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Computing problems and generating functions
Q. Find the generating function for the sequence $\{a_n\}$, where $a_n$ is the number of solutions to the equation: $a+b+c=k$ when $a, b, c$ are non-negative integers such that $a\ge2, 0\le b\le3$ and $2\le c\le5.$
As far as I have understood, here, $a_n$ must be the coefficient of $x^k$ in the expansion of $(x^2+x^3+x^4+\cdots)(1+x+x^2+x^3)(x^2+x^3+x^4+x^5).$
This follows because we obtain the term $x^k$ in the product by picking a term $x^a$ in the first sum, $x^b$ in the second and $x^c$ in the third, where the exponents $a, b, c$ satisfy the equation $a+b+c=k$ along with the given constraints.
But since the positive constant $k$ is not specified here so I'm not aware of how to obtain the desired coefficient in the above product. Does there exist any closed form for the same? Please suggest.. Give me some insights. It would be so helpful. Thanks in advance...
| Let's do a few simplifications:
*
*$(x^{2} + x^{3} + \cdots) = x^{2}(1 + x + x^{2} + \cdots) = x^{2} \cdots \dfrac{1}{1-x}$
*$1 + x + x^{2} + x^{3} = \dfrac{1-x^{4}}{1-x}$ (by the sum of a geometric series)
*$x^{2} + x^{3} + x^{4} + x^{5} = x^{2}(1 + x + x^{2} + x^{3}) = x^{2} \cdot \dfrac{1-x^{4}}{1-x}$
So really, you have the expression:
$$x^{4} \cdot (1-x^{4})^{2} \cdot \frac{1}{(1-x)^{2}}$$
It suffices to look for the coefficient of $x^{k-4}$ in the expansion of:
$$(1-x^{4})^{2} \cdot \frac{1}{(1-x)^{2}}$$
Now using the Binomial and (Derivative of a) Geometric Series Expansions, we have that:
\begin{align*}
(1-x^{4})^{2} \cdot \frac{1}{(1-x)^{2}} &= \left( 1 -2x^{4} + x^{8} \right) \cdot \left( \sum_{i=0}^{\infty} \binom{i+2-1}{i} x^{i} \right)
\end{align*}
By expanding things out, we have that the coefficient of $x^{k-4}$ is:
\begin{align*}
\binom{(k-4)+2-1}{k-4} - 2\binom{(k-8)+2-1}{k-8} + \binom{(k-12) + 2 - 1}{k-8}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4016659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Showing the sequence $\sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, \dots$ tends to $2$ using the Epsilon-Neighbourhood definition I am given the sequence
$$a_{n+1} = \sqrt{2 a_n}, \quad a_1 = \sqrt{2}.$$
That is, the sequence
$$ \sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, \dots$$
I am aware of using the recursive method to find the limit: setting $x = \sqrt{2x}$ and getting $x = 2$, rejecting $x = 0$ due to the sequence being monotone and positive.
I now want to show that the limit of this sequence is $2$ using the epsilon-neighbourhood definition for a convergent series:
Let $(a_n)$ be a sequence that converges to a real number $a$. Then, for every number $\epsilon > 0$, there exists a number $N \in \mathbb{N}$ such that for all $n \geq N$, $|a_n - a| < \epsilon$.
I take the logarithm on both sides and arrive at
$$ \left| \left( \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} \right) \ln 2 - \ln 2 \right| < \epsilon. $$
Using triangle inequality, $|x - y| \geq |x| - |y|$, I arrive at
$$ \left| \left( \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} \right) \right| - |\ln 2| < \epsilon. $$
Add $\ln 2$ to both sides
$$ \left| \left( \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} \right) \right| < \epsilon + \ln 2. $$
The left hand side can be rewritten as a sum of $n$ terms in a geometric series with first term $\frac{1}{2}$ and common ratio $\frac{1}{2}$,
$$ \frac{\frac{1}{2} \left( 1 - \left( \frac{1}{2} \right)^n \right)}{1 - \frac{1}{2}} = 1 - \frac{1}{2^n} < \epsilon + \ln 2. $$
I am unsure of how to proceed beyond this point.
| Hint:
For $n>1$, we can write
$$\frac{a_n}2=2^{-2^{-n}}>2^{-\frac1n}>1-\frac1n$$ because $2^n>n$ and because by the binomial theorem
$$\left(1-\frac1n\right)^{-n}>\left(1+\frac1n\right)^n=1+\frac nn+\frac{n(n-1)}{2n^2}+\cdots+1>2.$$
Now it suffices to prove convergence of $1-\dfrac1n$ to $1$, which is easy, and will allow to write
$$n>N\implies\left|2-\frac2n\right|<\epsilon\implies\left|2^{1-2^{-n}}-2\right|<\epsilon.$$
The figure illustrates the first inequalities.
| {
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"url": "https://math.stackexchange.com/questions/4018603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Show that $\exists \delta > 0, \forall x \in ]0,\pi[, \exists n \in \Bbb N, |\sin(xk^n)|\ge \delta$.
Let $k \ge 2$, $k \in \Bbb N$. Show that $\exists \delta > 0, \forall x \in ]0,\pi[, \exists n \in \Bbb N, |\sin(xk^n)|\ge \delta$.
My intuition tells me that we can pick $\delta=1/2$. I tried to study $u_n = \sin(xk^n)=U_{k^n}(\sin(x))=U_k(u_{n-1})$. I thought of introducing a variant of Chebyshev polynomials $U_n(\sin(x))=\sin(nx)$. But it didn't help much. Does someone have a hint? Thanks. (here $\Bbb N = \{0,1,2,\ldots\}$.)
I needed this result for solving this exercice (which is a oral exam that was posed during the competitive exams of Polytechnique, #1 engineering school in France).
Show that for $f(z)=z^k$ with $k \ge 2$
$$
\exists \delta >0, \forall x,y \in \Bbb U, x \neq y \implies \exists n \in \Bbb N, |f^{(n)}(x)-f^{(n)}(y)| \ge \delta.
$$
Suppose $x \neq 0$ and $y/x=e^{i \theta}$,
\begin{align}
| f^{(n)}(x) - f^{(n)}(y) | &= |x^{k^n} - y^{k^n}|
=
| 1- e^{i \theta k^n}|\\
&= | e^{-i \theta k^n/2} - e^{i \theta k^n/2}|
= |2i \sin(\theta k^n/2)| = 2|\sin (\theta k^n/2)|
\end{align}
So
$$
|f^{(n)}(x)-f^{(n)}(y)| \ge \delta \iff |\sin (\theta k^n/2)| \ge \delta/2
$$
| First solution of the problem
We prove that for each $x \in(0,\pi)$, there exists $n,y\in \mathbb{N}$ such that
$$\pi y + \frac{\pi}{k^2} \le x k^n\le \pi y +\pi - \frac{\pi}{k^2} \iff y +\frac{1}{k^2}\le k^n\frac{x}{\pi} \le y+1-\frac{1}{k^2} \tag{1}$$
because $$(1) \iff |\sin(xk^n)| \ge \sin \left(\frac{\pi}{k^2} \right)$$
We write $\frac{x}{\pi}$ in base-k numeral system (aka: radix system)
as follows
$$\frac{x}{\pi} =\overline{0,s_1s_2s_3...s_n...}^{(\mathbf{k})} = \sum_{i=1}^{+\infty}\frac{s_i}{k^i}=\frac{s_1}{k}+\frac{s_2}{k^2}+\frac{s_3}{k^3}+...+\frac{s_n}{k^n}+...$$
and then
$$k^n\frac{x}{\pi} =\overline{s_1s_2s_3...s_n,s_{n+1}s_{n+2}...}^{(\mathbf{k})}$$
The integer part $\left[k^n\frac{x}{\pi}\right]$ and the fractional part $\{k^n\frac{x}{\pi}\}$ of $k^n\frac{x}{\pi}$ are defined as
$$a_n =\left[k^n\frac{x}{\pi}\right] = \overline{s_1s_2s_3...s_n}^{(\mathbf{k})}$$
$$\epsilon_n=\left\{k^n\frac{x}{\pi}\right\} = \overline{0,s_{n+1}s_{n+2}...}^{(\mathbf{k})}$$
We notice that $0 \le \epsilon_n < 1$ and $k^n\frac{x}{\pi} = a_n + \epsilon_n $.
Case 1: $\frac{x}{\pi}$ is a terminating decimal in the base-k. Suppose $\frac{x}{\pi}$ has $N$ numbers after the $0$. Then, $\frac{x}{\pi}$ can be written as
$$\frac{x}{\pi} =\overline{0,s_1s_2s_3...s_N}^{(\mathbf{k})} = \frac{s_1}{k}+\frac{s_2}{k^2}+\frac{s_3}{k^3}+...+\frac{s_N}{k^N}$$
with $ 1 \le s_N \le (k-1)$.
Choose $y = a_{N-1}$ then $\epsilon_{N-1} = \overline{0,s_N}^{(\mathbf{k})} = \frac{s_N}{k}$. We can verify easily that $(y,n) = (a_{N-1},N-1)$ satisfy (1). Indeed, we have
$$a_{N-1}+\frac{1}{k^2} \le a_{N-1}+\frac{1}{k} \le a_{N-1}+ \frac{s_{N-1}}{k} = a_{N-1}+ \epsilon_{N-1} = k^{N-1} \frac{x}{\pi}$$
and
$$k^{N-1} \frac{x}{\pi} = a_{N-1}+ \epsilon_{N-1} = a_{N-1}+ \frac{s_{N-1}}{k} \le a_{N-1}+ \frac{k-1}{k} \le a_{N-1}+ 1- \frac{1}{k^2}$$
Case 2-a: $\frac{x}{\pi}$ is a non-terminating decimal (aka repeating decimal) in the base-k and there exists an $n$ such that $1 \le s_{n+1} \le (k-2)$.
Then, $\epsilon_n = \overline{0,s_{n+1}s_{n+2}...}^{(\mathbf{k})}$ satisfies $\frac{1}{k} \le \epsilon_n \le \frac{k-1}{k}$. Choose $y=a_n$, we have
$$a_n +\frac{1}{k} \le k^n\frac{x}{k\pi} \le a_n +\frac{k-1}{k} $$
$$\iff a_n +\frac{1}{k} \le a_n+ \epsilon_n \le a_n +\frac{k-1}{k} \tag{2}$$
From (2), we can deduce that (1) holds true. Indeed, we have
$$a_n+\frac{1}{k^2 } \le a_n+\frac{1}{k} $$
and
$$a_N+ \frac{k-1}{k} \le a_N+ 1- \frac{1}{k^2}$$
Case 2-b: $\frac{x}{\pi}$ is a non-terminating decimal (aka repeating decimal) in the base-k and there doesn't exist an $n$ such that $1 \le s_{n+1} \le (k-2)$.
So for all $i=1,...,+\infty$ we have $s_i = 0$ or $s_i=(k-1)$
Let $u$ is the first value such that $s_u = 0$ (if $u$ doesn't existe, then $s_u = (k-1)$ for all $u$. As we suppose $\frac{x}{\pi}$ is a non-terminating decimal, we have $\frac{x}{\pi} =\sum_{i=1}^{+\infty}\frac{(k-1)}{k^i} \rightarrow 1$ which is a contradiction.).
Let $v$ is the first value such that $v>u$ and $s_v = (k-1)$. (If $v$ doesn't existe, then $s_v = 0$ for all $v>u$. Then $\frac{x}{\pi}$ is not a non-terminating decimal $\implies$ contradiction).
Hence, $\frac{x}{\pi}$ can be written as
$$\frac{x}{\pi} = \overline{0,s_{1}s_{2}...s_{u-1}00...00(k-1)s_{v+1}....}^{(\mathbf{k})}$$
In particular,
$$k^{v-1}\frac{x}{\pi} = \overline{..s_{v-2},0(k-1)s_{v+1}..}^{(\mathbf{k})}$$
Let's $y = a_{v-1}$ and $n = v-1$, we have
$$k^n\frac{x}{\pi} = a_{v-1} + \overline{0,0(k-1)s_{v+1}..}^{(\mathbf{k})}$$
and
$$\overline{0,0(k-1)s_{v+1}..}^{(\mathbf{k})} = \frac{k-1}{k^2} + ... \in \left(
\frac{1}{k^2}, 1-\frac{1}{k^2} \right)$$
$$\implies a_n + \frac{1}{k^2} \le k^n\frac{x}{\pi} \le a_n + 1-\frac{1}{k^2}$$
So, (1) holds true.
Conclusion
Hence, we have for all $x\in (0,\pi), \exists n\in \mathbb{N}$ such that $|\sin(xk^n)|\ge \sin(\frac{\pi}{k^2})$.
Note: We can prove $ \forall x \in ]0,\pi[, \exists n \in \Bbb N, |\cos(xk^n)|\ge \cos \left( \frac{\pi}{k} \right)$ by using the same method. In fact, for the $|\cos(xk^n)|$, the proof is less difficult, it suffices to prove there exists $y,n\in\mathbb{N}$ such that $y \le k^n\frac{x}{\pi} \le y+1-\frac{1}{k}$.
Hence, we have that $\forall x \in ]0,\pi[, \exists n,m \in \Bbb N$ such that $$|\sin(xk^n)|\ge \sin\left( \frac{\pi}{k^2} \right)$$
$$|\sin(xk^m)|\le \sin\left( \frac{\pi}{k} \right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4019502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
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Showing $\cos A\cos B\cos C=\frac{s^2-(2R+r)^2}{4R^2}$ and $\cos A+\cos B+\cos C=1+\frac rR$ in $\triangle ABC$
In a triangle with vertices $A$, $B$, $C$, semiperimeter $s$, inradius $r$ and circumradius $R$, prove that $$\cos A\cos B\cos C=\frac{s^2-(2R+r)^2}{4R^2}$$ and $$\cos A+\cos B+\cos C=1+\frac rR$$
(note: we can also discover the value of $\cos A\cos B+\cos B\cos C+\cos C\cos A$ using the identity $\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1$)
Since the last time I've posted this question (the original thread is now deleted), I've reflected a bit on the suggestions of several users. First, I included relevant informations and defintion and second I did try to use the cosine law, but It did not give me help.
I was referred by a friend to the identities
$$\begin{align}
a+b+c &= 2s \tag{1} \\[4pt]
ab+ac+bc &= s^2+r^2+4rR \tag{2} \\[4pt]
abc &= 4Rrs \tag{3}
\end{align}$$ The first and third facts are obvious, while the second I do not know for sure to be true (although it probably is) and appears to model the numerator of the first identity in $\cos A\cos B\cos C$.
Any other idea?
| Use the result $1+\frac rR=\cos A+\cos B+\cos C$ in the derivation below
\begin{align}
& \frac{s^2-(2R+r)^2}{4R^2}\\
= &\frac14\left(\frac{a+b+c}{2R}\right)^2-\frac14\left(2+\frac rR\right)^2\\
=& \frac14\left[ (\sin A +\sin B +\sin C)^2 -(1+ \cos A +\cos B +\cos C)^2 \right]\\
=& \frac14\left[(\sin^2A-\cos^2 A)+(\sin^2 B-\cos^2 B) + (\sin^2 C-\cos^2 C)\right.\\
& \left.-1 -2( \cos A +\cos B +\cos C)-2(\cos (A+B)+\cos (B+C) +\cos (C+A)) \right]\\
= &\frac14( -\cos2A - \cos2B -\cos2C-1)\\
= &\frac14( -2\cos(A+B)\cos(A-B) -2\cos^2C +1 -1)\\
= &\frac12\cos C(\cos(A-B)+\cos(A+B))\\
= &\>\cos A \cos B\cos C\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4019680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Calculate the integral $ \iint_R\sqrt{y/x}\,e^{\sqrt{xy}}\,dA$ The $R$ is the region of integration, described in the following image.
To solve this integral we make a change of coordinates with $u=\sqrt{xy}$ and $v=\sqrt{\frac{y}{x}}$. Furthermore, in the region of integration we have $x>0$ and $y>0$. Then $u^2=xy$ y $v^2=\frac{y}{x}$, so we have:
\begin{align*}
v^2 &=\frac{y}{x}\\
x^2v^2 &= xy\\
x^2v^2 &= u^2.
\end{align*}
\begin{align*}
x^2 &= \frac{u^2}{v^2}\\
x &= \frac{u}{v}.
\end{align*}
Observation: $x=-\frac{u}{v}$ is not taken, because $x>0$ in the $R$ region.
With the above we can solve for $y$.
\begin{align*}
v^2& = \frac{y}{x}\\
v^2& = \frac{y}{\frac{u}{v}}\\
uv & = y.
\end{align*}
On the other hand, in the $R$ region we have the following equations $xy = 1$, $y = x$ and $y = 2$. Now with all of the above, let's pass these equations to the coordinates $u$ and $v$.
\begin{align*}
xy = 1 &\Rightarrow u = \sqrt{1} \Rightarrow u = 1\\
y = x &\Rightarrow uv = \frac{u}{v} \Rightarrow v^2 = 1 \Rightarrow v = 1\\
y = 2 &\Rightarrow 2 = uv.
\end{align*}
Observation: By definition of $u$ and $v$, we have to be positive about the values taken by the $R$ region, therefore $v = -1$ is not taken.
Using the above equations, we have that the region $R$ is transformed into the region $S$, shown below.
Now using $x = \frac{u}{v}$ and $y = uv$ we calculate the Jacobian to use the change of variable in the original integral.
\begin{align*}
\frac{\partial x}{\partial u}&=\frac{1}{v} &\frac{\partial y}{\partial u} =v\\
\frac{\partial x}{\partial v}&=-\frac{u}{v^2} &\frac{\partial y}{\partial v} =u
\end{align*}
Thus, the Jacobian is:
\begin{align*}
\begin{vmatrix}
\frac{1}{v} &v\\
-\frac{u}{v^2} &u
\end{vmatrix}
=\frac{u}{v}+\frac{u}{v}=\frac{2u}{v}.
\end{align*}
So using all the information developed we have:
\begin{align*}
\iint\limits_R\sqrt{\frac{y}{x}}e^{\sqrt{xy}}dA &= \iint\limits_Sve^u\left(\frac{2u}{v}\right)dA\\
&= 2\int_{1}^{2}\int_{1}^{2/u}ue^u\,dv\,du\\
&= 2\int_{1}^{2}ue^uv\biggr\rvert_{1}^{2/u}\,du\\
&= 2\int_{1}^{2}ue^u\left(\frac{2}{u}-1\right)du\\
&= 2\int_{1}^{2}e^u(2-u)du=2(e^u(3-u))\biggr\rvert_{1}^{2}=2e(e-2).
\end{align*}
I think this is the correct solution, I await your comments. If anyone has a different solution or correction of my work I will be grateful.
| Your solution is correct. Still we can do without changing variables:
\begin{align*}
[\ldots]&=\int_1^2\int_{1/y}^y\sqrt\frac{y}{x}e^{\sqrt{xy}}\,dx\,dy
\\&=\int_1^2 2e^{\sqrt{xy}}\Bigg|_{x=1/y}^{x=y}\,dy
\\&=2\int_1^2(e^y-e)\,dy
\\&=2(e^y-ey)\Bigg|_{y=1}^{y=2}=2e(e-2).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4021177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Calculate $\int_0^{\infty} \frac{\cos 3x}{x^4+x^2+1} dx$ Calculate $$\int_0^{\infty} \frac{\cos 3x}{x^4+x^2+1} dx$$
I think that firstly I should use Taylor's theorem, so I have:$$\int_0^\infty \frac{1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots}{(x^2+1)^2}dx$$
However I don't know what I can do the next.
| Let $f(x)=\frac{\cos(3x)}{x^4+x^2+1}$. Inasmuch as $f$ is an even function, we can write
$$\begin{align}
\int_0^\infty \frac{\cos(3x)}{x^4+x^2+1}\,dx&=\frac12 \int_{-\infty}^\infty \frac{e^{i3x}}{x^4+x^2+1}\,dx\\\\
&=\pi i \left(\text{Res}\left(\frac{e^{i3z}}{z^4+z^2+1}, z=e^{i\pi/3}\right)+\text{Res}\left(\frac{e^{i3z}}{z^4+z^2+1}, z=e^{i2\pi/3}\right)\right)
\end{align}$$
where we have used the Residue Theorem.
Can you finish now?
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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Given a recursion $a_{n+ 1}= \sqrt{na_{n}+ 2n+ 1}$ with $a_{1}\geq 1.$ Prove that $a_{n}\sim n\,{\rm as}\,n\rightarrow\infty$
Given a recursion $a_{n+ 1}= \sqrt{na_{n}+ 2n+ 1}$ with $a_{1}\geq 1.$ Prove that
$$a_{n}\sim n\,{\rm as}\,n\rightarrow\infty$$
Let $b_{n}= \dfrac{a_{n}}{n},$ we need to prove $\lim b_{n}= 1,$ then we find a value $\beta\in\left ( 0, 1 \right )$ satisfying $\left | b_{n+ 1}- 1 \right |\leq\beta\left | b_{n}- 1 \right |$ so that
$$\beta\rightarrow 0\,{\rm aut}\,n\rightarrow\infty$$
I can't predict the relationship among them, I need the the help to go to the induction, thank you.
| You have
$$\begin{aligned}
a_{n+1} &=\sqrt{na_n +2n+1}\\
&= \sqrt{n^2+2n+1+(n a_n -n^2)}\\
&=(n+1) \sqrt{1+\left(\frac{n}{n+1}\right)^2\left(\frac{a_n}{n}-1\right)}
\end{aligned}$$
Hence denoting $b_n =\frac{a_n}{n}$
$$\begin{aligned}
b_{n+1} -1
&=\sqrt{1+\left(\frac{n}{n+1}\right)^2\left(b_n-1\right)}-1
\end{aligned}$$
As for $x\gt 0$
$$\begin{aligned}
\sqrt{1+x}-1 &=\frac{x}{\sqrt{1+x}+1}\\
&\le \frac{1}{2}x
\end{aligned}$$
you get
$$\begin{aligned}0 &\le b_{n+1}-1\\
&\le \frac{1}{2} \left(\frac{n}{n+1}\right)^2 (b_n-1)\\
&\le \frac{1}{2}(b_n-1)
\end{aligned}$$ and can conclude to the desired result.
| {
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"url": "https://math.stackexchange.com/questions/4024146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Prove that $\frac{x^2+y^2}{z}+\frac{y^2+z^2}{x}+\frac{z^2+x^2}{y}\geq 2(x+y+z)$ $x,y,z \in \mathbb R^+$ different than $0$, Prove that :
$$\frac{x^2+y^2}{z}+\frac{x^2+z^2}{y}+\frac{y^2+z^2}{x}\geq 2(x+y+z)$$
My attempt:
First, we should prove that :
$$2\left(\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x} \right)\geq 2(x+y+z)$$
I will consider it as proven, we gonna start by :
$$(x-y)^2 \geq 0 \iff x^2+y^2 \geq 2xy $$
by the same method we have :
$x^2+z^2 \geq 2xz$ and $y^2+z^2 \geq 2yz$
so we can replace $2xy$ by $x^2+y^2$ in this inequality
$$2\left(\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x} \right)\geq 2(x+y+z)$$
so :
$$\frac{x^2+y^2}{z}+\frac{x^2+z^2}{y}+\frac{y^2+z^2}{x}\geq 2(x+y+z)$$
I'm new to AM-GM inequality, if it exists a better proof using it please post it.
| Probably the following is shorter and easier.
Firstly, note that for positive reals $a$ and $b$ the inequality
$$
\frac{a^2}{b}\ge 2a-b
$$
holds (since it's equivalent to $\frac{(a-b)^2}{b}\ge 0$). Applying this observation six times gives
$$
\frac{x^2+y^2}{z}+\frac{x^2+z^2}{y}+\frac{y^2+z^2}{x}\ge(2x-z+2y-z)+(2x-y+2z-y)+(2y-x+2z-x)=
\\
=2(x+y+z),
$$
as desired.
Comment. One can generalize this trick, namely the inequality
$$
\frac{a^x}{b^y}\ge\frac{xa^{x-y}-yb^{x-y}}{x-y}
$$
holds for $a,b,x,y>0$, $x\neq y$. This is often helpful when we need to estimate some fractions using polynomial expressions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4024476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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} |
If $x+2y=8$, find the minumum of $x+y+\frac{3}{x}+\frac{9}{2y}$ (x, y ∈ R+) Question : If $x+2y=8$, find the minumum of $x+y+\dfrac{3}{x}+\dfrac{9}{2y}$ $(x, y \in \mathbb{R}^+)$
I tried to use $x=8-2y$, and I thought I can plug in. But I think I'm not doing it well. I thought about AM-GM, and how can I expand this using $x+2y=8$?
| Knowing the answer from taking the derivative, I'd cheat by "observing"
$$x+y+\frac 3x + \frac{9}{2y}=\left(\frac x2+y\right) + \left(\frac x2 + \frac 2x\right)+\left(\frac 1x+\frac {9}{2y}\right)$$
$$\ge 4+2\sqrt{\frac x2 \cdot \frac 2x}+\frac{(x+2y)(\frac 1x + \frac{9}{2y})}{x+2y} \tag{AM-GM}$$
$$\ge 4+2+\frac{(1+3)^2}{8}=8 \tag{C-S}$$
Without knowing the answer and barred from using calculus, I'd do
$$x+y+\frac 3x + \frac{9}{2y}=\left(\frac x2+y\right) + \left(\frac x2 + \frac \alpha x\right)+\left(\frac {3-\alpha}{ x}+\frac {9}{2y}\right)\\
$$
and solve for $\alpha$ and $x_0$ such that
$$\frac {x_0} {2}=\frac{\alpha}{x_0} \tag1$$
$$\frac{x_0^2}{3-\alpha} = \frac{(8-x_0)^2}{9}\tag2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4024960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
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} |
Question on solving one algebra relation to obtain another If $a,b,c\in \mathbb R$ and,
$$\left(\frac{a-b}{a+b}\right)\left(\frac{b-c}{b+c}\right)\left(\frac{c-a}{c+a}\right)=-27$$
Evaluate $$\frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}$$
I noticed that you can reduce the question to the following:
If $(1-2x)(1-2y)(1-2z)=-27$, then evaluate $x+y+z$ where $x=\frac{b}{a+b}, \ y=\frac{c}{b+c}, \ z=\frac{a}{c+a}$
I'm not sure how to solve this. I think it may use inequalities or pure algebra.
Please help me out here. Thanks a lot :)
| Let's call $\begin{cases} U=\frac a{a+b}+\frac b{b+c}+\frac c{c+a}\\V=\frac b{a+b}+\frac c{b+c}+\frac a{c+a}\end{cases}$
We have trivially $U+V=3$
Now $U-V=\frac {a-b}{a+b}+\frac {b-c}{b+c}+\frac {c-a}{c+a}=\cdots=-\frac {a-b}{a+b} \times\frac {b-c}{b+c}\times\frac {c-a}{c+a}=27$
Show the identity $\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}=-\frac{a-b}{a+b}\cdot\frac{b-c}{b+c}\cdot\frac{c-a}{c+a}$
Therefore $U=15$ and $V=-12$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integrating $\frac1{2\sigma\sqrt{2\pi}}\int_{-\infty}^\infty(1+erf(\frac{z-\mu_i}{\sigma_i\sqrt2}))\exp( -\frac12(\frac{z-\mu}\sigma)^2)dz$ I am interested in the integral of
$$\frac{1}{2\sigma \sqrt{2\pi}}\int_{-\infty}^\infty \left( 1+\operatorname{erf}\left(\frac{z-\mu_i}{\sigma_i\sqrt{2}}\right)\right)\exp\left( -\frac{1}{2}\left( \frac{z-\mu}{\sigma}\right)^2\right)dz$$
where $\operatorname{erf}$ is the error function defined as the CDF of the standard normal (up to a constant).
In this case $\sigma, \mu, \mu_i, \sigma_i$ are constants.
I truly have no clue how to even start something like this. Is Mathematica even capable of this type of integral?
For those curious, this is evaluated when considering $Pr(f_i \leq f \leq f_j)$ where $f_j, f_i, f$ follow a normal distribution with different parameters. The full integral is
$$\frac{1}{2\sigma \sqrt{2\pi}}\int_{-\infty}^\infty \left( 1+erf\left(\frac{z-\mu_i}{\sigma_i\sqrt{2}}\right)\right)\exp\left( -\frac{1}{2}\left( \frac{z-\mu}{\sigma}\right)^2\right)dz \\
- \frac{1}{4 \sigma \sqrt{2 \pi}} \int_{-\infty}^\infty \left( 1+erf \left(\frac{z-\mu_i}{\sigma_i \sqrt{2}}\right)\right)\left(1+erf\left(\frac{z-\mu_j}{\sigma_j \sqrt{2}}\right)\right)\exp\left(-\frac{1}{2}\left(\frac{z-\mu}{\sigma}\right)^2\right)dz$$
But I thought that might be inappropriate ;)
| Integral of one error function and Gaussian
Mathematica does not evaluate the integral. Surprisingly, there is an exact result. By changing variables, $z-\mu=x$, your integral of $\operatorname{erf} \times \text{Gaussian}$ may be written as
$$
I(b)=\int\limits_{-\infty}^\infty dx \ \operatorname{erf}(ax+b) \exp \left(-cx^2 \right)
$$
Since $\partial_t \operatorname{erf}(t)= \frac{2}{\sqrt{\pi}}e^{-t^2}$, differentiating with respect to $b$ produces
$$
\partial_b I(b)=\int\limits_{-\infty}^\infty dx \ \frac{2}{\sqrt{\pi}} \exp\left( -(ax+b)^2-cx^2 \right)
$$
This is just a Gaussian integral over $x$, the result is
$$
\partial_b I(b)= \frac{2}{\sqrt{a^2+c}} \exp \left( -\frac{b^2 c}{a^2 +c} \right)
$$
Now we integrate to recover $I(b)$
$$
I(b)-I(0)=\int\limits_0^b \ dt \frac{2}{\sqrt{a^2+c}} \exp \left( - t^2\frac{ c}{a^2 +c} \right)
$$
$I(0)=\int dx \ ( \text{odd function})=0$, and the integral on the right is simply the definition of $\operatorname{erf}$, so we have
$$
I(b)=\sqrt{\frac{\pi}{c}} \operatorname{erf} \left( b \sqrt{\frac{c}{a^2+c}} \right)
$$
$$
\tag*{$\blacksquare$}
$$
Integral of two error functions and Gaussian
We may use a similar method to find that
$$
\int\limits_{-\infty}^\infty dx \ \operatorname{erf}(a_1 x) \operatorname{erf}(a_2 x) e^{-cx^2} = \frac{2}{\sqrt{c\pi}} \tan^{-1}\left( \frac{a_1 a_2}{ \sqrt{c(a_1^2+a_2^2+c)}} \right)
$$
However, once we introduce $b\neq 0$, we meet the more complicated integral in an intermediate step
$$
\int\limits_0^b dt \ \operatorname{erf}(\alpha t + \beta) e^{-\gamma t^2}
$$
Update! Using the result from this post, we can make progress. Let
$$
J(\beta)=\int\limits_{-\infty}^\infty dx \ \operatorname{erfc}(\alpha x+\beta) \operatorname{erf}(a x+b) \exp \left(-cx^2 \right)
$$
Then $J(\infty)=0$. Differentiating
$$
\partial_{\beta} J(\beta)=-\frac{2}{\sqrt{\pi}} \int\limits_{-\infty}^\infty dx \ \operatorname{erf}(a x+b) \exp \left(-cx^2 - (\alpha x +\beta)^2 \right)
$$
Which we can evaluate using our first result
$$
\partial_{\beta} J(\beta)=-\frac{2}{\sqrt{\pi}} \sqrt{\frac{\pi}{c+\alpha^2}} \operatorname{erf}\left( \left(b-\frac{a \alpha \beta}{c+\alpha^2} \right) \sqrt{\frac{c+\alpha^2}{c+\alpha^2+a^2}} \right) \exp \left( -\beta^2 \left( 1+\frac{\alpha^2}{c+\alpha^2}\right) \right)
$$
With $\Lambda=\frac{2}{\sqrt{\pi}} \sqrt{\frac{\pi}{c+\alpha^2}}$, $A=-\frac{a \alpha }{c+\alpha^2} \sqrt{\frac{c+\alpha^2}{c+\alpha^2+a^2}}$, $B=b\sqrt{\frac{c+\alpha^2}{c+\alpha^2+a^2}}$, and $C=\left( 1-\frac{\alpha^2}{c+\alpha^2}\right)$. Now we integrate
$$
J(\infty)-J(\beta)=-\int\limits_\beta^\infty \ dt \Lambda \operatorname{erf}(At+B)e^{-Ct^2}
$$
$$
J(\beta)=\int\limits_\beta^\infty dt \ \Lambda \operatorname{erf}(At+B)e^{-Ct^2}= \frac{\Lambda}{\sqrt{2C}} \int\limits_{\sqrt{2C}\beta}^\infty du \ \operatorname{erf}(Au/\sqrt{2C}+B)e^{-u^2/2}
$$
Which may be written in terms of the 'Generalized Owen-T' function
$$
J(\beta)=\frac{2 \sqrt{\pi} \Lambda}{\sqrt{C}}T\left( \sqrt{2C}\beta, A/\sqrt{C},\sqrt{2}B\right)
$$
Using identity ii we have
$$
J(\beta)=\frac{2 \sqrt{\pi} \Lambda}{\sqrt{C}} \left[ \frac{1}{2\pi} \left( \tan^{-1}(A/\sqrt{C}) -\tan^{-1}(A/\sqrt{C} + B/\beta\sqrt{C}) \\ -\tan^{-1}(B^{-1}(\beta \sqrt{C} +AB/\sqrt{C} + A^2 \beta /\sqrt{C} )) \right) +\frac{1}{4} \operatorname{erf}(B (1+A^2/C)^{-1/2}) \\ +T(\beta \sqrt{2C},(A\beta^{-1}+B)/\beta \sqrt{C})+T(\sqrt{2}B (1+A^2/C)^{-1/2}, B^{-1}(\beta \sqrt{C} +AB/\sqrt{C} + A^2 \beta /\sqrt{C} ) \right]
$$
Where $T$ is the Owen T function. I've checked it numerically, I hope there are no typos here.
Approximate integral of two or more error functions with Gaussian
We can approximate the integral using Laplace's method. I'll demonstrate with two error functions, but in principle you could have more. Let
$$
I(c)=\int\limits_{-\infty}^\infty dx \ \operatorname{erf}(a_1 x+ b_1) \operatorname{erf}(a_2 x+ b_2) e^{-cx^2}
$$
Expand the product of $\operatorname{erf}$s around the Gaussian peak: $x=0$
$$
\operatorname{erf}(a_1 x+ b_1) \operatorname{erf}(a_2 x+ b_2) = \operatorname{erf}( b_1) \operatorname{erf}( b_2) + x \left( \text{junk} \right) + x^2 \frac{2 e^{-b_1^2-b_2^2}}{\pi} \left( 2 a_1 a_2 -a_2^2 b_2 e^{b_1^2} \sqrt{\pi} \operatorname{erf}(b_1) - a_1^2b_1 e^{b_2^2} \sqrt{\pi} \operatorname{erf}(b_2) \right) + \mathcal{O}(x^3)
$$
Term by term integration produces the asymptotic series. The $(\text{junk})$ and all $x^{\text{odd}}$ terms integrate to zero. Noting that
$$
\int\limits_{-\infty}^\infty dx \ x^{2n} e^{-cx^2} = c^{-n-1/2} \Gamma(n+1/2)
$$
For fixed $a_i, \ b_i$ we have
$$
I(c) \sim \sqrt{\frac{\pi}{c}} \operatorname{erf}( b_1) \operatorname{erf}( b_2) \\ +\frac{1}{c^{3/2}} \frac{ e^{-b_1^2-b_2^2}}{\sqrt{\pi}} \left( 2 a_1 a_2 -a_2^2 b_2 e^{b_1^2} \sqrt{\pi} \operatorname{erf}(b_1) - a_1^2b_1 e^{b_2^2} \sqrt{\pi} \operatorname{erf}(b_2) \right) \ \ , \ \ c \to \infty
$$
Which is the two term approximation.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}<2$ Prove that for every $a,b,c \in \mathbb{R}^{+}$ We have $$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}<2$$
Unfortunately i can just prove that :
$$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{b^2}{b^2+ac} <3$$
like this :
$$a^2+bc>a^2 \iff \frac{a^2}{a^2+bc}<1$$
and by the same method we have :
$$\frac{b^2}{b^2+ac}<1,\frac{b^2}{b^2+ac}<1$$
Adding them together will give us the desired inequality.
and please don't use any $\sum_{cyc}$ because I get confused with it.
| From
$$\frac{a^2}{a^2+bc}+\frac{b^2}{b^2+ac}+\frac{c^2}{c^2+ab}<2 \iff\\
\frac{a^2+bc-bc}{a^2+bc}+\frac{b^2+ac-ac}{b^2+ac}+\frac{c^2+ab-ab}{c^2+ab}<2 \iff\\
3-\frac{bc}{a^2+bc}-\frac{ac}{b^2+ac}-\frac{ab}{c^2+ab}<2 \iff\\
\frac{bc}{a^2+bc}+\frac{ac}{b^2+ac}+\frac{ab}{c^2+ab}>1$$
and applying Titu's Lemma
$$\frac{bc}{a^2+bc}+\frac{ac}{b^2+ac}+\frac{ab}{c^2+ab}=\\
\frac{b^2c^2}{a^2bc+b^2c^2}+\frac{a^2c^2}{b^2ac+a^2c^2}+\frac{a^2b^2}{c^2ab+a^2b^2}\geq\\
\frac{(bc+ac+ab)^2}{b^2c^2+a^2c^2+a^2b^2+a^2bc+b^2ac+c^2ab}=\\
\frac{\color{green}{b^2c^2+a^2c^2+a^2b^2}+\color{red}{2a^2bc+2b^2ac+2c^2ab}}{\color{green}{b^2c^2+a^2c^2+a^2b^2}+\color{red}{a^2bc+b^2ac+c^2ab}}>1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4029795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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How to show that this $2n \times n^2$ matrix has rank $2n-1$? The matrix is fairly messy to present, but quite easy to understand. When $n=3$, the matrix is
\begin{bmatrix}
1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0\\
0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0\\
0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1\\
1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1
\end{bmatrix}
So, basically, it splits into two parts:
*
*the upper part (from row $1$ to row $n$) has $n$ identity matrices $I_n$.
*the lower part is $O_n^1, \ldots, O_n^n$, where each $O_n^i$ is an $n \times n$ matrix whose $i$-th row is all one whereas the other entries are zero.
By some examples and intuition, I am aware that the rank of such a matrix is $2n-1$, but how should I rigorously prove it?
| Things will perhaps be clearer if we write these $2n$ rows from $\Bbb{R}^{n^2}$ as
$n \times n$ matrices.
For $n=3$ we are looking at
$$\begin{bmatrix} 1 & 0 & 0 \\ 1 & 0 & 0 \\ 1 & 0 & 0\end{bmatrix},\begin{bmatrix} 0 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 0\end{bmatrix},\begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 1\end{bmatrix}, \begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix},\begin{bmatrix} 0 & 0 & 0 \\ 1 & 1 & 1\\ 0 & 0 & 0\end{bmatrix},\begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0 \\ 1 & 1 & 1\end{bmatrix}.$$
Call them $A_1, \ldots, A_{2n}$. We will prove that $\{A_1, \ldots, A_{2n-1}\}$ is linearly inependent and $A_{2n}$ is in its span.
Assume $$0 = \sum_{i=1}^{2n-1}\alpha_iA_i = \begin{bmatrix} \alpha_{1}+\alpha_{n+1} & \alpha_{2}+\alpha_{n+1} & \cdots & \alpha_{n}+\alpha_{n+1} \\
\alpha_{1}+\alpha_{n+2} & \alpha_{2}+\alpha_{n+2} & \cdots & \alpha_{n}+\alpha_{n+2}\\
\vdots & \vdots & \ddots & \vdots \\
\alpha_{1}+\alpha_{2n-1} & \alpha_{2}+\alpha_{2n-1} & \cdots & \alpha_{n}+\alpha_{2n-1}\\
\alpha_{1} & \alpha_{2}& \cdots & \alpha_{n}\end{bmatrix}$$
so first we get $\alpha_1=\cdots=\alpha_n = 0$ and then $\alpha_{n+1}=\cdots=\alpha_{2n-1}=0$.
On the other hand, we have
$$A_{2n}= \sum_{i=1}^n A_i - \sum_{i=n+1}^{2n-1}A_i.$$
Therefore $\dim\operatorname{span}\{A_1, \ldots, A_{2n}\} = 2n-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4034693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluate an absolute monster integral $\int\limits_{0}^{1} \frac{\log(1-x+x^2)}{\sqrt{x}(1+x)}\mathrm{d}x.$ I want to figure out a way to evaluate $$\int\limits_{0}^{1} \frac{\log(1-x+x^2)}{\sqrt{x}(1+x)}\mathrm{d}x.$$
I tried to substitute $x = u^2$ and cancel the square root in the denominator, getting $$2\int\limits_{0}^{1} \frac{\log(1-u^2+u^4)}{1+u^2}\mathrm{d}u.$$But now I am stuck again.
| Note
$$I=\int\limits_{0}^{1} \frac{\ln(1-x+x^2)}{\sqrt{x}(1+x)}{d}x
\overset{t=\sqrt x} = 2\int\limits_{0}^{1} \frac{\ln(1-t^2+t^4)}{1+t^2}dt
$$
Let $J(a) = \int_0^1 \frac{\ln(\frac14(1+t^2)^2 \sec^2a -t^2)}{1+t^2}dt
$.
Then, $J(0) = \int_0^1 \frac{2\ln\frac{1-t^2}2}{1+t^2}dt=
-2G$
$$J’(a)= \int_0^1 \left(\frac{\tan a}{t^2+2t\cos a+1} + \frac{\tan a}{t^2-2t\cos a+1} \right)dt=\frac\pi2 \sec a
$$
$$J(\frac\pi6) = \int\limits_{0}^{1} \frac{\ln(1-t^2+t^4)-\ln3}{1+t^2}dt=\frac12I-\frac\pi4\ln3
$$
Thus
\begin{align}
I= \frac\pi2\ln3+2J(\frac\pi6)
= \frac\pi2\ln3+2J(0) +2\int_0^{\pi/6} J’(a)da
=\pi \ln3-4G
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4035977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How to Evaluate $\int_{0}^{1} \frac{x^2 \ln(1-x^4)}{1+x^4}dx$? How to evaluate
$$ \int_{0}^{1} \frac{x^2 \ln(1-x^4)}{1+x^4} \,dx \approx -0.162858 \tag{1}$$
The integral arises in the computation of
$$\left( \sum_{n=1}^{\infty} \frac{(-1)^n}{n}\right)\left(\sum_{n=1}^{\infty} \frac{(-1)^n}{4n-1}\right)$$
as
$$ \scriptsize{\frac{\pi \ln(2)}{4\sqrt{2}} + \frac{\ln(2) \ln(3-2\sqrt{2})}{4\sqrt{2}} -3\ln(2) + \frac{\pi}{2}= \int_{0}^{1} \frac{x^2 \ln(1-x^4)}{1+x^4} \,dx + \int_{0}^{1} \frac{x^{1/4}}{2(1+x)}\left(\tan^{-1}(x^{1/4}) - \tanh^{-1}(x^{1/4}) \right)}dx $$
A similar Integral
$$ \int_{0}^{1}\left( \frac{x^2 \ln(2)}{x^4-1} - \frac{x^2 \ln(1+x^4)}{x^4-1}\right)dx = C-\frac{\pi^2}{16}+\frac{\ln^2(\sqrt{2}-1)}{4}+\frac{\pi \ln (\sqrt{2}-1)}{4} \tag{2} $$
Where $ C $ = Catalan Constant
Unfortunately the same techniques I used to evaluate $(2)$ have not worked for $(1)$.
I know only for integration - By parts, U-Sub, and using Taylor Series as well as Mathematica.
Q = Is there a closed form for Integral $(1)$?
EDIT
$$ (1) = \frac{-\pi^2}{8\sqrt{2}} + \frac{\pi \ln(8)}{4\sqrt{2}} +\frac{\ln(8) \ln(3-2\sqrt{2})}{4\sqrt{2}} -\frac{\pi \ln(3-2\sqrt{2})}{8\sqrt{2}} + 4\sum_{k=1}^{\infty} \frac{(-1)^k}{4k-1}\sum_{n=1}^{k} \frac{1}{4n-1} $$
$$\sum_{k=1}^{\infty} \frac{(-1)^k}{4k-1}\sum_{n=1}^{k} \frac{1}{4n-1} = \frac{1}{64}\left(\psi^{(1)}\left(\frac{7}{8}\right)-\psi^{(1)}\left(\frac{3}{8}\right)\right) + W $$
Where $W$ is some value.
| This may be counterproductive but we can obtain another form from @yarchik's solution.
$$
\begin{aligned}
S%
&=-\sum _{n=0}^{\infty } \frac{(-1)^n }{4n+3}H_{n+\frac{3}{4}},\\
&=-\frac{1}{4}\sum _{n=0}^{\infty } \frac{(-1)^n }{n+3/4}H_{n+\frac{3}{4}},\\
&=-\frac{1}{4}\int_0^1x^{-1/4}\sum _{n=0}^{\infty } (-x)^nH_{n+\frac{3}{4}}\,\mathrm dx,\\
\end{aligned}
$$
The remaining series can be solved exactly yielding
$$
\begin{aligned}
S%
&=\frac{1}{7}\int_0^1\frac{x^{3/4}}{1+x}F\left({1,7/4\atop 11/4};-x\right)\,\mathrm dx-\frac{H_{3/4}}{4}\int_0^1\frac{x^{-1/4}}{1+x}\,\mathrm dx,\\
&=\frac{1}{7}\int_0^1\frac{x^{3/4}}{1+x}F\left({1,7/4\atop 11/4};-x\right)\,\mathrm dx-\frac{H_{3/4}(-1)^{1/4}}{2}(\operatorname{arctan}((-1)^{1/4})- \operatorname{arctanh}((-1)^{1/4})),
\end{aligned}
$$
The hypergeometric function in the integral is zero-balanced and may reduce further.
| {
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Express $S =\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\ldots+\frac{1}{(n-2)\cdot(n-1)}+\frac{1}{n\cdot(n+1)}$ in terms of $n$.
Express $S =\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\ldots+\frac{1}{(n-2)\cdot(n-1)}+\frac{1}{n\cdot(n+1)}$ in terms of $n$.
Here's what I have done so far:
$$\begin{align*}
S &=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{n-2}-\frac{1}{n-1}\right)+\left(\frac{1}{n}-\frac{1}{n+1}\right)\\[5pt]
&= \left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\ldots+\frac{1}{n}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\ldots+\frac{1}{n+1}\right)\\[5pt]
&= \sum_{k=1}^n \frac{1}{2k-1} - \sum_{k=1}^{n+1} \frac{1}{2k} \\[5pt]
\end{align*}$$
How do I continue the problem from here? Is it even possible?
| Let $ n\in\mathbb{N}^{*} $ be an odd integer, we have : \begin{aligned}\frac{1}{1\times 2}+\frac{1}{3\times 4}+\cdots +\frac{1}{n\left(n+1\right)}=\sum_{k=1}^{\frac{n+1}{2}}{\frac{1}{2k\left(2k-1\right)}}&=\sum_{k=1}^{\frac{n+1}{2}}{\frac{1}{2k-1}}-\sum_{k=1}^{\frac{n+1}{2}}{\frac{1}{2k}}\\ &=\sum_{k=1}^{n}{\frac{1}{k}}-\sum_{k=1}^{\frac{n-1}{2}}{\frac{1}{2k}}-\sum_{k=1}^{\frac{n+1}{2}}{\frac{1}{2k}}\\ \sum_{k=1}^{\frac{n+1}{2}}{\frac{1}{2k\left(2k-1\right)}}&=H_{n}-\frac{H_{\frac{n-1}{2}}+H_{\frac{n+1}{2}}}{2}\end{aligned}
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 1
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Calculating this series integral $\int (\frac{x^{2n-2}}{(2n-2)!})\cdot \frac{x^{2n-1}}{(2n-1)!}$ I've calculated this integral and would like the communities feedback and support towards my solution:
$$\int \left(1 \cdot \frac{x^2}{2!} \cdot \frac{x^4}{4!} \cdots\frac{x^{2n-2}}{(2n-2)!}\right)\cdot \left(x \cdot \frac{x^3}{3!} \cdot \frac{x^5}{5!}\cdots\frac{x^{2n-1}}{(2n-1)!}\right)\,\mathrm{d}x$$
My attempt:
I performed substitution where $u = \left(x \cdot \frac{x^3}{3!} \cdot \frac{x^5}{5!}\cdot\frac{x^{2n-1}}{(2n-1)!}\right)$.
Then taking the derivative of both sides $\mathrm{d}u = \left(1 \cdot \frac{x^2}{2!} \cdot \frac{x^4}{4!} \cdots\frac{x^{2n-2}}{(2n-2)!}\right)\mathrm{d}x$.
This yields $$\int u\space \mathrm{d}u = \frac{u^2}{2}+C.$$
Then substituting back in the variable in $u$:
$$\frac{1}{2}\left(x \cdot \frac{x^3}{3!} \cdot \frac{x^5}{5!}\frac{x^{2n-1}}{(2n-1)!}\right)^2$$
Although I'm not entirely sure how to distrbute the $2$ properly. Do I multiply the single $x$ onto $\frac{x^3}{2!}$, then distribute the power, and multiply the $2$ on the numerator using this fraction rule like this
$$\left(\frac{2(x^6)}{3!} \cdot \frac{2(x^7)}{5!}\frac{2(x^{2n+1)}}{(2n-1)!}\right) ?$$
| Please keep in mind: the derivative of a product is not the product of the derivatives!
In this case you should simply multiply both factors obtaining
$$\int\prod_{k=0}^{2n-1}\frac {x^k}{k!}dx=\frac1 {\prod_{k=0}^{2n-1} k!}\int x^{\sum_{k=0}^{2n-1}k}dx=\frac1 {\prod_{k=0}^{2n-1} k!}\int x^{n (2n-1)}dx=
\frac{x^{n (2n-1)+1}}{(n (2n-1)+1)\prod_{k=0}^{2n-1} k!}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4039963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove $ \int_{5\pi/36}^{7\pi/36} \ln (\cot t )dt +\int_{\pi/36}^{3\pi/36} \ln (\cot t )dt = \frac49G $ I have the conjecture for the integral
$$ \int_{\frac{5\pi}{36}}^{\frac{7\pi}{36}} \ln (\cot t )\>dt +\int_{\frac{\pi}{36}}^{\frac{3\pi}{36}} \ln (\cot t )\>dt = \frac49G $$
where $G$ is the Catalan constant, following some heuristic effort. But, I am unable to derive it formally despite having tried for some time. The derivation seems tougher than the tools in my toolbox.
I stick to real methods, though, trying to work it out mainly with elementary approaches. The standard procedures do not help that much. For instance, the substitution $u=\tan t $ is not viable given the inordinate limits; integration-by-parts coverts the integrand to $\frac t{\sin 2t}$, which is not revealing either. The proof for a much simpler integral
$$\int_0^\frac{\pi}{12}\ln(\cot x)=\frac23G
$$
is known, yet perhaps with limited relevance. Maybe, there is no avoidance of resorting to infinite series, complex methods, etc.
Edit:
Taking the cue from the comments below, the integral can be equivalently expressed as
$$\int_{\frac{\pi}{12}}^{\frac{5\pi}{36}} \ln\left( \cot (t+\frac\pi{36} )\cot (t-\frac\pi{36} )\right)=\frac49G$$
which may not reduce the difficulty, albeit appearing compact.
| Using the third of these equations we can express your integrals in terms of the Clausen function:
\begin{align}
I &\equiv \int \limits_{\frac{\pi}{36}}^{\frac{3\pi}{36}} \log (\cot(t))\, \mathrm{d} t + \int \limits_{\frac{5\pi}{36}}^{\frac{7\pi}{36}} \log (\cot(t))\, \mathrm{d} t \\
&= \frac{1}{2} \left[- \operatorname{Cl}_2 \left(\vphantom{\operatorname{Cl}_2 \left(\frac{11\pi}{18}\right)}\frac{\pi}{18}\right) + \operatorname{Cl}_2 \left(\frac{3\pi}{18}\right) - \operatorname{Cl}_2 \left(\frac{5\pi}{18}\right) + \operatorname{Cl}_2 \left(\frac{7\pi}{18}\right) \right. \\
&\phantom{= \frac{1}{2} \left[ \vphantom{\operatorname{Cl}_2 \left(\frac{11\pi}{18}\right)} \right.} \left.+ \operatorname{Cl}_2 \left(\frac{11\pi}{18}\right) - \operatorname{Cl}_2 \left(\frac{13\pi}{18}\right) + \operatorname{Cl}_2 \left(\frac{15\pi}{18}\right) - \operatorname{Cl}_2 \left(\frac{17\pi}{18}\right)\right] \\
&= \frac{1}{2} \left[\operatorname{Cl}_2 \left(\frac{\pi}{2}\right) - \sum \limits_{k=0}^{8} (-1)^k \operatorname{Cl}_2 \left(\frac{(2k+1)\pi}{18}\right) \right] \, .
\end{align}
The Fourier series of $\operatorname{Cl}_2$ and its special value $\operatorname{Cl}_2 \left(\frac{\pi}{2}\right) = \mathrm{G}$ allow us to write this result as
$$ I = \frac{1}{2} \left[\mathrm{G} - \sum \limits_{n=1}^\infty \frac{\alpha(n)}{n^2}\right]$$
with $\alpha \colon \mathbb{N} \to \mathbb{R}$ given by
$$\alpha (n) = \sum \limits_{k=0}^8 (-1)^k \sin \left(\frac{(2k+1)n\pi}{18}\right) = \operatorname{Im} \left[\mathrm{e}^{\mathrm{i} \pi \frac{n}{18}} \sum \limits_{k=0}^{8} \mathrm{e}^{\mathrm{i} \pi \left(1+\frac{n}{9}\right) k} \right] \, .$$
Clearly, we have
$$\alpha (9 + 18 m) = \operatorname{Im} \left[\mathrm{i} (-1)^m \sum \limits_{k=0}^{8} 1 \right] = 9 (-1)^m $$
for $m \in \mathbb{N}_0$, while for all other $n \in \mathbb{N}$ we find
$$ \alpha (n) = \operatorname{Im} \left[\mathrm{e}^{\mathrm{i} \pi \frac{n}{18}} \frac{1 - \mathrm{e}^{\mathrm{i} \pi (9+n)}}{1 - \mathrm{e}^{\mathrm{i} \pi \left(1+\frac{n}{9}\right)}} \right] = \operatorname{Im} \left[2 \frac{1 + (-1)^n}{\left \lvert 1 + \mathrm{e}^{\mathrm{i} \pi n/9} \right \rvert^2} \cos \left(\frac{n \pi}{18}\right)\right] = 0 \, .$$
Therefore,
$$ I = \frac{1}{2} \left[\mathrm{G} - \sum \limits_{m=0}^\infty \frac{9 (-1)^m}{(9 + 18 m)^2}\right] = \frac{1}{2} \left[\mathrm{G} - \frac{1}{9}\sum \limits_{m=0}^\infty \frac{(-1)^m}{(2m+1)^2}\right] = \frac{4}{9} \mathrm{G} \, .$$
The same method leads to the more general result
\begin{align}
\sum \limits_{\mu = 0}^{\nu - 1} (-1)^{\nu - 1 - \mu} \int \limits_0^{\frac{2\mu +1}{2\nu+1} \frac{\pi}{4}} \log(\cot(t)) \, \mathrm{d} t &= \frac{1}{2} \left[\mathrm{G} + (-1)^{\nu - 1} \sum \limits_{k=0}^{2\nu} (-1)^k \operatorname{Cl}_2 \left(\frac{2k+1}{2\nu+1} \frac{\pi}{2}\right)\right] \\
&= \frac{\nu + 1_{2\mathbb{N} - 1} (\nu)}{2\nu+1} \mathrm{G}
\end{align}
for $\nu \in \mathbb{N}$ ($1_{2\mathbb{N}-1}$ is the indicator function of the odd positive integers). The first three cases are
\begin{align}
\int \limits_0^{\frac{\pi}{12}} \log(\cot(t)) \, \mathrm{d} t &= \frac{2}{3} \mathrm{G} \, , \\
\int \limits_{\frac{\pi}{20}}^{\frac{3\pi}{20}} \log(\cot(t)) \, \mathrm{d} t &= \frac{2}{5} \mathrm{G} \, , \\
\int \limits_0^{\frac{\pi}{28}} \log(\cot(t)) \, \mathrm{d} t + \int \limits_{\frac{3\pi}{28}}^{\frac{5\pi}{28}} \log(\cot(t)) \, \mathrm{d} t &= \frac{4}{7} \mathrm{G}
\end{align}
and $\nu = 4$ corresponds to the original question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4040125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 0
} |
How to simplify $\frac{\sqrt{2}+2(\cos 20^\circ+\cos 25^\circ)}{\sin \left(90-\frac{45}{2}\right)\sin 55^\circ \sin 57.5^\circ}$? The problem is as follows:
Simplify the following expression:
$B=\frac{\sqrt{2}+2(\cos 20^\circ+\cos 25^\circ)}{\sin \left(90-\frac{45}{2}\right)\sin 55^\circ \sin 57.5^\circ}$
The alternatives given in my book are as follows:
$\begin{array}{ll}
1.&7.5\\
2.&6\\
3.&8\\
4.&5\\
\end{array}$
I'm not sure how to proceed here because the division seems kind of complicated to simplify.
But I could spot that suspicioulsy $\sqrt{2}= \csc 45^\circ$
and also $2= \csc 30^\circ$
But I don't know if these would come into play in solving this problem as it is challenging. Perhaps does it exist a way to solve this without much fuss?.
I could also spot that:
$\sin \left(90-\frac{45}{2}\right)= \cos \frac{45}{2}$
$\sin 55^\circ = \cos 35^\circ$
and
$\sin 57.5^\circ= \cos 32.5^\circ = \sin \frac{65}{2}^\circ$
The rest I presume that involves the simplifcation of the expresion using sum to product formulas. But I got stuck with those. Can someone help me here?.
| The denominator is $\sin(55^\circ) \sin(57.5^\circ) \sin(67.5^\circ)$ and note that the angles add up to $180^\circ$.
Using the identity $\sin \alpha \cdot \sin \beta \cdot \sin \gamma = \frac14(\sin 2\alpha + \sin 2 \beta + \sin 2\gamma)$ listed in Wikipedia Further "conditional" identities for the case α + β + γ = 180° that is only valid if the angles sum to $180^\circ$, the denominator becomes
$$\frac14(\sin(110^\circ)+\sin(115^\circ)+\sin(135^\circ)) = \frac14(\sin(70^\circ)+\sin(65^\circ)+\sin(45^\circ))$$
since $\sin(180^\circ-\theta)=\sin(\theta)$. Then use the value for $\sin45^\circ$ and the fact that $\sin(90^\circ-\theta)=\cos(\theta)$ to obtain
$$\frac{1}{4}(\frac{1}{\sqrt2}+\cos(20^\circ)+\cos(25^\circ))$$
for the denominator. Writing the entire fraction,
$$\frac{\sqrt{2}+2\cos(20^\circ)+2\cos(25^\circ)}{\frac{1}{4}(\frac{1}{\sqrt2}+\cos(20^\circ)+\cos(25^\circ))} = \frac{8(\sqrt{2}+2\cos(20^\circ)+2\cos(25^\circ))}{2(\frac{1}{\sqrt2}+\cos(20^\circ)+\cos(25^\circ))} = 8$$
is the desired result.
Edited to add:
Here is a quick proof of the identity mentioned previously.
$$\sin\alpha\sin\beta\sin\gamma = \sin\alpha\sin\beta\sin(180^\circ-(\alpha+\beta))=\sin\alpha\sin\beta\sin(\alpha+\beta)$$
since $\alpha+\beta+\gamma=180^\circ$ and $\sin(180^\circ-\theta)=\sin(\theta)$. Use the angle sum identity for sine to obtain
$$\sin\alpha\sin\beta(\sin\alpha\cos\beta+\cos\alpha\sin\beta) = \sin^2\alpha\sin\beta\cos\beta+\sin^2\beta\sin\alpha\cos\alpha$$
which is transformed using the double- and half-angle formulas for sine
$$\frac12(\sin^2\alpha\sin2\beta+\sin^2\beta\sin2\alpha) = \frac14((1-\cos2\alpha)\sin2\beta+(1-\cos2\beta)\sin2\alpha)$$
$$=\frac14(\sin2\alpha+\sin2\beta-(\cos2\alpha\sin2\beta+\cos2\beta\sin2\alpha))=\frac14(\sin2\alpha+\sin2\beta-\sin(2\alpha+2\beta))$$
which is the desired result if the last term is equal to $\sin2\gamma$. That is easily shown since
$$\sin2\gamma=\sin(2(180^\circ-(\alpha+\beta)))=\sin(360^\circ-2(\alpha+\beta))$$
$$=\sin360^\circ\cos(2(\alpha+\beta))-\cos(360^\circ)\sin(2(\alpha+\beta)) = -\sin(2\alpha+2\beta)$$
completes the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4040272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve determinants with $|AB| = |A||B|$ How should I figure out the following determinants?
It is required to use $|AB| = |A||B|$ to figure them out.
(1) $D_1 = \begin{vmatrix}
1+x_1y_1 & 1+x_1y_2 & \dots & 1+x_1y_n \\
1+x_2y_1 & 1+x_2y_2 & \dots & 1+x_2y_n \\
\vdots & \vdots & \ddots & \vdots \\
1+x_ny_1 & 1+x_ny_2 & \dots & 1+x_ny_n
\end{vmatrix}$
(2) $D_2 = \begin{vmatrix}
1 & \cos(a_1-a_2) & \cos(a_1-a_3) & \dots & \cos(a_1-a_n) \\
\cos(a_1-a_2) & 1 & \cos(a_2-a_3) & \dots & \cos(a_2-a_n) \\
\cos(a_1-a_3) & \cos(a_2-a_3) & 1 & \dots & \cos(a_3-a_n) \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\cos(a_1-a_n) & \cos(a_2-a_n) & \cos(a_3-a_n) & \dots & 1
\end{vmatrix}$
(3) $D_3 = \begin{vmatrix}
a & a & a & a \\
a & a & -a & -a \\
a & -a & a & -a \\
a & -a & -a & a
\end{vmatrix}$
(4) Let $s_k=a_1^k+a_2^k+a_3^k+a_4^k \quad (k=1,2,3,4,5,6)$,
$$D_4 = \begin{vmatrix}
4 & s_1 & s_2 & s_3 \\
s_1 & s_2 & s_3 & s_4 \\
s_2 & s_3 & s_4 & s_5 \\
s_3 & s_4 & s_5 & s_6
\end{vmatrix}$$
My Attempt:
(1) I noticed for any $i,j$,
$1+x_iy_j = \begin{bmatrix} 1 & x_i \end{bmatrix}
\begin{bmatrix} 1 \\ y_j \end{bmatrix}$.
So, the matrix corresponds $D_1$ equals $\begin{bmatrix}
1 & x_1 \\ 1 & x_2 \\ \vdots & \vdots \\ 1 & x_n
\end{bmatrix}_{n \times 2}
\begin{bmatrix}
1 & 1 & \dots & 1 \\
y_1 & y_2 & \dots & y_n
\end{bmatrix}_{2 \times n}$.
But it's not helpful at all. :(
(2) It's obvious that
$a_{ij} = \cos(a_i - a_j) = \cos a_i \cos a_j + \sin a_i \sin a_j$.
(3) I've no idea about this problem at all. All I came up with, is
$$D_3 = a^4 \begin{vmatrix}
1 & 1 & 1 & 1 \\
1 & 1 & -1 & -1 \\
1 & -1 & 1 & -1 \\
1 & -1 & -1 & 1
\end{vmatrix}$$
(4) I noticed that $a_{ij} = s_{i+j-2}$. But still, not helpful.
Plz give me some hints. Thx in advance.
| for D_3 case;I think LU decomposition can help to rewrite as $AB$ and use $|AB|=|A||B|$
$$\left(\begin{matrix}
1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1
\end{matrix}\right).\left(\begin{matrix}
a & a & a & a \\
a & a & -a & -a \\
a & -a & a & -a \\
a & -a & -a & a
\end{matrix}\right)=\\
\left(\begin{matrix}
1 & 0 & 0 & 0 \\
1 & 1 & 0 & 0 \\
1 & 0 & 1 & 0 \\
1 & 1 & 1 & 1
\end{matrix}\right).\left(\begin{matrix}
a & a & a & a \\
0 & -2*a & 0 & -2*a \\
0 & 0 & -2*a & -2*a \\
0 & 0 & 0 & 4*a
\end{matrix}\right)$$ so det=$$(1.1.1.1).(a.(-2a).(-2a).(4a))=16a^4$$so det$D_3 = \begin{vmatrix}
a & a & a & a \\
a & a & -a & -a \\
a & -a & a & -a \\
a & -a & -a & a
\end{vmatrix}=-16a^4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4041667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
There is a rectangle in 3-dimensional space. If I am only given the $x$ and $y$ coordinates, how do I find the $z$ coordinates? We are given the $x$ and $y$ coordinates of the four vertices of a rectangle.
How do I find the $z$-coordinates?
I know that this 3 dimensional rectangle must obey all the usual conditions of rectangle.
Opposite sides are equal.
The angles between two adjacent sides are 90 degrees.
| (Rewritten after missing an entire degree of freedom!)
Let's say the vertices are in (clockwise or counterclockwise order) $\vec{A} = (x_A, y_A, z_A)$, $\vec{B} = (x_B, y_B, z_B)$, $\vec{C} = (x_C, y_C, z_C)$, and $\vec{D} = (x_D, y_D, z_D)$.
In order for the quadrilateral to be a parallelogram, pairs of opposing sides must be equal:
$$\left\lbrace\begin{aligned}
\vec{B} - \vec{A} &= \vec{C} - \vec{D} \\
\vec{D} - \vec{A} &= \vec{C} - \vec{B} \\
\end{aligned}\right. \quad \iff \left\lbrace\begin{aligned}
x_A - x_B + x_C - x_D &= 0 \\
y_A - y_B + y_C - y_D &= 0 \\
z_A - z_B + z_C - z_D &= 0 \\
\end{aligned}\right. \tag{1}\label{G1}$$
which means we can ignore $\vec{C}$ completely, because it is defined exactly by the three other points (via $\vec{C} = \vec{B} + \vec{D} - \vec{A}$), if it is a parallelogram. (And rectangles and squares are always also parallelograms.)
First, check if the coordinates are valid for any rectangle that is perpendicular to the $z$ axis. If
$$(x_D - x_A)(x_B - x_A) + (y_D - y_A)(y_B - y_A) = 0 \tag{2a}\label{G2a}$$
then all real $z_A = z_B = z_C = z_D$ are valid solutions, as are $z_A = z_B \ne z_C = z_D$ and $z_A = z_D \ne z_B = z_C$. (The latter are solutions because changing $z_A \to z_A + h$ and $z_B \to z_B + h$, or $z_A \to z_A + h$ and $z_D \to z_D + h$ is still a rectangle, increasing $h$ in magnitude just makes the rectangle taller or wider.)
For nontrivial solutions, we need all four angles of the parallelogram to be right angles:
$$\left\lbrace\begin{aligned}
\bigl( \vec{D} - \vec{A} \bigr) \cdot \bigl( \vec{B} - \vec{A} \bigr) &= 0 \\
\bigl( \vec{A} - \vec{B} \bigr) \cdot \bigl( \vec{C} - \vec{B} \bigr) &= 0 \\
\bigl( \vec{B} - \vec{C} \bigr) \cdot \bigl( \vec{D} - \vec{C} \bigr) &= 0 \\
\bigl( \vec{C} - \vec{D} \bigr) \cdot \bigl( \vec{A} - \vec{D} \bigr) &= 0 \\
\end{aligned}\right.$$
Because $\vec{C} = \vec{D} + \vec{B} - \vec{A}$, these all simplify to the same rule,
$$(x_B - x_A)(x_D - x_A) + (y_B - y_A)(y_D - y_A) + (z_B - z_A)(z_D - z_A) = 0$$
i.e.
$$(z_B - z_A)(z_D - z_A) = - (x_B - x_A)(x_D - x_A) - (y_B - y_A)(y_D - y_A)$$
which we can solve for $z_D$ (noting that we have already covered the $z_A = z_B$ cases in $\eqref{G2a}$):
$$z_D = z_A - \frac{ (x_B - x_A)(x_D - x_A) + (y_B - y_A)(y_D - y_A) }{z_B - z_A}, \quad z_A \ne z_B \tag{2b}\label{G2b}$$
meaning we can choose any pair of reals $z_A \ne z_B$, and $\eqref{G2b}$ and
$$z_C = z_D + z_B - z_A$$
will yield a valid rectangle.
Let's test this with
$$\vec{A} = \left[\begin{matrix} 3 \\ 3 \\ z_A \end{matrix}\right], \quad
\vec{B} = \left[\begin{matrix} 10 \\ 5 \\ z_B \end{matrix}\right], \quad
\vec{C} = \left[\begin{matrix} 11 \\ 8 \\ z_B + z_D - z_A \end{matrix}\right], \quad
\vec{D} = \left[\begin{matrix} 4 \\ 6 \\ z_D \end{matrix}\right]$$
(which yields $13 \ne 0$ for $\eqref{G2a}$, so $z_A=z_B=z_C=z_D$ are not valid rectangles).
Rule $\eqref{G2b}$ simplifies to
$$(z_D - z_A)(z_B - z_A) = -13 \quad \iff \quad z_D = z_A - \frac{13}{z_B - z_A}, z_A \ne z_B$$
Here are some randomly generated solutions:
$$\begin{array}{cc|cccc}
z_A & z_B & \vec{A} & \vec{B} & \vec{C} & \vec{D} \\
\hline
0 & 1 & (3, 3, 0) & (10, 5, 1) & (11, 8, -12) & (4, 6, -13) \\
1 & 0 & (3, 3, 1) & (10, 5, 0) & (11, 8, 13) & (4, 6, 14) \\
1 & -1 & (3, 3, 1) & (10, 5, -1) & (11, 8, 5.5) & (4, 6, 7.5) \\
-3 & 10 & (3, 3, -3) & (10, 5, 10) & (11, 8, 9) & (4, 6, -4) \\
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4044245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If we place $1$ to $n^2$ in an $n\times n$ table, what is the smallest $s$ where $s$ is the max of $a+b$ where $a,b$ are numbers in adjacent cells? Primary question : Let $n$ be an integer greater than 1. Numbers $1,2,...,n^2$ are placed in an $n\times n$ table, one number per cell. For each such configuration, let $s$ be the maximum value of $a+b$ where $a$ and $b$ are numbers in adjacent cells (sharing sides). What is the minimum possible value of $s$?
Write $S=S_n$ for the minimum value of $s$. Because $n>1$ we obviously have $S\ge n^2+2$ because $n^2$ is adjacent to at least two cells. It is clear also that when $n>3$, $S>n^2+2$ because either $n^2-1$ is adjacent to $\ge 3$ cells or both $n^2$ and $n^2-1$ are in some corners, so one of the $4$ neighbours of $n^2$ and $n^2-1$ is at least $4$. Hence $S\ge n^2+3$ for $n>3$. I know $S_2=6$, $S_3=11$, $S_4=19$. I am not seeing a pattern here.
I was asked for $S_{10}$, so an answer for $n=10$ is already good for me. But it'd be great if $S_n$ is known for an arbitrary $n$.
If you know where this problem came from, plz let me know. The same friend from another question asked me this question but didnt know if this problem is a contest problem from somewhere.
Modified questions :
(1) what is the smallest value of $s$ if we define 'adjacent' to mean 'sharing at least one vertex'? If $S'_n$ is the smallest number of the modified question, it is easy to see that $S'_n\ge n^2+5$ for $n>3$. We have $S'_2=5$, $S_3'=12$. I'd like to know the answer to this question too.
(2) The same question as the primary question but we are filling $1,2,...,mn$ in an $m\times n$ table.
(3) The same question as the modified question (1) but we are filling $1,2,...,mn$ in an $m\times n$ table.
| Indeed, a minimum of $n^2 + \lfloor \frac n 2\rfloor + 1$ can be arranged for any $n \geq 3$, as follows, where $m = \lfloor \frac n 2 \rfloor$:
\begin{matrix}
n^2, & 1, & n^2 - 1, & 2, & n^2 - 2, & \dots \\
m + 1, & n^2 - n + m, & m + 2, & n^2 - n + m + 1, & m + 3, &\dots \\
n^2 - n, & n + 1, & n^2 - n - 1, & n + 2, & n^2 - n - 2, & \dots \\
n + m + 1, & n^2 - 2n + m, & n + m + 2, & n^2 - 2n + m + 1, & n + m + 3, & \dots \\
\dots & \dots & \dots & \dots & \dots & \dots
\end{matrix}
As for the optimality, it seems that the following is true:
On a $2k \times 2k$ grid, if we mark $k^2$ cells such that no two are adjacent to each other, then the number of cells that are adjacent to at least one marked cell is at least $k^2 + k$.
If this statement is true, then the above arrangement for $n = 2k$ is optimal: we consider the largest $k^2$ numbers $4k^2, 4k^2 - 1, \dots, 3k^2 + 1$. If two of them are adjacent, then the sum is at least $6k^2 + 3$; otherwise, by the above statement, there are at least $k^2 + k$ numbers adjacent to them, among which the largest one must be at least $k^2 + k$. Since this number is adjacent to a number at least $3k^2 + 1$, their sum is at least $4k^2 + k + 1$.
However I'm not able to prove the above statement, although it seems very likely to be true.
| {
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"url": "https://math.stackexchange.com/questions/4046654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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What is the probability that the area of the triangle with vertices $(0,0)$, $(3,0)$ and $P$ is greater than 2? Let $P$ be a point chosen at random on the line segment between the points $(0,1)$ and $(3,4)$ on the coordinate plane. What is the probability that the area of the triangle with vertices $(0,0)$, $(3,0)$ and $P$ is greater than 2?
I started off by noticing that the base of the triangle is 3, so the height must be greater than $\frac{4}{3}$ if the area is greater than 2 because $\frac{1}{2}(3)(h)>2 \rightarrow h >\frac{4}{3}.$
Since $y>\frac{4}{3}, x > \frac{1}{3}$ because the line is $y=x+1$.
So for the area of the triangle to be greater than 2, $P$ can be any point from $(\frac{1}{3}, \frac{4}{3})$ to $(3,4)$ on the line $y=x+1$.
Therefore the probability of the triangle area being greater than 2 is $\dfrac{\text{length of segment (1/3, 4/3) to (3,4)}}{\text{length of segment (0,1) to (3,4)}}=\dfrac{2/3\sqrt{17}}{3\sqrt{2}}=\dfrac{\sqrt{34}}{9}.$
However the solution says the answer is 8/9, so what am I doing wrong here? I'm also pretty sure I overcomplicated a lot of things, and I was also wondering if there's a simpler solution. Thanks!
|
The point, $Z$ on the dark green line has coordinates
$x = x(P)$ and $y=area(\triangle PQR)$
So the ratio you are looking for is
$\dfrac{6-2}{6-1.5}=\dfrac{4}{4.5}=\dfrac 89$
| {
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"url": "https://math.stackexchange.com/questions/4051787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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studying the function $f(x)= \sqrt{x^2+x}-x$ I'm studying the function $$f(x)= \sqrt{x^2+x}-x$$
The domain is $(- \infty, -1] \cup [0, + \infty)$.
In particular I'm making some mistakes about the sign of the
first derivative $$ \frac{2x+1-2 \sqrt{x^2+x}}{2 \sqrt{x^2+x}}$$
The first derivative is never zero.
$ \frac{2x+1-2 \sqrt{x^2+x}}{2 \sqrt{x^2+x}}>0 \Rightarrow 2x+1-2 \sqrt{x^2+x} >0 \Rightarrow 2x+1>2 \sqrt{x^2+x} \Rightarrow 1>0 $ . So it seems that the derivative is always positive in the domain.
But comparing with the result in wolfram alpha the function is decreasing in $(-\infty, -1]$
| Note that the domain of the derivative
$$f^\prime(x)\:=\: \frac{2x+1}{2 \sqrt{x^2+x}}-1$$
is not equal to the domain of $f$ since the denominator zeros $\{-1,0\}\,$ have to be excluded.
The sign analysis is quickly achieved using the AGM inequality:
If $0<x$ then
$$\sqrt{x(x+1)}\:<\:\frac{2x+1}{2}\quad\implies\; 0\;<\;\frac{2x+1}{2 \sqrt{x^2+x}}-1$$
For $x<-1$ one obtains
$$\sqrt{(-x)(-(x+1))}\:<\:\frac{-x-(x+1)}2 \;=\; -\frac{2x+1}{2}\quad\implies\;
f^\prime(x)<-2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$(x^2-9)^{(3x+5)}=(x-3)^{(x-1)}(x+3)^{(x-1)}$ What is $x$? $(x^2-9)^{(3x+5)}=(x-3)^{(x-1)}(x+3)^{(x-1)}$
$(x^2-9)^{(3x+5)}=(x^2-9)^{(x-1)}$
$3x+5=x-1$
$x=-3$
But when I try to use WolfarmAlpha the integer solution is $3$ instead of $-3$. The numerical solution is $x\approx 3.1622776600...$
So I tried to do other things, such as making sure the base is larger than $0$.
$x^2-9>0$
$x<-3$ or $x>3$
But I still didn't obtain the $x$.
So what is $x$?
| The equation:
$$(x^2-9)^{(3x+5)}=(x^2-9)^{(x-1)}
$$
obviously holds if $x^2-9=0$ provided that the exponents are non-negative. This is the case for $x=3$.
If $x^2-9\ne0$ we can divide both sides by $(x^2-9)^{(x-1)}$ obtaining:
$$(x^2-9)^{(2x+6)}=1.$$
This equation holds if $2x+6=0$ or $x^2-9=1$. The roots of the equations are $x=-3$ and $x=\pm\sqrt{10}$, respectively.
The substitution in the original equation shows that
$x=3,\sqrt{10}$ satisfy the equation whereas the roots $x=-3,-\sqrt{10}$ do not suit it.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show limit exists if and only if $m+n>2$. Let $m,n \in \mathbb{N}$.
Show that the limit $\lim_{x \to (0,0)} \frac{x^ny^m}{x^2+y^2}$ exists if an only if $n+m>2$.
Attempt:
$(\Leftarrow)$. If $n+m>2$ without loss of generality assume $n \geq m$. Then $|\frac{x^ny^m}{x^2+y^2}|\leq |\frac{x^ny^m}{x^2}|=|x^{n-2}y^m|\leq (\sqrt{x^2+y^2})^{n-2}(\sqrt{x^2+y^2})^m=(\sqrt{x^2+y^2})^{m+n-2}$.
So for $\epsilon>0$, choose $\delta=\epsilon^{\frac{1}{m+n-2}}$, then $0<\sqrt{x^2+y^2}<\delta \implies |\frac{x^ny^m}{x^2+y^2}|<\epsilon$.
$(\Rightarrow)$.If $n+m\leq 2$. $f(\frac{1}{k},\frac{1}{k})=(\frac{1}{2})(\frac{1}{k})^{n+m-2}$ which does not approach zero as $k \rightarrow 0$, while $f(\frac{1}{k},0)=0$ approaches $0$ as $k \rightarrow 0$. Since these limits are different, the limit does not exist in this case.
| Problem like this is fun to solve for sure. Let's do the backward direction: If $m+n > 2$, then either $m \ge 2$ or $n \ge 2$. Let's say $m \ge 2$, then rewrite it as $\dfrac{x^2y}{x^2+y^2}\cdot x^{m-2}y^{n-1}$. Observe that $x^{m-2}y^{n-1} \to 0$ or $1$, and $0 \le \left|\dfrac{x^2y}{x^2+y^2}\right| \le |y|$, thus it goes to $0$, and the whole thing goes to $0$. Conversely, if the limit exists, say $L < \infty$ and if $m+n \le 2 \implies m = n = 1\implies \dfrac{x^my^n}{x^2+y^2} = \dfrac{xy}{x^2+y^2}$ does not have a limit as you can choose $2$ paths to $(0,0)$, contrary to the existence of $L$, thus it is the case that $m+n > 2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Closed form of $\sum_{m=1}^{\infty} \frac{(-1)^mH_{\frac{3m}{4}}}{3m}$ I've been working on an integral, namely: $${\displaystyle \int_0^1 \frac{x^2}{1 + x^3}\ln(1 - x^4)dx}$$
Which I managed to narrow down to the following expression: $$\sum_{m=1}^{\infty} \frac{(-1)^mH_{\frac{3m}{4}}}{3m}$$
Where $H_n$ is the n-th harmonic number. I managed to get here after converting the integrand into a double summation, recognizing the digamma function hidden inside, and summing back to a harmonic number, but couldn't get past that.
Is there another way to solve the integral, or to solve the summation? I've tried breaking the sum down to its parts, but the best I could do was to find a few terms and was left with
$$\sum_{m=1}^{\infty} \frac{(-1)^m\psi(\frac{3m}{4})}{3m}$$
Where $\psi(x)$ is the digamma function, and still couldn't solve that.
| Maple gives a rather unpleasant expression involving complex dilogarithm functions:
$$-\frac{19 \pi^{2}}{144}+\frac{7 \ln \! \left(2\right)^{2}}{12}-\frac{\ln \! \left(2-\sqrt{3}\right)^{2}}{6}+\frac{\mathit{dilog}\! \left(\frac{1}{2}+\frac{i \sqrt{3}}{6}\right)}{3}-\frac{\mathit{dilog}\! \left(\frac{3}{2}-\frac{i}{2}+\left(-\frac{1}{2}+\frac{i}{2}\right) \sqrt{3}\right)}{3}-\frac{\mathit{dilog}\! \left(\frac{3}{2}-\frac{i}{2}+\left(\frac{1}{2}-\frac{i}{2}\right) \sqrt{3}\right)}{3}-\frac{\mathit{dilog}\! \left(\frac{3}{2}+\frac{i}{2}+\left(\frac{1}{2}+\frac{i}{2}\right) \sqrt{3}\right)}{3}-\frac{\mathit{dilog}\! \left(1-\frac{i \sqrt{3}}{3}\right)}{3}+\frac{\mathit{dilog}\! \left(\frac{1}{2}-\frac{i \sqrt{3}}{6}\right)}{3}-\frac{\mathit{dilog}\! \left(1+\frac{i \sqrt{3}}{3}\right)}{3}-\frac{\mathit{dilog}\! \left(\frac{3}{2}+\frac{i}{2}-\left(\frac{1}{2}+\frac{i}{2}\right) \sqrt{3}\right)}{3}+\frac{\mathit{dilog}\! \left(\frac{1}{2}+\frac{i \sqrt{3}}{2}\right)}{3}+\frac{\mathit{dilog}\! \left(\frac{1}{2}-\frac{i \sqrt{3}}{2}\right)}{3}
$$
I suspect this comes from writing the rational function $x^2/(1+x^3)$ in partial fractions and also writing $\ln(1-x^4) = \ln(1-x) + \ln(1+x) + \ln(1-ix) + \ln(1+ix)$.
| {
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Algebra question with sequences. I dont know how to solve it. Can somebody help but also tell me how to approach these kind of questions.
Terms of the sequence $a_1, a_2, \ldots, a_{2020}$ are obtained by the rule
$$
a_2 = \frac{1+a_1}{1-a_1}, \qquad
a_3 = \frac{1+a_2}{1-a_2}, \qquad \ldots, \qquad
a_{2020} = \frac{1+a_{2019}}{1-a_{2019}}
$$
If $a_{2020} = \frac{1}{5}$, find $a_1$.
| Given equatiobn is:
$$A_{n+1}=\frac{1+A_n}{1-A_n} \implies A_{n+1} (1-A_n)=1+A_n~~~(1)$$
Let $$1-A_n=\frac{B_n}{B_{n-1}} \implies A_{n+1}=1-\frac{B_{n+1}}{B_n}$$Then (1) becomes
$$\frac{B_n-B_{n+1}}{B_n}\frac{B_n}{B_{n-1}}=2-\frac{B_n}{B_{n-1}}$$
$$\implies B_n-B_{n+1}=2B_{n-1}-B_n \implies B_{n+1}+2B_{n-1}-2B_n=0$$
Let $B_n=x^n$, then $x^2-2x+2=0 \implies x=(1\pm i)=a,b$ $$\implies B_n=P a^n +Q b^n\implies A_n=\frac{B_{n-1}-B_n}{B_n}=\frac{P a^{n-1}(1-a)+Qb^{n-1}(1-b)}{Pa^{n-1}+Qb^{n-1}}$$
So $$\implies A_n=-i\frac{a^{n-1}-Rb^{n-1}}{a^{n-1}+Rb^{1-n}}=-i\frac{1-Re^{i\pi(1-n)/2}}{1+Re^{i\pi(1-n)/2}}$$
$$A_{2020}=-i\frac{1-iR}{1+iR}=\frac{1}{5}\implies R=\frac{-1-5i}{i+5} \implies A_1=-i\frac{1-R}{1+R}=\frac{3}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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For which values of $a$ does this system of equations $\mathbf{{not}}$ have a unique solution? Here's my system of linear equations:
$\begin{cases}
x + 2y + 2z = 1\\x + ay + 3z = 3\\x + 11y +az = 0\\
\end{cases}$
Thus I have the augmented matrix $\left[\begin{array}{ccc|c}1&2&2&1\\1&a&3&3\\1&11&a&0\end{array}\right]$
By row reduction, I obtain:
$\left[\begin{array}
{ccc|c}1&2&2&1\\0&a-2&1&2\\0&9&a-2&-1\end{array}\right]$
Unfortunately, I am stuck at this stage. I have tried swapping rows around but I didn't have much luck.
Update: I have managed to solve this with the use of the determinant.
Matrix of minors:
$\left[\begin{array}
{ccc}a^2-33&a-3&11-a\\2a-22&a-2&9\\6-2a&1&a-2\end{array}\right]$
Matrix of cofactors:
$\left[\begin{array}
{ccc}a^2-33&3-a&11-a\\22-a&9&a-2\\6-2a&-1&a-2\end{array}\right]$
Adjugate matrix:
$\left[\begin{array}
{ccc}a^2-33&22-2a&6-2a\\3-a&a-2&-1\\11-a&-9&a-2\end{array}\right]$
Det(A) = $1(a^2 - 33) + 2(3 - a) + 2(11 - a) = a^2 - 4a - 5$
$(a - 5)(a + 1) = 0$
Thank you all for your help!
| Well if $a = 2$ you have $z=2; y=-\frac 19$ and $x = whatever$.
But if $a-2\ne 0$ keep at it.
Multiply row 2 by $\frac 9{a-2}$ and subtract getting rows.
$\begin{array}\\
1&2&2&|\,\,1\\
0&a-2&1&|\,\,2\\
0&0&(a-2)-\frac 9{a-2}&|-\frac 9{a-2}\\
\end{array}$
Now if $(a-2)-\frac 9{a-2} = 0$ we have a case with not solutions.
That is if $(a-2)^2 = 9$ of $a = 5,-1$.
If we assume $a\ne 2,5,-1$ then we can reduce
$\begin{array}\\
1&2&2&|\,\,1\\
0&1&\frac 1{a-2}&|\,\,\frac 2{a-2}\\
0&0&1&|-\frac 9{a-2}\cdot\frac 1{(a-2)-\frac 9{a-2}}=-\frac{9}{(a-2)^2-9}\\
\end{array}$
..... and that's in the whatchamacallit form "upper triangle"?) so we know this results in a unique solution.
So $a = 5$ or $1$ are the cases without solutions.
(because we'd get $((a-2)-\frac 9{a-2})z=-\frac 9{a-2}$ which if $a = 5$ is
$(3-3)z = -3$ or $0 = 3$ and if $a=-1$ we get $(-3 + 3)z = 3$ or $0 = 3$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Conclude integral is zero without calculation I have the following two integrals:
$$\int_0^{2\pi} \frac{\cos{\theta}d\theta}{\sqrt{a^2\cos^2{\theta}+b^2\sin^2{\theta}+c^2}},$$
$$\int_0^{2\pi} \frac{\sin{\theta}d\theta}{\sqrt{a^2\cos^2{\theta}+b^2\sin^2{\theta}+c^2}},$$
and I know both are zero. I can argue why the second one is so:
$$\int_0^{2\pi} \frac{\sin{\theta}d\theta}{\sqrt{a^2\cos^2{\theta}+b^2\sin^2{\theta}+c^2}}=\int_{-\pi}^{\pi} \frac{\sin{\theta}d\theta}{\sqrt{a^2\cos^2{\theta}+b^2\sin^2{\theta}+c^2}}.$$
Therefore,
$$\int_0^{2\pi} \frac{\sin{\theta}d\theta}{\sqrt{a^2\cos^2{\theta}+b^2\sin^2{\theta}+c^2}}=\int_{-\pi}^{0} \frac{\sin{\theta}d\theta}{\sqrt{a^2\cos^2{\theta}+b^2\sin^2{\theta}+c^2}}+\int_{0}^{\pi} \frac{\sin{\theta}d\theta}{\sqrt{a^2\cos^2{\theta}+b^2\sin^2{\theta}+c^2}}.$$
With the variable change $\phi=-\theta$ I get $\sin{\theta}d\theta=\sin{\phi}d\phi$ because sine is an odd function (also, the denominator does not change), so the first term becomes
$$\int_{-\pi}^{0} \frac{\sin{\theta}d\theta}{\sqrt{a^2\cos^2{\theta}+b^2\sin^2{\theta}+c^2}}=\int_{\pi}^{0} \frac{\sin{\phi}d\phi}{\sqrt{a^2\cos^2{\phi}+b^2\sin^2{\phi}+c^2}}=-\int_{0}^{\pi} \frac{\sin{\phi}d\phi}{\sqrt{a^2\cos^2{\phi}+b^2\sin^2{\phi}+c^2}},$$
and it cancels the second term so the sine integral is zero.
Is there a similar procedure for the cosine function? Since it is an even funcion the argument can't be exactly the same. Any help will be appreciated!
| Hint: Use the substitution $\phi = \pi/2 - \theta$.
| {
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Proving that $a \cos A+b\cos B+c \cos C$ is the semi-perimeter of $\triangle ABC$
With each symbol having its usual meaning, prove that in $\triangle ABC$,
$$a \cos A+b\cos B+c \cos C=s$$
where $s$ is the semi-perimeter.
I started by applying the rule of sines:
*
*$a=k\sin A$
*$b=k\sin B$
*$c=k \sin C$
where $k$ is a constant
so, we have:
$k(\sin A\cos A+\sin B\cos B+\sin C\cos C)$
manipulating this, we get:
$\frac{k}{2}(\sin 2A+\sin 2B+\sin 2C)$
$\frac{k}{2}(2\sin(A+B)\cos(A-B) + \sin 2C)$
$\frac{k}{2}(2\sin C\cos(A-B)+\sin 2C)$
$\frac{k}{2}2\sin C(\cos(A-B)+\cos C)$
But this doesn't seem to be yielding desired results. Could someone help me out with this? Or maybe tell me a better approach to this problem?
| What is true is the relation $a\cos A+b \cos B+c\cos C=\dfrac{2rs}{R}$ where $r$ is the inradius, $R$ is the circumradius and $s$ is the semi perimeter of the triangle.
By the sine rule, we can write the length of the sides as $a=2R\sin A,b=2R\sin B,c=2R\sin C$ therefore we have $$a\cos A+b \cos B+c\cos C=R(\sin 2A+\sin 2B+\sin 2C)$$
It is well known that in a triangle, we have $\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B \sin C$
So we have $a\cos A+b \cos B+c\cos C=4R\sin A\sin B \sin C$
Again, by the sine rule, re writing $\sin A=\dfrac{a}{2R}$ and similarly for other sides, we obtain $a\cos A+b \cos B+c\cos C=\dfrac{abc}{2R^2}$
It is also well known that $abc$ and $\Delta$ are related by the relation $\Delta=\dfrac{abc}{4R}$, so substituting the value into the equation we obtain the value as $\dfrac{2\Delta}{R}$, but since $\Delta=rs$, we have $$a\cos A+b \cos B+c\cos C=\dfrac{2rs}{R}$$
Helpful: Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle
| {
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"timestamp": "2023-03-29T00:00:00",
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A question about limit Can anyone help this questions?
Find the limit of
$$1+\sqrt{2+\sqrt[3]{3+\sqrt[4]{4+\sqrt[5]{5+....+\sqrt[n]{n}}}}}$$
I can only solve that this formula is less than 3 but can not find the exact answer for this.
| It can be shown that $1.9<a_0<2$:
We use following inequalities:
$\sqrt[k]{k+1}>1; \sqrt[k]{k}<\sqrt[k]{k+2}<2$
Seris of $a_n$ is increasing and from first inequality we have:
$a_n=\sqrt{2+\sqrt[3]{3+\sqrt[4]4+ \cdot\cdot\cdot+\sqrt[n]n}}<\sqrt{2+\sqrt[3]{3+\sqrt[4]4+ \cdot\cdot\cdot+\sqrt[n-1]{n-1+2}}}<\cdot\cdot\cdot<\sqrt{2+\sqrt[3]5}$
Therefore the series has a limit like $a_n $ such that :
$a_0\leq\sqrt{2+\sqrt[3]5}<2$
Now using first inequality we find:
$a_n>\sqrt{2+\sqrt[3]{3+\cdot\cdot\cdot+\sqrt[n-1]n}}>\cdot\cdot\cdot>\sqrt{2+\sqrt[3]{3+\sqrt[4]5}}$
Therefore:
$a_0>\sqrt{2+\sqrt[3]3+\sqrt[4]5}$
It can easily be seen rhat:
$\sqrt{2+\sqrt[3]{3+\sqrt[4]5}}>1,9$
Therefore:
$1.9<a_0<2$
| {
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Prove inequality consisting of sum using mathematical induction I wish to seek help for a question from the Finnish matriculation exam in mathematics. There are multiple methods to solve the exercise, but I am ONLY interested in solving it using mathematical induction. The question is as follows:
$$\text{let}\ n\ \text{be a positive integer}, n\ge1$$
$$\text{Show that:}$$
$$\frac{127}{7}\cdot(n-1)^7\le\sum_{k=n}^{2n-1}k^6\le\frac{127}{7}\cdot\ n^7$$
I have tried hard to solve this using induction, but I cannot get very far at all. Any help would be appreciated.
EDIT: What I have done so far.
*
*Test with n = 1, which gives:
$$0\le1\le\frac{127}{7}, \text{which is true}$$
*Assumption:
$$\frac{127}{7}\cdot(n-1)^7\le\sum_{k=n}^{2n-1}k^6\le\frac{127}{7}\cdot\ n^7$$
*Induction step:
$$\frac{127}{7}\cdot\ n^7\le\sum_{k=n+1}^{2n+1}k^6\le\frac{127}{7}\cdot\ (n+1)^7$$
But I don't know how to prove this in terms of the assumption. I would assume that there is some way to remove the sum from this inequality. It seems to be possible to expand it using my CAS-calculator. But I just can't wrap my head around it.
I hope this enough explanation of what I have done so far.
(I might have used the wrong mathematical terms here, since English is not my first language)
| Thanks to the help of @maxmilgram, I've managed to figure out the solution.
SOLUTION:
*
*Test with n = 1, which gives:
$$0\le1\le\frac{127}{7}, \text{which is true}$$
*Assumption:
$$\frac{127}{7}\cdot(n-1)^7\le\sum_{k=n}^{2n-1}k^6\le\frac{127}{7}\cdot\ n^7$$
*Induction step:
$$\frac{127}{7}\cdot\ n^7\le\sum_{k=n+1}^{2n+1}k^6\le\frac{127}{7}\cdot\ (n+1)^7$$
And since:
$$\sum_{k=n+1}^{2n+1}k^6=\sum_{k=n}^{2n-1}k^6-n^6+(2n)^6+(2n+1)^6$$
we can write the induction step in terms of the assumption:
$$\frac{127}{7}n^7\le\sum_{k=n}^{2n-1}k^6-n^6+\left(2n\right)^6+\left(2n+1\right)^6\le\frac{127}{7}\left(n+7\right)^7$$
$$\frac{127}{7}n^7+n^6-\left(2n\right)^6-\left(2n+1\right)^6\le \sum _{k=n}^{2n-1}k^6\le \frac{127}{7}\left(n+7\right)^7+n^6-\left(2n\right)^6-\left(2n+1\right)^6$$
In other words, if we could prove that:
$$\frac{127}{7}n^7+n^6-\left(2n\right)^6-\left(2n+1\right)^6\le \frac{127}{7}\left(n-1\right)^7$$ and
$$\frac{127}{7}\left(n+7\right)^7+n^6-\left(2n\right)^6-\left(2n+1\right)^6\ge \frac{127}{7}n^7$$
we would know that the induction step is indeed true. After some simplification we end up with:
$$-573n^5+395n^4-795n^3+321n^2-139n+\frac{120}{7}\le 0, \text{which is true since } n\ge 1$$
$$189n^5+395n^4+475n^3+321n^2+115n+\frac{120}{7}\ge 0, \text{which is true since } n \ge 1$$
In other words, we have shown that the induction step is true if the induction assumption also is true. Therefore, the statement is true for all positive integers n.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4069342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Problem with system of equations $(x,y,z)$ System of equations:
$$\begin{cases}\frac{x}{yz}+\frac{y}{zx}+\frac{z}{xy}=\frac{3}{xyz}\\xy+yz+zx=3\end{cases}$$
Solution:
$$\begin{cases}\frac{x}{yz}+\frac{y}{zx}+\frac{z}{xy}=\frac{3}{xyz}\\xy+yz+zx=3\end{cases}\Rightarrow\begin{cases}x^2+y^2+z^2=3\\xy+yz+zx=3\end{cases}$$
$x^{2}+y^{2}+z^{2}=xy+yz+zx$
$(x+y+z)^{2}-2(xy+yz+zx)=xy+yz+zx$
$(x+y+z)^{2}-2*3=3$
$(x+y+z)^{2}=9$
What should I do next?
| For the system $\begin{cases}x^2+y^2+z^2=3\\x+y+z=3\end{cases}\quad$ just offset the variables $\begin{cases}x=1+X\\y=1+Y\\z=1+Z\end{cases}$
Then you get $\begin{cases}3+X+Y+Z=3\\3+2X+2Y+2Z+X^2+Y^2+Z^2=3\end{cases}\implies X^2+Y^2+Z^2=0$
Which in the real domain has only the $(0,0,0)$ solution.
For $x+y+z=-3$ proceed similarly with $x=X-1,y=Y-1,z=Z-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4070632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the following limit? How can I calculate the following limit?
What do I miss here? What am I doing wrong?
$$\begin{align*}
\lim_{x\rightarrow0}\frac{6x\cdot\sin x-6\cdot\sin\left(x^{2}\right)+x^{4}}{x^{5}\left(e^{x}-1\right)}= & \lim_{x\rightarrow0}\frac{6x\cdot\sin x}{x^{5}\left(e^{x}-1\right)}-\lim_{x\rightarrow0}\frac{6\cdot\sin\left(x^{2}\right)}{x^{5}\left(e^{x}-1\right)}+\lim_{x\rightarrow0}\frac{x^{4}}{x^{5}\left(e^{x}-1\right)}=\\
= & \lim_{x\rightarrow0}\frac{6}{x^{3}\left(e^{x}-1\right)}\cdot\underbrace{\frac{\sin x}{x}}_{\rightarrow1}-\lim_{x\rightarrow0}\frac{6}{x^{3}\left(e^{x}-1\right)}\cdot\underbrace{\frac{\sin\left(x^{2}\right)}{x^{2}}}_{\rightarrow1}+\lim_{x\rightarrow0}\frac{1}{x\left(e^{x}-1\right)}=\\
= & \lim_{x\rightarrow0}\frac{1}{x\left(e^{x}-1\right)}=\infty\neq\frac{21}{20}.
\end{align*}$$
According to WolframAlpha $21/20$ is the solution. What am I doing wrong?
| Using the limit $\lim_{x\to 0}(e^x-1)/x=1$ the denominator of the expression under limit can be safely replaced by $x^6$. Next we add and subtract $6x^2$ in numerator and express it as $$6(x^2-\sin x^2)+6x\sin x-6x^2+x^4$$ This allows us to split the desired limit as a sum of two limits the first one of which is $$6\lim_{x\to 0}\frac{x^2-\sin x^2}{x^6}=6\lim_{t\to 0}\frac{t-\sin t} {t^3}=1$$ via L'Hospital's Rule (just applying once) or Taylor series.
The other limit we need to evaluate is $$\lim_{x\to 0}\frac{6\sin x - 6x+x^3}{x^5}$$ We can apply L'Hospital's Rule once to get the expression $$\frac{3x^2-6(1-\cos x)} {5x^4}$$ and putting $x=2t$ this transforms into $$\frac {3}{20}\cdot\frac{t^2-\sin^2t}{t^4}=\frac{3}{20}\cdot\frac{t+\sin t} {t} \cdot\frac{t-\sin t} {t^3}$$ and this tends to $(3/20)(2)(1/6)=1/20$.
The desired limit is thus $1+(1/20)=21/20$. You should see that the limit has been evaluated using just two applications of L'Hospital's Rule.
Also observe that the the term $6x^2$ was added and subtracted in numerator by observing the term $-6\sin x^2$ and knowing that this could lead to a split as $6(x^2-\sin x^2)$. Whenever you see that an expression can be split into multiple terms always try to ensure that at least one of the terms after split has a finite limit. Then that particular term can be handled without knowing anything about limiting behavior of other terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4076365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Showing that $\frac{x^{2x}}{(x+1)^{x+1}}\rightarrow +\infty$ as $x\rightarrow +\infty$ I am trying to show that $$\frac{x^{2x}}{(x+1)^{x+1}}\rightarrow +\infty \ \ \text{as} \ \ x\rightarrow +\infty.$$
My attempt is as follows:
\begin{align}
\frac{x^{2x}}{(x+1)^{x+1}}&=\frac{x^{x}}{x+1}\left(\frac{x^x}{(x+1)^x}\right) \\
&=\frac{x^{x}}{x+1}\left(\frac{x}{x+1}\right)^x \\
&=\frac{x^{x}}{x+1}\left(\frac{1}{(1+1/x)^x}\right).
\end{align}
I can see that the second fraction will converge to $1/e$, but I am unsure of how to approach the first fraction.
| Hint:$$\frac{x^{2x}}{(x+1)^{x+1}}=x^{2x-(x+1)}\frac{1}{\left(1+ \frac{1}{x} \right)^{x+1}} \sim\frac{x^{x-1}}{e} $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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find the roots $(2z+3)^3=\frac{1}{64}$ There are 3 roots 1 real and 2 imaginary i found one z by doing $\frac{\frac{1}{4}-3}{2}$ so $z=\frac{-11}{8}$ however there are two more complex roots which are $z=\frac{-25+i√3}{16}$ and $z=\frac{-25-i√3}{16}$ but i dont know how to get to it any help is much appricated
. thank you
| $$(2z+3)^3=\frac{1}{64} \implies 2z+3=\frac{1}{4}, \frac{\omega}{4}, \frac{\omega^2}{4}$$
$$\implies z=\frac{-11}{8}, \frac{\omega -12}{8}, \frac{\omega^2-12}{8}.$$ Here $\omega$ is cube-root of unity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4076926",
"timestamp": "2023-03-29T00:00:00",
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Integral of Euler. Here is an integration problem I found in an old book:
Integrate
$$\int\frac{1+x^2}{1-x^2}\frac{dx}{\sqrt{1+x^4}}$$
The integral is attributed to Euler.
My solution is
let $$y=\frac{1+x^2}{1-x^2}$$ then get
$$\frac{1}{\sqrt{2}} \int \frac{ydy}{\sqrt{y^4-1}}$$
After another substitution $z=y^2$ get
$$\frac{1}{2\sqrt{2}}\int\frac{dz}{\sqrt{z^2-1}}=\frac{1}{2\sqrt{2}}\cosh^{-1}z$$ so my answer is
$$\frac{1}{2\sqrt{2}}\cosh^{-1}\left(\frac{1+x^2}{1-x^2}\right)^2$$
However the answer in the back is
$$\frac{1}{\sqrt{2}} \sinh^{-1}\frac{\sqrt{2}x}{1-x^2}$$
After much work (!!!) I have shown that the two forms are equal to a constant.
My question is: How can one solve the original integral to get the alternative answer directly. What substitution to use ?
| Knowing the form of the answer, it is natural to apply the substitution $t= \frac{\sqrt2 x}{1-x^2}$, which leads to
$$1+t^2 = \frac{1+x^4}{(1-x^2)^2},\>\>\>\>\>
dt = \frac{\sqrt2 (1+x^2)}{(1-x^2)^2}dx$$
Then
$$\int\frac{1+x^2}{1-x^2}\frac{1}{\sqrt{1+x^4}}dx
=\int \frac{1+x^2}{1-x^2}\frac{\frac{(1-x^2)^2}{\sqrt2(1+x^2)}dt }{(1-x^2)\sqrt{1+t^2}}\\=\frac1{\sqrt2}\int \frac{1}{\sqrt{1+t^2}}dt
= \frac1{\sqrt2}\sinh^{-1}t= \frac{1}{\sqrt{2}} \sinh^{-1}\frac{\sqrt{2}x}{1-x^2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4077788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Solving $\frac{3^x+2^x}{3^x-2^x}=7$: More than one answer? How to solve for all $x$?
I am trying to solve this problem (math for fun):
$$\frac{3^x+2^x}{3^x-2^x}=7$$
Step 1. Let$\:a=3^x\:and\:b=2^x$
Step 2. $\frac{a+b}{a-b}=7$
Step 3. $a+b=7\left(a-b\right)=7a-7b$
Step 4. $6a-8b=0$
Step 5. $6a=8b=3\cdot 2\cdot a=4\cdot 2\cdot b$
Step 6. $3a=4b$
Step 7. $3\cdot 3^x=2\cdot 2\cdot 2^x=3^{x+1}=2^{x+2}$
Step 8. Log both sides
Step 9. $\ln \left(3\right)+x\ln 3=x\ln2+\ln 4$
Step 10. $\ln \left(3\right)-\ln 4=x\:\left(\ln 2-\ln 3\right)$
Step 11. $\frac{\:\left(\ln 3-\ln 4\right)}{\left(\ln 2-\ln 3\right)\:}=x≈0.71$
SOLVED^
| $$3^x+2^x=7\cdot3^x-7\cdot2^x$$
$$8\cdot2^x=6\cdot3^x$$
$$\left(\frac32\right)^x=\frac 43$$
and
$$x=\log_{3/2}\frac43.$$
As every steps are equivalent, this is the only solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4081887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Shorthand limits of matrices? Is there any relatively easy way to solve limits with matrices?
$$
\lim_{h\rightarrow \infty}
\begin{pmatrix}
1 & 0 & 0\\
\frac{1}{r} & 0 & \frac{-1}{r}\\
0 & \frac{1}{hc} & 1
\end{pmatrix}^h
=
\begin{pmatrix}
1 & 0 & 0\\
\frac{1}{r} & 0 & \frac{-1}{r}\\
\frac{1}{rc} & 0 & 1-\frac{1}{rc}
\end{pmatrix}
$$
I have painstakingly identified the solution for this particular input matrix by identifying numbers with gigantic values of h in matlab. I would prefer to not square a matrix about a hundred times and manually identify the numbers. It's like calculating the derivative of a function with limits, I rather figure out the tricks than always going through the limit.
The 4th power of the input matrix looks like this and I suppose you can make certain deductions, such as $\frac{1}{h}$ and $\frac{1}{h^2}=$ 0 and $\frac{3}{h}$ = 1.
$$
\begin{pmatrix}
1 & 0 & 0\\
\frac{1}{r} & 0 & \frac{-1}{r}\\
0 & \frac{1}{hc} & 1
\end{pmatrix}^4
=
\begin{pmatrix}
1 & 0 & 0\\
\frac{1}{r} & \frac{-1}{rhc} & \frac{-1}{r}\\
\frac{1}{rhc} & \frac{1}{hc} & 1-\frac{1}{rhc}\\
\end{pmatrix}^2
=
\begin{pmatrix}
1 & 0 & 0\\
\frac{1}{r}-\frac{1}{(rhc)^2}-\frac{1}{r^2hc} & \frac{1}{(rhc)^2}-\frac{1}{rhc} & \frac{-1}{r}+\frac{2}{r^2hc}\\
\frac{3}{rhc}-\frac{1}{(rhc)^2} & \frac{1}{hc}-\frac{2}{r(hc)^2} & 1-\frac{3}{rhc}+\frac{1}{(rhc)^2}\\
\end{pmatrix}
$$
I would like to understand how to go from the input matrix straight to the solution, I can reach the solution if I take the 4th power of the input matrix and look for the number 3. But that will become very time consuming for big matrices, on the order of 100 x 100.
Edit
The $\frac{1}{rc}$ part in the previous limit result had to receive another limit so here's the solution to that.
$$
\lim_{h\rightarrow \infty}
\begin{pmatrix}
1 & 0 & 0\\
\frac{1}{r} & 0 & \frac{-1}{r}\\
\frac{1}{rhc} & 0 & 1-\frac{1}{rhc}
\end{pmatrix}^h
=
\begin{pmatrix}
1 & 0 & 0\\
\frac{1}{r} & 0 & \frac{-1}{r}\\
1-e^\frac{-1}{rc} & 0 & e^\frac{-1}{rc}
\end{pmatrix}
$$
Again, I found the solution by squaring many times and seeing error patterns and using wolframalpha to turn error sums to e functions.
Maybe there is no simple way to find limits to matrices.
| The problem here is the power $h$. But if the matrix is diagonalizable, there is an easy way to handle it. Let
$$H :=\begin{pmatrix}1 & 0 & 0\\
\frac{1}{r} & 0 & \frac{-1}{r}\\
0 & \frac{1}{hc} & 1
\end{pmatrix}$$
And suppose there is some matrix $Q$ with $Q^{-1}DQ = H$ where $$D =\begin{pmatrix}d_1&0&0\\0&d_2&0\\0&0&d_3\end{pmatrix}$$ is diagonal. Then
$$H^h = (Q^{-1}DQ)^h = Q^{-1}DQQ^{-1}DQQ^{-1}DQ\cdots Q^{-1}DQ = Q^{-1}D^hQ.$$
$$\lim_{h \to \infty} H^h = \lim_{h\to \infty} Q^{-1}D^hQ$$
Because $h$ also appears in $H$, $Q$ will depend on $h$ as well. And $$D^h = \begin{pmatrix}d_1^h&0&0\\0&d_2^h&0\\0&0&d_3^h\end{pmatrix}$$
If you can find $\lim_{h \to \infty}Q$, then $\lim_{h \to \infty}Q^{-1}$ will be its inverse. And if you can also find $\lim_{h \to \infty}d_i^h$ for each $i$, then you have $\lim_{h\to\infty} D^h$, and $$\lim_{h\to \infty} H^h = \lim_{h \to \infty}Q \lim_{h\to\infty} D^h\lim_{h \to \infty}Q^{-1}$$
Which you can then compare to the other side.
| {
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"source": "stackexchange",
"question_score": "4",
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Limit $\underset{x\to 0}{\text{lim}}\frac{\sqrt[3]{a x+b} - 2}{x}$ equals to $\frac{5}{12}$ A friend of mine asked this question to me. It seems it's from Stewart.
Find the values of a and b such that
$\underset{x\to 0}{\text{lim}}\frac{\sqrt[3]{a x+b} - 2}{x} = \frac{5}{12}$
This is what I tried with better results.
For $b$:
$$\frac{\sqrt[3]{a x+b} - 2}{x} = \frac{5}{12} $$
$$\sqrt[3]{a x+b} - 2 = x\frac{5}{12} $$
$$\underset{x\to 0}{\text{lim}} \sqrt[3]{a x+b} - 2 = \underset{x\to 0}{\text{lim}} x\frac{5}{12} $$
$$\sqrt[3]{b} - 2 = 0 $$
$$\sqrt[3]{b} = 2 $$
$$ b = 8$$
For $a$:
$$a x+8 = (x\frac{5}{12} + 2)^3$$
$$a x+8 = \frac{125 x^3}{1728}+\frac{25 x^2}{24}+5 x+8$$
$$a x = \frac{125 x^3}{1728}+\frac{25 x^2}{24}+5 x$$
$$a = \frac{125 x^2}{1728}+\frac{25 x}{24}+5$$
$$\underset{x\to 0}{\text{lim}} a =\underset{x\to 0}{\text{lim}} \frac{125 x^2}{1728}+\frac{25 x}{24}+5$$
$$a = 5 $$
But the limit $$\underset{x\to 0}{\text{lim}}\frac{\sqrt[3]{5x+8} - 2}{x}$$ doens't go to $\frac {5}{12}$.
May someone help.
| Actually, the limit
$\lim\limits_{x\to 0}\frac{(5x+8)^{1/3}-2}{x}$ (of the form $\frac{0}{0}$)
$=\lim\limits_{x\to 0}\frac{5.\frac{1}{3}(5x+8)^{-2/3}}{1}$, by L' Hospital's rule (since the limiting value of both numerator and denominator is 0 and the first derivative of the denominator w.r.t. $x$ is non-zero, the rule is applicable).
$=\frac{5}{12}$
Also, letting $y^3=ax+b$, we have the limit $L=\lim\limits_{y\to b^{1/3}}a.\frac{y-2}{y^3-b}$, the denominator tends to $0$, the limit exists iff numerator tends to $0$ too, $\implies b=8$.
Now, $L=\lim\limits_{y\to 2} \frac{a}{y^2+2y+4}=\frac{a}{12}=\frac{5}{12}$ (given) $\implies a=5$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Areas in triangle
In the above isosceles right triangle ABC, with its two sides $AB = AC = 1$ unit, we take a random point D on the hypotenuse and draw perpendicular lines to the sides AB and AC, which intersect them at points E and F respectively.
Show that the maximum of the three areas AFDE, EBD and CDF is always $\geq \frac {2}{9}$.
If we set $EB = x$ then, area $EBD = \frac {x^2}{2}$,
$AFDE = (1-x)*x$,
$CDF = \frac {(1-x)^2}{2}$.
Also $AFDE + EBD + CDF = \frac {1}{2}$.
Clearly the 3 areas can't be equal.
We can only have 2 of them equal, when $x=0.5$ or $x=0.33$ or $x=0.66$.
When $x=0.33 = \frac {1}{3}$ then
$EBD = \frac {x^2}{2} = \frac{1}{18}$ and $AFDE = CDF = \frac{1}{2}*(\frac{9}{18}-\frac{1}{18}) = \frac{2}{9}$.
But I don't know if this is a sufficient proof.
| Start with D as the mid point of BC. WLOG, imagine D moves up by a vertical distance of $\Delta x$ .
\begin{align}S_{AEDF}=\frac{1}{4}-\Delta x^2, S_{\triangle DFC}=\frac{1}{2}(\frac{1}{2}+\Delta x)^2\end{align}
\begin{align}S_{AEDF}\geq \frac{2}{9}(\Delta x \leq\frac{1}{6})\\
S_{\triangle DFC}\geq \frac{2}{9}(\Delta x\geq \frac{1}{6})\end{align}
Similar conclusion can be obtained if D moves down by $\Delta x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4089455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Compute $\int \frac {dx} {(a + b \cos x)^2} $ for $a > b.$ Let $a>b$ be real numbers. As the title suggests, I would like to compute $$\int \frac 1 {(a + b \cos x)^2} \, dx.$$
My attempt consisted of converting the cosine to half-angle tangents, substituting the half-angle tangent for $t$, and simplifying to $$\int \frac{2(1+t^2)}{[(a+b)+(a-b)t^2]^2} \, dt.$$
From here, I multiplied and divided the integrand by $(a-b)$, added an $(a+b)-2b$ to the numerator, and simplified it to $$\frac{4b}{a-b}\int {\left(\frac 1 {[(a+b)+(a-b)t^2]^2} + \frac{2t}{a-b} \right)} \, dt.$$
Now, just looking at the integral left, I substituted $t$ for $\sqrt{\frac{a+b}{a-b}} \tan\theta$, did some simplifying, converted the resulting $\cos^2\!\theta$ in the numerator to $1+\cos 2 \theta$, and wrapped up the integral, finishing up with $$\frac{2\tan \frac{x}{2}}{a-b}-\frac{2b}{(a^2-b^2)^{\frac{3}{2}}}{\left(\arctan \sqrt{\frac{a+b}{a-b}}\tan \frac{x}{2}+ \frac{\sqrt{ \frac{a+b}{a-b}}\tan
\frac{x}{2}}{1+{ \frac{a-b}{a+b}\tan^2 \frac{x}{2}}} \right)}$$
This is apparently wrong. Can someone help me with why?
| Proceed as follows after half-angle substitution $t=\tan\frac{x}2$
\begin{align}
& \int \frac{2(1+t^2)}{[(a+b)+(a-b)t^2]^2} \, dt\\
=& \frac2{a^2-b^2} \int \frac{a[(a-b)t^2+(a+b)]+ b[(a-b)t^2-(a+b)] }{[(a+b)+(a-b)t^2]^2} \, dt \\
=& \frac2{a^2-b^2} \int \frac{a[(a-b)+\frac{a+b}{t^2} ]+ [b[(a-b)-\frac{a+b}{t^2}]}{[(a-b)t+\frac{a+b}t]^2}dt\\
=& \frac2{a^2-b^2} \left(a \int \frac{d[(a-b)t-\frac{a+b}{t}]}{[(a-b)t-\frac{a+b}t]^2+4(a^2-b^2)}
+b\int \frac{d[(a-b)t+\frac{a+b}{t}] }{[(a-b)t +\frac{a+b}t]^2} \, \right)\\
=& \frac{a}{(a^2-b^2)^{3/2}} \tan^{-1} \frac{(a-b)t-\frac{a+b}{t}}{2\sqrt{a^2-b^2}}- \frac{2b}{a^2-b^2}
\frac{1}{(a-b)t +\frac{a+b}t}\\
=& \frac{a}{(a^2-b^2)^{3/2}} \tan^{-1} \frac{(a-b)\tan^2\frac{x}2-(a+b)}{ 2\sqrt{a^2-b^2}\tan\frac{x}2 }- \frac{2b}{a^2-b^2}
\frac{\tan\frac{x}2}{(a-b) \tan^2\frac{x}2 +(a+b)}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Double Integration in region $A=\{ (x,y)\in \mathbb{R}^{2} : x^{2}+y^{2}\le 1, x+y \ge 1, y \le x\}$.
Calculate
$$\iint _{A} (x^{2}+y^{2})^{-3/2} \,dx\,dy\,,$$
where $A=\{ (x,y)\in \mathbb{R}^{2} : x^{2}+y^{2}\le 1, x+y \ge 1, y \le x\}$.
I first found the intersection points that are $(1,0)$, $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ and $\left(\frac{1}{2}, \frac{1}{2}\right)$. And the determined the regions
\begin{split}
D & =\left\{(x,y)\in \mathbb{R}^{2}: x^{2}+y^{2}, y \le x\} = \{(r,\theta)\in \mathbb{R}^{2}: 0 \le r \le 1, 0 \le \theta \le \frac{\pi}{4} \right\}\\
B & =\left\{(x,y)\in \mathbb{R}^{2}: 0 \le x \le \frac{1}{\sqrt{2}}, 0 \le y \le x\right\}\\
C & =\left\{(x,y)\in \mathbb{R}^{2}: \frac{1}{\sqrt{2}} \le x \le 1, \frac{1}{\sqrt{2}} \le y \le 1-x \right\}
\end{split}
Then,
$$\iint _{A} (x^{2}+y^{2})^{-3/2} \,dx\,dy = \iint_{D} (x^{2}+y^{2})^{-3/2} \,dx\,dy - \iint_{B} (x^{2}+y^{2})^{-3/2} \,dx\,dy - \iint_{C} (x^{2}+y^{2})^{-3/2} \,dx\,dy$$
Is this right? Or is there any other easiest way?
| In polar coordinates, the integration region is enclosed by $\theta \in (0,\frac\pi4)$ and $ r\in (\frac1{ \sin \theta + \cos\theta},1)$. Then, the integral becomes
$$I=\int_0^{\pi4}\int_{\frac1{\sin \theta + \cos\theta}}^1 \frac1{r^2}drd\theta= \int_0^{\pi4}(\sin \theta +\cos \theta -1 )d\theta=1-\frac\pi4
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Trapezium from two diagonals and two non-parallel sides Is it possible to construct (and to calculate) a trapezium from it's two non-parallel sides and it's two diagonals, with other words $b,d,e,f$ are given:
I read out the equations system
$\begin{array}{|l l}
(1) & a =c+p+q \\
(2) & h^2 =e^2 -(a-q)^2 \\
(3) & h^2 =f^2 -(a-p)^2 \\
(4) & b^2 =h^2+q^2 \\
(5) & d^2 =h^2+p^2 \\
\end{array}$
How can I solve that for $a,c,h,p,q$ ?
| Per the cosine rule
$$a^2 = b^2+e^2 -2be \cos C= d^2 +f^2 -2df \cos D\tag1
$$
and note that the triangles ADB and ACB are equal in area
$$\frac12 be \sin C =\frac12df \sin D\implies b^2e^2 (1-\cos^2 C )= d^2f^2(1-\cos^2 D)\tag2
$$
Substitute (1) into (2) to eliminate $\cos C$ and $\cos D$, yielding
$$a^2= \frac{(b^2-e^2)^2- (d^2-f^2)^2}{2(b^2+e^2 -d^2-f^2)}
$$
Similarly
$$c^2= \frac{(b^2-f^2)^2- (d^2-e^2)^2}{2(b^2+f^2 -d^2-e^2)}
$$
Depending on the given values of $b$, $d$, $e$ and $f$, the trapezium is either unique or impossible since the RHS’s of above expressions have to be positive. In the special case of $b=d$ and $e=f$, the trapezium is possible yet not unique.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Hypergeometric functions for $\sum_{k} {{a+b} \choose{a+k}}{{b+c} \choose{b+k}}{{c+a} \choose{c+k}}(-1)^k$ with $a,b,c\geq 0$ The sum is: $\sum_{k} {{a+b} \choose{a+k}}{{b+c} \choose{b+k}}{{c+a} \choose{c+k}}(-1)^k$ with $a,b,c \geq 0$.
The equivalent hypergeometric function can be found in page 214, Concrete Mathemtics, which is:
$$_3F_2\left(1-a-2n,1-b-2n,-2n;a,b;1\right) (*)$$
This is my approach:
Let $n = max(-a,-b,-c)$.
$$\sum_{k} {{a+b} \choose{a+k}}{{b+c} \choose{b+k}}{{c+a} \choose{c+k}}(-1)^k = \sum_{k\geq n} {{a+b} \choose{a+k}}{{b+c} \choose{b+k}}{{c+a} \choose{c+k}}(-1)^k = \sum_{k\geq0} {{a+b} \choose{a+k+n}}{{b+c} \choose{b+k+n}}{{c+a} \choose{c+k+n}}(-1)^{k+n}$$
Now let $$t_k = {{a+b} \choose{a+k+n}} {{b+c} \choose{b+k+n}}{{c+a} \choose{c+k+n}}(-1)^{k+n} = \frac{(a+b)!}{(a+k+n)!(b-k-n)!}\frac{(b+c)!}{(b+k+n)!(c-k-n)!}\frac{(c+a)!}{(c+k+n)!(a-k-n)!}(-1)^{k+n}$$
I would like to evaluate $$\frac{t_{k+1}}{t_k}$$,which is:
$$\frac{(a-k-n)(b-k-n)(c-k-n)(-1)}{(a+k+n+1)(b+k+n+1)(c+k+n+1)}(-1)^n = \frac{(k-a+n)(k-b+n)(k-c+n)}{(a+k+n+1)(b+k+n+1)(c+k+n+1)}(-1)^n $$
The above fraction doesn't have the form of:
$$ \frac{(k+a_1)(k+a_2)...(k+a_m)z}{(k+b_1)(k+b_2)...(k+b_n)(k+1)}$$
Even if we let n be even, the answer would be:
$$_4F_3\left(n-a,n-b,n-c,1;n+1+a,b+1+n,c+1+n;1\right)$$, not matching $(*)$. Apart from this way, i can't find any other approaches to this problem. Any help would be greatly appreciated.
*Edit: We might assume the biggest number of three is c. So $a \leq b \leq c$. This could work because we will have $k!$ in $t_k$. (9/4/2021)
| The dumbest mistake is in this $(-1)^n$ in:
$$\frac{(a-k-n)(b-k-n)(c-k-n)(-1)}{(a+k+n+1)(b+k+n+1)(c+k+n+1)}(-1)^n $$
Clearly, it should be canceled as well! And the idea in the comment works!
Let we assume $a \leq b \leq c$. So, we have $n = -c$!
Now, $$\sum_{k} {{a+b} \choose{a+k}}{{b+c} \choose{b+k}}{{c+a} \choose{c+k}}(-1)^k = \sum_{k \geq0} {{a+b} \choose{a+k-c}}{{b+c} \choose{b+k-c}}{{c+a} \choose{k}}(-1)^k$$
Repeating the steps, we have:
$$\frac{t_{k+1}}{t_k} = \frac{(b+c-k)(c+a-k)(2c-k)(-1)}{(a-c+k+1)(b-c+k+1)(k+1)} = \frac{(k-(b+c))(k-(c+a))(k-2c)}{(a-c+k+1)(b-c+k+1)(k+1)}$$
And it is: $$_{3} F_{2}\left(\begin{matrix} -b-c,-c-a,-2c\\a-c+1,b-c+1 \end{matrix} ;1\right)$$
Finally, we simply let $a_0 = a-c+1$, $b_0 = b-c+1$ and $c=n$. So, $a=a_0+c-1=$ and $b=b_0+c-1$.
$$_{3} F_{2}\left(\begin{matrix} -b_0-c+1-c,-c-a_0-c+1,-2c\\a_0,b_0 \end{matrix} ;1\right)$$ = $$_{3} F_{2}\left(\begin{matrix} -b_0-2n+1,-a_0-2n+1,-2n\\a_0,b_0 \end{matrix} ;1\right)$$, we're done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Uncertain how to solve this trigonometric equation I am currently attempting to discover how to find the general solutions to $$\sqrt 3\tan^2x=2\tan x+\sqrt 3$$
The given solutions are $x= \frac{\pi}{3}+ \pi k$ , $\frac{5\pi}{6} + \pi k$
To solve this equation I removed the square root from $\sqrt 3\tan^2x$ leaving me with $\tan x$.
Then I subtracted $2\tan x$ from $3\tan x$ leaving me with $\tan x=\sqrt 3$.
Should I then not solve $\tan x = \sqrt 3$ for the generalized solutions?
This would give $x=\frac{\pi}{3}$ and $x=\frac{2\pi}{3}$ which would then be generalized to $\frac{\pi}{3} + \pi k$ and $\frac{2\pi}{3} + \pi k$, which doesn't agree with the given answers.
| Elaborating on player3236's comment: Let $u = \tan x$. Then your equation is $\sqrt{3} u^2 = 2u + \sqrt{3}$. Solve this quadratic equation, and then use $u = \tan x$ to solve for $x$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How can we prove $\gcd(3n^2+1, 2n-3)\mid 31$ I want to show that $\gcd(3n^2+1, 2n-3)$ divides 31 $\forall n\in\mathbb{N}$.
I have tried to begin by eliminating the $3n^2$ factor on the left by adding and subtracting multiples and powers of $2n-3$ but I still haven't figured out how. Can someone give me some hint?
| Note if $\gcd(k,a) = 1$ then $\gcd(a, b) = \gcd(a, bk)$. THis is because $k$ having no factors in common with $a$ then multiplying $b$ by $k$ will not add any common factors of $a$.
And as $3|3n^2$ then $3\not \mid 3n^2 + 1$. And as $2|2n$ then $2\not \mid 2n-3$. And as $3,2$ are prime $\gcd(3n^w+ 1, 3) = 1$ and $\gcd(2n-3, 2) = 1$.
SO $\gcd(3n^2 +1, 2n-3) = \gcd(2\cdot (3n^2 + 1), 3\cdot (2n-3))=$
$\gcd (6n^2 + 2, 6n -9)= $
$\gcd([6n^2 + 2] - n[6n-9], 6n-9) = \gcd(9n + 2, 6n - 9)=\gcd(9n+ 2, 2n-3)$
Againd $3\not \mid 9n + 2$ and $2\not \mid 2n-3$ so
$\gcd(9n + 2, 2n-3) = \gcd(2(9n+2), 9(2n-3))=\gcd(18n +4 , 18n - 27)=$
$\gcd([18n + 4] - [18n-27], 18n - 27) = \gcd(31, 18n-27)$.
Now $31$ is prime. So either $31 \mid 18n -27$ or $31$ has no factors in common.
So either $\gcd (31, 18n-27)=\begin{cases} 31 &\text{if } 31|18n-27\\ 1&\text{if } 31 \not \mid 18n-27\end{cases}$.
And as $1|31$ and $31|31$ we have $\gcd(3n^2 + 1, 2n -3 )$ is a divisors of $31$.
....
Actually if those multiplying terms by $3$ and $2$ rub you the wrong way (they rub me the wrong way)
Consider Bezout's lemma
If we can find $M (3n^2 + 1) + N(2n-3) = 31$ where $N$ and $M$ are integers (and any polynomial with integer coefficients in terms of $n$ will be evaluated as integers for all $n$) the we have $\gcd(3n^2 + 1, 2n-3)|31$.
And we have $2(3n^2 + 1) - 3n(2n -3) = 9n +2$
$4(3n^2 + 1) - 6n(2n-3) = 18n + 4 = 9(2n-3) + 31$
$4(3n^2 + 1) - (6n+9)(2n-3) = 31$.
We are done.
We are still multiplying in extra terms but... well, its clearer to me that it is allowable
| {
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"timestamp": "2023-03-29T00:00:00",
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Converting a radical to a fractional exponent I want to understand how to convert a radical to a fractional exponent. Given the following equation:
$\sqrt[3]{(x)^6\cdot x^9}=\sqrt[3]{x^{24}\cdot x^9}=\sqrt[3]{x^{33}}=x^{\frac{33}3}=x^{11}$
How does: $\sqrt[3]{(x)^6\cdot x^9} = \sqrt[3]{x^{24}\cdot x^9}\;\;$??
| Power rules of exponent :
*
*$(a^m)^n=a^{mn}$ $\to \sqrt[3]{(x^4)^6\cdot x^9} = \sqrt[3]{x^{24}\cdot x^9}$
*$a^ma^n=a^{m+n}$ $\to \sqrt[3]{x^{24}\cdot x^9}$$=\sqrt[3]{x^{33}}$
*$\sqrt[n]{a}=a^{\frac{1}{n}},\sqrt[n]{a^m}=a^{\frac{m}{n}} \to \sqrt[3]{x^{33}}$$=x^{\frac{33}{3}}=x^{11}\;$
Or,
$\sqrt[3]{(x^4)^6\cdot x^9} = \sqrt[3]{x^{24}\cdot x^9}$(2)$=\sqrt[3]{x^{33}}$$=x^{\frac{33}{3}}=x^{11}\;$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The functional equation $ f \left( x ^ 2 y + f ( y ) \right) = x ^ 2 f ( y ) + f \big( f ( y ) \big) $
Find all functions $ f : \mathbb R \to \mathbb R $ such that
$$ f \left( x ^ 2 y + f ( y ) \right) = x ^ 2 f ( y ) + f \big( f ( y ) \big) $$
for all $ x , y \in \mathbb R $.
It's straightforward to check that for any constant $ a \in \mathbb R $, the function $ f : \mathbb R \to \mathbb R $ defined by $ f ( x ) = a x $ for all $ x \in \mathbb R $ is a solution. I suspect that those are the only solutions.
Setting $ x = 1 $ and $ y = 0 $, we can see that $ f ( 0 ) = 0 $. Letting $ a = f ( 1 ) $ and defining $ g : \mathbb R \to \mathbb R $ with $ g ( x ) = f ( x ) - a x $ for all $ x \in \mathbb R $, we can see that $ g ( 0 ) = g ( 1 ) = 0 $, and the functional equation becomes
$$ g \Big( \left( x ^ 2 + a \right) y + g ( y ) \Big) = x ^ 2 g ( y ) + g \big( a y + g ( y ) \big) $$
for all $ x , y \in \mathbb R $. Setting $ y = 1 $, we find out that $ g ( x ) = g ( a ) $ for all $ x \ge a $. In case $ a \le 1 $, this directly gives $ g ( a ) = 0 $, as we have $ g ( 1 ) = 0 $. In case $ a > 0 $ one can set $ y = a $ and choose $ x $ large enough so that $ \left( x ^ 2 + a \right) y + g ( y ) \ge a $, and get $ x ^ 2 g ( a ) = g ( a ) - g \big( a ^ 2 + g ( a ) \big) $, and since this is true for all large enough $ x $, we must have $ g ( a ) = 0 $. So in any case, we have $ g ( x ) = 0 $ for all $ x \ge a $.
It remains to prove that $ g $ is zero everywhere, which will show that linear functions are the only possible solutions.
Source:
Number six on the list at the end of this page.
| In fact, it's not true that linear functions are the only solutions. For any $ a , b \in \mathbb R _ { 0 + } $, the function $ f : \mathbb R \to \mathbb R $ defined with
$$ f ( x ) =
\begin {cases}
a x & x \ge 0 \\
b x & x < 0
\end {cases} $$
is another solution. This can be verified easily, using the fact that $ y $, $ a y $, $ b y $, $ \left( x ^ 2 + a \right) y $ and $ \left( x ^ 2 + b \right) y $ will always have the same sign (in the nonstrict sense), because $ a , b \ge 0 $. Considering these solutions together with linear functions, one can show that there is no other solution, which we will prove.
We continue the argument put forward by the OP, except that we only prove $ g ( y ) = 0 $ for $ y > 0 $, as it may not be true for $ x < 0 $ in the case of the above solutions. Fixing $ y > 0 $, we can choose $ x $ large enough, namely $ x \ge \sqrt { \frac { | a - a y - g ( y ) | } y } $, so that we have $ \left( x ^ 2 + a \right) y + g ( y ) \ge a $, and hence $ g \Big( \left( x ^ 2 + a \right) y + g ( y ) \Big) = 0 $. Then we can use the functional equation for $ g $ to get $ x ^ 2 g ( y ) = - g \big( a y + g ( y ) \big) $, which shows that $ g ( y ) = 0 $, as this is true for all large enough $ x $.
For the negative points, we can take a similar path. Letting $ b = - f ( - 1 ) $ and defining $ h : \mathbb R \to \mathbb R $ with $ h ( x ) = f ( x ) - b x $ for all $ x \in \mathbb R $, we have $ h ( 0 ) = h ( - 1 ) = 0 $ and the functional equation
$$ h \Big( \left( x ^ 2 + b \right) y + h ( y ) \Big) = x ^ 2 h ( y ) + h \big( b y + h ( y ) \big) $$
for all $ x , y \in \mathbb R $. Setting $ y = - 1 $, we get $ h ( x ) = h ( - b ) $ for all $ x \le - b $. In case $ b \le 1 $, we have $ - 1 \le - b $ and since $ h ( - 1 ) = 0 $, $ h ( - b ) = 0 $. In case $ b > 0 $, letting $ y = - b $ and choosing $ x $ large enough so that $ \left( x ^ 2 + b \right) y + h ( y ) \le - b $, we get $ x ^ 2 h ( - b ) = h ( - b ) - h \big( - b ^ 2 + h ( - b ) \big) $, which gives $ h ( - b ) = 0 $ since it's true for all large enough $ x $. So in any case, we have $ h ( x ) = 0 $ for all $ x \le - b $. Again, for any $ y < 0 $, choosing $ x $ large enough so that $ \left( x ^ 2 + b \right) y + h ( y ) \le - b $, we get $ x ^ 2 h ( y ) = - h \big( b y + h ( y ) \big) $, which shows that $ h ( y ) = 0 $, as that is true for all large enough $ x $.
So we know that $ g ( x ) = 0 $ for all $ x \ge \min ( a , 0 ) $ and that $ h ( x ) = 0 $ for all $ x \le \max ( - b , 0 ) $. Note that by definition of $ g $ and $ h $ we have $ g ( x ) - h ( x ) = ( a - b ) x $ for all $ x \in \mathbb R $. This implies that if $ a < 0 $, we can choose $ x $ so that $ a < x < 0 $, and thus have $ g ( x ) = h ( x ) = 0 $, which gives $ a = b $. Similarly, if $ b < 0 $, choosing $ x $ with $ 0 < x < - b $ we get $ a = b $. Therefore, if $ a $ is not equal to $ b $ then $ a , b \ge 0 $.
By definition of $ g $ and $ h $ and the previous results, we have $ f ( x ) = a x $ for $ x \ge 0 $ and $ f ( x ) = b x $ for $ x < 0 $. If $ f $ is not linear, then $ a $ is not equal to $ b $, and thus we must have $ a , b \ge 0 $. We already know that whatever the value of $ a $ and $ b $, we'll get a solution in such a case. Thus there is no solution other than linear ones and those of this form, and we're done.
| {
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"timestamp": "2023-03-29T00:00:00",
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maximum of $x^2-3xy-2y^2$ subjects to $x^2+xy+y^2=1$ without Lagrange multiplier What is the maximum of $x^2-3xy-2y^2$ subjects to $x^2+xy+y^2=1$? I wonder there is a precalculus method, without using the Lagrange multiplier.
| $$x^2+xy+y^2=1\implies y_\pm=-\frac{1}{2} \left(x \pm\sqrt{4-3 x^2}\right)$$
Using $y_-$ gives
$$f=x^2-3xy_m-2y_m^2=\frac{7 x^2}{2}+\frac{1}{2} \sqrt{4-3 x^2} x-2$$ Compute the derivative, set it equal to $0$ and solve for $x$. You should get
$$x_1=\sqrt{\frac{1}{39} \left(26+7 \sqrt{13}\right)} \qquad x_2=-\sqrt{\frac{1}{39} \left(26-7 \sqrt{13}\right)}$$ Compute the coreesponding $y_1$ and $y_2$.
Do the same with $y_+$.
| {
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"timestamp": "2023-03-29T00:00:00",
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$I_n= \int_{0}^{ + \infty} \dfrac{ x^{4n-2} }{ ( 1+x^4)^n}$
*
*$x \geq 0$
*$n \geq 0$
*$f_n(x)= \dfrac{ x^{4n-2} }{ ( 1+x^4)^n}$
*$I_n= \int_{0}^{ + \infty}f_n(x) $
*What is the limit of $I_n$ ?
*What is an equivalent of $I_n$ ?
$
\begin{align*}
u'&= (-n+1) 4x^3 (1+x^4)^{-n-1}\\
u &=x^{4n-1}\\
v &= x^{4n-1} \\
v' &= (4n-1) x^{4n-2} \\
I_{n+1}&= \int_{0}^{ + \infty} \dfrac{-4n x^3 x^{4n-1} }{ 4 (1+x^4)^{n+1} (-n) }\\
I_{n+1}&= \left[ \dfrac{ x^{4n-1}}{-4n (1+x^4)^n } \right] + \int_{0}^{ + \infty} \dfrac{(4n-1) x^{4n-2}}{4n (1+x^4)^n } \\
I_{n+1} &= \dfrac{4n-1}{4n}I_n \\
I_2 &=\dfrac{4-1}{4 \times 1}I_1 \\
I_{n+1} &= \dfrac{\prod_{k=1}^{n} (4k-1) }{4^n n!} I_1 \\
I_1&= \int_{0}^{ + \infty} \dfrac{ x^{2}}{ (1+x^4) } dx \\
\end{align*}
$
Quanto's answer for $I_1$ :
$$ \int_0^{\infty} \dfrac{ x^{2}}{ 1+x^4 } dx
\overset{x\to \frac1x}=\frac12\int_0^{\infty} \dfrac{ 1+x^2}{ 1+x^4 } dx = \frac12 \int_0^{\infty} \dfrac{ 1+\frac1{x^2}}{ x^2+\frac1{x^2} } dx \\
= \frac12 \int_0^{\infty} \dfrac{ d(x-\frac1{x})}{ (x-\frac1{x})^2+2 } dx =\frac\pi{2\sqrt2}
$$
$I_{n+1} = \dfrac{ \prod_{k=1}^n (4k-1) }{ 4^n n!} I_1 =\prod_{k=1}^n \dfrac{4k-1}{4k} I_1=\prod_{k=1}^n (1- \dfrac{1}{4k}) I_1$
What is the limit ?
$\ln( 1 - u) \sim u$ donc $\ln (1 - \dfrac{1}{4k}) \sim - \dfrac{1}{4k}$
and $- \sum \dfrac{1}{k} = - \infty$ so $I_n \to 0$ ?
| If you like the gaussian hypergeometric function
$$J_n= \int \dfrac{ x^{4n-2} }{ ( 1+x^4)^n}\,dx=\frac{x^{4 n-1} }{4 n-1}\,\,_2F_1\left(n-\frac{1}{4},n;n+\frac{3}{4};-x^4\right)$$ Assuming that $n \gt \frac 14$, for $x=0$ the rhs tends to $0$ and
$$I_n= \int_{0}^{ + \infty} \dfrac{ x^{4n-2} }{ ( 1+x^4)^n}=\frac{\Gamma \left(\frac{1}{4}\right) \Gamma \left(n+\frac{3}{4}\right)}{(4
n-1) \Gamma (n)}=\frac{\Gamma \left(\frac{5}{4}\right) \Gamma \left(n-\frac{1}{4}\right)}{\Gamma(n)}$$ which the same as @Svyatoslav's result in comments.
Now, consider
$$y=\frac{\Gamma \left(n-\frac{1}{4}\right)}{\Gamma(n)}\implies \log(y)=\log \left(\Gamma \left(n-\frac{1}{4}\right)\right)-\log \left(\Gamma \left(n\right)\right)$$ Apply Stirling approximation twice and continue with Taylor expansion
$$\log(y)=\frac{1}{4} \log \left(\frac{1}{n}\right)-\frac{3}{32 n}+\frac{1}{128 n^2}+O\left(\frac{1}{n^3}\right)$$
$$y\sim\frac 1 {n^{\frac 14}} \exp\Big[-\frac{3}{32 n}+\frac{1}{128 n^2}\Big]$$ and then the approximation
$$I_n\sim\frac {\Gamma \left(\frac{5}{4}\right) } {n^{\frac 14}} \exp\Big[-\frac{3}{32 n}+\frac{1}{128 n^2}\Big]$$
Trying for $n=10^4$, the exact value is $0.0906417$ while the approximation gives $0.0906394$.
| {
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$\frac{1}{a} = \frac{1}{b} + \frac{1}{c} - \frac{1}{abc}$ and $a^2 + b^2 = c^2$ I have found this in an Romanian magazine. We have to solve for natural numbers:
$$\frac{1}{a} = \frac{1}{b} + \frac{1}{c} - \frac{1}{abc}$$
$$a ^ 2 + b ^ 2 = c ^ 2$$
After some elementary calculations, we get that $bc + 1 = a(b + c)$. I tried using some divisibility things, hoping to get $b + c$ dividing some constant, but, in the end, got $b + c \ | \ (b + c) ^ 2$. Then, $a^2(b + c) ^ 2 = (bc + 1) ^ 2$, so $(c^2 - b^2)(b + c)^2 =(bc + 1) ^ 2$ and tried solving this, but, again, couldn't. Can you help me? I doubt there are natural solutions, but are there any integers?
| (This is not a solution. I made an error.)
Show that (If you get stuck, show your work.)
*
*$(a, b, c)$ is a primitive pythagorean triplet.
*If $(a, b, c ) = (m^2 - n^2, 2mn, m^2+n^2)$ with $m > n > 0$, then there are no solutions.
Substituting in, we get $n^4 = m^4 - 4mn^3$.
View this as a quartic in $ x = \frac{n}{m}$, we get $ x^4 + 4x^3$ - 1 = 0, which has no rational solutions.
Hence, there are no integer solutions to $(m, n)$.
*If $(a, b, c) = ( 2mn, m^2 - n^2, m^2 + n^2)$ with $m > n > 0$, then the only solution is $(m, n) = (4, 1) $ which leads to $ (a, b, c) = (8, 15, 17)$.
(Hm, I thought I had a solution to this part, but it turns out that I made an error.)
Substituting in, we have $ m^4 - 4m^3 n= n^4 - 1$.
If $ n = 1$, we have $ m = 4n = 4$.
Otherwise, writing it as $ m^3 (m-4n) = n^4 - 1 > 0 \Rightarrow m -1 \geq 4n$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $\frac{7}{2^{1/2} + 2^{1/4} + 1} = A + B*2^{1/4} + C*2^{1/2} + D*2^{3/4},$ find $A,B,C,D$
If $\frac{7}{2^{1/2} + 2^{1/4} + 1} = A + B*2^{1/4} + C*2^{1/2} + D*2^{3/4},$ find $A,B,C,D$ .
What I Tried: I wrote :-
$$\rightarrow \frac{7}{2^{1/2} + 2^{1/4} + 1} = \frac{2^2 + 2 + 1}{2^{1/2} + 2^{1/4} + 1}$$
Then I rationalized the denominator to get :-
$$\rightarrow\frac{(2^2 + 2 + 1)(2^{1/2} + 2^{1/4} - 1)}{(2^{1/2} + 2^{1/4} + 1)(2^{1/2} + 2^{1/4} -1)}$$
$$\rightarrow \frac{2^{5/2}+2^{9/4}-2^2+2^{3/2}+2^{5/4}-2+2^{1/2}+2^{1/4}-1}{(2^{1/2} + 2^{1/4})^2 - 1}$$
$$\rightarrow \frac{2^{5/2}+2^{9/4}-2^2+2^{3/2}+2^{5/4}-2+2^{1/2}+2^{1/4}-1}{( 2^{5/4}+2^{1/2} + 1)}$$
I am stuck here. I probably cannot proceed further unless I find a way to factorise the numerator, but I have not find a way to do that for now. Can anyone help me with this problem?
| Let $2^{1/4}=p$
So, we need
$$\dfrac{1+p^4+p^8}{1+p+p^2}$$
Now $1+p^4+p^8=(1+p^4)^2-(p^2)^2=?$
and $1+p^2+p^4=(1+p^2)^2-(p)^2=?$
| {
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How to calculate the limit of the following sum as it approaches a divergence I am interested in the behaviour of summations like the following:
$$
\lim_{\alpha\to 0}\sum_{n=1}^{\infty}\frac{e^{in\alpha}}{\sqrt{\beta+n^{2}}}
$$
where $\beta$ is a constant. Just putting $\alpha=0$ doesn't help because the sum doesn't converge (it has a logarithmic singularity). I know that the case $\beta=0$ is easily calculated because the summation can be done exactly:
$$
\sum_{n=1}^{\infty}\frac{e^{in\alpha}}{n}=-ln(1-e^{i\alpha})\sim\frac{\pi i}{2}-ln(\alpha)+O(\alpha)
$$
but I don't know how to extend this to the case when $\beta$ is non-zero.
Thanks in advance for any help.
| I will assume that $\alpha>0$ and $\beta \geq 0$. Let us write
$$
\sum\limits_{n = 1}^\infty {\frac{{e^{in\alpha } }}{{\sqrt {\beta + n^2 } }}} = \sum\limits_{n = 1}^\infty {\frac{{e^{in\alpha } }}{n}} + \sum\limits_{n = 1}^\infty {\left( {\frac{1}{{\sqrt {\beta + n^2 } }} - \frac{1}{n}} \right)e^{in\alpha } } .
$$
As you noted,
$$
\sum\limits_{n = 1}^\infty {\frac{{e^{in\alpha } }}{n}} = - \log \alpha + \frac{\pi }{2}i + \mathcal{O}(\alpha )
$$
as $\alpha\to 0+$. Now,
\begin{align*}
\sum\limits_{n = 1}^\infty\! {\left( {\frac{1}{{\sqrt {\beta + n^2 } }} - \frac{1}{n}} \right)e^{in\alpha } } =\; & \sum\limits_{n = 1}^\infty \! {\left( {\frac{1}{{\sqrt {\beta + n^2 } }} - \frac{1}{n}} \right)} \\ & + \sum\limits_{n = 1}^\infty {(e^{in\alpha } - 1)\!\left( {\frac{1}{{\sqrt {\beta + n^2 } }} - \frac{1}{n}} \right)} .
\end{align*}
It is easy to see that
$$
0 < \frac{1}{n} - \frac{1}{{\sqrt {\beta + n^2 } }} = \frac{1}{2}\int_0^\beta {\frac{{dx}}{{(x + n^2 )^{3/2} }}}
\le \frac{\beta }{{2n^3 }},
$$
whence
\begin{align*}
\left| {\sum\limits_{n = 1}^\infty {(e^{in\alpha } - 1)\left( {\frac{1}{{\sqrt {\beta + n^2 } }} - \frac{1}{n}} \right)} } \right| & \le \sum\limits_{n = 1}^\infty {\left| {\frac{1}{{\sqrt {\beta + n^2 } }} - \frac{1}{n}} \right|\left| {e^{in\alpha } - 1} \right|}\\ & \le \beta \sum\limits_{n = 1}^\infty {\frac{1}{{n^3 }}\left| {\sin \left( {\frac{{n\alpha }}{2}} \right)} \right|} \le \beta \alpha \frac{{\pi ^2 }}{{12}}
\end{align*}
and
$$\left|\sum\limits_{n = 1}^\infty {\left( {\frac{1}{{\sqrt {\beta + n^2 } }} - \frac{1}{n}} \right)}\right|<+\infty.$$
In summary,
$$
\sum\limits_{n = 1}^\infty {\frac{{e^{in\alpha } }}{{\sqrt {\beta + n^2 } }}} = - \log \alpha + C_\beta + \frac{\pi }{2}i + \mathcal{O}(\alpha ) + \mathcal{O}(\alpha \beta )
$$
as $\alpha \to 0+$, uniformly in $\beta\geq 0$ with
$$
C_\beta = \sum\limits_{n = 1}^\infty {\left( {\frac{1}{{\sqrt {\beta + n^2 } }} - \frac{1}{n}} \right)} .
$$
A simplified result which is useful for larger values of $\beta$ may be obtained as follows. It can be shown using the Euler–Maclaurin formula that
$$
C_\beta = - \log \sqrt{\beta} - \gamma + \log 2 + \mathcal{O}\!\left( {\frac{1}{{\sqrt {\beta+1} }}} \right).
$$
Therefore,
$$
\sum\limits_{n = 1}^\infty {\frac{{e^{in\alpha } }}{{\sqrt {\beta + n^2 } }}} = - \log (\alpha\sqrt{\beta}) - \gamma + \log 2 + \frac{\pi }{2}i + \mathcal{O}(\alpha ) + \mathcal{O}(\alpha \beta ) + \mathcal{O}\!\left( {\frac{1}{{\sqrt {\beta+1} }}} \right)
$$
as $\alpha \to 0+$, uniformly in $\beta\geq 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Length of cardiod $r=1+\sin\left(\theta\right)$ I'm pretty sure this is correct since I'm able to justify each of the steps but I'm just wondering if there's a more efficient way that isn't just skipping steps? Because I had to do 2 substitutions, one of which converted sine into cosine.
\begin{align*}
\text{Length }= & \ 2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{2 + 2\sin\left(\theta\right)} d\theta \\
= & \ 2 \sqrt{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{1 + \sin\left(\theta\right)} d\theta\\
= & \ - 2\sqrt{2} \int_{\pi}^{0} \sqrt{1+ \sin\left(\frac{\pi}{2}-x\right) } dx \\
= & \ - 2\sqrt{2} \int_{\pi}^{0} \sqrt{1+ \cos\left(x\right) } dx \\
= & \ - 2\sqrt{2} \int_{\pi}^{0} \sqrt{2 \cos^2\left(\frac{x}{2}\right)} dx \\
= & \ - 4 \int_{\pi}^{0} \left|\cos\left(\frac{x}{2}\right)\right| dx \\
= & \ 4 \int_{0}^{\pi} \cos\left(\frac{x}{2}\right) dx \\
= & \ 4 \cdot 2 \int_{0}^{\frac{\pi}{2}} \cos\left(u\right) du \\
= & \ 8 \left[\sin\left(u\right)\right]_{u=0}^{u=\frac{\pi}{2}} \\
= & \ 8 \left[\sin\left(\frac{\pi}{2}\right) - \sin\left(0\right) \right] \\
= & \ 8 \\
\end{align*}
edit: admittedly, I do know of other answers but I have no idea how one would have come across such trigonometric identities, or reason out how such an identity would be beneficial.
| Instead
\begin{align*}
\text{L }= & \ 2 \sqrt2\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{1 +\sin\theta}\> d\theta
= \ 2 \sqrt{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{\left(\sin\frac{\theta}2+ \cos\frac{\theta}2 \right)^2} \>d\theta\\
= &\ 2\sqrt{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(\sin\frac{\theta}2+ \cos\frac{\theta}2 \right) d\theta = 8 \\
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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simplify the expression ${\sqrt{x - 2\sqrt{x -1}} +\sqrt{x + 2\sqrt{x -1}}}$
Simplify the expression ${\sqrt{x - 2\sqrt{x -1}} +\sqrt{x + 2\sqrt{x -1}}}$
The problem is from a not so well renowned book for calculus in India - Concepts of Functions & Calculus - Vikas Rahi, ISBN 9780070080805. The answer for this question is given as
$$\lvert \sqrt{x-1} -1 \rvert + \lvert \sqrt{x-1}+1 \rvert$$
My efforts weren't great at all, I tried rationalizing them but the denominator becomes very similar to the question, which is totally not helpful for simplification.
| $$\begin{align}{\sqrt{x - 2\sqrt{x -1}} +\sqrt{x + 2\sqrt{x -1}}}&=T(x)>0&\end{align}$$
$$T^2(x)=2x+2 \sqrt{x^2-4x+4}$$
$$T^2(x)=2x+2|x-2|$$
$$\begin{align}\color {gold}{\boxed {\color{black}{T(x)=\sqrt{2x+2|x-2|}}}}\end{align}$$
*
*If $x≥2$, then $\begin{align}T(x)&=\sqrt{2x+2x-4}\\
&=2\sqrt{x-1}\end{align}$
*If $x<2$, then $\begin{align}T(x)=\sqrt{2x+4-2x}=2.\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4115583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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integrate $\int \frac{x^{(2n-1)}}{(1+x^2)^{(n+3)}} dx$ So, I need to integrate $\int \frac{x^{(2n-1)}}{(1+x^2)^{(n+3)}} dx, n \in \mathbb{N}$.
I've tried partial fractions, but I get nowhere.
Then I tried using substitution, using the substitution $t = 1+x^2$, but I get:
$$
\int \frac{x^{(2n-1)}}{(1+x^2)^{(n+3)}}dx = \int \frac{x^{(2n-1)}}{(t)^{(n+3)}} \frac{1}{2x}dt =2\int int\frac{x^{(2n-2)}}{(t)^{(n+3)}}dt = 2\int\frac{(t-1)^{(n-1)}}{(t)^{(n+3)}}dt $$
And then I don't know what to do.
Should I use another approach, or am I on the right path?
Thanks in advance.
| Denote
$$
I_n = \int \frac{x^{2n-1}}{(1+x^2)^{n+3}} dx.
$$
Using integration by parts it's easy to see that
$$
I_n = \int \frac{x^{2n-1}}{(1+x^2)^{n+3}} dx = \begin{bmatrix} u = \frac{1}{(1+x^2)^{n+3}} & dv = x^{2n-1} dx \\ du = - \frac{2(n+3)x}{(1+x^2)^{n+4}} dx & v =\frac{x^{2n}}{2n}\end{bmatrix} =
$$
$$
= \frac{x^{2n}}{2n (1+x^2)^{n+3}} + \frac{n+3}{n} \int \frac{x^{2n+1}}{(1+x^2)^{n+4}} = \frac{x^{2n}}{2n (1+x^2)^{n+3}} + \frac{n+3}{n} I_{n+1}.
$$
Therefore,
$$
I_{n+1} = \frac{n}{n+3} \left(I_n - \frac{x^{2n}}{2n (1+x^2)^{n+3}} \right)
$$
and one can compute $I_n$ for any $n>1$ knowing $I_1$.
| {
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"url": "https://math.stackexchange.com/questions/4120043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the equation $\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x+2}+\frac{1}{x+3}+\frac{1}{x+4}=0$ Solve the equation $$\dfrac{1}{x}+\dfrac{1}{x+1}+\dfrac{1}{x+2}+\dfrac{1}{x+3}+\dfrac{1}{x+4}=0.$$
For $x\ne -4;-3;-2;-1;0$ we have $$(x+1)(x+2)(x+3)(x+4)+x(x+2)(x+3)(x+4)+x(x+1)(x+3)(x+4)+\text{...}=0$$
Most likely that's not the author's intention. I have tried to substitute $t=x+2$ to get $$\dfrac{1}{t-2}+\dfrac{1}{t-1}+\dfrac{1}{t}+\dfrac{1}{t+1}+\dfrac{1}{t+2}=0$$ which actually isn't easier to work with than the original problem.
|
Most likely that's not the author's intention
It could very well be. If we replace $y = x + 2$ we have
$\frac 1{y-2} + \frac 1{y-1} +\frac 1y + \frac 1{y+1}+\frac 1{y+2} = 0$ so
$(y-1)y(y+1)(y+2) + (y-2)y(y+1)(y+2) + (y-2)(y-1)(y+1)(y+2) + (y-2)(y-1)y(y+2) + (y-2)(y-1)y(y-1) =0$
$(y^2-1)(y^2+2y) + (y^2-4)(y^2+ y) + (y^2-4)(y^2 - 1) + (y^2 -4)(y^2-y) + (y^2-1)(y^2 - y)=0$
$(y^2-1)[(y^2 + 2y)+(y^2 - 2y)] + (y^2-4)[(y^2 + y)+(y^2-y)] + (y^2 -4)(y^2 -1) = 0$
$(y^2-1)2y^2 + (y^2-4)2y^2 + (y^2 -4)(y^2 -1) = 0$
$2y^2(2y^2 - 5) + (y^4 -5y^2 +4) = 0$
$5y^4 - 15y^2 + 4 = 0$
$y^2 = \frac {15 \pm \sqrt{15^2-80}}{10}= \frac {15\pm \sqrt {145}}{10}$
$y = \pm\frac {15\pm \sqrt {145}}{10}$
$x = \pm \frac {15\pm \sqrt {145}}{10}-2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4120143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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A little bit of trouble computing the characteristic polynomial of a matrix? I'm reading Smirnov's Linear Algebra and Group Theory, here:
He says that is the characteristic equation but when I compute it on Mathematica, I obtained: $-(\lambda -2)^2 \left(\lambda ^3-27 \lambda -26\right)$. For reference, I computed it as follows:
I think the different result may come from Mathematica computing it in some different number domain perhaps? Or this is some typo in the book?
| I assume there is a typo in the matrix $A$ and the issue is that $a_{41}$ should be negative, so lets use this and find the Jordan canonical form. We have
$$A = \begin{pmatrix}
-2 & -1 & -1 & 3 & 2 \\
-4 & 1 & -1 & 3 & 2 \\
1 & 1 & 0 & -3 & -2 \\
-4 & -2 & -1 & 5 & 1 \\
4 & 1 & 1 & -3 & 0 \\
\end{pmatrix}$$
The eigenvalues are $\lambda = 2$ (a triple root) and $\lambda_2 = -1$ (a double root).
This is a deficient matrix with some generalized eigenvectors.
We find
$$P = \begin{pmatrix}
-1 & 0 & 1 & -1 & 1 \\
-1 & 0 & 0 & -1 & 0 \\
1 & 1 & -1 & 1 & -1 \\
-1 & 0 & 1 & -2 & 0 \\
1 & 0 & 0 & 1 & 1 \\
\end{pmatrix}, ~~~J = \begin{pmatrix}
-1 & 1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 & 0 \\
0 & 0 & 2 & 0 & 0 \\
0 & 0 & 0 & 2 & 1 \\
0 & 0 & 0 & 0 & 2 \\
\end{pmatrix}$$
This does not appear to match the authors results, but we can validate both.
Using the above, we verify
$$ A = P J P^{-1} = \begin{pmatrix}
-2 & -1 & -1 & 3 & 2 \\
-4 & 1 & -1 & 3 & 2 \\
1 & 1 & 0 & -3 & -2 \\
-4 & -2 & -1 & 5 & 1 \\
4 & 1 & 1 & -3 & 0 \\
\end{pmatrix}~~~\Large\color{\green}{\unicode{10004}}$$
Using the authors' results, we can verify
$$A = V \Lambda V^{-1} =
\begin{pmatrix}
0 & 1 & 0 & 0 & 1 \\
1 & 0 & 0 & -1 & 0 \\
1 & 0 & 0 & 0 & 1 \\
1 & 0 & 1 & 0 & 0 \\
0 & 0 & 1 & 1 & 1 \\
\end{pmatrix} \begin{pmatrix}
2 & 0 & 0 & 0 & 0 \\
1 & 2 & 0 & 0 & 0 \\
0 & 0 & 2 & 0 & 0 \\
0 & 0 & 0 & -1 & 0 \\
0 & 0 & 0 & 1 & -1 \\
\end{pmatrix} \begin{pmatrix}
0 & -1 & 1 & 1 & -1 \\
1 & -1 & 0 & 1 & -1 \\
0 & 1 & -1 & 0 & 1 \\
0 & -2 & 1 & 1 & -1 \\
0 & 1 & 0 & -1 & 1 \\
\end{pmatrix}~~~\Large\color{\green}{\unicode{10004}}$$
A couple of items to note:
$1.$ Eigenvectors are not unique.
$2.$ The Jordan canonical form is unique up to order of permutation.
So, these answers are equivalent and we have proven using the authors' calculations that indeed there is a typo in the book.
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplify convolution integral $\int_{-\infty}^\infty\frac{1}{(a^2+t^2)(a^2+(x-t)^2) }dt$ I am trying to show that $$\frac{1}{a^2+x^2} \ast \frac{1}{a^2+x^2}=\int_{-\infty}^\infty\frac{1}{(a^2+t^2)(a^2+(x-t)^2) }dt = \frac{2\pi}{a(4a^2+x^2)}$$ I wrote out the integral for the convolution and it becomes a big exercise in partial fractions and integration of rational functions.
$$ \int(\frac{1}{a^2+t^2}) (\frac{1}{a^2+(x-t)^2}) dt=
\int\frac{x + 2 t}{x (4 a^2 + x^2) (a^2 + t^2)} + \frac{3 x - 2 t}{x (4 a^2 + x^2) (a^2 + x^2 - 2 xt + t^2)} dt =$$
$$=\frac{a (log(a^2 + t^2) - log(a^2 + (x - t)^2)) + x \cdot tan^{-1}(\frac{t}{a}) + x \cdot tan^{-1}(\frac{t - x}{a})}
{a x (4 a^2 + x^2)}
$$
Plugging in the bounds of plus/minus infinity makes the logs cancel out, and gives the 2$\pi x$ from the arctan terms.
My question: Is there a way to simplify / make this any easier using the properties of the convolution? Or are these problems just 'plug and chug' like back in my engineering days?
| Symmetrize the integrand with $y= t-\frac x2$
\begin{align}
& \int_{-\infty}^\infty\frac{1}{(a^2+t^2)(a^2+(x-t)^2) }dt\\
= & \int_{-\infty}^\infty\frac{1}{(a^2+(y+\frac x2)^2)(a^2+(y-\frac x2)^2)}dy\\
=&\
\frac2{x(4a^2+x^2)}\int_{-\infty}^\infty
\frac{{\frac x2 +(y+\frac x2)}}{a^2+(y+\frac x2)^2 }
+ \frac{{\frac x2 -(y-\frac x2)}}{a^2+(y-\frac x2)^2 } \overset{y\pm\frac x2\to y}{dy}\\
= &\ \frac2{4a^2+x^2}\int_{-\infty}^\infty\frac{1}{ a^2+y^2}dy
= \frac{2\pi}{a(4a^2+x^2)}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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how to prove that $1^x+2^x+3^x+4^x+\cdots+N^x$ will never sum to a prime number except $1^x+2^x$? I am a a web developer programming in PhP which is limited to large calculations, but running a quick script shows that $1^x+2^x+3^x+4^x+\cdots+N^x$ can never sum to a prime number unless in the case of $1^x+2^x$, such as in the cases of $x=1$ and $x=2$ where $1^1+2^1=3$ and $1^2+2^2=5$.
As a self learner, I am currently refreshing my learning in Algebra 2 (before moving on), and sometimes my mind wonders into questions that I just can't find the answers to (mostly because I am not familiar with the concerned topics). I tried finding an answer but if this is a duplicate with a relevant answer, please close and refer me to it.
How to prove (if possible) that $1^x+2^x+3^x+4^x+\cdots+N^x$ will never be the sum of a prime number, unless in the case of $1^x+2^x$, such as in the cases of $x=1$ and $x=2$ where $1^1+2^1=3$ and $1^2+2^2=5$?
Edit: $x$ and $N$ are positive integers
I appreciate any answers even if it is just a hint or a reference.
| This is also only a partial solution that may provide more insight.
We can split into two cases, $x$ even and odd.
If $x$ is odd, then note that $i^x \equiv -(n-i)^x \mod n$. Then, consider $1^x+ 2^x...+n^x$ If $n$ is odd, we can pair up $1$ and $n-1$, etc. Each pair’s sum is divisible by $n$, $n^n$ is divisible by $n$ and the total is greater than $n$ and divisible by it and so composite. If $n$ is even, we can pair up 1 and $n$, etc and get that the sum is divisible by $n+1$ and divisible by it and is greater than it when either n or x are greater than 2.
If $x$ is even, then it’s trickier. By considering Vieta’s formula, it’s possible to show that $p| 1^x +2^x ... + (p-1)^x$ whenever $p-1$ does not divide $x$. This means that if $p$ divides $N$ and $p-1$ doesn’t divide $x$ then you can break up the sum into groups of size $p$ each divisible by $p$. Similarly if $p$ divides $N+1$, you can do the same thing by adding the $N+1$ term which is divisible by $p$. Similarly, you can double the sum and get terms from $1$ to $2N$ module $2N+1$. Thus, if $p$ doesn’t divide the sum and is a divisor of either $x$, $x+1$, or $2x+1$, then $p-1$ divides $N$.
Additionally, if $p^2$ divides $x$,$x+1$, or $2x+1$, then since each of the aforementioned groupings is the same module $p$ and their are now a multiple of $p$ of them, $p$ would divide the sum. This, in order for $p$ to not divide the sum, $x$, $x+1$, and $2x+1$ must not be divisible by its square.
In particular, this means for a given $n$, all of the primes of $x$, $x+1$, $2x+1$ are of the finite set of those for which $p-1$ divides $N$ and they are squarefree, so there are only a finite number of cases to check. Also, note that $x$,$x+1$, and $2x+1$ are relatively prime, so their primes are disjoint.
One other interesting fact is that if $p-1$ divides $n$, then everything is 0 or 1 mod p, so p divides the sum iff p divides $x - \lfloor x/p \rfloor$
It’s possible to use these facts to work through the small cases and quickly narrow things down to fairly large numbers.
In particular, if x is 2, the primes 2,3,5 divide $x$,$x+1$,$2x+1$, so 1,2,4 divides $N$, so $N$ is a multiple of 4. In fact, all the prime examples are the Fermat primes.
$x$ can’t be 3 since $x+1=4$ is not squarefree.
$x$ can’t be 4 since it’s squarefree.
If $x$ is 5, then to eliminate 5,2*3,11, we need $\operatorname{lcm}(5-1,2-1,3-1,11-1)=20$ to divide N. The smallest such example $1^{20}+2^{20}+3^{20}+4^{20}+5^{20}$ is divisible by 137. The fact that there can be a factor significantly larger than both $x$ and $n$ suggests that this kind of brute force approach with small factors is unlikely to completely solve the problem.
If $x$ is 6, then $2x+1$ is 13, so 12 divides N. Therefore $4=5-1$ divides N and $6=5+\lfloor 6/5 \rfloor$, so 5 will divide the sum of powers.
We can continue in the vein and eliminate many small numbers. In particular, I find that if $x$ is 10, then N must be a multiple of 60. If x is 21, then $N$ is a multiple of 210. If $x$ is 29, N is a multiple of $812=4*7*29$. Etc.
I’m not sure how to proceed from here. Even the simplest exceptional case here $1^{20}+2^{20}+3^{20}+4^{20}+5^{20}$ seems difficult to figure out that it’s composite and a multiple of 137 without brute force.
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"url": "https://math.stackexchange.com/questions/4122803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Are these the only "intersections" between the following series, and why all of them are multiples of $10$ As a web developer that programs in PhP, I enjoy running some scripts to see some of math wonders, however, PhP is limited to large calculations.
I wanted to see if there are any intersections for the following $2$ series:
A) $1 + 2 + 3+ 4 + 5 + \ldots$
B) $(1) + (1 + 2) + (1 + 2 + 3) +(1 + 2 + 3 + 4) + (1 + 2 + 3 + 4 + 5) + \ldots$
Out of the first $116410911$ positive integers, I have "found" the following "intersections" between A) and B):
At the number $10$:
$10 - (1+2+3+4)=0$
$10 - ((1)+(1+2)+(1+2+3))=0$
At the number $120$:
$120 - (1+2+3+4+5+6+7+8+9+10+11+12+13+14+15) = 0$
$120 - ((1)+(1+2)+(1+2+3)+(1+2+3+4)+(1+2+3+4+5)+(1+2+3+4+5+6)+(1+2+3+4+5+6+7)+(1+2+3+4+5+6+7+8)) = 0$
And I am not going to list the other procedures, but also:
At the numbers $1540$ and $7140$.
Can anyone confirm if there are any more "intersections"? In other words, am I to expect a finite case or an infinite case? Also, is it a coincidence that all four of them are numbers that are multiples of $10$?
| I think you are trying to find tuples $(a,b) $that satisfy the equation
$$\sum_{i = 1}^{a} i = \sum_{i = 1}^{b}\sum_{j =1}^{i}j$$
The left hand side is
$$\sum_{i = 1}^{a}i = \frac{a(a + 1)}{2}$$
and the right-hand side is
$$\sum_{i = 1}^{b}\sum_{j = 1}^{i}j = \frac{b(b + 1)(b + 2)}{6}.$$
Then, the equation becomes \begin{align*}\frac{a(a + 1)}{2} &= \frac{b(b + 1)(b + 2)}{6} \\ 3a(a + 1) &= b(b + 1)(b + 2)\end{align*}
This seems to be a Diophantine equation, but since I have no knowledge about it, someone could continue solving this.
To give a shot, I'll try to solve this by noticing that both sides have three factors, then we can equate the factors and see if it works.
We can let \begin{align*}3 &= b \\ a &= b + 1 \\ a + 1 &= b + 2.\end{align*} Notice that the second equation implies the third equation and $a = 4$ by solving for $a$ from the first equation. This is the first case that you gave which is for $10$.
The other cases is then left as an exercise.
By the way, I have no idea why you found tuples $(a,b)$ that are equal to a sum that is a multiple of 10. Someone might just answer this by continuing my solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4123915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Binary $[15,5,d]$ cyclic code My goal is to find a generator polynomial for a binary $[15,5,d]$ cyclic code that corrects all errors with weight less than or equal to $3$.
I have to find a degree $10$ monic polynomial that divides $x^{15}-1$ ($\bmod 2$) and has weight $7$. If $15$ and $5$ were coprime I would know how to factor $x^{15}-1$ using cyclotomic theory. But that is not the case.
Is my reasoning ok? How do I factor $x^{15}-1\bmod 2$? Thanks in advance!
| If you try systematically dividing by quartics you will see that more polynomials than $x^4+x^3+x^2+x+1$ factor into $x^{15}+1$ over $\mathbb F_2$:
$$x^{15}+1=(x+1)(x^2+x+1)(x^4+x^3+x^2+x+1)(x^4+x^3+1)(x^4+x+1)$$
Then
$$(x^2+x+1)(x^4+x^3+1)(x^4+x^3+x^2+x+1)=x^{10}+x^9+x^8+x^6+x^5+x^2+1$$
is the desired weight-$7$ polynomial.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4124386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Find the determinant of nth order $$
\begin{vmatrix}
2 & 2 & ... & 2 & 2 & 1 \\
2 & 2 & ... & 2 & 2 & 2 \\
2 & 2 & ... & 3 & 2 & 2 \\
... & ... & ... & ... & ... & ... \\
2 & n-1 & ... & 2 & 2 & 2 \\
n & 2 & ... & 2 & 2 & 2
\end{vmatrix}
$$
I got this in my linear algebra homework. In the task, it is required to find the determinant of a matrix by the method of representing the sum of determinants. By that I mean this property of determinants:
$$
\begin{vmatrix}
a & b+e \\
c & d+f \\
\end{vmatrix} = \begin{vmatrix}
a & b \\
c & d \\
\end{vmatrix} + \begin{vmatrix}
a & e \\
c & f \\
\end{vmatrix}
$$
What I have tried:
*
*Add the i-th with the (i-1)-th;
*Add the last row to all, getting two determinants, one of which is 0 (because of the (n-1)-st column);
*Tried to get n on the diagonal, this is what I got:
$$
\begin{vmatrix}
2 & 3 & ... & n-1 & n & n \\
2 & 3 & ... & n-1 & n & n+1 \\
2 & 3 & ... & n & n & n+1 \\
... & ... & ... & ... & ... & ... \\
2 & n & ... & n-1 & n & n+1 \\
n & 3 & ... & n-1 & n & n+1
\end{vmatrix}
$$
The closest one to mine from StackExchange was this one. But I didn't manage to link these two determinants.
No matter how I transform it, nothing worked for me. Any ideas?
| Let's do some elementary operations on rows and columns:
$$\begin{align}
\begin{vmatrix}
2 & 2 & ... & 2 & 2 & 1 \\
2 & 2 & ... & 2 & 2 & 2 \\
2 & 2 & ... & 3 & 2 & 2 \\
... & ... & ... & ... & ... & ... \\
2 & n-1 & ... & 2 & 2 & 2 \\
n & 2 & ... & 2 & 2 & 2
\end{vmatrix}
&\overset{1}=
\begin{vmatrix}
2 & 2 & ... & 2 & \color{red}2\downarrow & 1 \\
0 & 0 & ... & 0 & 0 & 1 \\
0 & 0 & ... & 1 & 0 & 1 \\
... & ... & ... & ... & ... & ... \\
0 & n-3 & ... & 0 & 0 & 1 \\
n-2 & 0 & ... & 0 & 0 & 1
\end{vmatrix} \\[2mm]
&\overset{2}= 2\begin{vmatrix}
0 & 0 & ... & 0 & \leftarrow\color{red}1 \\
0 & 0 & ... & 1 & 1 \\
... & ... & ... ... & ... & ... \\
0 & n-3 & ... & 0 & 1 \\
n-2 & 0 & ... & 0 & 1
\end{vmatrix} \\[2mm]
&\overset{3}= (-1)^{n-1}2\begin{vmatrix}
0 & 0 & ... & 1 \\
... & ... & ... ... & ... \\
0 & n-3 & ... & 0 \\
n-2 & 0 & ... & 0
\end{vmatrix} \\[2mm]
&\overset{4}= (-1)^{n-1 + \frac{(n-2)(n-3)}{2}}2\cdot (n-2)!
\end{align}$$
Explanation:
$1)$ Subtract the first row from all others
$2)$ Expand through $(n-1)$-th column
$3)$ Expand through the first row
$4)$ Calculate the anti-diagonal determinant
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4126432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
For positive integer $n$, why is $\lfloor \log_{10}(2^n)\rfloor + \lfloor \log_{10}(5^n)\rfloor + 2 = n+1$?
For positive integer $n$, why is $\lfloor \log_{10}(2^n)\rfloor + \lfloor \log_{10}(5^n)\rfloor + 2 = n+1$?
This question comes from counting the number of digits of $10^n$ in terms of the number of digits of $2^n$ and $5^n$. Number of digits of $10^n$ is $n+1$ which equals to the sum of the number of digits of $2^n$ and $5^n$. I know that the number of digits of a positive integer $x$ is $\lfloor \log_{10}(x)\rfloor + 1$.
Using programming, I've checked that this is true for $n$ less than $100$.
| $\lfloor{\log_{10}(2^n)}\rfloor = \lfloor{\log_{10}(10^n) - \log_{10}(5^n)}\rfloor = n + \lfloor{ - \log_{10}(5^n)}\rfloor = n - \lfloor{\log_{10}(5^n)}\rfloor - 1$
So:
$\lfloor{\log_{10}(2^n)}\rfloor + \lfloor{\log_{10}(5^n)}\rfloor + 2 = n - \lfloor{\log_{10}(5^n)}\rfloor - 1 + \lfloor{\log_{10}(5^n)}\rfloor + 2 = n + 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4130146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
$25$ people will be divided into $5$ groups ,each group have $5$ individuals . What is the probability that $25$ people will be divided into $5$ groups ,each group have $5$ individuals . What is the probability that
a-) Dennis , John and Jack are in the same group.
b-)Dennis , John and Jack are in different groups.
c-)Dennis and John are in the same group but not Jack
My attempt :
a-) If they are in same group , then there are $5$ ways to select this group. Moreover , we should select $2$ people for the group by $C(22,2)$ . Then $5 \times \frac{C(22,2) \times C(20,5) \times C(15,5) \times C(10,5) \times C(5,5)}{C(25,5) \times C(20,5) \times C(15,5) \times C(10,5) \times C(5,5)}$
b-)If they are in different groups , we can disribute them by $P(5,3)=60$ ways.Then , $60 \times \frac{C(24,4) \times C(19,4) \times C(14,4) \times C(10,5) \times C(5,5)}{C(25,5) \times C(20,5) \times C(15,5) \times C(10,5) \times C(5,5)}$
c-)We can choose $5$ groups for Dennis and John , 4 groups for Jack ,so $P(5,2)=20$ ways. Then ; $20 \times \frac{C(23,3) \times C(19,4) \times C(15,5) \times C(10,5) \times C(5,5)}{C(25,5) \times C(20,5) \times C(15,5) \times C(10,5) \times C(5,5)}$
Is my solution way correct ? If not ,can you help..
| Part b should have been $\frac{60 \times C(22,4) \times C(18,4) \times C(14,4) \times C(10,5) \times C(5,5)}{ C(25,5) \times C(20,5) \times C(15,5) \times C(10,5) \times C(5,5)}$
part c should have been $\frac{20 \times C(22,3) \times C(19,4) \times C(15,5) \times C(10,5) \times C(5,5)}{ C(25,5) \times C(20,5) \times C(15,5) \times C(10,5) \times C(5,5)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4130868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Prove convexity of $f(x_1,x_2,x_3)=\frac{x_1^2+x_1+1}{x_2+x_3}$ over $\{(x_1,x_2,x_3) : x_2,x_3\gt0\}$
Prove convexity of $$f(x_1,x_2,x_3)=\frac{x_1^2+x_1+1}{x_2+x_3}$$ over $\{(x_1,x_2,x_3) : x_2,x_3\gt0\}$
I am looking for an easy (or relatively easy) way to show $f$ is convex.
I tried to split the function into:
$$\frac{x_1^2}{x_2+x_3}+\frac{1}{x_2+x_3}+\frac{x_1}{x_2+x_3}$$
The first and the second functions are convex:
*
*The first is convex because it's quadratic-over-linear which is convex;
*The second is convex because it's linear change of variables of $\frac{1}{x}$ which is convex over $x>0$.
I stuck proving the third one. I found it's Hessian matrix. Let $g(x)=\frac{x_1}{x_2+x_3}$.
$$\nabla ^2g(x)=
\begin{bmatrix}
0 & \frac{-1}{(x_2+x_3)^2} & \frac{-1}{(x_2+x_3)^2} \\
\frac{-1}{(x_2+x_3)^2} & \frac{2x_1}{(x_2+x_3)^3} & \frac{2x_1}{(x_2+x_3)^3} \\
\frac{-1}{(x_2+x_3)^2} & \frac{2x_1}{(x_2+x_3)^3} & \frac{2x_1}{(x_2+x_3)^3}
\end{bmatrix}
$$
It suffices to show $\nabla g(x)\succeq0$, but I failed to do so. Last hope is calculating eigenvalues, but it seems too difficult. By the way, I can't use level-sets.
Maybe there's an eaier way to show its convexity.
Please advise.
Thank you.
| $g$ is not convex, as $(2,2)$-entry of $\nabla^2 g(x)$ is negative when $x_1$ is negative.
We have $\nabla f(x) = \begin{bmatrix} \frac{2x_1+1}{x_2+x_3} \\ - \frac{x_1^2+x_1+1}{(x_2+x_3)^2} \\ - \frac{x_1^2+x_1+1}{(x_2+x_3)^2}\end{bmatrix}$.
$$\nabla^2 f(x) = \begin{bmatrix} \frac{2}{x_2+x_3} & - \frac{2x_1+1}{(x_2+x_3)^2} & - \frac{2x_1+1}{(x_2+x_3)^2} \\ - \frac{2x_1+1}{(x_2+x_3)^2} & \frac{2(x_1^2+x_1+1)}{(x_2+x_3)^3} & \frac{2(x_1^2+x_1+1)}{(x_2+x_3)^3}\\ - \frac{2x_1+1}{(x_2+x_3)^2} & \frac{2(x_1^2+x_1+1)}{(x_2+x_3)^3} & \frac{2(x_1^2+x_1+1)}{(x_2+x_3)^3}\end{bmatrix}$$
One possible way to check positive semidefiniteness is by examine the determinant of each principal minor is nonnegative.
Since the second row and the third row is identitcal, the determinant is $0$. Also $|a_{23}|=0$.
Notice that $|a_{11}|>0$. Since $x_1^2+x_1+1 \ge \frac34> 0$, we have $|a_{22}|=|a_{33}|>0$.
$$|a_{12}|=|a_{13}|=\frac1{(x_2+x_3)^4}[4(x_1^2+x_1+1)-(2x_1+1)^2]=\frac{3}{(x_2+x_3)^4}>0.$$
Hence, it is convex.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4132168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that $f(x)=\frac{1+11x-6x^2}{(x-3)(1-2x)}$ is a decreasing function How do I prove that $f(x)$ is a decrasing function
$$f(x)=\frac{1+11x-6x^2}{(x-3)(1-2x)},x>3$$
I did:
$$f{'(x)}=-\frac{4}{(x-3)^2}+\frac{2}{(1-2x)^2}$$
$$f^{'}(x)<0$$
then I got
$$-\frac{7}{9}<x<1$$
So where do I go from here
| \begin{align}f(x) &= \frac{1+11x-6x^2}{(x-3)(1-2x)}\\&=\frac{3(-2x^2+7x-3)+11x-21x+1+9}{(x-3)(1-2x)}\\
&=3 + \frac{10(1-x)}{(x-3)(1-2x)}\\
&=3 + \frac{4}{x-3}- \frac{2}{1-2x}\\
&= 3 + \frac4{x-3}+\frac{1}{x-0.5}\end{align}
For $x>3$, as $x$ increases, we can see that $f(x)$ decreases.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4136384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find and solve a recurrence relation In a casino, there are $k(\geq 2)$ colored poker chips, one of which is blue. Find and solve a recurrence relation for the number of ways Joseph can stack n of these chips without consecutive blue chips.
Attempt
Let $a_n$ be the number of such arrangements of $n$ poker chips. If we look at the chip, the one underneath it must be of $k-1$ different colours, so we have $(k-1)a_{n-2}$ ways to arrange the rest of the stack. If it's one the other $k-1$ chips on top, we have $a_{n-1}$ ways to arrange the rest of the stack.
Therefore, we're trying to solve the following recurrence relation:
\begin{align*}
a_n=(k-1)a_{n-1}+(k-1)a_{n-2}, n\geq 2
\end{align*}
With $a_0=1$ y $a_1=k$.
Is this correct?
| If the reccurence relation is
$$a_n=(k-1)(a_{n-1}+a_{n-2})$$ the solution is given by
$$a_n=\frac {2^{-(n+1)}} K \Big[(K-k-1) (k-K-1)^n+(K+k+1) (k+K-1)^n\big]$$ where $K=\sqrt{(k-1)(k+3)}$.
This generates the sequence
$$\left(
\begin{array}{cc}
0 & 1 \\
1 & k \\
2 & (k-1) (k+1) \\
3 & (k-1) \left(k^2+k-1\right) \\
4 & (k-1)^2 k (k+2) \\
5 & (k-1)^2 \left(k^3+2 k^2-k-1\right) \\
6 & (k-1)^3 (k+1) \left(k^2+2 k-1\right) \\
7 & (k-1)^3 k \left(k^3+3 k^2-3\right) \\
8 & (k-1)^4 \left(k^2+k-1\right) \left(k^2+3 k+1\right) \\
9 & (k-1)^4 \left(k^5+4 k^4+2 k^3-5 k^2-2 k+1\right) \\
10 & (k-1)^5 k (k+1) (k+2) \left(k^2+2 k-2\right)
\end{array}
\right)$$ where interesting patterns can be found.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4137053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Find Two Rank 1 Matrices I'm looking for rank one matrices B and C such that A = B + C and BC = 0. Where
$$A = \begin{bmatrix}
0 & 2 & 2\\
2 & 4 & 2\\
2 & 2 & 0
\end{bmatrix}$$
with eigenvalues 0, -2, 6 and eigenvectors \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix} \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}
Sorry, I don't really know how to format
| What about $B=\left(\begin{array} & 1 & 2 & 1 \\ 2 & 4 & 2 \\ 1 & 2 & 1 \end{array}\right)$ and $C = \left(\begin{array} & -1 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & -1 \end{array}\right)$?
Both matrices have rank 1, their sum is $A = B+C$ and
$BC = \left(\begin{array} & 0 & 0 & 0 \\ 0 & 0 & 0 \\0 & 0 & 0\end{array}\right)$.
The reason why you can do this is the following:
At first, note that all these eigenvectors are orthogonal to each other, meaning that $u^Tv=0$, so using them in $B$ and $C$ will give a result of $BC=0$.
$B$ and $C$ each consist of multiples of the eigenvectors corresponding to eigenvalues $-2$ and $6$. This means we get:
$A\cdot B_i = 6 \cdot B_i$ and $A\cdot C_i = -2 \cdot C_i$ where $B_i$ and $C_i$ denote the $i$-th column. Define $b$ to be the eigenvector to $6$ and $c$ to be the eigenvector to $-2$. As we have rank $1$ matrices, each column has to be a multiple of the corresponding eigen vector. That means our matrices look like this:
$B=\left(\begin{array} & & & \\ \lambda_1 b & \lambda_2 b & \lambda_3 b \\ & & \end{array}\right)$ as well as $C=\left(\begin{array} & & & \\ \mu_1 c & \mu_2 c & \mu_3 c \\ & & \end{array}\right)$
We can use the information that if a $3\times 3$ matrix has distinct eigenvectors and -values which generate $\mathbb{R}^3$ (which is the case), this matrix can be reconstructed unambiguously given only the eigenvectors and eigenvalues.
So let's look what values for $\lambda_i,\mu_j$ are needed to get the same eigen values as $A$:
First $(B+C)\cdot b = 6\cdot b$. We have $B \cdot b = (\lambda_1+\lambda_2+\lambda_3) b$ so we get $\lambda_1+\lambda_2+\lambda_3 = 6$. Analogously from $(B+C)\cdot c = -2 \cdot c$ we can conclude that $\mu_1+\mu_2+\mu_3 = -2$.
But we also want to have $C\cdot b = B\cdot c = 0$. To make this true the vector $\lambda = \left(\begin{array} & \lambda_1 & \lambda_2 & \lambda_3 \end{array}\right)$ and
$\mu = \left(\begin{array} & \mu_1 & \mu_2 & \mu_3 \end{array}\right)$ must be a multiple of $b$ and $c$. Together with above equations we get our final $B$ and $C$.
Faster approach
As $A$ is symmetric we see that $A = A^T = (B+C)^T = B^T + C^T = B+C$ which means that both $B$ and $C$ should be symmetric. As we already have their eigenvectors, we can just construct $B$ and $C$ by using $B = \lambda b\cdot b^T$ and $C = \mu c\cdot c^T$ and simple experimentation gives $\lambda=\mu=1$
Remark: We cannot use $d$ in our matrix because $d^Tb = d^Tc=d^Td=0$. In addition $(d+b)^Tc=0$ and $(d+c)^Tb=0$ which means that $b$ and $c$ cannot be the correct eigenvectors for the matrices $D+B$ or $D+C$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4138465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Solve $\int e^{x^2+x}(4x^3+4x^2+5x+1)dx$ One of the solutions tried to write the integral a
$$\int e^{x^2+x}((2x+1)p(x) + p’(x))dx$$
Where $p(x)=2x^2+bx+c$
I have no idea why they chose the lead coefficient to be $2$.
Going with this, $b=1$ and $c=0$
So
$$\int e^{x^2+x} ((2x+1)(2x^2+x)+4x+1)dx $$
Now even after substituting $x^2+x=t$, the integration still seems to be mess, not in the form of the desired $\int e^t(f(t)+f’(t))dt$ form.
Is there a better solution or a better explanation for this one?
| $$\begin{align}
\int e^{x^2+x}(4x^3+4x^2+5x+1)dx &= \int e^{x^2+x}(4x^3 + 2x^2 + 2x^2 + x + 4x + 1)dx\\
&= \int e^{x^2+x}(2x+1)(2x^2 + x) + e^{x^2+x}(4x+1)dx\\
&= \int e^{f(x)}(f'(x)g(x) + g'(x)) dx
\end{align}$$
where $f(x) = x^2 + x$ and $g(x) = 2x^2 + x$. You should be able to see (integration by parts is one way) that
$$\int e^{f(x)}(f'(x)g(x) + g'(x)) dx = e^{f(x)} g(x) + C$$
To summarize,
$$\int e^{x^2+x}(4x^3+4x^2+5x+1)dx = e^{x^2 + x} (2x^2 + x) + C$$
Comments on your solution:
Great work, you almost completed it! It's worth noting that instead of $\int e^{t} (f(t) + f'(t)) dt$, you really want a polynomial in the exponent. Take $f(x) = x^2 + x$ (the obvious choice) and see where you can get with differentiating $e^{f(x)}g(x)$? That's how I would reverse engineer the solution. Once you know $g(x)$, the game is over.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Condition that satisfies inequality involving fractions and the floor function I want to solve the following inequality for $a$ :
$$\left \lfloor \frac{a}{b} \right \rfloor > \frac{a}{b+1}$$
Where both $a$ and $b$ are strictly positive integers.
My end goal with this is, for a given $b$, to be able to find the smallest $a$ of the form $6 \times 10^{7+p}$ ($p \in \mathbb{Z}$)
If you would like to known more about the context :
I stumbled upon this problem while coding a musical file format converter.
In one file format, the BPM of the song is stored as some integer version of $$\frac{6 \times 10^7}{\textrm{BPM}}$$ which I call the tempo value.
My code internally handles decimal BPMs, so it needs to convert back and forth between the two "representations". I wanted this conversion to be both "idempotent" and "minimal" in some way, "idempotent" as in : if I convert a tempo value to a BPM and back, I get the same value, and "minimal" meaning that the intermediate BPM is stored with the minimum number of decimal places needed to recover the exact tempo value if I just floor the "raw" conversion on the way back.
if $v$ is the tempo value, the "raw" BPM can be computed as $$\textrm{BPM}_\textrm{raw} = \frac{6 \times 10^7}{v}$$ I then need to choose some number of decimal places $p$ to truncate it to, which (thanks wikipedia) can be expressed this way : $$\textrm{BPM}_\textrm{trunc} = \textrm{floor(}\textrm{BPM}_\textrm{raw}, p\textrm{)} = \frac{\left \lfloor 10^p \cdot \textrm{BPM}_\textrm{raw} \right \rfloor}{10^p}$$ And then to convert it back to an integer tempo value, I reuse the formula then floor the result again : $$v_\textrm{recovered} = \left \lfloor \frac{6 \times 10^7}{\textrm{BPM}_\textrm{trunc}} \right \rfloor$$ Plugging every formula together, I want to find a condition on p that satisfies : $$ \begin{align} v &= v_\textrm{recovered} \\ v &= \left \lfloor \frac{6 \times 10^7}{\textrm{BPM}_\textrm{trunc}} \right \rfloor \\ v &= \left \lfloor \frac{6 \times 10^7}{\frac{\left \lfloor 10^p \cdot \textrm{BPM}_\textrm{raw} \right \rfloor}{10^p}} \right \rfloor \\ v &= \left \lfloor \frac{6 \times 10^7}{\frac{\left \lfloor 10^p \cdot \frac{6 \times 10^7}{v} \right \rfloor}{10^p}} \right \rfloor \\ v &= \left \lfloor \frac{6 \times 10^{7+p}}{\left \lfloor \frac{6 \times 10^{7+p}}{v} \right \rfloor} \right \rfloor \\ \end{align} $$
If we rename $v$ to $b$ and $6 \times 10^{7+p}$ to $a$ we need to find a condition on $a$ so that $$b = \left \lfloor \frac{a}{\left \lfloor \frac{a}{b}\right \rfloor}\right \rfloor$$
$b$ is equal to the floor of that fraction if and only if the value of that fraction sits between $b$ and $b+1$ exclusive, which translates to $$ b \leq \frac{a}{\left \lfloor \frac{a}{b}\right \rfloor} < b+1$$ The left part is true, this comes from the fact that $\left \lfloor x \right \rfloor \leq x$ for any $x$, plugging in $\frac{a}{b}$ gives $$\left \lfloor \frac{a}{b} \right \rfloor \leq \frac{a}{b} \Leftrightarrow \frac{a}{\frac{a}{b}} \leq \frac{a}{\left \lfloor \frac{a}{b} \right \rfloor} \Leftrightarrow b \leq \frac{a}{\left \lfloor \frac{a}{b} \right \rfloor}$$ So I just need to find a condition for the right side, which can be rewritten a bit : $$\frac{a}{\left \lfloor \frac{a}{b}\right \rfloor} < b+1 \Leftrightarrow \frac{a}{b+1} < \left \lfloor \frac{a}{b}\right \rfloor$$ And here we are ...
| *
*The floor function obeys$$\mathrm{ bn\le a<b(n+1),y=b, n\in\Bbb N}$$ Graph,
*Using the identity above: $$\mathrm{y_1=b=\left\lfloor\frac ab\right\rfloor>y_2=\frac a{b+1}}$$
*Therefore when we let a vary:$$\mathrm{a>b(b+1)=b^2+b, bn\le a<b(n+1),y=b, a,b,n\in\Bbb N}$$
*Graphical proof
But this only gives the bound when the two graphs no longer cross and $a\ge \left\lfloor \frac ab \right \rfloor$. Modifying this a little bit after doing some pattern analysis of the inequalities satisfying b=1,2,... gives:
$$\mathrm{a>n,b(n+1)\le a <(b+1)(n+1)=bn+b+n+1, a,b,n \in \Bbb N^0}$$
or simply: $$\mathrm{a>n,b(n+1)\le a <(b+1)(n+1)=bn+b+n+1, a,b,n \in \Bbb N^0, n\le b,\ if\ n=b,\ then \ a>b^2,\ or\ otherwise\ undefined}$$
Here is graphical proof of this one.
All in all:$$\mathrm{p>log_{10}(n)-7,0\le a-b(n+1)=10^{7+p}-bn-b<n+1⇔log_{10}(b)+log_{10}(n+1)-7 \le p < log_{10}(b+1)+log_{10}(n+1)-7, a,b,n\in \Bbb N^0=0,1,2,...,p\in\Bbb Z}$$
I will leave it to you to find other forms and edit my solutions.
These formulas even work with $a=x^x$ and other functions.
Please correct me and give me feedback!
| {
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$\int_{1}^\infty(1-\frac{1}{x})^{x^2}dx$ converges or diverges Determine if the integral $\int_{1}^\infty(1-\frac{1}{x})^{x^2}dx$ converges / diverges:
My attempt:
Notice that $\lim_{x \to \infty}(1+\frac{-1}{x})^x=e^{-1}$, so for every $\varepsilon>0$ there exists $M>0$ such that for every $x>M$ we have $|(1+\frac{-1}{x})^x-\frac{1}{e}|<\frac{1}{2}-\frac{1}{e}$ so we have $(1+\frac{-1}{x})^x<\frac{1}{2}$.
So for every $x \ge M$ we have $(1+\frac{-1}{x})^{x^2} \le \frac{1}{2^x}$,
So $\int_{1}^\infty(1-\frac{1}{x})^{x^2}dx \le \int_{1}^\infty\frac{1}{2^x}dx < \infty$, so it converges.
Is my proof correct?
Thanks a lot!
| A continuation of @Gary's observation:
As you have showed, we know that for every $x \ge M$ (for some $M\geq 1$) we have $(1+\frac{-1}{x})^{x^2} \le \frac{1}{2^x}$. Therefore
$$\int_{1}^\infty(1-\frac{1}{x})^{x^2}dx =\int_{1}^M(1-\frac{1}{x})^{x^2}dx+\int_{M}^\infty(1-\frac{1}{x})^{x^2}dx\le \int_{1}^M(1-\frac{1}{x})^{x^2}dx+ \int_{M}^\infty\frac{1}{2^x}dx $$
Since
$$\int_{M}^\infty\frac{1}{2^x}dx < \infty$$
holds, it suffices to show
$$\int_{1}^M(1-\frac{1}{x})^{x^2}dx<\infty.$$
Because $[1,M]$ is closed and bounded, it suffices to show that there exists a $C\geq 0$ with
$$(1-\frac{1}{x})^{x^2}\le C,\quad x\in[1,M].$$
| {
"language": "en",
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Estimate of $\int_y^{+∞}\frac{[2-({\rm erf}(x+a)+{\rm erf}(x-a))]^2}{e^{-(x+a)^2}+e^{-(x-a)^2}}\,\mathrm dx$ $\DeclareMathOperator{\erf}{erf}\DeclareMathOperator{\erfc}{erfc}$I am trying to find either a closed form expression or an $y$-dependent analytical estimate for
$$\int_y^{+\infty}\frac{[2-(\erf(x+a)+\erf(x-a))]^2}{e^{-(x+a)^2}+e^{-(x-a)^2}}\,\mathrm dx.$$
This appeared in my research. It is related to the diffusion equation under a specific potential. All I could come up with is to define
$$f(x)=\erfc(x+a)+\erfc(x−a)$$
and then I see that this integral is
$$-\frac{2}{\sqrt{\pi}}\int_{y}^{+\infty}\frac{f(x)^{2}}{\frac{\mathrm df(x)}{\mathrm dx}}\,\mathrm dx.$$
Any suggestions are welcome!
| $\DeclareMathOperator{\erf}{erf}\def\d{\mathrm{d}}\def\e{\mathrm{e}}\def\paren#1{\left(#1\right)}$Assume $a > 0$. For any $ν \in \mathbb{R}$, by l'Hospital's rule,\begin{gather*}
\lim_{x → +∞} \frac{\displaystyle\int_x^{+∞} t^ν \exp(-t^2) \,\d t}{x^{ν - 1} \exp(-x^2)} = \lim_{x → +∞} \frac{-x^ν \exp(-x^2)}{-(2x^ν - (ν - 1) x^{ν - 2}) \exp(-x^2)}\\
= \lim_{x → +∞} \frac{x^2}{2x^2 - (ν - 1)} = \frac{1}{2},
\end{gather*}
thus when $x → +∞$,\begin{gather*}
1 - \erf(x) = \frac{2}{\sqrt{π}} \int_x^{+∞} \exp(-t^2) \,\d t = \frac{1}{\sqrt{π} x} \exp(-x^2) + o\paren{ \frac{1}{x} \exp(-x^2) }, \tag{1}\\
\int_x^{+∞} \frac{1}{t^2} \exp(-t^2) \,\d t \sim \frac{1}{2x^3} \exp(-x^2). \tag{2}
\end{gather*}
Note that $\exp(-(x + a)^2) = o(\exp(-(x - a)^2))$ ($x → +∞$). Therefore when $x → +∞$,\begin{align*}
&\mathrel{\phantom=} 2 - \erf(x + a) - \erf(x - a)\\
&= (1 - \erf(x - a)) + (1 - \erf(x + a))\\
&\stackrel{(1)}{=} \frac{\exp(-(x - a)^2)}{\sqrt{π} (x - a)} + o\paren{ \frac{\exp(-(x - a)^2)}{x - a} } + \frac{\exp(-(x + a)^2)}{\sqrt{π} (x + a)} + o\paren{ \frac{\exp(-(x + a)^2)}{x + a} }\\
&= \frac{\exp(-(x - a)^2)}{\sqrt{π} (x - a)} + o\paren{ \frac{\exp(-(x - a)^2)}{x - a} } \sim \frac{\exp(-(x - a)^2)}{\sqrt{π} (x - a)},
\end{align*}
which implies that$$
\frac{(2 - \erf(x + a) - \erf(x - a))^2}{\exp(-(x + a)^2) + \exp(-(x - a)^2)} \sim \frac{\paren{ \dfrac{\exp(-(x - a)^2)}{\sqrt{π} (x - a)} }^2}{\exp(-(x - a)^2)} = \frac{\exp(-(x - a)^2)}{π(x - a)^2}.
$$
Therefore when $y → +∞$,\begin{gather*}
\int_y^{+∞} \frac{(2 - \erf(x + a) - \erf(x - a))^2}{\exp(-(x + a)^2) + \exp(-(x - a)^2)} \,\d x\\
\sim \int_y^{+∞} \frac{\exp(-(x - a)^2)}{π(x - a)^2} \,\d x = \frac{1}{π} \int_{y - a}^{+∞} \frac{1}{t^2} \exp(-t^2) \,\d t \stackrel{(2)}{\sim} \frac{\exp(-(y - a)^2)}{2π(y - a)^3}.
\end{gather*}
| {
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Finding Taylor's series of the function: $\frac{e^{a \sin^{-1}x}}{\sqrt{1-x^2}}$ Show that $$\frac{e^{a \sin^{-1}x}}{\sqrt{1-x^2}}=1+\frac{ax}{1!}+\frac{(a^2+1^2)x^2}{2!}+\frac{a(a^2+2^2)x^3}{3!}+\frac{(a^2+1^2)(a^2+3^2)x^4}{4!}+\cdots$$
My attempt:
I integrated the function and got $\frac{e^{a \sin^{-1}x}}{a}$ then I wrote the series of $e^{a \sin^{-1}x}$ but it contained terms like $(\sin^{-1}x)^2$, $(\sin^{-1}x)^3$ and so on so I could not find the series. My idea was to find the series of the anti derivative of the function and then to derivate the obtained series. Any other way to do it?
| We can use your idea of finding the antiderivative. We have:
\begin{align}
e^{a\sin^{-1}(x)}& =1+a\sin^{-1}x+\dfrac{(a\sin^{-1}x)^2}{2!}+...+\dfrac{(a\sin^{-1}x)^5}{5!}+...\\
&=1+a\left(x+\dfrac{x^3}{6}+\dfrac{3x^5}{40}+...\right)+\dfrac{a^2}{2!}\left(x+\dfrac{x^3}{6}+...\right)^2+\dfrac{a^3}{3!}\left(x+\dfrac{x^3}{6}+...\right)^3+\dfrac{a^4}{4!}x^4+\dfrac{a^5}{5!}x^5+...\\
&=1+a\left(x+\dfrac{x^3}{6}+\dfrac{3x^5}{40}+...\right)+\dfrac{a^2}{2!}\left(x^2+\dfrac{x^4}{3}+...\right)+\dfrac{a^3}{3!}\left(x^3+\dfrac{x^5}{2}+...\right)+\dfrac{a^4}{4!}x^4+\dfrac{a^5}{5!}x^5+...\\
&=1+ax+\dfrac{a^2x^2}{2!}+\left(\dfrac{ax^3}{3!}+\dfrac{a^3x^3}{3!}\right)+\left(\dfrac{a^2x^4}{3!}+\dfrac{a^4x^4}{4!}\right)+\left(\dfrac{3ax^5}{40}+\dfrac{a^3x^5}{2\cdot3!}+\dfrac{a^5x^5}{5!}\right)+...\\
&=1+ax+\dfrac{a^2x^2}{2!}+\dfrac{(a+a^3)x^3}{3!}+\dfrac{(4a^2+a^4)x^4}{4!}+\dfrac{(9a+10a^3+a^5)x^5}{5!}+...\\
\end{align}
Thus, $\dfrac{1}{a}e^{a\sin^{-1}(x)}=\dfrac{1}{a}+x+\dfrac{ax^2}{2!}+\dfrac{(1+a^2)x^3}{3!}+\dfrac{(4a+a^3)x^4}{4!}+\dfrac{(9+10a^2+a^4)x^5}{5!}+...$
Finally, take the derivative on both sides and we are done.
| {
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Generating functions in discrete mathematics - How to find coefficient from polynomial with multiplication So I encountered a certain problem when studying generating functions that I seem to be unable to fully solve.
The problem is as follows:
Carol is collecting money from her cousins to have a party for her aunt. If eight of the cousins promise to give 2, 3, 4, or 5 dollars each, and two others each give 5 or 10 dollars, what is the total number of ways Carol can collect exactly 40 dollars?
The solution is the coefficient of the $x^{40}$ term in the generating function
$(x^2 + x^3 + x^4 + x^5)^8(x^5+x^{10})^2$
which may be simplified to finding the coefficient of the $x^{14}$ term in
$(1+x+x^2+x^3)^8(1+x^5) = (\frac{1-x^4}{1-x})^8(1+2x^5+x^{10}) = (1-x^4)^8(1-x)^{-8}(1+2x^5+x^{10})$
This is where I am stuck. To find the coefficient I know that I am suppose to find all possible combinations of how $x^{14}$ can be created by evaluating the polynomial. This would be done quite easily if only the first term $(1-x^4)^8(1-x)^{-8}$ would exist but how I am suppose to do this with the additional term $(1+2x^5+x^{10})$ present? I.e how do I evaluate the different combinations $x^{14}$ may be created and the find the coefficient with the second term present?
There is a solution given in the book, but I do not understand the principle behind it.
| It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in a series. With $A(x)$ a generating function we can write for instance
\begin{align*}
[x^{14}]\left(1+2x^5+x^{10}\right)A(x)=\left([x^{14}]+2[x^9]+[x^{4}]\right)A(x)\tag{1}
\end{align*}
In (1) we apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$. In the current situation we obtain
\begin{align*}
\color{blue}{[x^{14}]}&\color{blue}{\left(1+2x^5+x^{10}\right)\left(1-x^4\right)^8(1-x)^{-8}}\\
&=\left([x^{14}]+2[x^9]+[x^{4}]\right)\left(1-x^4\right)^8(1-x)^{-8}\tag{2}\\
&=[x^{14}]\left(1-\binom{8}{1}x^4+\binom{8}{2}x^8-\binom{8}{3}x^{12}\right)(1-x)^{-8}\\
&\qquad+2[x^{9}]\left(1-\binom{8}{1}x^4+\binom{8}{2}x^8\right)(1-x)^{-8}\\
&\qquad+[x^4]\left(1-\binom{8}{1}x^4\right)(1-x)^{-8}\tag{3}\\
&=\left([x^{14}]-\binom{8}{1}[x^{10}]+\binom{8}{2}[x^6]-\binom{8}{3}[x^{2}]\right)(1-x)^{-8}\\
&\qquad+\left(2[x^9]-2\binom{8}{1}[x^5]+2\binom{8}{2}[x^1]\right)(1-x)^{-8}\\
&\qquad+\left([x^4]-\binom{8}{1}[x^0]\right)(1-x)^{-8}\tag{4}\\
&\,\,\color{blue}{=\binom{-8}{14}(-1)^{14}-\binom{8}{1}\binom{-8}{10}(-1)^{10}}\\
&\qquad\qquad\color{blue}{+\binom{8}{2}\binom{-8}{6}(-1)^6-\binom{8}{3}\binom{-8}{2}(-1)^2}\\
&\qquad\color{blue}{+2\binom{-8}{9}(-1)^9-2\binom{8}{1}\binom{-8}{5}(-1)^5}\\
&\qquad\qquad\color{blue}{+\binom{8}{2}\binom{-8}{1}(-1)^1}\\
&\qquad\color{blue}{+\binom{-8}{4}(-1)^4-\binom{8}{1}\binom{-8}{0}(-1)^0}\tag{5}\\
&=\cdots
\end{align*}
Comment:
*
*In (2) we use the rule we already applied in (1).
*In (3) we expand $\left(1-x^4\right)^8$ up to terms which give a contribution when calculating the coefficient of operator. If for instance $[x^q]$ is given, it is sufficient to expand $\left(1-x^4\right)^8$ up to terms with powers of $x$ less or equal to $q$.
*In (4) we use again the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$ and the linearity of the coefficient of operator.
*In (5) we select the coefficient of $(1-x)^{-8}$ accordingly.
| {
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Line integral , potential function and Gradient theorem
Given the field $F=(\frac {x+y}{x^2+y^2},\frac {-x+y}{x^2+y^2}$ , calculate $\int_C Pdx+Qdy$ where $C$ is the line $y=\frac{x+25}{7}$ that begins at point $(3,4)$ and ends at $(-4,3)$
What I did was:
First I checked if $P_y=Q_x$ and got $P_y=\frac {x^2-y^2-2xy}{(x^2+y^2)^2}$ and $Q_x=\frac {x^2-y^2-2xy}{(x^2+y^2)^2}$ , since they are equal I can look for the potential function so we need ($\phi(x,y)=(P,Q)$
I integrated $\phi$ with respect to $x$ and got $\int \frac {x+y}{x^2+y^2}$=$\frac {\ln(x^2+y^2)}{2}+\arctan(\frac{x}{y})+C(y)$
then I did the derivative of $\phi$ with respect to $y$ and got $\phi_y=\frac{y-x}{x^2+y^2}+C'(y)$
the I comapred $\phi_y=Q$ , and the result was $\frac{y-x}{x^2+y^2}+C'(y)=\frac{-x+y}{x^2+y^2}$ so the answer is $C'(y)=0$ which means $C(y)=C$ just a constant.
according to the gradient theorem $W=\int_c F\cdot dr=\phi(x_1,y_1)-\phi(x_0,y_0)$
$\phi(x_1,y_1)$=$\phi(-4,3)=\frac{\ln(9+16)}{2}+\arctan(\frac{-4}{3})$
$\phi(x_0,y_0)$=$\phi(3,4)=\frac{\ln(9+16)}{2}+\arctan(\frac{3}{4})$
$W=0.6821-1.256=-0.6435$ which is a wrong answer , It should be $\frac{-\pi}{2}$
what am I doing wrong? Thanks for any help and tips!
| Your working is correct but you have some calculation mistakes in the end.
$\phi(-4,3)=\frac{\ln(9+16)}{2}+\arctan(\frac{-4}{3})$
$\phi(3,4)=\frac{\ln(9+16)}{2}+\arctan(\frac{3}{4})$
Line integral is $\phi(-4,3) - \phi(3,4) = \arctan(-\frac{4}{3}) - \arctan(\frac{3}{4})$
$ = - \arctan(\frac{4}{3}) - \arctan(\frac{3}{4}) = - (\frac{\pi}{2} - \arctan(\frac{3}{4})) - \arctan(\frac{3}{4}) = - \frac{\pi}{2}$
| {
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If $p^k m^2$ is an odd perfect number with special prime $p$, then which is larger: $D(p^k)$ or $D(m^2)$? Denote the classical sum of divisors of the positive integer $x$ to be $\sigma(x)=\sigma_1(x)$. Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$. Finally, denote the deficiency of $x$ by $D(x)=2x-\sigma(x)$.
The topic of odd perfect numbers likely needs no introduction.
Let $p^k m^2$ be an odd perfect number with special prime $p$, satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.
Here is my question:
If $p^k m^2$ is an odd perfect number with special prime $p$, then which is larger: $D(p^k)$ or $D(m^2)$?
MY ATTEMPT
Since $p$ is prime, we have
$$1<I(p^k)=\frac{\sigma(p^k)}{p^k}=\frac{p^{k+1}-1}{p^k (p - 1)}<\frac{p^{k+1}}{p^k (p - 1)}=\frac{p}{p-1}.$$
Now, because $p$ is a prime satisfying $p \equiv 1 \pmod 4$, we have the lower bound $p \geq 5$, whereupon we obtain the upper bound
$$I(p^k)<\frac{p}{p-1} \leq \frac{5}{4}.$$
Note that, since $p^k m^2$ is perfect and $\gcd(p,m)=1$, then we have
$$2=I(p^k m^2)=I(p^k)I(m^2) \iff I(m^2) = \frac{2}{I(p^k)}.$$
This implies that we have the lower bound
$$I(m^2) > \frac{2}{(5/4)} = \frac{8}{5},$$
from which we finally get
$$1 < I(p^k) < \frac{5}{4} < \frac{8}{5} < I(m^2) < 2,$$
since $m^2$ is a proper factor of the perfect number $p^k m^2$, and is therefore deficient.
In particular, we have
$$0 < 2 - I(m^2) < 2 - I(p^k) < 1$$
$$0 < \frac{D(m^2)}{m^2} < \frac{D(p^k)}{p^k} < 1$$
$$0 < D(m^2) < \frac{m^2}{p^k}\cdot{D(p^k)} < m^2.$$
Alas, this is where I get stuck.
| Too long to comment :
One can prove that
$$D(m^2)\lt D(p^k)\iff m^2\lt \frac{(p^{k+1}-1)(p^{k+1}-2p^k+1)}{2(p-1)(p^k-1)}$$
Proof :
We have
$$\begin{align}&D(p^k)-D(m^2)
\\\\&=2p^k-\sigma(p^k)-2m^2+\sigma(m^2)
\\\\&=2p^k-\sigma(p^k)-2m^2+\frac{2p^km^2}{\sigma(p^k)}
\\\\&=2p^k-\frac{p^{k+1}-1}{p-1}-2m^2+\frac{2p^km^2(p-1)}{p^{k+1}-1}
\\\\&=\frac{(p-1)(p^{k+1}-1)(2p^k-2m^2)-(p^{k+1}-1)^2+2p^km^2(p-1)^2}{(p-1)(p^{k+1}-1)}
\\\\&=\frac{ p^{2 k + 2}- 2 p^{2 k + 1} + 2 p^k - 1-2(p^{k + 1} - p^k - p +1)m^2}{(p-1)(p^{k+1}-1)}
\\\\&=\frac{(p^{k+1}-1)(p^{k+1}-2p^k+1)-2(p-1)(p^k-1)m^2}{(p-1)(p^{k+1}-1)}\end{align}$$
from which we get
$$\begin{align}D(m^2)\lt D(p^k)&\iff \frac{(p^{k+1}-1)(p^{k+1}-2p^k+1)-2(p-1)(p^k-1)m^2}{(p-1)(p^{k+1}-1)}\gt 0
\\\\&\iff (p^{k+1}-1)(p^{k+1}-2p^k+1)\gt 2(p-1)(p^k-1)m^2
\\\\&\iff m^2\lt \frac{(p^{k+1}-1)(p^{k+1}-2p^k+1)}{2(p-1)(p^k-1)}\end{align}$$
| {
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synthetic divide polynomial $(2x^3+7x^2-13x-3) \div (2x-3)$ I need to figure out how synthetic division works. The problem is $$ (2x^3+7x^2-13X-3) \div (2x-3)$$ I can do the long division.
$$\require{enclose}
\begin{array}{r}
x^2+5x+1 \\
(2x-3) \enclose{longdiv}{2x^3+7x^2-13x-3}\\ \underline{-2x^3+3x^2} \phantom{100000000} \\ 10x-13x \phantom{1000} \\\underline{-10x+15x} \phantom{1000} \\ 2x-3 \\\underline{-2x+3} \\ 0
\end{array}$$
But when I do synthetic division my answer is slightly different;
$$ 3/2 \left[ \begin{array}{r}{2 \phantom{10}7 \phantom{2}-13\phantom{2}-3} \\ \underline{ \phantom{100}3 \phantom{100} 15 \phantom{1000} 3} \\ 2 \phantom{1}10\phantom{1000} 2 \phantom{1000} 0\end{array} \right] = 2x^2 +10x +2$$
I just realized that I could factor out the two from the polynomial after I synthetic divide. The answer then would be; $$2(x^2+5x+1)$$ The only problem I have is the two graphs would be different. The second graph would stretch along the y axis. If I put a 1 in for x then y =14 a
fter synthetic division. Y would = 7 if I put a 1 in for x after I use long division. I am not sure what to do next.
| Since you have taken $3/2$ while doing synthetic division, we can write
$$ 2x^3 +7x^2-13x -3 = (x- \frac32)(2x^2 + 10x + 2) $$
Now rearranging terms in RHS we get
$$2x^3 +7x^2-13x -3 = (2x- 3)(x^2 + 5x + 1)$$ which is the same answer you got when you did the long division.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4164545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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If $a - b = 3$, $a + b + x = 2$, then find the value of $(a - b)[x^3 - 2ax^2 + a^2x - (a + b)b^2]$ After solving the question I got to the conclusion that it has something to do with $(a-b)$ being a factor of the whole polynomial. And since the value of $(a-b)$ is already given we can further solve the equations to get the value of $a$,$b$ and $x$ and substitute the values obtained to find the answer. But I am stuck on finding the value of x using the fact that $(a-b)$ is a factor.
If anyone knows the answer please share it. Thank you
| Let $f(x)=(a-b)[x^3-2ax^2+a^2x-(a+b)b^2]$. Now using the Remainder Theorem, which tells us that if you divide a polynomial $f(x)$ by $(x-y)$ then the remainder is $f(y)$. Now, we know that $(a-b)$ is a factor of $f(x)$ so $x=3$ is a root. Substituting it in $a+b+x=2$ we get $$a+b+3=2$$ $$\implies a+b=-1$$ Now add $a-b=3$ and $a+b=-1$ getting $$a-b+a+b=3-1$$ $$\implies 2a=2$$ $$\implies a = 1$$. Now substitute the value of $a$ in $a-b=3$ we get $b=-2$. Now putting all these three values in $f(x)$, we get
$$f(x)=(1-(-2))[3^3-2\times 1 \times 3^2 + 1^2 \times 3 - (1-2)(-2)^2]$$
$$=(1+2)[27-18+3-(-4)]$$
$$=3(27-18+3+4)$$
$$=3(34-18)$$
$$=3 \times 16$$
$$=48$$
Hence the value of $f(x)$ is $48$
Hope it helps
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4165087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Use Implicit Differentiation to find $\frac{d^2y}{dx^2}$? Given a system of equation,
\begin{align*}
x &= t^2 + 2t \\
y &= 3t^4 + 4t^3
\end{align*}
I want to find $\frac{d^2 y}{dx^2}$ at $(x,y) = (8, 80)$. Then, $\partial_x(y) = \frac{d y}{dt} \frac{dt}{dx}$. By chain rule,
\begin{align*}
\partial_x^2(y) &= \partial_x \left(\frac{d y}{dt}\right)\frac{dt}{dx} + \frac{dy}{dt} \partial_x \left(\frac{dt}{dx}\right) \\
&= \frac{d^2 y}{dt^2}\left(\frac{dt}{dx}\right)^2 + \frac{dy}{dt}\frac{d^2t}{dx^2}
\end{align*}
Here, how do I find $\frac{d^2 t}{dx^2}$?
| Just for your information since you will need it quite often with differential equation problem
$$\large\frac{d^2 t}{dx^2}=-\frac {\frac{d^2 x}{dt^2} } {\Big[\frac{d x}{dt} \Big]^3}$$
Just checking
$$x=\tan(t) \implies \frac{d x}{dt}= \sec ^2(t)\implies \frac{d^2 x}{dt^2}=2 \tan (t) \sec ^2(t)\implies \frac{d^2 t}{dx^2}=-2 \sin (t) \cos ^3(t)$$
$$t=\tan^{-1}(x) \implies \frac{d t}{dx}=\frac{1}{1+x^2}\implies \frac{d^2 t}{dx^2}=-\frac{2 x}{\left(1+x^2\right)^2}=-2 \sin (t) \cos ^3(t)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4166548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Lower bound on probability of choosing odd integer out of $2m$ integers
The probability of choosing an integer ‘$k$’ out of $2m$ integers
$1,2,3,…2m$ is inversely proportional to $k^4$.If $P_1$ is the probability that chosen number is odd and $P_2$
is the probability that chosen number is even, then
A. $P_1 < \frac{2}{3}$
B. $P_1 > \frac{1}{2}$
C. $P_1 > \frac{2}{3}$
D. $P_2 < \frac{1}{2}$
It is obvious that B and D are true, since probability of choosing odd number is greater than choosing even number. After trying on for some values of $m$, I feel C is also true, although the official answer is BD. The value of $P_1$ will be $\sum_{r=1} ^ m \frac{x}{(2r-1)^4}$ where $x= \frac{1}{\sum_{k=1}^{2m} k^{-4}}$.
I tried to prove C by taking $\displaystyle{\lim_{m \to \infty}} P_1$ but couldn't compute it. Wolfram Alpha says that it's equal to $\frac{15}{16}$. Is there some other way to prove option C?
| For $m=1$ you have $P_1 = \frac{\frac1{1^4}}{\frac1{1^4}+\frac1{2^4}} = \frac{16}{17}> \frac23$
Now consider the position as $m \to \infty$.
Let $S_\infty= \sum\limits_{k=1}^\infty \frac{1}{k^4}$ so $\frac{S_\infty}{2^4} = \sum\limits_{k=1}^\infty \frac{1}{(2k)^4}$ and $P_1 \to \frac{S_\infty-\frac{S_\infty}{2^4}}{S_\infty} = \frac{15}{16}> \frac23$ as $n$ increases
This suggests that $C$ is true. For a proof, you can adjust the previous part:
Let $S_m= \sum\limits_{k=1}^{2m} \frac{1}{k^4}$ so $\frac{S_m}{2^4} = \sum\limits_{k=1}^{2m} \frac{1}{(2k)^4} > \sum\limits_{k=1}^{m} \frac{1}{(2k)^4}$ so for all $m$ we have $$P_1 = \frac{\sum\limits_{k=1}^{2m} \frac{1}{k^4}- \sum\limits_{k=1}^{m} \frac{1}{(2k)^4} }{\sum\limits_{k=1}^{2m} \frac{1}{k^4}} > \frac{S_m-\frac{S_m}{2^4}}{S_m} = \frac{15}{16}> \frac23.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $a$ and $b$ such that $x^3-4ax^2-8bx-3+a+b$ has a triple root $$P(x)=x^3-4ax^2-8bx-3+a+b$$
NOTE: a & b are both real and the solution is not necessarily unique
So far i've done this:
If $P(x)$ is a $3$rd degree polynomial and has a triple root, then it has the form:
$$
P(x)=(x-n)^3=x^3-3x^2 n+3xn^2-n^3.
$$
Then
$$
x^3-3x^2 n+3xn^2-n^3=x^3-4ax^2-8bx+(-3+a+b).
$$
And from this, I got three equations:
$$-3n=-4a$$
$$3n^2=-8b$$
$$-n^3=(-3+a+b)$$
I don't know what to do from here. Is my reasoning correct?
| Doing the same as @Soumya Patel, the cubic equation
$$8k^3−3k^2+6k−24=0$$ has only one real root (the discriminant being negative).
Using the hyperbolic method for one real root, we have
$$k=\frac{1}{8} \left(1+2 \sqrt{15} \sinh \left(\frac{1}{3} \sinh
^{-1}\left(\frac{149}{3 \sqrt{15}}\right)\right)\right)$$
Now,
$$a=\frac 3 4 k \quad \implies \quad a=\frac{3}{32} \left(1+2 \sqrt{15} \sinh \left(\frac{1}{3} \sinh
^{-1}\left(\frac{149}{3 \sqrt{15}}\right)\right)\right)$$
$$b=-\frac{3 }{8}k^2 \quad \implies \quad b=-\frac{3}{512} \left(1+2 \sqrt{15} \sinh \left(\frac{1}{3} \sinh
^{-1}\left(\frac{149}{3 \sqrt{15}}\right)\right)\right)^2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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If $x^{19}+x^{17}+x^{13}+x^{11}+x^{7}+x^{5}+x^{3}$ is divided by $(x^2 +1)$, then find the remainder
If the polynomial $x^{19}+x^{17}+x^{13}+x^{11}+x^{7}+x^{5}+x^{3}$
is divided by $(x^
2
+1)$, then the remainder is:
How Do I solve this question without the tedious long division?
Using remainder theorem , we can take $x^3$ common and put $x^2 =-1$ although $x$ is not a real number. By this method, I got the right answer as $-x$. Is it the right way? Because $x$ comes out to be $i$ which is not real.
Also , can I apply remainder theorem to quadratic divisor polynomials in this way?
| With $y:=x^2$,$$P(x)=x^{19} +x^{17} +x^ {13} +x^ {11} +x^ 7 +x^ 5 +x^ 3=(y^9+y^8+y^6+y^5+y^3+y^2+y)x$$ and you are asked the remainder of the division by the polynomial $x^2+1=y+1$.
You can write
$$P(x)=(Q(y)(y+1)+R(y))x$$ where $R$ is of degree $0$, i.e. a constant (which you know how to compute by division by a linear binomial).
From this
$$P(x)=Q(x^2)(x^2+1)x+Rx$$ and the remainder is $Rx$.
Note that at no time do you have to deal with the solutions of $x^2+1=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4182764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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If ($ACBD$) $= -1$, show that $DA \cdot DB = DC \cdot DO$, where O is the midpoint of the segment $AB$.
If ($ACBD$) $= -1$, show that $DA \cdot DB = DC \cdot DO$, where O is the midpoint of the segment $AB$.
My attempt to proof:
Suppose ($ACBD$) $= -1$ then $\frac{AC}{CB} = \frac{AD}{BD}$ then $AC \cdot BD = CB \cdot AD$. Also, since $O$ is a midpoint for $AB$ then $AO = OB$.
\begin{align*}
AC \cdot BD &= CB \cdot AD \\
\Longrightarrow (AD + DC) \cdot BD &= (CD + DB) \cdot(AO + OD) \\
\Longrightarrow AD \cdot BD + BD \cdot DC &= CD \cdot AO + DB \cdot OD + DB \cdot AO + CD \cdot OD \\
\Longrightarrow AD \cdot BD - CD \cdot OD &= CD \cdot AO + DB \cdot DC + DB \cdot OD + DB \cdot AO \\
&= CD \cdot AO + DB \cdot DC + DB \cdot AD \\
&= CD \cdot AO + DB \cdot AC \\
&= (CA + AD) \cdot AO + (DA + AB) \cdot AC \\
&= CA \cdot AO + AD \cdot AO + DA \cdot AC + AB \cdot AC \\
&= AC \cdot (AB - AO) + DA \cdot (AC - AO) \\
&= AC(OB) + DA(OC)
\end{align*}
I'm stuck in this. Is my proof wrong? how to make the right hand side $0$?
| Rephrasing of problem statement
Let $ A C = a, CB = b, BD = c$.
We are given that $ac = (a+b+c)b$.
WTS $(a+b+c)c = (b+c)(c+ (a+b)/2)$.
Taking the difference of the 2 sides, and multiplying by 2
WTS $ -ab + ac - b^2 - bc = 0$.
Notice that this is exactly the condition $ac = (a+b+c)b$. Hence we are done.
Note: If so desired, you can rephrase this in terms of $AC, CB, BD$, and use only those line segments, to make it clear what we should be focusing on.
Introducing line segments like $OA, OC$ etc makes it harder to manipulate, ad we should just replace them by $OA = (AC+CB)/2$, etc.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$\frac{1-a^3}{a}=\frac{1-b^3}{b}=\frac{1-c^3}{c}, a \neq b \neq c$; What is $a^3+b^3+c^3$?
$\frac{1-a^3}{a}=\frac{1-b^3}{b}=\frac{1-c^3}{c}\\a \neq b \neq c\\a^3+b^3+c^3=$
Since $\frac{1-a^3}{a}=\frac{1-b^3}{b}$, $a^3=b^2a-\frac{a}{b}-1$.
$\therefore a^3+b^3+c^3=ab^2+bc^2+ca^2-\frac{ab+bc+ca}{abc}-1$
Since $a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc$,
$ab^2+bc^2+ca^2-\frac{ab+bc+ca}{abc}-1=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc$
(stuck here)
| It looks like you don't need to do any calculations.
We can construct a new cubic equation with roots $a^3,b^3,c^3.$
$$a^3=x_1,~ b^3=x_2,~ c^3=x_3$$
$$\begin{align}r=\frac{1-a^3}{a}=\frac{1-b^3}{
b}=\frac{1-c^3}{c}\end{align}$$
$$\begin{align}&\implies1-a^3=ar \\
&\implies(1-a^3)^3=a^3r^3 \\
&\implies(1-x)^3=xr^3 \\
&\implies1-3x+3x^2-x^3-xr^3=0 \\
&\implies x^3-3x^2+x(r^3+3)-1=0 \\
&\implies x_1+x_2+x_3=3 \\
&\implies a^3+b^3+c^3=3.\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4188295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Polynomial Inequality Holds for All Real Numbers x Problem: The quadratic polynomial $P(x),$ with real coefficients, satisfies
$$P(x^3 + x) \ge P(x^2 + 1)$$ for all real numbers $x.$ Find the sum of the roots of $P(x).$
Work: Since $P(x)$ is a quadratic polynomial, let $P(x)=ax^2+bx+c$. Then, we have $$a(x^3+x)^2+b(x^3+x)+c \ge a(x^2+1)^2+b(x^2+1)+c \implies a(x^3+x)^2+b(x^3+x) \ge a(x^2+1)^2+b(x^2+1).$$ Since we need to find the sum of the roots of $P(x)$, we need the value $-\frac{b}{a}$ by Vieta's Formulas. I then tried to manipulate using that form, but got nothing. Can somebody finish off this solution, or take another path of their own to solve this problem? Thanks!
| You have
$$(x^3 - x^2 + x - 1) (a(x^3 + x^2 + x + 1) +b) \ge 0.$$
So, for any $x > 1$, $a(x^3+x^2+x+1)+b \ge 0$ and for any $x < 1$, $a(x^3+x^2+x+1)+b \le 0$.
It follows from continuity that $a(1^3+1^2+1+1) + b = 0$, that is, $-b/a = 4$.
You can further check that $x^2-4x$ satisfies the inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4189822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding $a$ and $b$ such that the graph of $y=-2\sin3x+1$ passes through $(-\frac\pi{12},a)$ and $(b,1)$ Please look at the attached image through the link. I already figured out $b$, which I'm pretty sure is $2\pi/3$. However, I'm confused on figuring out $a$. This is what I tried:
\begin{align*}
y & = -2\sin(3x)+1\\
y & = -2\sin\left(3 \cdot -\frac{\pi}{12}\right)+1\\
y & = -2\sin\left(-\frac{\pi}{4}\right)+1
\end{align*}
I used the unit circle to find $\sin\left(-\frac{\pi}{4}\right)$.
\begin{align*}
y & = -2\left(-\frac{\sqrt{2}}{2}\right)+1\\
y & = \sqrt{2} + 1\\
y & = 2.41421356
\end{align*}
My answer isn't correct, so could anyone please explain how I should solve for $b$? Thanks!
| To find b simply let $1 = -2\sin(3x) +1$ and it is when $\sin3x = 0$ for the some $\frac{\pi}{4} < x < \frac{\pi}{2}$
Since $\arcsin(0) \in (0, \pi, 2\pi...) $ you may obtain $x = 0, \frac{\pi}{3}, \frac{2\pi}{3}...$ from which clearly $\frac{\pi}{3}$ is the desired value
To find a is a similiar method, let $a = 1 - 2\sin(-\frac{\pi}{4})$ such that $ a = 1 + \sqrt{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Showing $\int_{0}^{2\pi}\cos(x)\cos(2x)\cos(3x)\,dx = \frac\pi2$ An integral from MIT Integration Bee:
Show that
$$I = \int_{0}^{2\pi}\cos(x)\cos(2x)
\cos(3x)\,dx = \frac\pi2$$
This integral appeared in the 2019 paper. Below is my own solution:
$$\begin{align}
I &= \int_{0}^{2\pi}\cos(x)\cos(2x)(\cos x\cos2x-\sin x\sin2x)\,dx \\[6pt]
&= \int_{0}^{2\pi}\cos^2(x)\cos^2(2x) \,dx -\int_{0}^{2\pi}\cos(x)\cos(2x)\sin(x)\sin(2x)\, dx
\end{align}$$
Replacing $\cos^2(x)= \frac{1+\cos(2x)}{2}$ for the first integral and $\sin(x)\cos(x)= \sin(2x)/2 $ for the second, we get
$$\begin{align}
&\int_{0}^{2\pi}\frac{1+\cos(2x)}{2}\cos^2(2x) dx-\frac{1}{2}\int_{0}^{2\pi}\cos(2x)\sin^2(2x) dx \\[6pt]
=\; &\frac{1}{2}\left(\int_{0}^{2\pi}\cos^2(2x)dx \, + \int_{0}^{2\pi}\cos(2x)\cos(4x)dx \right) \\[6pt]
=\; &\frac{1}{2} \left( \pi + 0\right) \qquad \text{$\because$ the orthogonality of $\cos(mx)$} \\[6pt]
=\; &\frac\pi2
\end{align}$$
This solution is rather awkward, and I'm sure there's a better and faster approach to this integral. Could anyone provide a more elegant solution(or a sketch of it)? Thanks.
| Another method:
$$I=\int_0^{2\pi} \cos x\cos 2x(\cos 3x+\cos x-\cos x)\, dx=\\
\int_0^{2\pi} \cos x \cos 2x (2\cos 2x\cos x-\cos x)\, dx=\\
\int_0^{2\pi} \cos^2x\cos 2x(2\cos2x-1)\,dx=\\
\int_0^{2\pi} 2\cdot \frac{1+\cos 2x}{2}\cdot \frac{1+\cos 4x}{2}\,dx- \frac{1+\cos 2x}{2}\cdot \cos 2x\,dx=\\
\int_0^{2\pi} \frac12+\frac12\cos 4x+\frac12\cos 2x\cos 4x-\frac12\cdot \frac{1+\cos 4x}{2}\,dx=\\
\pi+0+0-\frac{\pi}{2}-0=\frac{\pi}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 7,
"answer_id": 6
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Question on Series Transformations from "The Harmonic Series and Logarithmic Intragrals" This is from an example in Section 1.7 of the book, "The Harmonic Series and Logarithmic Integrals" by, Ali Shadhar Olaikhan
$$\sum_{n=0}^\infty \frac{-1^n}{(2n+1)^3} = \mathcal{J}\sum_{n=1}^\infty \frac{i^n}{n^3}$$
I don't understand how the left side of this example has any Complex terms. It appears to be entirely Real. What am I not understanding, please?
| \begin{align*}
\Im \sum_{n=1}^\infty \frac{\mathrm{i}^n}{n^3}
&= \Im \left( \frac{\mathrm{i}^1}{1^3} + \frac{\mathrm{i}^2}{2^3} + \frac{\mathrm{i}^3}{3^3} + \frac{\mathrm{i}^4}{4^3} + \cdots \right) \\
&= \Im \frac{\mathrm{i}^1}{1^3} + \Im \frac{\mathrm{i}^2}{2^3} + \Im \frac{\mathrm{i}^3}{3^3} + \Im \frac{\mathrm{i}^4}{4^3} + \cdots \\
&= \frac{1}{1^3} + 0 + \frac{-1}{3^3} + 0 + \cdots \\
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4192060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\ln x\ln(1-x)\le \ln^2 2$. Define $f(x)=\ln x\ln(1-x)$ where $x \in (0,1)$. Note that $f(x)$ is symmetric with respect to $x=\frac{1}{2}$.Thus we may only study the range of $f(x)$ over $\left(0,\frac{1}{2}\right]$.
Now, by differentiating we obtain $$f'(x)=\frac{(1-x)\ln(1-x)-x\ln x}{x(1-x)}.$$
Obviously, $f'(x)$ has a zero at $x=\frac{1}{2}$. But can we conclude that this is unique？
| My second solution:
It suffices to prove that, for all $x \in (0, 1/2]$,
$$\ln x \ln (1 - x) \le \ln^2 2.$$
Let $f(x) = \ln x \ln (1 - x) - \ln^2 2$. We have
$$f'(x) = \frac{(1 - x)\ln (1 - x) - x\ln x}{x(1 - x)}.$$
Since $u\mapsto \ln u$ is concave on $(0, \infty)$, we have (see the remarks)
$$\frac{1 - 2x}{1 - x}\ln \frac{1}{x} + \frac{x}{1 - x}\ln 1 \le \ln\left(\frac{1 - 2x}{1 - x} \cdot \frac{1}{x} + \frac{x}{1 - x} \cdot 1\right) = \ln\frac{1 - x}{x}$$
which results in $(1 - x)\ln(1 - x) - x\ln x \ge 0$.
Thus, $f'(x) \ge 0$. Also, $f(1/2) = 0$.
Thus, $f(x) \le 0$ for all $x\in (0, 1/2]$.
We are done.
Remarks: i) Here we used @trancelocation's nice idea in How do I show this inequality that $\frac{\ln({x+1})}{\ln({x})} \leq \frac{x}{x-1}$ for $x>1$?.
ii) Alternatively, using @Quanto's nice idea in the link above, we have
\begin{align*}
(1 - x)\ln(1 - x) - x\ln x
= x\int_x^{1/2} \frac{1}{u^2}\int_0^u \frac{1 - 2v}{1 - v} \mathrm{d} v \mathrm{d}u \ge 0.
\end{align*}
| {
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"url": "https://math.stackexchange.com/questions/4192684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 3
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A question about solving quotients all. I ran into this problem and was wondering where my logic failed. I was solving an absolute value involving a quotient and went about it the following way:
$$
\begin{eqnarray}
\lvert \frac{x+1}{x-2} \rvert &<& 3\\
\Rightarrow \frac{x+1}{x-2} &>& -3 \hspace{1mm}and\hspace{1mm}\frac{x+1}{x-2} < 3\\
x+1&>&-3(x-2)\hspace{2cm}\text{(Solving for the left inequality)}\\
x+1&>&-3x+6\\
4x&>&5\\
x&>&\frac{5}{4}
\end{eqnarray}
$$
which cannot be true, as $x$ can equal $2$ with these restrictions which is obviously not allowed. Doing the left side the correct way:
$$
\begin{eqnarray}
\frac{x+1}{x-2} &>& -3\\
\frac{x+1}{x-2} + 3 &>& 0\\
\frac{x+1}{x-2} + \frac{3x-6}{x-2} &>& 0\\
\frac{4x-5}{x-2} &>& 0\\
\Rightarrow x < \frac{5}{4}\lor x>2\\
\end{eqnarray}
$$
leads to the solution set $(-\infty,\frac{5}{4})\cup(2,\infty)$. My question is, what happened with the first method where I failed to come up with a solution for $(2,\infty)$, (and why the signage for $x>\frac{5}{4}$ is backwards/incorrect in the first example, does the multiplication by -3 reverse the inequality even if I'm not introducing a new negative term to one side?). I assume since you cannot divide by zero, you're not allowed to multiply the denominator to the other side without restrictions, but I am not quite sure. Thank you, I greatly appreciate it.
| HINT
Since both sides are nonnegative, you can square them in order to obtain an equivalent inequation:
\begin{align*}
\left|\frac{x+1}{x-2}\right| < 3 & \Longleftrightarrow \left(\frac{x+1}{x-2}\right)^{2} < 9\\\\
& \Longleftrightarrow \frac{(x^{2} + 2x + 1) - 9(x^{2} - 4x + 4)}{(x-2)^{2}} < 0\\\\
& \Longleftrightarrow \frac{-8x^{2} + 38x - 35}{(x-2)^{2}} < 0\\\\
& \Longleftrightarrow
\begin{cases}
8x^{2} - 38x + 35 > 0\\\\
x\neq 2
\end{cases}
\end{align*}
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If A is the 4 by 4 matrix of ones, find the eigenvalues and the determinant of A−I So I want to find the eigen values and eigen vectors of a matrix with all 1's
\begin{bmatrix}1&1&1&1\\1&1&1&1\\1&1&1&1\\1&1&1&1\end{bmatrix}
Only 1 independent would be left,
\begin{bmatrix}1&1&1&1\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix}
Now, Let's assume λ= 1.
A-λI would give me,
\begin{bmatrix}0&1&1&1\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix}
With eigen values as, -1,-1,-1 and 3.
Ultimately, λ1=3 and λ2=-1.
But this is something I have assumed, how can I get eigen values and its vectors by a method? Or what steps should I take ahead? Thanks in advance
| HINT
To begin with, obtain the characteristic polynomial:
\begin{align*}
\det(A - \lambda I) & =
\begin{vmatrix}
1 - \lambda & 1 & 1 & 1\\
1 & 1 - \lambda & 1 & 1\\
1 & 1 & 1 - \lambda & 1\\
1 & 1 & 1 & 1 - \lambda
\end{vmatrix}\\\\
& =
\begin{vmatrix}
4 - \lambda & 1 & 1 & 1\\
4 - \lambda & 1 - \lambda & 1 & 1\\
4 - \lambda & 1 & 1 - \lambda & 1\\
4 - \lambda & 1 & 1 & 1 - \lambda
\end{vmatrix}\\\\
& =
(4-\lambda)\begin{vmatrix}
1 & 1 & 1 & 1\\
1 & 1 - \lambda & 1 & 1\\
1 & 1 & 1 - \lambda & 1\\
1 & 1 & 1 & 1 - \lambda
\end{vmatrix}\\\\
& =
(4-\lambda)\begin{vmatrix}
1 & 0 & 0 & 0\\
1 & -\lambda & 0 & 0\\
1 & 0 & -\lambda & 0\\
1 & 0 & 0 & - \lambda
\end{vmatrix}\\\\
& = (\lambda - 4)\lambda^{3}
\end{align*}
whence we get the eigenvalues $\lambda\in\{0,4\}$.
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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In a right triangle, the perimeter is equal to 30. How many integer values can the hypotenuse take? (Answer:2)
I did:
$a+b+h = 30\rightarrow a+b = h-30\\
a^2+b^2 = h^2 \rightarrow h = \sqrt{a^2+b^2}\\
a-b<h<a+b \rightarrow a-b < h < \sqrt{a^2+b^2}-30...\\
\text{I didn't find other relationships...}
$
| There are indeed not enough restrictions. We can write $\cos(\theta) = \frac a h$, $\sin(\theta) = \frac b h$, hence $h\cos(\theta)+h\sin(\theta)+h=30$, or $\cos(\theta)+\sin(\theta)+1 = \frac{30}h$.
$\cos(\theta)+\sin(\theta)$ is equal to $\sqrt 2\sin(\theta +\frac{\pi}4)$, so we get $\sqrt 2 \sin(\theta +\frac{\pi}4) = \frac2 {\sqrt 2}(\frac{30}h-1)$ or $\theta = \sin^{-1}(\frac2 {\sqrt 2}(\frac{30}h-1))-\frac{pi}4$, which has solutions for more than two values of $h$.
If we have the requirement that all sides are positive integers, there's only one solution.
All Pythogorean triples are of the form $2kmn$, $k(m^2-n^2)$, $k(m^2+n^2)$ for some positive integers $k,n,m$. $k$ has to be a factor of $30$, so that gives $1,2,3,5,6,10,30$. The smallest Pythogorean triplet is $3,4,5$, which adds up to $12$, for $k=2.5$, which doesn't work because $k$ needs to be an integer. All larger $k$ are prohibited (a larger $k$ would means that the reduced triples is smaller, but $3,4,5$ is smallest). The only smaller $k$ is $1$, so we now know that any solution must be of the form $2mn$, $m^2-n^2$, $m^2+n^2$. Adding those together, we get $2m^2+2mn=2m(m+n)$, thus $m(m+n)=15$. The only factorizations of $15$ are $15\cdot1$ and $5\cdot3$. The factorization $15\cdot1$ would require $m=1$, which doesn't work ($m>n$, so if $m=1$, $n=0$, but $m$ and $n$ both have to be positive), so that leaves $5\cdot3$. Since $m+n>m$, the $3$ must be $m$ and the $5$ is $m+n$, so $n=2$. That gives $5,12,13$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\ \ \frac{\ 9}{\left(a+b+c\right)^2}$.
Let $a,b,c$ be positive real numbers. Prove that $$\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\ \ \frac{\ 9}{\left(a+b+c\right)^2}.$$
I want to prove the inequality with elementary inequalities ( Mean inequalities, Cauchy-Schwarz etc.). I tried to use Cauchy-Schwarz, but I got $$\frac{\ 1}{a^2+b^2+ab}+\frac{\ 1}{b^2+c^2+bc}+\ \frac{\ 1}{c^2+a^2+ca}\ge\frac{9}{2(a^2+b^2+c^2)+ab+bc+ca}.$$
Then we have to show that $$2(a^2+b^2+c^2)+ab+bc+ca\leq (a+b+c)^2 $$
which is not true.
Multiplying the numerators of the fractions in LHS by $a^2,b^2,c^2$ also makes the problem more difficult.
So, how to solve the problem with elementary inequalities?
| Multiplying the numerators of the fractions by $a^2$, $b^2$ and $c^2$ gives a wrong inequality:
$$\sum_{cyc}\frac{1}{b^2+c^2+bc}=\sum_{cyc}\frac{a^2}{a^2(b^2+c^2+bc)}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(2a^2b^2+a^2bc)}$$ and we need to prove that:
$$(a+b+c)^4\geq9\sum\limits_{cyc}(2a^2b^2+a^2bc),$$ which is wrong for $c\rightarrow0^+$ and $a=b=1$.
But the following C-S helps already:
$$\sum_{cyc}\frac{1}{b^2+c^2+bc}=\sum_{cyc}\frac{(3a+b+c)^2}{(3a+b+c)^2(b^2+c^2+bc)}\geq\frac{25(a+b+c)^2}{\sum\limits_{cyc}(3a+b+c)^2(b^2+c^2+bc)}$$ and it's enough to prove that:
$$25(a+b+c)^4\geq9\sum\limits_{cyc}(3a+b+c)^2(b^2+c^2+bc)$$ or
$$\sum_{cyc}(7a^4+19a^3b+19a^3c-48a^2b^2+3a^2bc)\geq0$$ or
$$3\sum_{cyc}(a^4-a^3b-a^3c+a^2bc)+\sum_{cyc}(4a^4+22a^3b+22a^3c-48a^2b^2)\geq0,$$ which is true by Schur and Muirhead.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Determine whether the series absolutely converge, conditionally converge or diverge The series is $\sum_{n=1}^{\infty}\frac{n^2+1}{n^5-n^4+3n}$.
In my book I just see examples and exercises for determining whether the series absolutely converge, conditional converge or diverge in alternating series.
This series is not alternating. So I want to make sure my analysis is right.
I have these rules:
The series $\sum a_n$ is:
*
*Absolutely convergent if $\sum |a_n|$ converges.
*Conditionally convergent if $\sum |a_n|$ diverges but $\sum a_n$ converges.
*Divergent if $\sum |a_n|$ diverges but $\sum a_n$ also diverges.
Let me know if the resume of these rules is OK.
Following these rules then:
$\sum_{n=1}^{\infty}|\frac{n^2+1}{n^5-n^4+3n}|=\sum_{n=1}^{\infty}\frac{n^2+1}{n^5-n^4+3n}$
I'm using the Direct Comparison Test:
$\frac{n^2+1}{n^5-n^4+3n}<\frac{n^2}{n^5}=\frac{1}{n^3}$
Then:
$\sum_{n=1}^{\infty}\frac{1}{n^3}$ converges because $p=3>1$ by $p$-series.
Therefore by the Direct Comparison test $\sum \frac{n^2+1}{n^5-n^4+3n}$ also converges.
Therefore $\sum \frac{n^2+1}{n^5-n^4+3n}$ is absolutely convergent.
Am I right?
Thanks in advance for your time.
| First off, your series isn't alternating, so you cannot use the Alternating Series Test. The terms "absolutely\conditionally converge" ONLY apply to alternating series. If your series is not alternating, it either just converges or diverges. These terms don't apply. You could say it's absolutely convergent, because the absolute value doesn't affect the value of the sum, but this is rather unnecessary.
Notice you have a polynomial of degree $2$ on the numerator, and a polynomial of degree $5$ in the denominator. So, as $n$ gets very large, this fraction looks more and more like $\frac{n^2}{n^5} = \frac{1}{n^3}$. Here is a graph with $\frac{1}{n^3}$ in red and $\frac{n^2+1}{n^5-n^4+n}$ in blue. Notice the functions converge to zero together as $n$ gets large. This is the idea behind the Limit Comparison Test, which goes like this:
Suppose we have two series $\sum a_n$ and $\sum b_n$, and let $L = \lim_{n\to\infty} \frac{a_n}{b_n}$. If $L$ is finite and positive, then $a_n$ and $b_n$ either BOTH converge or BOTH diverge. Otherwise, we cannot come to any conclusion.
So, take $a_n$ = $\frac{1}{n^3}$ and $b_n = \frac{n^2+1}{n^5-n^4+n}.$ Then,
$$\frac{a_n}{b_n} = \frac{\frac{1}{n^3}}{\frac{n^2+1}{n^5-n^4+n}} = \frac{n^5 - n^4 + 1}{n^5+n^3}$$
And, when we take the limit, we see that
$$L = \lim_{n\to\infty} \frac{a_n}{b_n} = \lim_{n\to\infty} \frac{n^5 - n^4 + 1}{n^5+n^3} = 1$$
Because $L=1$, it is positive and finite, and we conclude that BOTH $\sum\frac{1}{n^3}$ and $\sum\frac{n^2+1}{n^5-n^4+n}$ converge, or both diverge. Because $\sum\frac{1}{n^3}$ converges, we know that our given sum, $\sum\frac{n^2+1}{n^5-n^4+n}$, must also converge.
| {
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"source": "stackexchange",
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Given a function, find values of when it is equal to $100$. Question: Let $f(n) = 2n - [\frac{(1 + \sqrt{8n – 7})}{2}]$ for each positive integer, where $[x]$ is denoted for the largest integer less than or equal to $x$ for any real number $x$. Find the sum of all possible value(s) $n$ such that $f(n) = 100$.
My solution:
If $f(n) = 100$ then $2n - [\frac{1 + \sqrt{8n – 7}}{2}] = 100$. It is also known that $[x] = \lfloor{x}\rfloor$, so I will be using $\lfloor{x}\rfloor$ throughout the rest of my answer.
First of all, $2n - [\frac{1 + \sqrt{8n – 7}}{2}] = 2n - \lfloor(2 + \frac{\sqrt{8n – 7}}{2})\rfloor$.
So, since $2n - \lfloor(2 + \frac{\sqrt{8n – 7}}{2})\rfloor = 100$, and $\lfloor{x}\rfloor \approx x$, we want $2n - (2 + \frac{\sqrt{8n – 7}}{2}) \approx 100$
Solving this equation, we get $2 + \frac{\sqrt{8n - 7}}{2} \approx 100 - 2n$. So, $\sqrt{8n - 7} \approx 196 - 4n.$ Since $\sqrt{8n - 7}$ is positive, $196 - 4n$ is also positive. So, $n > 49$.
Testing values of $n$, we see that $n = 55$ works. If $n > 55$, then $f(n) > 100$. This is because $2n$ grows much bigger than $\sqrt{8n - 7}$ when $n > 55$.
Hence, there is only one value of $n$, which is $55$, so the answer is $\boxed{55}$.
Could anyone review my solution for a better solution or if there are any mistakes in my solution? Feel free to write a better answer in the answers section and thanks in advance!
| $$2n- \left\lfloor \frac{1+\sqrt{8n-7}}2\right\rfloor= 100$$
$$2n-100 =\left\lfloor \frac{1+\sqrt{8n-7}}2\right\rfloor $$
$$ \frac{\sqrt{8n-7}-1}2\le 2n-100 \le \frac{1+\sqrt{8n-7}}{2}$$
$$4n-200-1 \le \sqrt{8n-7} \le 4n-200 + 1$$
$$(4n-201)^2 \le 8n-7 \le (4n-199)^2$$
The first inequality is satisfied for $$46 \le n \le 55$$
and the second inequality is satisfied by $$n \le 45 \lor n \ge 55$$
$n=55$ is the only solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that if $a,b,c$ are sides of a triangle and $s$ is the semi-perimeter, then $a^2+b^2+c^2 \geq \dfrac{36}{35}\bigg(s^2 + \dfrac{abc}{s}\bigg)$
Prove that, if $a,b,c$ are the sides of a triangle and $s$ is the semi-perimeter, then $a^2+b^2+c^2 \geq \dfrac{36}{35}\bigg(s^2 + \dfrac{abc}{s}\bigg)$.
What I Tried:- Nothing special really came in my mind. I did not find a way to use Triangle Inequality. What I did was, by AM-GM :-
$$a^2 + b^2 + c^2 \geq \frac{36}{35}\bigg(s^2 + \dfrac{abc}{s}\bigg) > \frac{72}{35}(\sqrt{sabc}).$$
But I couldn't proceed from this.
Another Idea I had was :- $a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca)$.
$\rightarrow a^2 + b^2 + c^2 = 4s^2 - 2(ab + bc + ca).$
But I did not know how to use this here, and would make the calculations a bit messy, especially of the $\dfrac{36}{35}$ part present there.
Can Anyone Help me? Thank You.
| Here is another approach.
By AM-GM,
$(a+b+c)(a^2+b^2+c^2) \geq 3 (abc)^{2/3} \cdot 3 (abc)^{1/3} = 9abc$
$4 (a^2+b^2+c^2) \geq \cfrac{18abc}{s}$
$13 (a^2+b^2+c^2) \geq 9(a^2+b^2+c^2) + \cfrac{18abc}{s}$
Using $a^2+b^2+c^2 \geq ab+bc+ca$,
$13 (a^2+b^2+c^2) \geq 9(ab+bc+ca) + \cfrac{18abc}{s}$
$26 (a^2+b^2+c^2) \geq 18(ab+bc+ca) + \cfrac{36abc}{s}$
$35 (a^2+b^2+c^2) \geq 9(a^2+b^2+c^2) + 18(ab+bc+ca) + \cfrac{36abc}{s}$
$35 (a^2 + b^2 + c^2) \geq 36 s^2 + \cfrac{36abc}{s}$
$a^2 + b^2 + c^2 \geq \cfrac{36}{35} \left(s^2 + \cfrac{abc}{s} \right)$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the most complete binary quadratic form table? Is there a free, easily accessible quadratic form resource — maybe even just a big web-accessible table — which shows “all known results” [reasonably speaking] about numbers of the form $mx^2+ny^2$ for various $m,n$?
Immediate motivation: Given integers $a$ and $x$ with $a^3 \mid (x^2+1)(3x^2+5)$, I’m trying to characterize $a$. I know that if [odd] prime $\phi \mid (x^2+1)$, then $\phi \equiv 1\!\pmod{4}$… but what about $\phi \mid (3x^2+5)$?
Ultimately, of course, this kind of question really comes down to characterizing the primes that can divide a given quadratic form. I know about scattered resources like this Mathworld page and the (very short!) table on this page — I just can’t seem to find comprehensive information on that subject presented in a nice compact form.
| Particular case where $(x, 3)=1$
For $3x^2+5$ we may write:
$x^2\equiv 1\bmod 3\Rightarrow 3x^2+5\equiv 8\equiv 2\bmod 3$
We may conclude that primes of the form $p=3m+2$ can divide $3x^2+5$ , for example:
$x=1\rightarrow 3x^2+5=8, k=0\rightarrow p=2; 2|8$,
$x=6\rightarrow 3x^2+5=113=3\times 37+2$
$x=4\rightarrow 3x^2+5=53=3\times 17+2$
$x=8\rightarrow 3\times 64+5=197=3\times 65+2$
$x=17\rightarrow 3x^2+5=872=2\times 436; 2=3\times 0+2$
$x=18\rightarrow 3x^2+5=977=3\times 325+2$
For $mx^2+ny^2$ we may use the same method, suppose $(m, n, 3)=1$ then:
$mx^2+ny^2\equiv (m+n)\bmod 3$
| {
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.