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How to prove that for all natural $n$, $133|11^{n+2}+12^{2n+1}$
How to prove that for every natural $n$, the number
$$11^{n+2} +12^{2n+1}$$
is divisible by $133$?
I tried Induction method, so assuming that
$$
n=k \implies
A = (11^{k+2} +12^{2k+1} ) \pmod{133} \equiv 0
$$
Then trying to prove that
$$
n=k+1 \implies
B = (11^{k+3} +12^{2k+3} ) \pmod{133} \equiv 0
$$
For this I wanted to split $B$ into multiplication with at least one multiple of $A$. But no luck.
Any Ideas?
| Given that $11^{k+2}+12^{2k+1}=133\lambda$, you can infer that $11^{k+2}=133\lambda-12^{2k+1}$.
Thus you have that $$11^{k+3}+12^{2k+3}= 11^{k+2}\cdot 11 +144 \cdot 12^{2k+1}=11\cdot 133\lambda -11\cdot 12^{2k+1} +144 \cdot 12^{2k+1} =11\cdot 133\lambda +12^{2k+1} \cdot (144-11)=11\cdot 133\lambda +133\cdot 12^{k+1} $$
which is divisible by 133
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Solving $\frac{x^2-2}{x^2+2} \leq \frac{x}{x+4}$. Solve the inequality
$\frac{x^2-2}{x^2+2} \leq \frac{x}{x+4} \Leftrightarrow$ $\frac{x^2-2}{x^2+2} - \frac{x}{x+4} \leq 0 \Rightarrow$ $\frac{x^3+4x^2-2x-8-x^3-2x}{(x^2+2)(x+4)} \geq 0 \Leftrightarrow$ $4x^2-4x-8 \geq 0 \Rightarrow$ $x \geq \frac{1}{2} \pm \sqrt{2.25} \Rightarrow$ $x_1 \geq -1, \; x_2 \geq 2$
We notice that $x_2 \geq 2$ is a false root and testing implies that the solutions of the inequality lies within the interval $-1 \leq x \leq 2$.
Problem: But $x<-4$ also solves the inequality, so I must have omitted or done something wrong? And also, am I using implication and equivalence symbols correctly when doing the calculations?
Thank you for your help!
| It might be useful to simplify the inequality before trying to solve:
$$\frac{x^2-2}{x^2+2} \leq \frac{x}{x+4} \Leftrightarrow x^2-2 \leq \frac{x^3+2x}{x+4}$$
Now, note that (using polynomial division)
*
*$\frac{x^3+2x}{x+4} = x^2 - 4x+18 -\frac{72}{x+4}$
So, the inequality turns into
$$x^2-2 \leq x^2 - 4x+18 -\frac{72}{x+4} \Leftrightarrow \frac{72}{x+4} \leq -4x+20 = 4(5-x) \Leftrightarrow \boxed{\frac{18}{x+4} \leq 5-x}$$
Taking into consideration that the inequality sign flips when multiplying by a negative number you get two cases:
$$\frac{18}{x+4} \leq 5-x \Leftrightarrow \begin{cases} 18 \leq (5-x)(x+4) & \mbox{for } x >-4 \\ 18 \geq (5-x)(x+4) & \mbox{for }x<-4\end{cases} \Leftrightarrow \boxed{\begin{cases} (x+1)(x-2) \leq 0 & \mbox{for } x >-4 \\ (x+1)(x-2) \geq 0 & \mbox{for }x<-4\end{cases}}$$
Note that $(x+1)(x-2)$ corresponds to an upwards open parabola with zeros at $x=-1$ and $x=2$. So, you get
$$\begin{cases} (x+1)(x-2) \leq 0 & \mbox{for } x >-4 \\ (x+1)(x-2) \geq 0 & \mbox{for }x<-4\end{cases} \Leftrightarrow \begin{cases} x \in [-1,2] \cap (-4, \infty)\\ x \in ((-\infty,-1] \cup [2,\infty)) \cap ( -\infty,-4) \end{cases} \Leftrightarrow \boxed{x\in ([-1,2] \cup ( -\infty,-4))}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Prove that for $f(z)=\sum_{n=0}^{\infty} \frac{z^{n}}{2^n+1}$
Prove that for $f(z)=\sum_{n=0}^{\infty} \frac{z^{n}}{2^n+1}$,
$$\lim_{z \to2} f(z) (2-z) = 2$$
My approach :
I was thinking of applying Abel's limit theorem to compute the limit but since it is valid only for $z \to 1 $ I am totally clueless
I also tried by this approach:
$$ \lim_{\epsilon \to 0 } f(2-\epsilon)(\epsilon)= \sum_{0}^{\infty} \frac{(2-\epsilon)^n}{2^n +1}(\epsilon)$$
but again i am not sure of how to proceed.
| Just note that for any $z \in (-2,2)$, because of absolute convergence in $[-\frac{2+z}{2},\frac{2+z}{2}]$, we have $$f(z)(2-z) = \sum\limits_{n=0}^{+\infty} \frac{2z^n}{2^n+1} - \sum\limits_{n=0}^{+\infty} \frac{z^{n+1}}{2^n+1} = 1 + \sum\limits_{n=0}^{+\infty} z^{n+1} \cdot \frac{2(2^n+1)-(2^{n+1}+1)}{(2^n+1)(2^{n+1}+1)} = 1 + \sum\limits_{n=0}^{+\infty} z^{n+1} \cdot \frac{1}{(2^n+1)(2^{n+1}+1)}$$
So, since this has radius $4$, it is continuous in $2$ (in $2^-$).
Thus $$\lim\limits_{z \to 2^-} f(z)(2-z) = 1 + \sum\limits_{n=0}^{+\infty} 2^{n+1} \cdot \frac{1}{(2^n+1)(2^{n+1}+1)} = 1 + 2\sum\limits_{n=0}^{+\infty} \frac{1}{(2^n+1)} - \frac{1}{(2^{n+1}+1)}$$
Therefore the result is $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2908741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$n^3+n<3^n$ for $n \geq4$ by induction. To prove that $n^3+n<3^n$ for $n \geq4$ by induction.
I have proved the fact but it became very long and I have to use two more induction proof within the proof.
Can someone give a better solution by induction?
Thank You.
| $n^3 + n < 3^n \tag 1$
certainly holds for $n = 4$; so this will be the base case; assuming (1) binds for some $n$, then
$3(n^3 + n) < 3^{n + 1}; \tag 2$
now,
$(n + 1)^3 + (n + 1) = n^3 + 3n^2 + 3n + 1 + n + 1 = n^3 + 3n^2 + 4n + 2, \tag 3$
and
$3(n^3 + n) = 3n^3 + 3n, \tag 4$
so what we need is
$n^3 + 3n^2 + 4n + 2 \le 3n^3 + 3n, \; n \ge 4; \tag 5$
this is true if
$3n^3 + 3n - n^3 - 3n^2 - 4n - 2 = 2n^3 - 3n^2 -n - 2 \ge 0, \; n \ge 4; \tag 6$
now, as our colleague xbh has observed,
$2n^3 - 3n^2 - n - 2 = (n - 1)(2n^2 - n - 2) - 4; \tag 7$
we have
$n \ge 4 \Longrightarrow n - 1 \ge 3; \tag 8$
also we have
$n \ge 4 \Longrightarrow 2n^2 - n - 2 = n^2 + n(n - 1) - 2 \ge 16 + 4 \cdot 3 - 2 = 26; \tag 9$
thus,
$n \ge 4 \Longrightarrow 2n^3 - 3n^2 - n - 2 \ge 3 \cdot 26 - 4 = 74; \tag{10}$
we see that (6) binds, thus does (5), hence the induction is complete and (1) holds for all $n \ge 4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 3
} |
Given that $m$ is a real number not less than $-1$ The question is:
Given that $m$ is a real number not less than $-1$, such that the equation in $x$ is $x^2+2(m-2)x+m^2-3m+3=0$ has two distinct roots $r$ and $s$. If $r^2+s^2=6$, find the value of $m$.
Here's what I've tried:.
Using Vieta's formulas:
$rs = m^2-3m+3$
$r+s = -2m+4$
Then I set the equation as such; $(r+s)^2 - 2rs = 6$. I got $2m^2-10m+4=0$. Using the quadratic equation, I got $\frac {5+\sqrt {17}}2$ and $\frac {5-\sqrt {17}}2$. Is there anything wrong with my solution?
| The posted solution is correct up to that point, though it remains to be verified that $\,r \ne s\,$ for those values of $\,m\,$.
Alt. hint: $\color{blue}{r+s}=-2m+4$, and each root satisfies $\,x^2= -\big(2(m-2)x+m^2-3m+3\big)\,$, so:
$$
\begin{align}
6 = r^2+s^2 &= -2(m-2)(\color{blue}{r+s})- 2(m^2-3m+3) \\
&= 2(m-2)(2m-4)- 2(m^2-3m+3) \\
&= 2 (m^2 - 5 m + 5)
\end{align}
$$
Then what remains to solve is $\,m^2-5m+5=3\,$ for solutions $\,\ge -1\,$ (and, again, verify that they yield distinct roots $\,r,s\,$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Using Gram-Schmidt to find the QR decomposition I'm having problems doing a QR decomposition of a matrix...
Let $A=\begin{bmatrix}
{1} & {1} & {0} \\
{0} &{1} &{1} \\
{1} & {0} &{1}
\end{bmatrix}$
Find the QR decomposition for A
Here's what I've been doing:
I choose this basis, $B=\left \{(1,0,1), (1,1,0), (0,1,1)\right \}$ (the columns of the matrix).
Now I use the Gram-Schmidt process (and this is where I'm having trouble)
$u_{1}=(1,0,1)$
$u_{2}=(1,1,0)$ (cuz $<(1,0,1), (1,1,0)>=0$)
$u_{3}=(0,1,1)-\frac{<(0,1,1), (1,1,0)>}{<(1,1,0), (1,1,0)>}(1,1,0)-\frac{<(0,1,1), (1,0,1)>}{<(1,0,1), (1,0,1)>}(1,0,1)=$ $(0,1,1)-1/2(1,1,0)-1/2(1,0,1)=(-1, 1/2, 1/2)$
And now I find the norm for all the three vectors:
$||u_{1}||=||u_{2}||=||u_{3}||=\sqrt{2}$
So the orthonormal basis must be $B'= \left \{(\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}), (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0), (\frac{-1}{\sqrt{2}}, \frac{1}{2\sqrt{2}}, \frac{1}{2\sqrt{2}})\right \}$ (Which it isn't orthonormal)
So $Q=\begin{bmatrix}
{\frac{1}{\sqrt{2}}} & {\frac{1}{\sqrt{2}}} & {\frac{-1}{\sqrt{2}}} \\
{0} &{\frac{1}{\sqrt{2}}} &{\frac{1}{2\sqrt{2}}} \\
{\frac{0}{\sqrt{2}}} & {0} &{\frac{1}{2\sqrt{2}}}
\end{bmatrix}$
Which $Q^{t}Q \neq I$ ($I$ being the identity matrix), so all I did was wrong...
Where's my mistake?
| $u_1 $ is correct, your mistake is in $u_2$
\begin{equation}
u_2 =(1,1,0) - \frac{<(1,0,1), (1,1,0)>}{<(1,0,1), (1,0,1)>}(1,0,1)
\end{equation}
which is
\begin{equation}
u_2 = (1,1,0) - (\frac{1}{2})(1,0,1)
=
\frac{1}{2}(1,1,-1)
=
(\frac{1}{2}, \frac{1}{2},-\frac{1}{2})
\end{equation}
Now,
\begin{equation}
u_3 =(0,1,1) - \frac{<(1,0,1), (0,1,1)>}{<(1,0,1), (1,0,1)>}(1,0,1)
-
\frac{<(\frac{1}{2}, \frac{1}{2},-\frac{1}{2}), (0,1,1)>}{<(\frac{1}{2}, \frac{1}{2},-\frac{1}{2}), (\frac{1}{2}, \frac{1}{2},-\frac{1}{2})>}(\frac{1}{2}, \frac{1}{2},-\frac{1}{2})
\end{equation}
that is
\begin{equation}
u_3 = (0,1,1) - \frac{1}{2}(1,0,1)
-
\frac{0}{\frac{3}{4}}(\frac{1}{2}, \frac{1}{2},-\frac{1}{2})
=
(-\frac{1}{2},1,\frac{1}{2})
\end{equation}
This is an orthogonal basis, i.e.
\begin{equation}
O
=
\begin{bmatrix}
u_1 & u_2 & u_3
\end{bmatrix}
\end{equation}
where $O^T O $ is diagonal. To have an orthonormal one, just take (as you also mention)
\begin{align}
l_1 &= \frac{u_1}{\Vert u_1 \Vert}\\
l_2 &= \frac{u_2}{\Vert u_2 \Vert}\\
l_3 &= \frac{u_3}{\Vert u_3 \Vert}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Area of ellipse using double integral I am trying to find the area of a quadrant of an ellipse by double integrating polar coordinates but the answer I'm getting is incorrect.
ellipse : $ x^2/a^2 + y^2/b^2 =1 $
Any point on ellipse : $ ( a\cos(\theta), b\sin(\theta)) $
At $ \theta$, taking $ d\theta $ segment, Thus $ r^2 = a^2\cos^2(\theta) + b^2\sin^2(\theta) $ [Using pythagoras theorem]
$$ Area = \int_{0}^{\pi/2} \int_{0}^{\sqrt{a^2\cos^2(\theta) + b^2\sin^2(\theta)}} rdrd\theta $$
$$ = 1/2 \int_{0}^{\pi/2} r^2 \Big|_{0}^{\sqrt{a^2\cos^2(\theta)+b^2\sin^2(\theta)}} d\theta $$
$$ = 1/2 \int_{0}^{\pi/2} (a^2\cos^2(\theta)+b^2\sin^2(\theta)) d\theta $$
$$ = 1/2 \int_{0}^{\pi/2} ((a^2 - b^2)\cos^2(\theta)+b^2) d\theta $$
$$ = 1/4 \int_{0}^{\pi/2} (a^2 - b^2)(1+ \cos(2\theta)) d\theta +2b^2 d\theta $$
I am getting $$ \pi/8 (a^2 + b^2).$$ But the correct answer is $ \pi ab/4 $
| You can also use green's theorem.
Set x=acos($\theta$), y=bsin($\theta$). Now we can set up the intergal.
$A=\frac{1}{2}\int_{c} xdy-ydx. \\
dy=bcos(\theta), dx=-asin(\theta) \\
A=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}abcos^2(\theta)+absin^2(\theta)\text{ }d\theta=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}ab \text{ }d\theta=\frac{\pi ab}{4}. \\
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2915515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Evaluating $\iiint_R\left(1-\frac{x}{\sqrt{(x^2+y^2)}}\right)\,dx\,dy\,dz$ given the region $R$
Evaluate $$\iiint_R\left(1-\frac{x}{\sqrt{(x^2+y^2)}}\right)\,dx\,dy\,dz$$, where $R$ is the region bounded by $z=x^2+y^2$ and $z=1-x^2-y^2$.
Here is what I did:
Let: $x=r\cos\theta$, $y=r\sin\theta$, $z=z$.
First, consider $z=x^2+y^2$.
$r\in(0,\frac{\sqrt2}{2})$, $\theta\in(0,2\pi)$, $z\in(r^2,\frac{1}{2})$
$$\iiint\left(1-\frac{x}{\sqrt{(x^2+y^2)}}\right)\,dx\,dy\,dz=\int_{0}^{\frac{\sqrt2}{2}}\int_{0}^{2\pi}\int_{r^2}^{\frac{1}{2}}r(1-\cos\theta)\,dz\,d\theta\, dr=\frac{\pi}{8}$$
Then consider $z=1-x^2-y^2$
For $r\in(0,\frac{\sqrt2}{2})$, $\theta\in(0,2\pi)$, $z\in(1-r^2,1)$ $$\iiint\left(1-\frac{x}{\sqrt{(x^2+y^2)}}\right)\,dx\,dy\,dz=\int_{0}^{\frac{\sqrt2}{2}}\int_{0}^{2\pi}\int_{1-r^2}^{1}r(1-\cos\theta)\,dz\,d\theta\, dr=\frac{\pi}{8}$$
For the final answer I added those up and I got $\frac{\pi}{8}+\frac{\pi}{8}=\frac{\pi}{4}$
Where did I do wrong?
| Your subsitution is fine, cylindrical coordinates. But you messed up with the limits. Obviously $\theta \in (0,2\pi)$ . We have that z goes from the bottom surface to the top one, so you have to first identify them. In this case $z=r^2$ is the one below and $z=1-r^2$ the one above, so $z \in (r^2,1-r^2)$. Then we have to define the limits for the radius. The radius goes from $0$ to the interception between both surfaces:
$$ r^2=1-r^2 \Rightarrow r=\frac{\sqrt{2}}{2} \text{ ,because }r>0 $$
Considering the jacobian is $r$, the integral becomes:
$$ \int_0^{2\pi} \int_0^\frac{\sqrt{2}}{2} \int_{r^2}^{1-r^2} (1-\cos{\theta})\space r \space dz\space dr \space d\theta $$
The cosine part is $0$ because of the symmetry, and then we have that:
$$ \text{Integral} =2\pi \int_0^{\frac{\sqrt{2}}{2}} (1-2r^2) \space rdr = \frac{\pi}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2915731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Find the generating function for sequence $1,2,4,0,8,24,120,184,312,56,568,1592,...$ I'm having trouble finding a generating function for the sequence that has a closed form. The sequence can be deduced using two powers, with alternating negative and multiples of three as shown:
first term:
$1=1$
second term:
$2=1+2^{0}$
third term:
$4=1+2^{0}+2^{1}$
fourth term:
$0=1+2^{0}+2^{1}-2^{2}$
...
n-th term:
$n = 1+2^{0}+2^{1}-2^{2}+2^{3}+2^{4}+3(2^{5})+2^{6}+2^{7}-2^{8}+...$
Therefore, the -1 appears at every $4(mod6)$ term and the 3 appears at every $1(mod6)$ term greater than 1.
So I can construct the n-th ($n>1$) term as follows:
$f(n) = 1+\sum_{i=2}^{n}2^{i-2}u(i)$
where
$u(i)=-1$, if $i\equiv4(mod6)$
$u(i)=3$, if $i\equiv1(mod6)$
$u(i)=1$, otherwise
Is there a way to get a closed form of the function $f(n)$?
| Begin by finding the generating function for the sequence $2^0, 2^1, -2^2, 2^3, 2^4, 3 \cdot 2^5, 2^6, 2^7, -2^8, \dots$. This is the sum of
\begin{align}
\frac{1}{1-2x} &= 2^0 + 2^1x + 2^2x^2 + 2^3x^3 + 2^4x^4 + 2^5x^5 + \dots \\
-2\cdot\frac{(2x)^2}{1-(2x)^6} &= -2 \cdot 2^2x^2 - 2 \cdot 2^8x^8 - 2\cdot 2^{14}x^{14} - \dots \\
+2\cdot\frac{(2x)^5}{1-(2x)^6} &= 2 \cdot 2^5 x^5 + 2\cdot 2^{11}x^{11} + 2\cdot 2^{17}x^{17} + \dots.
\end{align}
Call this $g(x)$.
Then $1 + x \cdot g(x)$ will add the $1$ term at the beginning, which doesn't follow the pattern. Finally, multiplying by $\frac{1}{1-x}$ to get $$\frac{1+x \cdot g(x)}{1-x}$$ will give the generating function for this sequence's partial sums, which is what you want.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2916267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Determinant of Symmetric Matrix $\mathbf{G}=a\mathbf{I}+b\boldsymbol{ee}^T$ To give the close-form of $\det(\mathbf{G})$, where $\mathbf{G}$ is
\begin{align}
\mathbf{G}=a\mathbf{I}+b\boldsymbol{ee}^T
\end{align}
in which $a$ and $b$ are constant, and $\boldsymbol{e}$ is a column vector with all elements being $1$. In addition, $(\cdot)^T$ is transposition operation. $\mathbf{G}$ is $u\times u$. We use $\mathbf{G}_u$ to underline the dimension of $\mathbf{G}$.
The question is to determine $\det(\mathbf{G})$.
As I know: We rewrite $\mathbf{G}_u$ as
\begin{align}
\mathbf{G}_u=\left[\begin{array}{ccc}
\mathbf{G}_{u-1} & b\\
b & a+b
\end{array}
\right]
\end{align}
Using the determinant of block matrix lemma
\begin{align}
\det\left[\begin{array}{ccc}
\mathbf{A} & \mathbf{B}\\
\mathbf{C} & \mathbf{D}
\end{array}
\right]=\det(\mathbf{A})\det(\mathbf{D}-\mathbf{C}\mathbf{A}^{-1}\mathbf{B})
\end{align}
We then have
\begin{align}
\det(\mathbf{G}_u)=\det(\mathbf{G}_{u-1})\det(a+b-b^2\boldsymbol{e}^T\mathbf{G}_{u-1}^{-1}\boldsymbol{e})
\end{align}
It still needs to get $\mathbf{G}_{u-1}^{-1}$ via matrix inverse lemma
\begin{align}
(\mathbf{A}+\mathbf{BC})^{-1}=\mathbf{A}^{-1}-\mathbf{A}^{-1}\mathbf{B}(\mathbf{I}+\mathbf{CA}^{-1}\mathbf{B})^{-1}\mathbf{CA}^{-1}
\end{align}
We then have
\begin{align}
\mathbf{G}_{u}
&=\frac{1}{a}\mathbf{I}-\frac{1}{a^2}b\boldsymbol{e}\left(1+\frac{b}{a}\boldsymbol{e}^T\boldsymbol{e}\right)^{-1}\boldsymbol{e}^T\\
&=\frac{1}{a}\mathbf{I}-\frac{b}{a(a+bu)}\boldsymbol{ee}^T
\end{align}
where $\boldsymbol{e}^T\boldsymbol{e}=u$ is used. Plugging it into
\begin{align}
&\det(a+b-b^2\boldsymbol{e}^T\mathbf{G}_{u-1}^{-1}\boldsymbol{e})\\
=&a+b-b^2\left({\frac{u-1}{a}-\frac{b(u-1)^2}{a[a+b(u-1)]}}\right)
\end{align}
Then
\begin{align}
\det(\mathbf{G}_u)=\det(\mathbf{G}_{u-1})\left[{a+b-b^2\left({\frac{u-1}{a}-\frac{b(u-1)^2}{a[a+b(u-1)]}}\right)}\right]
\end{align}
Although, the connection between $\mathbf{G}_u$ and $\mathbf{G}_{u-1}$ is found, I can't give the expression of $\mathbf{G}_u$. Please give me hand, thanks a lot!
| Note that $G$ is a circulant matrix, so we know the eigenvectors are $(1,\zeta,\zeta^2,\dots,\zeta^{u-1})$ where $\zeta^u=1$. Its eigenvalue is
$$ a+b\sum_{j=0}^{u-1} \zeta^j=
\begin{cases}
a & \text{if }\zeta\neq 1\\
a+bu & \text{if }\zeta=1
\end{cases} $$
and so the determinant is $a^{u-1}(a+bu)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2916569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
} |
Is the sequence $a_{n} = \frac {1}{2^2} + \frac{2}{3^2} + .... +\frac{n}{(n+1)^2}$ Cauchy? First:
I tried substituting natural numbers for $n$ to calculate the consecutive terms of the sequence and then see the difference between their values and I found that the difference is decreasing for large values of $n$ (not very large because I am calculating by my hand)so I concluded that it is a Cauchy sequence but unfortunately when I looked at the hint for solving this problem I found it to be:
"Prove that $a_{2n} - a_{n} \geq n * \frac{2n}{(2n +1)^2} \geq \frac{2}{9}$" so I concluded that it is not Cauchy. So can anyone tell me please why testing by numbers sometimes lead to wrong values and when it is preferable to use this test?
Second:
If $a_{n} = \frac {1}{2^2} + \frac{2}{3^2} + .... +\frac{n}{(n+1)^2}$, Does $a_{2n} = \frac {1}{2^2} + \frac{2}{3^2} + .... +\frac{2n}{(2n+1)^2}$? because I am confused what it equals.
I hope my question fulfills the requirements of a good question, if no please let me know.
| "I tried substituting natural numbers for n to calculate the consecutive terms of the sequence and ...found that the difference is decreasing"
It doesn't matter if the difference between consecutive terms get smaller (and converge to zero). The terms $a_n$ and $a_{n+1}$ may get minisculely close together but in all the terms further down $a_{10^{\text{500 hundred quadrillion zillion oogleplex}}}$ can still be huge.
What does matter is that as $N$ gets large then the difference between any two terms (not just consecutive terms). To be Cauchy $a_n$ and $a_{10^{\text{500 hundred quadrillion zillion oogleplex}}}$ must be close together if $n, 10^{\text{500 hundred quadrillion zillion oogleplex}}$ are both greater than $N$.
So it doesn't matter of $a_{n+1} - a_n = \frac {n+1}{(n+2)^2} \to 0$. We need to have as $a_m - a_n = \frac {n+1}{(n+2)^2} + \frac {n+2}{(n+3)^2}+ \frac {n+3}{(n+4)^2}+ ....... + \frac {m}{(m+1)^2}\to 0$ as $n\to \infty$. As $a_m - a_n = \frac {n+1}{(n+2)^2} + \frac {n+2}{(n+3)^2}+ \frac {n+3}{(n+4)^2}+ ....... + \frac {m}{(m+1)^2}$ may have many, many,many terms this is a very tall order.
And it doesn't hold true.
$a_{2n} - a_n = \frac {n+1}{(n+2)^2} + \frac {n+2}{(n+3)^2} + ..... + \frac {2n}{(2n+1)^2} >$
$\frac {2n}{(2n+1)}^2 +\frac {2n}{(2n+1)}^2+.... + \frac {2n}{(2n+1)}^2 =$
$n*\frac {2n}{(2n+1)}^2 = \frac {2n^2}{4n^2 + 4n + 1} >$
$\frac {2n^2}{4n^2 + 4n^2 + n^2} =\frac {2n^2}{9n^2}= \frac 29$.
This means you will always be able to find to terms that are at least $\frac 29$ apart. No matter how large we take $N$ we can always find $n > N$ and $2n > N$ so that $a_{2n} - a_n > \frac 29$. So it is not true that the difference of any two terms goes to $0$.
So the sequence is not Cauchy.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A ring such that $(a+b)^2=a^2+b^2$ and $(a+b)^3=a^3+b^3$
Let $(A,+,\cdot)$ be a ring such that there are $a,b \in A$ which satisfy $$(a+b)^2=a^2+b^2, \quad (a+b)^3=a^3+b^3$$
Prove that $(a+b)^n=a^n+b^n,$ for all positive integers $n.$
I have found the following solution, but I am not quite satisfied with it.
From the hypothesis we get $ab+ba=0$ and $ab^2+ba^2=0.$ We will prove the identity using induction. Suppose that it is true for $1,2,...,n-1, \: n \geq 4.$ We can write $$(a+b)^n=(a+b)^{n-1}(a+b)=(a^{n-1}+b^{n-1})(a+b)=a^n+a^{n-1}b+b^{n-1}a+b^n$$
It is left to prove that $a^{n-1}b+b^{n-1}a=0.$ We can write $$a^{n-1}b+b^{n-1}a=a^{n-2}ab+b^{n-2}ba=-a^{n-2}ba-b^{n-2}ab \quad (*)$$
But $(a+b)^{n-1}=(a+b)^{n-2}(a+b)=(a^{n-2}+b^{n-2})(a+b)=a^{n-1}+b^{n-1},$ so $$a^{n-2}b+b^{n-2}a=0 \Rightarrow b^{n-2}a=-a^{n-2}b$$
Plugging this back in $(*)$ gives $$a^{n-1}b+b^{n-1}a=-a^{n-2}ba+a^{n-2}b^2=a^{n-2}(-ba+b^2)=a^{n-3}ab(-a+b)$$
But $0=ab^2+ba^2=ab^2-aba=ab(b-a),$ so $$a^{n-1}b+b^{n-1}a=a^{n-3}\cdot 0 = 0$$ and this completes the indution.
Is there any other solution, maybe quicker or more beautiful?
| This is the same, but maybe rearranged to see an idea of norming (non-commutative) monomials in $a,b$. As noticed in the OP, from $a^2+b^2=(a+b)^2=(a+b)(a+b)=a^2+ab+ba+b^2$ we get the "supercommutativity relation"
$$
ab= -ba\ .
$$
The other relation, extended as $a^3+b^3=(a+b)^3=(a+b)^2(a+b)=(a^2+b^2)(a+b)=a^3+a^2b+b^2a+b^3$ gives as in the OP
$$
a^2b=-b^2a\ .
$$
We start now a "new" proof. (Essentially the same.) The strategy being to write each monomial in $a,b$ in a "normalized form". First, using supercommutativity we can write any monomial $aa\dots abb\dots baa\dots a bb\dots b\dots$ in the form $\pm aa\dots a\ aa\dots a\ \dots bb\dots b\ bb\dots b\dots$ by pushing all $a$'s in the front and changing signs. Using the second rule, we can reduce the $a$ powers in front of the $b$'s to get a "normalized" monomial of the shape $\pm b^?a$. Let us show inductively the relation $a^nb=-b^na$, starting from the given one for $n=2$. For $n\ge 2$ we have
$$
\begin{aligned}
a^{n+1}b
&=
aa^n b\\
&=
-ab^n a&&\text{(by induction)}\\
&=
-(-1)^n aa b^n&&\text{(by supercommutativity)}\\
&=-(-1)^n a^2 \underbrace{bbb\dots b}_{n\text{ times}}\\
&=-(-1)^n(- b^2 a)\underbrace{bb\dots b}_{n-1\text{ times}}\\
&=+(-1)^n b^2\ (a\underbrace{bb\dots b}_{n-1\text{ times}})\\
&=+(-1)^n b^2\ (-1)^{n-1}\underbrace{bb\dots b}_{n-1\text{ times}}a&&\text{(by supercommutativity)}\\
&=-b^{n+1}a\ .
\end{aligned}
$$
The wanted relation now follows also inductively,
$$
(a+b)^{n+1}=(a^n+b^n)(a+b)=a^{n+1}+\underbrace{a^nb+b^na}_{=0\text{ shown above}}+b^{n+1}
=
a^{n+1}+b^{n+1}
\ .$$
| {
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"url": "https://math.stackexchange.com/questions/2920135",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "24",
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} |
Coefficient of a generating function I have the following generating function,
$$f(x)=(1+x+x^2+x^3)^4$$
I want to find the $[x^5]$ coefficient, to do so I wrote,
$$[x^5]f(x)=[x^5](1+x+x^2+x^3)^4=[x^5]\left(\frac{1-x^4}{1-x}\right)^4=
[x^5](1-x^4)^4(1-x)^{-4}
$$
I am wondering how to proceed from here?
| Write the two binomials in terms of series:
$$[x^5](1-x^4)^4(1-x)^{-4}=[x^5]\sum_{i=0}^{4} {4\choose i}(-x^4)^i \cdot \sum_{j=0}^{\infty}{-4\choose j}(-x)^j=\\
[x^5]{4\choose 0}(-x^4)^\color{green}{0}\cdot \color{red}{{-4\choose 5}}(-x)^\color{green}5+{4\choose 1}(-x^\color{green}4)^\color{green}1\cdot \color{blue}{{-4\choose 1}}(-x)^\color{green}1=\\
[x^5]\color{red}{{4+5-1\choose 5}(-1)^5}(-x)^5-4x^4\color{blue}{{4+1-1\choose 1}(-1)^1}(-x)^1=\\
{8\choose 5}-4{4\choose 1}=40,$$
where it was used negative binomial series:
$$(x+a)^{-n} = \sum_{k=0}^{\infty} {-n\choose k}x^ka^{-n-k};\\
{-n\choose k}={n+k-1\choose k}(-1)^k.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Existence proof of infinite solutions of a diophantine equation Prove by construction that $x(x+1)=2 y^2$ has an infinite number of solutions where $x$ and $y$ are both positive whole numbers. Do not use an iterative procedure like that for solutions of Pell's equation in your proof.
Background: I came up with a proof of this years ago in which $x$ was a multiple of $32$ and a quadratic or quartic polynomial of an integer parameter. Given that, $x(x+1)/2$ was guaranteed to be a perfect square. I've since encountered another, probably better proof the above but efforts to find my old one have been in vain.
Proof using Pell:
1) Complete the square of the $x$ Side of the equation to get $x^2+x+1/4=2y^2+1/4$.
2) This is equal to $(x+1/2)^2=2y^2+1/4$.
3) Multiply both sides by $4$ to get: $(2x+1)^2=8y^2+1$.
4) Let $m=2x+1$ and $n=2y$. Then we have $m^2-2n^2=1$, Pell's Equation.
5) Given solution ($m,n$), construct a symmetric $2x2$ matrix with ($m,n \, \sqrt{2}$) as the first row.
6) The square of a symmetric matrix is symmetric. The determinant of the square of a matrix is the square of that matrix's determinant. We thus end up with two new integers in the first row ($m^2+2n^2, 2 \, m \, n$) which are again solutions of Pell's equation.
7) This can be iterated to generate a sequence of solutions.
8) $x$ and $y$ can be represented in terms of $m$ and $n$, so we thus have an infinite number of solutions for $x$ and $y$.
| If you have one solution to
$x^2-dy^2 = 1$,
by using
$\begin{array}\\
(x^2-dy^2)(u^2-dv^2)
&=x^2u^2-d(x^2v^2+y^2u^2)+d^2y^2v^2\\
&=x^2u^2+d^2y^2v^2\pm 2dxuyv-d(x^2v^2+y^2u^2\pm 2xuyv)\\
&=(xu\pm dyv)^2-d(xv\pm yu)^2\\
\end{array}
$
you can construct
an arbitrary number of them.
When $d=8$,
since
$3^2-8\cdot1^2 = 1$,
from this first solution
you can get
$u^2-8v^2
=(3u\pm 8v)^2-8(3v\pm u)^2
$.
Choosing "+"
gives Will Jagy's iteration.
Wow, is this not original.
| {
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"timestamp": "2023-03-29T00:00:00",
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Does the fact that $x^2=(x-1)(x+1)+1$ have a name? Just curious about this pattern
$$x^2 = (x-1)(x+1) +1$$
So:
$$\begin{align}
1^2 &= \phantom{1}0\cdot\phantom{1}2+1 = 1 \\
2^2 &= \phantom{1}1\cdot\phantom{1}3+1 = 4 \\
3^2 &= \phantom{1}2\cdot\phantom{1}4+1 = 9 \\
4^2 &= \phantom{1}3\cdot\phantom{1}5+1 = 16 \\
9^2 &= \phantom{1}8\cdot10+1 = 81 \\
15^2 &= 14\cdot16+1 = 225
\end{align}$$
and so on.
Then, to know any number raised to the power of $2$, you can multiply the previous number $(x-1)$ by the next one $(x+1)$, and add $1$.
So, to know the squared root of a number like $64$, you have to substract $1$ ($63$) and look for two numbers multiplied are $63$ and subtracted are $2$:
$$x \cdot y = 63 \qquad x - y = 2$$
Solving the equation you get $9$ and $7$ (or $-7$ and $-9$). The number between these is the square root ($8$).
I don't know if this apply for power of $3$.
Does this fact/theorem/relation has a name or something?
| Trickery;
$x^2= x^2 -1 +1 = (x^2-1) +1=$
$ (x+1)(x-1)+1$.
Used: $x^2-1=(x+1)(x-1)$.
P.S. Example :
$19^2= (20)(18)+1 = 360 +1=$
$ 361$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Evaluate $\ \int_{|z|=2}\frac{z^2+1}{(z-3)(z^2-1)}\ dz$
I am trying to solve $$\ \int_{|z|=2}\frac{z^2+1}{(z-3)(z^2-1)}\ dz,$$
using Cauchy's Integral Formula.
My attempt:
$$I=\int_{|z|=2}\frac{z^2+1}{(z-3)(z^2-1)}\ dz=\int_{|z|=2}\frac{z^2+1}{(z-3)(z-1)(z+1)}\ dz$$
Now, $I$ has singularities $\pm 1, 3$, but only $\pm 1$ lie inside the circle $|z|=2$. Now if we draw the following
We see that
$f(z)=\frac{z^2+1}{(z-3)(z-1)}$ and $g(z)=\frac{z^2+1}{(z-3)(z+1)}$ are holomorphic outside the green and blue ball respectively.
Now, i've been told that we can use a corollary of the Cauchy-Goursat theorem to show that $$I=\int_{a} \frac{f(z)}{z+1}\ dz+\int_{b} \frac{g(z)}{z-1}\ dz,$$ where $a$ and $b$ are the contours of the green and blue cirlces respectively. But I don't understand why this is? I can solve beyond this point.
| *
*Method 1: Using Partial fractions:
$$ \int_{|z|=2}\frac{z^2+1}{(z-3)(z^2-1)} dz = \int_{|z|=2}\frac{-\frac12}{z-1} + \frac{\frac14}{z+1} + \frac{\frac54}{z-3} dz$$
$$ = \int_{|z|=2}\frac{-\frac12}{z-1}dz + \int_{|z|=2}\frac{\frac14}{z+1}dz + \int_{|z|=2}\frac{\frac54}{z-3} dz$$
$$ = -\frac12 (2\pi i) + \frac14 (2\pi i) + \frac54 (0) = \frac{-\pi i}{2}$$
*Method 2: Boundaries of the balls, $\gamma_1$ and $\gamma_2$, respectively that you drew:
$$ \int_{|z|=2}\frac{z^2+1}{(z-3)(z^2-1)} dz$$
$$ = \int_{\gamma_1}\frac{z^2+1}{(z-3)(z^2-1)} dz + \int_{\gamma_2}\frac{z^2+1}{(z-3)(z^2-1)} dz + \int_{\text{the boundary of the region inside} \ |z|=2 \ \text{but outside} \ \gamma_1 \ \text{and} \ \gamma_2}\frac{z^2+1}{(z-3)(z^2-1)} dz$$
$$ = \int_{\gamma_1}\frac{z^2+1}{(z-3)(z^2-1)} dz + \int_{\gamma_2}\frac{z^2+1}{(z-3)(z^2-1)} dz + 0 \tag{1}$$
$$ = \int_{\gamma_1} \frac{\frac{z^2+1}{(z-3)(z-1)}}{(z+1)} dz + \int_{\gamma_2} \frac{\frac{z^2+1}{(z-3)(z+1)}}{(z-1)} dz$$
$$ = \frac{z^2+1}{(z-3)(z-1)}|_{z=-1} (2\pi i) + \frac{z^2+1}{(z-3)(z+1)}|_{z=1} (2\pi i)$$
$$ = \frac{2}{(-4)(-2)} (2\pi i) + \frac{2}{(-2)(2)} (2\pi i) = \frac{-\pi i}{2}$$
*Method 3: Using Residue:
$$ \int_{|z|=2}\frac{z^2+1}{(z-3)(z^2-1)} dz = (2\pi i) \text{Res}_{\{z=1\}}(\frac{z^2+1}{(z-3)(z^2-1)}) + (2\pi i) \text{Res}_{\{z=-1\}}(\frac{z^2+1}{(z-3)(z^2-1)})$$
$$ = (2\pi i) \lim_{z \to 1} (z-1)(\frac{z^2+1}{(z-3)(z^2-1)}) + (2\pi i) \lim_{z \to -1} (z+1)(\frac{z^2+1}{(z-3)(z^2-1)}) = \frac{-\pi i}{2}$$
$(1)$ is because "a corollary of the Cauchy-Goursat theorem" states that
$$0= \int_{\gamma} f(z) dz$$ if $f$ is holomorphic in the region bounded by $\gamma$ under suitable conditions of $\gamma$, where conditions on the region and $\gamma$ may vary by book.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving $\sqrt{8-x^2}-\sqrt{25-x^2}\geq x$
I would like to find the solution of
$$\sqrt{8-x^2}-\sqrt{25-x^2}\geq x.$$
My try:
First I used the hint of this answer.
$$ \frac{8-x^2-25+x^2}{\sqrt{8-x^2}+\sqrt{25-x^2}}\geq x \leftrightarrow \frac{-17}{\sqrt{8-x^2}+\sqrt{25-x^2}}\geq x.$$
Then the solution can be found by
$$\left(-17\right)^2\geq \left(x\sqrt{8-x^2}+x\sqrt{25-x^2}\right)^2.$$
But I think this is not the best approach.
| Unfortunately the approach I did there is not useful here. However, here is another approach:
Write the inequality as
$$\sqrt{8-x^2}-x\geq\sqrt{25-x^2}. \tag{1}$$
Now use the Cauchy–Bunyakovsky–Schwarz inequality,
$$(x_1y_1+x_2y_2+\cdots+x_ny_n)^2\leq\left(x_1^2+x_2^2+\cdots+x_n^2\right)\left(y_1^2+y_2^2+\cdots+y_n^2\right),$$
to evaluate the left-side of $(1)$. Using this formula, one gets
$$\left(\sqrt{8-x^2}-x\right)^2\leq\left(1+1\right)\left(8-x^2+x^2\right)=16. $$
Notice we must have $-2\sqrt{2}\leq x \leq 2\sqrt{2}$ since, see Doug M's answer, $>$ and $<$ are not defined concepts over the complex numbers (sorry for that). Then, the right-side of $(1)$ takes
$$\sqrt{25-x^2}\geq \sqrt{25-8}=\sqrt{17}.$$
Futher, notice also that the left-side of $(1)$ is less than or equal to $4$, and that the right-side is greater than or equal to $4$. Therefore, there's no real solution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Positive Definite Matrix (Block Matrix) Let $B$ be an $m\times n$ matrix. Is $$ A=\begin{pmatrix} I & B \\ B^T & I+B^TB \end{pmatrix} $$ positive definite?
Attempt:
Let $\mathbf{z}=\begin{pmatrix} \mathbf{x} \\ \mathbf{y} \end{pmatrix}$. To show that $A$ is positive definite, $\mathbf{z}^TA\mathbf{z}>0$. Expanding $\mathbf{z}^TA\mathbf{z}>0$ gives $\mathbf{x}^T\mathbf{x}+\mathbf{y}^T\mathbf{y}+(B\mathbf{y})^T(B\mathbf{y})+2\mathbf{x}^TB\mathbf{y}$. The first three terms are positive, but what can be concluded about the $2\mathbf{x}^TB\mathbf{y}$ term?
| One hint could be to try to rewrite with sum of self-outer products:
$$A = \begin{bmatrix}I&B\\B^T&I+B^TB\end{bmatrix}= \cdots\\=\begin{bmatrix}I\\B^T\end{bmatrix} \begin{bmatrix}I&B\end{bmatrix} + \begin{bmatrix}0\\I\end{bmatrix} \begin{bmatrix}0&I\end{bmatrix}$$
What can we say about it now?
| {
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"timestamp": "2023-03-29T00:00:00",
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First order differential equation solution check. I have an equation:
$\frac{dy}{y}=(0.0001x^2+0.005x)dx$
$y=ce^{(0.00003x+0.0025)x^2}$
Now I try to substitute $y$ in the euqation and I am getting
$(\frac{1}{ce^{(0.00003x+0.0025)x^2}})'=0.0001x^2+0.005x$
$(\frac{1}{ce^{(0.00003x+0.0025)x^2}})'=(\frac{1}{c}e^{(-0.00003x-0.0025)x^2})'=(\frac{1}{c}e^{(-0.00003x-0.0025)x^2})(-0.0001x^2-0.005x)$
But
$(\frac{1}{c}e^{(-0.00003x-0.0025)x^2})(-0.0001x^2-0.005x) \neq 0.0001x^2+0.005x$
What am I missing here while checking the solution?
| The equation is of the form:
$$ \frac{\mathrm{d}y}{y} = (ax^2+bx)\mathrm{d}x $$
where $a=0.0001$ and $b=0.005$. Integrating both sides gives:
$$ \int\frac{\mathrm{d}y}{y} = \int(ax^2+bx)\mathrm{d}x $$
The solution is
$$ \ln y = \frac{a}{3}x^3+\frac{b}{2}x^2+c$$
Therefore, in terms of $y$:
$$ y = C\mathrm{e}^{(\frac{a}{3}x+\frac{b}{2})x^2}, \label{eq1} \tag{1}$$
where $C=\mathrm{e}^c$. We can test the solution in the following way:
$$\begin{align}
\frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\mathrm{d}}{\mathrm{d}x}\left [ C\mathrm{e}^{(\frac{a}{3}x+\frac{b}{2})x^2} \right]\\
&=\frac{\mathrm{d}}{\mathrm{d}x}\left [ \frac{a}{3}x^3+\frac{b}{2}x^2 \right]C\mathrm{e}^{(\frac{a}{3}x+\frac{b}{2})x^2} \\
&=\left ( ax^2+bx^2 \right)C\mathrm{e}^{(\frac{a}{3}x+\frac{b}{2})x^2}
\end{align}.$$
Using $\eqref{eq1}$:
$$
\frac{\mathrm{d}y}{\mathrm{d}x} =\left ( ax^2+bx^2 \right)y
$$
Therefore:
$$
\frac{\mathrm{d}y}{y} =\left ( ax^2+bx^2 \right) \mathrm{d}x
$$
Regarding the numeric values: $a=0.0001=\frac{1}{10000}$, then $\frac{a}{3}=\frac{1}{30000}\neq 0.00003$. On the other hand,
$$
\frac{\mathrm{d}y}{y} \neq \frac{1}{y'}
$$
The right way to view the equation is the following: The differential of a function $y(x)$ is given by
$$
\mathrm{d}y=y'(x)\mathrm{d}x
$$
Divide each side of the equation by $y(x)$
$$
\frac{\mathrm{d}y}{y}=\frac{y'(x)}{y}\mathrm{d}x.
$$
Then $\frac{y'(x)}{y}$ should give you $0.0001x^2+0.005x$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Does the elliptic curve $y^2 = 4 x^3 -6075$ have any integer points? Let $E$ be the elliptic curve $y^2 = 4 x^3 -6075$. I ran the following Mathematica code, which searches naively for integer solutions to $E$ but it did not find any solutions $(x, y) \in E(\mathbb{Z})$ satisfying $0 \leq x \leq 10^6$.
T = Table[ z = 4 x^3 - 6075;
If[ IntegerQ[Sqrt[z]],
{x, Sqrt[z]}
, 0]
, {x, 0, 1000000}];
DeleteCases[T, 0]
Is $E(\mathbb{Z})$ empty? Also, what is the easiest database on Mordell curves to access? I have had some trouble installing the Pari database in the past.
| No, $y^2=4x^3-6075$ has no integer solution.
An elementary observation: $3\nmid x$: because $$3\mid x \implies 27\mid y^2 \implies 9\mid y \implies
9\mid x$$ but $y^2+6075$ has no solution in $\mathbb{Z}/3^6\mathbb{Z}$.
The ring of integer of $\mathbb{Q}(\sqrt{-3})$ is a UFD. The equation can be written as
$$\left( {\frac{{y + 45\sqrt { - 3} }}{2}} \right)\left( {\frac{{y - 45\sqrt { - 3} }}{2}} \right) = {x^3}$$
elements inside brackets, denoted by $\alpha$ and $\beta$, are integral over $\mathbb{Z}$ since $y$ is odd.
I claim that $\alpha,\beta$ are relatively prime. If a prime $\pi$ divides both, then $\pi\mid 45\sqrt{-3}$, so $\pi = \sqrt{-3}$ or $5$. If $\pi = \sqrt{-3}$, then $3\mid x$, contradiction. If $\pi = 5$,
let $v_5$ denote valuation at $5$, normalized so that $v_5(5)=1$, note that $v_5(\alpha) \in \mathbb{Z}$ as $v_5$ is unramified. $$0< v_5(\alpha)+v_5(\beta) = 2v_5(\alpha)= 3v_5(x) $$
this says $v_5(x)$ is even, hence $5^6 \mid (y^2+6075)$, but $y^2+6075=0$ has no solution in $\mathbb{Z}/5^6\mathbb{Z}$, ruling out $\pi = 5$. Valuation can be saved by noting that $5^3 \mid (y^2+6075)$ is already impossible, but it seems difficult to deduce the stronger $5^6 \mid (y^2+6075)$ from consideration in $\mathbb{Z}$ alone.
Since $\alpha,\beta$ are relatively prime, they are both cube, say
$$\frac{{y + 45\sqrt { - 3} }}{2} = {(\frac{{a + b\sqrt { - 3} }}{2})^3} \qquad \text{ or } \qquad \left( {\frac{{1 + \sqrt { - 3} }}{2}} \right){\left( {\frac{{a + b\sqrt { - 3} }}{2}} \right)^3}$$
with $a,b$ both even or odd. The first case gives $60= {a^2}b - {b^3}$, so $b$ has only a few possible values, checking them gives no integer solutions to original equation. The second case gives $$\frac{{45}}{2} = \frac{{{a^3} + 3{a^2}b - 9a{b^2} - 3{b^3}}}{{16}}$$ this says $v_3(a) \geq 1$, thus $v_3(a^3+3a^2b-9ab^2)\geq 3$ but $v_3(45/2) = 2$, so $v_3(3b^3) = 2$, absurd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2935205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Find the relationship of the length of triangle's sides.
Denote the three sides of $\triangle ABC$ to be $a,b,c$. And they satisfy $$a^2+b+|\sqrt{c-1}-2|=10a+2\sqrt{b-4}-22 $$
Now determine what kind of triangle $\triangle ABC$ is.
A.Isosceles triangle which its leg and base is not equal.
B.equilateral triangle
C.Right triangle
D.Isosceles Right triangle
The only information I got is from the number in the radical need to be greater than $0$. Then $b\ge4$ and $c\ge 1$. Also $10a+2\sqrt{b-4}-22\ge0 $. But they are all inequalities. What we need is some equalities. It would be great to have some hints.
| We have
$$|\sqrt{c-1}-2|=10a+2\sqrt{b-4}-22-a^2-b\tag1$$
from which
$$10a+2\sqrt{b-4}-22-a^2-b\ge 0\tag2$$
follows.
$(2)$ is equivalent to
$$a^2-10a+b-2\sqrt{b-4}+22\le 0,$$
i.e.
$$(a-5)^2-25+(b-4)+4-2\sqrt{b-4}+22\le 0,$$
i.e.
$$(a-5)^2+(\sqrt{b-4}-1)^2\le 0$$
from which
$$a-5=\sqrt{b-4}-1=0$$
i.e.
$$a=b=5$$
follows.
Now you can get $c$ from $(1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2935522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Divergence of infinite series $(\frac{3k-2}{4k+2})^{2k-3}$ I would like to ask if my solution for testing the divergence of the infinite series below is correct.
$$\sum_{k=1}^{\infty} \left(\frac{3k-2}{4k+2}\right)^{2k-3}$$
I used the Cauchy ratio test.
$$
\lim_{k \to \infty} \left(\frac{3k-2}{4k+2}\right)^{2k-3} =
\lim_{k \to \infty} \left(\frac{3k-2}{4k+2}\right)^{2k} \cdot \left(\frac{3k-2}{4k+2}\right)^{-3} =
\lim_{k \to \infty} \left(\frac{\left(3k-2\right)^{2}}{\left(4k+2\right)^{2}}\right)^{k} \cdot \left(\frac{3k-2}{4k+2}\right)^{-3} =
\lim_{k \to \infty}\left(\frac{9k^{2}-12k+4}{16k^2+16k+4}\right)^{k} \cdot \left(\frac{k\left(3-\frac{2}{k}\right)}{k\left(4+\frac{2}{k}\right)}\right)^{-3} =
\lim_{k \to \infty} \left(\frac{k^{2}\left(9-\frac{12}{k}+\frac{4}{k^{2}}\right)}{k^{2}\left(16+\frac{16}{k}+\frac{4}{k^{2}}\right)}\right)^{k} \cdot \left(\frac{3}{4}\right)^{-3} =
\lim_{k \to \infty} \frac{64}{27} \cdot \left(\frac{9}{16}\right)^{k}
$$
Therefore:
$$
\left\lvert \frac{\frac{64}{27} \cdot \left(\frac{9}{16}\right)^{k}}{\frac{64}{27} \cdot \left(\frac{9}{16}\right)^{k+1}} \right\rvert =
\left\lvert \frac{\frac{9^{k}}{16^{k}}}{\frac{9^{k+1}}{16^{k+1}}} \right\rvert =
\left\lvert \frac{9^{k}\cdot 16^{k+1}}{16^{k}\cdot 9^{k+1}} \right\rvert =
\left\lvert \frac{9^{k}}{9^{k+1}} \cdot \frac{16^{k+1}}{16^{k}} \right\rvert =
\left\lvert 9^{k-k+1}\cdot16^{k+1-k} \right\rvert =
\left\lvert 9\cdot16 \right\rvert =
144
$$
As $144 > 1$, the series is divergent.
Is this a correct solution?
| For this problem, the ratio test requires careful calculations with which you made some mistakes. The easiest approach is to note the $k$th term is positive but less than $(3/4)^{2k-3}$, so the series converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2936436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Determine $(n-1)$ eigenvectors of $f$ associated to $0$?
$$A=\begin{pmatrix}1&1&\cdots&1\\1&1&\cdots&1\\\vdots&\vdots&\ddots&1\\1&1&\cdots&1\end{pmatrix}\in\mathcal M_n(\mathbb R).$$
*
*Prove that $0$ and $n$ are eigenvalue of $A$.
*Determine the characteristic polynomial. (Hint: You can use the basis $(e_1, e_2 - e_1, \cdots, e_n - e_1)$)
*Determine the eigenvalues of $A$.
*Prove that $f(e_1)$ is an eigenvector.
*Determine $(n-1)$ eigenvectors of $f$ associated to $0$.
*Prove that there exists a basis $\mathcal{B_2}$ in which the matrix is of the form:
$$D = \begin{pmatrix}
n & 0 & \cdots & 0 \\
0 & 0 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & 0
\end{pmatrix}$$
*
*Taking the vector $ v =
\begin{pmatrix}
1 \\
1 \\
\vdots \\
1
\end{pmatrix}$
We get: $Av = nv$.
But I could not see how $0$ can be an eigenvalue.
*$$P_A(X) = \begin{vmatrix}
1 - X & 1 & \cdots & 1 \\
1 & 1 - X & \cdots & 1 \\
\vdots & \vdots & \ddots & 1 \\
1 & 1 & \cdots & 1 - X
\end{vmatrix}
=
(1 - X)\begin{vmatrix}
1 & 1 & \cdots & 1 \\
1 & 1 - X & \cdots & 1 \\
\vdots & \vdots & \ddots & 1 \\
1 & 1 & \cdots & 1 - X
\end{vmatrix}$$
I could not proceed to get an expression of $P_f(X)$ as I could not see how to use the hint.
*Computing $ Ae_1 = \begin{pmatrix}
1 \\
1 \\
\vdots \\
1
\end{pmatrix}$ which is not related to the eigenvalues $0$ or $n$.
*I am stuck in this question too.
| Let $f_1 = e_1$, $f_2 = e_2 - e_1$, ..., $f_n = e_n - e_1$ as in the hint. Note that $\sum f_i = (\sum e_i) - (n-1) e_1$.
Then $A$ annihilates $f_2$, $f_3$, ..., and $f_n$. Meanwhile,
\begin{align*}
Af_1 &= \sum_i {e_i} \\
&= \sum_i f_i + (n-1) e_1 \\
&= nf_1 + \sum_{i \neq 1} f_i,
\end{align*}
so the matrix for $A$ in the new basis is
$$
\begin{pmatrix}
n & 0 & 0 & \ldots & 0 \\
1 & 0 & 0 & \ldots & 0 \\
1 & 0 & 0 & \ddots & 0 \\
1 & 0 & 0 & \ldots & 0 \\
\end{pmatrix}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2941464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the slant asymptote for $y=\frac{x^2\arctan\left(x\right)}{3x+3}$ I got stuck at the end and wonder if you can give some guide on how to proceed to find the slant asymptote?
$$y=\frac{x^2\arctan\left(x\right)}{3x+3}$$
$$y= kx+m\\
\frac{f(x)}{x} \rightarrow k\\
f(x) - kx \rightarrow m $$
Solution:
$$\lim _{x\to \infty }\left(\frac{\frac{x^2\arctan\left(x\right)}{3x+3}}{x}\:\right)=\lim _{x\to \infty }\left(\frac{x^2\arctan\left(x\right)}{x(3x+3)}\:\right)=\lim _{x\to \infty }\left(\frac{x^2\arctan\left(x\right)}{(3x^{2\:\:}+3x)}\:\right)\\
=\lim _{x\to \infty }\left(\frac{x^2}{x^2} \cdot\frac{\arctan\left(x\right)}{(3+\frac{3}{x})}\:\right)=\frac{\pi}{6}\Longrightarrow k=\frac{\pi}{6}\\
f(x)-kx \implies \frac{x^2\arctan\left(x\right)}{3x+3} - \frac{\pi}{6} \cdot {x}\\
\frac{6\cdot x^2\arctan -3\pi{x^2} - 3\pi {x}}{18(x+1)}$$
Searching for $m$.
| hint
Put $y=\frac 1x$ and use the fact that
for $y>0$, we have
$$\arctan(\frac 1y)=\frac {\pi}{2}-\arctan(y)$$
use L'Hopital rule to compute the limit when $y\to 0^+$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2943785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
What is $15^{15} + 16^{16} + 17^{17} + 18^{18} + 19^{19} + 20^{20} \pmod{7}$? I am trying to evaluate $15^{15} + 16^{16} + 17^{17} + 18^{18} + 19^{19} + 20^{20} \pmod{7}$. I have found that $15^{15} \equiv 1 \pmod{7}$ and that $16^{16} \equiv 2 \pmod{7}$.
To evaluate $15^{15} \pmod{17}$, I did the following:
$$15 = 2 \times 7 + 1 \equiv 1 \pmod{7}$$
$$15^{15} \equiv 1 \pmod{7}$$
Then, to evaluate $16^{16}$, I wrote:
$$16 = 15 + 1 \equiv 1 + 1 = 2 \pmod{7}$$
$$16^{16} \equiv 2^{16} \pmod{7}$$
$$2^{3} = 8 = 7+1 \equiv 1 \pmod{7}$$
$$2^{16} = 2^{3} \times 2^{13} \equiv 2^{13} = 2^{3} \times 2^{10} \equiv 2^{10} \equiv \dots \equiv 2 \pmod{7}$$
Nevertheless, I have not managed to figure out how to evaluate $17^{17}$. How should I go about this and is my overall approach for evaluating the sum in question a good one?
| You can go with the same process I think:
$$17^{17} \equiv 3^{17} \mod{7}$$
and we have
$$3^1 \equiv 3 \mod{7}$$
$$3^2 \equiv 2 \mod{7}$$
$$3^3 \equiv 6 \mod{7}$$
$$3^4 \equiv 4 \mod{7}$$
$$3^5 \equiv 5 \mod{7}$$
$$3^6 \equiv 1 \mod{7} \text{ (We could have this by using Fermat's Little Theorem as well)}$$
From here, we can conclude that $3^{17} \equiv 5 \mod{7}$.
Then, notice that
$$18 \equiv 4 \equiv -3 \mod{7}$$
$$19 \equiv 5 \equiv -2 \mod{7}$$
$$20 \equiv 6 \equiv -1 \mod{7}$$
which means you can find other terms by using your previous results for $15,16$ and $17$.
But as suggested on comments, a better way is to use Fermat's Little Theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2944232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
placing balls into compartments
Suppose we have $n$ balls that are randomly distributed into $N$
compartments. Find the probability that $m$ balls will fall into the
first two compartments. Assume all $N^n$ arrangements are equally
likely.
Try.
First, we select $m$ balls out of $n$, this we can do in ${n \choose m}$ ways and we want to put them in the first two compartments. That is, we have ${n \choose m}$ choices for both compartments. Therefore, we can do this in ${n \choose m}^2$ ways. Now, the remaining $n-m$ balls must be placed in the remaining $N-2$ compartments. This we can do in $(N-2)^{n-m}$ ways. So probability must be
$$ \frac{ {n \choose m}^2 \cdot (N-2)^{n-m} }{N^n } $$
However, I have some doubts.
If we assume that $m$ are to be distributed to any $2$ compartments, not just the first $2$, then we would have an extra ${N \choose 2}$ term in the above, is this correct?
| There are $\binom{n}{m}$ ways to select which $m$ of the $n$ balls will be in the first two compartment and $2^m$ ways to distribute them to those two compartments. The remaining $n - m$ balls can be distributed to the remaining $N - 2$ compartments in $(N - 2)^{n - m}$ ways. Hence, the number of favorable cases is
$$\binom{n}{m}2^m(N - 2)^{n - m}$$
Since there are $N^n$ ways to distribute the balls, the probability that exactly $m$ of the $n$ balls land in the first two of the $N$ compartments is
$$\frac{\binom{n}{m}2^m(N - 2)^{n - m}}{N^n}$$
Sanity Check: Suppose $N = 5$ and $n = 3$. Then we should have $$\Pr(m = 0) + \Pr(m = 1) + \Pr(m = 2) + \Pr(m = 3) = 1$$
Our formula gives
\begin{align*}
\Pr(m = 0) & = \frac{\binom{3}{0}2^0(5 - 2)^{3 - 0}}{5^3} = \frac{1 \cdot 1 \cdot 27}{125} = \frac{27}{125}\\
\Pr(m = 1) & = \frac{\binom{3}{1}2^1(5 - 2)^{3 - 1}}{5^3} = \frac{3 \cdot 2 \cdot 9}{125} = \frac{54}{125}\\
\Pr(m = 2) & = \frac{\binom{3}{2}2^2(5 - 2)^{3 - 2}}{5^3} = \frac{3 \cdot 4 \cdot 3}{125} = \frac{36}{125}\\
\Pr(m = 0) & = \frac{\binom{3}{3}2^3(5 - 2)^{3 - 3}}{5^3} = \frac{1 \cdot 8 \cdot 1}{125} = \frac{8}{125}
\end{align*}
so
$$\Pr(m = 0) + \Pr(m = 1) + \Pr(m = 2) + \Pr(m = 3) = \frac{27 + 54 + 36 + 8}{125} = 1$$
as expected.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2946710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Simplify the Dot Product in terms of $a$ and $b$ Where $a$ and $b$ are arbitrary vectors
$(a+2b) \cdot (2a-b)$
$$a\cdot(2a-b)+2b\cdot(2a-b) = 2(a\cdot a)-a\cdot b+1(b\cdot a)-2(b\cdot b)$$
$$=2(a)-ab+4ab-2(b)^2$$
$$=2a^2-2b^2$$
$$=2(a^2-b^2)$$
Where did i go wrong in simplifying this?
| I marked the mistakes in red and corrected these
\begin{align}
(a+2b) * (2a-b)& =a*(2a-b)+2b*(2a-b)\\
& = 2(a*a)-a*b+\color{red}{4}(b*a)-2(b*b)\\
& = 2(a)\color{red}{^2}\underbrace{-ab+4ab}_{\color{red}{3ab}}-2(b)^2\\
& = 2a^2-2b^2+\color{red}{3ab}\\
& = 2(a^2-b^2)+\color{red}{3ab}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2948870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $7|13\cdot 6^{n}+8\cdot 13^{n}$ for all natural numbers. I would like to verify whether my proof is correct. The answer sheet used a much more intuitive and logical approach but I think mine is correct also.
To prove: $7|13\cdot 6^{n}+8\cdot 13^{n}$ for all natural numbers
Proof: We proceed by induction and show the base case holds.
$13+8=21 $. Since 7 divides 21 the base case holds.
We assume $7|13\cdot 6^{m}+8\cdot 13^{m}$ and we need to show $7|13\cdot6^{m+1}+8\cdot 13^{m+1}$.
Since, $7|13\cdot 6^{m}+8\cdot 13^{m}$, we have, $13\cdot 6^{m}+8\cdot 13^{m} =7x, \space x \in \mathbb{N}$.
We rewrite $13\cdot 6^{m+1}+8\cdot 13^{m+1}$ as $6(13\cdot 6^m)+13(8\cdot 13^m)$ and notice, $13\cdot 6^m=7x-8\cdot13^m$.
We substitute and find,
$6(7x-8\cdot13^m)+13(8\cdot13^m)=42x-48\cdot13^m+104\cdot13^m = 42x+56\cdot13^m=7(6x+8\cdot13^m)$
Since $6x+8\cdot13^m \space \in \mathbb{N}$
$\\ \therefore $ By the principle of mathematical induction, $7|13\cdot 6^{n}+8\cdot 13^{n}$ for all natural numbers $n \space \blacksquare$.
| It is correct, but:
*
*When you wrote that “We rewrite $13\cdot 6^{m}+8\cdot 13^{m}$ as $6(13\cdot 6^m)+13(8\cdot 13^m)$”, what you meant was “We rewrite $13\cdot 6^{m+1}+8\cdot 13^{m+1}$ as $6(13\cdot 6^m)+13(8\cdot 13^m)$”.
*You don't need to compute $6\times8$ and $13\times8$. Note that\begin{align}6(7x-8\times13^m)+13(8\times13^m)&=7\times(6x)+(-6+13)\times(8\times13^m)\\&=7\times(6x+8\times13^m).\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2949124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Maclaurin series of $\frac{1}{1-x}\ln\frac{1}{1-x}$ Find the Maclaurin series expansion (power law series) of the function:
$$f(x) = \frac{1}{1-x}\ln\frac{1}{1-x}.$$
I could find the expansion terms, up to the 5-th term, and is:
$$F(x) = x + \frac{3}{2!}x^2 + \frac{11}{3!}x^3 + \frac{50}{4!}x^4 + \frac{274}{5!}x^5 + \dots$$
and so on...
But I could not find a closed formula for it. Does any exist?
I've tried manipulating the power law series for $\dfrac{1}{1-x}$ and $\ln(1+x)$, but without success...
| $$ 1/(1-x)ln(1/(1-x)) = $$
$$ (1+x+x^2+x^3+x^4+...)(x+x^2/2+x^3/3+x^4/4+...) = $$
$$ x+(1+1/2)x^2+(1+1/2+1/3)x^3+(1+1/2+1/3+1/4)x^4+...= $$
$$ x + 3x^2/2 + 11x^3/6+25x^4/12+...=$$
$$ \sum_{a=1..\infty } (\sum_{b=1..a} 1/b) x^a =$$
$$ \sum_{a=1..\infty} H_a x^a$$
Numerators of harmonic numbers $H_n$ : https://oeis.org/A001008
Denominators of harmonic numbers $H_n$ : https://oeis.org/A002805
https://en.wikipedia.org/wiki/Harmonic_number
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Evaluating $\int_0^1\ln(1+x^2)\ln(x^2+x^3)\frac{dx}{1+x^2}$
How to evaluate $$I=\int_0^1\ln(1+x^2)\ln(x^2+x^3)\frac{dx}{1+x^2}?$$
It equals $\frac5{64}\pi^3-\frac92G\ln2+\frac14\pi\ln^22$ according to Mathematica, where $G$ denotes Catalan's constant.
Attempt
$$I=\frac d{ds}\int_0^1\ln(x^2+x^3)\frac{dx}{(1+x^2)^{1-s}}$$
or, $$I=\int_0^{\pi/4}2\ln\sec t\ln(\tan^2t(1+\tan t))dt$$
$$=2\int_0^{\pi/4}\left(\ln2+\sum_{n=1}^\infty\frac{(-1)^n\cos(2nx)}n\right)\left(-2\sum_{n=1}^\infty\frac{\cos(4n-2)x}{2n-1}+\ln(1+\tan x)\right)dx$$
$$=-4G\ln2+\frac14\pi\ln^22+2\sum_{n=1}^\infty\frac{(-1)^n}n\int_0^{\pi/4}\cos(2nx)\ln(\tan^2 x+\tan^3x)dx$$
| $$I=\int_0^1\frac{\ln(1+x^2)(2\ln(x)+\ln(1+x))}{1+x^2}dx$$
$$=2\int_0^1\frac{\ln(x)\ln(1+x^2)}{1+x^2}dx+\int_0^1\frac{\ln(1+x)\ln(1+x^2)}{1+x^2}dx$$
$$=2I_1+I_2$$
$I_1$ is calculated here:
$$\boxed{I_1=-2\,\Im\operatorname{Li_3}(1+i)+\frac{3\pi^3}{32}+\frac{\pi}8\ln^2(2)-\ln(2)G}$$
For $I_2$, let $x\to (1-x)/(1+x)$
$$I_2=\int_0^1\frac{\ln\left(\frac{2}{1+x}\right)\ln\left(\frac{2(1+x^2)}{(1+x)^2}\right)}{1+x^2}dx$$
$$=\ln(2)\underbrace{\int_0^1\frac{\ln\left(\frac{2(1+x^2)}{(1+x)^2}\right)}{1+x^2}dx}_{x\to (1-x)/(1+x)}-\ln(2)\int_0^1\frac{\ln(1+x)}{1+x^2}dx+2\int_0^1\frac{\ln^2(1+x)}{1+x^2}dx-I_2$$
$$=\ln(2)\int_0^1\frac{\ln(1+x^2)}{1+x^2}dx-\ln(2)\int_0^1\frac{\ln(1+x)}{1+x^2}dx+2\int_0^1\frac{\ln^2(1+x)}{1+x^2}dx-I_2$$
$$=\ln(2)\left(\frac{\pi}{2}\ln(2)-G\right)-\ln(2)\left(\frac{\pi}{8}\ln(2)\right)$$
$$+2\left(4\,\mathfrak{J}\operatorname{Li}_3(1+i)-\frac{7\pi^3}{64}-\frac{3\pi}{16}\ln^2(2)-2\ln(2)G\right)-I_2$$
$$\Longrightarrow \boxed{I_2=4\,\mathfrak{J}\operatorname{Li}_3(1+i)-\frac{7\pi^3}{64}-\frac52\ln(2)G}$$
$$\Longrightarrow I=2I_1+I_2=\frac{5\pi^3}{64}+\frac{\pi}{4}\ln^2(2)-\frac92\ln(2)G.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2953815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
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Is my particular solution of $ {(3xy + y^2)}\ dx + {(x^2 + xy)}\ dy = 0 $ for $x=1$ and $y=1$ correct?
Find the particular solution to
$$
{(3xy + y^2)}\ dx + {(x^2 + xy)}\ dy = 0
$$
for $x=1$ and $y=1$.
My Solution
$$
\frac{dy}{dx}=-\frac{3xy+y^2}{x^2+xy}
$$
Substituting $y=vx$ , $\frac{dy}{dx}=v+x\frac{dv}{dx}$
$$
v+x\frac{dv}{dx}=-\frac{3vx^2+x^2v^2}{x^2+vx^2}
$$
$$
v+s\frac{dv}{dx}=-\frac{3v+v^2}{1+v}
$$
$$
x\frac{dv}{dx}=-\frac{2v^2+4v}{v+1}
$$
$$
\frac{v+1}{2v^2+4v}=-\frac{dx}{x}
$$
$$
\frac{(2v+2)}{v^2+2v}=-4\frac{dx}{x}
$$
$$
\int\frac{(2v+2)}{v^2+2v}=-4\int\frac{dx}{x}
$$
$$
\log |v^2+2v| = -4\log|x|+\log C
$$
$$
\log |v^2+2v| = \log\frac{C}{x^4}
$$
$$
|v^2+2v|=\frac{C}{x^4}
$$
$$
|y^2+2xy|=\frac{C}{x^2}
$$
Substituting $x=1$ and $y=1$, $C=3$
The answer is
$$
|{(y^2 + 2xy)}|= \frac{3}{x^2}
$$
The answer in the book is left till here only,but what I did after this was as follows
$$
(y^2 + 2x y) = \pm\frac{3}{x^2}
$$
And then rejecting
$$
(y^2 + 2x y) = -\frac{3}{x^2}
$$
As it does not satisfy $x=1$, $y=1$
So According to me the particular solution is
$$
(y^2 + 2x y) = \frac{3}{x^2}
$$
Whose answer is correct my or book's?
| Neither you nor the book is correct. Both are partially correct, but not correct entirely. The false step is the belief that
$$\int\,\frac{1}{t}\,\text{d}t=\ln|t|+\text{constant}\,.$$
Since $t=0$ is not in the domain of the function $t\mapsto\dfrac1t$, what is correct is saying that
$$\int\,\frac1t\,\text{d}t=\ln|t|+c(t)\,,$$
where $c:(-\infty,0)\cup(0,\infty)\to\mathbb{C}$ is a locally constant function. That is, for some constants $c_-$ and $c_+$ which are not necessarily the same, we have
$$c(t)=\begin{cases}c_-&\text{if }t<0\,,\\c_+&\text{if }t>0\,.\end{cases}$$
Therefore, what is safe to say is that
$$|v^2+2v|=\frac{\tilde{\Gamma}(x,v)}{x^4}$$
for some locally constant function $\tilde{\Gamma}$ in two variables with domain $\Omega:=\big(\mathbb{R}\setminus\{0\} \big)\times \big(\mathbb{R}\setminus\{-2,0\}\big)$. You can remove the absolute value and say
$$v^2+2v=\frac{\Gamma(x,v)}{x^4}$$
for some locally constant $\Gamma:\Omega\to\mathbb{C}$.
From the initial condition $(x,y)=(1,1)$ (whence $(x,v)=1$) that you are dealing with the connected component $(0,\infty)\times (0,\infty)$ of $\Omega$. You can take $\Gamma(x,v)$ to be a constant $C$ there, which you found out that $C=3$. That is,
$$y^2+2xy=x^2(v^2+2v)=\frac{C}{x^2}=\frac{3}{x^2}\text{ for }x>0\,.$$
However, you cannot jump the boundary and deduce that $y^2+2xy=\dfrac{3}{x^2}$ for $x<0$ too.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2954913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How evaluate $ \sum\limits_{n=1}^{\infty}\frac{1}{n}\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)^{2}$ How prove
$$ \sum\limits_{n=1}^{\infty}\frac{1}{n}\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)^{2}=\frac{\pi^2\ln2}{6}-\frac{\ln^32}{3}-\frac{3\zeta(3)}{4} $$
$\mathbf {My\,Attempt:}$
I put $$\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right) = \int_0^1 \frac{x^n}{1+x}dx$$ and hence
$$\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)^2=\int_0^1\int_0^1 \frac{(xy)^n}{(1+x)(1+y)}dxdy$$
So, the sum equals $$\int_0^1\int_0^1 \frac{1}{(1+x)(1+y)}\left(\sum\limits_{n=1}^{\infty} \frac{(xy)^n}{n}\right)dxdy=-\int_0^1\int_0^1 \frac{\ln(1-xy)}{(1+x)(1+y)}dxdy$$
The inner integral is
$$\int_0^1 \frac{\ln(1-xy)}{1+y}dy=\mathrm{Li}_2 \left(\frac{1-x}{1+x}\right)-\mathrm{Li}_2\left(\frac{1}{1+x}\right)+\ln(1-x)\ln\left(\frac{2x}{1+x}\right)$$
The remaining stuff is a lot of calculations.
$\text{Any hint for a better method or idea?}$
| The second round of integration is completed with use of
\begin{align}
I_{1} &= \int_{0}^{1} \mathrm{Li}_2 \left(\frac{1-x}{1+x}\right) \, \frac{dx}{1+x} = \zeta(2) \, \ln 2 - \frac{5}{8} \, \zeta(3) \\
I_{2} &= \int_{0}^{1} \mathrm{Li}_2 \left(\frac{1}{1+x}\right) \, \frac{dx}{1+x} = \frac{\zeta(3)}{8} + \frac{\zeta(2) \, \ln 2}{2} - \frac{\ln^{3}2}{6} \\
I_{3} &= \int_{0}^{1} \ln(1-x) \, \ln\left(\frac{2 x}{1+x}\right) \, \frac{dx}{1+x} =
\frac{3 \, \zeta(3)}{2} + \frac{\ln^{3}2}{6} - \frac{3}{2} \, \zeta(2) \, \ln 2.
\end{align}
With these values then:
\begin{align}
S &= \sum_{n=1}^{\infty} \frac{1}{n} \, \left( \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k+n} \right)^{2} \\
&= \sum_{n=1}^{\infty} \frac{1}{n} \, \int_{0}^{1} \int_{0}^{1} \frac{(u t)^{n} \, du \, dt}{(1+u) (1+t)} \\
&= - \int_{0}^{1} \int_{0}^{1} \frac{\ln(1- u t) \, du \, dt}{(1+u)(1+t)} \\
&= - \int_{0}^{1} \left( \mathrm{Li}_2 \left(\frac{1-t}{1+t}\right)-\mathrm{Li}_2\left(\frac{1}{1+t}\right)+\ln(1-t)\ln\left(\frac{2t}{1+t}\right) \right) \, \frac{dt}{1+t} \\
&= - \left[ \left(\zeta(2) \, \ln 2 - \frac{5}{8} \, \zeta(3) \right) - \left(\frac{\zeta(3)}{8} + \frac{\zeta(2) \, \ln 2}{2} - \frac{\ln^{3}2}{6} \right) \right. \\
& \hspace{20mm} \left. + \left(\frac{3 \, \zeta(3)}{2} + \frac{\ln^{3}2}{6} - \frac{3}{2} \, \zeta(2) \, \ln 2 \right) \right] \\
&= \zeta(2) \, \ln 2 - \frac{3 \, \zeta(3)}{4} - \frac{\ln^{3}2}{3}.
\end{align}
This leads to saying
$$ \sum\limits_{n=1}^{\infty}\frac{1}{n}\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)^{2}= \zeta(2)\ln 2 -\frac{3\zeta(3)}{4} - \frac{\ln^{3}2}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2955340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Euclidean GCD calculation and mod
Calculate $6/87 \pmod{137}$
I do not understand the Euclidean GCD algorithm. If someone can please explain the overall logic of this it would be much appreciated.I have posted what the solution is supposed to be below:
| $$ \frac{ 137 }{ 87 } = 1 + \frac{ 50 }{ 87 } $$
$$ \frac{ 87 }{ 50 } = 1 + \frac{ 37 }{ 50 } $$
$$ \frac{ 50 }{ 37 } = 1 + \frac{ 13 }{ 37 } $$
$$ \frac{ 37 }{ 13 } = 2 + \frac{ 11 }{ 13 } $$
$$ \frac{ 13 }{ 11 } = 1 + \frac{ 2 }{ 11 } $$
$$ \frac{ 11 }{ 2 } = 5 + \frac{ 1 }{ 2 } $$
$$ \frac{ 2 }{ 1 } = 2 + \frac{ 0 }{ 1 } $$
Simple continued fraction tableau:
$$
\begin{array}{cccccccccccccccc}
& & 1 & & 1 & & 1 & & 2 & & 1 & & 5 & & 2 & \\
\frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 1 }{ 1 } & & \frac{ 2 }{ 1 } & & \frac{ 3 }{ 2 } & & \frac{ 8 }{ 5 } & & \frac{ 11 }{ 7 } & & \frac{ 63 }{ 40 } & & \frac{ 137 }{ 87 }
\end{array}
$$
$$ $$
$$ 137 \cdot 40 - 87 \cdot 63 = -1 $$
..
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2957154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $x>0, y>0,x+y=\frac{\pi}{3}$ then maximum value of $\tan x\tan y$
If $x>0, y>0$ and $x+y=\frac{\pi}{3}$, then find the maximum value of $\tan x\tan y$
My Attempt
$x>0, y>0, x+y=\frac{\pi}{3}\implies x, y$ in $1^\text{st}$ quadrant.
$\tan x, \tan y>0, \tan(x+y)=\sqrt{3}, \tan x\tan y>0$
$$
\tan x+\tan y\geq2\sqrt{\tan x.\tan y}\implies \tan^2x+\tan^2y+2\tan x\tan y\geq 4\tan x\tan y\\
2\tan x\tan y\leq\tan^2x+\tan^2y\implies \color{red}{?}
$$
or
$$
1-\tan x\tan y=\frac{\tan x+\tan y}{\tan(x+y)}\implies\tan x\tan y=1-\frac{\tan x+\frac{\sqrt{3}-\tan x}{1+\sqrt{3}\tan x}}{\sqrt{3}}\\
\tan x\tan y=1-\frac{\frac{\tan x+\sqrt{3}\tan^2x+\sqrt{3}-\tan x}{1+\sqrt{3}\tan x}}{\sqrt{3}}=1-\frac{1+\tan^2x}{1+\sqrt{3}\tan x}\leq\color{red}{?}\\
=\frac{1+\sqrt{3}\tan x-1-\tan^2x}{1+\sqrt{3}\tan x}=\frac{\tan x(\sqrt{3}-\tan x)}{1+\sqrt{3}\tan x}=\color{red}{?}
$$
Note: I prefer not to do differentiation
| Hint: One way is finding the extrimum of $f(x,y)=\tan x\tan y$ with condition $x+y=\dfrac{\pi}{3}$, then with
$$F(x,y)=\tan x\tan y-\lambda\left(x+y-\dfrac{\pi}{3}\right)$$
we have
$$\tan y=\lambda\cos^2x~~~,~~~\tan x=\lambda\cos^2y$$
then
$$\tan y\lambda\cos^2y=\tan x\lambda\cos^2x$$
shows $\sin x\cos x=\sin y\cos y$ or $\color{blue}{x=y}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2958959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How did I obtain incorrect results for $P_{B\leftarrow C}$ and $P_{C\leftarrow B}$ geometrically? I'm given two bases in $\mathbb{R}^2$:
$B = \{\begin{pmatrix} 1 \\ 1 \end{pmatrix}, \begin{pmatrix} -1 \\ 2 \end{pmatrix}\}$ and $C = \{\begin{pmatrix} -4 \\ 2 \end{pmatrix}, \begin{pmatrix} 2 \\ 5 \end{pmatrix}\}$ and I'm asked to find $P_{B\leftarrow C}$ and $P_{C\leftarrow B}$ completely geometrically.
I started out by finding them mechanically using the fact that $P_{B\leftarrow C} = (P_{S\leftarrow B})^{-1}(P_{S\leftarrow C})$, and $P_{C\leftarrow B} = (P_{S\leftarrow C})^{-1}(P_{S\leftarrow B})$ where $S$ is the standard basis in $\mathbb{R}^2$.
$P_{B\leftarrow C} = \begin{pmatrix} 1 & -1 \\ 1 & 2 \end{pmatrix}^{-1} \begin{pmatrix} -4 & 2 \\ 2 & 5 \end{pmatrix} = \begin{pmatrix} -2 & 3 \\ 2 & 1 \end{pmatrix}$.
And: $P_{C\leftarrow B} = \begin{pmatrix} -4 & 2 \\ 2 & 5 \end{pmatrix}^{-1} \begin{pmatrix} 1 & -1 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} -\frac{1}{8} & \frac{3}{8} \\ \frac{1}{4} & \frac{1}{4} \end{pmatrix}$.
Then, this is the work I did attempting to find these same results geometrically:
To find $P_{B\leftarrow C}$, where do the vectors $\begin{pmatrix} -4 \\ 2 \end{pmatrix}$ and $\begin{pmatrix} 2 \\ 5 \end{pmatrix}$ go? $\begin{pmatrix} -4 \\ 2 \end{pmatrix} \rightarrow \begin{pmatrix} -1 \\ 2 \end{pmatrix}$, which is a clockwise rotation by angle $\alpha_1$.
$\cos{\alpha_1} = \frac{\begin{pmatrix} -4 \\ 2 \end{pmatrix} \cdot \begin{pmatrix} -1 \\ 2 \end{pmatrix}}{\sqrt{20}\sqrt{5}} = \frac{8}{10} = \frac{4}{5}$. Therefore, $\sin{\alpha_1} = \frac{-3}{5}$.
Additionally, $\begin{pmatrix} 2 \\ 5 \end{pmatrix} \rightarrow \begin{pmatrix} 1 \\ 1 \end{pmatrix}$ has a clockwise rotation by angle $\alpha_2$.
$\cos{\alpha_2} = \frac{\begin{pmatrix} 2 \\ 5 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 1 \end{pmatrix}}{\sqrt{2}\sqrt{29}} = \frac{7}{\sqrt{58}}$. Therefore, $\sin{\alpha_2} = \frac{-3}{\sqrt{58}}$. Now, we must multiply the rotated $\begin{pmatrix} -4 \\ 2 \end{pmatrix}$ by a factor. $\frac{\sqrt{5}}{\sqrt{20}} = \sqrt{\frac{5}{20}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$. And we must multiply the rotated $\begin{pmatrix} 2 \\ 5 \end{pmatrix}$ by a factor $\frac{\sqrt{2}}{\sqrt{29}} = \frac{\sqrt{58}}{29}$.
Thus, $P_{B\leftarrow C} = \begin{pmatrix} \frac{2}{5} & \frac{3}{29} \\ -\frac{3}{10} & \frac{7}{29} \end{pmatrix}$.
I then followed this same methodology, obtaining: $P_{C\leftarrow B} = \begin{pmatrix} \frac{7}{2} & -\frac{6}{5} \\ \frac{3}{2} & \frac{8}{5} \end{pmatrix}$.
Obviously, these results don't match the results I obtained with mechanical calculations, so something is wrong. I feel that I may have a major misunderstanding in the geometry of change-of-basis. Perhaps I shouldn't be calculating the change-of-basis matrices by sending one set of basis vectors to the other because that may not reflect what is happening for any arbitrary vectors switching between those bases. Also, I have a feeling that my algebra may be wrong somewhere in determining the rotation matrices and then applying scalars to them. It is strange because we are trying to essentially use one transformation matrix to rotate two vectors two different angles, which I feel may be the wrong approach.
What am I missing?
| First a bit of review. The coordinates of a vector are the coefficients of its unique expression as a linear combination of basis vectors. That is, if we have the ordered basis $\mathcal B = \left(\mathbf b_1,\mathbf b_2\right)$, then $[\mathbf v]_{\mathcal B} = (v_1,v_2)^T$ means that $\mathbf v = v_1\mathbf b_1+v_2\mathbf b_2$. Now suppose that $\mathbf b_1 = p_{11}\mathbf c_1+p_{21}\mathbf c_2$ and $\mathbf b_2 = p_{12}\mathbf c_1+p_{22}\mathbf c_2$ for another ordered basis $\mathcal C = (\mathbf c_1,\mathbf c_2)$. Then $$\mathbf v = v_1(p_{11}\mathbf c_1+p_{21}\mathbf c_2)+v_2(p_{12}\mathbf c_1+p_{22}\mathbf c_2) = (p_{11}v_1+p_{12}v_2)\,\mathbf c_1 + (p_{21}v_1+p_{22}v_2)\,\mathbf c_2,$$ or, in matrix form, $$[\mathbf v]_{\mathcal C} = \begin{bmatrix} p_{11}v_1+p_{12}v_2 \\ p_{21}v_1+p_{22}v_2 \end{bmatrix} = \begin{bmatrix}p_{11}&p_{12}\\p_{21}&p_{22}\end{bmatrix} \begin{bmatrix}v_1\\v_2\end{bmatrix} = P_{\mathcal C\leftarrow\mathcal B}[\mathbf v]_{\mathcal B}.$$ So, the columns of the change-of-basis matrix $P_{\mathcal C\leftarrow\mathcal B}$ are the coordinates of the elements of $\mathcal B$ expressed in the basis $\mathcal C$.
To find these coordinates geometrically, we can use the paralellogram rule for vector addition. For instance, to find $[\mathbf c_1]_{\mathcal B}$, draw a paralellogram with sides parallel to $\mathbf b_1$ and $\mathbf b_2$ that has $\mathbf c_1$ as a diagonal. The ratios of its side lengths to the lengths of the corresponding basis vectors (taking care to get the signs right) are the coordinates of $\mathbf c_1$ relative to $\mathcal B$.
The endpoints of the other diagonal, which are the vectors we’re looking for, can be computed by several methods such as intersecting pairs of lines, but in this case we can just read them off the diagram: they are $(-2,-2)^T = -2\mathbf b_1$ and $(-2,4)^T = 2\mathbf b_2$, so $[\mathbf c_1]_{\mathcal B} = (-2,2)^T$. This is the first column of $P_{\mathcal B\leftarrow\mathcal C}$, which matches what you got with a mechanical computation. I’ll leave working out the others to you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2959669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\frac{1+\sin\theta + i\cos\theta}{1+\sin\theta-i\cos\theta}=\sin\theta+i\cos\theta.$ Prove that $$\frac{1+\sin\theta + i\cos\theta}{1+\sin\theta-i\cos\theta}=\sin\theta+i\cos\theta.$$
Hence, show that $$(1+\sin\frac{\pi}{5}+i\cos\frac{\pi}{5})^5+i(1+\sin\frac{\pi}{5}-i\cos\frac{\pi}{5})^5=0.$$
In the first part, I tried realising LHS so I got $$LHS=\frac{(1+\sin\theta+i\cos\theta)^2}{(1+\sin\theta)^2+\cos^2\theta}=\frac{1+2\sin\theta+2i\cos\theta+2i\cos\theta\sin\theta+\sin^2\theta+i^2\cos^2\theta}{2+2\sin\theta}.$$
but now I am stuck :( . Any help would be greatly appreciated, thanks!
| I write
$i\cos \theta + \sin \theta = i(\cos \theta -i\sin \theta) = i(\cos (-\theta) + i\sin (-\theta)) = ie^{-i\theta}; \tag 1$
then
$\sin \theta -i\cos \theta = -i(\cos \theta + i\sin \theta) = -ie^{i \theta}; \tag 2$
we have
$(1 + \sin \theta - i\cos \theta)(\sin \theta + i\cos \theta) = (1 - ie^{i \theta})ie^{-i\theta} = ie^{-i\theta} + 1 = 1 + \sin \theta + i\cos \theta, \tag 3$
simply a slightly altered form equivalent to the desired result; some minor re-arrangemens yield
$\dfrac{ 1 + \sin \theta + i\cos \theta}{ 1 + \sin \theta - i\cos \theta } = \dfrac{1 + ie^{-i\theta}}{1 - ie^{i \theta}} = ie^{-i\theta} = \sin \theta + i\cos \theta, \tag 4$
the equation in the form given.
We also wish to show that
$\left (1+\sin\dfrac{\pi}{5}+i\cos\dfrac{\pi}{5} \right )^5+i \left ( 1+\sin\dfrac{\pi}{5}-i\cos\dfrac{\pi}{5} \right )^5=0; \tag 5$
if we set
$\theta = \dfrac{\pi}{5}, \tag 6$
then it follows that
$\left ( \dfrac{ 1 + \sin \theta + i\cos \theta}{ 1 + \sin \theta - i\cos \theta } \right )^5 = (\sin \theta + i \cos \theta)^5$
$= (ie^{-i\theta})^5 = i^5 e^{-i 5 \theta} = ii^4 e^{-i 5(\pi / 5)} = ie^{-i \pi} = -i; \tag 7$
therefore,
$\dfrac{( 1 + \sin \theta + i\cos \theta)^5}{( 1 + \sin \theta - i\cos \theta )^5} = \left ( \dfrac{ 1 + \sin \theta + i\cos \theta}{ 1 + \sin \theta - i\cos \theta } \right )^5 = (\sin \theta + i \cos \theta)^5 = -i, \tag 8$
whence
$( 1 + \sin \theta + i\cos \theta)^5 + i ( 1 + \sin \theta - i\cos \theta )^5 = 0; \tag 9$
if we substitute (6) into this equation we find that in concrete terms
$\left ( 1 + \sin \dfrac{\pi}{5} + i\cos \dfrac{\pi}{5} \right )^5 + i \left ( 1 + \sin \dfrac{\pi}{5} - i\cos \dfrac{\pi}{5} \right )^5 = 0, \tag{10}$
as per request.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2961689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Simplifying $\sin^{-1}(\cos x)$ I have a question on $$\sin^{-1}(\cos x)$$
Since $\cos(x)=\sin(\frac \pi 2 \pm x)$, the above expression can simplify to either $\frac \pi 2 + x$ or $\frac \pi 2 - x$. This seems like a contradiction.
What's the problem here?
| Let $x=\pm\frac{\pi}{2}\mp(y+2\pi n),y\in [-\frac{\pi}{2},\frac{\pi}{2}]$.
Then:
$$\begin{align}\arcsin(\cos x)&=\arcsin(\cos [\pm\frac{\pi}{2}\mp(y+2\pi n)])=\\
&=\arcsin(\sin (y+2\pi n))=\\
&=\arcsin(\sin y)=\\
&=y.\end{align}$$
Example 1: If $x=300^\circ$, then $300^\circ=-90^\circ+(30^\circ+360^\circ)$ and:
$$\arcsin(\cos 300^\circ)=30^\circ.$$
Example 2: If $x=-300^\circ$, then $-300^\circ=90^\circ-(30^\circ+360^\circ)$ and:
$$\arcsin(\cos (-300^\circ))=30^\circ.$$
Example 3: If $x=180^\circ$, then $180^\circ=90^\circ-(-90^\circ+0^\circ)$ and:
$$\arcsin(\cos (180^\circ))=-90^\circ.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2961789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
find $\cos\theta$ if $\sin\theta=\frac{3}{4}$ and $\tan\theta=\frac{9}{2}$
If $\sin\theta=\frac{3}{4}$ and $\tan\theta=\frac{9}{2}$ then find $\cos\theta$
The solution is given in my reference as $\frac{1}{6}$.
$$
\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\frac{9}{16}}=\sqrt{\frac{7}{16}}=\frac{\sqrt{7}}{4}
$$
$$
\cos\theta=\frac{1}{\sec\theta}=\frac{1}{\sqrt{1+\tan^2\theta}}=\frac{1}{\sqrt{1+\frac{81}{4}}}=\frac{2}{\sqrt{85}}
$$
$$
\cos\theta=\frac{\sin\theta}{\tan\theta}=\frac{\frac{3}{4}}{\frac{9}{2}}=\frac{3}{4}.\frac{2}{9}=\frac{1}{6}
$$
Why is it getting confused here ?
| The problem is actually incorrect. Here is why:
Consider a right triangle and angle $\theta$ in that triangle such that $\sin\theta=\dfrac{3}{4}.$
That means that the right triangle has one leg $3$ and hypotenuse $4$. This implies that the other leg is $\sqrt{4^2-3^2}=\sqrt{7}.$
This implies that $\cos \theta=\dfrac{\sqrt{7}}4.$ Now also note that $\tan \theta \neq \frac 92$ as stated in your problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2961982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Sum of series $\sum_{n=1}^{\infty}\sin(\frac{1}{2^n})\cos(\frac{3}{2^n})$ How can we get sum of this series? $$\sum_{n=1}^{\infty}\sin(\frac{1}{2^n})\cos(\frac{3}{2^n})$$
I think we must apply this theorem.
if in $\sum_{n=1}^\infty a_{n}$, $a_n = b_{n+1}-b_n$ and $\lim_{n\to\infty} b_n = b$; $ \implies $ $\sum_{n=1}^\infty a_{n}=b-b_1$
But how?
| The identity $\sin x\cos 3x=\frac{1}{2}(\sin 4x-\sin 2x)$ gives$$\sum_{n\ge 1}\sin\frac{1}{2^n}\cos\frac{4}{2^n}=\frac{1}{2}\sum_{n\ge 1}(\sin\frac{4}{2^n}-\sin\frac{2}{2^n})=\frac{1}{2}\sin 2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2963068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Showing that two random variables are independent I have the following problem:
Given two independent standard normal random variables, call them $X$ and $Y$, how can I show that $Z = X^2 + Y^2$ and $W=\frac{X}{Y}$ are also independent?
I know that because $X$ and $Y$ are standard normal I can write their distributions as $$f_X(x) = \frac{1}{2\pi}e^\frac{-(x)^2}{2}$$ $$f_Y(y) = \frac{1}{2\pi}e^\frac{-(y)^2}{2}$$
Then because they are independent I can write their joint probability distribution as $$f_{X,Y}(x,y) = \frac{1}{2\pi}e^\frac{-(x^+y^2)}{2}$$
Now I know that to show $Z$ and $W$ are independent, I need to show that their joint probability distribution is equal to the product of the marginal distribution functions. But I'm not sure how to find
$f_{Z,W}(z,w)$, $f_Z(z)$ and $f_W(w)$.
Any help is much appreciated.
| The joint distribution for $(X^2,Y^2)$ is
$$\frac{x^{-1/2}e^{-x/2}}{\sqrt{2}\Gamma(1/2)}1_{(0,+\infty)}(x)\times \frac{y^{-1/2}e^{-y/2}}{\sqrt{2}\Gamma(1/2)}1_{(0,+\infty)}(x)=\frac{(xy)^{-1/2}e^{-(x+y)/2}}{2(\Gamma(1/2))^2}1_{(0,+\infty)}(x)1_{(0,+\infty)}(y)$$
Since $X^2=Z-\frac{Z}{W^2+1}=\frac{ZW^2}{W^2+1},Y^2=\frac{Z}{W^2+1}$, the Jacobian matrix is $$
\left|\begin{array}{cc}
\frac{W^2}{W^2+1}& \frac{1}{W^2+1}\\
\frac{2ZW}{(W^2+1)^2} & -\frac{2ZW}{(W^2+1)^2}
\end{array}\right| =\frac{2ZW}{(W^2+1)^2}
$$
Therefore, the joint pdf of $(Z,W)$ is
$$\frac{(\frac{ZW^2}{W^2+1}\frac{Z}{W^2+1})^{-1/2}e^{-(\frac{ZW^2}{W^2+1}+\frac{Z}{W^2+1})/2}}{2(\Gamma(1/2))^2}\times \frac{2ZW}{(W^2+1)^2}1_{(0,+\infty)}(z)=\frac{1}{w^2+1}\frac{e^{-z/2}}{\pi}1_{(0,+\infty)}(z).$$
Hence, the pdf of $W$ and $Z$ are $\frac{1}{\pi (1+w^2)}$ and $\frac{1}{2}e^{-z/2}1_{(0,+\infty)}(z)$.
For now, one can see the independence of $W$ and $Z$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that X$^3$+X+1|X$^7$+1 This is a part of the question. Which states Let K= $\mathbb{Z}$/2$\mathbb{Z}$[X]/(d)$\mathbb{Z}$/2$\mathbb{Z}$[X], where
d=X$^3$+X+1 and let a be the class of X modulo d. This is the only part that i do not know how to solve. I have tried calculating the quotient and the remainder but i do not know if it has any meaningful use. And since it has a remainder i cannot find a q(x) that satisfies f(x)=g(x)*q(x). Any tips?
| Here are the steps for polynomial long division $\mod 2$, where the notation
$\dfrac{p(X)}{q(X)} \longrightarrow X^k \tag 0$
indicates the quotient of the leading terms is $X^k$:
$\dfrac{X^7 + 1}{X^3 + X + 1} \longrightarrow X^4; \tag 1$
$X^7 + 1 - X^4(X^3 + X + 1) = X^7 + 1 - X^7 - X^5 - X^4 = X^5 + X^4 + 1; \tag 2$
$\dfrac{X^5 + X^4 + 1}{X^3 + X + 1} \longrightarrow X^2; \tag 3$
$X^5 + X^4 + 1 - X^2(X^3 + X + 1) = X^5 + X^4 + 1 - X^5 - X^3 - X^2 = X^4 + X^3 + X^2 + 1; \tag 4$
$\dfrac{X^4 + X^3 + X^2 + 1}{X^3 + X + 1} \longrightarrow X; \tag 5$
$X^4 + X^3 + X^2 + 1 - X(X^3 + X + 1) = X^4 + X^3 + X^2 + 1 - X^4 - X^2 - X = X^3 + X + 1; \tag 6$
$\dfrac{X^3 + X + 1}{X^3 + X + 1} \longrightarrow 1, \tag 7$
and there is clearly no remainder; we gather the terms of the quotient:
$\dfrac{X^7 + 1}{X^3 + X + 1} = X^4 + X^2 + X + 1, \tag 8$
which is easily checked:
$(X^3 + X + 1)( X^4 + X^2 + X + 1)$
$= X^7 + X^5 + X^4 + X^3 + X^5 + X^3 + X^2 + X + X^4 + X^2 + X + 1$
$X^7 + (X^5 + X^5) + (X^4 + X^4) + (X^3 + X^3) +(X^2 + X^2) + (X + X) + 1$
$= X^7 + 1; \tag 9$
thus we see that
$X^3 + X + 1 \mid X^7 + 1. \tag{10}$
| {
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"source": "stackexchange",
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How do I evaluate the integral $\int_0^1\frac{x^2+x+1}{x^4+x^3+x^2+x+1}dx$. stuck on this integral
$$\int_0^1\dfrac{(x^2+x+1)}{(x^4+x^3+x^2+x+1)}\ dx$$
I was attempting to evaluate the infinity sum S = $ 1- \frac{1}{4} + \frac {1}{6} - \frac {1}{9} + \frac {1}{11} -\frac {1}{14}+ ........ $
what I then did was define the S to be equal to $$ \int_0^1 (1-x^3 +x^5-x^8+x^{10}-x^{13}+........) dx $$
I simplified this and got the above integral I tried to do partial fraction but did not succeed.
| The hint:
Use $$x^4+x^3+x^2+x+1=\left(x^2+\frac{1}{2}x+1\right)^2-\left(\frac{\sqrt5}{2}x\right)^2=$$
$$=\left(x^2+\frac{1-\sqrt5}{2}x+1\right)\left(x^2+\frac{1+\sqrt5}{2}x+1\right).$$
Can you end it now?
| {
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"source": "stackexchange",
"question_score": "2",
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Prove $\sum_{1}^{\infty} \frac{1}{\sqrt{n}( n + \sqrt{n})} \lt 2 $
Prove that $$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}( n + \sqrt{n})} \lt 2$$
I have found that $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}( n + \sqrt{n})} < \pi / 2 $ with integrating from $1$ to infinity but integral isn’t allowed.
| Let us go with creative telescoping. As $n\to +\infty$,
$$ \frac{1}{n+n\sqrt{n}}=\frac{1}{n\sqrt{n}}-\frac{1}{n^2}+\frac{1}{n^2\sqrt{n}}+O\left(\frac{1}{n^3}\right) $$
while
$$ \frac{2}{\sqrt{n-\frac{7}{6}}}-\frac{2}{\sqrt{n-\frac{1}{6}}}=\frac{1}{n\sqrt{n}}+\frac{1}{n^2\sqrt{n}}+\frac{95}{96 n^2\sqrt{n}}+o\left(\frac{1}{n^3}\right) $$
$$ \frac{1}{n-1}-\frac{1}{n} = \frac{1}{n^2}+\frac{1}{n^3}+o\left(\frac{1}{n^3}\right) $$
hence by setting $f(n)=\frac{2}{\sqrt{n-\frac{7}{6}}}-\frac{1}{n-1}$ we have that $f(n)-f(n+1)$ is an excellent approximation of $\frac{1}{n+n\sqrt{n}}$ for any $n\geq 4$, such that
$$ \sum_{n\geq 1}\frac{1}{n+n\sqrt{n}}\approx \sum_{n=1}^{3}\frac{1}{n+n\sqrt{n}}+f(4) =1.6839589\ldots$$
is accurate up to the second figure. In particular $\sum_{n\geq 1}\frac{1}{n+n\sqrt{n}}<1.7$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding minimum value of $ \frac{x^2 +y^2}{y} $
Finding the minimum value of $\displaystyle \frac{x^2 +y^2}{y}.$ where $x,y$ are
real
numbers satisfying $7x^2 + 3xy + 3y^2 = 1$
Try: Equation $7x^2+3xy+3y^2=1$ represent Ellipse
with center is at origin.
So substitute $x=r\cos \alpha $ and $y=r\sin \alpha$
in $7x^2+3xy+3y^2=1$
$$3r^2+4r^2\cos^2 \alpha+3r^2\sin \alpha \cos \alpha =1$$
$$3r^2+2r^2(1+\cos 2 \alpha)+\frac{3r^2}{2}\sin 2 \alpha =1$$
$$8r^2+r^2(4\cos 2 \alpha+3\sin \alpha)=2$$
So $$r^2=\frac{2}{8+(4\cos 2 \alpha+3\sin \alpha)}$$
$$\frac{2}{8+5}=\frac{2}{13}\leq r^2\leq \frac{2}{8-5}=\frac{2}{3}$$
we have to find minimum of $$\frac{x^2+y^2}{y}=\frac{r}{\sin \alpha}$$
How can i find it, could some help me
| Lagrange function is
$$L=k\, \left( 3 {{y}^{2}}+3 x y+7 {{x}^{2}}-1\right) +\frac{{{y}^{2}}+{{x}^{2}}}{y}$$
Solve system:
$$L'_x=0,\quad L'_y=0,\quad L'_k=0.$$
We get two solutions
$$k=1/4,y=-2/5,x=-1/5$$ $$k=-1/4,y=2/5,x=1/5$$
Answer:
$$f_{min}=f\left(-\frac{1}{5},-\frac{2}{5}\right)=-\frac{1}{2}$$
with CAS Maxima:
| {
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How do I use the definition of partial derivative to get $f_x(x,y) = \frac{y(y^2-x^2)}{(x^2 + y^2)^2}$? $\\f(x,y) =
\begin{cases}
\frac{xy}{x^2 + y^2}, & \text{if $(x,y) \ne (0,0)$} \\
0, & \text{if $(x,y) = (0,0)$}
\end{cases}$
Using the definition of partial derivatives:
\begin{align*}
f_x(x,y) &= \lim_{h\to 0} \frac{f(x+h,y) - f(x,y)}{h}
\\&= \lim_{h\to 0} \frac{\frac{(x+h)y}{(x+h)^2 + y^2}-\frac{xy}{x^2 + y^2}}{h}\\&= \lim_{h\to 0} \frac{(xy + hy)(x^2 +y^2)-xy(x^2 +2hx+ h^2 + y^2)}{h(x^2 + 2hx + h^2 + y^2)(x^2 + y^2)}
\end{align*}
I'm not sure how to get rid of the variable $h$. I wanted to know exactly how I could further simplify in order to get $\frac{y(y^2-x^2)}{(x^2 + y^2)^2}$?
| Continuing your calculation, we get
$$f_x(x,y)=\lim_{h\rightarrow 0}\frac{hy(y^2-x^2-hx)}{h(x^2+2hx+h^2+y^2)(x^2+y^2)}.$$
Or
$$f_x(x,y)=\lim_{h\rightarrow 0}\frac{y(y^2-x^2-hx)}{(x^2+2hx+h^2+y^2)(x^2+y^2)}.$$
Can you finish from here?
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the $2n$ th derivative of $\frac{1}{1+x^2y^2}$ with repect to $x$ We know that with repect to $x$,
\begin{align}
\left( \frac{1}{1-x^2 y^2} \right)^{(2n)} =
\frac{(2n)!y^{2n}}{2 } \left( \frac{1}{(1+xy)^{2n+1}} +\frac{1}{(1-xy)^{2n+1}} \right)
\end{align}
What about
\begin{align}
\left( \frac{1}{1+x^2 y^2} \right)^{(2n)}
\end{align}
| The first thing to do is make this substitution $y = ik$, we can do this freely because the derivative is with respect to x not to y, and it is only a contsant.Then after this substitution we get
$$\left( \frac{1}{1+x^2 y^2} \right)^{(2n)}=\left( \frac{1}{1-x^2 k^2} \right)^{(2n)}
$$
Which, as you said, is equal to
$$\frac{(2n)!k^{2n}}{2 } \left( \frac{1}{(1+xk)^{2n+1}} +\frac{1}{(1-xk)^{2n+1}} \right)
$$
Finally we just substitute $\frac{y}{i}=k$ and we obtain that
for the derivative of a number with the form 4k we have
$$\frac{(2n)!y^{2n}}{2 } \left( \frac{1}{(1+\frac{xy}{i})^{2n+1}} +\frac{1}{(1-\frac{xy}{i})^{2n+1}} \right)
$$
and for the derivative of a number with the form 4k+2 we have
$$\frac{-(2n)!y^{2n}}{2 } \left( \frac{1}{(1+\frac{xy}{i})^{2n+1}} +\frac{1}{(1-\frac{xy}{i})^{2n+1}} \right)
$$
then if we do not want i, we only have to add the two terms and apply the binomial theorem.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help with induction proof I need help with the following induction proof which I am not sure if I am doing correctly.
$$\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{(2n-1)(2n+1)}=\frac{n}{2n+1}$$
I check for $n=1$ (Base case)
$$\frac{1}{1\cdot3}=\frac13$$
$$\frac{1}{2\cdot1+1}=\frac13$$
Now, is this the correct next step in my proof?
$$\frac{k}{2\cdot k+1}+\frac{1}{(2\cdot (k+1)-1)(2\cdot(k+1)+1)}=\frac{k+1}{2\cdot(k+1)+1}$$
I we assume it is correct for $n=k$ then it is also true for $n=k+1$ which means that the RHS must be equal to the LHS.
| $$
\\\frac{1}{1\cdot3}+\frac{1}{1\cdot5}+...+\frac{1}{(2n-1)(2n+1)}=
\\\frac{1}{2}\cdot((\frac{1}{1}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{5})+\cdots+(\frac{1}{2n-1}-\frac{1}{2n+1}))=
\\\frac{1}{2}\cdot(1-\frac{1}{2n+1})=\frac{n}{2n+1}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $\int_0^{2 \pi} \dfrac{d \theta}{5+4 \sin\theta}=\dfrac{2 \pi}{3}.$ By change of variable and letting $C_0$ be the unit circle about the origin, we have:
\begin{align*}
\int_0^{2 \pi} \dfrac{d \theta}{5+4 \sin\theta} &= \int_{C_0} \dfrac{dz/zi}{5+4(z-z^{-1})/2i}\\
& = \int_{C_0} \dfrac{dz}{2z^2+5iz-2}\\
& = \int_{C_0} \dfrac{dz}{(z+i/2)(z+2i)}\\
& = 2 \pi i (-i/2+2i)^{-1}\\
& = \dfrac{4 \pi}{3}
\end{align*}
But wolframalpha gives $\dfrac{2 \pi}{3}.$ Is what I wrote valid?
| $\oint_{|z|=1} \frac {1}{i(2z^2 + 5iz -2} \ dz\\
\oint_{|z|=1} \frac {1}{2(z +2i)(z + \frac 12i)} \ dz$
You dropped a factor of $\frac 12$
$2\pi i\frac {1}{2(-\frac 12 i + 2i)} = \frac {2\pi}{3}$
| {
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Sequence Inequality question from RMO 2018 Define a sequence {$a_n$} of real numbers by $a_1 = 2$ and $a_{n+1} = \dfrac{a_n^2+1}{2}$, for $n\ge 1$,
Prove that for every natural number $N, \sum_{j=1}^{N} \frac{1}{1+a_j} \lt 1$
I tried mathematical induction after coming to the step where
$Sum_n = \frac{1}{2}\left(\frac{a_1-1}{a_2-1} + \frac{a_2-1}{a_3-1} +\cdots \frac{a_n-1}{a_{n+1}-1}\right)$.
Having gotten this how would go about proving it using induction?
| To get $\dfrac{1}{a_n+1}$, Firse subtract $1$ from both sides
$$a_{n+1}-1=\frac{a_{n}^2-1}{2}=\frac{(a_n+1)(a_n-1)}{2}.$$
Then take the multiplicative inverse ( assume $a_n\neq \pm1$, as we'll see later)
$$\frac{1}{a_{n+1}-1}=\frac{2}{(a_n+1)(a_n-1)}$$
where the right side can be written as
$$\frac{1}{a_n-1}-\frac{1}{a_n+1}.$$
Thus
$$\frac{1}{a_n+1} = \frac{1}{a_n-1}-\frac{1}{a_{n+1}-1}$$
Sum up both sides from $1$ to $N$ we get
$$\sum_{k=1}^{N}\frac{1}{a_k+1}=\frac{1}{a_1-1}-\frac{1}{a_{N+1}-1}=1-\frac{1}{a_{N+1}-1}.$$
Now we just need to prove $a_n>1$ for all $n\geqslant2$.
Notice that $$a_{n+1}-a_{n}=\frac{a_{n}^2+1}{2}-a_n=\frac{(a_n-1)^2}{2}>0.$$
Then $\{a_n\}$ is incresing with $a_1=2$. Hence the proof is done.
| {
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Prove that $10|n+3n^3+7n^7+9n^9$ Prove that $10|n+3n^3+7n^7+9n^9$ for every $n\in \mathbb N$
Only what i see that 10=5*2 and both number is free numbers, and if I show that $5|n+3n^3+7n^7+9n^9$ and $2|n+3n^3+7n37+9n^9$ that I prove, since 5 and 2 is free numbers i can use Fermat's little theorem such that $5|n^5-n$ and $2|n^2-n$ but i can not see that somehow help me, do you have some idea?
| $\begin{align} \bmod 10\!:\ f_n & = \ n+9n^9 +\ 3n^3\,+ 7n^7\\
&\equiv n(1\!-\!n^8) + 3n^3(1\!-\!n^4)\ \ \ {\rm by}\ \ \ 9\equiv -1,\,\ 7\equiv -3\\
&\equiv \color{#c00}{n(1\!-\!n^4)}\,g_n
\end{align}$
Notice that $\,2,5\mid \color{#c00}{n\,-\,n^5}\ $ by Fermat.
Or, equivalently, $\,\color{#c00}{n^5\equiv n}\,\Rightarrow\, n^7\equiv n^3,\, n^9\equiv n\ $ so $\,f_n\equiv (1\!+\!9)n+(3\!+\!7)n^3\equiv 0$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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What does the minimum x that satify $x=24n+12=15m+6=11k+2$? My attempt so far was: let $x=24n+12=15m+6=11k+2$
find $x$ as the form $x=24*15*a+24*11*b+15*11*c$
$$2\equiv 24\cdot 15\cdot a\quad (mod\quad 11)\quad \Rightarrow 1\equiv 4\cdot a\Rightarrow \quad a=3\\ 6\equiv 24\cdot 11\cdot b\quad (mod\quad 15)\quad \Rightarrow \quad 2\equiv 8\cdot 11\cdot b\quad (mod\quad 5)\quad \Rightarrow 2\equiv 3\cdot b\quad (mod\quad 5)\quad \quad $$ and I stack here after finding $a$
| $\!x\equiv 12\pmod{\!24}\!\iff x/3\,\equiv\ \ \ 4\,\pmod{\! 8}$
$\!\!\left.\begin{align}
&x\equiv \ 6\!\!\pmod{\!15}\!\iff x/3\,\equiv\ \ \ 2\!\!\!\pmod{ 5}\\
&x\equiv \ 2\!\!\pmod{\!11}\!\iff x/3\,\equiv -3\!\!\!\pmod{\!\!11}\\
\end{align}\right\}\!\!\!\iff\!\dfrac{x}3\equiv -3\pmod{\!55}\!\iff\! \dfrac{x}3 = \color{#0a0}{-3\!+\!55j}$
$\!\!\!\bmod \color{#c00}8\!:\, 4\equiv \dfrac{x}3\equiv \color{#0a0}{-3\!+\!55}\color{#c00}j\equiv 5\!-\!j\!\!\iff\! \color{#c00}{j\equiv 1}\!\iff\! \dfrac{x}3=-3\!+\!55(\color{#c00}{1\!+\!8i})\!\iff\!\! x\equiv 156\!+\!1320i $
| {
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ّFind $x$ such that $ \frac{1}{x^2} + \frac{1}{(3-x)^2} = \frac{104}{25}$.
ّFind $x$ such that
$$ \frac{1}{x^2} + \frac{1}{(3-x)^2} = \frac{104}{25}\,.$$
My attempt:
After clearing the denominators, I obtain this quartic equation
$$104 x^{4} -624 x^{3} +886 x^{2} +150x-225=0.$$
I don't know how to proceed from here.
| Let $u=x$ and $v=3-x$. Then we have
$$
u+v = 3,
\qquad
\frac{1}{u^2} + \frac{1}{v^2} = \frac{104}{25}
$$
But
$$
\frac{1}{u^2} + \frac{1}{v^2}
=
\frac{u^2+v^2}{(uv)^2}
=
\frac{(u+v)^2-2uv}{(uv)^2}
=
\frac{9-2uv}{(uv)^2}
$$
Thus, $uv$ is a root of
$$
\frac{9-2z}{z^2} = \frac{104}{25}
$$
a quadratic equation. Once you know $uv$ and $u+v$, you know $u$ and $v$ by solving another quadratic equation.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\int^{\infty}_{-\infty}\frac{1}{(x^2+x+1)^3}dx=\frac{4\pi}{3\sqrt3}$. I have no idea how to do this question.
I'm given $\int^{\infty}_{-\infty}\frac{1}{x^2+2ax+b^2}dx=\frac{\pi}{\sqrt{b^2-a^2}}$ if $b>|a|$ and I'm asked to prove $\int^{\infty}_{-\infty}\frac{1}{(x^2+x+1)^3}dx=\frac{4\pi}{3\sqrt3}$.
What I've tried:
$\int^{\infty}_{-\infty}\frac{1}{(x^2+x+1)}dx=\frac{2\pi}{\sqrt3}$. I've tried setting a variable as the power, ie: $I(r)=\int^{\infty}_{-\infty}\frac{1}{(x^2+x+1)^r}dx$ but differenciating the inside w.r.t to $r$ doesn't seem to form any differential equation that can be solved.
Answers and hints appreciated!
| Differentiate with respect to $b$ gives
$$\dfrac{d}{db}\int^{\infty}_{-\infty}\frac{1}{x^2+2ax+b^2}dx=\dfrac{d}{db}\frac{\pi}{\sqrt{b^2-a^2}}$$
$$\int^{\infty}_{-\infty}\frac{-2b}{(x^2+2ax+b^2)^2}dx=\frac{-2b\pi}{2\sqrt{(b^2-a^2)^3}}$$
or
$$\int^{\infty}_{-\infty}\frac{1}{(x^2+2ax+b^2)^2}dx=\frac{\pi}{2\sqrt{(b^2-a^2)^3}}$$
and one another derivative gives following result
$$\int^{\infty}_{-\infty}\frac{1}{(x^2+2ax+b^2)^3}dx=\frac{3\pi}{8\sqrt{(b^2-a^2)^5}}$$
now you have the answer with $a=\dfrac12$ and $b=1$.
| {
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Show $x_{n+1} = {1\over 2}x_n^2 - 1$ is bounded below and unbounded above and $x_n$ is increasing.
Let:
$$
\begin{cases}
x_{n+1} = {1\over 2}x_n^2 - 1\\
x_1 = 3\\
n\in \mathbb N
\end{cases}
$$
Show that the sequence $x_n$ is bounded only below and is increasing.
I've started with the following:
$$
x_1 = 3 \\
x_2 = 3.5 \\
x_2 > x_1
$$
Suppose $x_{n+1} > x_n$. Consider the following equation:
$$
x_{n+2} - x_{n+1} = \frac{1}{2}x_{n+1}^2 - \frac{1}{2}x_n^2
$$
By initial assumption:
$$
x_{n+1} > x_n \implies x_{n+1}^2 > x_n^2 \implies \frac{1}{2} x_{n+1}^2 > \frac{1}{2} x_n^2 \implies x_{n+2} > x_{n+1}
$$
Thus $x_n$ sequence is increasing.
But now how do I show the lower bound exist and upper does not? Intuitively for the sequence to be unbounded we need the following condition to be satisfied:
$$
{1\over 2}x_n^2 > 1 \iff x_n > \sqrt2
$$
I'm kindly asking to verify whether i've correctly shown that $x_n$ is monotonically increasing and help with showing the bounds.
| $x_{n+1} - x_n = \dfrac{x_n^2}{2}-1 - x_n= \dfrac{x_n^2 - 2x_n - 1}{2}=\dfrac{(x_n-1)^2-2}{2}$. Thus we need to show: $x_n \ge 3, \forall n \ge 1$. We show this by induction on $n\ge 1$. We have $x_1 = 3$, assume $x_n \ge 3$, we have $x_{n+1} = \dfrac{x_n^2}{2} - 1 \ge \dfrac{3^2}{2}-1= 4.5-1 = 3.5 > 3$. Thus $x_{n+1} - x_n \ge \dfrac{(3-1)^2-2}{2}=1 > 0$. Thus $x_n$ is a strictly increasing sequence, and clearly bounded below by $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2997083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to eliminate vector from equation This is probably a super simple question.
Let's say I have an equation
$$
\mathbf{v}^T(\mathbf{a} + \mathbf{b}) = \mathbf{v}^T(\mathbf{c}).
$$
Does this imply that
$$
(\mathbf{a} + \mathbf{b}) = \mathbf{c}
$$?
Intuitively, I would say that it does. On the other hand, there is no left hand side operation I can come up with that eliminates $\mathbf{v}$. I tried with:
$$
\mathbf{v} = \begin{pmatrix}v_1 \\ v_2\end{pmatrix}
$$
$$
\mathbf{x} = \begin{pmatrix}1/v_1\\1/v_2\end{pmatrix}
$$
And then
$$\mathbf{x}\mathbf{v}^T(\mathbf{a} + \mathbf{b}) = \mathbf{x}\mathbf{v}^T(\mathbf{c})
$$
But this doesn't result in the identity matrix on the left side as I had hoped:
$$
\begin{pmatrix}1/v1\\1/v2\end{pmatrix}\begin{pmatrix}v1\ v2\end{pmatrix} = \begin{pmatrix}1 \ \ v_2/v_1 \\ v_1/v_2 \ 1 \end{pmatrix}
$$
| Consider $\mathbf{v} = \begin{bmatrix}1\\0\\0\end{bmatrix}$, $\mathbf{x} = \begin{bmatrix}0\\1\\0\end{bmatrix}$, $\mathbf{y} = \begin{bmatrix}0\\0\\1\end{bmatrix}$.
Then $\mathbf{v}^T\mathbf{x} = \mathbf{v}^T\mathbf{y} = 0$ but $\mathbf{x} \neq \mathbf{y}$.
In general, $\mathbf{u}^T\mathbf{v}$ is the dot product of $\mathbf{u}$ and $\mathbf{v}$, which measures the angle between them:
$$\mathbf{u}\cdot\mathbf{v} = \lVert \mathbf{u} \rVert \lVert \mathbf{v} \rVert \cos \theta$$
so if $\mathbf{u}\cdot\mathbf{v} = \mathbf{u}\cdot\mathbf{w}$, then all you can conclude is that $\mathbf{v}$ and $\mathbf{w}$ are the same angle from $\mathbf{u}$, provided they have the same length.
| {
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"source": "stackexchange",
"question_score": "1",
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If $a$, $b$, $c$ are positive real numbers such that $a^2 + b^2 + c^2 = 27$. Find the least value of $a^3 + b^3 + c^3$? If $a$, $b$, $c$ are positive real numbers such that $a^2 + b^2 + c^2 = 27$. Find the least value of $a^3 + b^3 + c^3$ ?
I tried with Tchebyshev inequality on sets $\{a, b, c\}$ and $\{a^2, b^2 , c^2\}$ but was blocked. I also tried with mean of $m-$the power with $m=3$ but again could not proceed further.
I only know elementary inequalities like Cauchy-Schwarz and AM-GM apart from above mentioned inequalities. Please help.
| Okay, the question requires knowledge of the AM-GM inequality, which you do have.
You need to apply it twice; first on $a^2,b^2,c^2$ and obtain an inequality on the product $abc$; then on $a^3,b^3,c^3$ and substitute $abc$.
The answer is $81$.
Hope this helps.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solution of complex differential equation involving $2$ variables
Solve the differential equation
$\displaystyle \frac{dy}{dx} = \sqrt{\frac{1}{2}+\int^{-\sin^4
\theta}_{\cos^4 \theta}\frac{\sqrt{f(\phi)}d\phi}{\sqrt{f(\theta)}+\sqrt{f(\cos 2 \theta-\phi)}}},$
where $\theta = x+y$ and $\displaystyle x+y\in \bigg(\frac{\pi}{4}\;,\frac{3\pi}{4}\bigg)$
Try: First we will sove $$I = \int^{-\sin^4
\theta}_{\cos^4 \theta}\frac{\sqrt{f(\phi)}d\phi}{\sqrt{f(\theta)}+\sqrt{f(\cos 2 \theta-\phi)}}\cdots \cdots (1)$$
Replace $\phi = \cos^4 \theta-\sin^4 \theta-\phi = \cos (2\theta)-\phi.$
So $$I = \int^{-\sin^4
\theta}_{\cos^4 \theta}\frac{\sqrt{f(\cos 2 \theta -\phi)}d\phi}{\sqrt{f(\theta)}+\sqrt{f(\cos 2 \theta-\phi)}}\cdots (2)$$
So $$2I = \int^{-\sin^4 \theta}_{\cos^4 \theta}d\phi = \phi\bigg|^{-\sin^4 \theta}_{\cos^4 \theta}$$
$$I = \frac{1}{2}\bigg[\sin^4 \theta +\cos ^4 \theta\bigg] = \frac{1}{2}-\sin^2 \theta \cos^2 \theta$$
So $$\frac{dy}{dx} = \sqrt{1-\sin^2 \theta \cos^2 \theta}$$
could some help me how to solve further, i don,t get any clue
thanks
answer given as $\displaystyle \frac{2}{\sqrt{3}}\tan^{-1}\bigg(\frac{2\tan(x+y)+1}{\sqrt{3}}\bigg)=x+c.$
| I'll assume that you meant to have $f(\phi)$ and not $f(\theta)$ in the denominator of the integral
You made a sign error
$$ I = \frac12 \left(-\sin^4 \theta - \cos^4 \theta \right) = -\frac12 +\sin^2\theta\cos^2\theta $$
Then
$$ \frac{dy}{dx} = \sqrt{\frac12 + I} = \big\vert\sin\theta\cos\theta\vert $$
You also have
$$ \frac{d\theta}{dx} = 1 + \frac{dy}{dx} = 1 + \big\vert\sin\theta\cos\theta\big\vert
= \begin{cases}
1 + \sin\theta\cos\theta, & \theta \in \left(\frac{\pi}{4}, \frac{\pi}{2} \right) \\
1 - \sin\theta\cos\theta, & \theta \in \left( \frac{\pi}{2}, \frac{3\pi}{4} \right)
\end{cases} $$
Separating leads to the integral
$$ \int \frac{1}{1+\sin\theta\cos\theta}d\theta = \int \frac{\sec^2\theta}{\sec^2\theta+\tan\theta}d\theta = \int\frac{1}{u^2+u+1}du $$
where $u = \tan\theta$. Completing the square on the denominator gives
$$ \int\frac{1}{u^2+u+1}du = \int\frac{1}{\left(u+\frac12\right)^2+\frac34}du = \frac{2}{\sqrt{3}}\arctan\left(\frac{2u+1}{\sqrt{3}}\right) + c $$
| {
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Infinite sequence $2^{n}-3 (n=2,3,...)$ contains no term divisible by 65 Show that the infinite sequence $2^{n}-3 (n=2,3,...)$ contains infinitely many terms which are divisible by $5$ and infinitely many terms which are divisible by $13$, but no terms which are divisible by $65$
My attempt at this:-
By Fermat's theorem,
$$2^{4}\equiv 1\pmod5$$
Raising to the power k we get,
$$2^{4k}\equiv 1\pmod5$$
$$2^{4k+3}\equiv 8\pmod5$$
$$2^{4k+3}\equiv 3\pmod5$$
So, $5\mid 2^{n}-3\quad\forall \quad n=4k+3$ where $k$ is any non-negative integer.
Similarly, by Fermat's theorem
$$2^{12}\equiv 1\pmod{13}$$
$$2^{12k}\equiv 1\pmod{13}$$
$$2^{12k+4}\equiv 16\pmod{13}$$
$$2^{12k+4}\equiv 3\pmod{13}$$
Therefore, $13\mid 2^{n}-3\quad\forall \quad n=12k+4$
How do I show that it contains no term which is divisible by 65?
Thank you!
| Hint:
Write $12k+4$ as $12n+4$ which needs to be equal $4k+3$ which is odd unlike the former and $k,n$ are arbitrary integers
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $k$ in Maclaurin series expansion of $\frac{dy}{dx}=-\frac{1}{2}+\frac{1}{4}x+kx^2+...$ where $y=\ln\Bigl(\frac{e^{-x}+1}{2}\Bigl)$
Given that $y=\ln\Bigl(\frac{e^{-x}+1}{2}\Bigl)$, show $\frac{dy}{dx}=\frac{1}{2}e^{-y}-1$. Show that the series expansion of $\frac{dy}{dx}$ in ascending powers of $x$, up to and including the term in $x^2$ is $-\frac{1}{2}+\frac{1}{4}x+kx^2+\dots$ , where $k$ is to be determined.
I'm able to solve this question but I'm unsure if my $k$ value is correct. My $k$ is $0$. Am I correct? Otherwise, I might have made a mistake somewhere.
My work
1) $f(0)=y=0$
2) $\frac{dy}{dx}=\frac{1}{2}e^{-y}-1\ $
therefore $f'(0)=-\frac{1}{2}$
3) $\frac{d^2y}{dx^2}=-\frac{1}{2}e^{-y}\cdot\Big(\frac{dy}{dx}\Bigl)\ $
therefore $f''(0)=\frac{1}{4}$
4) $\frac{d^3y}{dx^3}=\frac{1}{2}e^{-y}\cdot\Bigl(\frac{dy}{dx}\Bigl)^2-\frac{1}{2}e^{-y}\cdot\Bigl(\frac{d^2y}{dx^2}\Bigl)\ $
therefore $f'''(0)=0$
Hence
$$\frac{dy}{dx}=f'(0)+f''(0)x+\frac{f'''(x)}{2}x^2+\dots=-\frac{1}{2}+\frac{1}{4}x+\dots$$
| Yes, you are correct $k=0$. This is an alternative solution where we use the expansions of $e^t$ and $(1+t)^{-1}$ at $t=0$:
$$\begin{align}
\frac{dy}{dx}&=\frac{2}{e^{-x}+1}\cdot \frac{-e^{-x}}{2} =-\frac{1}{1+e^x}\\
&=-\frac{1}{1+1+x+\frac{x^2}{2}+o(x^2)}\\
&=-\frac{1}{2}\left(1+\frac{x}{2}+\frac{x^2}{4}+o(x^2)\right)^{-1}\\
&=-\frac{1}{2}\left(1-\left(\frac{x}{2}+\frac{x^2}{4}\right)+\left(\frac{x}{2}+o(x)\right)^2+o(x^2)\right)\\
&=-\frac{1}{2}+\frac{x}{4}+\underbrace{\left(-\frac{1}{4}+\frac{1}{4}\right)}_{=0}\cdot x^2+ o(x^2).
\end{align}$$
| {
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General method to find sum of binomial Specifically I want to ask the method to solve
$$\sum_{k=0}^n \binom{4n+b}{4k};\ b=[0,1,2,3]$$
And how to solve series of type
$$\sum_{k=0}^n \binom{an+b}{ak};\ a=[1,2,3,...],\ b=[0,1,2,...,a-1]$$
| I am sure that you will enjoy the result !
If $$S_n=\sum_{k=0}^n \binom{4n+b}{4k}$$ then a CAS gives
$$4S_n=2^{\frac{4 n+b}{2} } \left((-1)^b\, 2^{\frac{4 n+b}{2} } \cos (\pi b)+\cos
\left(\frac{ (4 n+b)\pi}{4} \right)+(-1)^b \cos \left(\frac{3(4
n+b)\pi}{4} \right)\right)-$$ $$4 \binom{4 n+b}{4 n+4} \,
_5F_4\left(1,\frac{4-b}{4},\frac{5-b}{4},\frac{6-b}{4},\frac
{7-b}{4};\frac{4n+5}{4},\frac{4n+6}{4},\frac{4n+7}{4},\frac{4n+8}{4};1\right)$$ where appears the generalized hypergeometric function.
For the first values of $b$, the sequences can be found in $OEIS$.
$$\left(
\begin{array}{cc}
b & \text{OEIS sequence} \\
0 & \text{A070775} \\
1 & \text{A090407} \\
2 & \text{A001025} \\
3 & \text{A090408}
\end{array}
\right)$$
Be sure that I did not make this by myself.
| {
"language": "en",
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Proving $(n+1)^2+(n+2)^2+...+(2n)^2= \frac{n(2n+1)(7n+1)}{6}$ by induction My question: $(n+1)^2+(n+2)^2+(n+3)^2+...+(2n)^2= \frac{n(2n+1)(7n+1)}{6}$
My workings
*
*LHS=$2^2$ =$4$ RHS= $\frac{24}{6} =4 $
*$(k+1)^2+(k+2)^2+(k+3)^2+...+(2k)^2= \frac{k(2k+1)(7k+1)}{6}$
*LHS (subsituting $n= k+1$)----> $(k+2)^2+(k+3)^2+(k+4)^2+...+(2k)^2+(2k+1)^2$
Now, this is where my problem has started. When I substitute $n=k+1$ in the third step, I do not have the $(k+1)^2$ anymore. So I cannot use the statement in second step. So my question is what should i do next?
| Assuming the step for $n=k$ $\:$For
$n=k+1$ $(k+2)^2+\cdots+(2k+2)^2=\dfrac{k(2k+1)(7k+1)}{6}-(k+1)^2+(2k+1)^2+(2k+2)^2$ $=\dfrac{k(2k+1)(7k+1)}{6}+3(k+1)^2+(2k+1)^2=\dfrac{k(2k+1)(7k+1)}{6}+7k^2+10k+4=\dfrac{(k+1)(2k+3)(7k+8)}{6}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3011450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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$\lim_{x\to -\infty} x+\sqrt{x^2-3x}$ Hey so I'm having a bit of a hard time understanding this one.
$\lim_{x\to -\infty} x+\sqrt{x^2-3x}$
1) $x+\sqrt{x^2-3x}$ * $(\frac{x-\sqrt{x^2-3x}}{x-\sqrt{x^2-3x}})$
2) $\frac{x^2-(x^2-3x)}{x-\sqrt{x^2-3x}}$
3) $\frac{3x}{x-\sqrt{x^2(1-\frac{3}{x}})}$
4) $\frac{3x}{x-\sqrt{x^2}*\sqrt{1-\frac{3}{x}}}$
5) $\frac{3x}{x-x(\sqrt{1-\frac{3}{x}})}$
6) $\frac{3}{1-(\sqrt{1-\frac{3}{x}})}$
Now I would just take the limit, it would result in $\frac{3}{1-1}$ which would be undefined. For some reason, the $x$ in the denominator of step 5 should turn into $-(-x)$ which in turn would be positive and therefore be $\frac{3}{1+\sqrt{1=\frac{3}{x}}}$ which would equal $\frac{3}{2}$.
I really don't get it. Apparently the $-\infty$ would mean that $\sqrt{x^2}$ = $-x$. We didn't even evaluate the limit yet.. how does that turn into $-x$, just because we know the limit is negative does not mean we evaluated it yet..., why not simplify until there is no more simplification to be done, which is what I did in my steps, which would evaluate to undefined?
Would love some help, thanks!
| We are near $-\infty$, so $x<0$ and $|x|=-x$.
thus
$$x+\sqrt{x^2-3x}=x+\sqrt{x^2(1-\frac 3x)}$$
$$=x+|x|\sqrt{1-\frac 3x}$$
$$=x(1-\sqrt{1-\frac 3x})$$
$$=x\frac{\frac 3x}{1+\sqrt{1-\frac 3x}}$$
the limit is $\frac 32$.
| {
"language": "en",
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"question_score": "3",
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"answer_id": 3
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Use $\cos 5\theta$ to find the roots of $x(16x^4 - 20x^2 + 5) = 0$ I used $\cos(3\theta + 2\theta)$ to prove the first part, but I don't know how to the $2$nd part.
Show that $\cos 5\theta=16\cos^5\theta-20\cos^3\theta+5\cos\theta,$ and hence show that
$$\text{the roots of }x(16x^4 - 20x^2 + 5) \text{ are: } 0,\cos\frac{\pi}{10}, \cos\frac{3\pi}{10},\cos\frac{7\pi}{10}, \cos\frac{9\pi}{10}$$
|
Here the value of cosine will repeat after point.
Final values wil be only 5
| {
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How to get from $x^{p-1}-1$ to $(x-1)(x^{p-2}+x^{p-3}+\cdots+x+1)$? How would I get from $x^{p-1}-1$ to $(x-1)(x^{p-2}+x^{p-3}+\cdots+x+1)$?
It make sense to me logically. When one multiplies it out, it would condense to $x^{p-1}-1$. But it's just not clicking. What is the arithmetic between these steps?
| I'm not sure if this fully answers your question, however, when I am factoring higher degree polynomials (degree greater than 2) I often like to use the following trick that I will demonstrate below:
On polynomials like the one you have there, it is easy to see that $x=1$ is a zero of the polynomial $x^{p-1}-1$. Then you want to use the following trick where you kind of 'reverse' expand the polynomial:
\begin{align}
x^{p-1} - 1 &= x^{p-1} - x^{p-2} + x^{p-2} - x^{p-3} + x^{p-3} -...- x + x - 1 \\
&= (x^{p-1} + x^{p-2} + x^{p-3} + ... + x) - (x^{p-2} + x^{p-3} + x^{p-4} +...+ 1) \\
&= x(x^{p-2} + x^{p-3} + x^{p-4} +...+ 1) - (x^{p-2} + x^{p-3} + x^{p-4} +...+ 1) \\
&= (x-1)(x^{p-2} + x^{p-3} + x^{p-4} +...+ 1) \\
\end{align}
So you basically add and subtract terms without changing the polynomial so it is easy to pull out the $(x-1)$ factor. Just as another example to better illustrate the technique on a polynomial you don't know the factors of:
Lets say we wanted to factor the polynomial $x^3 -9x^2 +26x -24$. With some quick trial and error you can find that $x=2$ is a zero of the polynomial. Then, as before, you want to rearrange the polynomial so that it easy to pull out the $(x-2)$ factor. We do it as follows:
\begin{align}
x^3 -9x^2 +26x -24 &= x^3 - 2x^2 - 7x^2 + 14x + 12x - 24 \\
&= (x^3 - 7x^2 + 12x) - (2x^2 - 14x +24) \\
&= x(x^2 -7x +12) - 2(x^2 - 7x + 12) \\
&= (x-2)(x^2 -7x +12)\\
\end{align}
So all we have done is leave the highest/lowest degree terms alone and split the middle terms into two parts, one a multiple of $x$ and the other a multiple of -2. It is then easy to rearrange the polynomial and factor out the $(x-2)$ factor. You can apply this technique to any polynomial that you already know a zero of, however, as you can imagine it gets messy if the zeroes of the polynomial aren't 'small' integers.
| {
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Produce formula Let $a_{i,j}$ any positive integers wher $1\leq i,j \leq 3$.
I want to write closed form for following summation:
$$a_{1,3}(a_{2,1}+a_{2, 2}+a_{2,3})(a_{3,1}+a_{3, 2}+a_{3,3})+a_{2,3}(a_{1,1}+a_{1, 2})(a_{3,1}+a_{3, 2}+a_{3,3})+a_{3,3}(a_{1,1}+a_{1, 2})(a_{2,1}+a_{2, 2})$$
$\textbf{My attempt:}$
\begin{equation*}
\sum_{j=1}^3 a_{j,3}\left[\prod_{\substack{i=1 \\ i\neq j}}^3\left(\sum_{\substack{k=1 \\ k<j}}^{3 \ \ \ \text{or} \ \ 3-1} a_{i,k} \right) \right]
\end{equation*}
We will take $3-1$ when $k<j$.
Actually, I want to generalize this. My attempt so complicated to me.
If we consider all of $a_{i,j}$ as a matrix:
\begin{pmatrix}
a_{1,1} & a_{1,2} & a_{1,3} \\
a_{2,1} & a_{2,2} & a_{2,3} \\
a_{3,1} & a_{3,2} & a_{3,3} \\
\end{pmatrix}
then are there any computer program to make formula?
$$a_{1,3}(a_{2,1}+a_{2, 2}+ a_{2,3})(a_{3,1}+a_{3, 2}+a_{3,3})\cdots (a_{r,1}+a_{r,2}+a_{r,3})+a_{2,3}(a_{1,1}+a_{1, 2})(a_{3,1}+a_{3, 2}+a_{3,3})+\cdots (a_{r,1}+a_{r,2}+a_{r,3})+a_{3,3}(a_{1,1}+a_{1, 2})(a_{2,1}+a_{2, 2})(a_{4,1}+a_{4,2}+a_{4,3}+\cdots (a_{r,1}+a_{r,2}+a_{r,3}) \\ \vdots \\
+a_{r,3}(a_{1,1}+a_{1,2})(a_{2,1}+a_{2,2})\cdots (a_{r-1,1}+a_{r-1,2})
$$
| $$a_{1,3}(a_{2,1}+a_{2, 2}+a_{2,3})(a_{3,1}+a_{3, 2}+a_{3,3})+a_{2,3}(a_{1,1}+a_{1, 2})(a_{3,1}+a_{3, 2}+a_{3,3})+a_{3,3}(a_{1,1}+a_{1, 2})(a_{2,1}+a_{2, 2})$$
$$=(a_{1,1}+a_{1,2}+a_{1,3})(a_{2,1}+a_{2,2}+a_{2,3})(a_{3,1}+a_{3,2}+a_{3,3})-(a_{1,1}+a_{1,2})(a_{2,1}+a_{2,2})(a_{3,1}+a_{3,2})$$
$$=\prod_{i=1}^3\sum_{j=1}^3a_{i,j}-\prod_{i=1}^3\sum_{j=1}^2a_{i,j}.$$
More generally,
$$\sum_{i=1}^ra_{i,m}\left(\prod_{h=1}^{i-1}\sum_{j=1}^{m-1}a_{h,j}\right)\left(\prod_{h=i+1}^r\sum_{j=1}^ma_{h,j}\right)=\prod_{i=1}^r\sum_{j=1}^ma_{i,j}-\prod_{i=1}^r\sum_{j=1}^{m-1}a_{i,j}$$
$$=\sum_{m\in\{j_1,j_2,\dots,j_r\}}a_{1,j_1}a_{2,j_2}\cdots a_{r,j_r}.$$
| {
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Why is the inequality $\sum_{n=1}^{\infty} \frac{1}{n^2} \leq 1 + \int_1^{\infty} \frac{1}{x^2}$ true? $$\sum_{n=1}^{\infty} \frac{1}{n^2} \leq 1 + \int_1^{\infty} \frac{1}{x^2}dx$$
I'm having trouble figuring out why the inequality above is true. I understand the following inequality:
$$\int_1^{\infty} \frac{1}{x^2}dx \leq \sum_{n=1}^{\infty} \frac{1}{n^2}$$
It makes sense because the rectangles formed from the right side of the inequality have pieces that go over-top of the function like so:
So let's say I rewrite $\sum_{n=1}^{\infty} \frac{1}{n^2}$ as $1 + \sum_{n=2}^{\infty} \frac{1}{n^2}$ since they are equivalent.
Why is that less than $1 + \int_1^{\infty} \frac{1}{x^2}dx$?
If I picture it in my head it sounds like its saying that the first term is greater than all the little pieces that form above the curve in the picture above.
My claims come specifically from page 60 of this webpage from Dartmouth
| The right endpoint sums for the integral have the form:
$$
\sum_{n = 2}^\infty \frac{1}{n^2}
$$
and we know:
$$
\sum_{n = 2}^\infty \frac{1}{n^2} \leq \int_1^\infty \frac{1}{x^2} \leq \sum_{n = 1}^\infty \frac{1}{n^2}
$$
Subtracting the RHS, we have:
$$
-1 \leq \int_1^\infty \frac{1}{x^2} - \sum_{n = 1}^\infty
\frac{1}{n^2}$$
Multiplying by $-1$, we have:
$$
1 \geq \sum_{n = 1}^\infty
\frac{1}{n^2} -\int_1^\infty \frac{1}{x^2}
$$
as we wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3016140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Number of solutions in quadratic congruence I use an example to explain my question:
How many solutions are there to $x^2+3x+18\equiv 0$ (mod $28$).
Usually I will encounter these kind of problems. My first step must transform this equation to become 2 equations:
$$x^2+3x+18\equiv 0 \quad (\text{mod }7) $$
$$x^2+3x+18\equiv 0\quad (\text{mod }4) $$
Then, I just plug in the numbers $0$ to $6$ to eq 1 to find out there is one solution $x\equiv 2$ mod $7$.
For equation 2, I do the same thing, plug in $0$ to $3$ and there are two solutions:$x\equiv 2$ and $x\equiv 3$ mod $4$.
Then how should I continue?? Using Chinese Remainder Theorem? Because I have made the original equation into a two simultaneous equations, maybe something like:
*
*$x\equiv 2$ mod $7$ and $x\equiv 2$ mod $4$
*$x\equiv 2$ mod $7$ and $x\equiv 3$ mod $4$
Is it the general way to do it? But is there a faster way to determine the number of solutions without performing CRT?
| $x\equiv2\pmod4,x\equiv2\pmod7\implies$lcm$(4,7)|(x-2)\implies x\equiv2\pmod{28}$
For the second, $$7a+2=4b+3\iff7a=4b+8-7\iff\dfrac{7(a+1)}4=b+2$$ which is an integer
$\implies4|7(a+1)\iff4|(a+1)$ as $(4,7)=1$
$\implies a+1=4c$
$\implies x=7a+2=7(4c-1)+2=28c-5\equiv-5\pmod{28}\equiv-5+28$
| {
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"url": "https://math.stackexchange.com/questions/3019723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving the Diophantine equation $y^2 = 4x^3 + 1$ for $x,y \in \mathbb{Z}$
I want to solve the Diophantine equation $y^2 = 4x^3 + 1$ for $x,y \in \mathbb{Z}$.
Note that $y$ is odd, since $y$ even would give a contradiction $\mod{2}$. Hence $\frac{y-1}{2}, \frac{y+1}{2} \in \mathbb{Z}$.
So we can rewrite the equation to: $\frac{y-1}{2} \frac{y+1}{2} = x^3$.
Claim $\frac{y-1}{2}$ and $\frac{y+1}{2}$ are coprime.
Proof: Let $d = \gcd(\frac{y-1}{2},\frac{y+1}{2})$. Let $\frac{y-1}{2} = a d$ and $\frac{y+1}{2} = b d$ for some $a,b \in \mathbb{Z}$. Then $y + 1 = bd \,2$ and $y-1 = a d\, 2$, so $y + 1 = ad\,2+2$, it follows that $bd = ad + 1$, so $d(b-a)=1$, so $d =\pm1$. So we conclude that indeed $\frac{y-1}{2}$ and $\frac{y+1}{2}$ are coprime.
Since they are coprime we can perform a descent. That gives $\frac{y-1}{2} = e^3$ and $\frac{y+1}{2} = f^3$ for some coprime $e,f \in \mathbb{Z}$. After adding and subtracting these two equations we find that $y = e^3+f^3$ and $1 = f^3 - e^3$.
Am I now correct to say that the only solutions are $f = 1, e=0$ and $f = 0, e=-1$?
This gives $y=\pm1$. In the above equation we see that in both cases $x = 0$.
We conclude by saying that the only solutions are $(x,y) = (0,\pm1)$.
Did I make any mistakes?
Are there more efficient methods?
| Of course you can, from $f^3-e^3=1$ we get $$(f-e)(f^2+fe+e^2)=1$$
so $f-e=1$ and $f^2+fe+e^2 = 1$ or $f-e=-1$ and $f^2+fe+e^2 = -1$
So you have 2 cases. You can express $f$ with $e$ and plug in the second equation...
Else, I would introduce $y= 2k+1$ for some $k$, then you would have $k(k+1)=x^3$ which is a little easer to handle. One thing you can notice faster is that $k$ and $k+1$ are coprime.
Nicely done anyway.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Laurent Series expansion about the point $z_0 = i$ of $\frac{z}{z^2+1}$ I am trying to construct the Laurent series expansion of $f(z) = \frac{z}{z^2+1}$ about $z_0 = i$ in the region $\{z \in \mathbb{C}: 0 < |z - i| < 2\}$ but I am stuck.
We can re-write $f(z) = \dfrac{z}{z^2+1} = \dfrac{z}{(z-i)(z+i)}$
This allows us to see more easily that there are first order poles at $\pm i$. We can use partial fractions to obtain:
$$\frac{1}{(z+i)(z-i)} = \frac{i}{2(z+i)} - \frac{i}{2(z-i)}$$
We want to re-write everything in terms of $z-i$:
$$\frac{1}{z+i} = \frac{1}{2i + (z - i)} = \frac{1}{2i}\left(\frac{1}{1+\frac{1}{2i}(z-i)}\right)$$
Using the geometric series this gives:
$$ = \frac{-i}{2} \left(\sum_{n=0}^{\infty} (-1)^n \frac{(z-i)^n}{(2i)^n}\right)$$
Then we have that:
$$f(z) = z\left(\frac{i}{2}\right)\left(\left(\frac{-i}{2}\sum_{n=0}^{\infty} (-1)^n \frac{(z-i)^n}{(2i)^n}\right) - \frac{1}{z-i}\right)$$
But the problem is, I still have a term in the form of $z$ rather than $z-i$. How do I proceed from here?
| \begin{align}&f(z)=\frac{z}{z^2+1}=\frac{A}{z-i}+\frac{B}{z+i}\\&=\frac{1}{2}\frac{1}{z-i}+\frac{1}{2}\frac{1}{z+i}\end{align}
\begin{align}&A=Res[f,i]=\lim\limits_{z\rightarrow i}\frac{z}{z+i}=\frac{1}{2}
\\&B=Res[f,-i]=\lim\limits_{z\rightarrow -i}\frac{z}{z-i}=\frac{1}{2}\end{align}\begin{align}f(z)&= \frac{1}{2}\frac{1}{z-i}+\frac{1}{2}\frac{1}{2i\left(1+\frac{z-i}{2i}\right)}\\&=\frac{1}{2}\frac{1}{z-i}+\frac{1}{2}\sum_{n=0}^{\infty}\frac{(-1)^n (z-i)^n}{(2i)^{n+1}}\end{align}\begin{align}=&\frac{1}{2}\frac{1}{z-i}-\frac{i}{4}+\frac{1}{8}(z-i)\\&+\frac{i}{16}(z-i)^2+\cdots\end{align}
in the region $\{z\in\mathbb{C}:0<|z-i|<2\}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3023007",
"timestamp": "2023-03-29T00:00:00",
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Solve $\sqrt{2}\sec x+\tan x=1$
Solve $\sqrt{2}\sec x+\tan x=1$
I understand it can be very easily solved by expanding in terms of $\sin x$ and $\cos x$, gives $x=2n\pi-\frac{\pi}{4}$. But, what if I do the following:
$$
\sqrt{2}\sec x+\tan x=1\\
\text{Differentiating}\implies\sqrt{2}\sec x\tan x+\sec^2 x=0\implies\sqrt{2}\tan x+\sec x=0
$$
Step 1
$$
\sec x=-\sqrt{2}\tan x=\frac{1-\tan x}{\sqrt{2}}\implies2\tan x=\tan x-1\implies\tan x=-1\\
\boxed{x=n\pi-\frac{\pi}{4}}
$$
Step 2
$$
\tan x=1-\sqrt{2}\sec x=\frac{-\sec x}{\sqrt{2}}\implies2\sec x-\sqrt{2}=\sec x\\
\implies\sec x=\sqrt{2}\implies\boxed{x=2n\pi\pm\frac{\pi}{4}}
$$
$$
x=n\pi-\frac{\pi}{4}\quad\&\quad x=2n\pi\pm\frac{\pi}{4}\\\implies \bigg[x=2n\pi-\frac{\pi}{4}\text{ or }x=2n\pi+\frac{3\pi}{4}\bigg]\quad\&\quad x=2n\pi\pm\frac{\pi}{4}\\
\implies \boxed{x=2n\pi-\frac{\pi}{4}}
$$
In my attempt why do I need Step 2 to get the complete solution ?
Can someone give a proper explanation to my attempt ?
| HINT
As noticed we can't use differentiation to obtain the result indeed in general
$$f'(x)=g'(x) \not \Rightarrow f(x)=g(x)$$
consider for example the simple case
$$2x+3=1 \to 2=0$$
I suggest to use tangent half-angle identities by $t = \tan \frac x2$ to obtain
$$\sqrt{2}\sec x+\tan x=1 \iff \sqrt{2}\frac{1+t^2}{1-t^2}+\frac{2t}{1-t^2}=1$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Radius of convergence of $\sum\limits_{n=1}^\infty((\frac{1}{4})^n+(\frac{1}{3})^n)x^n$ according to wolfram alpha the radius is 3. I'm struggeling with the proof and would be very glad if someone could take a short look. Here my approach using the root test:
$a_n:=\sqrt[n]{(\frac{1}{4})^n+(\frac{1}{3})^n}<\sqrt[n]{(\frac{1}{3})^n+(\frac{1}{3})^n}=\sqrt[n]{2(\frac{1}{3})^n}=\sqrt[n]2(\frac{1}{3})=:b_n$
$\implies \limsup\limits_{n\rightarrow\infty}a_n<\limsup\limits_{n\rightarrow\infty}b_n=\frac{1}{3}$
$\left(\implies \forall x \in \mathbb{R}: |x|<3 \implies \exists \alpha \in \mathbb{R}: \lim\limits_{n\rightarrow\infty}\sum_{n=1}^\infty b_nx^n=\alpha \right)$
But that does not necessarily mean that $\lim\limits_{n\rightarrow\infty}\sum_{n=1}^\infty a_nx^n$ has the same radius of convergence, does it? The radius could still be smaller, right?
Hints to a better proof are greatly appreciated! :)
| We have that
$$|x|\cdot\sqrt[n]{\left(\frac{1}{4}\right)^n+\left(\frac{1}{3}\right)^n}=|x|\cdot\frac13\sqrt[n]{\left(\frac{3}{4}\right)^n+1} \to|x|\cdot\frac13\cdot 1 = |x|\cdot\frac13<1$$
and therefore the radius of convergence is $3$.
Moreover note that
*
*for $x=3 \implies \left(\left(\frac{1}{4}\right)^n+\left(\frac{1}{3}\right)^n\right)x^n=\left(\frac{3}{4}\right)^n+1$
*for $x=-3 \implies \left(\left(\frac{1}{4}\right)^n+\left(\frac{1}{3}\right)^n\right)x^n=\left(-\frac{3}{4}\right)^n+(-1)^n$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Seeking Methods to solve $\int_{0}^{\frac{\pi}{2}} \ln\left|\sec^2(x) + \tan^4(x) \right|\:dx $ After weeks of going back and forth I've been able to solve the following definite integral:
$$I = \int_{0}^{\frac{\pi}{2}} \ln\left|\sec^2(x) + \tan^4(x) \right|\:dx $$
To solve this I employ Feynman's Trick with Glasser's Master Theorom but I'm excited to learn of other methods that can be employed. Are there any other 'tricks' that can be used? or alternatively series based solutions? or transformations? (or anything for that matter).
For those who may be interested my process was:
(1) First make the substitution: $u = \tan(x)$
$$I = \int_{0}^{\infty} \frac{\ln\left|u^2 + 1 + u^4 \right|}{u^2 + 1}\:du = \int_{0}^{\infty} \frac{\ln\left|1 + u^2\left(u^2 + 1\right) \right|}{u^2 + 1}\:du$$
(2) Now employ Feynman's Trick by introducing a new parameter:
$$I(t) = \int_{0}^{\infty} \frac{\ln\left|1 + t^2u^2\left(u^2 + 1\right) \right|}{u^2 + 1}\:du$$
Note here that $I = I(1)$ and $I(0) = 0$
(3) Take the derivative w.r.t 't'
$$I'(t) = \int_{0}^{\infty} \frac{2tu^2\left(u^2 + 1\right)}{1 + t^2u^2\left(u^2 + 1\right)}\frac{1}{u^2 + 1}\:du = \frac{1}{t} \int_{-\infty}^{\infty} \frac{1}{\left(u - \frac{1}{tu}\right)^2 + \frac{2}{
t} + 1}\:du$$
(4) Employ Glasser's Master Theorem:
$$I'(t) = \frac{1}{t} \int_{-\infty}^{\infty} \frac{1}{\left(u - \frac{1}{tu}\right)^2 + \frac{2}{t} + 1} \:du= \frac{1}{t}\int_{-\infty}^{\infty}\frac{1}{u^2 + \frac{2}{t} + 1} \:du$$
As: $\frac{2}{t} + 1 > 0 $ we arrive at
$$I'(t) = \frac{1}{t}\left[\frac{1}{\sqrt{\frac{2}{t} + 1}}\arctan\left(\frac{u}{\frac{2}{t} + 1}\right)\right]_{-\infty}^{\infty}= \frac{\pi}{\sqrt{t\left(t + 2\right)}}$$
(5) We now integrate w.r.t 't'
$$I(t) = \int \frac{\pi}{\sqrt{t\left(t + 2\right)}}\:dt = 2\pi\sinh^{-1}\left(\frac{t}{\sqrt{2}} \right) + C$$
Where $C$ is the constant of integration. As above $I(0) = 0 \rightarrow C = 0$ and so, our final solution is given by:
$$I = I(1) = 2\pi\sinh^{-1}\left(\frac{1}{\sqrt{2}} \right)$$
| $$I = \int_{0}^{\frac{\pi}{2}} \ln\left(\sec^2(x) + \tan^4(x) \right)dx=\int_0^\infty \frac{\ln(1+x^2+x^4)}{1+x^2}dx$$Consider: $$I(a)=\int_0^\infty \frac{\ln((1+x^2)a+x^4)}{1+x^2}dx$$
Derivating under the integral sign with respect to $a$ gives: $$I'(a)=\int_0^\infty \frac{1+x^2}{(1+x^2)a+x^4}\frac{dx}{1+x^2}=\int_0^\infty \frac{1}{x^4+ax^2+a}dx\overset{\large{x=\frac{\sqrt a}{t}}}=\int_0^\infty \frac{\frac{t^2}{\sqrt a}}{t^4+at^2+a}dt$$
$$2I'(a)=\int_0^\infty \frac{\frac{t^2}{\sqrt a}+1}{t^4+at^2+a}dt\Rightarrow I'(a)=\frac{1}{2\sqrt a}\int_0^\infty \frac{t^2+\sqrt a}{t^4+at^2+a}dt$$
$$=\frac{1}{2\sqrt a}\int_0^\infty \frac{1+\frac{\sqrt a}{t^2}}{\left(t-\frac{\sqrt a}{t}\right)^2+a+2\sqrt a}dt=\frac{1}{2\sqrt a}\int_0^\infty \frac{d\left(t-\frac{\sqrt a}{t}\right)}{\left(t-\frac{\sqrt a}{t}\right)^2+\left(\sqrt{a+2\sqrt a}\,\right)^2}$$
$$=\frac{1}{2\sqrt a}\frac{1}{\sqrt{a+2\sqrt a}}\arctan\left(\frac{t-\frac{\sqrt a}{t}}{\sqrt{a+2\sqrt a}}\right)\bigg|_0^\infty \Rightarrow I'(a)=\frac{\pi}{2\sqrt a}\frac{1}{\sqrt{a+2\sqrt a}}$$
And noticing that $I(0)=4\int_0^\infty \frac{\ln x}{1+x^2} dx=0$. By the fundamental theorem of Calculus we have: $$I=I(1)-I(0)=\int_0^1 I'(a)da=\frac{\pi}{2}\int_0^1 \frac{1}{\sqrt a \sqrt {a+2\sqrt a}}da$$ Finally setting $\sqrt a =x$ gives:
$$I=\pi \int_0^1 \frac{1}{\sqrt{(x+1)^2-1}}dx=\pi\ln(2+\sqrt 3)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
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Limit Question: doubting about my answer! A limit where $x \to \infty$ Hey guys so I have this limit:
$$\lim_{x \to ∞} f(x) = \frac{(3x+1)^3(x-1)}{(x-2)^4}$$
and I got $27$ as final answer, just wondering if you guys can check;
I expanded the numerator and denominator and then divided everything by $x^4$
| Note that
$$\frac{(3x+1)^3(x-1)}{(x-2)^4}=27\frac{(x+1/3)^3(x-1)}{(x-2)^4}=27\frac{\frac{(x+1/3)^3}{x^3}\frac{(x-1)}x}{\frac{(x-2)^4}{x^4}}\\=27\frac{(1+1/x)^3(1-1/x)}{(1-2/x)^4}\to27\frac{(1+0)^3(1-0)}{(1-0)^4}=27$$
as $x\to\infty$, so your result is correct.
| {
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"timestamp": "2023-03-29T00:00:00",
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Joint PDF P[X+Y<=0.5] I am having some difficulty with this probability that I need to compute, and I could use some help from other eyes to see where I am messing up:
Joint PDF: $x+y$ for $0 < x < 1$, $0 < y < 1$.
I need to find the $P(x+y\leq 0.5)$.
For the double integration, I have the following bounds:
*
*Outer bound is respectively to $x$ and is from $0$ to $0.5$.
*Inner bound is respective to $y$ and is from $0$ to $0.5 - x$.
Here is the initial set up:
$\int_0^.5 \int_0^{.5-x}(x +y) dydx$.
My steps of integration:
1: $\int_0^.5 \left(xy + \frac{y^2}{2}\right)\Big|_0^{.5-x}dx$.
2: $\int_0^.5$ $\left(\frac{x}{2} -x^2+\frac{(0.5-x)^2}{2}\right) dx $.
3: $(\frac{x^2}{4} - \frac{x^3}{3} + \frac{(0.5-x)^3}{6})\Big|_0^{0.5}$.
I am getting a negative number based on this final equation, and the answer should be $1/24$. Can someone help me figure out where I went wrong?
| Your made the sign error in step 2 (as indicated by KaviRamaMurthy).
As an alternative, you can simplify in step 2 before integrating:
$$\int_0^{0.5}\left(\frac{x}{2} -x^2+\frac{(0.5-x)^2}{2}\right) dx=
\int_0^{0.5}\left(\require{cancel}\cancel{\frac{x}{2}} -x^2+\frac{1}{8}-\cancel{\frac x2}+\frac{x^2}{2}\right) dx=\\
\int_0^{0.5}\left(-\frac{x^2}{2}+\frac18\right) dx=
\left(-\frac{x^3}{6}+\frac x8\right)|_0^{0.5}=
-\frac{1}{48}+\frac{1}{16}=
\frac 1{24}.$$
Answering your comment, yes, for the upper limit $0.5$ the term will be $0$, but for the lower limit $0$, the term will be $\frac1{48}$:
$$(\frac{x^2}{4} - \frac{x^3}{3} \color{red}{-} \frac{(0.5-x)^3}{6})\Big|_0^{0.5}=\left(\frac{1}{16}-\frac1{24}-0\right)-\left(0-0-\frac1{48}\right)=\frac1{24}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integer equations I have $2$ following problems. Find integer roots of
$$\begin{align}
&1)~\frac{x+y}{x^2-xy+y^2}=\frac3z \\
&2)~x^3y^3-4xy^3+y^2+x^2-2y-3=0
\end{align}$$
I have no idea to solve them. I try to guess roots of the second, they are $\left( -2, 1\right), \left( 0, -1\right), \left( 0, 3\right), \left( 2, 1\right) $. Please help me. Thank you very much.
| Question 1:
\begin{align}
& \frac{x+y}{x^2-xy+y^2} = \frac 3z \\
\iff & \frac{3(x^2-xy+y^2)}{x+y} = z\\
\iff & 3(x+y) - \frac{9xy}{x+y} = z \\
\iff & \frac{9xy}{x+y} = 3x+3y-z \\
\implies & \frac{9}{\frac 1x + \frac 1y} = 3x+3y-z
\end{align}
From the LHS we see that $\frac 1x + \frac 1y$ must be equal to $\pm 1, \pm 3, \pm 9$ since the RHS is an integer. But of course, since $x,y$ are integers, $\frac 1x + \frac 1y \in [-2,2]$ and it follows that $\frac 1x + \frac 1y = \pm 1$.
Moreover, the only way this can happen is if $x = y = \pm 2$.
Hence, the only solutions are $(x,y,z) = (2,2,3)$ and $(x,y,z) = (-2,-2,-3)$.
EDIT:
As pointed out in the comments, I have made the mistake of dividing by $0$, so I have changed the last $\iff$ to an $\implies$.
If $x=0$, then $z=3y$ so we get the solutions $(x,y,z) = (0,t,3t)$
Similarly when $y=0$ we get the solutions $(x,y,z) = (t,0,3t)$ for any $t \in \Bbb Z$ with $t \neq 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the Jordan form and basis for a matrix I have this matrix $A=\left[ {\begin{array}{ccc}
-3&3&-2\\
-7&6&-3\\
1&-1&2\\
\end{array} } \right]$
I computed the characteristic polynomial $C_A(t)=-(t-2)^2(t-1)$
When I go to try and find the eigenvectors and generalized eigenvectors I compute $A-2I=\left[ {\begin{array}{ccc}
-5&3&-2\\
-7&4&-3\\
1&-1&0\\
\end{array} } \right]$
Using row reduction $\left[ {\begin{array}{ccc}
1&-1&0\\
0&1&1\\
0&0&0\\
\end{array} } \right]x=0$ gives me eigenvectors having the form $e=r\left[ {\begin{array}{c}
1\\
1\\
-1\\
\end{array} } \right]$
$v=\left[ {\begin{array}{c}
1\\
1\\
-1\\
\end{array} } \right]$
Is the only eigenvector up to a constant. So I know there is 1 generalized eigenvector and I simply chose this eigenvector to be my initial vector in the cycle.
At this point I know the Jordan form is $J=\left[ {\begin{array}{ccc}
2&1&0\\
0&2&0\\
0&0&1\\
\end{array} } \right]$
Now to find the generalized eigenvector I can take an augmented matrix
$\left(\begin{array}{ccc|c}
5&3&-2&1\\
-7&4&1&1\\
1&-1&0&-1
\end{array}\right)$
I row reduced this to the form
$\left(\begin{array}{ccc|c}
1 & -1 & 0&-1\\
0 & 1 & 1&2\\
0&0&0&0
\end{array}\right)$
So I need a vector that satisfies $x_1-x_2=-1$ and $x_2+x_3=2$ I let $x_3=1$
Which gives me a vector $x=\left(\begin{array}{c}
0\\
1\\
1
\end{array}\right)$ which seems to work, so I have a basis for the generalized eigenspace for $\lambda=2$
then I need to find a solution for $(A-I)x=0$ for $\lambda=1$.
I row reduce $A-I=\left[\begin{array}{ccc}
-4&3&-2\\
-7&5&1\\
1&-1&1
\end{array}\right]$ to get $\left[\begin{array}{ccc}
1&0&-1\\
0&1&-2\\
0&0&0
\end{array}\right]$ and using $(A-I)x=0$ found that all eigenvectors for $\lambda=1$ have the form $v=s\left(\begin{array}{c}
1\\
2\\
1
\end{array}\right)$
So I can make a matrix $Q=\left[\begin{array}{ccc}
1&1&0\\
2&1&1\\
1&-1&1
\end{array}\right]$ s.t $A=QJQ^{-1}$ where the columns of $Q$ are a jordan basis for $J$
| Because you are looking for a Jordan canonical basis, $Ax = 2x+v$, where $x$ is the generalized eigenvector you're looking for. Equivalently, $(A-2I)x = v$. Applying $A-2I$ one more time to each side will give you the following statement: $(A-2I)^{2}x = 0$.
So, your generalized eigenvector is in the nullspace of $(A-2I)^2$, and it is not in the nullspace of $(A-2I)$. Can you finish from there?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3031391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve for positive relatively prime numbers? When finding the relative prime for large numbers like 360 (that has a prime factorization with multiples of the same prime factor), how would it be solved?
ie. How many numbers between 1 and 360 are relatively prime to 360?
I would think either
*
*N = 360 [prime factorization: 2, 2, 2, 5, 9]
*$N(A_1) = \frac{360}{2}$
*$N(A_2) = \frac{360}{5}$
*$N(A_3) = \frac{360}{9}$
*$N(A_1 \cup A_2) = ...$
*$N(A_1 \cup A_3) = ...$
*$N(A_2 \cup A_3) = ...$
...
or
*
*N = 360 [prime factorization: 2, 2, 2, 5, 9]
*$N(A_1) = \frac{360}{2}$
*$N(A_2) = \frac{360}{2}$
*$N(A_3) = \frac{360}{2}$
*$N(A_4) = \frac{360}{5}$
*$N(A_5) = \frac{360}{9}$
*$N(A_1 \cup A_2) = ...$
...
| What primes do you need to worry about? The primes that divide $360$ are $2,3,5$, so you really want to find those numbers which are not divisible by $2, 3$, or $5$.
Now
*
*$360/2 = 180$ numbers up to $360$ are divisible by $2$.
*$360/3 = 120$ numbers up to $360$ are divisible by $3$.
*$360/5 = 72$ numbers up to $360$ are divisible by $5$.
But some numbers are divisible by both $2$ and $3$, or by $3$ and $5$, or $2$ and $5$. So we've overcounted. (This is an inclusion-exclusion argument in progress).
*
*There are $360/6 = 60$ divisible by $2$ and $3$.
*There are $360/15 = 24$ divisible by $3$ and $5$.
*There are $360/10 = 36$ divisible by $2$ and $5$.
But we've overcounted how much we've overcounted! There are also numbers divisible by $2$, $3$, and $5$.
*
*There are $360/(2 \cdot 3 \cdot 5) =12$ numbers up to $360$ that are divisible by $2\cdot3\cdot5 = 30$.
Thus in total there will be
$$ 360 - (180 + 120 + 72 - (60 + 24 + 36 - (12))) = 96.$$
In fact, this is one way of understanding the expression for $\varphi(n)$ given by
$$ \varphi(n) = n\prod_{p \mid n} \left( 1 - \frac{1}{p} \right),$$
which encodes this inclusion-exclusion argument within it.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A functional equation (another) I would like to find a continuous concave function from $[1/2,1]$ to $[0,1]$ such that $f(1)=1$ and for all $x\in [1/2,1]$
$$f(x)= \frac{1}{2} + \frac{1}{4}f\left(\frac{2x}{1+x}\right).$$
I am actually not sure about existence of a solution.
| $f(x)=\dfrac{1}{2}+\dfrac{1}{4}f\left(\dfrac{2x}{1+x}\right)$
$f\left(\dfrac{2^x}{2^x-1}\right)=\dfrac{1}{2}+\dfrac{1}{4}f\left(\dfrac{2\times\dfrac{2^x}{2^x-1}}{1+\dfrac{2^x}{2^x-1}}\right)$
$f\left(\dfrac{2^x}{2^x-1}\right)=\dfrac{1}{2}+\dfrac{1}{4}f\left(\dfrac{\dfrac{2^{x+1}}{2^x-1}}{\dfrac{2^{x+1}-1}{2^x-1}}\right)$
$f\left(\dfrac{2^x}{2^x-1}\right)=\dfrac{1}{2}+\dfrac{1}{4}f\left(\dfrac{2^{x+1}}{2^{x+1}-1}\right)$
$f\left(\dfrac{2^{x+1}}{2^{x+1}-1}\right)-4f\left(\dfrac{2^x}{2^x-1}\right)=-2$
$f\left(\dfrac{2^x}{2^x-1}\right)=4^x\Theta(x)-2x+1$ , where $\Theta(x)$ is an arbitrary periodic function with unit period (according to http://eqworld.ipmnet.ru/en/solutions/fe/fe1102.pdf)
$f(x)=\dfrac{x^2}{(x-1)^2}\Theta\left(\log_2\dfrac{x}{x-1}\right)-2\log_2\dfrac{x}{x-1}+1$ , where $\Theta(x)$ is an arbitrary periodic function with unit period
| {
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"source": "stackexchange",
"question_score": "2",
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} |
Solution for $(a\cdot x+b)\cdot \sin(x)+c\cdot \cos(x)=0$? As part of an engineering problem, I've been trying for a generic solution for $(a\cdot x+b)\cdot \sin(x)+c\cdot \cos(x)=0$
Here's something I tried:
$(a⋅x+b)⋅\tan(x)+c=0$, when $\cos(x)\neq 0$
$$\tan(x)=\frac{-c}{a⋅x+b}$$
It sort of looking like this, when, say, $a=2, b=3, c=5$:
Any chance that can be solved without using numerical methods?
EDIT: I was looking for solutions in the form of $x \approx f(a,b,c) $, preferably finding solution in any range, because $a$, $b$, and $c$ can take wierd values. It's used for a program where it will not be trivial to solve problem procedurally.
| Besides numerical solutions for particular parameter values, you could try series solutions. Thus when $c$ is small compared to $a$ and $b$, the solution near $x=0$ is
$$ -\frac{b}{a} + \cot\left(\frac{b}{a}\right) \frac{c}{a} + \cot\left(\frac{b}{a}\right) \csc^2 \left(\frac{b}{a}\right) \frac{c^2}{a^2}
+ \left(\cot^3\left(\frac{b}{a}\right) + \cot\left(\frac{b}{a}\right) \sec^4 \left(\frac{b}{a}\right)\right) \frac{c^3}{a^3} + \ldots $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $x=y+\sqrt{y^2+1}$
Then how to write $y$ in terms of $x$? If $$x=y+\sqrt{y^2+1}$$
Then how to write $y$ in terms of $x$? Please tell me some hints
| I assume that
$x, y \in \Bbb R. \tag 0$
We are given
$x = y + \sqrt{y^2 + 1}. \tag 1$
We observe that this equation implies
$x > 0, \forall y \in \Bbb R, \tag{1.2}$
since
$y^2 < y^2 + 1 \Longrightarrow \vert y \vert < \sqrt{y^2 + 1}; \tag{1.5}$
we proceed:
$x - y = \sqrt{y^2 + 1}; \tag 2$
$x^2 - 2xy + y^2 = (x - y)^2 = y^2 + 1; \tag 3$
$x^2 - 2xy = 1; \tag 4$
$2xy = x^2 - 1; \tag 5$
by virtue of (1.2), we may divide by $2x$:
$y = \dfrac{x^2 - 1}{2x}. \tag 6$
CHECK:
From (6),
$y^2 = \dfrac{(x^2 - 1)^2}{4x^2} = \dfrac{1 - 2x^2 + x^4}{4x^2}; \tag 7$
$y^2 + 1 = \dfrac{1 - 2x^2 + x^4}{4x^2} + 1 = \dfrac{1 - 2x^2 + x^4}{4x^2} + \dfrac{4x^2}{4x^2} = \dfrac{1 + 2x^2 + x^4}{4x^2} = \dfrac{(1 + x^2)^2}{4x^2}; \tag 8$
and again, by virtue of (1.2), we may apply $\sqrt \cdot$ to both sides and find
$\sqrt{y^2 + 1} = \dfrac{1 + x^2}{2x}; \tag 9$
$y + \sqrt{y^2 + 1} = \dfrac{x^2 - 1}{2x} + \dfrac{1 + x^2}{2x} = \dfrac{2x^2}{2x} = x. \tag{10}$
| {
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"source": "stackexchange",
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An inequality for positive definite matrix with trace 1 Given a positive definite matrix $A \in \mathbb R^{n \times n}$. If $\operatorname{trace}(A) = 1$, for $n \geq 3$, prove that $$\text{det}(A) \leq \frac{n^n}{(n-1)^{2n}} \text{det}(I -A)^2$$ and the equation only holds when $A = \frac{1}{n}I$.
What I have tried:
For any positive definite matrix $P \in \mathbb{R}^{n \times n}$, there exists an invertible matrix $\Omega$, s.t.
\begin{align}
&P = \Omega^{-1} \Sigma \Omega \\
&\Sigma = \text{diag}(x_1, \cdots, x_n) \\
&x_1, \cdots, x_n > 0
\end{align}
We have $\text{det}(P) = \text{det}(\Sigma) =\prod_{i = 1}^n x_i$ and $\text{det}(I - P) = \text{det}(I - \Sigma) = \prod_{i = 1}^n (1 - x_i)$.
Therefore, we only need to prove $\forall x_1, \cdots, x_n$ with $x_1, \cdots, x_n > 0$ and $\sum_{i=1}^n x_i=1$, the inequality holds:
\begin{align}
\prod_{i = 1}^n (1 - x_i)^2 \geq \frac{(n - 1)^{2n}}{n^n} \prod_{i = 1}^n x_i
\end{align}
Assume $f(x_1, \cdots, x_n) = \prod_{i = 1}^n (1 - x_i)^2 / \prod_{i = 1}^n x_i$ and we then find the minimum of $f$.
| In order to investigate behavior of the function $f$, at the first we consider the following auxiliary problem.
Let $0<x,y,a$ and $x+y+a=1$. If $a$ is fixed then
$$h(x,y)=\frac{(1-x)^2(1-y)^2}{xy}=\frac{(a+xy)^2}{xy}=\frac{a^2}{xy}+2a+xy.$$
A function $\frac{a^2}{t}+t$ has a derivative $-\frac{a^2}{t^2}+1$, so it decreases when $0<t<a$ and increases when $a<t<1$. By AM-GM inequality, $xy\le \frac{(1-a)^2}4$ and the equality is attained iff $x=y$. In particular, if $\frac{(1-a)^2}4\le a$ then $h(x,y)$ attains its minimum at $(x,y)$ iff $x=y$. It is easy to check that $\frac{(1-a)^2}4\le a$ iff $$a\ge 3-2\sqrt{2}=0.1715\dots.$$
Consider the function $f(x_1,\dots, x_n)$. Fixing the smallest $x_l$ we restrict domain of $f$ to a compact set $C$ consisting of $x$ such that $\{x_j\ge x_l\}$ for each $j$ and $\sum x_j=1$. Since $f$ on $C$ is continuous, it attains its minimum at some point $x$. Let $x_i$ be the largest coordinate of $x$. Then $$x_j+x_k\le \frac 23(x_i+x_j+x_k)\le \frac 23<1-(3-2\sqrt{2})$$ for each remaining distinct $j$ and $k$, so the minimality of $f(x)$ on $C$ and the properties of the function $h$ imply that $x_j=x_k=t$ for each remaining $j$ and $k$. Then $x_i=1-(n-1)t$ and
$$f(x)=r(t)=\frac{((n-1)t)^2}{1-(n-1)t}\left(\frac{(1-t)^2}{t}\right)^{n-1}=\frac{(n-1)^2(1-t)^{2n-2}}{(1-(n-1)t)t^{n-3}}.$$
If $t$ tends to $0$ or to $\frac 1{n-1}$ then $r(t)$ tends to infinity. So $r(t)$ attains its minimum at a compact subset of an interval $(0, \frac 1{n-1})$ in some its point $s$. Then $r’(s)=0$. This easily implies that
$$\left((1-s)^{2n-2}\right)’ (1-(n-1)s)s^{n-3}=(1-s)^{2n-2}\left((1-(n-1)s)s^{n-3}\right)'$$
$$(2-2n)(1-(n-1)s)s^{n-3}=(1-s)\left((1-(n-1)s)s^{n-3}\right)'$$
The following cases are possible
1) $n=3$. Then $-4(1-2s)=(1-s)\left(1-2s\right)'$ and $s=\frac 13$.
2) $n>3$. Then
$$(2-2n)(1-(n-1)s)s^{n-3}=(1-s)\left((n-3)s^{n-4}-(n-1)(n-2)s^{n-3}\right)$$
$$(2-2n)(1-(n-1)s)s=(1-s)\left((n-3)-(n-1)(n-2)s\right)$$
$$s^2(n^2-n)+s(n^2-4n+1)-n+3=0$$
$$s=\frac 1n\mbox{ or }s=-\frac {n-3}{n-1}<0.$$
Thus anyway $s=\frac 1n $. Then all $x_j$ equal to $\frac 1n$ and $f(x)=\frac{(n - 1)^{2n}}{n^n}$.
| {
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Calculate the area of trapezoid. In an isosceles trapezoid $ABCD$ the leg $AB$ and the smaller base $BC$ are 2 cm long, and $BD$ is perpendicular to $AB$.Calculate the area of trapezoid.
Let $\angle BAD=\theta,$
$BD=h$,
$\angle ABD=90^\circ$
$\angle CBD=90^\circ-\theta$
$CD=2$because trapezoid is isosceles
Apply cosine law in triangle BDC,
$\cos(90-\theta)=\frac{h^2+2^2-2^2}{2\times 2\times h}=\frac{h}{4}$
$\sin\theta=\frac{h}{4}..(1)$
In right triangle $ABD,\sin \theta=\frac{h}{\sqrt{h^2+4}}..(2)$
From $(1)$ and $(2)$,$h=2\sqrt3$
Area of $ABCD=\frac{1}{2}\times 2\times h+\frac{1}{2}\times 2\times 2\times \sin2\theta=3\sqrt3$
But the answer given is $2\sqrt2(\sqrt{5}+1)$
| you are correct. The answer should be $3\sqrt{3}$
| {
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Solving $\int\limits_{-\infty}^\infty \frac{1}{x^8+1}dx$ through Glasser's Master Theorem Trying to find a way to solve
$$\int_{-\infty}^\infty \frac{1}{x^8+1}dx$$
through Glasser's Master Theorem, more specifically the Cauchy–Schlömilch substitution. Preferably, I'm looking for the closed form solution, and I am already aware of how to attain this through contour integration.
Solution: $$\frac{\pi}{4\sin(\frac{\pi}{8})}$$
Link to general closed form solution: solutions to $\int_{-\infty}^\infty \frac{1}{x^n+1}dx$ for even $n$
| I don't know what is GMT. I looked it up on Wikipedia and it states $u = x - 1/x$ as Cauchy–Schlömilch substitution. Here is my solution that splits the integral into 4 integrals, each one of them uses $u = x - 1/x$. I will happily delete the answer if it is useless or wrong :)
$$\int^{\infty}_{-\infty} \dfrac{1}{1+ x^8} dx = -\dfrac{1}{2 \sqrt{2}}\int^{\infty}_{-\infty} \dfrac{x^2 - \sqrt{2}}{x^4 - \sqrt 2 x^2 + 1} + \dfrac{1}{2 \sqrt{2}}\int^{\infty}_{-\infty}\dfrac{x^2 + \sqrt{2}}{x^4 + \sqrt2 x^2 + 1} \\= -\dfrac{1}{2 \sqrt{2}}\int^{\infty}_{-\infty} \dfrac{x^2 + 1}{x^4 - \sqrt 2 x^2 + 1} +\dfrac{\sqrt{2} +1}{2 \sqrt{2}} \int^{\infty}_{-\infty} \dfrac{1}{x^4 - \sqrt 2 x^2 + 1} \\+ \dfrac{1}{2 \sqrt{2}}\int^{\infty}_{-\infty}\dfrac{x^2 + 1 }{x^4 + \sqrt2 x^2 + 1}+\dfrac{\sqrt2 -1}{2 \sqrt{2}}\int^{\infty}_{-\infty}\dfrac{1}{x^4 + \sqrt2 x^2 + 1} $$
$$\int^{\infty}_{-\infty} \dfrac{x^2 + 1}{x^4 - \sqrt 2 x^2 + 1} = \int^{\infty}_{-\infty} \dfrac{1 + 1/x^2}{ (x -1/x)^2 + 2 - \sqrt 2}$$
$u = x - 1/x, du = 1 + 1/x^2$
$$\int^{\infty}_{-\infty} \dfrac{1}{u^2 + (2 - \sqrt 2 )} du $$
$$J = \int^{\infty}_{-\infty} \dfrac{1}{x^4 - \sqrt 2 x^2 + 1} = 2 \int^{\infty}_{0} \dfrac{1}{x^4 - \sqrt 2 x^2 + 1}$$
$x = 1/t, dx = -1/t^2 dt$
$$J = 2 \int^{\infty}_{0} \dfrac{t^2}{t^4 - \sqrt 2 t^2 + 1} dt$$
Adding the two equations,
$$J = \int^{\infty}_{0} \dfrac{t^2 + 1}{t^4 - \sqrt 2 t^2 + 1} dt$$
This can be solved with $ u = t - 1/t$ like the previous equation.
$$\int^{\infty}_{-\infty}\dfrac{x^2 + 1 }{x^4 + \sqrt2 x^2 + 1} = \int^{\infty}_{-\infty}\dfrac{1 + 1/x^2 }{(x - 1/x)^2 + \sqrt2 + 2}$$
Similar to first integral, now.
$$I = \int^{\infty}_{-\infty}\dfrac{1}{x^4 + \sqrt2 x^2 + 1} = 2\int^{\infty}_{0}\dfrac{1}{x^4 + \sqrt2 x^2 + 1}$$
$x = 1/t, dx = -1/t^2 dt$
$$ I = \int^{\infty}_{0}\dfrac{t^2 + 1}{t^4 + \sqrt2 t^2 + 1} dt $$
Can be solved with $u = t - 1/t$ like other integrals here.
| {
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"source": "stackexchange",
"question_score": "9",
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Can any improvements be made to my proof that "$\sqrt{3} $ is irrational"? Suppose, for contradiction that $\sqrt{3}$ is rational. Then there exists $a,b \in \mathbb{Z}$ such that $$\frac{a}{b}= \sqrt{3},$$
where $a/b$ is in its simplest form. Then the above equation implies $$a^2=3b^2.$$ If $b$ is even, then $a$ is even, which is a contradiction since $a/b$ is therefore not in its simplest form.
Now, consider $b$ to be odd, then a is odd. Then for $m,n \in \mathbb{Z}$, we have $$(2m+1)^2=3(2n+1)^2\\ 4m^2+4m+1=12n^2+12n+3\\
2(2m^2+2m)=2(6n^2+6n+1)\\2(m^2+m)=2(3n^2+3n)+1.$$
The LHS is even since $m^2+m \in \mathbb{Z}$ and the RHS is odd since $ 3n^2+3n\in \mathbb{Z}$. This is a contradiction, and we therefore conclude that $\sqrt{3}$ is irrational.
| Your proof looks correct to me. Instead of doing the algebra at the end, you could reduce the equation $a^2 = 3b^2$ modulo $4$. If $a$ and $b$ are both odd, then $$a^2 \equiv 1 \mod 4,$$
and
\begin{align*}
b^2 &\equiv 1\mod 4\\
3b^2 &\equiv 3\mod 4,
\end{align*}
contradiction.
Another approach is to note that in the prime factorization of $a^2 = 3b^2$, the power of $3$ dividing the left hand side is even, while the power of $3$ dividing the right hand side is odd.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the differential equation $\frac{dy}{dx}=\frac{y+2y^5}{4x+y^4}$ Solve the differential equation $$\frac{dy}{dx}=\frac{y+2y^5}{4x+y^4}$$
My try:
we can write the equation as:
$$\frac{dy}{dx}=\frac{1}{y^3}\frac{\left(1+2y^4\right)}{1+\frac{4x}{y^4}}$$
Multiplying both sides with $\frac{1}{y^5}$ we get:
$$\frac{1}{y^5}\frac{dy}{dx}=\frac{1}{y^8}\frac{y^4(2+\frac{1}{y^4})}{1+\frac{4x}{y^4}}=\frac{1}{y^4}\frac{(2+\frac{1}{y^4})}{1+\frac{4x}{y^4}}$$
Now letting $$\frac{1}{y^4}=t$$ we get
$$\frac{-1}{4}\frac{dt}{dx}=\frac{t^2+2t}{4tx+1}$$
Any way further to convert in to variable separable?
| This is an "almost" exact equation. Write it in $A(x,y)\text dx + B(x,y)\text dy=0$ form
$$(-y-2y^5)\text dx + (4x+y^4)\text dy = 0$$
Then, look for a function $\mu(y)$ such that
$$ (-y-2y^5)\mu(y)\text dx + (4x+y^4)\mu(y)\text dy = 0 $$
Let $\mu(y) = y^{-5}$. Then, the equation becomes
$$ (-y^{-4}-2)\text dx + (4xy^{-5}+y^{-1})\text dy = 0 \\
\text d\left[ -xy^{-4} - 2x + \log y \right] = 0$$
Integrating, we get
$$ -xy^{-4} - 2x + \log y = C $$
Solving for $x$ gives us
$$ x = \frac{y^4}{2y^4+1}\log(cy) $$
Solving for $y$ in terms of the function $W(z)$ such that $W(z)\exp(W(z)) = W(z\exp(z)) = z$,
$$y = \left( \frac{4x}{W(kx\exp(-8x))} \right)^\frac{1}{4}$$
| {
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"source": "stackexchange",
"question_score": "6",
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Solving $\Bigl\{\begin{smallmatrix}x+\frac{3x-y}{x^2+y^2}=3\\y-\frac{x+3y}{x^2+y^2}=0\end{smallmatrix}$ in $\mathbb R$
$$\begin{cases}
x+\dfrac{3x-y}{x^2+y^2}=3 \\
y-\dfrac{x+3y}{x^2+y^2}=0
\end{cases}$$
Solve in the set of real numbers.
The furthest I have got is summing the equations, and I got
$$x^3+(y-3)x^2+(y^2+2)x+y^3-3y^2-4y=0.$$
But I have no idea how to solve this problem. How can I solve it?
| Make the substitution $x = r\cos(\theta)$ and $y = r\sin(\theta)$ in order to get
\begin{cases}
r^{2}\cos(\theta) + 3\cos(\theta) - \sin(\theta) = 3r\\
r^{2}\sin(\theta) - \cos(\theta) - 3\sin(\theta) = 0\\
\end{cases}
From the second equation, it results that $\cos(\theta) = (r^{2}-3)\sin(\theta)$. Therefore we get
\begin{align*}
(r^{2} + 3)\cos(\theta) - \sin(\theta) & = 3r \Longleftrightarrow (r^{2}+3)(r^{2}-3)\sin(\theta) - \sin(\theta) = 3r \Longleftrightarrow\\
(r^{4}- 10)\sin(\theta) & = 3r \Longleftrightarrow\sin(\theta) = \frac{3r}{r^{4}-10}
\end{align*}
Consequently, we have
\begin{align*}
&\cos^{2}(\theta) + \sin^{2}(\theta) = 1 \Leftrightarrow\left[\frac{3r(r^{2}-3)}{r^{4}-10}\right]^{2} + \left[\frac{3r}{r^{4}-10}\right]^{2} = 1 \Leftrightarrow\\\\
& 9r^{2}(r^{4}-6r^{2}+9) + 9r^{2} = (r^{4}-10)^{2} \Leftrightarrow 9r^{6}-54r^{4} + 90r^{2} = r^{8} - 20r^{4} + 100 \Leftrightarrow\\\\
& r^{8} - 9r^{6} + 34r^{4} - 90r^{2} + 100 = 0 \Leftrightarrow (r^{8} - 5r^{6}) - (4r^{6} - 20r^{4}) + (14r^{4} - 70r^{2}) - (20r^{2} - 100) = 0 \Leftrightarrow\\\\
& r^{6}(r^{2}-5) - 4r^{4}(r^{2}-5) + 14r^{2}(r^{2}-5) - 20(r^{2}-5) = (r^{6} - 4r^{4} + 14r^{2} - 20)(r^{2}-5) = 0 \Leftrightarrow\\\\
& [(r^{6} - 2r^{4}) - (2r^{4} - 4r^{2}) + (10r^{2}-20)](r^{2}-5) = 0 \Leftrightarrow (r^{4}-2r^{2}+10)(r^{2}-2)(r^{2}-5) = 0 \Leftrightarrow\\\\
& (r^{2} - 2)(r^{2} - 5) = 0 \Leftrightarrow r\in\{\sqrt{2},\sqrt{5}\}\quad\text{since}\quad r\geq 0
\end{align*}
Finally, for each value of $r$ there corresponds a solution. More precisely, the solution set is described by $$S = \{(1,-1),(2,1)\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Finding $\iiint x^2\,{\rm d}x{\rm d}y{\rm d}z$ over the volume bounded by $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2 }= 1$
Finding the value of $$\iiint x^2\,\mathrm dx\mathrm dy\mathrm dz$$ over the volume bounded by the ellipsoid $$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}= 1.$$
I am taking limits as follow:
$$-a \leq x \leq a$$
$$-\frac{b}{a} (a^2-x^2)^{1/2} \leq y \leq \frac{b}{a} (a^2-x^2)^{1/2}$$
$$-\frac{c}{ab} ((ab)^2 - (bx)^2 - (ay)^2)^{1/2} \leq z \leq \frac{c}{ab} ((ab)^2 - (bx)^2 - (ay)^2)^{1/2}$$
And solving the integral as $\mathrm dz$ first then $\mathrm dy$ and then $\mathrm dx$. But it is just becoming harder and harder with each step.
I do not have a tutor. So, I highly need help of you masters. Please help me.
| First, rewrite the integral as
$$\int_{-a}^a x^2 \int\int_{E_x}1\ dy\ dz\ dx $$
where $E_x$ is the ellipse in the $yz$-plane satisfying $\frac{y^2}{b^2} + \frac{z^2}{c^2} = (1-\frac{x^2}{a^2})$. By arranging the formula of the ellipse into the more familiar form $\frac{y^2}{b^2\left(1-\frac{x^2}{a^2}\right)} + \frac{z^2}{c^2\left(1-\frac{x^2}{a^2}\right)}=1$, we see that the area of said ellipse is $\pi bc(1-\frac{x^2}{a^2})$ (e.g. see Wikipedia for area of ellipse formula).
Thus, the triple integral simplifies to the single-variable integral
$$\int_{-a}^a x^2 \pi bc(1-\frac{x^2}{a^2})\ dx= \frac{4}{15}\pi a^3 bc$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3049891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Proof verification. $\{x_n\}$ is a sequence such that $|x_{n+1} - x_n| \le C\alpha^n$ for $\alpha\in (0, 1), n\in\Bbb N$. Prove $x_n$ converges.
Let $\{x_n\}, n\in \Bbb N$ denote a sequence such that:
$$
\begin{cases}
|x_{n+1} - x_n| \le C\alpha^n \\
0 < \alpha < 1
\end{cases}
$$
Prove $\{x_n\}$ converges.
Given the fact $|x_{n+1} - x_n| \le C\alpha^n$ consider the following inequalities:
$$
|x_{n+1} - x_n| \le C\alpha^n \\
|x_{n+2} - x_{n+1}| \le C\alpha^{n+1} \\
|x_{n+3} - x_{n+2}| \le C\alpha^{n+2} \\
\dots \\
|x_{n+p+1} - x_{n+p}| \le C\alpha^{n+p} \\
$$
Consider the sum of the inequalities:
$$
|x_{n+1} - x_{n}| + |x_{n+2} - x_{n+1}| + |x_{n+3} - x_{n+2}| + \cdots + |x_{n+p+1} - x_{n+p}| \\
\le \sum_{k=0}^p C\alpha^{n+k} = C \sum_{k=0}^p \alpha^{n+k} \tag1
$$
By geometric sum:
$$
C \sum_{k=0}^p \alpha^{n+k} = C \cdot \frac{\alpha^n(1-\alpha^{p + 1})}{1-\alpha} \le C \cdot \frac{\alpha^n}{1-\alpha}
$$
Lets fix some $\epsilon > 0$, and $N\in \Bbb N$ such that:
$$
C\cdot \frac{\alpha^{N}}{1-\alpha} < \epsilon
$$
Rewrite $\alpha$ as:
$$
\alpha = \frac{1}{1+r},\ r \in \Bbb R_{>0}
$$
Thus:
$$
C\cdot \frac{1}{(1-\alpha)(1+r)^N} < \epsilon \\
(1+r)^N > {C\over (1-\alpha)\epsilon} \\
N > \log_{1+r} {C\over (1-\alpha)\epsilon}
$$
Returning to $(1)$ we have by triangular inequality:
$$
|x_{n}- x_{n+1} + x_{n+1} - x_{n+2} + \cdots + x_{n+p} - x_{n+p+1}| \\ \le |x_{n+1} - x_{n}| + |x_{n+2} - x_{n+1}| + \cdots + |x_{n+p+1} - x_{n+p}|
$$
Since the values are telescoping we obtain:
$$
|x_{n} - x_{n+p+1}| < C \cdot \frac{\alpha^n}{1-\alpha} < \epsilon
$$
Now if we choose:
$$
n > N > \log_{1+r} {C\over (1-\alpha)\epsilon}
$$
we obtain a regular definition of the Cauchy criteria, which means $\{x_n\}$ is a convergent sequence.
Could you please verify the reasoning above? Thank you!
| Seems fine.
We have $C \ge 0$.
In the event that $C=0$, we have a constant sequence and hence it converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3051454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
Compute $\sum\limits_{j = 0}^{m - 1} \left(c_j + 1\right)\ln\left(c_j + 1\right)$ where $c_j = \cos\left(\frac{\pi}{2m}\left(1 + 2j\right) \right)$ As part of solving:
\begin{equation}
I_m = \int_0^1 \ln\left(1 + x^{2m}\right)\:dx.
\end{equation}
where $m \in \mathbb{N}$. I found an unresolved component that I'm unsure how to start:
\begin{equation}
G_m = \sum_{j = 0}^{m - 1} \left(c_j + 1\right)\ln\left(c_j + 1\right),
\end{equation}
where $c_j = \cos\left(\frac{\pi}{2m}\left(1 + 2j\right) \right)$
I'm just looking for a starting point. Any tips would be greatly appreciated.
By the way, I was able to show (and this was part of the solution too) :
\begin{equation}
\sum_{j = 0}^{m - 1} c_j = 0
\end{equation}
Edit: For those that may be interested. In collaboration with clathratus, we found that for $n > 1$
\begin{equation}
\int_{0}^{1} \frac{1}{t^n + 1}\:dt = \frac{1}{n}\left[\frac{\pi}{\sin\left(\frac{\pi}{n} \right)}- B\left(1 - \frac{1}{n}, \frac{1}{n}, \frac{1}{2}\right)\right]
\end{equation}
Or for any positive upper bound $x$:
\begin{align}
I_n(x) &= \int_{0}^{x} \frac{1}{t^n + 1}\:dt = \frac{1}{n}\left[\Gamma\left(1 - \frac{1}{n} \right)\Gamma\left(\frac{1}{n} \right)- B\left(1 - \frac{1}{n}, \frac{1}{n}, \frac{1}{x^n + 1}\right)\right]
\end{align}
Here though, I was curious to investigate when $n$ was an even integer. This is my work:
Here we will consider $r = 2m$ where $m \in \mathbb{N}$. In doing so, we observe that the roots of the function are $m$ pairs of complex roots $(z, c(z))$ where $c(z)$ is the conjugate of $z$. To verify this:
\begin{align}
x^{2m} + 1 = 0 \rightarrow x^{2m} = e^{\pi i}
\end{align}
By De Moivre's formula, we observe that:
\begin{align}
x = \exp\left({\frac{\pi + 2\pi j}{2m} i} \right) \mbox{ for } j = 0\dots 2m - 1,
\end{align}
which we can express as the set
\begin{align}
S &= \Bigg\{ \exp\left({\frac{\pi + 2\pi \cdot 0}{2m} i} \right) , \:\exp\left({\frac{\pi + 2\pi \cdot 1}{2m} i} \right),\dots,\:\exp\left({\frac{\pi + 2\pi \cdot (2m - 2)}{2m} i} \right)\\
&\qquad\:\exp\left({\frac{\pi + 2\pi \cdot (2m - 1)}{2m} i} \right)\Bigg\},
\end{align}
which can be expressed as the set of $2$-tuples
\begin{align}
S &= \left\{ \left( \exp\left({\frac{\pi + 2\pi j}{2m} i} \right) , \:\exp\left({\frac{\pi + 2\pi(2m - 1 - j )}{2m} i} \right)\right)\: \bigg|\: j = 0 \dots m - 1\right\}\\
& = \left\{ (z_j, c\left(z_j\right)\:|\: j = 0 \dots m - 1 \right\}
\end{align}
From here, we can factor $x^{2m} + 1$ into the form
\begin{align}
x^{2m} + 1 &= \prod_{r \in S} \left(x + r_j\right)\left(x + c(r_j)\right) \\
&= \prod_{i = 0}^{m - 1} \left(x^2 + \left(r_j + c(r_j)\right)x + r_j c(r_j)\right) \\
&= \prod_{i = 0}^{m - 1} \left(x^2 + 2\Re\left(r_j\right)x + \left|r_j \right|^2\right)
\end{align}
For our case here $\left|r_j \right|^2 = 1$ and $\Re\left(r_j\right) = \cos\left(\frac{\pi + 2\pi j}{2m} \right)= \cos\left(\frac{\pi}{2m}\left(1 + 2j\right)\right) = c_j$
\begin{align}
\int_0^1 \log\left( x^{2m} + 1\right)\:dx &= \int_0^1 \log\left(\prod_{r \in S} \left(x^2 + 2c_jx+ \left|r_j \right|^2\right)\right)\\
&= \sum_{j = 0}^{m - 1} \int_0^1 \log\left(x^2 + 2c_jx + 1 \right)\\
&= \sum_{j = 0}^{m - 1} \left[2\sqrt{1 - c_j^2}\arctan\left(\frac{x + c_j}{\sqrt{1 - c_j^2}}\right) + \left(x + c_j\right)\log\left(x^2 + 2c_jx + 1\right) - 2x \right]_0^1 \\
&= \sum_{j = 0}^{m - 1} \left[ 2\sqrt{1 - c_j^2}\arctan\left(\sqrt{\frac{1 - c_j}{1 + c_j}} \right) + \log(2)c_j + \left(\log(2) - 2\right) + \left(c_j + 1\right)\log\left(c_j + 1\right) \right] \\
&= 2\sum_{j = 0}^{m - 1}\sqrt{1 - c_j^2}\arctan\left(\sqrt{\frac{1 - c_j}{1 + c_j}} \right) + \log(2)\sum_{j = 0}^{m - 1} c_j + m\left(\log(2) - 2\right)\\
&\qquad+ \sum_{j = 0}^{m - 1}\left(c_j + 1\right)\log\left(c_j + 1\right)
\end{align}
Thus,
\begin{align}
\int_0^1 \log\left( x^{2m} + 1\right)\:dx &=\sum_{j = 0}^{m - 1}c_j\sin\left(\frac{\pi}{2m}\left(1 + 2j\right)\right) + \log(2)\sum_{j = 0}^{m - 1} c_j + m\left(\log(2) - 2\right)\\
&\qquad+ \sum_{j = 0}^{m - 1}\left(c_j + 1\right)\log\left(c_j + 1\right)
\end{align}
| This does not answer the question as asked in the post.
Consider
$$J_m=\int \log(1+x^{2m})\,dx$$ One integration by parts gives
$$J_m=x \log \left(1+x^{2 m}\right)-2m\int \frac{ x^{2 m}+1-1}{x^{2 m}+1}\,dx=x \log \left(1+x^{2 m}\right)-2mx+2m\int \frac{dx}{x^{2 m}+1}$$ and
$$\int \frac{dx}{x^{2 m}+1}=x \, _2F_1\left(1,\frac{1}{2 m};1+\frac{1}{2 m};-x^{2 m}\right)$$ where appears the Gaussian or ordinary hypergeometric function.
So
$$K_m=\int_0^a \log(1+x^{2m})\,dx=a \log \left(1+a^{2 m}\right)-2ma+2ma \, _2F_1\left(1,\frac{1}{2 m};1+\frac{1}{2 m};-a^{2 m}\right)$$ and, if $a=1$,
$$I_m=\int_0^1 \log(1+x^{2m})\,dx= \log \left(2\right)-2m+2m \, _2F_1\left(1,\frac{1}{2 m};1+\frac{1}{2 m};-1\right)$$ which can write
$$I_m=\log (2)-\Phi \left(-1,1,1+\frac{1}{2 m}\right)$$
where appears the Lerch transcendent function.
Now, (this is something I never looked at), a few (ugly) expressions for $f_m=\Phi \left(-1,1,1+\frac{1}{2 m}\right)$ before any simplification
$$f(1)=\frac{\pi }{2}-2$$
$$f(2)=\frac{1}{4} \left(\pi \tan \left(\frac{\pi }{8}\right)+\pi \cot \left(\frac{\pi
}{8}\right)-4 \sqrt{2} \log \left(\sin \left(\frac{\pi }{8}\right)\right)+4
\sqrt{2} \log \left(\cos \left(\frac{\pi }{8}\right)\right)\right)-4$$
$$f(3)=\frac{2 \left(\pi -\sqrt{3} \log \left(\sqrt{3}-1\right)+\sqrt{3} \log
\left(1+\sqrt{3}\right)\right)}{\left(\sqrt{3}-1\right)
\left(1+\sqrt{3}\right)}-6$$
$$f(4)=\frac{1}{4} \left(\pi \tan \left(\frac{\pi }{16}\right)+\pi \cot \left(\frac{\pi
}{16}\right)-8 \sin \left(\frac{\pi }{8}\right) \log \left(\sin \left(\frac{3
\pi }{16}\right)\right)+8 \cos \left(\frac{\pi }{8}\right) \log \left(\cos
\left(\frac{\pi }{16}\right)\right)-8 \cos \left(\frac{\pi }{8}\right) \log
\left(\sin \left(\frac{\pi }{16}\right)\right)+8 \sin \left(\frac{\pi
}{8}\right) \log \left(\cos \left(\frac{3 \pi }{16}\right)\right)\right)-8$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3053596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Factoring $3^6-3^3 +1$ $3^6-3^3 +1$ factors?, 37 and 19, but how to do it using factoring, $3^3(3^3-1)+1$, can't somehow put the 1 inside
| $x^2-x+1$ factorises as $(x-\omega)(x+\omega^2)$ where $\omega^3=-1$.
Here $x=27$ and working modulo $27$ the cubes are $1,8,0,10,17,0,19,26, 0, 1, 8 \dots$ so $8^3\equiv -1$, and we can take $\omega = 8, \omega^2=64\equiv 10$ and obtain the factorisation $$(27-8)(27+10)=19\times 37$$
Since we have a cube root involved and the modulus is a power of $3$ there are some curiosities about the factorisation, but it checks out all the same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3053975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
A Series For the Golden Ratio
Question: Can we show that $$\phi=\frac{1}{2}+\frac{11}{2}\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2} $$; where $\phi={1+\sqrt{5} \above 1.5pt 2}$ is the golden ratio ?
Some background and motivation:
Wikipedia only provides one series for the golden ratio - see also the link in the comment by @Zacky. I became curious if I could construct another series for the Golden Ratio based on a slight modification to a known series representation of the $\sqrt{2}.$ At first I considered $$\sqrt{2}=\sum_{n=0}^\infty(-1)^{n+1}\frac{(2n+1)!!}{(2n)!!}$$; which series can be accelerated via an Euler transform to yield $$\sqrt{2}=\sum_{n=0}^\infty(-1)^{n+1}\frac{(2n+1)!!}{2^{3n+1}(n!)^2}$$ This last series became the impetus to try and and get to the Golden ratio. Through trial and error I stumbled upon
$$\frac{\sqrt{5}}{11}=\sum_{n=0}^\infty\frac{(2n)!}{5^{3n+1}(n!)^2}$$
| Using Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $,
$$\dfrac{(2n)!}{5^{3n+1}(n!)^2}=\dfrac{2^n\cdot1\cdot3\cdot5\cdots(2n-3)(2n-1)}{5^{3n+1}n!}=\dfrac{-\dfrac12\left(-\dfrac12-1\right)\cdots\left(-\dfrac12-(n-1)\right)}{n!\cdot5}\left(-\dfrac4{5^3}\right)^n$$
$$\implies5\sum_{n=0}^\infty\dfrac{(2n)!}{5^{3n+1}(n!)^2}=\left(1-\dfrac4{5^3}\right)^{-1/2}=?$$
Alternatively using Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $,
Using Generalized Binomial Expansion, $$(1+x)^m=1+mx+\frac{m(m-1)}{2!}x^2+\frac{m(m-1)(m-2)}{3!}x^3+\cdots$$ given the converge holds
$mx=\dfrac{2!}{5^3(1!)^2},\dfrac{m(m-1)}2x^2=\dfrac{4!}{5^6(2!)^2}$
$\implies m=-\dfrac12,x=-\dfrac4{5^3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
Calculate $\lim\limits_{ x\to + \infty}x\cdot \sin(\sqrt{x^{2}+3}-\sqrt{x^{2}+2})$ I know that $$\lim\limits_{ x\to + \infty}x\cdot \sin(\sqrt{x^{2}+3}-\sqrt{x^{2}+2})\\=\lim\limits_{ x\to + \infty}x\cdot \sin\left(\frac{1}{\sqrt{x^{2}+3}+\sqrt{x^{2}+2}}\right).$$ If $x \rightarrow + \infty$, then $\sin\left(\frac{1}{\sqrt{x^{2}+3}+\sqrt{x^{2}+2}}\right)\rightarrow \sin0 $. However I have also $x$ before $\sin x$ and I don't know how to calculate it.
| \begin{align}
x \sin \left( \sqrt{x^2+3}-\sqrt{x^2+2}\right)
&=x \sin \left( \frac{1}{\sqrt{x^2+3}+\sqrt{x^2+2}}\right)\\
&=\left(\dfrac{x}{\sqrt{x^2+3}+\sqrt{x^2+2}}\right)
\dfrac{\sin \left( \dfrac{1}{\sqrt{x^2+3}+\sqrt{x^2+2}}\right)}
{\left(\dfrac{1}{\sqrt{x^2+3}+\sqrt{x^2+2}}\right)}\\
&=\left(\dfrac{1}{\sqrt{1+\frac{3}{x^2}}+\sqrt{1+\frac{2}{x^2}}}\right)
\dfrac{\sin \left( \dfrac{1}{\sqrt{x^2+3}+\sqrt{x^2+2}}\right)}
{\left(\dfrac{1}{\sqrt{x^2+3}+\sqrt{x^2+2}}\right)}\\
&\to \dfrac 12 \cdot 1 \ \text{as $x \to \infty$}\\
&\to \dfrac 12 \ \text{as $x \to \infty$}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3057710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $\sum_{r=2}^{\infty} \frac{2-r}{r(r+1)(r+2)}$
Evaluate
$$\sum_{r=2}^{\infty} \frac{2-r}{r(r+1)(r+2)}$$
So in a previous part of the question I calculated that
$$\sum_{r=1}^{n} \frac{2-r}{r(r+1)(r+2)} = \sum_{r=1}^{n}\left( \frac{1}{r}-\frac{3}{r+1}+\frac{2}{r+2}\right)=\frac{n}{(n+1)(n+2)}$$
So my question is then why does $$\sum_{r=2}^{\infty} \frac{2-r}{r(r+1)(r+2)} = -\frac{1}{6}$$
As I though that it would be $\frac{1}{2}$.
| If $H_{n}$ is the $n$-th harmonic number, we have $$\sum_{r=2}^{n}\left(\frac{1}{r}-\frac{3}{r+1}+\frac{2}{r+2}\right)=\left(H_{n}-1\right)-3\left(H_{n}-1-\frac{1}{2}\right)+2\left(H_{n}-1-\frac{1}{2}-\frac{1}{3}\right)$$ $$=-1+\frac{3}{2}-\frac{2}{3}=-\frac{1}{6}.$$ Now take the limit as $n\rightarrow+\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3057821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
Laplace transform of square root of a trigonometric function Need help with this question from my university paper.
My question : Find Laplace Transform of the following: $\sqrt{1 + \sin(4t)}$
I do know how to solve $\sqrt{1 + \sin(t)}$
By taking $1 = \sin^2 \left(\frac{t}{2}\right) + \cos^2 \left(\frac{t}{2}\right)$
Further taking $\sin(t) = 2\sin \left(\frac{t}{2}\right) \cos \left(\frac{t}{2}\right)$
But I'm a bit confused about the above question.
Thanks in advance.
| The function $\sqrt{1+\sin at}$ is periodic with period $T = 2\pi/a$ and the first zero at $3\pi/2a$. Following your steps, we can write this as
$$ \sqrt{1+\sin at} = \sqrt{\cos^{2}\frac{at}{2} + \sin^{2}\frac{at}{2} + 2\sin\frac{at}{2}\cos\frac{at}{2}} = \left|\cos\frac{at}{2}+\sin\frac{at}{2}\right|. $$
Using the Laplace transform of a periodic function, we split the integral into two pieces. This is necessary because of the absolute value sign. To get rid of it in the region $[3\pi/2a, 2\pi/a]$, we have to take the negative of the function.
$$\begin{aligned} I &= \int_{0}^{\infty}\sqrt{1+\sin at}\,e^{-st}\,\mathrm{d}t = \frac{1}{1-e^{-2\pi s/a}}\int_{0}^{2\pi/a}\left|\cos\frac{at}{2} + \sin\frac{at}{2}\right|e^{-st}\,\mathrm{d}t \\
&= \frac{1}{1-e^{-2\pi s/a}}\left[\int_{0}^{3\pi/2a}\left(\cos\frac{at}{2} + \sin\frac{at}{2}\right)e^{-st}\,\mathrm{d}t - \int_{3\pi/2a}^{2\pi/a}\left(\cos\frac{at}{2} + \sin\frac{at}{2}\right)e^{-st}\,\mathrm{d}t \right]\end{aligned}$$
These integrals can be done by considering the integral
$$ \int_{0}^{3\pi/2a}e^{iat/2}e^{-st}\,\mathrm{d}t $$
and summing the real and imaginary contributions. One should then obtain
$$ \mathcal{L}[\sqrt{1+\sin at}] = \frac{1}{s^{2}+a^{2}/4}\left(s + \frac{a}{2} + \frac{\sqrt{2}\,ae^{-3\pi s/2a}}{1-e^{-2\pi s/a}}\right). $$
For $a = 4$, we have
$$ \boxed{\mathcal{L}[\sqrt{1+\sin 4t}] = \frac{1}{s^{2}+4}\left(s + 2 + \frac{4\sqrt{2}\,e^{-3\pi s/8}}{1-e^{-\pi s/2}}\right)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3058396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Determining the missing digits of $15! \equiv 1\square0767436\square000$ without actually calculating the factorial
$$15! \equiv 1\cdot 2\cdot 3\cdot\,\cdots\,\cdot 15 \equiv 1\square0767436\square000$$
Using a calculator, I know that the answer is $3$ and $8$, but I know that the answer can be calculated by hand.
How to calculate the missing digits? I know that large factorials can be estimated using Stirling's approximation:
$$15! \approx \sqrt{2\pi\cdot 15}
\cdot \left(\frac{15}{e}\right)^{15}$$
which is not feasible to calculate by hand.
The resulting number must be divisible by 9 which means that the digit sum must add up to 9 and is also divisible by 11 which means that the alternating digit sum must be divisible by 11:
$1+ d_0 + 0 + 7 +6 +7 +4 +3+6+d_1+0+0+0 \mod \phantom{1}9 \equiv \,34 + d_0 + d_1 \mod \phantom{1}9 \equiv 0 $
$-1+ d_0 - 0 + 7 -6 +7 -4 +3-6+d_1-0+0-0 \mod 11 \equiv d_0 + d_1 \mod 11 \equiv 0 $
The digits $3$ and $8$, or $7$ and $4$, fulfill both of the requirements.
| $15!=2^{11}\cdot 5^3\cdot 7^2\cdot 11\cdot 13=(1000)X$ where $X=2^8\cdot 3^6\cdot 7^2\cdot 11\cdot 13.$
The last digits of $2^8(=16^2), 3^6 (=9^3),7^2, 11,13$ are, respectively $6,9,9,1,3 .$
Modulo $10$ we have $6\cdot 9\cdot 9 \cdot 1\cdot 3\equiv 6\cdot(-1)^2\cdot 3\equiv 18\equiv 8$. So the last digit of $X$ is an $8$.
Therefore the 2nd digit of $15!$ must be a $3$ in order for the sum of all its digits to be divisible by $9$.
| {
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Show that if $a \equiv b\pmod{kn}$, then $a^k\equiv b^k\pmod{k^2n}$
Problem
Show that if $a \equiv b\pmod{2n}$, then $a^2\equiv b^2\pmod{4n}$. More generally, show that if $a \equiv b\pmod{kn}$, then $a^k\equiv b^k\pmod{k^2n}$. [Introduction to Higher Arithmetic, ex. 2.1]
Proof of the first statement
Note that $2n \mid a - b\implies 2n (a + b)\mid a^2 - b^2$ but the $\pmod{2n}$ congruence implies that $a$ and $b$ are of the same parity and so $2\mid (a + b)$ which implies that $2n \cdot 2\mid 2n (a + b)\mid a^2 - b^2$ or $a^2 = b^2\pmod{4n}$.
Attempt at proof of the general case
Multiplying $a \equiv b\pmod{kn}$ by itself $k$ times, we get $a^k \equiv b^k\pmod{kn}$, but it is still needed to prove that $k$ divides $a^k - b^k$. It maybe stands reasonable to consider the following expansion: $$(a - b) ^ k = a^k - {k \choose 1} a^{k-1}b^1 + {k \choose 2} a^{k-2}b^2 - \cdots \mp {k \choose 1} a^1b^{k-1} \pm b^k$$ Under $\pmod{kn}$, we can replace any $b$ with $a$ and vice versa, which might be useful. Maybe the trick is to sum binomial coefficients to a number (or two numbers) which are divisible by $k$. But sum over all of coefficients is $2^k$, which is not necessarily divisible by $k$.
So, I got stuck here.
| Note that $a\equiv b\pmod{kn}\implies a=b+kln$ for some integer $l$. Then $$a^k-b^k=(b+kln)^k-b^k=\binom k1b^{k-1}kln+\binom k2b^{k-2}(kln)^2+\cdots+(kln)^k$$ Every term has a factor of $k^2n$ from $(kln)^t$ for $1<t\le k$ but $\dbinom k1=k$ so $k^2n\mid (a^k-b^k)$.
| {
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For integers $a,b$ and prime $p$, are there infinitely many solutions to $p = (a-b)(a+b)$? I was thinking on this problem: suppose, for a prime $p$ and integers $a,b$, we have
$$p = (a-b)(a+b)$$
This implies immediately that $p \mid (a-b)$ or $p \mid (a+b)$. Then, since $p$ is prime and thus its only factors are $1$ and itself, this means $a+b = p$ and $a-b = 1$.
From that, $a=b+1.$ It's easy to see that, say, if $a$ ends in $3$ and $b$ ends in $2$, or if $a$ ends in $8$ and $b$ ends in $7$, that do not necessarily satisfy the original equation, just $a=b+1$. They don't necessarily satisfy $a+b=p$, however.
So my question is:
For integers $a,b$ and prime $p$, are there infinitely many solutions to the equation below?
$$p = (a-b)(a+b)$$
| Of course not!
If $p$ is prime so its only factors are $\pm 1$ and $\pm p$ so so if $p = (a-b)(a+b)$ then $a+b$ and $a-b$ can only be $\pm 1$ and/or $\pm p$.
So either $a-b = 1$ and $a+b = p$ so $a=\frac {p+1}2; b = \frac {p-1}2 = a-1$.
Or $a-b = p$ and $a + b = 1$ so $a= \frac {p+1}2; b = -\frac {p-1}2= 1-a$
Or $a-b = - 1$ and $a+b = -p$ so $a =-\frac {p+1}2; b=-\frac {p-1}2=a+1$
or $a-b = -p$ and $a + b = -1$ so $a=-\frac{p+1}2; b = \frac {p-1}2 = -1-a$
| {
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Solving integral using residues I'm trying to find the value of the integral $\int_0^{2\pi}\frac{\cos^2u}{2-\sin u}du$ using the substitution: $\cos u =\frac{1}{2}(z+\frac{1}{z})$ and $\sin u = \frac{1}{2i}(z-\frac{1}{z})$. Making the substitution, we obtain $\int_C(z^2+1)^2/(2z^2(z^2+4iz-1)dz$ where C denotes the unit circle. I leave out the steps inbetween because I do not believe the error was made there.
Next, I found the roots of the polynomial in the denominator, which are the poles of the integrand. I found these to be $z_0=0$, $z_1=i(-2+\sqrt3)$, $z_2=i(-2-\sqrt3)$. Only $z_0$ and $z_1$ lie inside the unit circle, so the value of the integral will be $2\pi i[Res(0)+Res(i(-2+\sqrt3))]$. $Res(0)$ is easily obtained. $Res(i(2-\sqrt3))$ should be too, since $z_1$ is a first order pole, so I can just take $lim_{z\to z_1}[f(z)(z-z_1)]$. The results are $$Res(0)= lim_{z\to 0}\frac{d}{dz}z^2f(z)=...=-2i$$ and $$Res(i(\sqrt3-2))=lim_{z\to z_1}(z-z_1)f(z)=...=-3/2+\sqrt3$$
What mistake did I make? My answer turns out to be $2\pi i(-2i-3/2+\sqrt3)$, the solution is $-2\pi(\sqrt3-2)$.
| Let $\tan\frac{u}{2}=x$.
Thus, $$dx=\frac{1}{2\cos^2\frac{u}{2}}du=\frac{1}{2}(1+x^2)du,$$ which gives
$$du=\frac{2}{1+x^2}dx.$$
Hence, $$\int\frac{\cos^2u}{2-\sin{u}}du=\int\left(2+\sin{u}-\frac{3}{2-\sin{u}}\right)du=$$
$$=2u-\cos{u}-3\int\frac{1}{2-\frac{2x}{1+x^2}}\frac{2}{1+x^2}dx=$$
$$=2u-\cos{u}-3\int\frac{1}{1-x+x^2}dx=2u-\cos{u}-3\int\frac{1}{\left(x-\frac{1}{2}\right)^2+\left(\frac{\sqrt3}{2}\right)^2}dx=$$
$$=2u-\cos{u}-2\sqrt3\arctan\frac{2\tan\frac{u}{2}-1}{\sqrt3}+C.$$
Can you end it now?
| {
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On Basak's "Bounds On Factors Of Odd Perfect Numbers" Let $N = q^k n^2$ be an odd perfect number given in Eulerian form, i.e. $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. In what follows, we denote the abundancy index of $x \in \mathbb{N}$ by $I(x)=\sigma(x)/x$, where $\sigma(x)$ is the sum of the divisors of $x$.
In Case 1 under Remark 3.1 on page 4 of Basak's Bounds On Factors Of Odd Perfect Numbers, it is proven that
$$\frac{16}{7\zeta(3)} < \frac{16q^3}{7\zeta(3)(q^3 - 1)} < \bigg(1 + \frac{1}{q}\bigg)\prod_{p \mid n, \hspace{0.05in} p \neq q}\bigg(1 + \frac{1}{p} + \frac{1}{p^2}\bigg).$$
But we also have
$$\bigg(1 + \frac{1}{q}\bigg)\prod_{p \mid n, \hspace{0.05in} p \neq q}\bigg(1 + \frac{1}{p} + \frac{1}{p^2}\bigg) < I(q)I(n^2) \leq \frac{6I(n^2)}{5},$$
since $q$ is prime with $q \equiv 1 \pmod 4$ implies that $q \geq 5$.
(Note that this last inequality is unconditional on the truth of the Descartes-Frenicle-Sorli Conjecture that $k=1$.)
This implies that
$$\frac{16}{7\zeta(3)} < \frac{6I(n^2)}{5}$$
from which it follows that
$$I(n^2) > \frac{5}{6}\cdot\frac{16}{7\zeta(3)} = \frac{40}{21\zeta(3)} \approx 1.58458547158229994034881195966.$$
But then this resulting numerical lower bound for $I(n^2)$ is trivial, as it is known that
$$I(q^k) < \frac{q}{q - 1} \leq \frac{5}{4} < \frac{8}{5} \leq \frac{2(q - 1)}{q} < I(n^2),$$
so that we already know, unconditionally, that $I(n^2) > 1.6$.
Here is my question:
Would it be possible to tweak Basak's argument in order to come up with an improved lower bound for
$$\bigg(1 + \frac{1}{q}\bigg)\prod_{p \mid n, \hspace{0.05in} p \neq q}\bigg(1 + \frac{1}{p} + \frac{1}{p^2}\bigg)?$$
Update (May 1, 2020 - 13:45 PM Manila time)
I re-read Basak's argument in Case 1 under Remark 3.1 on page 4 of the hyperlinked paper in arXiv, it appears that Basak does make use of the assumption that $k=1$.
| We get $$\frac{16q^3}{7\zeta(3)(q^3 - 1)} < I(q)I(n^2)$$ from which we have
$$I(n^2)\gt\frac{16}{7\zeta(3)}f(q)$$
where
$$f(q)=\frac{q^4}{(q^3-1)(q+1)}$$
Now, we have, for $q\ge 5$,
$$f'(q)=\frac{ q^3 (q(q^2-3)-4)}{(q + 1)^2 (q^3 - 1)^2}\gt 0$$
from which we have
$$I(n^2)\gt\frac{16}{7\zeta(3)}f(q)\ge \frac{16}{7\zeta(3)}f(5)=\frac{1250}{651\zeta(3)}\approx 1.597364$$
which is better than $1.584585$, but still smaller than $1.6$.
| {
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Find minimum value of $a^2+b^2$ Given that $a$ and $b$ are real constants and that the equation $x^4+ax^3+2x^2+bx+1=0$ has at least one real root, find the minimum possible value of $a^2+b^2$.
I began this way: Let the polynomial be factorized as $(x^2+\alpha x + 1)(x^2+\beta x +1)$. Then expanding and comparing coefficients we get $\alpha\beta=0$, meaning either $\alpha=0$ or $\beta=0$. Suppose $\alpha=0$. Then we see that $(x^2+\beta x+1)$ should have real roots, from which we get $\beta^2 \geq 4$. But we get $a=b=\beta$ from the comparison above. So $a^2+b^2 = 2\beta^2 \geq 8$.
Is it correct? Or is there any mistake? Any other solution is also welcome.
| I would write first $(x^2+cx+d) (x^2+ex+f) $ and then find $c, d, e, f$ in terms of $a, b$.
| {
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