I do that here, I still get something complicated: $$ ... = \frac{|x - a| |a|}{ (1+x^2)(1+a^2)} + \frac{|x-a| |x|}{(1+x^2)(1+a^2)} \leq \frac{|x-a|}{ 2 (1+x^2)} + \frac{|x-a|}{2(1+a^2)}. $$ I still got my RHS in terms of $x$, while I only want it in terms of $|x-a|$ and $a$ so than I can pick my $\delta$.
Any help please?
| In general, to show $f$ is continuous at $a$, it suffices to bound $|f(x)-f(a)|\le M|x-a|$ when $x$ closed to $a$,where $M$ is a constant depends on $a$.
For example, let $a \in \mathbb{R}$ be arbitrary. Then for every $x \in \mathbb{R}$ we have
$$ | f(x) - f(a) | = \bigg| \frac{1}{1 + x^2} - \frac{1}{1 + a^2} \bigg| \le \frac{| x - a| (|a| + |x|)}{(1+x^2)(1+a^2)}\le | x - a| (|a| + |x|). $$
Hence given $\epsilon>0$, we let $\delta=\min\{1,\frac{\epsilon}{2(2|a|+1)}\}$, then for all $|x-a|<\delta\le 1$, we will have $|x|\le |a|+1$, hence
$| f(x) - f(a) |\le (2|a|+1)|x-a|\le \frac{\epsilon}{2(2|a|+1)}(2|a|+1)= \frac{\epsilon}{2}<\epsilon$
| {
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Expressing $(-8)^{\frac13}$ in polar form I want to express $(-8)^{\frac{1}{3}}$ in polar and cartesian coordinates.
What I did was to solve the equation $-8 = r^3e^{3i\theta}= r^3(\cos(3\theta)+i\sin(3\theta))$ which implies that I must solve the equations
$$
r^3\cos(3\theta) = -8
$$
and the equation
$$
r^3\sin(3\theta)= 0
$$
The latter gives me $\theta = 0, \frac{\pi}{3}, \frac{2\pi}{3},\dotsc$ and so using $\theta = 0$, using the equation $r^3\cos(3\theta) = -8$, I get $r = -2$ and in this case the polar and cartesian form is $-2$, then using $\theta = \frac{\pi}{3}.$ I get $r =2$ and so the polar form is $2e^{\frac{\pi}{3}}$ and the cartesian form is $1+i \sqrt{3}$ and finally using $\theta = \frac{2\pi}{3}$ I get $r=-2$ and in polar form the answer is $-2e^{\frac{2\pi}{3}}$ and in cartesian coordinates the answer is $-1 + i \sqrt{3}$. However, the answers in my book are different mostly by signs, like the first answer is $2$ instead of $-2$. Can anyone explain why I am getting this sign errors?
| You're using $r=-2$, which is incorrect: the polar form is $re^{i\theta}$, with $r>0$.
You should write
$$
-8=r^3(\cos3\theta+i\sin3\theta)
$$
so $r=2$ and
$$
\cos3\theta+i\sin3\theta=-1
$$
This gives
$$
3\theta=\pi+2k\pi
$$
or
$$
\theta=\frac{\pi}{3}+\frac{2}{3}k\pi
$$
and the principal arguments are
$$
\frac{\pi}{3},\quad \pi,\quad \frac{5\pi}{3}
$$
so the solutions are
$$
2e^{i\pi/3}=1+i\sqrt{3},\quad 2e^{i\pi}=-2,\quad 2e^{5\pi/3}=1-i\sqrt{3}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Total number of values of $a$ such that the equation $x^2+ax+a+1=0$ has integral roots Find the total number of values of $a$ such that the equation $x^2+ax+a+1=0$ has integral roots
The equation can also be written as $$\left(x+\frac{a}{2}\right)^2=\frac{a^2-4a-4}{4}$$
So $a$ is a multiple of $2$ and $a^2-4a-4$ is a multiple of $4$ and $\frac{a^2-4a-4}{4}$ is a perfect square.
How can I find $a$ using the above conditions?
| The sum $r+s$ of the roots is $-a$ and their product $rs$ is $a+1$. So, as in your calculation, we want
$$(r-s)^2=(r+s)^2-4rs=a^2-4(a+1)$$
to be a perfect square. So we want $(a-2)^2-8$ to be a perfect square.
There are only two perfect squares that differ by $8$, so $(a-2)^2=9$, and now we can find the possible values of $a$. It is easy to verify that they both work.
| {
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Square 3x3 filled with numbers with every 2x2 subsquare having equal sum. You are given 3x3 square filled with numbers from 1 to 9 without repetition. Every subsquare 2x2 has equal sum. What are the possible sums of those subsquares? I think that all values between 16 and 24 are possible, but I don't know how to prove it. I can check all the possibilites, but I am looking for an elegant and as simple as possible proof.
| As a modest complement to Ross Millikan's answer,
I have tested the $9!$ matrices and all the sums from $16$ to $24$ appear,
with the following frequencies:
$$
\begin{array}{c|ccccccccc}
s & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 & 24 \\\hline
\sharp & 16 & 40 & 40 & 64 & 56 & 64 & 40 & 40 & 16 \\
\end{array}
$$
And here are examples for each sum
$$
\begin{array}{ccc}
\left(
\begin{array}{ccc}
5 & 3 & 8 \\
7 & 1 & 4 \\
6 & 2 & 9 \\
\end{array}
\right) &\left(
\begin{array}{ccc}
3 & 6 & 8 \\
7 & 1 & 2 \\
4 & 5 & 9 \\
\end{array}
\right) &\left(
\begin{array}{ccc}
2 & 6 & 7 \\
9 & 1 & 4 \\
3 & 5 & 8 \\
\end{array}
\right)\\[3mm]
\left(
\begin{array}{ccc}
1 & 7 & 4 \\
8 & 3 & 5 \\
6 & 2 & 9 \\
\end{array}
\right) &\left(
\begin{array}{ccc}
1 & 6 & 7 \\
8 & 5 & 2 \\
3 & 4 & 9 \\
\end{array}
\right) &\left(
\begin{array}{ccc}
1 & 5 & 6 \\
8 & 7 & 3 \\
4 & 2 & 9 \\
\end{array}
\right)\\[3mm]
\left(
\begin{array}{ccc}
2 & 3 & 5 \\
9 & 8 & 6 \\
4 & 1 & 7 \\
\end{array}
\right) &\left(
\begin{array}{ccc}
1 & 5 & 3 \\
9 & 8 & 7 \\
4 & 2 & 6 \\
\end{array}
\right) & \left(
\begin{array}{ccc}
1 & 6 & 2 \\
8 & 9 & 7 \\
4 & 3 & 5 \\
\end{array}
\right)
\end{array}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding a solution to this logarithm equation. How is:
$ \log_{(xyz)^{1/3}} \left(\frac{yz}{x^{3}}\right)^{2}$
expressed in terms of a and b when:
$ a = \log_{y} x $
$ b = \log_{z} x $
EDIT: It is z in the numerator, apologies i posted by memory. Also reframed the question.
| $$\log _{ (xyz)^{ \frac { 1 }{ 3 } } } \left( \frac { yz }{ x^{ 3 } } \right) ^{ 2 }=\frac { 2 }{ \frac { 1 }{ 3 } } \log _{ (xyz) } \left( \frac { yz }{ x^{ 3 } } \right) =6\frac { \log _{ x }{ \left( \frac { yz }{ x^{ 3 } } \right) } }{ \log _{ x }{ \left( xyz \right) } } =6\frac { \log _{ x }{ \left( y \right) +\log _{ x }{ \left( z \right) } -3\log _{ x }{ x } } }{ \log _{ x }{ \left( x \right) +\log _{ x }{ \left( y \right) } +\log _{ x }{ \left( z \right) } } } =\\ =6\frac { \log _{ x }{ \left( y \right) +\log _{ x }{ \left( z \right) } -3 } }{ 1+\log _{ x }{ \left( y \right) } +\log _{ x }{ \left( z \right) } } =6\frac { \frac { 1 }{ \log _{ y }{ \left( x \right) } } +\frac { 1 }{ \log _{ z }{ \left( x \right) } } -3 }{ 1+\frac { 1 }{ \log _{ y }{ \left( x \right) } } +\frac { 1 }{ \log _{ z }{ \left( x \right) } } } =6\frac { \frac { 1 }{ a } +\frac { 1 }{ b } -3 }{ 1+\frac { 1 }{ a } +\frac { 1 }{ b } } =6\frac { \frac { a+b-3ba }{ ba } }{ \frac { ab+a+b }{ ab } } =\\ =\frac { 6a+6b-18ba }{ ab+a+b } $$
| {
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How to find $\angle$ b? How to find $\angle$ b ?
The vertices of the triangle are on the foci of the ellipse and on the ellipse.
$\angle$ a, the major axis and eccentricity are known.
| Let $\alpha, \beta$ be the angles, $2a>0$ the length of the major axis, $P = (-ae, 0)$, $Q = (ae, 0)$ the foci, and $R = (x,y)$ the point on the ellipse. I assume for simplicity that $-ae < x < ae$ and $y >0$, since the calculations are analogous in other cases.
Note that $x,y$ are such that
\begin{align}
\frac{x^2}{a^2} + \frac{y^2}{(1-e^2)a^2} = 1
\end{align}
and
\begin{align*}
y = \tan \alpha (x+ae).
\end{align*}
We have
\begin{align*}
\tan \beta = \frac{y}{ae - x} = \frac{ae + x}{ae - x}\tan \alpha.
\end{align*}
Let $t = (ae+x)/(ae-x)$, so that $x = ae(t-1)/(t+1)$ and $y = ae\tan \alpha (2t)/(t+1)$. Then the ellipse equation for $t$ is
\begin{align*}
e^2\frac{(t-1)^2}{(t+1)^2} + 4\tan^2 \alpha \frac{e^2}{1-e^2}\frac{t^2}{(1+t)^2} = 1.
\end{align*}
If my calculations are correct, from this one gets
\begin{align*}
t = \frac{1-e^2}{4e^2\tan^2 \alpha - (1-e^2)^2}\left ( 1 + e^2 + \frac{2e}{\cos \alpha} \right ).
\end{align*}
This times $\tan \alpha$ gives $\tan \beta$. Of course the value of $\alpha$ such that $2e\tan \alpha = 1-e^2$ is the one for which $x = ae$, since $t \to \infty$ gives $x \to ae$. Note also that for $\tan^2 \alpha = (1-e^2)/e^2$, i.e. when $R = (0, b)$, the expression simplifies to $t = 1$, as expected.
Unfortunately I cannot write the solution in a better form than this... Hope it can help anyway!
| {
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Find the three positive values of p for which the equation $px^2-4x+1=0$ will have rational roots
Question: Find the three positive values of p for which the equation $$px^2-4x+1=0$$
will have rational roots.
My attempt (Algebraically):
Usually if it has to have rational roots then the discriminant must equal zero so
$$b^2 - 4ac = 0$$
$$ (-4)^4-4(p)(1)=0$$
$$ 16-4p = 0$$
$$ p = 4 $$
But the answers given are $3 , \frac{7}{4} , \frac{15}{4}$ how do they get that algebraically?
| Given the original and solving for x, we get
$$px^2-4x+1=0$$
$$x=\frac{4\pm\sqrt{16-4(p)(1)}}{2p}=\frac{2\pm\sqrt{4-p}}{p}$$
From this is appears that there are many answers where $4-p$ is a perfect square but one is bogus.
$$p=0\implies x=\frac{2+2}{0}\not\in \mathbb{N}$$
$$p=3\implies x=\frac{2\pm1}{3}=\big\{1, \frac{1}{3}\big\}\quad \land\quad p=4\implies x=\frac{2+0}
{4}=\frac{1}{2}\qquad$$
$$p=-5\implies x=\frac{2+\pm3}{-5}=\big\{-1,\frac{1}{5}\big\}\quad \land\quad p=-12\implies x=\frac{2\pm4}{-12}=\big\{\frac{-1}{2},\frac{1}{6}\big\}$$
$$p=-21\implies x=\frac{2\pm5}{-21}=\big\{\frac{-1}{3},\frac{1}{7}\big\}\quad \land\quad p=-32\implies x=\frac{2\pm6}{-32}=\big\{\frac{-1}{4},\frac{1}{8}\big\}$$
Solving for p: we get $$p=\frac{4x-1}{x^2}$$
$$x=1\implies p=\frac{3}{1}\qquad x=2\implies p=\frac{7}{4}\qquad x=3\implies p=\frac{11}{9}\qquad x=4\implies p=\frac{15}{16}$$
It looks like there are infinite answers with $x$ as the independent variable but I don't know if they will fit back into the original solution for $x$.
| {
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Find the value of $(a - b)^2$ given $a + b = 2$ and $a^2 + b^2 = 6$
$$a + b = 2\\
a^2 + b^2 = 6$$
Find the value of $(a-b)^2 $
My workings till I got stuck -
$$(a-b)^2 = a^2 - 2ab + b^2 \\
= a^2 + b^2 - 2ab\\
= 6 - 2 ab $$
I'm stuck at how to find $ab$ . Can I get hints on how to find $ab$? Thanks a lot.
| We will solve for the values of $a$ and $b$. $a=2-b$ so $a^2+b^2=6 \iff (2-b)^2 + b^2 +6 $
This means $2b^2-4b-2=0 \iff b^2 -2b -1 +0 $. The solutions of this equation are $b=1-\sqrt{2}$ and $b=1+\sqrt{2}$. Then $a=1+\sqrt{2}$ or $a=1-\sqrt{2}$
Then, in any case $(a-b)^2=8$
| {
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Prove that $\lim _{x\to \infty \:}(1+\frac{x^x}{x!})^{\frac{1}{x}} = e$ Using a graphing calculator, it seems that $\lim _{x\to \infty \:}(1+\frac{x^x}{x!})^{\frac{1}{x}} = e$. How can this be proven?
| Since
$n! > (n/e)^n$,
for integer $x$,
$1+\frac{x^x}{x!}
<1+\frac{x^x}{(x/e)^x}
=1+e^x
$.
Therefore,
for integer $x$,
$(1+\frac{x^x}{x!})^{\frac{1}{x}}
< (1+e^x)^{1/x}
=e (1+e^{-x})^{1/x}
=e (1+\frac{e^{-x}}{x})
=e +e\frac{e^{-x}}{x}
$.
Since
$n! < (n/e)^{n+1}$,
for integer $x$,
$1+\frac{x^x}{x!}
\gt 1+\frac{x^x}{(x/e)^{x+1}}
=1+\frac{e^{x+1}}{x}
$.
Therefore,
for integer $x$,
$(1+\frac{x^x}{x!})^{\frac{1}{x}}
> (1+\frac{e^{x+1}}{x})^{1/x}
= e (e^{-x}+\frac{e}{x})^{1/x}
= e \left(\frac{e}{x}\right)^{1/x}(1+e^{-x}\frac{x}{e})^{1/x}
\gt e +\left(\frac{e}{x}\right)^{1/x}
$.
Therefore,
for integer $x$,
$e \left(\frac{e}{x}\right)^{1/x}
<(1+\frac{x^x}{x!})^{\frac{1}{x}}
<e +e\frac{e^{-x}}{x}
$
and the limits of both sides
are $e$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1675262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 1
} |
Which is greater, |$\vec a + \vec b$| or |$\vec a$|+|$\vec b$|? Pardon me if this has been asked before, I have encountered this question and its many variations in many physics mock papers.
I want to know if there is some sort of way to logically attempt these types of questions, I am not in requirement of an answer but a method which will help in solving questions of this form...
| \begin{align*}
|\mathbf{a}+\mathbf{b}|^{2} &=
(\mathbf{a}+\mathbf{b}) \cdot (\mathbf{a}+\mathbf{b}) \\
&= \mathbf{a} \cdot \mathbf{a}+\mathbf{a} \cdot \mathbf{b}+
\mathbf{b} \cdot \mathbf{a}+\mathbf{b} \cdot \mathbf{b} \\
&= |\mathbf{a}|^{2}+2\, \mathbf{a} \cdot \mathbf{b}+|\mathbf{b}|^{2} \\
& \leq |\mathbf{a}|^{2}+2|\mathbf{a}||\mathbf{b}|+|\mathbf{b}|^{2}
\quad (\because \: \mathbf{a} \cdot \mathbf{b} \leq
|\mathbf{a}||\mathbf{b}|) \\
& = (|\mathbf{a}|+|\mathbf{b}|)^{2} \\
\therefore \quad |\mathbf{a}+\mathbf{b}|
& \leq |\mathbf{a}|+|\mathbf{b}|
\quad (\because \: |\mathbf{a}+\mathbf{b}|, |\mathbf{a}|,
|\mathbf{b}| \geq 0)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1675452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integral using partial fractions I am having trouble using partial fractions to evaluate
$$\int \frac{6x}{x^3-8} dx.$$ I can find the denominator but using equations to find the numerator is difficult.
|
Partial fractions for: $\frac{6x}{x^3-8}$
$$\frac{6x}{(x-2)(x^2+2x+4)}=\frac{C+Bx}{x^2+2x+4}+\frac{A}{x-2}$$
$$6x=A(x^2+2x+4)+(x-2)(C+Bx)$$
$$6x=4A-2C+(A+B)x^2+(2A-2B+C)x$$
$$\begin{cases}
4A-2C=0\\
2A-2B+C=6\\
A+B=0
\end{cases}$$
Solve
$$\int \frac{6x}{x^3-8}dx$$
$$=6\int\frac{dx}{(x-2)(x^2+2x+4)}\overset{\text{partial fractios}}{=}\int\frac{2-x}{x^2+2x+4}dx+\int\frac{dx}{x-2}$$
$$=-\frac 1 2\int\frac{2x+2}{x^2+2x+4}dx+3\int\frac{dx}{x^2+2x+4}+\int\frac{dx}{x-2}$$
Set $t=x^2+2x+4$ and $dt=(2x+2)dx$
$$=-\frac 1 2 \int \frac 1 t+3\int\frac{dx}{x^2+2x+4}+\ln|x-2|$$
$$\overset{\text{complete square}}{=}-\frac{\ln|t|}{2}+\ln|x-2|+3\int\frac{dx}{(x+1)^2+3}$$
$$=\color{red}{-\frac{\ln|x^2+2x+4|}{2}+\ln|x-2|+\sqrt 3\arctan\left(\frac{x+1}{\sqrt 3}\right)+\mathcal C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1676225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Solving $\int {\frac{1}{(-x^{2}+6x-5)^{1/2}}}\;dx$ such that $(1\lt x\lt5)$ using trigonometric substitutions and pythagorean identities First post so I'll get right to the question;
$$\int {\frac{1}{(-x^{2}+6x-5)^{1/2}}}\;dx,\qquad(1\lt x\lt5)$$
To begin with I completed the square which yields: $$-(x-3)^{2}+4$$
substituting this completed square into the integrand yields:
$$\int {\frac{1}{(-(x-3)^{2}+4)^{1/2}}}\,\,dx$$
solving by substituting the trig substitution:
let $$x = 3+2\tan u$$
hence: $$\frac{dx}{du} = 2\sec^{2}u$$
thus:
$$dx = (2\sec^{2}u) du$$
Rearranging $$x = 3+2\tan u$$
yields $$u=\tan^{-1}(\frac{x-3}{2})$$
Now rewriting the integral using the value of x yields:
$$\int {\frac{1}{(-(3+2\tan u-3)^{2}+4)^{1/2}}}\,\,\,(2\sec^{2}u) du$$
which yields:$$\int {\frac{1}{(-(2\tan u)^{2}+4)^{1/2}}}\,\,(2\sec^{2}u) du$$
which is the same as saying: $$\int {\frac{1}{(-4\tan^{2}u+4)^{1/2}}}\,\,\,(2\sec^{2}u) du$$
which can be rearranged to give: $$\int {\frac{1}{(4-4\tan^{2}u)^{1/2}}}\,\,\,(2\sec^{2}u) du$$
Applying the constant multiple rule to the integral yields:
$$\frac 12\int {\frac{1}{(1-\tan^{2}u)^{1/2}}}\,\,\,(\sec^{2}u) du$$
Now we know that pythagorean identity: $$\tan^{2}u=\sec^{2}u-1$$
so substituting I end up with an indefinite integral that looks like this:
$$\frac 12\int {\frac{1}{(1-\sec^{2}u+1)^{1/2}}}\,\,\,(\sec^{2}u) du$$
which gives:
$$\frac 12\int {\frac{1}{(2-\sec^{2}u)^{1/2}}}\,\,\,(\sec^{2}u) du$$
However this is obviously wrong because I think it is unsolvable. So where did I veer from the right path?
| $$\int\sqrt{\frac{1}{-x^2+6x-5}}\space\text{d}x=\int\frac{1}{\sqrt{4-(x-3)^2}}\space\text{d}x=$$
Substitute $u=x-3$ and $\text{d}u=\text{d}x$:
$$\int\frac{1}{\sqrt{4-u^2}}\space\text{d}u=\int\frac{1}{2\sqrt{1-\frac{u^2}{4}}}\space\text{d}u=\frac{1}{2}\int\frac{1}{\sqrt{1-\frac{u^2}{4}}}\space\text{d}u=$$
Substitute $s=\frac{u}{2}$ and $\text{d}s=\frac{1}{2}\space\text{d}u$:
$$\int\frac{1}{\sqrt{1-s^2}}\space\text{d}s=\arcsin\left(s\right)+\text{C}=\arcsin\left(\frac{u}{2}\right)+\text{C}=\arcsin\left(\frac{x-3}{2}\right)+\text{C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1677308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
Multiple answers to $\int \sqrt{4t - t^2} \, \textrm{d} t$ I'm trying to understand why I'm getting different answers when taking different approaches to integrating
$$
\int \sqrt{4t - t^2} \, \textrm{d} t
$$
First, I tried substituting $\sqrt t = 2 \sin \theta$:
$$
\begin{eqnarray}
\int \sqrt{t}\sqrt{4 - t} \, \textrm{d} t
&=& \int 2 \sin \theta \cdot \sqrt{4 - 4 \sin^2 \theta} \cdot 8 \sin \theta \cos \theta \, \textrm{d} \theta \\
&=& 32 \int \sin^2 \theta \cos^2 \theta \, \textrm{d} \theta \\
&=& 4 \int 1 - \cos 4 \theta \, \textrm{d} \theta \\
&=& 4 \theta - \sin 4 \theta + C \\
&=& 4 \theta - 4 \sin \theta \cos^3 \theta + 4 \sin^3 \theta \cos \theta + C \\
&=& 4 \arcsin \frac{\sqrt{t}}{2} + \frac{1}{2} (t - 2)\sqrt{4t - t^2} + C \\
\end{eqnarray}
$$
Second, I tried completing the square and substituting $t - 2 = 2 \sin \theta$:
$$
\begin{eqnarray}
\int \sqrt{4 - (t^2 - 4t + 4)} \, \textrm{d} t
&=& \int \sqrt{4 - (t - 2)^2} \, \textrm{d} t \\
&=& \int \sqrt{4 - 4 \sin^2 \theta} \cdot 2 \cos \theta \, \textrm{d} \theta \\
&=& 4 \int \cos^2 \theta \, \textrm{d} \theta \\
&=& 2 \int 1 - \cos 2 \theta \, \textrm{d} \theta \\
&=& 2 \theta + \sin 2 \theta + C \\
&=& 2 \theta + 2 \sin \theta \cos \theta + C \\
&=& 2 \arcsin \left(\frac{t - 2}{2}\right) + \frac{1}{2}(t - 2)\sqrt{4t - t^2} + C \\
\end{eqnarray}
$$
The second answer is the same as in the book but I don't understand why the first approach gives the wrong answer.
| The answers differ by a constant. In fact,
$$\arcsin\left(\frac t2-1\right) = 2\left(\arcsin\frac{\sqrt t}2 -\frac\pi4\right).\tag{*}$$
Proof: Write $\theta:=\arcsin\frac{\sqrt t}2$. Then $\sin^2\theta=\frac t4$ and $\cos^2\theta=1-\frac t4$, and using the angle-difference identities,
$$
\sin\left(\theta-\frac\pi4\right)=\frac1{\sqrt 2}(\sin\theta-\cos\theta)\tag1
$$
while
$$\cos\left(\theta-\frac\pi4\right)=\frac1{\sqrt 2}(\sin\theta+\cos\theta).\tag2$$
Therefore
$$
\sin2\left(\theta-\frac\pi4\right)=2\sin\left(\theta-\frac\pi4\right)\cos\left(\theta-\frac\pi4\right)
\stackrel{(1),(2)}=\sin^2\theta-\cos^2\theta =\frac t4-\left(1-\frac t4\right)=\frac t2-1.$$
Therefore both sides of (*) have the same sine. Similarly you can show that both sides have the same cosine.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1679302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Minimum of $\frac{x}{1+y^2}+\frac{y}{1+x^2}$ on $x,y\ge 0$, $x+y=2$ let $x,y\ge 0$, and such $x+y=2$ find the minimum
$$\dfrac{x}{1+y^2}+\dfrac{y}{1+x^2}$$
I think $x=y=1$ is minimum of the value $1$,How can I prove?
| There is yet another way of proving this inequality.
$\dfrac{x}{1+y^2} =x(1-\dfrac{y^2}{1+y^2})=x-\dfrac{xy^2}{1+y^2} \geq x-\dfrac{xy^2}{2y}=x-\dfrac{xy}{2}$
Similarly, $\dfrac{y}{1+x^2} \geq y-\dfrac{xy}{2}$
Adding these up, we have $\dfrac{x}{1+y^2} + \dfrac{y}{1+x^2} \geq (x+y) -xy = 2-xy$
Also, $xy \leq \dfrac{1}{4} (x+y)^2=1 \Rightarrow -xy \geq -1$
Therefore, $\dfrac{x}{1+y^2} + \dfrac{y}{1+x^2} \geq 2-1=1$
Finally, it can be seen that the equality indeed holds when $x=y=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1679491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Find the area of the region bounded above by the curve... Find the area of the region bounded above by the curve $x^2+y^2=2$ and below by the curve $y=x^2$.
(Large Version)
Here's my attempt:
Since $y =x^2$, then $x^2+x^2=2$, which simplifies to give $x=1$ or $x=-1$. Since $y^2=2-x^2$, then $y=\sqrt( 2-x^2)$. Finally, I tried to integrate $\sqrt (2-x^2)-x^2$ and I got area $\pi/2 $ as my answer. Please correct me if I'm wrong, thanks!
| This is the calculation of the area between $\;y=x^2\;$ below and $\;x^2+y^2=2\;$ above:
Since both $\;y=\sqrt{2-x^2}\;,\;\;y=x^2\;$ are even functions, and since also
$\;y=\sqrt{2-x^2}=x^2\iff x=\pm1\;$ , the area is given by
$$A=2\int_0^1\left(\sqrt{2-x^2}-x^2\right)dx=2\sqrt2\int_0^1\sqrt{1-\left(\frac x{\sqrt2}\right)^2}\,dx-\left.\frac23x^3\right|_0^1$$
Substitute in the last integral $\sin u=\frac x{\sqrt2}\implies \sqrt2\,\cos u\,du=dx\;$ , and we get
$$A=4\int_0^{\pi/4}\cos^2u\,du-\frac23=\left.4\left(\frac{u+\sin u\cos u}2\right)\right|_0^{\pi/4}-\frac23=2\left(\frac\pi4+\frac12\right)-\frac23=\frac\pi2+\frac13$$
Now, if you want the area inside the disk $\;x^2+y^2\le2\;$ and below the parabola $\;y=x^2\;$, we have to substract the above from the disk's area:
$$2\pi-\left(\frac\pi2+\frac13\right)=\frac{3\pi}2-\frac13$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1682919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Showing that $xy \leq \frac{x^p}{p} + \frac{y^q}{q}$ Question:
Let $x \geq 0$ , $y \geq 0$ and $p > 0$, $q>0$ with $\frac{1}{p} + \frac{1}{q} = 1$.
Show that $$xy \leq \frac{x^p}{p} + \frac{y^q}{q} $$
[Suggestion: Without loss of generality suppose $xy = 1$].
Attempt: Let $f, \varphi : U \to \mathbb R$, $U = \{(x,y) \in \mathbb R^2; x > 0 , y > 0 \}$ given by $f(x,y) = \frac{x^p}{p} + \frac{y^q}{q}$ and $\varphi (x,y) = xy$. Then we have
$$\mathrm {grad}\, f(x,y) = (x^{p-1}, y^{q-1}) \,\,\,\text{and}\,\,\, \mathrm {grad} \,\varphi (x,y) = (y,x)$$
Then $1$ is a regular value of $\varphi$. Consider $M = \varphi^{-1} (1)$, the hyperbola $xy =1$. Now $(x,y) \in M$ is a critical point of $f|_M$ iff
$$\mathrm {grad}\, f(x,y) = \lambda\, \mathrm{grad} \, \varphi (x,y) \,\,\,\text{and}\,\,\, \varphi (x,y) = 1$$
As $x> 0 $ and $y>0$ we have
$$x^{p-1} = \lambda y \,\,\, , y ^{q-1} = \lambda x \,\,\,\text{and} \,\,\,xy = 1$$
Then $$\frac{x}{y} = \frac{y ^{q-1}}{x ^{p-1}} \implies x^p = y^q$$
This gives us $$\begin{align}\frac{x^p}{p} + \frac{y^q}{q} &= \frac{qy^q + py^q}{pq } = y^q \frac{p + q}{pq}\\&= y^q = y ^{1 + \frac{q}{p}}\\&=y^{\frac{q}{p}}\cdot y = x \cdot y\end{align}$$
Now $f$ is of class $C^{\infty}$ and its Hessian is given by
$$Hf(x,y) = \begin{pmatrix} (p-1)x^{p-2} & 0 \\ 0 & (q-1)y^{q-2}\end{pmatrix} $$
and it is positive, therefore $xy$ is a local minimum. It follows then $$xy \leq \frac{x^p}{p} + \frac{y^q}{q}$$
as we wanted.
The cases $x = y = 0$, $x = 0 $ and $y> 0$ were considered trivially true.
Note: This inequality is used to prove Hölder's Inequality.
| Use Jensen's inequality. The function $f(x)=-\log x$ is convex, and we have $\frac1p +\frac1q=1$,
so
$$f\left(\frac1p x^p +\frac 1q y^q\right)\le \frac1p f(x^p)+\frac1q f(y^q).$$
Plugging in $f$, this simplifies to $$-\log\left(\frac{x^p}p +\frac {y^q}q\right)\le -\log(xy).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1683150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Integrate $\int \frac{2x}{x^2 +2x+1}$ Here is my attempt:
$$ \frac{2x}{x^2 +2x+1}= \frac{2x}{(x+1)^2 } = \frac{2}{x+1}-\frac{2}{(x+1)^2 }$$
Then I tried to integrate it,I got $2\ln(x+1)+\frac{2}{x+1}+C$ as my answer. Am I right? please correct me if I'm wrong.
| Add and remove $2$ in the numerator, and you will obtain:
$$\frac{2x+2-2}{x^2+2x+1} = \frac{2x+2}{x^2+2x+1} - \frac{2}{x^2+2x+1}$$
Now you see that the numerator in the first fraction is nothing but the derivative of the denominator, id est you have a function of the form $\frac{f'(x)}{f(x)}$.
This hen integrated is simply
$$\ln(x^2+2x+1)$$
Now the other term
$$2\int\frac{1}{x^2 + 2x + 1}\ \text{d}x = 2\int\frac{1}{(x+1)^2}\ \text{d}x = -\frac{2}{(x+1)}$$
Obtained by simply using $y = (x+1)$.
Final result:
$$\ln(x^2+2x+1) -\frac{2}{(x+1)}$$
The logarithm can be written also as
$$\ln((x+1)^2) = 2\ln(x+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1684124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Rationalising factor of $a+b \sqrt{2}+c \sqrt{3} + d \sqrt{6}$ I am trying to express the inverse of $a+b \sqrt{2}+c \sqrt{3} + d \sqrt{6}$ (given $a, b, c, d \in \mathbb{Q}$) in the form $e+f \sqrt{2}+g\sqrt{3}+h\sqrt{6}$ (where $e, f, g, h \in \mathbb{Q}$).
I could come up with a long way to do this by solving 4 linear equations to find $e, f, g, h$ which we get by expading the following,
$(a+b \sqrt{2}+c \sqrt{3} + d \sqrt{6})(e+f \sqrt{2}+g\sqrt{3}+h\sqrt{6})=1$ and equating the rational term equal to 1 and the rest of the terms equal to $0$.
But, I am now trying to find a more creative (elegant, generalisable, insightful) way to do this.
For this I tried a few things which did not work (unecessary detail?).
I guessed, $(a+b \sqrt{2}+c\sqrt{3}+d\sqrt{6})(a+b \sqrt{2}+c\sqrt{3}-d\sqrt{6})(a+b \sqrt{2}-c\sqrt{3}+d\sqrt{6})(a-b \sqrt{2}+c\sqrt{3}+d\sqrt{6}) \cdots$
(all the terms with all possible signs), but this did not work as I checked with a special case.
I guessed that since a similar thing works for the rationalising factor of $\sqrt{a}+\sqrt{b}+\sqrt{c}$ which has $(\sqrt{a}+\sqrt{b}-\sqrt{c})(-\sqrt{a}+\sqrt{b}+\sqrt{c})(\sqrt{a}-\sqrt{b}+\sqrt{c})$ (I was motivated to this from heron's formula for area of triangle).
So, can anyone help me? Please.
By the way I am in high school so I don't know anything about field splittings and so on.
| The idea with multiplying the same expression with all possible signs is good, but requires a lot of multiplying, so it is easy to make a mistake. It turns out that it is sufficient to only multiply four such expressions (where sign next to $\sqrt{6}$ is a product of signs next to $\sqrt{2}$ and $\sqrt{3}$.):
$$(a+b\sqrt{2} + c\sqrt{3} + d\sqrt{6})(a+b\sqrt{2} - c\sqrt{3} - d\sqrt{6}) = a^2 + 2b^2 + 3c^2 + 6d^2 + (2ab - 6cd)\sqrt{2}$$
$$(a-b\sqrt{2} + c\sqrt{3} - d\sqrt{6})(a-b\sqrt{2} - c\sqrt{3} + d\sqrt{6}) = a^2 + 2b^2 + 3c^2 + 6d^2 - (2ab - 6cd)\sqrt{2}$$
So, product of all four expressions is $(a^2 + 2b^2 + 3c^2 + 6d^2)^2 - 2(2ab - 6cd)^2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1684506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Prove that $n(n+1)(n+5)$ is a multiple of $6$ I need to prove that $n(n+1)(n+5)$ is divisible by 6. where $n$ is a natural number. I have used the method of induction. But not successful
I got the expression $(k^3+6k^2+5k)+3k^2+15k+12$ when $n=k+1$.
The term inside the bracket is divisible by 6 since we have assumed that the result is true when $n=k$. If we can show that $3k^2+15k+12$ is also divisible by 6, then we are done. But how to proceed?
| $$
\begin{align}
n(n+1)(n+5)
&=n(n+1)(n+2)+3n(n+1)\\[6pt]
&=6\binom{n+2}{3}+6\binom{n+1}{2}
\end{align}
$$
Binomial Coefficient Basics
If, instead of Pascal's Triangle, we define the binomial coefficients as
$$
\binom{n}{k}=\frac{n(n-1)\cdots(n-k+1)}{k!}
$$
Then we have
$$
\begin{align}
\binom{n-1}{k}+\binom{n-1}{k-1}
&=\binom{n-1}{k-1}\frac{n-k}k+\binom{n-1}{k-1}\\
&=\binom{n-1}{k-1}\frac nk\\
&=\binom{n}{k}
\end{align}
$$
Thus, if $\binom{n-1}{k}\in\mathbb{Z}$ for all $k\in\mathbb{Z}$, then $\binom{n}{k}\in\mathbb{Z}$ for all $k\in\mathbb{Z}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1685246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 1
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Cannot find $\displaystyle \int_0^{\frac{\pi}{6}} \frac{1}{\sin x-\cos x} \, dx$ I am asked to find:
$$\int_0^{\frac{\pi}{6}} \frac{1}{\sin x-\cos x} \, dx$$
I have tried:
$$A=\int_0^{\frac{\pi}{6}} \frac{\sin x+\cos x}{\sin^2 x-\cos^2 x} \, dx$$
$$A=\int_0^{\frac{\pi}{6}} \frac{\sin x}{2\sin^2 x-1} \, dx + \int_0^{\frac{\pi}{6}} \frac{\cos x}{1-2\cos^2 x} \, dx$$
$$u=\sin x$$
$$du=\cos x\,dx$$
$$v=\cos x$$
$$dv=-\sin x \,dx$$
$$A=\int_0^{\frac{\pi}{6}} \frac{du}{2u^2-1} \, dx + \int_0^{\frac{\pi}{6}} \frac{dv}{2v^2-1} \, dx$$
But I am unable to move forward.
| \begin{align}
& \int_0^{\pi/6} \frac{\sin x}{2\sin^2 x-1} \, dx = \int_0^{\pi/6} \frac{\sin x}{1 - 2\cos^2 x} \, dx & & \text{since }\sin^2 x = 1-\cos^2 x \\[15pt]
= {} & \int_1^{\sqrt 3/2} \frac{-dw}{1-2w^2} = \int_1^{\sqrt 3/2} \frac{-dw}{\left( 1 - w\sqrt 2\right)\left( 1 + w\sqrt2\right)} & & \text{and then use partial fractions.}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1688907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
How to prove $\sin\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{bc}};\cos\frac{A}{2}=\sqrt{\frac{s(s-a)}{bc}}$ I have read these formulae in my book but i could not understand how these are proved.
$\sin\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{bc}};\sin\frac{B}{2}=\sqrt{\frac{(s-c)(s-a)}{ca}};\sin\frac{C}{2}=\sqrt{\frac{(s-a)(s-b)}{ab}}$ and
$\cos\frac{A}{2}=\sqrt{\frac{s(s-a)}{bc}};\cos\frac{B}{2}=\sqrt{\frac{s(s-b)}{ca}};\cos\frac{C}{2}=\sqrt{\frac{s(s-c)}{ab}}$
where $A,B,C$ are the angles of a triangle $ABC$ and $a,b,c$ are the sides opposite to the angles $A,B,C$ respectively.
I know that area of triangle$\Delta=\frac{1}{2}bc\sin A$
But area of triangle by Heron's formula is $\Delta=\sqrt{s(s-a)(s-b)(s-c)}$
$\sin A=\frac{2\Delta}{bc}$
$2\sin\frac{A}{2}\cos\frac{A}{2}=\frac{2\Delta}{bc}$
$\sin\frac{A}{2}\cos\frac{A}{2}=\frac{\sqrt{s(s-a)(s-b)(s-c)}}{bc}$
I do not know how to prove it further or there is some other method to prove it.
| \begin{aligned}
\cos A &= \frac{b^2+c^2-a^2}{2bc} \\
\cos A &= 2\cos^2\frac{A}{2} - 1 \\
2\cos^2\frac{A}{2} - 1 &= \frac{b^2+c^2-a^2}{2bc}\\
\cos^2 \frac{A}{2} &= \frac{b^2+c^2-a^2+2bc}{4bc} = \frac{(b+c-a)(b+c+a)}{4bc}\\
\cos \frac{A}{2} &= \sqrt{\frac{(2s-2a)(2s)}{4bc}} = \sqrt{\frac{s(s-a)}{bc}}
\end{aligned}
$\sin \frac{A}{2}$ can be solved in the same way
| {
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"url": "https://math.stackexchange.com/questions/1691711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Odds of two players meetings in an eight person single elimination tournament There is an $8$ person tournament. The odds of winning are $50\%$ for each player.
What is the probability of any $2$ players meetings at any point in the tournament?
I understand that there are $7$ matches, and there are are a total of $\binom 82$ ways to choose $2$ people out of $8$. So the probability is $\dfrac{7}{\frac{8!}{2!6!}} = 0.25$.
What I am stuck at is another way of approaching the problem. The question is asking what is the probability of $A$ and $B$ would meet. There are $3$ rounds of matches. In the first round, P($A$ meeting $B$) is $1/7$. In the second round, given that $A$ hasn't met $B$ in round $1 (p=6/7)$, and that both $A$ and $B$ proceeded into the second round $(p=0.5*0.5 = 1/4), A$ has a probability of $1/3$ to meet $B$. So the probability that they meet in round $2\;\; is\;\; 6/7*1/4*1/3 = 1/14$.
To make $1/7+1/14+P(A, B$ meet in round $3) = 0.25$, $P(A,B$ meet in round $3) = 1/28.$ But I could not figure out why the probability that $A$ and $B$ meet in round $3$ is $1/28.$ Any ideas?
| Here's another way of approaching the problem.
A will meet 1, 2 or 3 opponents, with probabilities $\frac{1}{2}$ (he gets knocked out in round 1 [R1]), $\frac{1}{4}$ (he wins in R1 and loses in R2), and $\frac{1}{4}$ (he wins in R1 and R2).
If he meets 1 opponent, there's $\frac{1}{7}$ chance it's B.
If he meets 2, there's $\frac{2}{7}$ chance that one of them is B.
If he meets 3, there's $\frac{3}{7}$ chance that one of them is B.
So the probability he meets B is $(\frac{1}{2}*\frac{1}{7})+(\frac{1}{4}*\frac{2}{7})+(\frac{1}{4}*\frac{3}{7})=\frac{1}{4}$.
And if he meets 3 opponents and one of them is B, then the probability that he meets B in any particular round is $\frac{1}{3}$. So the overall probability that he meets B in the 3rd round is
$(\frac{1}{4}*\frac{3}{7}) * \frac{1}{3} = \frac{1}{28}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluation of $1-\frac{1}{7}+\frac{1}{9}-\frac{1}{15}+\frac{1}{17}-\dotsb$
How can we calculate sum of following infinite series
$\displaystyle \bullet\; 1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\frac{1}{11}-\cdots$
$\displaystyle \bullet\; 1-\frac{1}{7}+\frac{1}{9}-\frac{1}{15}+\frac{1}{17}-\cdots$
$\textbf{My Try:}$ Let $$S = \int_0^1 (1+x^2-x^4-x^6+x^8+x^{10}+\cdots) \, dx$$
So we get $$S=\int_0^1 \left(1-x^4+x^8-\cdots\right)dx+x^2\int_0^1 (1-x^4+x^8-\cdots)\,dx$$
So we get $$S=\int_0^1 \frac{1+x^2}{1+x^4} \, dx= \frac{\pi}{2\sqrt{2}}$$ after that we can solve it
Again for second One, Let $$S=\int_0^1 (1-x^6+x^8-x^{14}+x^{16}+\cdots)$$
So we get $$S=\int_0^1 (1+x^8+x^{16}+\cdots) \, dx-\int_0^1 (x^6+x^{14}+
\cdots)\,dx$$
So we get $$S=\int_{0}^{1}\frac{1-x^6}{1-x^8}dx = \int_{0}^{1}\frac{x^4+x^2+1}{(x^2+1)(x^4+1)}dx$$
Now how can i solve after that, Help me
Thanks
| Start with partial fractions
$$\frac{x^4+x^2+1}{(x^2+1)(x^4+1)} = \frac{A}{x^2+1}+ \frac{B x^2+C}{x^4+1}$$
Thus,
$$A+B=1$$
$$B+C=1$$
$$A+C=1$$
or $A=B=C=1/2$. Also note that
$$x^4+1 = (x^2+\sqrt{2} x+1)(x^2-\sqrt{2} x+1) $$
so that
$$\frac{x^2+1}{x^4+1} = \frac{P}{x^2-\sqrt{2} x+1} + \frac{Q}{x^2+\sqrt{2} x+1}$$
where $P=Q=1/2$. Thus,
$$\frac{x^4+x^2+1}{(x^2+1)(x^4+1)} = \frac14 \left [2 \frac1{x^2+1} + \frac1{(x-\frac1{\sqrt{2}})^2+\frac12} + \frac1{(x+\frac1{\sqrt{2}})^2+\frac12} \right ]$$
And the integral is
$$\frac12 \frac{\pi}{4}+ \frac14 \sqrt{2} \left [\arctan{(\sqrt{2}-1)}-\arctan{(-1)} \right ] + \frac14 \sqrt{2} \left [\arctan{(\sqrt{2}+1)}-\arctan{(1)} \right ]= \frac{\pi}{8} (\sqrt{2}+1) $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to simplify the nested radical $\sqrt{1 - \frac{\sqrt{3}}{2}}$ by hand? I was solving a Mock Mathcounts Contest Mock contest (.pdf) written by a user on the Art of Problem Solving Forums. In problem #24 the only thing I couldn't do by hand was simplify the radical mentioned above. Note that the contest should involve math topics accessible to a high performing Mathcounts middle school student.
How do you simplify $$\sqrt{1 - \frac{\sqrt{3}}{2}}$$
| Proposed another way to collection
You can also use formulas
$$\sqrt{a\pm\sqrt b}=\sqrt{\frac{a + \sqrt{a^2-b}}{2}}\pm\sqrt{\frac{a - \sqrt{a^2-b}}{2}}$$
Then
$$\sqrt{1-\frac{\sqrt3}{2}}=\frac12\sqrt{4-\sqrt{12}}=\frac12 \cdot\left(\sqrt{\frac{4 + \sqrt{4^2-12}}{2}}-\sqrt{\frac{4 - \sqrt{4^2-12}}{2}} \right)=$$
$$=\frac12 \left(\sqrt{\frac{4 + \sqrt{4}}{2}}-\sqrt{\frac{4 - \sqrt{4}}{2}} \right)=\frac12 \left(\sqrt{\frac{6}{2}}-\sqrt{\frac{2}{2}} \right)=\frac 12(\sqrt3-1)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How to show convergence of a particular series from first principles I am refreshing my techniques for series and am currently stuck with the following exercise:
Show that for every real number $x > 1$ the series
$$
\sum_k \frac{2^k}{1 + x^{2^k}}
$$
converges.
*
*Hint: Add $\frac{1}{1-x}$.
This exercise appears right at the start of the chapter so nothing other than the bare definition of convergence of a series (via convergence of the partical sums) can be used -- not even the convergence of the geometric series as this has not been established yet. Though basic sequence results are ok to be applied to the sequence of partial sums.
I can find a solution, but not via the hint and I was wondering how I could use the hint!
P.S. Here is the argument that I have:
Using the fact that $x > 1$ and $2^k \ge 2k$ for each $k \in \mathbb N$ We have
\begin{align}
s_n = \sum^n_{k=1} \frac{2^k}{1 + x^{2^k}} \le \sum^n_{k=1} \frac{2^k}{(1 + x)^{2^k}} \le \sum^n_{k=1} \frac{2^k}{(1 + x)^{2k}} \le \sum^n_{k=1} \frac{2^k}{2^{2k}} = \sum^n_{k=1} \left(\frac{1}{2}\right)^k
\end{align}
and we know that the right hand side converges as $n \to \infty$. This approach does not use the hint, and rests on the knowledge that $\sum_k (\frac{1}{2})^k < \infty$.
| Hint Let's add $\frac{1}{1-x}$ to the first term and the result to the next term:
$$\begin{align}\frac{1}{1-x}+\frac{1}{1+x}&&=&\frac{2}{1-x^2}\\
\frac{2}{1-x^2}+\frac{2}{1+x^2} &&=&\frac{2^2}{1-x^{2^2}}\\
\frac{2^2}{1-x^4}+\frac{2^2}{1+x^4} &&=&\frac{2^3}{1-x^{2^3}} \end{align}$$
Can you see the telescoping formula:
$$\frac{1}{1-x}+\left(\frac{2^0}{1+x^{2^0}}+\frac{2^1}{1+x^{2^1}}+\frac{2^2}{1+x^{2^2}}\right)=\frac{8}{1-x^8} $$
Telescoping formula is : ( pass the mouse over to see it)
$$\frac{2^k}{1+x^{2^k}}=\frac{2^{k}}{x^{2^{k}}-1}-\frac{2^{k+1}}{x^{2^{k+1}}-1} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove this stronger inequality with $\frac{e^x}{x+1}-\frac{x-1}{\ln{x}}-\left(\frac{e-2}{2}\right)>0$ Let $x>1$ show that
$$\dfrac{e^x}{x+1}-\dfrac{x-1}{\ln{x}}-\left(\dfrac{e-2}{2}\right)>0$$
It seem this inequality can use derivatives solve it,But it is ugly.can you help？
$$\lim_{x\to 1}\left(\dfrac{e^x}{x+1}-\dfrac{x-1}{\ln{x}}-\left(\dfrac{e-2}{2}\right)\right)=0$$
and $f(x)=\dfrac{e^x}{x+1},g(x)=\dfrac{x-1}{\ln{x}}$,But $$f'(x)=\dfrac{xe^x}{(x+1)^2}>0, g'(x)=\dfrac{\ln{x}-\dfrac{x-1}{x}}{\ln^2{x}}>0$$
Unfortunately I don't know any nice expression for this 1-th derivative which could help. I'll be grateful for all useful suggestions.
| Set $$f(x)=\frac{e^x}{x+1}-\frac{ex}{4},\,\,g(x)=\frac{ex}{4}-\frac{x-1}{\ln x}.$$ Then $$f'(x)=\frac{xe^x}{(x+1)^2}-\frac{e}{4}=e\cdot\frac{4xe^{x-1}-(x+1)^2}{4(x+1)^2},$$ and since $e^{x-1}>x$ for all $x>1$, we obtain $$4xe^{x-1}-(x+1)^2>4x^2-(x+1)^2>0,$$ since $x>1$. This shows that $f$ is increasing. Moreover, $$g'(x)=\frac{x-1-x\ln x}{x\ln^2x}+\frac{e}{4}=\frac{x-1-x\ln x}{x\ln^2x}+\frac{1}{2}+\frac{e-2}{4}.$$ We will show that $g$ is also increasing for $x>1$; to do this, it is enough to show that $$x-1-x\ln x+\frac{1}{2}x\ln^2x>0,$$ or, equivalently, $$\ln^2x-2\ln x+2-\frac{2}{x}>0,$$ for $x>1$.
Set $h(x)=\ln^2x-2\ln x+2-\frac{2}{x}$, and suppose that $h$ has a root $x_0$, for some $x_0>1$. Since $h(1)=0$, there exists a root of $h'(x)=0$ in $(1,\infty)$. But, $$h'(x)=2\frac{\ln x}{x}-\frac{2}{x}+\frac{2}{x^2}=\frac{2}{x}\left(\ln x-1+\frac{1}{x}\right),$$ and the last function does not have roots greater than $1$, which is a contradiction. This shows that $h$ has a fixed sign on $(1,\infty)$, and since $h$ goes to $\infty$ as $x\to\infty$, we obtain that $h(x)>0$ for $x>1$.
So, we obtain that $g$ is increasing as well. Therefore, $$\frac{e^x}{x+1}-\frac{x-1}{\ln x}=f(x)+g(x)$$ is increasing as well, which shows the inequality.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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If $b$ is the geometric mean of $a$ and $c$, then $b^2$ is the geometric mean of $a^2$ and $c^2$ Can someone tell me if I proved this correctly? I'm new to proofs.
\begin{align}
b^2 &= ac \\
\sqrt{b^2} &= \sqrt{ac} \\
b &= \pm \sqrt{ac} \\
b^4 &= (\sqrt{ac})^4 \\
b^4 &= a^2 c^2 \\
\sqrt[4]{b^4} &= \sqrt[4]{a^2c^2} \\
b &= \pm \sqrt{ac}
\end{align}
| I would write the proof as follows:
If b is the geometric mean of a and c, then $b=\sqrt{ac}$. Therefore $b^2=ac=\sqrt{a^2c^2}$, so $b^2$ is the geometric mean of $a^2$ and $c^2$.
Note that the converse is not necessarily true. If $b^2=\sqrt{a^2c^2}$, b could be negative, or a or c could be negative, in which case b would not be the geometric mean of a and c.
| {
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"question_score": "1",
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Universal factoring method or list of methods (trinomials) I am a student in calculus II. I'm now failing tests solely because I cannot factor; I understand everything else. This is compounded by the fact it seems to exceedingly hard to find anything comprehensive online. This is a long-running problem I've been facing; I can't be the only one.
I know some methods, but I always run into more problems I can't factor. I know about synthetic/long division, binomial/perfect square trinomial factoring, difference and sum of cubes and squares, the grouping method, the quadratic formula (unsure how to use for factoring), guess and check, and some trinomial methods. I only know the difference/sum of squares/cubes well. For example, I can't factor $4x^2 - 4x - 3$.
It would be dandy if I could get a comprehensive list or a universal solution to all these factoring questions, but that's unlikely to happen (and would be too broad). I'll narrow this down to something more tangible: What are the methods I should know to factor trinomials?
| Before we consider how to factor $4x^2 - 4x - 3$, let's see what happens when we form the product of two linear monomials.
\begin{align*}
(6x - 7)(3x + 5) & = 6x(3x + 5) - 7(3x + 5)\\
& = \color{blue}{18}x^2 + \color{green}{30}x \color{green}{ - 21}{x} \color{blue}{- 35}\\
& = \color{blue}{18}x^2 - \color{green}{9}x - \color{blue}{35}
\end{align*}
To factor, what we wish to do is carry out the steps in reverse. The key observation is that in the expression $\color{blue}{18}x^2 + \color{green}{30}x \color{green}{ - 21}{x} \color{blue}{- 35}$, the product of the quadratic and constant coefficients is equal to the product of the two linear coefficients whose sum is the linear coefficient of the product. That is,
$$(\color{blue}{18})(\color{blue}{-35}) = (\color{green}{30})(\color{green}{-21}) = -630$$
This suggests that we can factor a quadratic polynomial $ax^2 + bx + c$ with rational coefficients with respect to the rationals if we can split the linear term if we can find two numbers with product $ac$ and sum $b$.
In your example of $4x^2 - 4x - 3$, we can split the linear term if we can find two numbers with product $(\color{blue}{4})(\color{blue}{-3}) = -12$ and sum $-4$. The possible factorizations of $-12$ are
\begin{align*}
-12 & = -1 \cdot 12 & -12 & = 1 \cdot -12\\
& = -2 \cdot 6 & & = \color{green}{2} \cdot \color{green}{-6}\\
& = -3 \cdot 4 & & = 3 \cdot -4
\end{align*}
By inspection, the only pair of factors of $-12$ with sum $-4$ are $2$ and $-6$. Hence,
\begin{align*}
4x^2 - 4x - 3 & = 4x^2 + 2x - 6x - 3 && \text{split the linear term}\\
& = 2x(2x + 1) - 3(2x + 1) && \text{factor by grouping}\\
& = (2x - 3)(2x + 1) && \text{extract the common factor}
\end{align*}
Another option is completing the square.
\begin{align*}
4x^2 - 4x - 3 & = 4(x^2 - x) - 3\\
& = 4\left(x^2 - x + \frac{1}{4}\right) - 1 - 3\\
& = 4\left(x - \frac{1}{2}\right)^2 - 4\\
& = 4\left[\left(x - \frac{1}{2}\right)^2 - 1\right]\\
& = 4\left[\left(x - \frac{1}{2}\right) + 1\right]\left[\left(x - \frac{1}{2}\right) - 1\right]\\
& = 4\left(x + \frac{1}{2}\right)\left(x - \frac{3}{2}\right)
\end{align*}
Since
\begin{align*}
4\left(x + \frac{1}{2}\right)\left(x - \frac{3}{2}\right) & = 2 \cdot 2 \left(x + \frac{1}{2}\right)\left(x - \frac{3}{2}\right)\\
& = \left[2\left(x + \frac{1}{2}\right)\right]\left[2\left(x - \frac{3}{2}\right)\right]\\
& = (2x + 1)(2x - 3)
\end{align*}
our two factorizations are equivalent.
For the general case,
\begin{align*}
ax^2 + bx + c & = a\left(x^2 + \frac{b}{a}x\right) + c\\
& = a\left(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}\right) - \frac{b^2}{4a} + c\\
& = a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2 - 4ac}{4a}\\
& = a\left[\left(x + \frac{b}{2a}\right)^2 - \frac{b^2 - 4ac}{4a^2}\right]\\
& = a\left[x + \frac{b}{2a} + \frac{\sqrt{b^2 - 4ac}}{2a}\right]\left[x + \frac{b}{2a} - \frac{\sqrt{b^2 - 4ac}}{2a}\right]
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality $abdc$ $\leq$ $3$ $a+b+c+d=6$
and
$a^2+b^2+c^2+d^2=12$.
and $a,b,c,d$ are reals.
Prove: $abcd$ $\leq$ $3$ without Lagrange multipliers, complex numbers or convexity help.
Using Cauchy–Schwarz inequality I found: $a,b,c,d \in [0,3]$.
How solve inequality?
| An altenative interpretation. By Vieta's formulas, we may assume that $a,b,c,d$ are four real roots of the polynomial
$$ q(t) = t^4-6t^3+12t^2+Kt+ p $$
and we have to prove that under such assumptions $p=abcd\geq 3$, so we have to study the cases for which
$$ f(t) = t(t-2)^3 = t^4-6t^3+12t^3-8t = Ct-p $$
has four real solutions. That is the same as requesting that
$$ g(t) = \frac{1}{4}(2t^2-6t+3)^2 = t^4-6t^3+12t^3-9t+\frac{9}{4} = Dt-\left(p-\frac{9}{4}\right) $$
has four real solutions. Since $g(t)\geq 0$, that may happen only if $D\geq 0$. Moreover, that may happen only if the value of $Dt-\left(p-\frac{9}{4}\right)$ at $t=0$ is less than the value of $g(0)$. That gives $abcd\geq 0$, for starters. The midpoint of the roots of $g'(t)$ is at $t=\frac{3}{2}$, so we must have that the value of $Dt-\left(p-\frac{9}{4}\right)$ at $t=\frac{3}{2}$ is between $0$ and $g\left(\frac{3}{2}\right)=\frac{9}{16}$ (VRC-very restrictive constraint). Now it is enough to compute the equations of the tangent line to the graph of $g(t)$ at its first inflection point $t=1$ to get $\left(p-\frac{9}{4}\right)\leq \frac{3}{4}$, from which $\color{red}{abcd\leq 3}$ follows.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the system of equations $\frac{x}{3} + \frac{y}{2} = \frac{2x}{3} - \frac{y}{6} = 7$ for $x$ and $y$. Here is the problem. The answer is suppose to be x = 12, y = 6
$$\frac{ x }{3} + \frac{y}{2} = \frac{ 2x }{3} - \frac{ y }{6} = 7$$
I've tried this. I'm having trouble figuring out what do on the last step I will describe.
$$6 • \frac{ x }{3} + \frac{y}{2} = \frac{ 2x }{3} - \frac{ y }{6} = 7$$
$$2x + 3y = 4x - y = 42$$
$$2x - 4y = 42$$
$$4y = 42 - 2x$$
$$y = \frac{ 42 - 2x}{4}$$
$$2x + 3[ \frac{ 42 - 2x }{4} ] = 7$$
| You are really given two equations here:
$$
\left\{\begin{align}
\frac{x}{3} + \frac{y}{2} &= 7\\[0.3cm]
\frac{2x}{3} - \frac{y}{6} &= 7
\end{align}
\right.
$$
Multiply the first equation by 6 to get $2x + 3y = 42$. Multiply the second equation by 6 to get $4x - y = 42$.
$$
\left\{
\begin{align} \displaystyle
2x + 3y &= 42\\
4x - y &= 42
\end{align}
\right.
$$
Multiply the second equation by $3$ and add it to the first equation to end up with $14x = 168$, and so $x = 12$. Then use either equation to get $y = 6$.
| {
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"timestamp": "2023-03-29T00:00:00",
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logarithm equation with scalar on right hand side $\log_7 (x^2-1) - \log_7 (x-1) = 2$
$\log_7 49 = 2$
=> $\log_7 (x^2-1) - \log_7 (x-1) = \log_7 49$
=> $\frac{(x^2-1)}{x-1} = 49$
=> $x^2 -1 = 49(x -1)$
=> $x^2 -1 = 49x -49$
=> $x^2 - 49x + 48 = 0$
The answer in the book is 48 so I'm obviously doing this wrong.
| We have that $x>1$ ($\log_7x$ is only defined if $x>0$) and
$$
\log_7(x^2-1)-\log_7(x-1)=\log_7\biggl(\frac{x^2-1}{x-1}\biggr)=\log_7\biggl(\frac{(x-1)(x+1)}{x-1}\biggr)=\log_7(x+1)=2.
$$
Hence, $x+1=49$ and $x=48$.
| {
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Laurent series expansion of $f(z)$ which converges for $\frac12 <|z|<1$
My task is to find the Laurent series expansion of $$f(z)=\frac{1}{2z^2-z-1},$$ which converges for $\displaystyle \frac12 <|z|<1$.
I proceeded by doing the following: $$\frac{1}{2z^2-z-1} = \frac{1}{(2z+1)(z-1)} = \frac13 \left( \frac{1}{z-1}-\frac{2}{2z+1} \right).$$
For the first term: $$\frac13 \frac{1}{z-1} = \frac{1}{3z} \frac{1}{1-\frac1z}=\frac{1}{3z}\sum_{n=0}^\infty \frac{1}{z^n},$$ for $\displaystyle \left| \frac1z \right|<1 \Rightarrow |z|>1.$ Hence, for $|z|<1$ (using geometric series properties): $$\frac13 \frac{1}{z-1} =-\frac{1}{3z}\sum_{n=1}^\infty z^n=-\frac13 \sum_{n=1}^\infty z^{n-1}.$$ For the second term: $$-\frac13\frac{2}{2z+1}=-\frac23\frac{1}{1+2z}=-\frac23\sum_{n=0}^\infty\left( -2z \right)^n,$$ for $\displaystyle |2z|<1 \Rightarrow |z|<\frac12$. Hence, for $\displaystyle |z|>\frac12$: $$-\frac13\frac{2}{2z+1}=\frac23 \sum_{n=1}^\infty\frac{1}{(-2z)^n}=\frac13 \sum_{n=1}^\infty(-1)^{-n}2^{1-n}z^{-n}.$$ Hence, $$\frac{1}{2z^2-z-1}=\frac13 \sum_{n=1}^\infty(-1)^{-n}2^{1-n}z^{-n}-\frac13 \sum_{n=1}^\infty z^{n-1}.$$
I think this is the correct approach. Some further guidance/clarification would be great!
| Note: This is just an affirmation of OPs calculation.
The function
\begin{align*}
f(z)&=\frac{1}{3}\frac{1}{1-z}-\frac{1}{3}\frac{1}{z+\frac{1}{2}}\\
\end{align*}
has two simple poles at $z=1$ and $z=-\frac{1}{2}$.
Since we want to find a Laurent expansion with center $0$, we look at these poles and see they determine three regions.
\begin{align*}
|z|<\frac{1}{2},\qquad\quad
\frac{1}{2}<|z|<1,\qquad\quad
1<|z|
\end{align*}
*
*The first region $ |z|<\frac{1}{2}$ is a disc with center $0$, radius $\frac{1}{2}$ and the pole $\frac{1}{2}$ at the boundary of the disc. In the interior of this disc all two fractions with poles $\frac{1}{2}$ and $1$ admit a representation as power series at $z=0$.
*The second region $\frac{1}{2}<|z|<1$ is the annulus with center $0$, inner radius $\frac{1}{2}$ and outer radius $1$. Here we have a representation of the fraction with pole $\frac{1}{2}$ as principal part of a Laurent series at $z=0$, while the fraction with pole at $1$ admits a representation as power series.
*The third region $|z|>1$ containing all points outside the disc with center $0$ and radius $1$ admits for all fractions a representation as principal part of a Laurent series at $z=0$.
A power series expansion of $\frac{1}{z+a}$ at $z=0$ is
\begin{align*}
\frac{1}{z+a}&=\frac{1}{a}\cdot\frac{1}{1+\frac{z}{a}}\\
&=\sum_{n=0}^{\infty}\frac{(-1)^n}{a^{n+1}}z^n
\end{align*}
The principal part of $\frac{1}{z+a}$ at $z=0$ is
\begin{align*}
\frac{1}{z+a}&=\frac{1}{z}\cdot\frac{1}{1+\frac{a}{z}}=\frac{1}{z}\sum_{n=0}^{\infty}\frac{(-a)^n}{z^n}\\
&=\sum_{n=1}^{\infty}\frac{(-a)^{n-1}}{z^n}
\end{align*}
OPs calculation corresponds to the representation in the second region and is correct. It is sufficient to calculate the power series part for $\frac{1}{1-z}$ and the principal part of the Laurent series for $\frac{2}{2z+1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1710433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Using Fourier Series to find the sum of a numerical series I have to use a Fourier series to compute the sum of the series $$\frac{1}{2} + \sum_{n=1}^\infty (-1)^n \frac{1}{n^2 + a^2}$$
My guesses are the Fourier series $$e^{ax} = \frac{e^{a\pi} - e^{-a\pi}}{\pi}[\frac{1}{2} + \sum_{n=1}^\infty (-1)^n \frac{1}{n^2 + a^2} (a\cos nx - n \sin nx)]$$
and
$$\cosh x = \frac{2}{\pi}\sinh a\pi [\frac{1}{2} + \sum_{n=1}^\infty (-1)^n \frac{a}{n^2 + a^2} (\cos nx)]$$
I try to simplify each Fourier series expansion to get a $1$ in the numerator, but I always seem to get something else - such as an unwanted $a$ or $n$. Any useful hints?
| Let $S$ be the series given by
$$S=\frac12 + \sum_{n=1}^\infty \frac{(-1)^n}{n^2 + a^2}\tag 1$$
Then, note that the Fourier series of $\cosh(ax)$ on the interval $[-\pi,\pi]$ is given by
$$\cosh(ax)=\frac{\sinh(a\pi)}{\pi a}\left(1+2a^2\sum_{n=1}^\infty \frac{(-1)^n}{n^2+a^2}\cos(nx)\right) \tag 2$$
Setting $x=0$ in $(2)$ and solving for the series $\sum_{n=1}^\infty \frac{(-1)^n}{n^2 + a^2}$ reveals
$$\sum_{n=1}^\infty \frac{(-1)^n}{n^2 + a^2}=\frac{\pi}{2a\sinh(\pi a)}-\frac1{2a^2} \tag 3$$
Comparing $(1)$ and $(3)$, we obtain
$$\bbox[5px,border:2px solid #C0A000]{S=\frac{a^2-1}{2a^2}+\frac{\pi}{2a\sinh(\pi a)}}$$
And we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1712431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simplifying the integral $\int\frac{dx}{(3 + 2\sin x - \cos x)}$ by an easy approach $$I=\displaystyle\int\frac{dx}{(3 + 2\sin x - \cos x)}$$
If $$\tan\left(\frac{x}{2}\right)=u$$
or $$x=2\cdot\tan^{-1}(u)$$
Then,
$$\sin{x}=\dfrac{2u}{1+u^2}$$
$$\cos{x}=\dfrac{1-u^2}{1+u^2}$$
$$dx=\dfrac{2}{1+u^2}$$
Substitute $$\tan\left(\dfrac{x}{2}\right)=u$$
Let us simplify the integrand before integrating
$$\dfrac{1}{3+2\sin{x}-\cos{x}}$$
$$=\dfrac{1}{3+2\frac{2u}{1+u^2}-\frac{1-u^2}{1+u^2}}$$
$$=\dfrac{1}{3+\frac{4u-1+u^2}{1+u^2}}$$
$$=\dfrac{1}{\frac{4u-1+u^2+3+3u^2}{1+u^2}}$$
$$=\dfrac{1+u^2}{4u^2+4u+2}$$
$$=\dfrac{1+u^2}{(2u+1)^2+1}$$
$$I=\displaystyle\int\dfrac{1+u^2}{(2u+1)^2+1}\cdot\dfrac{2}{1+u^2}\ du$$
$$=\displaystyle\int\dfrac{1}{(2u+1)^2+1}\ 2\,du$$
Now,
Take : $$v=2u+1$$
Therefore, $$dv=2\,du$$
$$I=\displaystyle\int\dfrac{1}{v^2+1}\ dv$$
$$I=\tan^{-1}(v)$$
Substitute everything back
$$I=\tan^{-1}(2u+1)$$
$$I=\tan^{-1}\left(2\tan\left(\frac{x}{2}\right)+1\right)$$
$$\boxed{\displaystyle\int\frac{dx}{(3 + 2\sin x - \cos x)} = \tan^{-1}\left(2\tan\left(\frac{x}{2}\right)+1\right)+C}$$
I know that my approach is also not so difficult but still I think there must be an relatively easy approach to this integral. I have tried many different things using trigonometric identities but nothing seems to bring to the solution easily. Kindly help me out.
| Generally, consider the real function
$$f(x)=\frac{1}{a+b\cos x+c\sin x}$$
With $a^2>b^2+c^2$ so that $f$ is defined on $\mathbb{R}$.
It is not hard to check that the derivative of
$$F(x)=\frac{x}{d}+\frac{2}{d}\arctan\left(\frac{c\cos x-b\sin x}{d+a+b\cos x+c\sin x}\right)$$
with $d=\sqrt{a^2-b^2-c^2}$, is $f(x)$. So $\int f(x)dx=F(x)+k$. The advantage
of this expression of $F$ is that it is also defined on $\mathbb{R}$. In particular,
$$\int \frac{1}{3-\cos x+2\sin x}=\frac{x}{2}+\arctan\left(\frac{2\cos x+\sin x}{5-\cos x+2\sin x}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1712481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Find $\lim_{x \to 0}\left(\frac{\cos x}{\cos(2x)}\right)^{\frac{1}{x^2}}$ $$\lim_{x \to 0}\left(\frac{\cos x}{\cos(2x)}\right)^{\frac{1}{x^2}}$$
What I have done is to take the $\ln$
$$e^{\lim_{x \to 0}\ln\left(\left(\frac{\cos x}{\cos(2x)}\right)^{\frac{1}{x^2}}\right)}$$
$$y={\lim_{x \to 0}{\frac{1}{x^2}}\cdot \ln\frac{\cos x}{\cos(2x)}}$$
Applying l'hopital to get
$$y={\lim_{x \to 0}{\frac{-2}{x^3}}\cdot \frac{\cos(2x)}{\cos x}\cdot \frac{-\sin x\cos(2x)+2\sin(2x)\cos x}{(\cos(2x)^2)}}={\lim_{x \to 0}\frac{-2}{x^3\cdot \cos x}\cdot \frac{-\sin x\cos(2x)+2\sin(2x)\cos x}{(\cos(2x))}}$$
Should I apply l'hopital again?
| If $L=\lim\ (\frac{\cos\ x}{\cos\ 2x})^\frac{1}{x^2}$ then
\begin{align} \ln\ L &= \lim \frac{\ln\ \cos\ x -\ln\ \cos\ 2x}{x^2}
\\
&=\lim \ \frac{ \frac{-\sin\ x}{\cos\ x} + \frac{2\sin\ 2x}{\cos\
2x}}{2x} \\&= \lim\ \frac{-\sin\ x +2\sin\ 2x}{2x}\\& = \lim\
\frac{-\cos\ x + 4\cos\ 2x}{2} =\frac{3}{2},\ L=e^\frac{3}{2}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1712973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Find the maximum and minimum value of the expression $\frac{x+y+z}3+\frac{2016}{\sqrt[3]{xyz}}$ Find the maximum and minimum value of the expression
$$\frac{x+y+z}3+\frac{2016}{\sqrt[3]{xyz}}$$
if $x,y,z \in [1,2016]$
My work so far:
Let $2016=a, x,y,z \in [1,a]$, $$\frac{x+y+z}3+\frac a{\sqrt[3]{xyz}}=A$$
$$m\le A \le M$$
Use $AM \ge GM:$
$$A=\frac{x+y+z}3+\frac a{\sqrt[3]{xyz}} \ge \sqrt[3]{xyz}+\frac a{\sqrt[3]{xyz}}\ge 2\sqrt a$$
So $A \ge m=2\sqrt a$, $(x=y=z=\sqrt a)$.
Help me find $M$.
| Well I'd just use calculus. We want to find the critical points of $f(x,y,z)=\frac{x+y+z}{3}+\frac{2016}{\sqrt[3]{xyz}}$. So we want to solve the system
$$\begin{align*}
\frac{\partial f}{\partial x}=\frac{1}{3}-\frac{672\cdot yz}{xyz\cdot\sqrt[3]{xyz}}=0 \\
\frac{\partial f}{\partial y}=\frac{1}{3}-\frac{672\cdot xz}{xyz\cdot\sqrt[3]{xyz}}=0 \\
\frac{\partial f}{\partial z}=\frac{1}{3}-\frac{672\cdot yx}{xyz\cdot\sqrt[3]{xyz}}=0 \\
\end{align*}$$
Doing a little manipulation we can get the solution to be
$$x=y=z=\sqrt{2016}$$
Now to find the minimum and maximum of the expression we just compare the value of $f$ at the point we got as well as at the endpoints of the allowed domain. What we have is
$$\begin{align*}
f(1,1,1)=f(2016,2016,2016)=2017 \\
f(2016,1,2016)=f(2016,2016,1)=f(1, 2016, 2016)\approx 1356.97 \\
f(1,1,2016)=f(2016,1,1)=f(1,2016,1)\approx 832.25 \\
f(\sqrt{2016},\sqrt{2016},\sqrt{2016})\approx 89.8
\end{align*}$$
So as you can see the expression is at a minimum when $(x,y,z)=(\sqrt{2016},\sqrt{2016},\sqrt{2016})$ and it is at a maximum when $(x,y,z)=(1,1,1),(2016,2016,2016)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1713148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find $(a,b)$ such that $\lim\limits_{x\to 0}\frac{f(x)}{x}=1$ implies $\lim\limits_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{(f(x))^3}=1$
Let $f$ denote any function such that $\lim\limits_{x\to 0}\frac{f(x)}{x}=1$. Find the value of $a$ and $b$, assuming that $\lim\limits_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{(f(x))^3}=1$.
My attempt:
$\lim_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{(f(x))^3}=1$
$\lim_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{x^3\frac{(f(x))^3}{x^3}}=1$
$\lim_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{x^3}=1$
$\lim_{x\to 0}\frac{1+a\cos x-b\frac{\sin x}{x}}{x^2}=1$
$\lim_{x\to 0}\frac{1+a\cos x-b}{x^2}=1$
As denominator is tending to zero,so numerato will also tend to zero.
$1+a-b=0............................(1)$
Applying L Hospital rule,
$\lim_{x\to 0}\frac{-a\sin x}{2x}=1$
So $a=-2$ and $b=-1$
But the answer given in my book is $a=\frac{-5}{2}$ and $b=-\frac{3}{2}$.
I do not understand where have i gone wrong?
| Since we can write
$$\lim_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{(f(x))^3}=\lim_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{x^3}\cdot\frac{1}{(f(x)/x)^3}$$
we have to have
$$\lim_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{x^3}=1$$
We can write
$$\lim_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{x^3}\tag1$$$$=\lim_{x\to 0}\frac{1+a\cos x-b\frac{\sin x}{x}}{x^2}\tag2$$
But we cannot write $(2)$ as
$$\lim_{x\to 0}\frac{1+a\cos x-b\cdot 1}{x^2}$$
From $(1)$, by L'Hôpital's rule,
$$(1)=\lim_{x\to 0}\frac{1+a\cos x-ax\sin x-b\cos x}{3x^2}\tag3$$
Here, we have to have
$$1+a-b=0\tag 4$$
Using L'Hôpital's rule several times,
$$\begin{align}(3)&=\lim_{x\to 0}\frac{1-ax\sin x-\cos x}{3x^2}\\&=\lim_{x\to 0}\frac{-a(\sin x+x\cos x)+\sin x}{6x}\\&=\lim_{x\to 0}\frac{-2a\cos x+ax\sin x+\cos x}{6}\\&=\frac{-2a+1}{6}\end{align}$$
and so
$$\frac{-2a+1}{6}=1\tag5$$
Now solve $(4)(5)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1713981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Complex number $z=cis(\frac{2k\pi}{5})$ for any integer $k$ such that $z \ne 1$? Consider the complex number $z=cis(\frac{2k\pi}{5})$ for any integer $k$ such that $z \ne 1$.
(a) Show that $z^n+\frac{1}{z^n}=2\cos(\frac{2nk\pi}{5})$ for any integer $n$.
(b) Show that $z^5=1$. Hence, or otherwise, show that $1+z+z^2+z^3+z^4=0$.
(c) Find the value of $b$, given that $(z+\frac{1}{z})^2+(z^2+\frac{1}{z^2})^2=b$.
My approach:
(a) This was easy, I rewrote this as $z^n+z^{-n}$, applied De Moivre's theorem, the sines cancelled out and that was all.
(b) I proved the first part that $z^5=1$, but have trouble with the second part ("hence, or otherwise...") - I tried rewritting a term, say the term $z$, as $cis(\frac{2k\pi}{5}))=cis(\frac{1}{5}*2k\pi)=(cis(2k\pi))^{1/5}=1^{1/5}=1$. This, I thought, will make all the terms be equal to 1. But the first sentence of the exercise says "for any integer $k$ such that $z\ne1$" - but how is it even possible? No matter what value of $k$ I substitute, it will always be a multiple of $2\pi$ so the $cis$ will be 1. What am I missing? How do I solve that?
(c) I think that I should use (a) here but I got stuck.
| Observe that
$$z^5=1\iff 0=z^5-1=(z-1)(z^4+z^3+z^2+z+1)$$
and since $\;z\neq1\;$ you get (b)
For (c) you can indeed use (b), and remember that all the roots of order $\;5\;$ are a cyclic (multiplicative) group, so
$$1+z+z^2+z^3+z^4=1+\overbrace{z}^{=z^{-4}}+z^2+z^{-2}+\overbrace{z^4}^{=z^{-1}}=\left(z+\frac1z\right)^2-1+\left(z^2+\frac1{z^2}\right)^2-2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1714093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the integral $\int\limits \frac{1}{(\sin x+\cos x)^3}\mathrm dx$ Using tangent half-angle substitution, the integral becomes $$\int\limits \frac{t^4+2t^2+1}{(-t^2+2t+1)^3}\mathrm dt$$
This integral can be solved using partial fractions,
$$\frac{t^4+2t^2+1}{(-t+\sqrt 2+1)^3(t+\sqrt 2-1)^3}=\frac{A}{-t+\sqrt 2+1}+\frac{B}{(-t+\sqrt 2+1)^2}+\frac{C}{(-t+\sqrt 2+1)^3}+\frac{D}{t+\sqrt 2-1}+\frac{E}{(t+\sqrt 2-1)^2}+\frac{F}{(t+\sqrt 2-1)^3}$$
What is the quicker way to solve this integral?
| HINT:
$$\sin x+\cos x=\sqrt2\cos\left(x-\dfrac\pi4\right)$$
Now use Finding $\int \sec^3x\,dx$
OR
Indefinite integral of secant cubed
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1715518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Linear algebra augmented matrix $$
\begin{bmatrix}
2 & h & 4 \\
3 & 6 & 7 \\
\end{bmatrix}
$$
Find $h$ such that this matrix is inconsistent
what I have done:
switched $R_1$ and $R_2$
$$
\begin{bmatrix}
3 & 6 & 7 \\
2 & h & 4 \\
\end{bmatrix}
$$
$R_2$ replaced by $(-3)R_2 + 2(R_1)$
$$
\begin{bmatrix}
3 & 6 & 7 \\
0 & 12-3h & 2 \\
\end{bmatrix}
$$
$R_2/(12-3h)$
$$
\begin{bmatrix}
3 & 6 & 7 \\
0 & 1 & 2/(12-3h) \\
\end{bmatrix}
$$
$R_1$ replaced by $(-6)R_2 + R_1$
$$
\begin{bmatrix}
3 & 0 & (4/h+4)+7 \\
0 & 1 & 2/(12-3h) \\
\end{bmatrix}
$$
$R_1/3$
$$
\begin{bmatrix}
1 & 0 & (4/(3(h+4))+7/3 \\
0 & 1 & 2/(12-3h) \\
\end{bmatrix}
$$
I feel like I am doing this right but I followed along on another example on the website and got lost near the end, can someone tell me where I went wrong?
| Another way to do this is to note that your system is $A\vec x=\vec b$ where
\begin{align*}
A&=
\left[\begin{array}{rr}
2 & h \\
3 & 6
\end{array}\right] &
\vec b &=
\begin{bmatrix}
4\\ 7
\end{bmatrix}
\end{align*}
Since $\det A=-3 \, h + 12$ we know that $A\vec x=\vec b$ has a solution if $h\neq 4$.
So, it remains to determine if the system given by
$$
\left[\begin{array}{rr|r}
2 & 4 & 4 \\
3 & 6 & 7
\end{array}\right]
$$
Row reducing gives
$$
\DeclareMathOperator{rref}{rref}\rref
\left[\begin{array}{rr|r}
2 & 4 & 4 \\
3 & 6 & 7
\end{array}\right]=
\left[\begin{array}{rr|r}
1 & 2 & 0 \\
0 & 0 & 1
\end{array}\right]
$$
Thus we see that the system is inconsistent when $h=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1719576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the solution to and limit of $a_{n+1} =\frac{v}{a_n+w} $ with $a_1>0, v > 0, w>0$ Find the solution to
and limit of
$a_{n+1}
=\frac{v}{a_n+w}
$
with
$a_1>0, v > 0, w>0$.
This was inspired by
my answer to
Converging sequence $a_{{n+1}}=6\, \left( a_{{n}}+1 \right) ^{-1}$.
I will give my answer
in two days.
| I was mystified by @Tnilk Imaniq's answer in the linked thread, but then I realized that what he was doing was to set $b_n=1/(a_n+c)$ and then get a linear difference equation for $b_n$. So we do the same:
$$b_n=\frac1{a_n+c};\,\,a_n=\frac1{b_n}-c$$
$$\frac1{b_{n+1}}-c=\frac v{\frac1{b_n}-c+w}=\frac{1-cb_{n+1}}{b_{n+1}}=\frac{vb_n}{1+(w-c)b_n}$$
Clearing denominators,
$$vb_nb_{n+1}=1-cb_{n+1}+(w-c)b_n-c(w-c)b_nb_{n+1}$$
Since all we know is linear difference equations, we demand $c^2-wc-v=0$.
$$c=\frac{w\pm\sqrt{w^2+4v}}2=\frac{w-\sqrt{w^2+4v}}2$$
because we anticipate that $$\lim_{n\rightarrow\infty}b_n=\infty$$
and we know that $a_n>0$, so we must have $c<0$. Then our linear difference equation is
$$cb_{n+1}+(c-w)b_n=\frac{w-\sqrt{w^2+4v}}2b_{n+1}+\frac{-w-\sqrt{w^2+4v}}2b_n=1$$
We seek a particular solution $b_{np}=A$, so
$$-A\sqrt{w^2+4v}=1,\,\,b_{np}=\frac{-1}{\sqrt{w^2+4v}}$$
Now we solve the homogeneous equation
$$\frac{w-\sqrt{w^2+4v}}2b_{n+1,h}+\frac{-w-\sqrt{w^2+4v}}2b_{nh}=0$$
This has a solution of the form $b_{nh}=c_1r^k$ if
$$r=\frac{\frac{w+\sqrt{w^2+4v}}2}{\frac{w-\sqrt{w^2+4v}}2}=\frac{\left(w+\sqrt{w^2+4v}\right)^2}{-4v}$$
so
$$b_n=b_{np}+b_{nh}=\frac{-1}{\sqrt{w^2+4v}}+c_1r^n$$
$$b_1=\frac1{a_1+\frac{w-\sqrt{w^2+4v}}2}=\frac{-1}{\sqrt{w^2+4v}}+c_1r$$
From which we obtain
$$c_1r=\frac{a_1+\frac{w+\sqrt{w^2+4v}}2}{\sqrt{w^2+4v}\left(a_1+\frac{w-\sqrt{w^2+4v}}2\right)}$$
So our solution is
$$a_n=\frac{\sqrt{w^2+4v}}{-1+\frac{a_1+\frac{w+\sqrt{w^2+4v}}2}{a_1+\frac{w-\sqrt{w^2+4v}}2}\left(\frac{\left(w+\sqrt{w^2+4v}\right)^2}{-4v}\right)^{n-1}}-\frac{w-\sqrt{w^2+4}}2$$
Since $|r|>1$, it follows that
$$\lim_{n\rightarrow\infty}a_n=\frac{\sqrt{w^2+4v}-w}2$$
I see that the OP has posted an answer while I was typing this up even though he said he would give us two days to reply. Oh well...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1720787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find the sum of the series: $\frac{1}{1*2} - \frac{1}{3*2^3} + \frac{1}{5*2^5} - \frac{1}{7*2^7}+\dots$? $$\frac{1}{1*2} - \frac{1}{3*2^3} + \frac{1}{5*2^5} - \frac{1}{7*2^7}+\dots$$
I made a series to get $$\sum_{n=0}^\inf \frac{(-1)^n}{(1+2n)*2^{1+2n}}$$ but what series can it manipulate and simplify to?
| The geometric series gives $$\frac{1}{1+x^2} = \sum_{n=0}^{\infty} (-1)^n x^{2n}, \,\,\,\,\,\,\,\, \text{when } \lvert x \rvert < 1.$$ Then $$\frac{1}{1+x^2} - 1 = \sum_{n=1}^{\infty} (-1)^n x^{2n},$$ for such $x$. Integrating from $x=0$ to $x=1/2$ (and integrating term-by-term in the sum, which is justifies by uniform convergence of the sum) gives $$\arctan\left( \tfrac 1 2\right) - \tfrac{1}{2} = \sum^\infty_{n=1} \frac{(-1)^n \left( \tfrac{1}{2}\right)^{2n+1}}{2n+1} = \sum^\infty_{n=1} \frac{(-1)^n}{(2n+1)2^{2n+1}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1722727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Power two of ordinal if $\omega$ is ordinal number of set $\mathbb{N}$ can I compute $(\omega + 1)^2$ like below
$\begin{align*}
(\omega + 1)^2 & = \ (\omega + 1).(\omega + 1)\\
& = \ ((\omega + 1).\omega)+(\omega + 1)\\
& = \ (\omega^2 + 1)+(\omega + 1)\\
& = \ \omega ^ 2 + (1 + \omega) + 1\\
& = \ \omega^2 + \omega + 1
\end{align*}$
is $(\omega+1).\omega = \omega^2 + 1$? how can I compute $(\omega+1).\omega$?
| Sometimes a picture can help, so I’m adding this addendum to Noah’s answer for the possible benefit of future readers. The first diagram shows the Cartesian product $\omega\times(\omega+1)$; $(\omega+1)\cdot\omega$ is the order type of this product under the lexicographic order.
$$\begin{array}{ccc}
\omega&\color{red}\omega&\color{blue}\omega&\color{green}\omega&\ldots\\
\vdots&\vdots&\vdots&\vdots&\\
4&4&4&4&\ldots\\
3&3&3&3&\ldots\\
2&2&2&2&\ldots\\
1&1&1&1&\ldots\\
0&0&0&0&\ldots
\end{array}$$
Now imagine taking each of the points in the top row and moving it to the bottom of the next column to the right; the result is shown below.
$$\begin{array}{ccc}
\vdots&\vdots&\vdots&\vdots&\vdots\\
5&4&4&4&4&\ldots\\
4&3&3&3&3&\ldots\\
3&2&2&2&2&\ldots\\
2&1&1&1&1&\ldots\\
1&0&0&0&0&\ldots\\
0&\omega&\color{red}\omega&\color{blue}\omega&\color{green}\omega&\ldots\\
\end{array}$$
The lexicographic order on this array is clearly isomorphic to that on the first array, but here it clearly has order type $\omega\cdot\omega=\omega^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1725608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Tangent line problems Problem 1
Find common tangent to the curve:
$y+x^2=-4$ and $x^2+y^2=4$.
My idea:
Let $t1... y=ax+b$ is a tangent line to the first curve.
Let $t2... y=cx+d$ is a tangent line to the second curve.
But we are seracing for common tangent so that mean that $a=c$ and $b=d$.
By using implicit differentiation we can find $a$ and $b$.
$$y+x^2=-4$$
$$y'+2x=0$$
$$y'=-2x$$
$$x^2+y^2=4$$
$$2x+2yy'=0$$
$$x+yy'=0$$
$$y'=\frac{-x}{y}$$
$$-2x=\frac{-x}{y} \to y=\frac{1}{2},$$$x$ is any real number
Problem 2
For which value of the coefficients $a$, $b$ and $c$ $\in$ R is the x-coordinate axis tangent to the curve?
$y=ax^2+bx+c$
My idea:
Curve and the tangent line must have just one common point. Our curve is a parabola and the number of dots on x axis is a number of solution of quadratic equation. So we need $a,b,c$ to be equal to $b^2-4ac=0$
Problem 3
Find a line that is tangent to the curve
$y = x^4 - 2x^3 - 3x^2 + 5x + 6$
in at least two points .
My idea:
Let that tangent be $t=ax+b$.
If we that line be a tangent line at two point to the curve equation $y(x)-t(x)=0$ must have at least two (not equal) solution.
$$x^4 - 2x^3 - 3x^2 + 5x + 6-(ax+b)=0$$
$$x^4 - 2x^3 - 3x^2 + 5x + 6=ax+b$$
$$a=5 , b=6$$
$$x^4 - 2x^3 - 3x^2=0$$ $$x_1=0, x_2=-1, x_3=3$$
Problem 4
I am trying to find the number of tangents to a curve that all pass through the origin. The curve's equation is $y=x^3+x^2−22x+20.$ I also need to find the equation of said tangents.
My work:
Let's use formula for tangent line:
$$y-y_0=y'(x_0)(x-x_0)$$
We know that $x=0$ and $y=0$.
Let's find $y'(x_0)$:
$y'(x_0)=3x_0^2+2x_0-22$
Plugin what we know:
$$-y_0=(3x_0^2+2x_0-22)*-x_0$$
We aslo know that $y_0=x_0^3+x_0^2-22x_0+20$
$$-x_0^3-x_0^2+22x_0-20=-3x_0^3-2x_0^2+22x_0$$
$$2x_0^3+x_0^2-20=0$$
Only real solution is 2. $x_0=2$ and $y_0=-12$.
Let tangent line be $y=kx+l$.
$k=y'(2)=-6$
$-12=-6*2+l\to l=0$
But we want l to be zero.
Our solution is $y=-6x$.
| For problem 1, note that the requirement is that two points(not necessarily the same) must share a tangent line(which means both $a=c$ and $b=d$). That they might not be the same point means you cannot assign them the same x and y values. So we have to simultaneously solve:
$$-2x_1=x_2/y_2$$
And:
$$b=d$$
To find expressions for b and d, we can use:
$$y=y_k'(x-x_k)-y_k$$
That is, $b=x_1y_1'-y_1$ and $d=x_2y_2'-y_2$.
You have to simultaneously solve:
$$-2t=-u/\sqrt{4-u^2}$$
And:
$$-t^2+4=-u^2/\sqrt{4-u^2}-\sqrt{4-u^2}$$
Wolfram is giving me a double surd solution to these equations. This seems pretty nasty to solve. Have you got the questions right?
Your answer to problem 2 is correct and well reasoned. =]
For problem 3, you need to find a line $y=ax+b$ such that $P(x)-y(x)=(x-\alpha)^2(x-\beta)^2$
$$x^4-2x^3-3^2+(5-a)x+(6-b)\equiv(x-\alpha)^2(x-\beta)^2$$
Expanding the right hand side:
$$x^4-2x^3-3^2+(5-a)x+(6-b)\equiv x^4-2(\alpha+\beta)x^3+(\alpha^2+\beta^2+4\alpha\beta)x^2-2(\alpha^2\beta+\alpha\beta^2)x+\alpha^2\beta^2$$
We can find $\alpha$ and $\beta$ by solving:
$$-2(\alpha+\beta)=-2$$
And:
$$\alpha^2+\beta^2+4\alpha\beta=-3$$
Giving us $\alpha=1$ and $\beta=-2$.
From there:
$$x^4-2x^3-3x^2+(5-a)x+(6-b)\equiv x^4-2x^3-3x^2+4x+4$$
That is, $5-a=4\to a=1$ and $6-b=4\to x=2$. Our tangent is $y=x+2$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Derivation and tangent problem set Problem 1
On the curve $y=\frac{1}{1+x^2}$ find a point in which tangent line is parallel to the horizontal axis.
My idea:
Let's find $y'$.
$$y'=\frac{-2x}{(1+x^2)^2}$$
If we want a tangent line to be parallel to x-axis it must look like this $t...y=kx+l$ such that $k=0$.
Let's find for which $x$ we have that situation.
$$y'=\frac{-2x}{(1+x^2)^2}=0$$
$$x=0\to y=1$$
$$(x,y)=(0,1)$$
Problem 2
Determine the angle of the curve
$y^2 - 3x^2 + x + 1 = 0$ and
$xy^2 - 1 = 0$
intersect in the first quadrant
My work:
$y^2 - 3x^2 + x + 1 =xy^2 - 1$
One solution for $x>0$ and $y>0$ is $x=y=1$.
Let's now find derivative of easch function at that point.
$$2xyy'=-y^2$$
$$y'=\frac{-y^2}{2xy}=\frac{-1}{2}$$
$$2yy'=6x-1$$
$$y'=\frac{6x-1}{2y}=\frac{5}{2}$$
And the angle is $\alpha =arctg|\frac{k_1-k_2}{1+k_1*k_2}|$ where $k_1=\frac{-1}{2}$ and $k_2=\frac{5}{2}$
Problem 3
Find the angle of intersection of the tangents to the curve
$3x^2 - 6x + 2y^2 - 3 = 0$
from the point $T( 1,3 )$?
My work:
First (1,3) isn't on the curve.
We will need that tangent equation:
$$y-y_0=y'(x_0)(x-x_0)$$
From the task we know that $x=1$ and $y=3$.
Let's try to find $x_0$ and $y_0$.
Using implict differentation we will find $y'$.
$$6x-6+4yy'=0$$
$$3x-3+2yy'=0$$
$$2yy'=-3x+3$$
$$y'=\frac{-3x+3}{2y}$$
Plugin that in equation:
$$3-y_0=\frac{-3x_0+3}{2y_0}(1-x_0)$$
$$3-y_0=\frac{-3x_0+3}{2y_0}+\frac{-3x_0^2-3x_0}{2y_0}$$
$$3-y_0=\frac{-3x_0^2+3-6x_0}{2y_0}$$
$$6y_0-2y_0^2=-3x_0^2+3-6x_0$$
But I can't get something nice...Is my problem 1 and 2 ok and please help me with 3
| Problem 1 is OK.
Problem 2 ;
First try to find intersection point $(x,y)$
$$ \frac1x - 3 x^2 + x +1 =0 $$
Factorize it
$$ -(x-1) (3 x^2 2 x +1) =0 $$
At $x=1, y= \dfrac{1}{\sqrt x } =1.\; i.e., (1,1) $ is the point of intersection. Other roots are neither in first quadrant nor real.
Now for the slope: $ y^2 = \frac 1x \rightarrow y'= \dfrac{-1}{ x^2 y} \rightarrow -1. $
Problem 3:
not answered.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1726094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finding a formula for a linear transformation Let $T: M_{22} \rightarrow\mathbb{R}^2$ be a linear transformation such that:
$$T\left(\begin{bmatrix} 1& 2\\3&4\end{bmatrix}\right) \rightarrow(5,5)$$
$$T\left(\begin{bmatrix} 1& 0\\0&1\end{bmatrix}\right) \rightarrow(2,0)$$
$$T\left(\begin{bmatrix} 0& 1\\1&0\end{bmatrix}\right) \rightarrow(0,2)$$
$$T\left(\begin{bmatrix} 4& 3\\2&2\end{bmatrix}\right) \rightarrow(6,5)$$
Give a formula for $T$ what is the kernel of $T$?
To be honest, I'm not completely sure where to start this one. Any hints to send me in the right direction would be vastly helpful. Thank you.
| You might find some clever trick as suggested in comments, but there is a standard way for describing the kernel.
Set
$$
E_1=\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix},\quad
E_2=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},\quad
E_3=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix},\quad
E_4=\begin{bmatrix} 4 & 3 \\ 2 & 2 \end{bmatrix},\quad
$$
You should have already proved that $\mathscr{B}=\{E_1,E_2,E_3,E_4\}$ is a basis for $M_{22}$. The matrix of the map $T$ with respect to this basis and the standard basis on $\mathbb{R}^2$ is
$$
A=\begin{bmatrix}
5 & 2 & 0 & 6 \\
5 & 0 & 2 & 5
\end{bmatrix}
$$
and an easy reduction to echelon form gives
$$
\begin{bmatrix}
1 & 0 & 2/5 & 1 \\
0 & 1 & -1 & 1/2
\end{bmatrix}
$$
A basis for the null space of $A$ is thus
$$
\left\{
v_1=\begin{bmatrix}-2\\5\\5\\0\end{bmatrix},
v_2=\begin{bmatrix}-2\\1\\0\\2\end{bmatrix}
\right\}
$$
This means that the kernel of $T$ is generated by
$$
-2E_1+5E_2+5E_3=
-2\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}
+5\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
+5\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}
=\begin{bmatrix}3 & 1 \\ -1 & -3\end{bmatrix}
$$
and
$$
-2E_1+E_2+2E_4=
-2\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}
+\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
+2\begin{bmatrix} 4 & 3 \\ 2 & 2 \end{bmatrix}
=\begin{bmatrix} 7 & 2 \\ -2 & -3\end{bmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1726294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to find if a sequence is converging or diverging? The question is as follows: Determine if the sequence:
$$\{1/1,\ 1/3,\ 1/2,\ 1/4,\ 1/3,\ 1/5,\ 1/4,\ 1/6,\ \ldots\}$$
diverges or converges. If it is convergent, find the limit.
I am not sure where to begin, I'd appreciate it if anybody could help me on this one.
Thankyou
| The approach of Bérénice is nice and looks quite reasonable, since the first step is to identify the elements of the sequence
\begin{align*}
\begin{array}{ccccccccc}
a_0&\ a_1&\ a_2&\ a_3&\ a_4&\ a_5&\ a_6&\ a_7&\ \cdots\\
\color{blue}{1}&\ \frac{1}{3}&\ \color{blue}{\frac{1}{2}}&\ \frac{1}{4}&\ \color{blue}{\frac{1}{3}}&\ \frac{1}{5}&\ \color{blue}{\frac{1}{4}}&\ \frac{1}{6}&\ \cdots\\
\\
\end{array}
\end{align*}
which typically results after some analysis in a separation of even and odd elements:
\begin{align*}
a_{2n}=\frac{1}{n+1}\qquad\qquad a_{2n+1}=\frac{1}{n+3}\qquad\qquad n\geq 0\tag{1}
\end{align*}
With (1) in mind we can continue along different lines. One of them is to see the elements are positive and essentially growing as $\frac{1}{n}$, So, we could think of squeezing the sequence between the constant sequence $(0)_{n\geq 0}$ and a sequence related with $\left(\frac{1}{n+1}\right)_{n\geq 0}$ where we use $n+1$ in the denominator to avoid division by zero when $n=0$.
Starting from (1) we obtain
\begin{align*}
\color{blue}{a_n}&=\frac{1}{\frac{n}{2}+1}\left(\frac{1+(-1)^n}{2}\right)+\frac{1}{\frac{n-1}{2}+3}\left(\frac{1-(-1)^n}{2}\right)\\
&=\frac{1}{n+2}\left(1+(-1)^n\right)+\frac{1}{n+5}\left(1-(-1)^n\right)\\
&\leq \frac{2}{n+2}+\frac{2}{n+5}\\
&\,\,\color{blue}{\leq \frac{4}{n+2}}
\end{align*}
We derive this way an upper bound $\frac{4}{n+2}$ for $a_n$ and can conclude from
\begin{align*}
0\leq a_n\leq \frac{4}{n+2}\quad\longrightarrow\quad 0
\end{align*}
the sequence $(a_n)_{n\geq 0}$ is a zero-sequence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1726398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
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Finding the Cartesian equation of an ellipse (Midpoints)
Question: The normal to the ellipse $ \frac{x^2}{25} + \frac{y^2}{9} = 1$ at a point $Q$ meets the coordinate axes at A and B respectively.
As $Q$ varies, the locus of the midpoint of $AB$ is another ellipse.
Find the Cartesian equation of this ellipse.
What I have done
Let $x = 5 \cos(\theta)$ and $y=3\sin(\theta)$ then
$$ \frac{dx}{d\theta} = -5 \sin(\theta) , \frac{dy}{d\theta}=3cos(\theta)$$
$$ \frac{dy}{dx} = \frac{dy}{d\theta} *\frac{d\theta}{dx} $$
$$ \frac{dy}{dx} = \frac{-3\cos(\theta)}{5 \sin(\theta)}$$
$$ m_{normal} \cdot m_{tgt} = -1 $$
$$ m_{normal} = \frac{5\sin(\theta)}{3\cos(\theta)}$$
For equation of normal
$$ y-y_1 = m(x-x_1) $$
$$ y - 3\sin(\theta)= \frac{5\sin(\theta)}{3\cos(\theta)}(x-5\cos(\theta))$$
$$ y - 3\sin(\theta) = \frac{5\sin(\theta)}{3\cos(\theta)}x - \frac{25}{3}\sin(\theta)$$
$$ y= \frac{5\sin(\theta)}{3\cos(\theta)}x - \frac{16}{3} \sin(\theta) $$
At
$$ y=0, x= \frac{16 \cos(\theta)}{5} $$
and
$$ x = 0 , y = - \frac{16}{3} \sin(\theta)$$
So we have 2 points which lie on the new ellipse
Hence the midpoint is given by $$\frac{1}{2} (X_1+X_2) $$
so
$$ \frac{1}{2}(0 + \frac{16 \cos(\theta)}{5}) $$
Midpoint is
$$ \frac{8\cos(\theta)}{5} $$
Now I am stuck , how should I continue?
| Equation of normal at $(a\cos t,b\sin t)$ is:
$$ax\sec t-by\csc t=a^2-b^2$$
The $x,y$-intercepts are $\frac{a^2-b^2}{a} \cos t$ and $\frac{a^2-b^2}{b} \sin t$ respectively.
Hence the locus is
$$\left( \frac{a^2-b^2}{2a} \cos t,\frac{a^2-b^2}{2b} \sin t \right)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $N$ with $N^2=10^4M+N$ I'm a high school student in France and I participated in a math olympiad, and there was a question which I found impossible to solve. Maybe they are here some people who can help me:
We take a number $N$ which can be written by numbers $a,b,c$ and $d$ in
the same order.
so $N=a(10^3)+b(10^2)+c10+d$. And $M\in \mathbb{Z}$ with
$N^2=M(10^4)+N$
Find $M$.
| $$10^4M=N^2-N=N(N-1)$$
Observe that $10^4$ divides $N(N-1)$ and $10^4>N\ge10^3$ as $N$ only has four digits. As their difference is 1, $N$ and $N-1$ are relatively prime to each other, meaning they don't have common divisors except $1$.
There are four cases
i) $10^4|N$, this cannot happen as $N$ is less than $10^4$.
ii) $2^4|N$ and $5^4|N-1$. Then, $N=k\cdot 5^4+1$ for some $k$. As $N$
has four digits, $2\le k\le 15$. $$5^4\equiv 1 \pmod{16}$$
So, $2^4=16|N$ implies $k=15, N=9376,M=8790$
iii) $5^4|N$ and $2^4|N-1$ Then, $N=k\cdot 5^4$ for some $k$. As $N$
has four digits, $2\le k\le 15$. $$5^4\equiv 1 \pmod{16}$$
So, $2^4=16|N-1$ implies $16|k-1$. However, no integer between $2$ and $15$ is equivelant to $1$ modulo $16$. Contradiction
iv) $10^4|N-1$, this cannot happen as $N-1$ is less than $10^4$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Prove that $n^4-n^2$ is divisible by $8$ if $n$ is an odd positive integer.
Prove that $n^4-n^2$ is divisible by $8$ if $n$ is an odd positive integer.
I'm supposed to use proof by induction, but I failed at it miserably. So far I have this:
$$(n^4) - (n^2) = (n^2)((n^2)-1) = n(n-1)n(n+1)$$
Let $n = 2k + 1$ because its always odd. Then
$$\begin{align}
&(2k+1)(2k)(2k+1)(2k+2)
\\\implies\quad &(2k+1)(2k)(2k+1)2(k+1)
\\\implies\quad &(2k+1)2(k)(2k+1)2(k+1)
\\\implies\quad &4(2k+1)(k)(2k+1)(k+1)
\\\implies\quad &4((2k+1)^2)(k)(k+1)
\\\implies\quad &4((2k+1)^2)((k^2)+k)
\end{align}$$
Now $(2k+1)^2$ is odd and $k^2 + k$ is always even because
$$\begin{cases}
\mathrm{odd}^2 + \mathrm{odd} = \mathrm{even} \\
\mathrm{even}^2 + \mathrm{even} = \mathrm{even}
\end{cases}$$
So that means I can take out a $2$ and write $((k^2)+k) = 2X$.
$$\begin{align}
&4(2X)((2k+1)^2)
\\\implies\quad &8X((2k+1)^2)
\end{align}$$
and because there is an $8$ I can say its always divisible by $8$, no matter what odd integer $k$ is? How can this be done using proof by induction?
| You are almost done in your first step. You showed
$$
(n^4)−(n^2)=(n^2)((n^2)−1)=n(n−1)n(n+1).
$$
Now $n$ is odd hence $n-1$ and $n+1$ are even, since every second even number is divisible by $4$ either $4k_1 =n-1$ and $2k_2=n+1$ or $4k_1=n+1$ and $2k_2=n-1$ for some $k_1,k_2\in\mathbb{N}$, this is
$$
(n^4)−(n^2) = n^2(4k_1)(2k_2).
$$
But if you want to prove it by induction you can try this: Let $n$ be odd such that $8|(n^4)−(n^2)$, we will show it for $n+2$.
First note that $8|(n^4)−(n^2)=n^2(n−1)(n+1)$ implies that $8|=(n−1)(n+1)$, hence $\exists k\in\mathbb{N}$ such that $8k=(n-1)(n+1)$, and compute for $n+2$
$$
((n+1)^4)−((n+2)^2) =(n+2)^2(n+1)(n+3)= (n+2)^2(n+1)((n-1)+4)=(n+2)^2(8k+(n+1)4)
$$
since $n$ is odd $n+1$ is even and hence $8|4(n+1)$ and we are done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Awkward terms in Differential equation Find the solution of the equation:
$(x^2+y^2+x)d x-(2x^2+2y^2-y)dy=0$
As a standard approach, I brought $dy/dx$ to the RHS of the equation. Then, I added and subtracted $x^2+y^2-y$ to the numerator of the LHS to find some symmetric relation between the numerator and denominator. Perfecting squares of the numerator or the denominator does not help either. Please advice.
| The equation can be written as
$$xdx +ydy + (x^2+y^2)(dx-2dy) = 0$$
Away from $(x,y) = (0,0)$ we can divide by $x^2+y^2$ to get
$$\frac{x}{x^2+y^2} dx + \frac{y}{x^2+y^2} dy + (dx-2dy) = 0,$$
which is the same as
$$d(\frac{1}{2}\ln({x^2+y^2)} + x-2y) = 0. $$
That leads to $\frac{1}{2}\ln{(x^2+y^2)} +x-2y = k$, as in @almagest's comment.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $f(x)=Ax^2+Bx+C$ and $2A,A+B,C$ are integers, prove that $f(x)$ is integer whenever $x$ is an integer
If $f(x)=Ax^2+Bx+C$ and $2A,A+B,C$ are integers, prove that $f(x)$ is integer whenever $x$ is an integer.
I found a way to prove the reverse statement. That is, I can prove that if $f(x)$ is integer whenever $x$ is integer, $2A,A+B,C$ are integers, by finding $f(0),f(1)$ and $f(-1)$.
How can I prove the actual statement?
| Using repeated differences on the values of $f(n)$ for $n=0,1,2,$ we get
$$
\begin{array}{lll}
C & A+B+C & 4A+2B+C \\
A+B & 3A+B \\
2A \\
\end{array}
$$
The last line repeats forever because $f$ is a polynomial of degree $2$.
Newton's interpolation formula (or direct verification) then gives
$$
Ax^2+Bx+C = C \binom{n}{0} + (A+B) \binom{n}{1} + 2A \binom{n}{2}
$$
from which the claims follows at once.
This solution tells us where the hypotheses come from.
| {
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"timestamp": "2023-03-29T00:00:00",
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Fewest steps to show $\sin^2(s) = \frac{1}{2} - \frac{1}{2} \cos(2s)$? Is there a method for getting from this statement
$
\sin^2(s)
$
to this statement
$
\frac{1}{2}\ -\ \frac{1}{2}\cos(2s)
$
in fewer steps than the following?
$\begin{align}
\cos(2s) &= \cos^2(s) - \sin^2(s) \\
\frac{1}{2} \cos(2s) &= \frac{1}{2} \cos^2(s) - \frac{1}{2} \sin^2(s) \\
-\frac{1}{2} \cos(2s) &= -\frac{1}{2} \cos^2(s) + \frac{1}{2} \sin^2(s) \\
\frac{1}{2} - \frac{1}{2} \cos(2s) &= \frac{1}{2} - \frac{1}{2} \cos^2(s) + \frac{1}{2} \sin^2(s) \\
\frac{1}{2} - \frac{1}{2} \cos(2s) &= \frac{1}{2} ( 1 - \cos^2(s) ) + \frac{1}{2} \sin^2(s) \\
\frac{1}{2} - \frac{1}{2} \cos(2s) &= \frac{1}{2} ( \sin^2(s) ) + \frac{1}{2} \sin^2(s) \\
\frac{1}{2} - \frac{1}{2} \cos(2s) &= \sin^2(s) \\
\end{align}$
| Write $$1=\sin^2(x)+\cos^2(x)$$
and $$\cos(2x) = \cos^2(x)-\sin^2(x).$$ Subtract these equalities $$1-\cos(2x) = 2\sin^2(x)$$ and divide by two $$\frac{1}{2}-\frac{1}{2}\cos(2x) = \sin^2(x).$$ Four lines, is it short enough ?
| {
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conjectured general continued fraction for the quotient of gamma functions Given complex numbers $a=x+iy$, $b=m+in$ and a gamma function $\Gamma(z)$ with $x\gt0$ and $m\gt0$, it is conjectured that the following general continued fraction which is symmetric on $a$ and $b$ is true
$$\frac{\displaystyle\Gamma\left(\frac{a+3b}{4(a+b)}\right)\Gamma\left(\frac{3a+b}{4(a+b)}\right)}{\displaystyle\Gamma\left(\frac{3a+5b}{4(a+b)}\right)\Gamma\left(\frac{5a+3b}{4(a+b)}\right)}=\cfrac{4(a+b)}{a+b+\cfrac{(2a)(2b)} {3(a+b)+\cfrac{(3a+b)(a+3b)}{5(a+b)+\cfrac{(4a+2b)(2a+4b)}{7(a+b)+\ddots}}}}\tag{1}$$
And can be further generalised to
$$\frac{1}{\sqrt{c}}\tan\left(\frac{b-a}{b+a}\tan^{-1}\left(\frac{\sqrt{c}}{d}\right)\right)=\cfrac{(b-a)}{d(a+b)+\cfrac{c(2a)(2b)} {3d(a+b)+\cfrac{c(3a+b)(a+3b)}{5d(a+b)+\cfrac{c(4a+2b)(2a+4b)}{7d(a+b)+\ddots}}}}\tag{2}$$
Corollaries:
(i) Specializing to $a=1/2$ and $b=2z/n+1/2$,we obtain
the continued fraction for $\tan\left(\frac{z\pi}{4z+2n}\right)$ found in this post after applying the functional equation of the gamma function
(ii) and specializing further to $a=-1/2$ and $b=2z/n+3/2$,we obtain immediately the continued fraction for $\cot\left(\frac{z\pi}{4z+2n}\right)$ found here after applying the functional equation of the gamma function.This continued fraction was proved by @GEdgar.
(iii) letting $2a=m-n$ and $2b=m+n$,we find $$\displaystyle\tan\left(\frac{\pi n}{4m}\right)=\cfrac{n}{m+\cfrac{m^2-n^2} {3m+\cfrac{4m^2-n^2}{5m+\cfrac{9m^2-n^2}{7m+\cfrac{16m^2-n^2}{9m+\ddots}}}}}$$
From which we obtain its hyperbolic companion $$\displaystyle\tanh\left(\frac{\pi n}{4m}\right)=\cfrac{n}{m+\cfrac{m^2+n^2} {3m+\cfrac{4m^2+n^2}{5m+\cfrac{9m^2+n^2}{7m+\cfrac{16m^2+n^2}{9m+\ddots}}}}}$$
(iv)and if $2a=-n$ and $2b=2m+n$,then it follows that
$$\displaystyle\tan\left(\frac{\pi(m+n)}{4m}\right)=\frac{1+\tan\Big(\frac{\pi n}{4m}\Big)}{1-\tan\Big(\frac{\pi n}{4m}\Big)}=\cfrac{m+n}{m+\cfrac{(-n)(2m+n)} {3m+\cfrac{(m-n)(3m+n)}{5m+\cfrac{(2m-n)(4m+n)}{7m+\cfrac{(3m-n)(5m+n)}{9m+\ddots}}}}}$$
Q: Is the conjectured general continued fraction true (for all complex numbers $a$,$b$ with $x\gt0$ and $m\gt0$)?
| If we consider the following symmetric q-continued fraction,where $q=ab$,$|q|\lt1$ see here and here
$$\cfrac{1}{1-ab+\cfrac{(ab)(1-b^2)(1-a^2)}{1-(ab)^3+\cfrac{(ab)^2(1-ab^3)(1-a^3b)}{1-(ab)^5+\cfrac{(ab)^3(1-a^2b^4)(1-a^4b^2)}{1-(ab)^7+\cfrac{(ab)^4(1-a^3b^5)(1-a^5b^3)}{1-(ab)^9+\ddots }}}}}=\frac{(a^5b^3;(ab)^4)_\infty\,(a^3b^5;(ab)^4)_\infty}
{(ab^3;(ab)^4)_\infty\,(a^3b;(ab)^4)_\infty}$$
and let $a\rightarrow q^a$ ,$b\rightarrow q^b$ then we have
$$\cfrac{1}{1-q^{(a+b)}+\cfrac{q^{(a+b)}(1-q^{2b})(1-q^{2a})}{1-q^{3(a+b)}+\cfrac{q^{2(a+b)}(1-q^{(a+3b)})(1-q^{(b+3a)})}{1-q^{5(a+b)}+\cfrac{q^{3(a+b)}(1-q^{(2a+4b)})(1-q^{(2b+4a)})}{1-q^{7(a+b)}+\cfrac{q^{4(a+b)}(1-q^{(3a+5b)})(1-q^{(5a+3b)})}{1-q^{9(a+b)}+\ddots }}}}}=\frac{(q^{(5a+3b)};q^{4(a+b)})_\infty\,(q^{3a+5b};q^{4(a+b)})_\infty}
{(q^{(a+3b)};q^{4(a+b)})_\infty\,(q^{(3a+b)};q^{4(a+b)})_\infty}$$
now after multiplying both sides by $1-q$ ,and performing equivalence transformation
$$\cfrac{1}{\frac{1-q^{(a+b)}}{1-q}+\cfrac{q^{(a+b)}\Big(\frac{1-q^{2b}}{1-q}\Big)\Big(\frac{1-q^{2a}}{1-q}\Big)}{\frac{1-q^{3(a+b)}}{1-q}+\cfrac{q^{2(a+b)}\Big(\frac{1-q^{(a+3b)}}{1-q}\Big)\Big(\frac{1-q^{(b+3a)}}{1-q}\Big)}{\frac{1-q^{5(a+b)}}{1-q}+\cfrac{q^{3(a+b)}\Big(\frac{1-q^{(2a+4b)}}{1-q}\Big)\Big(\frac{1-q^{(2b+4a)}}{1-q}\Big)}{\frac{1-q^{7(a+b)}}{1-q}+\cfrac{q^{4(a+b)}\Big(\frac{1-q^{(3a+5b)}}{1-q}\Big)\Big(\frac{1-q^{(5a+3b)}}{1-q}\Big)}{\frac{1-q^{9(a+b)}}{1-q}+\ddots }}}}}=(1-q)\frac{(q^{(5a+3b)};q^{4(a+b)})_\infty\,(q^{3a+5b};q^{4(a+b)})_\infty}
{(q^{(a+3b)};q^{4(a+b)})_\infty\,(q^{(3a+b)};q^{4(a+b)})_\infty}$$
then apply the equality $\lim_{q\rightarrow 1}\frac{1-q^x}{1-q}=x$,see q-analog and the q-gamma function $\Gamma_q(x)=(1-q)^{1-x}\frac{(q;q)_{\infty}}{(q^x;q)_{\infty}}$
which concludes the proof of continued fraction (1)
| {
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Maclaurin Expansion for $e^{e^{z}}$ at $z=0$ I need to find terms up to degree $5$ of $e^{e^{z}}$ at $z=0$.
I tried letting $\omega = e^{z} \approx 1+z+\frac{z^{2}}{2!}+\frac{z^{3}}{3!}+\cdots$, and then substituting these first few terms into the Taylor series expansion for $e^{z}$ as follows:
$e^{\omega} = \sum_{n=0}^{\infty}\frac{\omega^{n}}{n!} = 1 + \omega + \frac{\omega^{2}}{2!}+\frac{\omega^{3}}{3!}+\cdots = 1 + (1+z+\frac{z^{2}}{2!}+\frac{z^{3}}{3!}+\frac{z^{4}}{4!}+\cdots) + \frac{1}{2}\left(1+z+\frac{z^{2}}{2!}+\frac{z^{3}}{3!}+\frac{z^{4}}{4!}+\cdots \right)^{2}+ \frac{1}{3!}\left( 1+z+\frac{z^{2}}{2!}+\frac{z^{3}}{3!}+\frac{z^{4}}{4!}+\cdots\right)^{3}+\cdots$
But, then when I try to multiply this all out (using Maple), I find that going even further out than I have here, the terms of degree $\leq 5$ do not terminate. So, then I tried putting the command into Maple, and it came out to be something long and horrible with $e$ coefficients for each term.
I must be missing something here.
Could somebody please tell me how to finish this problem? I would ask for a complete solution, not hints or leading questions. I've been working on this for hours and am beyond frustrated. The most edifying thing for me at this point would be to see the correct, full way it's supposed to be done, work backwards, take it apart, and then be able to apply it to other situations.
Thank you.
| If you omit all terms with $x^6$ or higher you can compute
$$e^{e^x}= e^{1+x+x^2/2+x^3/6+x^4/24+x^5/120}$$
$$=e^{1}e^{x} e^{x^2/2} e^{x^3/6} e^{x^4/24}e ^{x^5/120}$$
$$=e(1+x+\tfrac{1}{2}x^2+\tfrac{1}{6}x^3+\tfrac{1}{24}x^4+\tfrac{1}{120}x^5) (1+\tfrac{1}{2}x^2+\tfrac{1}{8}x^4) (1+\tfrac{1}{6}x^3) (1+\tfrac{1}{24}x^4) (1+\tfrac{1}{120}x^5)$$
Distribute, omit $x^6\dots$ and get the Maclaurin expansion
$$e^{e^x}=e(1+x+x^2+\tfrac{5}{6}x^3+\tfrac{5}{8}x^4+\tfrac{13}{30}x^5)$$
| {
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Can someone explain how the cartesian equation is formed? Can someone explain how the cartesian equation of $r = 1 - \cos (\theta)$ is $x^4 + y^4 + 2x^2y^2 + 2x^3 + 2xy^2 - y^ 2 = 0$ ?
| It is slightly tricky. Multiply through by $r$ and put $r\cos\theta=x$, that gives $r^2=r-x$ or $x=r-r^2$. Squaring and rearranging gives $2r^3=r^4+y^2$. Squaring again gives $4r^6=r^8+2y^2r^4+y^4$. Substituting $r^2=x^2+y^2$ gives a degree 8 polynomial which factors: $(x^4+y^4+2x^2y^2+2x^3+2xy^2-y^2)(x^4+y^4+2x^2y^2-2x^3-2xy^2-y^2)=0$.
So the first question is whether the second factor is valid or whether it is an artefact of all that squaring. Try putting $y=0,x=2$. That fails to satisfy the first factor, but does satisfy the second. It also fails to satisfy the original equation (whereas $y=0,x=-2$ satisfies the first factor and the original equation but not the second factor).
Since the original equation obviously generates a continuous curve, we conclude that we have $x^4+y^4+2x^2y^2+2x^3+2xy^2-y^2=0$ as required.
| {
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Complete formalization of solutions to $a^2+b^2=c^2+k$ for fixed $k>0$ Is there a known complete formalization of solutions to $a^2 + b^2 = c^2 + k$ for a fixed constant $k>0$ similar to the one for primitive Pythagorean triples (i.e. $(a,b,c) = (m^2-n^2,2mn,m^2+n^2)$ )?
I saw on-line somewhere that Euler found all solutions to $a^2+b^2=c^2+d^2$ (I think it was $(a,b,c,d) = (xy-wz,xz+wy,xy+wz,xz-wy)$) so if $k$ is a square number $k=r^2$ we can find $x,y,w,z$ such that $xz-wy = r$. However, I was wondering if there is some known formulation having only two parameters (like we have for Pythagorean triples) given that $k$ is fixed.
| I do not know of a solution to your first part unless $k-0$ if you want $k$ to be the same across the board. I also don't know about Euler finding all solutions to $a^2+b^2=c^2+d^2$
because there are an infinit number of them found most easily with deriviatives of Euclid's formula shown here as
$$A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2\quad$$
By solving the C-function for k, we can find all triples with a given C by testing m-values to see which yield k-integers.
\begin{equation}
C=m^2+k^2\implies k=\sqrt{C-m^2}\qquad\text{for}\qquad \bigg\lfloor\frac{ 1+\sqrt{2C-1}}{2}\bigg\rfloor \le m \le \lfloor\sqrt{C-1}\rfloor
\end{equation}
The lower limit ensures $m>k$ and the upper limit ensures $k\in\mathbb{N}$.
$$C=65\implies \bigg\lfloor\frac{ 1+\sqrt{130-1}}{2}\bigg\rfloor=6 \le m \le \lfloor\sqrt{65-1}\rfloor=8\quad\land \quad m\in\{7,8\}\Rightarrow k\in\{4,1\}\\$$
$$F(7,4)=(33,56,65)\qquad \qquad F(8,1)=(63,16,65) $$
Here we have $33^2+56^2=63^2+16^2$ and there are infinite numbers of these that can by found by testing C-candidates of the form where $C=4n+1$.
| {
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False Proof that $\sqrt{4}$ is Irrational Everyone with any basic knowledge of number theory knows the classic proof of the irrationality of $\sqrt{2}$. Curious about generalizations using elementary methods, I looked up the irrationality of $\sqrt{3}$, and found the following:
Say $ \sqrt{3} $ is rational. Then $\sqrt{3}$ can be represented as $\frac{a}{b}$, where a and b have no common factors.
So $3 = \frac{a^2}{b^2}$ and $3b^2 = a^2$. Now $a^2$ must be divisible by $3$, but then so must $a $ (fundamental theorem of arithmetic). So we have $3b^2 = (3k)^2$ and $3b^2 = 9k^2$ or even $b^2 = 3k^2 $ and now we have a contradiction.
Such a proof follows the same basic logic as the proof for $\sqrt{2}$, except for using the fundamental theorem of arithmetic to replace and generalize the trivial fact that $n$ is even if $n^2$ is even. However, when I apply this proof format to $\sqrt{4} $ (which is clearly an integer and thus rational) I get the following:
Say $ \sqrt{4} $ is rational. Then $\sqrt{4}$ can be represented as $\frac{a}{b}$, where a and b have no common factors.
So $4 = \frac{a^2}{b^2}$ and $4b^2 = a^2$. Now $a^2$ must be divisible by $4$, but then so must $a $ (fundamental theorem of arithmetic). So we have $4b^2 = (4k)^2$ and $4b^2 = 16k^2$ or even $b^2 = 4k^2 $, which implies that $b=4n$ by the fundamental theorem. Now we have a contradiction (since can note that both $a$ and $b$ are divisible by $4$ and we assumed they were coprime)
This proof is clearly false, yet I fail to see where it differs. Where does it do so?
| Your error is stating that if $a^2$ is divisible by 4 so must $a$ be. The fundimental theorem states if a prime $p $ divides $ab $ then $p $ must divide $a $ or $p$ must divide $b $. That is true because $p $ is indivisable.
But if $p $ is composite it doesn't hold. $p$ could equal $jk $ and $j$ could divide $a $ and $k $ divide $b $. Example: 3 divides 4 times 9 so 3 either divides 4 or 3 divides 9 because 3 is prime. But 6 divides 4 times 9 but 6 neither divides 4 nor 9 but instead 3 divides 9 while 2 divides 4 so 6=2 times 3 divide 4 times 9.
So 4 divides $a^2$ means $2*2$ divides $a*a$ so 2 divides $a $ is all you can conclude with certainty. ... because 4 is not prime.
Actually because 4 is not square free. All of its prime factors must divide into $a$ but the square powers can be distributed among (and are) distributed among the square powers of $a$.
| {
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Solving $x^2+\frac{81x^2}{(9+x)^2}=40$
Solve the following equation:
$$x^2+\dfrac{81x^2}{(9+x)^2}=40$$
Unfortunately I have no ideas because after expanding I get an equation of 4 degree.
| Here is another way to solve it.
Fully expanding it gives:
$$x^2(9+x)^2 + 81x^2 = 40(9+x)^2$$
$$(x^2-40)(9+x)^2 + 81x^2 = 0$$
$$(x^2-40)(81+18x+x^2)+81x^2 = 0$$
$$x^4+18x^3+122x^2-720x-3240=0$$
Important step:
$$(x^2-2x-18)(x^2+20x+180)=0$$
The last step is tricky but if you assume that a factorization of the form exists:
$$(ax^2 + bx + c)(dx^2 + ex + f) = 0$$
then mapping the coeffeicients to the corresponding powers of x gives:
$$ (ad)x^4 + (ae + db)x^3 + (af + dc + be)x^2 + (bf + ec)x + fc = 0$$
You have the following equations to solve:
$$ (a*d) = 1 $$
a=d=1 (or -1) for simplicity
So we have 4 equations and 4 unknowns
$$ (e+b) = 18 $$
$$ (f + c + be) = 122 $$
$$ (b*f + e*c) = -720$$
$$ (f*c) = -3240 $$
These are tough and long to solve by hand but eventually you get:
$$a = 1, b = -2, c = -18, d = 1, e = 20, f = 180$$
| {
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Prove this inequality: $\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}+3$
Let $a$,$b$,$c$ be positive real numbers such that $abc=1$. Prove that
$$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}+3$$
I tried various methods. But, couldn't solve it. It'd be great if anyone can help.
| By AM-GM
$$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}
\ge \frac{2\sqrt{bc}}{\sqrt{a}}+\frac{\sqrt{ca}}{\sqrt{b}}+\frac{2\sqrt{ab}}{\sqrt{c}}
= \frac2a+\frac2b+\frac2c$$
For the last equality $abc=1$ was used.
Then again by AM-GM;
$$\frac1a+\frac1b+\frac1c \ge \frac3{\sqrt[3]{abc}}=3$$
and again by a well known inequality
$$\frac1a+\frac1b+\frac1c \ge \frac1{\sqrt{bc}}+\frac1{\sqrt{ca}}+\frac1{\sqrt{ab}}=\sqrt{a}+\sqrt{b}+\sqrt{c}$$
For the last equality $abc=1$ was used. Combine the last two inequalities to get the desired result.
| {
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Is this inequality true for all k ? $\sum_{n=k}^{n=+\infty} \frac{1}{n^4} \leq (\sum_{n=k}^{n=+\infty} \frac{1}{n^2})^3$ Can it be generalized for other powers ? Wolfram seems to say it is true for k below 20000.
I stumbled upon it randomly when trying to approximate $\sum_{n=1}^{n=+\infty} \frac{1}{n^4}$.
My reasoning was :
$$\left(\sum_{n=k}^{n=+\infty} \frac{1}{n^2}\right)^2=\sum_{n=k}^{n=+\infty} \frac{1}{n^4} + (\text{double products}) \geq\sum_{n=k}^{n=+\infty} \frac{1}{n^4}$$
So
$$\sum_{n=1}^{n=+\infty} \frac{1}{n^4} \leq \sum_{n=1}^{n=k-1} \frac{1}{n^4}+\left(\sum_{n=k}^{n=+\infty} \frac{1}{n^2}\right)^2 \leq \left(\sum_{n=1}^{n=k-1} \frac{1}{n^4}\right)+\left(\frac{1}{k-\frac{1}{2}}\right)^2$$
where the last inequality comes from An inequality: $1+\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt\frac53$.
Then I noticed that, perhaps, I could raise the last term to the power of 3 instead of just 2, making the inequality stronger.
| For $k > 1$,
$$\sum_{n=k}^\infty \dfrac{1}{n^4} < \int_{k-1}^\infty \dfrac{dx}{x^4} = \dfrac{1}{3(k-1)^3}$$
$$\left(\sum_{n=k}^\infty \dfrac{1}{n^2}\right)^3 > \left(\int_{k}^\infty \dfrac{dx}{x^2}\right)^3 = \dfrac{1}{k^3} $$
$\dfrac{1}{k^3} > \dfrac{1}{3(k-1)^3}$ when $3(k-1)^3 > k^3$, which
is true for $k > 3.2612$.
| {
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Prove that $(n+1)\int_0^1{\frac{x^n}{x+1}dx}\to\frac12$ Let $I_n=\int_0^1{\frac{x^n}{x+1}dx}$ , $n>0$
Show that $\lim_{n->\infty}{(n+1)I_n} = \frac{1}{2}$
All I could do was to show that the $I_n$ is decreasing.
| You may just integrate by parts,
$$
\begin{align}
I_n=\int_0^1\frac{x^n}{x+1}dx&=\left. \frac{x^{n+1}}{(n+1)}\frac{1}{(x+1)}\right|_0^1+\frac{1}{(n+1)}\int_0^1\frac{x^{n+1}}{(x+1)^2}\:dx\\\\
&=\frac1{2(n+1)}+\color{blue}{\frac{1}{(n+1)}\int_0^1\frac{x^{n+1}}{(x+1)^2}\:dx}
\tag1
\end{align}
$$ then observe that
$$
0\leq \color{blue}{\frac{1}{(n+1)}\int_0^1\frac{x^{n+1}}{(x+1)^2}\:dx}\leq \frac{1}{(n+1)}\frac{1}{(0+1)^2}\int_0^1x^{n+1}dx=\frac{1}{(n+1)(n+2)}\tag2
$$
Then using $(1)$ and $(2)$ gives easily
$$ \lim_{n \to +\infty}nI_n=\frac12.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to calculate this integral using dominated convergence theorem?
I want to calculate
$$\lim_{n\to\infty}\int_{n^{-1}}^n \frac{n^2xe^{-n^2x^2}}{1+x^2}\,dx\tag{1}$$
I have used two approach but get different answer.
(1) take $y=n^2x^2$, then
$$(1)=\lim_{n\to\infty}\int_1^{n^4}\frac{e^{-y}}{2(1+\frac{y}{n^2})}\,dy$$
$\frac{e^{-y}}{2(1+\frac{y}{n^2})}1_{[1,n^4]}(y)$ is dominated by $e^{-y}$ which is integrable, then by dominated convergence theorem we get $(1)=(2e)^{-1}$
(2) take $g(n)=\frac{n^2xe^{-n^2x^2}}{1+x^2}$, "differentiate" wrt $n$
$$g'(n)=\frac{2(1-n^2x^2)nxe^{-n^2x^2}}{1+x^2}\tag{3}$$
since $x\in[n^{-1},n]$, then $g'(n)<0$, so for all $x\in[n^{-1},n]$, $g(n)\le g(1)=\frac{xe^{-x^2}}{1+x^2}$ , then
$$\frac{n^2xe^{-n^2x^2}}{1+x^2}1_{[n^{-1},n]}(x)\le\frac{xe^{-x^2}}{1+x^2}\tag{2}$$
which is integrable, then by dominated convergence theorem, we get $(1)=0$
I am confused about what's wrong with my application of dominated convergence theorem, any help will be appreciated!
| The first approach is correct. However, there is a flaw in the second one.
Note that $g'(n)=\frac{2(1-n^2x^2)nxe^{-n^2x^2}}{1+x^2}=0$ when $n=1/x$. And at the maximum, $g(1/x)=\frac{e^{-1}}{x(1+x^2)}$, which is not integrable for $x\in [0,1]$.
The proposed dominating function, $\frac{xe^{-x^2}}{1+x^2}$ does not, in fact, dominate $g(n)=\frac{n^2xe^{-n^2x^2}}{1+x^2}$ for all $x\in [1/n,n]$.
At $x=1/n$, we have $g(n)=ne^{-1}> \frac1n e^{-1/n^2}$ for $n>1$.
One way around this difficulty is to write
$$\begin{align}
\int_{1/n}^n\frac{n^2xe^{-n^2x^2}}{1+x^2}\,dx&=\int_{1/n}^n n^2xe^{-n^2x^2}\,dx-\int_{1/n}^n\frac{n^2x^3e^{-n^2x^2}}{1+x^2}\,dx\\\\
&=\frac12 (e^{-1}-e^{-n^4})-\int_{1/n}^1\frac{n^2x^3e^{-n^2x^2}}{1+x^2}\,dx-\int_{1}^n\frac{n^2x^3e^{-n^2x^2}}{1+x^2}\,dx \tag 1
\end{align}$$
Now, let $h_n(x)=\frac{n^2x^3e^{-n^2x^2}}{1+x^2}$. Then, it is easy to show that $h_n(x)\le \frac{xe^{-1}}{1+x^2}$. So, the DCT guarantees that the first integral on the right-hand side of $(1)$ approaches $0$.
For the second integral, we note that for $x\ge 1$, $h_n(x)\le \frac{x^3e^{-x}}{1+x^2}$. Therefore, the DCT guarantees that the second integral on the right-hand side of $(1)$ also approaches $0$.
Hence, we have
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\int_{1/n}^n\frac{n^2xe^{-n^2x^2}}{1+x^2}\,dx=\frac{e^{-1}}{2}}$$
as expected!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Show that $x^2 + y^2$ and $x^2 - y^2$ cannot both be perfect squares at the same time where $x, y \in \mathbb{Z}^+$.
Show that $x^2 + y^2$ and $x^2 - y^2$ cannot both be perfect squares at the same time where $x, y \in \mathbb{Z}^+$.
I think that $x^2 + 2xy + y^2$ and $x^2 + y^2$ are not consecutive squares since the difference is even. I think it has some relation with other squares like $(x+y)^2$ and $(x-y)^2$.
How should I proceed? I would love some hints.
| Hint:
If$ x^2+y^2=c^2$ is a Pythagorean triple than there are two integers $m,n$ such that:
$$
x=m^2-n^2 \qquad y=2mn \qquad c=m^2+n^2
$$
so:
$$
x^2-y^2=(m^2-n^2)^2-4m^2n^2=m^4+n^4-6 m^2n^2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Prove that $6(\sin^{10}A+\cos^{10}A)-15(\sin^8A+\cos^8A)+10(\sin^6A+\cos^6A)-1=0$ Prove that
$$6(\sin^{10}A+\cos^{10}A)-15(\sin^8A+\cos^8A)+10(\sin^6A+\cos^6A)-1=0$$
Expression can be verified for different values of $A$ such as $\frac\pi4,\frac\pi2$ etc. But to prove it for general value of A?
| Hint: Try to use $\cos^2 A = 1-\sin^2 A $
Writing $s$ for $\sin^2 A$ we have $6(s^5 + (1-s)^5) -15(s^4+ (1-s)^4) + 10(s^3 + (1-s)^3) - 1 = 6s^5 + 6 - 30s + 60s^2 - 60s^3 + 30s^4 - 6s^5 - 15s^4 - 15 + 60s - 90s^2 + 60s^3 - 15s^4 + 10s^3 + 10 - 30s + 30s^2 - 10s^3 - 1 = 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the value of $ab+ 2cb+\sqrt3 ac$? Three positive real numbers $a,b,c$ satisfy the equations $a^2+\sqrt3 ab+b^2=25$, $b^2+c^2=9$ and $a^2+ac+c^2=16$ .Then find the value of $ab+ 2cb+\sqrt3 ac$?
Is there some way to find the desired value without actually finding values of $a,b,c$ or any other smart method to find $a,b,c$
| Here is a geometrical solution:
by rewriting the equations as
$$\left\{
\begin{aligned}
5^2&=a^2+b^2-2ab\cos\frac{5\pi}{6}\\
3^2&=b^2+c^2-2bc\cos\frac{\pi}{2}\\
4^2&=c^2+a^2-2ca\cos\frac{2\pi}{3}
\end{aligned}
\right.
$$
and evaluating
$$4\left(\frac{1}{2}ab\sin\frac{5\pi}{6}+\frac{1}{2}bc\sin\frac{\pi}{2}+\frac{1}{2}ca\sin\frac{2\pi}{3}\right)=4S_{\Delta ABC}=24.$$
Sketch of a existence proof:
Since $3^2+4^2=5^2$, $\Delta ABC$ exists. Consider $\overset{\mmlToken{mo}{⏜}}{HC}$, i.e., the segment that lies between $AB$ and $CA$ of a circle, for which $BC$ is a diameter. For any $P$ on the arc, $\angle BPC=\pi/2$ while $\angle CPA$ increases continuously from $\pi/2$ to $\pi$ (from $H$ to $C$), so there exists $P$ so that $\angle CPA=2\pi/3$.
In general, if such configuration does exist, the point where three segments meet is the intersection of three arcs. Each arc results the given opening angle with respect to each side of the triangle and they sum up to $2\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Evaluation of $ \int_{0}^{\pi}\ln(5-4\cos x)\,dx$
Evaluation of $\displaystyle \int_{0}^{\pi}\ln(5-4\cos x)dx = \int_{0}^{\pi}\ln(5+4\cos x)dx$
$\bf{My\; Try::}$ Let $\displaystyle I(a,b) = \int_{0}^{\pi}\ln(a+b\cos x)dx$
Then $$\frac{d}{db}(a,b) = \frac{d}{db}\left[\int_{0}^{\pi}\ln(a+b\cos x)dx\right]db$$
So $$I'(a,b) = \int_{0}^{\pi}\frac{\cos x}{a+b\cos x}dx = \frac{1}{b}\int_{0}^{\pi}\frac{(a+b\cos x)-a}{a+b\cos x}dx$$
So we get $$I'(a,b) = \frac{\pi}{b}-\frac{a}{b}\int_{0}^{\pi}\frac{1}{a+\cos x}dx$$
Using half angle formula $\displaystyle \tan x = \frac{1-\tan^2 x/2}{1+\tan^2 x/2}$
so we get $$I'(a,b) = \frac{\pi}{b}-\frac{a}{b}\int_{0}^{\pi}\frac{\sec^2 x/2}{(a+b)+(a-b)\tan^2 x/2}dx$$
Now Put $\tan x/2= t\; $ Then $\displaystyle \sec^2 \frac{x}{2}dx = 2dt$
So we get $$I'(a,b) = \frac{\pi}{b}-\int_{0}^{\infty}\frac{1}{(a+b)+(a-b)t^2}=\frac{\pi}{b}-\frac{a\pi}{2b\sqrt{a^2-b^2}}$$
So we get $$I(a,b) = \pi\ln|b|-\frac{\pi a}{2}\left[-\frac{\ln|b^2-a^2|+2\ln |b|}{2a^2}\right]$$
So we put $a = 5$ and $b=4$
We get $$I(5,4) = \pi\cdot \ln (5)-\frac{\pi}{2}\left[\frac{-\ln(9)+2\ln(5)}{2\cdot 5}\right]$$
So We get $$I(5,4) = \left[\frac{18\ln(5)+\ln(9)}{20}\right]\cdot \pi$$
I did not understand where i have done mistake, Help me
Thanks
| I get, assuming there's a $\mathrm{d}t$, $$
\int_0^\infty \frac{1}{(a+b)(a-b)t^2} \,\mathrm{d}t = \frac{\pi}{2 \sqrt{a^2-b^2}} \text{.}
$$
Integrating that with respect to $b$, I get $\displaystyle \frac{-\pi}{2}\arctan \frac{b\sqrt{a^2 - b^2}}{b^2 - a^2} $, which is not equivalent to your expression containing logarithms.
I haven't checked for further errors.
Edit: @user5713492 has (perhaps unintentionally) pointed out another error. Your second $\int_0^\infty$ has no $\frac{a}{b}$ prefactor. Assuming this was intended, I get $$
\frac{\partial I(a,b)}{\partial b} = \frac{\pi}{b} - \frac{a}{b} \int_0^\infty \frac{1}{(a+b)+(a-b)t^2} \,\mathrm{d}t = \frac{\pi}{b} - \frac{a \pi}{2b\sqrt{a^2 - b^2}} \text{,}
$$ which is what you get, so you probably dropped the prefactor.
Integrating that with respect to $b$, I get $$
\frac{\pi}{2}\ln b + \ln(a + \sqrt{a^2 - b^2}) + C(a) \text{,}
$$ where $C$ is an arbitrary function of $a$. This is not particularly close to your expression with logs and I see no path to equivalence, even ignoring the arbitrary function of integration.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1757448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Does an analytical form exist for the following integral I have an integral
$$f(n,a)=\int_0^{2\pi}\mathop{\mathrm{d}x}\frac{\cos(nx)\cos^2x}{1-a\cos^2x},$$
where $n$ is an even integer and $0<a<1$ is a real number. Does an analytical form exist for this integral?
Edit: Here is a partial answer that I discovered:
$$\int_0^{2\pi}\mathop{\mathrm{d}x}\frac{\cos(nx)}{1-a\cos^2x}=-\frac{2\pi}{\sqrt{1-a}}\left(\frac{1-\sqrt{1-a}}{\sqrt{a}}\right)^n,\qquad a<1, n>0.$$
$n$ is even.
| The answer is
$$f(n, a) = \frac{2\pi}{a\sqrt{1-a}}\left(\frac{1-\sqrt{1-a}}{\sqrt{a}}\right)^n \tag{1}$$
for $0 < a < 1$ and $n > 0$ an even integer; we also have $$f(0, a) = \frac{2\pi}{a \sqrt{1-a}} - \frac{2\pi}{a}$$
which (mysteriously) happens to equal what $\sqrt{a}f(1, a)$ would be if we put $n=1$ in formula $(1)$.
First, consider the related integral $$g(n, a) = \int_0^{2\pi}\frac{\cos nx}{1-a\cos^2 x}\,dx$$
which is useful because
$$g(n, a) - a f(n, a) = \int_0^{2\pi} \cos nx\,dx = \begin{cases}2\pi, & n = 0 \\ 0, & n \ne 0.\end{cases} \tag{2}$$
Express $g$ as a contour integral via $z = e^{ix}$,
noting that $\cos nx = \frac{1}{2}(z^n + z^{-n})$ and $dz = iz\,dx$.
After some simplification, we get:
$$g(n,a) = 2i \int_{|z|=1} \frac{z^n + z^{-n}}{az^4 + 2(a-2)z^2 + a} z\,dz. \tag{3}$$
The polynomial $Q_a(z)$ in the denominator has four roots,
but since it's biquadratic they are easy to determine explicitly.
The quadratic formula gives
$$\pm\sqrt{\frac{2}{a}\left(1\pm\sqrt{1-a}\right)-1}$$
and a stroke of luck denesting radicals gives
$$\pm\frac{1\pm\sqrt{1-a}}{\sqrt{a}}.$$
It isn't hard to see (either by elementary means or Rouché's theorem) that these four roots are real and distinct, and precisely two of them lie within the unit disk: $\alpha = (1-\sqrt{1-a})/\sqrt{a}$ and $-\alpha$. Let's call the other pair $\pm\beta$. From the coefficients of $Q_a$ we deduce that $\alpha^2 \beta^2 = 1$.
Write $P_n(z) = z(z^n + z^{-n})$. The integrand in $(3)$ has simple poles at $\alpha$ and $-\alpha$ with matching residues there: $$\frac{P(\pm\alpha)}{Q_a'(\pm\alpha)} = -\frac{1}{8\sqrt{1-a}}\left(\alpha^n + \beta^n\right).\tag{4}$$
For $n > 0$ there is also a pole at the origin of order $n-1$, and this is painful to deal with.
Here's what I did. (This may not be the best approach.)
Write
$$\frac{1}{aw^2 + 2(a-2)w + a} = \sum_{k=0}^\infty c_k w^k$$
as a (formal) power series.
Since this is a rational function,
the coefficients $c_k$ satisfy a recurrence with a closed form involving powers of the roots $\alpha^2$ and $\beta^2$. Cross multiplying, we see that
$$c_0 = \frac{1}{a},\quad c_1 = \frac{2(2-a)}{a^2},\quad c_k = a c_1 c_{k-1} - c_{k-2}.$$
Knowing that the general solution is of the form $c_k = \lambda (\alpha^2)^k + \mu (\beta^2)^k$, plug in the initial conditions and solve:
$$\lambda = \frac{a-2+2\sqrt{1-a}}{4a\sqrt{1-a}}$$
$$\mu = \frac{2-a+2\sqrt{1-a}}{4a\sqrt{1-a}} = \frac{1}{a} - \lambda.$$
We seek the coefficient of $z^{-1}$ in the integrand, which looks like
$$z(z^n + z^{-n})\sum_{k=0}^\infty c_k z^{2k}$$
around the origin, so we want $2k + 1 - n = -1$, or $k = \frac{n}{2}-1$.
Thus, the residue is $$\lambda\beta^2\alpha^n + \mu\alpha^2\beta^n\tag{5}$$
and combining $(5)$ with (the double contribution of) $(4)$ we obtain, by the residue theorem,
$$g(n, a) = 4\pi\left\{\left(\frac{1}{4\sqrt{1-a}}-\lambda\beta^2\right)\alpha^n + \left(\frac{1}{4\sqrt{1-a}}-\mu\alpha^2\right)\beta^n\right\}.$$
Miraculously, the coefficient of $\beta^n$ vanishes.
Simplifying, we conclude that
$$g(n, a) = \frac{2\pi}{\sqrt{1-a}}\left(\frac{1-\sqrt{1-a}}{\sqrt{a}}\right)^n$$
where, of course,
the sign of the root $\alpha$ does not matter as $n$ is even. This in $(2)$ yields $(1)$.
Update. A much simpler way to evaluate $g(n, a)$ has been brought to my attention.
$$g(n, a) = \int_0^{2\pi} \frac{e^{ix}}{1-a\cos^2 x}\,dx$$
because the imaginary part (the sine integral) vanishes (the integrand is odd);
now let $z = e^{ix}$. The integral becomes
$$4i \int_{|z|=1} \frac{z^{n+1}}{Q_a(z)}\,dz$$
which equals
$$(2\pi i)(4 i)\frac{2\alpha^n}{-8\sqrt{1-a}} = \frac{2\pi \alpha^n}{\sqrt{1-a}}$$
by the residue theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1760407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
$0\leq u_n\leq \frac {1}{n^2}\sum_{k=1}^nu_k\implies $ $\sum u_k$ converges.
Let $(u_n)_{n\geq 1}$ be a sequence of real numbers such that $\displaystyle \forall n\geq 1, \;\;\;\;0\leq u_n\leq \frac {1}{n^2}\sum_{k=1}^nu_k$.
Prove that $\sum u_k$ converges.
Let $S_n=\sum_{k=1}^nu_k$ and $A_n=\sum_{k=1}^n \frac 1{k^2}$.
It suffices to prove that $S_n$ converges.
Summing the inequalities $0\leq u_n\leq \frac {1}{n^2}\sum_{k=1}^nu_k$ from $n=1$ to $n=N$ yields $S_N\leq \sum_{k=1}^N \frac{S_k}{k^2}$
Summation by part gives $\displaystyle \sum_{k=1}^N \frac{S_k}{k^2}= S_NA_N-\sum_{k=1}^{N-1}u_{k+1}A_k\leq S_NA_N$
Hence the estimate $S_N\leq \sum_{k=1}^N \frac{S_k}{k^2}\leq S_NA_N$
As a consequence, $S_N$ converges if and only if the series $\displaystyle \sum_{k}\frac{S_k}{k^2}$ is convergent.
This is quite interesting, but it doesn't answer the problem...
| You can prove by induction that : $0 \leqslant u_n \leqslant \frac{2}{n(n+1)}u_1$.
Base case for $n=1$, that is okay
Inductive step Assume it is true for all $k < n$. Then :
$u_n \leqslant \frac{u_n}{n^2} + \frac{1}{n^2}\sum \limits_{k=1}^{n-1} u_k$, so $(1 - \frac{1}{n^2})u_n \leqslant \frac{1}{n^2}\sum \limits_{k=1}^{n-1} \frac{2}{k(k+1)}u_1 = \frac{1}{n^2}\sum \limits_{k=1}^{n-1} \Big( \frac{2}{k} - \frac{2}{k+1} \Big)u_1$.
Hence $\frac{n^2 - 1}{n^2} u_n \leqslant \Big( 2 - \frac{2}{n} \Big)u_1 = \frac{2(n-1)}{n} u_1$. We end up with $u_n \leqslant \frac{2}{n(n+1)}u_1$.
Finally you deduce that $\sum \limits_{k=1}^n u_k \leqslant 2 u_1$ (and all the $u_k$ are non-negative), so $\sum u_k$ converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1761305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$\int \frac{\cos(x)}{(1+\cos(x))^3} \, dx$? I'm having a bit of a trouble seeing how to factorize the result of substituting $t=\tan(x/2)$, $\cos(x)=\frac{1-t^2}{1+t^2}$ and $dx=\frac{2}{1+t^2} \, dt$ into
$$\int \frac{\cos(x)}{(1+\cos(x))^3} \, dx$$
The result is
$$\int\frac{\frac{1-t^2}{1+t^2}}{(1+\frac{1-t^2}{1+t^2})^3}\frac{2}{1+t^2} \, dt$$
that looks horrible.
| $$
\frac{\frac{1-t^2}{1+t^2}}{ \left(1+\frac{1-t^2}{1+t^2}\right)^3}\frac{2}{1+t^2}
$$
Start by multiplying the top and bottom of the expression above both by the common denominator $(1+t^2)^3$. The factor
$$
\frac{\frac{1-t^2}{1+t^2}}{1+t^2}
$$
becomes
$$
(1-t^2)(1+t^2).
$$
The factor
$$
\left( 1+ \frac{1-t^2}{1+t^2} \right)^3
$$
in the denominator becomes
$$
\Big( (1+t^2) + (1-t^2) \Big)
$$
and then simplifies to $2$. You end up with
$$
\int \frac{(1-t^2)(1+t^2)} 2 \, dt = \int (1-t^4)\,dt.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1765760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Solve the equation on the interval $\; 0 \le \theta \lt 2\pi $ Hey I have two questions for math exchange! Let me list them first and show you what I have tried. By the way one can not use a calculator on the test review!
Solve the equations on the interval $\; 0 \le \theta \lt 2\pi $:
*
*$2 \cos^2\theta - 3\cos \theta + 1 = 0 $
*$\cot \theta = 2 \cos \theta$
So for the first one I tried:
\begin{gather}
2 \cos^2\theta - 3\cos \theta + 1 = 0\\
(2\cos^2\theta - 1) \times (3\cos\theta - 1) = 0\\
\begin{aligned}
2 \cos^2 \theta -1 &= 0 &\text{ or }&& 3\cos\theta -1 =0\\
\cos^2\theta &= \dfrac{1}{2} &\text{ or }&& \theta = \cos^{-1}\left(\dfrac{1}{3}\right)
\end{aligned}
\end{gather}
After that I just got confused... I can't use a calculator so I do not know what to do next..
So The next problem was even more diffcult:
$$\cot \theta = 2 \cos \theta$$
And then immediately after that I got confused since there are two theta's, and I have absolutely no clue what to do next!
On the anwser key for test review, the correct anwsers are:
*
*$$2 \cos^2\theta - 3\cos \theta + 1 = 0$$
Answer:$\qquad\left\{ 0,\;\dfrac{\pi}{3} ,\; \dfrac{5\pi}{3} \right\}$
*$$\cot \theta = 2 \cos \theta$$
Answer : $\qquad\left\{\dfrac{\pi}{6} ,\; \dfrac{\pi}{2},\; \dfrac{5\pi}{6},\;\dfrac{3\pi}{2} \right\}$
Help would be greatly appreciated. Thankyou for reading!
| First question: let's call $x=\cos\theta$. Then you have a second order equation $2x^2-3x+1=0$. then you would get two solutions in general, $x_1$ and $x_2$ and the final solution will be $\theta=\arccos(x_1)+n\pi$ and $\theta=\arccos(x_2)+n\pi,\;n\in\mathbb{N}$
Second question, use the definition of $\cot\theta=\cos\theta/\sin\theta$. Then, the equation reads
$$\cos\theta\left(\frac{1}{\sin \theta}-2\right)=0$$
Then or $\cos\theta=0$, so $\theta=\pi/2+n\pi,\;n\in\mathbb{N}$ or $\frac{1}{\sin \theta}-2$, which leads to $\sin\theta=1/2$, or equivalently, $\theta=\pi/6+n\pi,5\pi/6+n\pi,\;n\in\mathbb{N}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1767127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Distance between incentre and orthocentre. I want to prove that the distance between incentre and orthocentre is $$\sqrt{2r^2-4R^2\cos A\cos B\cos C} $$here $r$ is inradius and $R$ is circumradius.
I considered $\triangle API$ ($P$ is orthocentre and $I$ is incentre). I could find $AP=2R\cos A$, $AI=4R\sin\frac{B}{2}\sin\frac{C}{2} $ and $\angle PAI=\angle \frac{B-C}{2}$.
So applying cosine rule I got $$PI^2=4R^2+16R^2\sin^2\frac{B}{2}\sin^2\frac{C}{2} -16R^2\cos A\sin\frac{B}{2}\sin\frac{C}{2}\Bigg(\cos\frac{B}{2}\cos\frac{C}{2}+\sin\frac{B}{2}\sin\frac{C}{2}\Bigg)$$
How to proceed form here ?
|
So applying cosine rule I got $$\small PI^2=4R^2+16R^2\sin^2\frac{B}{2}\sin^2\frac{C}{2} -16R^2\cos A\sin\frac{B}{2}\sin\frac{C}{2}\Bigg(\cos\frac{B}{2}\cos\frac{C}{2}+\sin\frac{B}{2}\sin\frac{C}{2}\Bigg)$$
I think that you have a typo (the red part) :
$$\begin{align}&\small PI^2=4R^2\color{red}{\cos^2A}+16R^2\sin^2\frac{B}{2}\sin^2\frac{C}{2} -16R^2\cos A\sin\frac{B}{2}\sin\frac{C}{2}\Bigg(\cos\frac{B}{2}\cos\frac{C}{2}+\sin\frac{B}{2}\sin\frac{C}{2}\Bigg)\\&\small=4R^2\left(\cos^2A+4\sin^2\frac{B}{2}\sin^2\frac{C}{2} -4\cos A\sin\frac{B}{2}\sin\frac{C}{2}\Bigg(\cos\frac{B}{2}\cos\frac{C}{2}+\sin\frac{B}{2}\sin\frac{C}{2}\Bigg)\right)\\&\small=4R^2\left(\cos^2A+4\sin^2\frac{B}{2}\sin^2\frac{C}{2}-\cos A\cdot \color{green}{2\sin\frac{B}{2}\cos\frac{B}{2}\cdot 2\sin\frac{C}{2}\cos\frac{C}{2}}-4\cos A\sin^2\frac{B}{2}\sin^2\frac{C}{2}\right)\\&\small=4R^2\left(\cos^2A+8\sin^2\frac{B}{2}\sin^2\frac{C}{2}\cdot \color{blue}{\frac 12\left(1-\cos A\right)}-\cos A\color{green}{\sin B\sin C}\right)\\&\small=4R^2\left(\cos^2A+8\color{blue}{\sin^2\frac A2}\sin^2\frac{B}{2}\sin^2\frac{C}{2}-\cos A\sin B\sin C\right)\\&\small=4R^2\left(8\sin^2\frac A2\sin^2\frac{B}{2}\sin^2\frac{C}{2}+\cos A(\cos A-\sin B\sin C)\right)\\&\small=4R^2\left(8\sin^2\frac A2\sin^2\frac{B}{2}\sin^2\frac{C}{2}+\cos A(\cos(180^\circ-(B+C))-\sin B\sin C)\right)\\&\small=4R^2\left(8\sin^2\frac A2\sin^2\frac{B}{2}\sin^2\frac{C}{2}-\cos A\cos B\cos C\right)\\&\small =2\left(4R\sin \frac A2\sin \frac B2\sin\frac C2\right)^2-4R^2\cos A\cos B\cos C\\&\small=2r^2-4R^2\cos A\cos B\cos C\end{align}$$
| {
"language": "en",
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"answer_count": 2,
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} |
Evaluation of $\sum^{\infty}_{n=1}\left(\frac{1}{3n+1}-\frac{1}{3n+2}\right)$
Evaluation of $$\sum^{\infty}_{n=1}\left(\frac{1}{3n+1}-\frac{1}{3n+2}\right)$$
$\bf{My\; Try::}$ Here I have solved it using Definite Integration,
Like $$\sum^{\infty}_{n=1}\left(\frac{1}{3n+1}-\frac{1}{3n+2}\right)=\sum^{\infty}_{n=1}\int_{0}^{1}\left(x^{3n}-x^{3n+1}\right)dx$$
So we get $$ = \int_{0}^{1}(1-x)\sum^{\infty}_{n=1}\left(x^{3n}\right)dx = \int_{0}^{1}\frac{(1-x)x^3}{1-x^3}dx=\int_{0}^{1}\frac{x^3}{x^2+x+1}dx$$
So we get Sum $$ = \sum^{\infty}_{n=1}\left(\frac{1}{3n+1}-\frac{1}{3n+2}\right) = \int_{0}^{1}\frac{x^3}{x^2+x+1}dx = \frac{1}{18}\left(2\pi\sqrt{3}-9\right)$$
My Question is can we solve above sum without Using DEfinite Integration, If yes
Then how can I solve it, Help required
Thanks
| Well you could rewrite your sum as :
$$\tag{1}S=-\frac 12+\frac 1{\sin(2\pi/3)}\;\sum_{k=1}^\infty\frac {\sin(2\pi k/3)}{k}$$
(the $-\dfrac 12$ is from your sum starting at $n=1$ and not $n=0$)
and use the Fourier series for the sawtooth wave to get your result
or consider $(1)$ as $-\frac 12$ plus the imaginary part of :
$$\frac 1{\sin(2\pi/3)}\;\sum_{k=1}^\infty\frac {\exp(2\pi k\,i/3)}{k}=\frac 1{\sin(2\pi/3)}\;\sum_{k=1}^\infty\frac {\left(\exp(2\pi \,i/3)\right)^{\;k}}{k}$$
that is
\begin{align}
\tag{2}S&=-\frac 12-\frac 2{\sqrt{3}}\;\Im\;\log(1-\exp(2\pi \,i/3))\\
&=-\frac 12-\frac 1{\sqrt{3}\;i}\;\log\frac{1-\exp(2\pi \,i/3)}{1-\exp(-2\pi \,i/3)}\\
&=-\frac 12-\frac 1{\sqrt{3}\;i}\;\log(\exp(-\pi \,i/3))\\
\end{align}
and thus
$$\tag{3}\boxed{S=\frac {\pi}{3\sqrt{3}}-\frac 12}$$
| {
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"answer_id": 1
} |
prove: $\lim\limits_{x \rightarrow a} \frac{f^2(x)- g^2(x)}{(f(x) -f(a))^2} = 1$ $f(x)$ and $g(x)$ both differentiable twice at $x = a$ and we know that $f''(a) =g''(a)+f(a)$, $f(a) = g(a) = f'(a) = g'(a) \not = 0$
(we don't know if $f(x)$ and $g(x)$ are differentiable anywhere else except $x=a$)
prove: $\lim\limits_{x \rightarrow a} \frac{f^2(x)- g^2(x)}{(f(x) -f(a))^2} = 1$
I tried to substitute $f^2(x)$ and $g^2(x)$ with a square of taylor expansion of degree 1 (with remainder of degree 2) for example:
$$f^2(x)= (f(a) + f'(a)x + \frac{f''(\xi)}{2}x^2)^2$$
same for $g^2(x)$.
Now because $f''(x)$ is continuous and since $x \rightarrow a$ we can conclude that $f(\xi) = f(a)$.
Now after simplifying the original expression i get:
$$\lim\limits_{x \rightarrow a} \frac{f^2(a)(3x^2+2x) + f(a)g''(a) \frac{x^4}{2}}{f^2(a) \cdot (x^2 + \frac{x^2}{2}) + f(a)(x+\frac{x^2}{2})g''(a)x^2 + g''}$$
| We can use Taylor series also (instead of L'Hospital's Rule). We have
\begin{align}
f(a + h) &= f(a) + hf'(a) + \frac{h^{2}}{2}f''(a) + o(h^{2})\notag\\
g(a + h) &= g(a) + hg'(a) + \frac{h^{2}}{2}g''(a) + o(h^{2})\notag
\end{align}
We can proceed as follows
\begin{align}
L &= \lim_{x \to a}\frac{f^{2}(x) - g^{2}(x)}{(f(x) - f(a))^{2}}\notag\\
&= \lim_{x \to a}\{f(x) + g(x)\}\cdot\frac{f(x) - g(x)}{(f(x) - f(a))^{2}}\notag\\
&= \{f(a) + g(a)\}\lim_{x \to a}\dfrac{f(x) - g(x)}{\left(\dfrac{f(x) - f(a)}{x - a}\right)^{2}(x - a)^{2}}\notag\\
&= \frac{2f(a)}{\{f'(a)\}^{2}}\lim_{x \to a}\frac{f(x) - g(x)}{(x - a)^{2}}\notag\\
&= \frac{2}{f(a)}\lim_{h \to 0}\frac{f(a + h) - g(a + h)}{h^{2}}\text{ (putting }x = a + h)\notag\\
&= \frac{2}{f(a)}\lim_{h \to 0}\frac{1}{h^{2}}\cdot\{f(a) + hf'(a) + \frac{h^{2}}{2}f''(a) - g(a) - hg'(a) - \frac{h^{2}}{2}g''(a) + o(h^{2})\}\notag\\
&= \frac{2}{f(a)}\cdot\frac{f''(a) - g''(a)}{2}\notag\\
&= \frac{2}{f(a)}\cdot\frac{f(a)}{2}\notag\\
&= 1\notag
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/1769611",
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"answer_id": 3
} |
How to Find $ \lim\limits_{x\to 0} \left(\frac {\tan x }{x} \right)^{\frac{1}{x^2}}$. Can someone help me with this limit? I'm working on it for hours and cant figure it out.
$$ \lim_{x\to 0} \left(\frac {\tan x }{x} \right)^{\frac{1}{x^2}}$$
I started transforming to the form
$ \lim_{x\to 0} e^{ {\frac{\ln \left(\frac {\tan x}{x} \right)}{x^2}} }$
and applied the l'Hopital rule (since indeterminated $\frac00$), getting:
$$ \lim_{x\to 0} \left( \frac{2x-\sin 2x }{2x^2\sin 2x} \right)$$
From here, I try continue with various forms of trigonometric substitutions, appling the l'Hopital rule again and again, but no luck for me. Can someone help me?
| It is possible to compute this limit only by making use of some basic limits:
\begin{eqnarray*}
\lim_{u\rightarrow 0}\frac{\ln (1+u)}{u} &=&1 \\
\lim_{x\rightarrow 0}\frac{\tan x-x}{x^{3}} &=&\frac{1}{3} \\
\lim_{x\rightarrow 0}\frac{\tan x}{x} &=&1.
\end{eqnarray*}
Take the logarithm
\begin{eqnarray*}
L &=&\ln \left( \frac{\tan x}{x}\right) ^{1/x^{2}} \\
&=&\frac{1}{x^{2}}\ln \frac{\tan x}{x} \\
&=&\frac{1}{x^{2}}\ln \left( 1+\frac{\tan x}{x}-1\right) \\
&=&\frac{\left[ \frac{\tan x}{x}-1\right] }{x^{2}}\cdot \frac{\ln \left( 1+%
\left[ \frac{\tan x}{x}-1\right] \right) }{\left[ \frac{\tan x}{x}-1\right] }
\\
&=&\frac{\tan x-x}{x^{3}}\cdot \frac{\ln (1+u(x))}{u(x)},\ with\ u(x)=\frac{%
\tan x}{x}-1
\end{eqnarray*}
since
\begin{equation*}
\lim_{x\rightarrow 0}u(x)=\lim_{x\rightarrow 0}(\frac{\tan x}{x}-1)=1-1=0
\end{equation*}
then
\begin{equation*}
\lim_{x\rightarrow 0}\frac{\ln (1+u(x))}{u(x)}=\lim_{u\rightarrow 0}\frac{%
\ln (1+u)}{u}=1,
\end{equation*}
and therefore
\begin{equation*}
\lim_{x\rightarrow 0}L(x)=\lim_{x\rightarrow 0}\frac{\tan x-x}{x^{3}}\cdot
\lim_{x\rightarrow 0}\frac{\ln (1+u(x))}{u(x)}=\frac{1}{3}\cdot 1=1.
\end{equation*}
Backward, one take the exponential, and using continuity arguments, obtains
\begin{equation*}
\lim_{x\rightarrow 0}\left( \frac{\tan x}{x}\right)
^{1/x^{2}}=\lim_{x\rightarrow 0}e^{L(x)}=e^{1/3}.
\end{equation*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 4
} |
Find the generating function for the recurrence $a_n=a_{n-1}-a_{n-2}$, with $a_0=0$ and $a_1=1.$ This was a test question and I felt confident about it but all he put on it was no and circled a problem and left it at that. My solution up until I messed up which was early was
$G_a(x) = \sum_{n=0}^\infty a_n x^n$
$G_a(x) = \sum_{n=0}^\infty (a_{n-1}-a_{n-2})x^n$
$G_a(x) = \sum_{n=1}^\infty a_{n-1}x^n+\sum_{n=2}^\infty a_{n-2}x^n + a_0+a_1$
in the second line he circled n=0 in the sum and and then in the third line he wrote n=2 instead of n=1 in the first sum. Can anyone explain and show me how to solve it the correct way.
My final answer was $G_a(x)=\frac{1}{x^2+x-1} $
| We can verify a certain periodicity in the terms without solving the recurrence $a_n$. In fact $$a_0=0\\a_1=1\\a_2=1\\a_3=0\\a_4=-1\\a_5=-1\\a_6=0\\a_7=1$$ Hence the sequence of values can be considered as beginning again from $a_6$ and $a_7$; an so on....
Consequently the asked generating function is
$$\color{red}{F(x)=(x+x^2-x^4-x^5)+(x^7+x^8-x^{10}-x^{11})+(x^{13}+x^{14}-x^{16}-x^{17}) +.....}$$
Considering finitely many terms, in order to get a closed form, we have $$F(x)=(x+x^2-x^4-x^5)(1+x^6+x^{12}+x^{18}+.....+x^{6n})$$
$$F(x)=(x+x^2-x^4-x^5)\left(\frac{x^{6(n+1)}-1}{x^6-1}\right)$$
Finally $$F(x)=(x-x^4)(x+1)\left(\frac{x^{6(n+1)}-1}{x^6-1}\right)=-x(x^3-1)(x+1)\left(\frac{x^{6(n+1)}-1}{x^6-1}\right)=x(x+1)\left(\frac{1-x^{6(n+1)}}{x^3+1}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1771409",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove that if $n \in \mathbb{N}$ and $n \ge 2$, then $2^{n + 1} \le 3^n$. Prove that if $n \in \mathbb{N}$ and $n \ge 2$, then $2^{n + 1} \le 3^n$.
My method:
If $n = 2$, $2^{n + 1} \le 3^n$ then $2^3 \le 3^2$ is $8 \le 9$, which holds for $n = 2$.
$2^{k + 1} \le 3^k$ then $2^{k + 2} \le 3^{k + 1}$. Then
$2 \cdot 2 \cdot 2^k \lt 3 \cdot 3^k$
$4 \cdot 2^k \lt 3 \cdot 3^k$
Therefore, $4 \cdot 2^k \lt 3 \cdot 3^k$. Hence, $2^{n + 1} \le 3^n$ for $n \ge 2$.
Is there a problem with: $2^{k + 1} \le 3^k$ then $2^{k + 2} \le 3^{k + 1}$ or any other part of this proof?
| The proof is easy:
*
*If $k=2$, we know that $2^{k+1} = 2^3 = 8 \leq 9 = 3^2 = 3^k$. Then, it is true.
*We assume that it is also true for $k=n$, then we assume that $2^{n+1} \leq 3^{n}$
*We want to see that it is true if $k = n+1$:
$2^{k+1} = 2^{n+2} = 2^{n+1} \cdot 2$
We assumed before $2^{n+1} \leq 3^{n}$, then
$2^{k+1} = 2^{n+1} \cdot 2 \leq 3^{n} \cdot 2 \leq 3^{n} \cdot 3 = 3^{n+1}=3^k \rightarrow 2^{k+1} \leq 3^k$
What we do is:
*
*We prove that it is true for $k=2$
*We prove that if it is true for $k=n$, then it is also true for $k=n+1$
Therefore, it is true for $k=3$ because it is true for $k=2$. It is true for $k=4$ because it is true for $k=3$...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1772201",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Let $f(x)=x^3-3x+1$.Find the number of distinct real roots of $f(f(f(x)))=3$ Let $f(x)=x^3-3x+1$.Find the number of distinct real roots of $f(f(f(x)))=3$
I have noticed that $f(x)=3$ has solutions $-1,-1,2$ But here how to find more roots ? in fact how to even say that they are roots of $f(f(f(x)))=3$?
If anyone has materials to share about these type of problems involving iterations, please do so.
| The exact solutions are the roots of the polynomial
? print(h)
x^27 - 27*x^25 + 9*x^24 + 324*x^23 - 216*x^22 - 2241*x^21 + 2268*x^20 + 9639*x^1
9 - 13584*x^18 - 25515*x^17 + 50598*x^16 + 35703*x^15 - 119232*x^14 - 1917*x^13
+ 170721*x^12 - 80352*x^11 - 127980*x^10 + 122958*x^9 + 22599*x^8 - 66339*x^7 +
20412*x^6 + 4374*x^5 - 2187*x^4
?
Its factorization is
? factor(h)
%33 =
[ x 4]
[ x^2 - 3 4]
[ x^3 - 3*x + 3 2]
[x^9 - 9*x^7 + 3*x^6 + 27*x^5 - 18*x^4 - 27*x^3 + 27*x^2 - 3 1]
?
The polynomial has $11$ real roots, $6$ negative and $4$ positive ones.
You can determine the roots of the $4$ polynomials numerically.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1772867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 6,
"answer_id": 5
} |
Evaluation of $\cos\left(\frac{\pi}{2n}\right)\cdot \cos\left(\frac{2\pi}{2n} \right)....\cos\left(\frac{(n-1)\pi}{2n}\right)$
Evaluation of $$\lim_{n\rightarrow \infty}\left(\tan \frac{\pi}{2n}\cdot \tan \frac{2\pi}{2n}\cdot \tan \frac{\pi}{3n}\cdot ...............\tan \frac{(n-1)\pi}{2n}\right)^{\frac{1}{n}} = $$ without using Limit as a sum.
$\bf{My\; Try::}$ Using the formula $$\displaystyle \sin\left(\frac{\pi}{n}\right)\cdot \sin\left(\frac{2\pi}{n} \right)....\sin\left(\frac{(n-1)\pi}{n}\right) = \frac{n}{2^{n-1}}$$
Replace $n\rightarrow 2n$
$$\displaystyle \sin\left(\frac{\pi}{2n}\right)\cdot \sin\left(\frac{2\pi}{2n} \right)....\sin\left(\frac{(2n-1)\pi}{2n}\right) = \frac{2n}{2^{2n-1}}$$
Now How can I calculate $$\displaystyle \sin\left(\frac{\pi}{2n}\right)\cdot \sin\left(\frac{2\pi}{2n} \right)....\sin\left(\frac{(n-1)\pi}{2n}\right)$$
and also How can I calculate $$\displaystyle \cos\left(\frac{\pi}{2n}\right)\cdot \cos\left(\frac{2\pi}{2n} \right)....\cos\left(\frac{(n-1)\pi}{2n}\right)$$
Help required, Thanks
| I will also share my thoughts on your formula.
In moving from
$$ \displaystyle \sin\left(\frac{\pi}{2n}\right)\cdot \sin\left(\frac{2\pi}{2n} \right)....\sin\left(\frac{(2n-1)\pi}{2n}\right) = \frac{2n}{2^{2n-1}} \ \ \ (1)$$
to
$$ \displaystyle \sin\left(\frac{\pi}{2n}\right)\cdot \sin\left(\frac{2\pi}{2n} \right)....\sin\left(\frac{(n-1)\pi}{2n}\right) \ \ \ (2)$$
Notice the symmetry of $\sin $ function around $\pi/2$, we can see that for all $k=1,2,...,n$
$$ \displaystyle \sin\frac{(n-k)\pi}{2n} = \sin(\frac{(n-k)\pi}{2n}+\frac{\pi}{2})=\sin\frac{(n+k)\pi}{2n}$$
So we see that (2) is equal to the square root of (1) -- almost! (Almost because depending on $n$ being odd or even there may or may not be a perfect match-up of $k$ less that $n$ with those bigger than $n$.)
For $\cos$ 's use the identity $$ \sin (\frac{\pi}{2}-x)=\cos (x)$$ to see that for any $k=1,2,...,n$
$$ =\cos (\frac{k\pi}{2n})=\sin (\frac{\pi}{2}-\frac{k\pi}{2n})=\sin (\frac{(n-k)\pi}{2})$$
But $n-k$ will range over the same integers.
Notice then that in your expression first numerator then will be cancelled from last denominator. Then 2nd numerator with the second to last denominator... Then your left with $1$! and the answer to your limit is $1$ -- which was corroborated by the alternative calculation.
| {
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"url": "https://math.stackexchange.com/questions/1774980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Splitting fractions with a linear denominator: $\frac{2x-1}{x+2}$ How can $$\frac{2x-1}{x+2}$$ be split to give $$A-\frac{B}{x+2}$$
where $A$ and $B$ are integers?
The solution is $$2-\frac{5}{x+2}.$$
| Observe that if you set
$$y=x+2,\quad x=y-2,$$ then you get
$$
\frac{2x-1}{x+2}=\frac{2(y-2)-1}{y}=\frac{2y-5}{y}=2-\frac{5}{y}=2-\frac{5}{x+2}
$$ as wanted.
| {
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Prove that $(a^2-bc)^3+(b^2-ac)^3+(c^2-ab)^3=3(a^2-bc)(b^2-ac)(c^2-ab)$
Let $a+b+c=0$. Prove that
$$(a^2-bc)^3+(b^2-ac)^3+(c^2-ab)^3=3(a^2-bc)(b^2-ac)(c^2-ab)$$
My work so far:
I used the formula
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$$
Then $$(a^2-bc)^3+(b^2-ac)^3+(c^2-ab)^3-3(a^2-bc)(b^2-ac)(c^2-ab)=$$
$$=(a^2+b^2+c^2-ab-bc-ca)\times$$
$$\times\left((a^2-bc)^2+(b^2-ac)^2+(c^2-ab)^2-(a^2-bc)(b^2-ac)-(b^2-ac)(c^2-ab)-(a^2-bc)(c^2-ab) \right)$$
But if $a\not=b$ or $a\not=b$ or $a\not=c$ then $a^2+b^2+c^2-ab-bc-ca>0$.
How do I prove
$$(a^2-bc)^2+(b^2-ac)^2+(c^2-ab)^2-(a^2-bc)(b^2-ac)-(b^2-ac)(c^2-ab)-(a^2-bc)(c^2-ab)=0?$$
| Let $a^2-bc=x$, $b^2-ca=y$, $c^2-ab=z$. Then note $$x-y=a^2-bc+ca-b^2=(a-b)(a+b+c)=0$$$$y-z=b^2-ca+ab-c^2=(b-c)(a+b+c)=0$$Thus we have $x=y=z$. We are done.
| {
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"source": "stackexchange",
"question_score": "2",
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Prove $1+a0,b>0,c>0$, $aSuppose:
$a>0,b>0,c>0$, $a<bc$ , $1+ a^{3} = b^{3} + c^{3} $.
Prove:
$$ 1+a<b+c $$
This inequality at the University of Toronto plan.
| This proof goes by showing the assumption $1+a \ge b+c$ leads to a contradiction. Since the variables are positive the assumption is equivalent on squaring the sides to
$$1+2a+a^2 \ge b^2+2bc+c^2.\tag{1}$$
Now from $a<bc$ follows $-3a>-3bc,$ which on adding to (1) gives the strict inequality
$$1-a+a^2>b^2-bc+c^2.\tag{2}$$
Together with the assumption $1+a \ge b+c,$ since one of the inequalities is strict, we get to
$$(1+a)(1-a+a^2)>(b+c)(b^2-bc+c^2),$$
that is, to $1+a^3>b^3+c^3,$ against the assumption $1+a^3=b^3+c^3$ of the problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1777322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Show that $\int_{0}^x \dfrac{1}{1+t^2}dt+\int_{0}^{1/x} \dfrac{1}{1+t^2}dt$ does not depend on $x$
Show that $\displaystyle \int_{0}^x \dfrac{1}{1+t^2}dt+\int_{0}^{1/x} \dfrac{1}{1+t^2}dt$ does not depend on $x$.
Attempt:
We have that $\displaystyle \int_{0}^x \dfrac{1}{1+t^2}dt+\int_{0}^{1/x} \dfrac{1}{1+t^2}dt = \tan^{-1}{x}+\tan^{-1}{\dfrac{1}{x}}$, which does depend $x$, so I am confused.
| So many answers here to choose from. Here is a slightly different approach.
Here you are:
$\tan^{-1} x + \tan^{-1} \frac{1}{x} + C\\
x = \tan y, -\frac{pi}{2}<y<\frac{pi}{2}\\
\tan^{-1} (\tan y) + \tan^{-1} (\frac{1}{\tan y}) + C\\
y + \tan^{-1} (\cot y) + C\\
y + \tan^{-1} (\tan (\frac{\pi}{2}- y) + C\\
\frac{\pi}{2} + C$
Actually,if $x<0,$ and hence $y<0$ then $\frac{\pi}{2} - y > \frac{\pi}{2}$ and $\tan^{-1} (\tan (\frac{\pi}{2} - y)) = -\frac{\pi}{2} - y$
hence $\tan^{-1} x + \tan^{-1} \frac{1}{x} =-\frac{\pi}{2},\frac{\pi}{2}$ depending on the sign of $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1778699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Partial fraction decompostion Solve the partial fraction.
Starting out with...
$${x^2+1\over x^3-1}={x^2+1\over (x-1)(x^2+x+1)}$$
Then the partial fraction formula part of $\displaystyle {A\over x-1}+{Bx+C\over x^2+x+1}$.
Solving with the "shortcut" method of substitution, here is what came up.
Multiplying the partial fraction part by the problem,
$${x^2+1\over x^3-1}={x^2+1\over (x-1)(x^2+x+1)}=
{A\over x-1}+{Bx+C\over x^2+x+1}
$$
is
$$x^2+1=A(x^2+x+1)+B(Bx+C)(x-1)$$
Then letting
$x=1$ then $2=3A$, $~~A=\frac23$
$x=0$ then $1=A-C$, then $C=-\frac{1}3$
$x=-1$ then
$$\eqalign{2&=A(-2(-B)+C)=A+2B-2C=\frac23+2B-2\left(-\frac13\right)=\frac43+2B\cr
2-\frac43&=2B\cr}
$$
$B= \frac13$
Then factoring, the answer is $\displaystyle{1\over3}\left({2\over x-1}+{x-1\over x^2+x+1}\right)$.
Is there any possible quicker way, or is this the fastest?
| Just to be clear, I think you meant that the partial fraction problem is reduced to finding the coefficients for
$x^2+1=A(x^2+x+1)+(Bx+C)(x-1)$
Solving with the method of substitution is indeed probably the quickest way. But note that the coefficient method reveals valuable information.
For example, since the $x^2$ on the LHS has coefficient 1, we expect that $A+B=1 \quad (Ax^2+Bx^2=x^2$)
An obvious substitution like you did would be at $x=1$. Then, $3A=2\Rightarrow A=2/3; B=1/3$
Another apparent coefficient is the last one. $1=A-C \Rightarrow C=-1/3$
Either methods work fine but the coefficient method is very powerful. In terms of efficiency, they will be around the same, with a slight advantage towards the coefficient method.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1778783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Integral of $\frac{\sqrt {x}}{x^2+x}$ I'm trying to find $$\int_{1/3}^3 \frac{\sqrt{x}}{x^2+x} dx.$$
I used a $u$ substitution where $u = \sqrt{x}$ to get $$2 \int_\sqrt{1/3}^\sqrt{3} \frac{u}{u^4+u^2} du.$$
Substituting $u = \tan v$, I get
$$2 \int \cot v dv
= \left. \ln |\sin v|\right|_{\arctan \sqrt{1/3}}^{\arctan \sqrt 3}
= 2\ln \left( \frac{\sqrt{3}}{2}\right)-2\ln(1/2),
$$ which is wrong. Where did I go wrong?
| Your mistake is in your first substitution. You replaced $dx$ with $2 du$ instead of $2u\,du$. The substitution, by the way, makes the integral trivial:
\begin{align}
\int_{1/3}^3 \frac{\sqrt{x}}{x^2+x} dx&=\int_{1/\sqrt 3}^{\sqrt3}\frac{2u^2}{u^4+u^2}\,du
=\int_{1/\sqrt 3}^{\sqrt3}\frac{2}{1+u^2}\,du\\ \ \\
&=2\left.\vphantom\int\arctan u\,\right|_{1/\sqrt3}^{\sqrt3}
=2\arctan\sqrt3-2\arctan1/\sqrt3\\ \ \\
&=\frac{2\pi}3-\frac{2\pi}6=\frac\pi3.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1780395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
solve the integral Could anyone help me on how to go about this integral.
Calculate: $$\int_0^{\pi/6}\left(\sin2x-\frac{\cos 3x}{3}+\frac{x^2}{2}\right)dx$$
This is how I've done and not coming up with right answer:
I get a primitive function
$$\left[-\frac{\cos 2x}{2}-\frac{1}{3}\frac{\sin3x}{3}+\frac{x^3}{6}\right]_0^{\frac{\pi}{6}}$$
and then I go ahead substituting in the respective $\pi/6$ och 0 boundary values which lead to $$\frac{1}{4}-\frac{1}{9\sqrt{2}}+\frac{\pi^3}{216*6}$$ and it's a wrong answer.
Any suggestions are highly appreciated!
| $$\int_{0}^{\frac{\pi}{6}}\left(\sin(2x)-\frac{\cos(3x)}{3}+\frac{x^2}{2}\right)\space\text{d}x=$$
$$\int_{0}^{\frac{\pi}{6}}\sin(2x)\space\text{d}x-\frac{1}{3}\int_{0}^{\frac{\pi}{6}}\cos(3x)\space\text{d}x+\frac{1}{2}\int_{0}^{\frac{\pi}{6}}x^2\space\text{d}x=$$
Substitute $u=2x$ and $\text{d}u=2\space\text{d}x$.
This gives a new lower bound $u=2\cdot0=0$ and upper bound $u=2\cdot\frac{\pi}{6}=\frac{\pi}{3}$:
$$\frac{1}{2}\int_{0}^{\frac{\pi}{3}}\sin(u)\space\text{d}u-\frac{1}{3}\int_{0}^{\frac{\pi}{6}}\cos(3x)\space\text{d}x+\frac{1}{2}\int_{0}^{\frac{\pi}{6}}x^2\space\text{d}x=$$
Substitute $s=3x$ and $\text{d}s=3\space\text{d}x$.
This gives a new lower bound $s=3\cdot0=0$ and upper bound $s=3\cdot\frac{\pi}{6}=\frac{\pi}{2}$:
$$\frac{1}{2}\int_{0}^{\frac{\pi}{3}}\sin(u)\space\text{d}u-\frac{1}{9}\int_{0}^{\frac{\pi}{2}}\cos(s)\space\text{d}s+\frac{1}{2}\int_{0}^{\frac{\pi}{6}}x^2\space\text{d}x=$$
$$\frac{1}{2}\left[-\cos\left(u\right)\right]_{0}^{\frac{\pi}{3}}-\frac{1}{9}\left[\sin\left(s\right)\right]_{0}^{\frac{\pi}{2}}+\frac{1}{2}\left[\frac{x^3}{3}\right]_{0}^{\frac{\pi}{6}}=$$
$$-\frac{1}{2}\left(\cos\left(\frac{\pi}{3}\right)-\cos\left(0\right)\right)-\frac{1}{9}\left(\sin\left(\frac{\pi}{2}\right)-\sin\left(0\right)\right)+\frac{1}{6}\left(\left(\frac{\pi}{6}\right)^3-0^3\right)=$$
$$-\frac{1}{2}\left(\frac{1}{2}-1\right)-\frac{1}{9}\left(1-0\right)+\frac{1}{6}\left(\frac{\pi^3}{216}-0\right)=\frac{5}{36}+\frac{\pi^3}{1296}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1780929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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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.