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Show that $ \int_0^a \frac{dx}{\sqrt{a^2-x^2}} \, [\arcsin{\frac{x}{a}} / \arccos{\frac{x}{a}}]= \frac{\pi^2}{8} $? I found this very interesting result:
$$ \int_0^a \frac{dx}{\sqrt{a^2-x^2}} \, \arcsin{\frac{x}{a}} = \int_0^a \frac{dx}{\sqrt{a^2-x^2}} \, \arccos{\frac{x}{a}} = \frac{\pi^2}{8}, $$
for $a>0.$ I wonder how one could prove this?
| We have that
$$\int_0^a \frac{dx}{\sqrt{a^2-x^2}} \,
\arcsin{\frac{x}{a}}=\int_0^1 \frac{dt}{\sqrt{1-t^2}} \, \arcsin{t}\\
=\int_0^1 \arcsin(t)\ d(\arcsin{t})=\frac{1}{2}\left[\arcsin^2{t}\right]_0^1=\frac{\pi^2}{8}.$$
For the other one you can integrate in a similar way or use the identity $\arcsin(t)+\arccos(t)=\frac{\pi}{2}$ for $t>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1893735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Let $z$ be a complex number $\ne 0$. What is the absolute value of $z\sqrt{z}$? $\color{red}{\mathbf{EDIT}}$ The question was misinterpreted - it was actually: 'what is the absolute value of $z/\bar{z}$?'; I'am grateful for the answers given on the original problem though and will keep this up as is in case someone else has a similar issue.
Exercise 3, page 379 of "Basic Mathematics" by S.Lang.
Problem: Let $z$ be a complex number $\ne 0$. What is the absolute value of $z\sqrt{z}$ $?
My approach to the question goes as follows:
Let $z = x + iy$ for real numbers $x, y$ different than $0$.
We have
\begin{align}
z\sqrt{z} & = (x + iy)\sqrt{z + iy} \\
& = x\sqrt{x+iy} + iy\sqrt{x+iy} &&\text{by distributivity}
\end{align}
We have
\begin{align}
|z\sqrt{z}| & = \sqrt{(x\sqrt{x + iy})^2 + (y\sqrt{x + iy})^2} &&\text{by definition} \\
& = \sqrt{x^2(x + iy)^2 + y^2(x + iy)^2} \\
& = \sqrt{(x^2 + y^2)(x + iy)} &&\text{by factoring} \\
& = \sqrt{(x^2 + y^2)} \sqrt{(x + iy)} \\
& = |z| \sqrt{z} &&\text{by definition} \\
\end{align}
The author's solution is $1$.
Thank you.
| https://ernstchan.com/b/src/1457375466-129.pdf
The question is actually "what is the value of $z /\overline{z}$.
Which is easy. By theorem 2:
$|z /\overline{z}| = |z|/|\overline{z}| = \sqrt{x^2 + y^2}/\sqrt{x^2 + (-y)^2} = 1$
;I don't see if Lang ever stated this in the text but it should be obvious $|z| = |\overline z|$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Examining convergence of $\sum_{n=1}^{\infty}\frac{1}{n}\sqrt{e^{\frac{1}{n}}-e^\frac{1}{n+1}}$ with mean value theorem $$\sum_{n=1}^{\infty}\frac{1}{n}\sqrt{e^{\frac{1}{n}}-e^\frac{1}{n+1}}$$
I would like to examine covergence of this series using mean value theorem.
I would like to check my solution and optionally alternative solution :)
Let $c_n \in (n, n+1)$.
$f(n)=e^{\frac{1}{n}}$.
$$\frac{e^{\frac1n}-e^{\frac1{n+1}}}{-1}=\left(e^{\frac{1}{c_n}}\right)'=-\frac{1}{c_n^2}e^{\frac1{c_n}}$$
$$\sqrt{e^{\frac1n}-e^{\frac{1}{n+1}}}=\sqrt{\frac{1}{c_n^2}e^{\frac{1}{c_n}}}=\frac{1}{c_n}\sqrt{e^{\frac{1}{c_n}}}$$
$$\sum_{n=1}^{\infty}\frac{1}{n}\sqrt{e^{\frac{1}{n}}-e^\frac{1}{n+1}}\le \sum_{n=1}^{\infty}\frac{1}{n^2}\sqrt{{e^{\frac{1}{n}}}}$$
$$\sum_{n=1}^{\infty}\frac{1}{n^2}\sqrt{{e^{\frac{1}{n}}}}\le \sum_{n=1}^{\infty}\frac{1}{n^2}{e^{\frac{1}{n}}}$$ Since, is $\sum_{n=1}^{\infty}\frac{1}{n^2}{e^{\frac{1}{n}}}$ covergent absolutely so our seires is also absolutely covergent.
| It is OK. Alternatively, you may write, by the Taylor series expansion, as $n \to \infty$,
$$
\begin{align}
\frac{1}{n}\sqrt{e^{\frac{1}{n}}-e^\frac{1}{n+1}}&=\frac{1}{n}\sqrt{\left(1+O\left(\frac{1}{n}\right)\right)-\left(1+O\left(\frac{1}{n+1}\right)\right)}
\\\\&=\frac{1}{n}\sqrt{O\left(\frac{1}{n}\right)}
\\\\&=O\left(\frac{1}{n^{3/2}}\right)
\end{align}
$$ giving the convergence of the initial series. We have used the fact that, as $n \to \infty$,
$$
O\left(\frac{1}{n+1}\right)=O\left(\frac1n\cdot\frac{1}{1+1/n}\right)=O\left(\frac1n\cdot(1-1/n)\right)=O\left(\frac{1}n\right).
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Is the series $\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots$ summable? Is the series $\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots$ summable? I think it diverges although:
$$ \sqrt{n+1} - \sqrt{n} \approx \frac{1}{2\sqrt{n}}$$
for example by the Mean Value Theorem $f(x+1)-f(x) \approx f'(x)$ and then I might argue:
$$ \sum_{n \geq 1} (-1)^{n+1} \sqrt{n} = \frac{1}{2}\sum_{m \geq 1} \frac{1}{\sqrt{2m}} = \infty $$
Are these Cesaro summable? For an even number of terms:
$$\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots - \sqrt{2n}
\approx - \frac{1}{2\sqrt{2}}\left( \frac{1}{\sqrt{1}} +
\frac{1}{\sqrt{2}} + \dots + \frac{1}{\sqrt{n}} \right)
\approx \sqrt{\frac{n}{2}}$$
so the Cesaro means tend to infinity. Does any more creative summation method work?
The result is from paper called "The Second Theorem of Consistency for Summable Series" in Vol 6 of the Collected Works of GH Hardy
the series $1 - 1 +1 - 1 \dots$ is summable $(1,k)$ for any $k$ but not summable $(e^n, k)$ for any value of $k$.
The series $\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots$ is summable $(n,1)$ but not $(e^{\sqrt{n}},1)$ and so on...
Here things like $(1,k), (n,1)$ refer to certain averaging procedures, IDK
| This sum can be done with some form of zeta function regularization. For $\Re s >1$, define:
$$\eta(s) = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} = (1-2^{1-s})\zeta(s) $$
Then, by analytic continuation, we can calculate:
$$\sum_{n=1}^\infty (-1)^{n-1} \sqrt{n} \to \eta\left(-\frac{1}{2}\right) = (1-2\sqrt{2})\zeta\left(-\frac{1}{2}\right) \approx .3801048$$
This is equal to the Abel sum $-\operatorname{Li}_{-\frac{1}{2}} (-1)$ from GEdgar's answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1896464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Smallest positive integer that gives remainder 5 when divided by 6, remainder 2 when divided by 11, and remainder 31 when divided by 35? What is the smallest positive integer that gives the remainder 5 when divided by 6, gives the remainder 2 when divided by 11 and 31 when divided by 35?
Also, are there any standard methods to solve problems like these?
| The system of equations
$$ \left\{\begin{array}{ll} n \equiv 5 &\pmod{6} \\ n \equiv 2&\pmod{11} \\ n\equiv 31 &\pmod{35}\end{array}\right. $$
is equivalent to $n\equiv m\pmod{2\cdot 3\cdot 5\cdot 7\cdot 11}$ by the Chinese remainder theorem.
If $m_6, m_{11}, m_{35}$ are the smallest integer solutions of the systems
$$ \left\{\begin{array}{ll} n \equiv 1 &\pmod{6} \\ n \equiv 0&\pmod{11} \\ n\equiv 0 &\pmod{35}\end{array}\right. \quad\left\{\begin{array}{ll} n \equiv 0 &\pmod{6} \\ n \equiv 1&\pmod{11} \\ n\equiv 0 &\pmod{35}\end{array}\right.\quad \left\{\begin{array}{ll} n \equiv 0 &\pmod{6} \\ n \equiv 0&\pmod{11} \\ n\equiv 1 &\pmod{35}\end{array}\right.$$
then we have $m\equiv 5m_6+2m_{11}+31 m_{35}\pmod{2310}$, and $m_6,m_{11},m_{35}$ are not difficult to compute. For instance, $m_6$ is a multiple of $11\cdot 35$ that is $\equiv 1\pmod{6}$, but
$$ k\cdot 11\cdot 35 \equiv 1\pmod{6} $$
is equivalent to $k\equiv 1\pmod{6}$, hence $m_6=11\cdot 35=385$. In a similar way you may find $m_{11}=210$ and $m_{35}=1716$, then re-combine them in order to find $m=\color{red}{101}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1897521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find large power of a non-diagonalisable matrix
If $A = \begin{bmatrix}1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$, then find $A^{30}$.
The problem here is that it has only two eigenvectors, $\begin{bmatrix}0\\1\\1\end{bmatrix}$ corresponding to eigenvalue $1$ and $\begin{bmatrix}0\\1\\-1\end{bmatrix}$ corresponding to eigenvalue $-1$. So, it is not diagonalizable.
Is there any other way to compute the power?
| Since the given matrix is rather simple, you could also compute a few powers:
\begin{align*}
A^1&=
\begin{pmatrix}
1 & 0 & 0 \\
1 & 0 & 1 \\
0 & 1 & 0
\end{pmatrix}\\
A^2&=
\begin{pmatrix}
1 & 0 & 0 \\
1 & 1 & 0 \\
1 & 0 & 1
\end{pmatrix}\\
A^3&=
\begin{pmatrix}
1 & 0 & 0 \\
2 & 0 & 1 \\
1 & 1 & 0
\end{pmatrix}\\
A^4&=
\begin{pmatrix}
1 & 0 & 0 \\
2 & 1 & 0 \\
2 & 0 & 1
\end{pmatrix}\\
A^5&=
\begin{pmatrix}
1 & 0 & 0 \\
3 & 0 & 1 \\
2 & 1 & 0
\end{pmatrix}\\
A^6&=
\begin{pmatrix}
1 & 0 & 0 \\
3 & 1 & 0 \\
3 & 0 & 1
\end{pmatrix}\\
\end{align*}
Can you notice some kind of pattern and prove it by induction? (Or, if you prefer, you can do the same thing with $B=A^2$ and try to find general form of $B^n$.)
Of course, this is "naive" approach - without using any theory you have learned about matrices. For more complicated matrices it would be rather difficult to spot some kind of pattern. That's why methods which work for arbitrary matrices are more useful.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 3
} |
Parabolic Representation The points $(-1/2, 0)$, $\left(0, \frac{\sqrt3}{2}\right)$, $(1/2, 0)$ are the vertices of an equilateral triangle. A parabolic equation that contains the three points is
$$y=
(-2 \sqrt3)x^2+\frac{\sqrt{3}}{2} \,.$$ How can the other two parabolas, that are congruent to the first parabola and contain the same three points, be represented? Can this be done with parametric equations or do I need a higher level of understanding?
| Since you are given the three points, there is a straight vectorial approach that you can follow.
Premised that
the parabola $y=x^2$, the parabola with vertex in $O=(0,0)$ and passing through
$U_{-1}=(-1,1)$ and $U_{1}=(1,1)$, can be written as:
$$
\mathop {OP}\limits^ \to \cdot \mathbf{u}_y = \left( {\mathop {OP}\limits^ \to \cdot \mathbf{u}_x } \right)^2
$$
where $\mathop {OP}\limits^ \to = (x,y)$ is the position vector, and
$\mathbf{u}_x = \left( {1,0} \right)\quad \mathbf{u}_y = \left( {0,1} \right)$
are the unit vectors on the axes, and that we have
$$
\mathop {OU_1 }\limits^ \to = \mathbf{u}_y + \mathbf{u}_x \quad \mathop {OU_{ - 1} }\limits^ \to = \mathbf{u}_y - \mathbf{u}_x
$$
then
in the case of the three symmetric points under consideration
$$
A = \left( { - 1/2,\;0} \right)\quad B = \left( {1/2,\;0} \right)\quad C = \left( {0,\;\sqrt 3 /2} \right)
$$
let's consider the vectors
$$
\begin{gathered}
\mathbf{v} = \frac{1}
{2}\left( {\mathop {AB}\limits^ \to - \mathop {AC}\limits^ \to } \right) = \frac{1}
{2}\left( {\left( {x_B - x_C } \right),\;\left( {y_B - y_C } \right)} \right) \hfill \\
\mathbf{w} = \frac{1}
{2}\left( {\mathop {AB}\limits^ \to + \mathop {AC}\limits^ \to } \right) = \frac{1}
{2}\left( {\left( {x_B + x_C - 2x_A } \right),\;\left( {y_B + y_C - 2y_A } \right)} \right) \hfill \\
\end{gathered}
$$
as shown in the figure
we want that
$$
\left\{ \begin{gathered}
\mathop {AB}\limits^ \to \cdot \mathbf{w} = \left\| \mathbf{w} \right\|^{\,2} \quad \Rightarrow \quad 1 \hfill \\
\mathop {AB}\limits^ \to \cdot \mathbf{v} = \left\| \mathbf{v} \right\|^{\,2} \quad \Rightarrow \quad 1 \hfill \\
\end{gathered} \right.
$$
therefore the parabola with vertex in $A$ e passing through $B$ and $C$ can be written as:
$$
\frac{{\mathop {AP}\limits^ \to \cdot \mathbf{w}}}
{{\left\| \mathbf{w} \right\|^{\,2} }} = \left( {\frac{{\mathop {AP}\limits^ \to \cdot \mathbf{v}}}
{{\left\| \mathbf{v} \right\|^{\,2} }}} \right)^2
$$
i.e.
$$
\left( {\left( {x + 1/2} \right),\;y} \right) \cdot \frac{1}
{2}\left( {\begin{array}{*{20}c}
{3/2} \\
{\sqrt 3 /2} \\
\end{array} } \right)\frac{4}
{3} = \left( {\left( {\left( {x + 1/2} \right),\;y} \right) \cdot \frac{1}
{2}\left( {\begin{array}{*{20}c}
{1/2} \\
{ - \sqrt 3 /2} \\
\end{array} } \right)4} \right)^2
$$
thus
$$
2\left( {6x + 3 + 2\sqrt 3 y} \right) = 3\left( {2x + 1 - 2\sqrt 3 y} \right)^2
$$
or
$$
12x^2 + 36y^2 - 24\sqrt 3 xy - 16\sqrt 3 y - 3 = 0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1899653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How can I prove that $\binom{3n}{n,n,n}$ divided by $6$ with no remainder for all $n>0$? How can I prove that $\binom{3n}{n,n,n}$ divided by $6$ with no remainder for all $n>0$?
| Referring to the standard result, the number of factors of $p$ in $\frac{(3n)!}{n!n!n!}$ is given by
$$
\sum_{i \ge 1} \left\lfloor \frac{3n}{p^i} \right\rfloor - 3 \sum_{i \ge 1} \left\lfloor \frac{n}{p^i} \right\rfloor
= \sum_{i \ge 1} \left( \left\lfloor \frac{3n}{p^i} \right\rfloor - 3 \left\lfloor \frac{n}{p^i} \right\rfloor \right).
$$
All of the terms of the sum are nonnegative. We have to prove that when $p =2$ or $3$, there is a term that is strictly positive.
*
*For $p = 3$, let $3^k$ be the highest power of $3$ dividing $n$. Then the term $i = 3^{k+1}$ works:
$$
\left\lfloor \frac{3n}{3^{k+1}} \right\rfloor - 3 \left\lfloor \frac{n}{3^{k+1}} \right\rfloor = 3 \left[\frac{n}{3^{k+1}} -\left\lfloor \frac{n}{3^{k+1}} \right\rfloor \right] > 0.
$$
*For $p = 2$, let $2^k$ be the highest power of $2$ dividing $n$, and again take the term $i = 2^{k+1}$:
\begin{align*}\left\lfloor \frac{3n}{2^{k+1}} \right\rfloor - 3 \left\lfloor \frac{n}{2^{k+1}} \right\rfloor
&= \left\lfloor \frac{2n}{2^{k+1}} + \frac{n}{2^{k+1}} \right\rfloor - 3 \left\lfloor \frac{n}{2^{k+1}} \right\rfloor \\
&= \frac{2n}{2^{k+1}} + \left\lfloor \frac{n}{2^{k+1}} \right\rfloor - 3 \left\lfloor \frac{n}{2^{k+1}} \right\rfloor \\
&= 2 \left[\frac{n}{2^{k+1}} - \left\lfloor \frac{n}{2^{k+1}} \right\rfloor\right] \\
&> 0.
\end{align*}
| {
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"question_score": "1",
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Help proving the product of any four consecutive integers is one less than a perfect square Apparently this is a true statement, but I cannot figure out how to prove this. I have tried setting
$$(m)(m + 1)(m + 2)(m + 3) = (m + 4)^2 - 1 $$
but to no avail. Could someone point me in the right direction?
| If you have trouble approaching this, try some examples: Note that (for instance)
$$
1 \times \color{red}{2 \times 3} \times 4 = 24 = 5^2-1
$$
$$
2 \times \color{red}{3 \times 4} \times 5 = 120 = 11^2-1
$$
$$
3 \times \color{red}{4 \times 5} \times 6 = 360 = 19^2-1
$$
and observe that
$$
5 = \color{red}{2 \times 3} - 1
$$
$$
11 = \color{red}{3 \times 4} - 1
$$
$$
19 = \color{red}{4 \times 5} - 1
$$
So try seeing if
$$
m\color{red}{(m+1)(m+2)}(m+3) = [\color{red}{(m+1)(m+2)}-1]^2-1
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Multi-index sum property Exercise 1.2.3.29 in Donald Knuth's The Art of Computer Programming (3e) states the following property of a multi-indexed sum:
$$
\sum_{i=0}^n \sum_{j=0}^i \sum_{k=0}^j a_ia_ja_k = \frac{1}{3}S_3 + \frac{1}{2}S_1S_2 + \frac{1}{6}S_1^3,
$$
where $S_r = \sum_{i=0}^n a_i^r$.
I tried to prove it and failed. What I found are the following identities:
$$
\sum_{i=0}^n \sum_{j=0}^i \sum_{k=0}^j a_ia_ja_k
= \frac{1}{2} \sum_{i=0}^n a_i \left( \left( \sum_{j=0}^i a_j \right)^2
+ \sum_{j=0}^i a_j^2 \right),
$$
$$
S_1S_2 = \sum_{i=0}^n \sum_{j=0}^n a_i a_j^2,
$$
$$
S_1^3 = \sum_{i=0}^n \sum_{j=0}^n \sum_{k=0}^n a_i a_j a_k.
$$
Can anybody help in completing the proof?
Based on grand_chat's answer, the identity can be proven inductively.
$$
\begin{align}
\sum_{i=0}^{n+1} \sum_{j=0}^i \sum_{k=0}^j a_ia_ja_k & =
\sum_{i=0}^n \sum_{j=0}^i \sum_{k=0}^j a_ia_ja_k
+ a_{n+1} \sum_{i=0}^{n+1} \sum_{j=0}^i a_ia_j
\\ & \stackrel{hyp}{=} \underbrace{\frac{1}{3}S_{n,3} + \frac{1}{2}S_{n,1}S_{n,2} + \frac{1}{6}S_{n,1}^3}_{=:S_n}
+ \frac{1}{2}
a_{n+1} \left( \left( \sum_{i=0}^{n+1} a_i \right)^2
+ \sum_{i=0}^{n+1} a_i^2 \right)
\\ & = S_n + \frac{1}{2} a_{n+1}
\left( \left( \sum_{i=0}^n a_i + a_{n+1} \right)^2
+ \sum_{i=0}^n a_i^2 + a_{n+1}^2 \right)
\\ & = S_n + \frac{1}{2} a_{n+1}
\left( \left( \sum_{i=0}^n a_i \right)^2 + 2 \sum_{i=0}^n a_i a_{n+1} + a_{n+1}^2
+ \sum_{i=0}^n a_i^2 + a_{n+1}^2 \right)
\\ & = S_n + a_{n+1}^3 + \sum_{i=0}^n a_i a_{n+1}^2
+ \frac{1}{2} \left( \sum_{i=0}^n a_i \right)^2 a_{n+1}
+ \frac{1}{2} \sum_{i=0}^n a_i^2 a_{n+1}
\\ & = S_n + a_{n+1}^3 \left( \frac{1}{3} + \frac{1}{2} + \frac{1}{6} \right)
+ \sum_{i=0}^n a_i a_{n+1}^2 \left( \frac{1}{2} + \frac{3}{6} \right)
+ \frac{3}{6} \left( \sum_{i=0}^n a_i \right)^2 a_{n+1}
+ \frac{1}{2} \sum_{i=0}^n a_i^2 a_{n+1}
\\ & = \frac{1}{3} \left( S_{n,3} + a_{n+1}^3 \right)
+ \frac{1}{2} \left(
S_{n,1}S_{n,2} + \sum_{i=0}^n a_i a_{n+1}^2 + \sum_{i=0}^n a_i^2 a_{n+1} + a_{n+1}^3
\right)
+ \frac{1}{6} \left(
S_{n,1}^3 + 3 \left( \sum_{i=0}^n a_i \right)^2 a_{n+1}
+ 3 \sum_{i=0}^n a_i a_{n+1}^2 + a_{n+1}^3
\right)
\\ & = \frac{1}{3}S_{n+1,3} + \frac{1}{2}S_{n+1,1}S_{n+1,2} + \frac{1}{6}S_{n+1,1}^3.
\end{align}
$$
The problem here is that the hypothesis is taken from thin air.
|
Here is a rather elementary approach based upon two aspects:
*
*Iterative calculation
We look at first at the simpler double sum
\begin{align*}
A_2&:=\sum_{i=0}^n\sum_{j=0}^ia_ia_j=\sum_{0\leq j\leq i\leq n}a_ia_j\\
&=\sum_{0\leq i\leq j\leq n}a_ia_j
\end{align*}
and obtain an expression in $S_1=\sum_{i=0}^na_i$ and $S_2=\sum_{i=0}^n a_i^2$. The result will then be used for calculating the triple sum.
*Manipulation of index range
We can represent the sum using a more compact index range notation and obtain after renaming the variables
\begin{align*}
A_3&:=\sum_{i=0}^n\sum_{j=0}^i\sum_{k=0}^ja_ia_ja_k
=\sum_{0\leq k\leq j\leq i\leq n}a_ia_ja_k\\
&=\sum_{0\leq i\leq j\leq k\leq n}a_ia_ja_k\\
\end{align*}
We will often rearrange the index range for our needs.
$$ $$
We obtain
\begin{align*}
A_2 &= \sum_{0\leq i\leq j\leq n}a_ia_j\\
&=\sum_{0\leq i,j\leq n}a_ia_j-\sum_{0\leq j<i\leq n}a_ia_j\tag{1}\\
&=\left(\sum_{0\leq i\leq n}a_i\right)\left(\sum_{0\leq j\leq n}a_j\right)
-\left(\sum_{0\leq j\leq i\leq n}a_ia_j-\sum_{0\leq j=i\leq n}a_ia_j\right)\\
&=S_1^2-\left(A_2-\sum_{i=0}^na_i^2\right)\\
&=S_1^2-A_2+S_2\\
\end{align*}
We conclude the double sum $A_2$ can be represented as
\begin{align*}
A_2&=\frac{1}{2}S_1^2+\frac{1}{2}S_2\tag{2}
\end{align*}
In (1) we represent the upper triangle range $0\leq i\leq j\leq n$ as square $0\leq i,j\leq n$ minus lower triangle $0\leq j< i\leq n$. In the following line we do some more rearrangements of the index range from which we can easily derive the representation in $S_1$ and $S_2$.
And now the triple sum. We show
the following is valid
\begin{align*}
A_3:=\sum_{i=0}^n\sum_{j=0}^i\sum_{k=0}^ja_ia_ja_k=\frac{1}{3}S_3+\frac{1}{2}S_1S_2+\frac{1}{6}S_1^3\tag{3}
\end{align*}
We start with the RHS
\begin{align*}
\frac{1}{3}S_3&+\frac{1}{2}S_1S_2+\frac{1}{6}S_1^3\\
&=\frac{1}{3}S_3+\frac{1}{2}S_1S_2+\frac{1}{6}S_1S_1^2\\
&=\frac{1}{3}S_3+\frac{1}{2}S_1S_2+\frac{1}{6}S_1\left(2A_2-S_2\right))\tag{4}\\
&=\frac{1}{3}\left(S_3+S_1S_2+S_1A_2\right)\\
&=\frac{1}{3}\left(\sum_{i=0}a_i^3+\sum_{i=0}^na_i\sum_{j=0}^na_j^2
+\sum_{i=0}^na_i\sum_{0\leq j\leq k\leq n} a_ja_k\right)\\
&=\frac{1}{3}\left(\sum_{0\leq i= j=k\leq n} a_ia_ja_k
+\sum_{0\leq i\leq n}\sum_{0\leq j=k\leq n}a_ia_ja_k
+\sum_{0\leq i\leq n}\sum_{0\leq j\leq k\leq n}a_ia_ja_k\right)\tag{5}
\end{align*}
In (4) we use the double sum representation (2). We show the last expression in brackets is equal to $3A_3$. At first we split the right-most triple sum.
*
*Right-most sum
Since $i$ in $0\leq i\leq n$ is either less or equal $j$ or greater $j$ and less or equal $k$ or greater $k$ we obtain three sums.
\begin{align*}
\sum_{0\leq i\leq n}\sum_{0\leq j\leq k\leq n}a_ia_ja_k
&=\sum_{0\leq i \leq j\leq k\leq n}a_ia_ja_k
+\sum_{0\leq j<i\leq k\leq n}a_ia_ja_k
+\sum_{0\leq j\leq k<i\leq n}a_ia_ja_k\\
&=A_3
+\sum_{0\leq j<i\leq k\leq n}a_ia_ja_k
+\sum_{0\leq j\leq k<i\leq n}a_ia_ja_k\\
&=A_3
+\sum_{0\leq i<j\leq k\leq n}a_ia_ja_k
+\sum_{0\leq i\leq j<k\leq n}a_ia_ja_k
\end{align*}
Next we focus on the middle sum and split the range conveniently
*
*Middle sum
We see the index $0\leq i\leq n$ in the middle sum is either less or equal $j(=k)$ or greater $j(=k)$ and we obtain two sums.
\begin{align*}
\sum_{0\leq i\leq n}\sum_{0\leq j=k\leq n}a_ia_ja_k
&=\sum_{0\leq i\leq j=k\leq n}a_ia_ja_k+\sum_{0\leq j=k<i\leq n}a_ia_ja_k\\
&=\sum_{0\leq i\leq j=k\leq n}a_ia_ja_k+\sum_{0\leq i=j<k\leq n}a_ia_ja_k
\end{align*}
We can now combine the left-most sum with the middle sum and the result with the right-most part.
*
*Combining sums
The colors $\color{red}{\text{red}}$, $\color{blue}{\text{blue}}$ and $\color{green}{\text{green}}$ indicate combined sums.
\begin{align*}
\sum_{0\leq i= \color{red}{j=k}\leq n}& a_ia_ja_k
+ \left(\sum_{0\leq i\leq j=k\leq n}a_ia_ja_k+\sum_{0\leq i=\color{red}{j<k}\leq n}a_ia_ja_k\right)\\
&\qquad +\left(A_3
+\sum_{0\leq i<j\leq k\leq n}a_ia_ja_k
+\sum_{0\leq i\leq j<k\leq n}a_ia_ja_k\right)\\
&=\sum_{0\leq i\leq \color{blue}{j=k}\leq n}a_ia_ja_k+\sum_{0\color{green}{\leq i=}\color{red}{j\leq k}\leq n}a_ia_ja_k\\
&\qquad +\left(A_3
+\sum_{0\leq \color{green}{i<j}\leq k\leq n}a_ia_ja_k
+\sum_{0\leq i\leq \color{blue}{j<k}\leq n}a_ia_ja_k\right)\\
&=A_3
+\sum_{0\leq \color{green}{i\leq j}\leq k\leq n}a_ia_ja_k
+\sum_{0\leq i\leq \color{blue}{j\leq k}\leq n}a_ia_ja_k\\
&=3A_3\tag{6}
\end{align*}
Finally we obtain from (5) and (6)
\begin{align*}
\frac{1}{3}S_3&+\frac{1}{2}S_1S_2+\frac{1}{6}S_1^3\\
&=\frac{1}{3}\left(\sum_{0\leq i= j=k\leq n} a_ia_ja_k
+\sum_{0\leq i\leq n}\sum_{0\leq j=k\leq n}a_ia_ja_k
+\sum_{0\leq i\leq n}\sum_{0\leq j\leq k\leq n}a_ia_ja_k\right)\\
&=\frac{1}{3}\left(3A_3\right)\\
&=A_3
\end{align*}
and the claim follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
What is $\int_0^1 \frac{\log \left(1-x^2\right) \sin ^{-1}(x)^2}{x^2} \, dx$? I encountered the integral $$\int_0^1 \frac{\log \left(1-x^2\right) \sin ^{-1}(x)^2}{x^2} \, dx = -0.9393323982...$$ while researching the evaluation of harmonic sums. Mathematica 11 is not able to evaluate this integral, and is not able to evaluate the underlying indefinite integral.
I considered using the Maclaurin series for the expression $\sin ^{-1}(x)^2$ to evaluate this integral. Using this Maclaurin series, it is easily seen that the problem of evaluating the above integral is equivalent to the problem of evaluating the series $$\sum _{n=1}^{\infty } \frac{2^{2 n-1} H_{n-\frac{1}{2}}}{(1-2 n) n^2 \binom{2 n}{n}}=-0.9393323982...$$ which Mathematica 11 is not able to evaluate directly.
I also considered using the Maclaurin series for the expression $\log \left(1-x^2\right)$ to evaluate the above integral. Using this Macluarin series, it is easily seen that this integral is equal to $$\frac{1}{4} \left(16 \pi C-\pi ^3-\pi ^2 \log (4)\right)+\sum _{n=1}^{\infty } \frac{\, _3F_2\left(\frac{1}{2},\frac{1}{2},1;\frac{3}{2},n+1;1\right)}{n^2-2 n^3}=-0.939332...,$$ but Mathematica is unable to evaluate the above infinite series.
What is a closed-form evaluation of $\int_0^1 \frac{\log \left(1-x^2\right) \sin ^{-1}(x)^2}{x^2} \, dx$? What techniques can be applied to evaluate this definite integral?
| As an addendum to the previous answer, it can be shown (through the FL-expansion of $\frac{\arcsin\sqrt{x}}{\sqrt{x}}$) that
$$\phantom{}_4 F_3\left(\tfrac{1}{2},\tfrac{1}{2},1,1;\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2};1\right)=4\text{ Im } \text{Li}_3\left(\frac{1+i}{2}\right)-\frac{\pi^3}{32}-\frac{\pi}{8}\log^2(2).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 1,
"answer_id": 0
} |
Craft Inequality Originally the inequality looks like :
$$
\frac{1}{4}(\sqrt{\sum_{cyc}\sin(u)\sin(v)} - \sum_{cyc}\sin(u)\sin(v))
\geq\sum_{cyc}\cos(u)-(\sum_{cyc}\sin(\frac{3u}{2}))
$$
with $$u+v+w=\pi$$
after many transformations the inequality looks like :
$$
\sqrt{\frac{1}{2}\left(\sum_{\mathrm{cyc}}2C+\sum_{\mathrm{cyc}}2AB\right)}-\frac{4}{2}\sum_{\mathrm{cyc}}C-\frac{1}{2}\sum_{\mathrm{cyc}}2AB\geq\frac{4}{1}\sum_{\mathrm{cyc}}\sqrt{\frac{1-C}{2}}(-2C-1)$$
with the constraint :
$$AB-\sqrt{(1-A²)}\sqrt{(1-B²)}=-C$$
and :
$$-1\le A\le B\le C\le 1$$
Furthermore we have
$$ 1\le A+B+C\le 1.5$$
$$ \frac{3}{4}\le A^2+B^2+C^2\le 3$$
$$ -1\le AB+BC+CA\le\frac{3}{4}$$
$$-3\le -\sum_{\mathrm{cyc}}\sqrt{\frac{1-C}{2}}(-2C-1)\le 3$$
and $$cos(u)=A$$
$$cos(v)=B$$
$$cos(w)=C$$
My try :
We have :
$$\frac{1}{\sqrt{2}}(A+B+C-1)\le \sqrt{\frac{1}{2}\left(\sum_{\mathrm{cyc}}2C+\sum_{\mathrm{cyc}}2AB\right)}$$
And
$$-(A^2+B^2+C^2)\le -\sum_{cyc}AB$$
We obtain :
$$
\sqrt{\frac{1}{2}\left(\sum_{\mathrm{cyc}}2C+\sum_{\mathrm{cyc}}2AB\right)}-\frac{4}{2}\sum_{\mathrm{cyc}}C-\frac{1}{2}\sum_{\mathrm{cyc}}2AB \geq \frac{1}{\sqrt{2}}(A+B+C-1)-\sum_{cyc}(2C+C^2) \geq \frac{4}{1}\sum_{\mathrm{cyc}}\sqrt{\frac{1-C}{2}}(-2C-1)$$
Or:
$$0\geq\frac{1}{\sqrt{2}}+\sum_{\mathrm{cyc}}\frac{4}{1}\sqrt{\frac{1-C}{2}}(-2C-1)+C^2+(2-\frac{1}{\sqrt{2}})C$$
Then I can't continue
| I begin with a transformation :
$$x^2=1-C$$
$$y^2=1-B$$
$$z^2=1-A$$
We obtain :
$$ \sqrt{\frac{1}{2}\left(\sum_{\mathrm{cyc}}2(1-x^2)+\sum_{\mathrm{cyc}}2(1-x^2)(1-y^2)\right)}-\frac{4}{2}\sum_{\mathrm{cyc}}((1-x^2))-\frac{1}{2}\sum_{\mathrm{cyc}}2((1-x^2)(1-y^2))\geq-\frac{1}{2}\sum_{\mathrm{cyc}}\sqrt{\frac{x^2}{2}}(-2(1-x^2)-1) $$
and the constraint wich was :
$$1=A^2+B^2+C^2+2ABC$$
is:
$$1=(1-z^2)^2+(1-y^2)^2+(1-x^2)^2+2(1-z^2)(1-x^2)(1-y^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1902307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $a$, $b$, and $c$ are sides of a triangle, then $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2$.
Let $a,b,c$ be the lengths of the sides of a triangle. Prove that
$$\sum_{\text{cyc}}\frac{a}{b+c}=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2\,.$$
Attempt. By clearing the denominators, the required inequality is equivalent to
$$a^2(b+c)+b^2(c+a)+c^2(a+b)>a^3+b^3+c^3\,.$$
Since $b+c>a$, $c+a>b$, and $a+b>c$, the inequality above is true. Is there a better, non-bruteforce way?
| As @user236182 pointed out, we can write $a=p+q$, $b=q+r$, and $c=r+p$, with $p,q,r>0$. Letting $s = p+q+r$, we have
$$\frac{a}{b+c} = \frac{s-r}{s+r} = 1 - \frac{2r}{s+r} $$
and hence
$$\sum\limits_{\text{cyc}}{\frac{a}{b+c}}< 2 \iff \sum\limits_{\text{cyc}}{\left(1-\frac{2r}{s+r}\right)}< 2 \iff \sum\limits_{\text{cyc}}{\frac{2r}{s+r}} > 1. $$
By Cauchy-Schwarz we have
$$\left(\sum\limits_{\text{cyc}}{\frac{r}{s+r}}\right)\left(\sum\limits_{\text{cyc}}{r(s+r)}\right)\ge\left(\sum\limits_{\text{cyc}}{r}\right)^2 = s^2. $$
Furthermore,
$$\sum\limits_{\text{cyc}}{r(s+r)} = s\sum\limits_{\text{cyc}}{r}+\sum\limits_{\text{cyc}}{r^2} = s^2 + (p^2+q^2+r^2) < s^2+(p+q+r)^2 = 2s^2. $$
Note that the inequality is strict as $p,q,r>0$. It follows that
$$ 2s^2\sum\limits_{\text{cyc}}{\frac{r}{s+r}} > \left(\sum\limits_{\text{cyc}}{\frac{r}{s+r}}\right)\left(\sum\limits_{\text{cyc}}{r(s+r)}\right) \ge s^2 $$
and hence $\sum\limits_{\text{cyc}}{\frac{2r}{s+r}} > 1$, as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1903775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Trouble understanding proof of the inequality - $(\frac{1}{a}+1)(\frac{1}{b}+1)(\frac{1}{c}+1) \ge 64 $, for $a,b,c > 0$ and $a+b+c = 1$ I was looking into this problem in a book discussing inequalities, However I found the proof quite hard to understand.The problem is as follows:
Let $a,b,c$ be positive numbers with $a+b+c=1$, prove that
$$\left(\frac{1}{a}+1\right)\left(\frac{1}{b}+1\right)\left(\frac{1}{c}+1\right) \ge 64$$
and the proof provided was the following:
Note that $$ abc \le (\frac{a+b+c}{3})^3 = \frac{1}{27} \tag{1}$$
by AM-GM inequality. Then
$$(\frac{1}{a}+1)(\frac{1}{b}+1)(\frac{1}{c}+1)=1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{1}{abc} \tag{2}$$
$$\ge 1+\frac{3}{\sqrt[3]{abc}}+\frac{3}{\sqrt[3]{(abc)^2}} +\frac
{1}{abc} \tag{3}$$
$$=(1+\frac{1}{\sqrt[3]{abc}})^3 \ge 4^3 \tag{4}$$
Steps 1 and 2 are easy for me to understand, but if someone could help me with steps 3 and 4 ,I would be very thankful.
| $$\frac{a + \frac 1 3 + \frac 1 3 + \frac 1 3}{4} \ge \sqrt[4]{a\cdot \frac 1 3 \cdot \frac 1 3 \cdot \frac 1 3}
$$
and similarly for b and c. So $(a+1)(b+1)(c+1) \ge 64\sqrt[4]{\frac{abc}{27^3}}$. You must show this is $\ge 64 abc$, which follows from your inequality (1).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1905278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
prove $abc \ge 8$ for $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1$ I am reading this book about inequalities and the chapter about AM-GM inequalities includes this problem:
Let $a,b,c$ be positive numbers for which $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1$, prove that
$$abc \ge 8$$
The book does not provide full solutions but only hints, and the one for this question is that it is similar to this problem:
Let $a,b,c$ be positive numbers with $a+b+c=1$, prove that
$$(\frac{1}{a}-1)(\frac{1}{b}-1)(\frac{1}{c}-1) \ge 8$$
again there's no solution provided but what I came up with is the following (you can correct me if it is wrong):
$$a+b=1-c$$
$$a+c=1-b$$
$$b+c=1-a$$
$$(\frac{1}{a}-1)(\frac{1}{b}-1)(\frac{1}{c}-1)=(\frac{1-a}{a})(\frac{1-b}{b})(\frac{1-c}{c})=(\frac{b+c}{a})(\frac{a+c}{b})(\frac{a+b}{c})=(\frac{b}{a}+\frac{c}{a})(\frac{a}{b}+\frac{c}{b})(\frac{a}{c}+\frac{b}{c})$$
By AM-GM we have for each term:
$$(\frac{b}{a}+\frac{c}{a}) \ge 2\sqrt{\frac{bc}{a^2}}$$
$$(\frac{a}{b}+\frac{c}{b}) \ge 2\sqrt{\frac{ac}{b^2}}$$
$$(\frac{a}{c}+\frac{b}{c}) \ge 2\sqrt{\frac{ab}{c^2}}$$
By multiplying the three inequalities we have
$$(\frac{b}{a}+\frac{c}{a})(\frac{a}{b}+\frac{c}{b})(\frac{a}{c}+\frac{b}{c}) \ge 8 \sqrt{\frac{a^2b^2c^2}{a^2b^2c^2}}=8$$
as desired.
Can someone please provide me a proof for the first problem as I cannot find any way around it, Thanks.
| Define
\begin{align*}
x&=\frac{1}{1+a}\\
y&=\frac{1}{1+b}\\
z&=\frac{1}{1+c}
\end{align*}
Then your problem transforms into: given $x+y+z=1$ prove $(\frac{1}{x}-1)(\frac{1}{y}-1)(\frac{1}{z}-1) \geq 8$. I believe you know the rest.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1906492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
If $a, b, c, d> 0$ then I try to prove the following inequality.. If $a, b, c, d> 0$ then prove that $$a^2 b^2 c^2 + b^2 c^2 d^2 + a^2 b^2 d^2 + a^2 c^2 d^2 \geq ab^2 c^2 d + ab^2 cd^2 + a^2 bcd^2 + a^2bc^2d$$
I'm using AM GM relation but I'm not getting the answer, probably due to a conceptual mistake?
| We may write $a^2b^2c^2 + b^2c^2d^2 + a^2b^2d^2 + a^2c^2d^2$ as
$$\frac{a^2b^2c^2 + b^2c^2d^2}{2} + \frac{b^2c^2d^2 + a^2b^2d^2}{2} + \frac{a^2b^2d^2 + a^2c^2d^2}{2} + \frac{a^2c^2d^2 + a^2b^2c^2}{2}$$
Apply the AM-GM inequality to each term to obtain the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1908678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How can it be proven that $\frac{x}{y}+\frac{y}{x}\geq2$, with $x$ and $y$ positive? So I realized that I have to prove it with the fact that $(x-y)^2+2xy=x^2+y^2$
So $\frac{(x+y)^2}{xy}+2=\frac{x}{y}+\frac{y}{x}$ $\Leftrightarrow$ $\frac{(x+y)^2}{xy}=\frac{x}{y}+\frac{y}{x}-2$
Due to the fact that $(x+y)^2$ is a square, it will be positive
$x>0$ and $y>0$ so $xy>0$
So $\frac{(x+y)^2}{xy}>0$
So $\frac{x}{y}+\frac{y}{x}>2$
But the problem is that I have proven that $\frac{x}{y}+\frac{y}{x}>2$, but in the case of $x=y$, it is equal, so it has to be $\geq$...
Could someone help?
| Let $a = x/y$, then this is equivalent to proving $a + 1/a \geq 2$. $a$ is positive so multiply both sides by $a$ we get
$$a^2 - 2a + 1 = (a - 1)^2 \geq 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1911037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Infinitude of solutions to Bézout's equation If $c=\gcd(a,b)$ for $a,b\in\Bbb Z$ it is possible to write $ax + by = c$ with $x,y\in\Bbb Z$. Is it possible to prove that there are infinitely many integer solutions for $x$ and $y$ without using Diophantine equations, and if so how?
(The particular problem I have is $1947x+264y=33$.)
| Writing $ax + by = c$, you get infinitely many solutions as follows: Let $k\in \mathbb Z$ be arbitrary, then
\begin{align*}
c &= ax + by = ax + \frac{ab}{c}\cdot k + by - \frac{ab}{c}\cdot k\\
&= a\cdot \left(x+\frac{b}{c}\cdot k\right) + b\cdot \left(y-\frac{a}{c}\cdot k\right) \tag{1}
\end{align*}
is another solution. Conversely, every solution is of this form: Take two solutions $ax + by = c$ and $au + bv = c$. Subtracting both equations, dividing by $c$ and rearranging yields
$$
\frac{a}{c}\cdot (x-u) = \frac{b}{c}\cdot (v-y). \tag{2}
$$
Since $\frac{a}{c}$ and $\frac{b}{c}$ are coprime, it follows that $\frac{b}{c}$ divides $x-u$ and, likewise, $\frac{a}{c}$ divides $v-y$. Hence, we find $n,m\in \mathbb Z$ with $\frac{a}{c}\cdot n = v-y$ and $\frac{b}{c}\cdot m = x-u$. Plugging both solutions into (2) gives
$$
\frac{a}{c}\cdot \frac{b}{c}\cdot m = \frac{b}{c}\cdot\frac{a}{c}\cdot n
$$
and hence $n=m$, because $\frac{ab}{c^2}\neq 0$. This shows that every solution is of the form (1).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1911772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
limit of product of $(a_1a_2.\dots a_n)^{\frac{1}{n}}$ How to calculate the following limit
$$ \lim_{n\rightarrow \infty} \left[ \left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)\cdots \left(1+\frac{n}{n}\right) \right]^\frac{1}{n} .$$
I was trying this by taking the $\log $ of the product and then limit but I am not getting the answer could anybody please help me. And, also is there any general rule for calculating the limit for $$ (a_1a_2.\dots a_n)^{\frac{1}{n}}.$$
Thanks.
| This answer uses Sterling's approximation to get the limit:
\begin{align}
l=\;&\left[ \left( 1+\frac { 1 }{ n } \right) \left( 1+\frac { 2 }{ n } \right) \cdots \left( 1+\frac { n }{ n } \right) \right]^{ \frac { 1 }{ n } } \\
=\; & \left[ \frac{1}{n^n}(n+1)(n+2)\dots(n+n) \right]^{ \frac { 1 }{ n } } \\
= \;& \frac{1}{n}\left[ \frac{(2n)!}{n!} \right]^{ \frac { 1 }{ n } } \\
\approx\; & \frac{1}{n}\left[\sqrt{2}\cdot 2^{2n}\left(\frac{n}{e} \right)^n\right]^{\frac{1}{n}} \Rightarrow \\
\lim_{n\rightarrow \infty}l= \;& \frac{4}{e}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1913595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 9,
"answer_id": 7
} |
Why isn't there an $x$ such that $\tan\left(\frac{\sin^{-1} x}{5}\right)=1$? If $\tan\left(\frac{\sin^{-1} x}{5}\right)=1$, then $x$ is equal to
$$\frac{-\pi}{2}\le \sin^{-1}x \le \frac{\pi}{2}$$
$$\frac{-\pi}{10}\le \frac{\sin^{-1}x}{5} \le \frac{\pi}{10}$$
This show that no value of such $x$ exist.
I want to know why this is happening.
Now in this approach,
$$\frac{\sin^{-1}x}{5}=\frac{\pi}{4}$$
$$\sin^{-1}x=\frac{5\pi}{4}$$
Why this is happening?
I think this is just happening because of principal value of $\sin^{-1}x$.
| $\sin^{-1} x = \frac{5 \pi}{4}$ gives $x=\sin \frac{5 \pi}{4} = \sin (\pi + \frac{\pi}{4})$, then $x = -\sin \frac{\pi}{4} = -\frac{1}{\sqrt 2}$. Your error is restricting $\sin^{-1} x$ to the values $-\frac{\pi}{2} \le \sin^{-1} x \le \frac{\pi}{2}$ (the principal value is within this range).
| {
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A faster way than tedious multiplying
Show that $(a+b+c)(a+b\epsilon +c\epsilon^2)(a+b\epsilon^2 + c\epsilon) = a^3 + b^3 + c^3 - 3abc$ If $$\epsilon^2 + \epsilon + 1 =0$$
The solution in the back of the book is given as
Proved by a direct check, taking into consideration that $\epsilon^2 = -\epsilon -1 \: \:$ and $\epsilon^3 = 1$
However the solutions still feels like tedious multiplication, doesn't it? Is there a faster, more elegant way to do this?
| One way would be by factorizing $a^3+b^3+c^3-3abc$.
$$a^3+b^3+c^3-3abc=(a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega)$$
where $\omega$ is the cube root of unity.
More on the factorization can be found here.
Then, clearly, $\epsilon=\omega$, and thus, $$\epsilon^3=1$$
Factorizing this gives,
$$(\epsilon-1)(\epsilon^2+\epsilon+1)=0$$
Thus,
$$\epsilon^2+\epsilon+1=0$$
| {
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$\left(\sum_{i=1}^n x_i\right) \cdot \left(\sum_{i=1}^n \frac{1}{x_i}\right) \ge n^2$, for all integers $n\ge 1$. Let $x_1,\ldots, x_n$ be positive integers. Use mathematical induction to prove that
$$\left(\sum_{i=1}^n x_i\right) \cdot \left(\sum_{i=1}^n \frac{1}{x_i}\right) \ge n^2$$ for all integers $n \ge 1$.
(Given Hint: For all positive integers $a$ and $b$, $\frac{a}{b}+\frac{b}{a} \ge 2$.)
Can anyone help? Thank you.
| Suppose the given inequality holds for an integer $n(\geq 1)$.
Then for $n+1$,
$$\left(\sum_{i=1}^{n+1} x_i\right)\left(\sum_{i=1}^{n+1} \frac{1}{x_i}\right)=\left(\sum_{i=1}^{n} x_i+x_{n+1}\right)\left(\sum_{i=1}^{n} \frac{1}{x_i}+\frac{1}{x_{n+1}}\right)$$
Expanding this gives us
$$=\left(\sum_{i=1}^{n} x_i\right)\left(\sum_{i=1}^{n} \frac{1}{x_i}\right)+x_{n+1}\sum_{i=1}^{n} \frac{1}{x_i}+\frac{1}{x_{n+1}}\sum_{i=1}^{n} x_i+1$$
Since the given inequality holds for the integer $n$, the first term is larger than $n^2$. So,
$$\geq n^2+x_{n+1}\sum_{i=1}^{n} \frac{1}{x_i}+\frac{1}{x_{n+1}}\sum_{i=1}^{n} x_i+1$$
Now we show that $x_{n+1}\sum_{i=1}^{n} \frac{1}{x_i}+\frac{1}{x_{n+1}}\sum_{i=1}^{n} x_i$ is larger than $2n$.
This is where we use the hint. Expanding the $\sum$ gives us
$$x_{n+1}\left(\frac{1}{x_1}+\frac{1}{x_2}+\cdots+\frac{1}{x_n}\right)+\frac{1}{x_{n+1}}\left(x_1+x_2+\cdots+x_n\right)$$
Rearranging the terms, we get
$$\left( \frac{x_{n+1}}{x_1}+\frac{x_1}{x_{n+1}} \right)+\left( \frac{x_{n+1}}{x_2}+\frac{x_2}{x_{n+1}} \right)+\cdots+\left( \frac{x_{n+1}}{x_n}+\frac{x_n}{x_{n+1}} \right)$$
and using the hint, this is larger than $2n$
Thus,
$$\left(\sum_{i=1}^{n+1} x_i\right)\left(\sum_{i=1}^{n+1} \frac{1}{x_i}\right)\geq n^2+2n+1=(n+1)^2$$
The inequality holds for $n+1$.
| {
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Mistake in basic algebra, I think? Problem prove $(2n+1)+(2n+3)+(2n+5)+...+(4n-1) =3n^2$
Induction proof: base case $n=1$ assume true for all $n$ prove for $n+1$.
The $n$th or last term becomes $(4(n+1)-1)=4n+3$.
We also sub $n+1$ in for all $n$ the $n-1$ term is $(4n-1)$ and the first term is $2(n+1)+1=2n+3$
The right side is $3(n+1)^2 = 3(n^2 + 2n +1 ) $
Next thing is I appear to be missing the first term in the sum $(2n+1)$ on the left.
Adding $(2n+1)$ to both sides and subtracting $4n+3$ from both sides we get the $n$ case that equals $ 3n^2 $ on the left and
$$3n^2 + 6n +3 + 2n+1 -4n -3= 3n^2 + 4n+1$$
Which leaves me with $(2n+1)+(2n+3)+(2n+5)+...+(4n-1) =3n^2 +4n +1$
Which is approximately $4n+1$ on the right bigger than what I started with and is exactly the same on the left algebraically I must of done something wrong.
Which leads me to my question, what was it?
| The left hand side is the sum of the odd numbers from $2n+1$ up to $4n-1$. Thus when going $n\to n+1$, the first summand $2n+1$ is dropped, and two summands $4n+1$ and $4n+3$ are appended a the end.
| {
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Prove: $\cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)}=\frac {3}{4} \cos{3A}$ Prove that:
$$\cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)}=\frac {3}{4} \cos{3A}$$
My Approach:
$$\mathrm{R.H.S.}=\frac {3}{4} \cos{3A}$$
$$=\frac {3}{4} (4 \cos^3{A}-3\cos{A})$$
$$=\frac {12\cos^3{A} - 9\cos{A}}{4}$$
Now, please help me to continue from here.
| First note that $\cos^3\theta=\frac34\cos\theta+\frac14\cos3\theta$. Observing that $$\cos(3A+3\cdot\tfrac23\pi)=\cos(3A+3\cdot\tfrac43\pi)=\cos3A,$$and adding the results of $\cos^3\theta$ for $\theta=A$, $\theta=A+\frac23\pi$, and $\theta=A+\frac43\pi$, using the fact that $\cos A+\cos(A+\frac43\pi)=2\cos(A+\frac23\pi)\cos\frac23\pi=-\cos(A+\frac23\pi)$, now gives the required form.
| {
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Proving this identity: $\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \sec^2\theta \csc^2\theta - 2$ I have tried solving this trig. identity, but I get stuck when it comes to the $-2$ part. Any suggestions?
$$\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \sec^2\theta \csc^2\theta - 2$$
| \begin{align*}
\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} &= \tan^2\theta + \cot^2\theta \\
&= \sec^2\theta - 1+ \csc^2\theta - 1\\
&= \frac{1}{\cos^2\theta} + \frac{1}{\sin^2\theta} - 2\\
&= \frac{\sin^2\theta + \cos^2\theta}{\sin^2\theta \cos^2\theta} -2\\
&= \sec^2\theta \csc^2\theta - 2
\end{align*}
| {
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Let $p,q$ be irrational numbers, such that, $p^2$ and $q^2$ are relatively prime. Show that $\sqrt{pq}$ is also irrational. Progress:
Since, $p,q$ are irrationals and $p^2$ and $q^2$ are relatively prime, thus, $p^2\cdot{q^2}$ cannot be a proper square, so, $pq$ is also irrational. Suppose, $pq=k$, then: $$\sqrt{k}\cdot\sqrt{k}=\sqrt{pq}\cdot\sqrt{pq}=k$$
Which implies that $\sqrt{pq}$ is irrational, since, $k$ is irrational and that $p^2\cdot{q^2}$ cannot be a proper square being $p^2$ and $q^2$ relatively prime and that concludes the proof.
The above lines are my attempt to prove the assertion. Is the proof correct? If not then how can I improve or disprove it.
Regrads
| Suppose $\sqrt{pq}$ is rational, say $\sqrt{pq} = \dfrac mn$ where $m$ and $m$ are positive, relatively prime integers. Then $p^2 q^2 = \dfrac{m^4}{n^4}$ where $p^2q^2$ is an integer. Since $m$ and $n$ are relatively prime integers, then $m^4$ and $n^4$ are also relatively prime integers. It follows that $n^4=1$ and hence $n=1$. So we find that $p^2q^2 = m^4$ for some positive integer $m$.
Since $p^2$ and $q^2$ are relatively prime integers,then any prime divisor of $m$ cannot be a common divisor of both $p^2$ and $q^2$. It follow that there must be relatively prime positive integers $a$ and $b$ such that $m = ab$, $\;p^2 = a^4$, and $q^2 = b^4$. Hence $p=a^2$ and $q=b^2$, which implies that $p$ and $q$ are rational (integers). By contradiction, $\sqrt{pq}$ must be irrational.
| {
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Show by induction that $n! > n \cdot F_n$ for all $n > 3$ Show by induction that $n! > n \cdot F_n$ for all $n> 3$, where $F_n$ is the $n$th fibonacci number.
For the basis step, I put $n=4$ and got:
\begin{align*}
&4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24 \\
&4 \cdot F_4 = 4 \cdot 3 = 12
\end{align*}
So the statement holds for $n=4$. For the inductive step, we assume the statement holds for a $k > 3$, and checks for $n = k+1$. We have:
\begin{align*}
&(k+1)! = (k+1)\cdot k \cdot (k-1) \cdot \; \dotsc \; \cdot 1 = (k+1 ) \cdot k! \\
&(k+1) \cdot F_{k+1} = (k+1)(F_k + F_{k-1}) = k \cdot F_k + k\cdot F_{k-1} + F_{k} + F_{k-1}
\end{align*}
By the induction assumption, we know that
\begin{align*}
k\cdot F_k + (k-1) \cdot F_{k-1} < k! + (k-1)! \leq k! + k! < 2k! <(k+1)k!,
\end{align*}
But we see that
\begin{align*}
k\cdot F_k + (k-1) \cdot F_{k-1} < (k+1) \cdot F_{k+1},
\end{align*}
so we can't really say if $(k+1) \cdot F_{k+1}$ is less than $(k+1)!$ or not.
Does this mean that the statement is not true, or is there just a better way of proving it?
| How about the following way?
It is sufficient to prove that $(n-1)!\gt F_n$ for $n\gt 3$.
Inductive step : we assume that $(k-1)!\gt F_k$.
Then, multiplying the both sides by $k$ gives
$$k!\gt kF_k\gt F_k+F_k\gt F_k+F_{k-1}=F_{k+1}$$
| {
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Partition of a prime $p$ into primes raised to a power to be dvisible by $p$ Find all the partitions of prime $p$ in primes less than $p$. Raise each term in each partition by some power $k > 1$ to see if the sum of these terms will be divisible by $p$. Of course, $k$ can differ for each partition. For example, for $5$, $2^3 + 3^3 = 35 = 5 \times 7$ and for $11$, $4 \times 2^4 + 3^4 = 209 = 11 \times 19$. Do you think all partitions will eventually have some least power $k$? Will it be less than $p$ itself?
| As McFry has already answered, the first question is answered by Fermat's little theorem: Since $a^p+b^p+\cdots+z^p\equiv a+b+\cdots+z$, there will always be a least power satisfying the OP's condition, and that least power will be no greater than $p$.
As for the second question, $p=7=2+2+3$ is an example where the least power is $p$. That is, $7$ does not divide any of the numbers $4+4+9=17$, $8+8+27=43$, $16+16+81=113$, $32+32+243=309$, or $64+64+729=857$.
It could be of interest to see how often $p$ is the least power for one of its partitions. To summarize the small examples (and someone should doublecheck this), the smallest powers for prime-only partitions of $7$ and $11$ are
$$\begin{align}
2^3+5^3&=7\cdot19\\
2^7+2^7+3^7&=7\cdot349\\
2^6+2^6+7^6&=11\cdot10707\\
3^6+3^6+5^6&=11\cdot1553\\
2^{11}+2^{11}+2^{11}+5^{11}&=11\cdot4439479\\
2^{11}+3^{11}+3^{11}+3^{11}&=11\cdot48499\\
2^4+2^2+2^4+2^4+3^4&=11\cdot19
\end{align}$$
(so $2$ out of $5$ prime-only partitions of $11$ have least power $11$).
| {
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Simplifying inverse trigonometric expressions such as $\cos^{-1}\left(\frac{x^2-1}{x^2+1}\right)+\tan^{-1}\left(\frac{2x}{x^2-1}\right)$ Okay so I'm just looking for a short cut method or a method that is not so long for simplifying expressions like this
$$\cos^{-1}\left(\frac{x^2-1}{x^2+1}\right)+\tan^{-1}\left(\frac{2x}{x^2-1}\right)$$
Here is how I do it: I take the first term of the equation and convert it to $\tan^{-1}$ form, and then I apply the formula of $\tan^{-1}a + \tan^{-1}b$. But this method is really time consuming; plus, it results in mistakes most of the times. Any alternatives or how you would do it, let me know, I'll be grateful. Thanking in anticipation.
Edit. Do I put $x= \tan y$ and solve? That should make it easy.
| Revised for the last part
In your specific case
$$
\cos ^{ - 1} \left( {\frac{{x^2 - 1}}
{{x^2 + 1}}} \right) + \tan ^{ - 1} \left( {\frac{{2x}}
{{x^2 - 1}}} \right)
$$
you may note that:
$$
\left\{ \begin{gathered}
x^2 - 1 = \operatorname{Re} \left( {\left( {x + i} \right)^2 } \right) \hfill \\
2x = \operatorname{Im} \left( {\left( {x + i} \right)^2 } \right) \hfill \\
\left( {x^2 + 1} \right) = \left| {\left( {x + i} \right)} \right|^2 = \left| {\left( {x + i} \right)^2 } \right| \hfill \\
\end{gathered} \right.
$$
so that
$$
\begin{gathered}
f(x)= \cos ^{ - 1} \left( {\frac{{x^2 - 1}}
{{x^2 + 1}}} \right) + \tan ^{ - 1} \left( {\frac{{2x}}
{{x^2 - 1}}} \right) = \hfill \\
= \cos ^{ - 1} \left( {\frac{{\operatorname{Re} \left( {\left( {x + i} \right)^2 } \right)}}
{{\left| {\left( {x + i} \right)^2 } \right|}}} \right) + \tan ^{ - 1} \left( {\frac{{\operatorname{Im} \left( {\left( {x + i} \right)^2 } \right)}}
{{\operatorname{Re} \left( {\left( {x + i} \right)^2 } \right)}}} \right) = \hfill \\
= \cos ^{ - 1} \left( {\cos \left( {2\arg \left( {x + i} \right)} \right)} \right) + \tan ^{ - 1} \left( {\tan \left( {2\arg \left( {x + i} \right)} \right)} \right) \hfill \\
\end{gathered}
$$
Here we have to stop and do some case distinction, because
of the codomain limitation in the definition of $arccos$ and $arctan$, which if overlooked will cause the error signalled by Jean-Claude Arbaut.
In fact
$$
\cos ^{ - 1} \left( {\cos \alpha } \right) = \left| {\alpha _ * } \right|\quad \quad \tan ^{ - 1} \left( {\tan \alpha } \right) = \alpha _{\, * * } \quad \left| {\;\alpha _{\, * * } \ne \pi /2} \right.
$$
where the star indices indicate the necessary reductions mod $2\pi$ and $\pi$, that is
$$
\begin{gathered}
\alpha _{\, * } = \pi - \bmod \left( {\pi - \alpha ,\;2\,\,\pi } \right)\quad \left| \; \right. - \pi < \alpha _{\, * } \leqslant \pi \hfill \\
\alpha _{\, * * } = \pi /2 - \bmod \left( {\pi /2 - \alpha ,\;\,\pi } \right)\quad \left| \; \right. - \,\pi /2 < \alpha _{\, * * } \leqslant \pi /2 \hfill \\
\end{gathered}
$$
So the development above continues as:
$$
\begin{gathered}
f(x) = \cos ^{ - 1} \left( {\frac{{x^2 - 1}}
{{x^2 + 1}}} \right) + \tan ^{ - 1} \left( {\frac{{2x}}
{{x^2 - 1}}} \right)\quad \left| {\;x \ne 0,\; \pm 1} \right.\quad = \hfill \\
= \cos ^{ - 1} \left( {\cos \left( {2\arg \left( {x + i} \right)} \right)} \right) + \tan ^{ - 1} \left( {\tan \left( {2\arg \left( {x + i} \right)} \right)} \right) = \hfill \\
= \left| {\left( {2\arg \left( {x + i} \right)} \right)_ * } \right| + \left( {2\arg \left( {x + i} \right)} \right)_{\, * * } = \hfill \\
= \left| {\left( {2\tan ^{ - 1} \left( {1/x} \right)} \right)_ * } \right| + \left( {2\tan ^{ - 1} \left( {1/x} \right)} \right)_{\, * * } \hfill \\
\end{gathered}
$$
for finally arriving at
$$
f(x) = \left\{ {\begin{array}{*{20}c}
0 & {x < - 1} \\
\pi & { - 1 < x \leqslant 0} \\
{4\tan ^{ - 1} \left( {1/x} \right) - \pi } & {0 < x < 1} \\
{4\tan ^{ - 1} \left( {1/x} \right)} & {1 < x} \\
\end{array} } \right.
$$
where the critical points have been filled according to the limit values
of $f(x)$.
| {
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Zeros of $a(1-x)x^b-(1-a)x(1-x)^b = 0$ What are the zeros of this equation?
$$a(1-x)x^b-(1-a)x(1-x)^b = 0$$
I was told that there were 3 zeros, however I've only been able to find 2 of them. Namely, $x=0$ and $x=1$. I can't think of any way to solve it algebraically and I'm not seeing any other obvious solutions. I also tried using Taylor expansion on the $(1-x)^b$ term, but it didn't prove to be too helpful. I graphed the equation with an arbitrary choice of a and b, but only noticed 2 zeros.
| The equation has at either $3$ or $4$ roots, assuming that $0 < a < 1$.
In particular, the third root they were probably looking for is
$$
x = \frac{1}{1 + \sqrt[(b-1)]{r}},
$$
where $r = \frac{a}{1-a}$.
This root is between $0$ and $1$.
If $b$ is odd and $a \ne \frac12$, there is a second root,
$$
x = \frac{1}{1 - \sqrt[(b-1)]{r}}.
$$
Let us write
$$
f(x) = x(1-x)\left( a x^{b-1} - (1-a) (1-x)^{b-1} \right).
$$
So in addition to the roots $x = 0$ and $x = 1$, we will have a root if
$$
a x^{b-1} = (1-a) (1-x)^{b-1}
$$
i.e.,
$$
\left(\frac{1}{x} - 1\right)^{b-1} = \frac{a}{1 - a}.
$$
Now, $\frac{a}{1-a}$ is some positive number $r$.
So it has a $(b-1)$th root, $\sqrt[(b-1)]{r} > 0$.
So $\frac{1}{x} = 1 + \sqrt[(b-1)]{r} > 1$,
and
$$
\boxed{x = \frac{1}{1 + \sqrt[(b-1)]{r}}}.
$$
If $b$ is odd, $b-1$ is even, so there is also a negative $(b-1)$th root of $r$, and thus we also have the solution
$$
x = \frac{1}{1 - \sqrt[(b-1)]{r}}.
$$
| {
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Show that $\{1, (1, 2)(3,4), (1, 3)(2,4), (1,4)(2, 3)\}$ is a subgroup of $S_4$ Show that {$1, (1, 2)(3,4), (1, 3)(2,4), (1,4)(2, 3)$} is a subgroup of $S_4$
My attempt:
Let $H = \{1, (1, 2)(3,4), (1, 3)(2,4), (1,4)(2, 3)\}$, where $1$ is the identity permutation. Then,
$1\circ(1, 2)(3,4) = (1, 2)(3,4) $
$1\circ(1, 3)(2,4) =(1, 3)(2,4) $
$1\circ(1,4)(2, 3)=(1,4)(2, 3) $
$(1, 2)(3,4)\circ1=(1, 2)(3,4) $
$(1, 3)(2,4) \circ1=(1, 3)(2,4) $
$(1,4)(2, 3)\circ1=(1,4)(2, 3) $
$(1,2)(3,4)\circ(1,3)(2,4)=(1,4)(2,3)$
$(1,3)(2,4)\circ(1,2)(3,4)=(1,4)(2,3)$
and so on.... (I will multiply all the elements with each other)
$H$ is finite and for all $a,b\in H$, $ab\in H$. Therefore, $H$ is a subgroup of $S_4$
Side note: What is
$(1,3)(2,4)\circ(1,3)(2,4)=?$
Do I need to include all the elements composed with itself?
| Let $S=\{1, (1, 2)(3,4), (1, 3)(2,4), (1,4)(2, 3)\}$, and relabel the elements $\sigma_1=1$, $\sigma_2=(1, 2)(3,4)$, $\sigma_3=(1, 3)(2,4)$, $\sigma_4=(1,4)(2, 3)$, where we note $\sigma_1=1=(1)(2)(3)(4)$ is the identity permutation.
Consider the product $(1,3)(2,4){\circ}(1,3)(2,4)$ in $S_4$. Remember when computing products one reads from right to left. So first look at the permutation $(1,3)$ on the RHS of $\circ$, this maps $1$ to $3$ (we can just ignore the permutation $(2,4)$ for the moment since $1$ and $3$ do not belong to it). Now consider the composition $(1,3){\circ}(1,3)$. Well the permutation $(1,3)$ on the RHS of $\circ$ maps $1$ to $3$, then the permutation $(1,3)$ on the LHS of $\circ$ maps $3$ back to $1$. Doing likewise with $3$ we see the mapping $3\mapsto1\mapsto3$, and so $(1,3){\circ}(1,3)=(1)(3)$, and $1$ and $3$ are fixed.
To work out $(1,3)(2,4){\circ}(1,3)(2,4)$ simply trace where each element on the RHS of $\circ$ is mapped to on the LHS of $\circ$:
$1\mapsto3\mapsto1$
so $1$ is fixed.
$3\mapsto1\mapsto3$
so $3$ is fixed.
$2\mapsto4\mapsto2$
so $2$ is fixed.
$4\mapsto2\mapsto4$
so $4$ is fixed.
Hence $(1,3)(2,4){\circ}(1,3)(2,4)=(1)(2)(3)(4)$, which is the identity permutation, and so the element $(1,3)(2,4)$ is self-inverse. In fact doing this with the other two nonidentity permutations reveals they are also self-inverse, and we can write $\sigma_i\circ\sigma_i^{-1}=1$, for $i=2,3,4$. Note to get the cycle decomposition of $\sigma^{-1}$ just write the numbers in each cycle back-to-front, e.g., $\sigma_2=(1, 2)(3,4)$, and $\sigma_2^{-1}=(2, 1)(4,3)$, this makes it clearer where things end up. You have already worked out $\sigma_2\circ\sigma_3=\sigma_3\circ\sigma_2=\sigma_4$: Check:
$\sigma_2\circ\sigma_4=\sigma_4\circ\sigma_2=\sigma_3$
and
$\sigma_3\circ\sigma_4=\sigma_4\circ\sigma_3=\sigma_2$.
This gives you a subgroup of $S_4$ isomorphic to the Klein $4$-group.
$$
\begin{array}{c|ccc}
\circ & \sigma_1 & \sigma_2 & \sigma_3 & \sigma_4 \\ \hline
\sigma_1 & \sigma_1 & \sigma_2 & \sigma_3 & \sigma_4 \\
\sigma_2 & \sigma_2 & \sigma_1 & \sigma_4 & \sigma_3 \\
\sigma_3 & \sigma_3 & \sigma_4 & \sigma_1 & \sigma_2\\
\sigma_4 & \sigma_4 & \sigma_3 & \sigma_2 & \sigma_1 \\
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1934863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to find all solutions of $x^2-y^2=x+y$? For this equation $$x^2-y^2=x+y$$ I know there are 2 functional solutions, $y=x-1$, and $y=-x$.
The second solution we can see immediately because both sides will always be $0$.
The first solution is also easy to get:$$(x-y)(x+y)=x+y\ \ ///\div(x+y)\ \ \ [y\neq-x]\\x-y=1\\y=x-1$$
But how do I solve this in a formal way? So that I'll find all solutions, and show that there aren't other solutions?
| $x^2 - y^2 = (x-y)(x+y) = x+y$
If $x+y=0$ then all $\{(x, -x)|x \in \mathbb R\}$ are solutions.
If $x+y\ne 0$ then $(x-y) = 1$ and $y = x -1$ so all $\{(x, x-1)|x \in \mathbb R\}$ are solutions.
So solutions are $\{(x, -x)|x \in \mathbb R\}\cup\{(x, x-1)|x \in \mathbb R\}$.
....
or $x^2 - y^2 = x+y$
$x^2 - y^2 -(x+y) = 0$
$(x+y)[(x-y) - 1] = 0$
so either $x+y = 0$ or $x-y-1 = 0$ so....
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1935369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Arc Circle Intersection in Euclidean 3D While programming on a GPS based game I came upon the following problem for 3D euclidean space:
Assume we have an arc (given by a sphere center coordinate C and two euclidean coordinates $A$ and $B$ on that sphere, with $AC = AB$ and $A !=B$) and another sphere (given by an arbitrary center coordinate $Q$ and radius $R > 0$). All points given in cartesian coordinates $(x, y, z)$. We will exclude the edge cases where $A, B, C$ are aligned.
Now we search for intersect points $X$ (trivial: $0 <= |X| <= 2$) between the arc and the sphere.
Is there a simple formula with acceptable numerical stability (we have GPS related precision limits anyway) for efficiently computing those two (or fewer) coordinates based on the given 5 parameters $A,B,C,Q,R$?
| I thought about it, and the second answer I proposed earlier is not all that bad. I'm going to treat the points $A$, $B$, $C$, $Q$, as coordinate triples, so that I can add and subtract them as vectors. I'm also going to write $r_2$ for $R$, the radius of the second sphere, and $r_1$ for the radius of the first. Here goes:
$$
\newcommand{\a}{\mathbf a}
\newcommand{\b}{\mathbf b}
\newcommand{\c}{\mathbf c}
\newcommand{\u}{\mathbf u}
\newcommand{\v}{\mathbf v}
\newcommand{\w}{\mathbf w}
$$
\begin{align}
r_1 &= \|A - C\| \\
r_2 & = R\\
\a &= (A - Q)/r_2 \\
\b &= (B - Q)/r_2\\
\c &= (C - Q)/r_2.
\end{align}
In doing this, we've changed to a coordinate system in which the second sphere is at the origin, and has radius 1; we'll convert back at the end.
Let
\begin{align}
\u &= \frac{r_2}{r_1}(\a - \c) \\
\v &= \frac{r_2}{r_1}(\b - \c) \\
\v' &= \v - (\u \cdot \v) \u \\
\w & = \v' / \| \v' \|
\end{align}
Note that $\|u\| = \|v \| = \|w\| = 1$. And $\u \cdot \w = 0$.
Now the arc from $\u$ to $\w$ passes through $\v$ at some angle $T$. What angle?
\begin{align}
T &= \arccos ({\v \cdot \u})
\end{align}
Let
\begin{align}
\gamma(t) = \c + r_1\cos(t) \u + r_1\sin(t) \w
\end{align}
for $0 \le t \le T$.
Then $\gamma$ parameterizes the arc from $\a$ to $\b$ on the (translated and scaled) first sphere. We seek points where
\begin{align}
\| \gamma(t) \|^2 &= 1.
\end{align}
because those are the ones that lie on the second sphere (in the new coordinate system).
Writing this out, we have
\begin{align}
1 &= \| \gamma(t) \|^2 \\
&= \gamma(t) \cdot \gamma(t)\\
&= (\c + r_1\cos(t) \u + r_1\sin(t) \w) \cdot (\c + r_1\cos(t) \u + r_1\sin(t) \w) \\
&= (\c\cdot \c) + 2r_1\cos(t) (\u \cdot \c) + 2 r_1\sin(t) (\w \cdot \c) + r_1^2\cos^2(t) (\u \cdot \u) + 2r_1^2\cos(t)\sin(t) (\u \cdot \w) + r_1^2\sin^2(t) (\w \cdot \w)
\end{align}
Fortunately, $\u$ and $\w$ are orthogonal, and each of $\u$ and $\w$ is a unit vector. So we have
\begin{align}
1 &= \| \gamma(t) \|^2 \\
&= (\c\cdot \c) + 2r_1\cos(t) (\u \cdot \c) + 2 r_1\sin(t) (\w \cdot \c) + r_1^2\cos^2(t) + r_1^2\sin^2(t) \\
&= (\c\cdot \c) + 2r_1\cos(t) (\u \cdot \c) + 2 r_1 \sin(t) (\w \cdot \c) + r_1^2 \\
\end{align}
Now we write $s = \tan(\frac{t}{2})$, which turns out to mean that $\sin t = \frac{2s}{1+s^2}$ and $\cos t = \frac{1-s^2}{1+s^2}$. So our equation becomes
\begin{align}
1
&= (\c\cdot \c) + 2r_1\frac{1-s^2}{1+s^2} (\u \cdot \c) + 2 r_1\frac{2s}{1+s^2} (\w \cdot \c) + r_1^2 \\
\end{align}
Multiplying through by $1+s^2$ gives
\begin{align}
1+s^2
&= (1+s^2)(\c\cdot \c) + 2r_1(1-s^2)(\u \cdot \c) + 4r_1s (\w \cdot \c) + r_1^2(1+s^2) \\
0
&= -1 -s^2 + (1+s^2)(\c\cdot \c) + 2r_1(1-s^2)(\u \cdot \c) + 4r_1s (\w \cdot \c) + r_1^2(1+s^2) \\
0 &= s^2 (\|c\|^2 - 1 - 2 r_1\u \cdot \c + r_1^2) + s 4r_1 (\w \cdot \c) + 2r_1(\u \cdot \c) + (\c \cdot \c) - 1 + r_1^2
\end{align}
which is a quadratic $As^2 + Bs + C = 0$, where the coefficients are
\begin{align}
A &= (\|c\|^2 - 1 - 2 r_1u \cdot c + r_1^2)\\
B &= 4r_1(w \cdot c)\\
C &= 2r_1(u \cdot c) + (c \cdot c) - 1 + r_1^2
\end{align}
(My apologies here for reusing the names $A, B, C$.) This quadratic can be solved with the quadratic formula to get solutions $s_1, s_2$. If both solutions have an imaginary part (which happens when $B^2 - 4AC < 0$), then the arc does not intersect the sphere. Otherwise, they are real, but may be identical. Let's see how to translate them back into an overall solution to the problem.
At this point, now that we've solved the quadratic, "A, B, C" go back to their original meanings as points in 3-space, OK?
From $s_i$ ($i = 1, 2$ henceforth), we can compute
$$
t_i = \arctan(2 s_i).
$$
If either value of $t_i$ is outside the interval $[0, T]$, we ignore it -- it does not correspond to a point of intersection of the arc and the sphere.
From this, we can compute $\cos t_i$ and $\sin t_i$, and thus compute
$$
\gamma_i = \c + r_1\cos t_i \u + r_1\sin t_i \w
$$
and then, going back to the original coordinate system, compute
$$
X_i = r_2\gamma_i + Q
$$
and the points $X_i$ are the intersections you seek.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1937019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
How to prove $\sin 10^\circ = \frac{-1+\sqrt{9-8\sin 50^\circ}}{4}$? How to prove this identity?
$$\sin 10^\circ = \frac{-1+\sqrt{9-8\sin 50^\circ}}{4}$$
Is this a particular case of a more general identity? Also, is it possible to give a geometric proof of this equality?
| The way you would derive this identity is as follows:
Note that $\sin A + \sin B = 2\sin\big(\frac{A+B}{2}\big)\cos\big(\frac{A-B}{2}\big)$
Putting $A=50$ and $B=10$,
$$
\sin 50 + \sin 10 = 2\sin\big(30\big)\cos\big(20\big) = \cos 20
$$
Now, we know that $\cos 20 = 1- 2\sin^2 10$. So we substitute:
$$
\sin 50 + \sin 10 = 1 - 2\sin^2 10
$$
Taking all the $\sin 10$ terms to one side to make a quadratic equation:
$$
2\sin^2 10 + \sin 10 + (\sin 50 - 1) = 0
$$
Solving for $\sin 10$ like a quadratic equation gives you your answer.
This answer contained two important points:
1) $\sin 30 = \frac{1}{2}$, helped us remove one sine factor from the sum.
2) $20 = 2*10$ allowed us to further reduce the equation to just the sines of $10$ and $50$.
I do not know of a general formula, but with respect to these constraints, $10$ and $50$ feel like unique choices, summing to $60$, twice of $30$ (whose sine is nice), and a difference of $40$, half of which is $20$, nicely expressible in terms of trigonometric functions of $10$. I cannot think of generalizations of this problem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Suppose $0< a,b,c < 1$ and $ab + bc + ca = 1$. Find the minimum value of $a + b + c + abc$. Suppose $0< a,b,c < 1$ and $ab + bc + ca = 1$. Find the minimum value of $a + b + c + abc$.
How can I use the first two equations to help solve the third? I'm stuck. Any solutions are greatly appreciated!
| When $a = b = c = \frac{1}{\sqrt3}$,
we have $ab + bc + ca = 1$ and $a + b + c + abc = \frac{10\sqrt3}{9}$.
Indeed, the minimum of $a + b + c + abc$ is $\frac{10\sqrt3}{9}$.
It suffices to prove that, for all $a, b, c\in (0, 1)$ with $ab + bc + ca = 1$,
$$a + b + c + abc\ge \frac{10\sqrt3}{9}.$$
Let $p = a + b + c, q = ab + bc + ca = 1, r = abc$.
We need to prove that
$$p + r \ge \frac{10\sqrt3}{9}. \tag{1}$$
Using $p^2 \ge 3q$, we have $p \ge \sqrt3$.
If $p \ge \frac{10\sqrt3}{9}$, clearly (1) is true.
If $\sqrt3 \le p < \frac{10\sqrt3}{9}$, using $r \ge \frac{4pq - p^3}{9}$ (3 degree Schur), we have
\begin{align*}
p + r - \frac{10\sqrt3}{9}
&\ge p + \frac{4pq - p^3}{9} - \frac{10\sqrt3}{9}\\
&= \frac{13}{9}p - \frac{1}{9}p^3
- \left(\frac{13}{9}\sqrt3 - \frac{1}{9}(\sqrt3)^3\right)\\
&= \frac19(p - \sqrt3)(10 - p^2 - p\sqrt 3)\\
&\ge 0.
\end{align*}
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1939973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Find limit of sequence sum Let us have sum of sequence (I'm not sure how this properly called in English): $$X(n) = \frac{1}{2} + \frac{3}{4}+\frac{5}{8}+...+\frac{2n-1}{2^n}$$
We need
$$\lim_{n \to\infty }X(n)$$
I have a solution, but was unable to find right answer or solution on the internet.
My idea:
This can be represented as $$ \frac{1}{2} + \frac{1}{4} + \frac{2}{4} + \frac{3}{8}+\frac{2}{8}+\frac{5}{16} + \frac{2}{16} ... + ...$$
Which is basically
$\frac{1}{2} + \frac{1}{2} +\frac{1}{4}+\frac{1}{8}$ - 1/2 + geometric progression + our initial sum divided by 2.
And then I thought: hey, so I can figure out one part of this sum, and second is twice smaller, and then it forms a cycle! (I suppose).
So it would be $B_1 = 1/2 + b_1/(1-1/2) = 3/2$
$$lim_{n \to\infty }X(n) = B_1/(1-1/2) = 3$$
Is this correct?
| $$\frac12+\frac34+\ldots+\frac{2n-1}{2^n}=\left(1-\frac12\right)+\left(1-\frac14\right)+\left(\frac34-\frac18\right)+\ldots\left(\frac n{2^{n-1}}-\frac1{2^n}\right)=$$
$$1+1+\frac34+\ldots+\frac n{2^{n-1}}-\frac12-\frac14-\ldots-\frac1{2^n}\xrightarrow[n\to\infty]{}$$
$$\to\sum_{k=1}^\infty\frac k{2^{k-1}}-\sum_{k=1}^\infty\frac1{2^k}$$
Taking into account that for $\;|x|<1\;$ we have
$$\frac1{1-x}=\sum_{n=0}^\infty x^n\implies\frac1{(1-x)^2}=\sum_{n=1}^\infty nx^{n-1}$$
Substitute $\;x=\cfrac12\;$ above , one in each sum, and get:
$$\sum_{n=1}^\infty\frac n{2^{n-1}}-\sum_{n=1}^\infty\frac1{2^n}=\frac1{\left(1-\frac12\right)^2}-\left(\frac1{1-\frac12}-1\right)=4-(2-1)=3$$
and your answer is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1947082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Weighted sum of two dice such that the result is a random integer between $0$ and $35$
The numbers shown by 2 dice are labelled $d$ and $e$. $A, B$ and $C$ are constants, giving a score $S=Ad + Be + C$. Find $A, B$ and $C$ such that the range of possible values for $S$ covers all integers from $0$ to $35$, with an equal probability of each score. [Oxford PAT exam, 2011]
I have tried to find equations in terms of $A, B, C$ for different values of $S$.
| The maximum value of $S$ (assuming $A,B\geqslant0$) is $6(A+B)+C$, and the minimum value $A+B+C$. This gives us the system of equations
\begin{align}
6A+6B+C &= 35\\
A+B+C &= 0.
\end{align}
Gaussian elimination yields $C=-7$, and therefore $A+B=7$. So $(A,B,C)=(6,1,-7)$ and $(1,6,-7)$. Taking $(A,B,C)=(6,1,-7)$ yields
$$S = 6d + e -7, $$
so that for any $j\in\{0,1,\ldots,35\}$, $$j= 6d_j+e_j-7 $$ where \begin{align}
d_j &= 1 +\left\lfloor \frac {j}6\right\rfloor\\
e_j &= j+7 -6d_j.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1949330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Why does $\sum_{n=0}^k \cos^{2k}\left(x + \frac{n \pi}{k+1}\right) = \frac{(k+1)\cdot(2k)!}{2^{2k} \cdot k!^2}$? In the paper "A Parametric Texture Model based on Joint Statistics of Complex Wavelet Coefficients", the authors use this equation for the angular part of the filter in polar coordinates:
$$\sum_{n=0}^k \cos^{2k}\left(x + \frac{n \pi}{k+1}\right)$$
My friend and I have tested many values of $k > 1$, and in each case this summation is equal to
$$\frac{(k+1)\cdot(2k)!}{2^{2k} \cdot k!^2}$$
The paper asserts this as well.
We are interested in having an analytic explanation of this equality, if it really holds. How can we derive this algebraically?
TL;DR
Is this true, and if so, why?
$$\sum_{n=0}^k \cos^{2k}\left(x + \frac{n \pi}{k+1}\right) = \frac{(k+1)\cdot(2k)!}{2^{2k} \cdot k!^2}$$
| First note that
$$
\left[\cos\left(x+\frac{n\pi}{k+1}\right)\right]^{2k} = \frac{1}{2^{2k}} \exp\left(-i2k\left(x+\frac{n\pi}{k+1}\right)\right)\sum_{j=0}^{2k} {2k \choose j} \exp\left(i2j\left(x+\frac{n\pi}{k+1}\right)\right)
$$
Then,
\begin{align}
\sum_{n=0}^k \left[\cos\left(x+\frac{n\pi}{k+1}\right)\right]^{2k} &= \frac{1}{2^{2k}} \sum_{n=0}^k \exp\left(-i2k\left(x+\frac{n\pi}{k+1}\right)\right)\sum_{j=0}^{2k} {2k \choose j} \exp\left(i2j\left(x+\frac{n\pi}{k+1}\right)\right) \\
&= \frac{1}{2^{2k}} \sum_{j=0}^{2k} {2k \choose j} \sum_{n=0}^k \exp\left(i2\left(x+\frac{n\pi}{k+1}\right)(j-k)\right) \tag{1}
\end{align}
Now, suppose that $j \neq k$. Then,
\begin{align}
(1) &= \frac{1}{2^{2k}} \sum_{j=0}^{2k} {2k \choose j} \exp(i2x(j-k)) \sum_{n=0}^k \left[\exp\left(i2\frac{\pi}{k+1}(j-k)\right)\right]^n \\
&= \frac{1}{2^{2k}} \sum_{j=0}^{2k} {2k \choose j} \exp(i2x(j-k)) \frac{1 - \left[\exp\left(i2\frac{\pi}{k+1}(j-k)\right)\right]^{k+1}}{1-\exp\left(i2\frac{\pi}{k+1}(j-k)\right)} \\
&= 0
\end{align}
Thus, considering $(1)$ only when $j=k$, we have the result.
| {
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"url": "https://math.stackexchange.com/questions/1951708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
$y'=$ ${-x+\sqrt{(x^2+4y)}}\over 2$ , $y(2)=-1$ Prove that the solutions $y_1$ and $y_2$ for the initial value problem
$y'=$${-x+\sqrt{(x^2+4y)}}\over 2$ , $y(2)=-1$
are:
$y_1=1-x$
$y_2={{-x^2}\over 4}$
And explain why the two solutions not contradiction with the theory of existence and uniqueness.
( sorry I don't speak English well )
My answer:
For the initial condition :
$y_1=1-x$ Satisfy the initial condition
$y(2)=-1$
$y_1=1-2=-1$
And Satisfy the differential equation
${y_1}^{\prime}=-1$
${{-x+\sqrt{x^2+4y}}\over 2}={{-x+\sqrt{x^2+4(1-x)}}\over 2}$
$={{-x+\sqrt{(x-2)^2}}\over 2}$
$={{-x+x-2}\over 2}=-1$
L.h.s.=R.h.s.
$y^{\prime}={{-x+\sqrt{x^2+4y}}\over 2}$
Also $y_2$ Satisfy the initial condition
$y_2={{-x^2}\over4}={-4\over4}=-1$
*
*And satisfy the differential equation.
$y_2^{\prime}={-x\over 2}$
${{-x+\sqrt{x^2+4y}}\over 2}={{-x+\sqrt{x^2+4({-x^2\over 4})}}\over 2}={-x\over 2}$
And for second part of exercise:
${\partial f\over \partial y}={1\over \sqrt{x^2+4y}}$
and this not defined when $x=2, \ y=-1$
So $f(x,y)$ no satisfy lipshtize condition in any rectangular has (2,-1), Lipshtize condition including uniqueness of solution, so no contradict with the existence and uniqueness.
True ?
If this also wrong I well be delet it , I answered by the same way of my teacher :(
| Let $u=\sqrt{x^2+4y}$ ,
Then $y=\dfrac{u^2-x^2}{4}$
$\dfrac{dy}{dx}=\dfrac{u}{2}\dfrac{du}{dx}-\dfrac{x}{2}$
$\therefore\dfrac{u}{2}\dfrac{du}{dx}-\dfrac{x}{2}=\dfrac{u-x}{2}$ with $u(2)=0$
$\dfrac{u}{2}\dfrac{du}{dx}-\dfrac{x}{2}=\dfrac{u-x}{2}$ with $u(2)=0$
$\dfrac{u}{2}\dfrac{du}{dx}=\dfrac{u}{2}$ with $u(2)=0$
$\dfrac{du}{dx}=1$ or $u=0$ with $u(2)=0$
$u=x-2$ or $u=0$
$\sqrt{x^2+4y}=x-2$ or $\sqrt{x^2+4y}=0$
$x^2+4y=x^2-4x+4$ or $x^2+4y=0$
$4y=-4x+4$ or $4y=-x^2$
$y=1-x$ or $y=-\dfrac{x^2}{4}$
| {
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"url": "https://math.stackexchange.com/questions/1952299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Greatest open ball in the unit $n$-cell
Let $n\geq2$ be an integer, let $C=[-1,1]^n$ and let $A$ be the set of all real numbers $r$ such that $r$ is the radius of some open ball $V$ such that $V$ is contained in $C$ and $V$ is disjoint to the open ball whose center is $\bf 0$ and radius is $1.$ Find $\sup A.$
I think the problem of finding $y=\sup A$ (without proof) isn't very difficult, because we can look at the case when $n=2$ and then "generalize" (graphically and using the pythagorean theorem and to generalize, we use the pythagorean theorem in $n$ dimensions). I found that $y=(\sqrt n-1)/(\sqrt n +1)$
I intituively see that a point $x$ (there are exactly $2^n$ such points) at which we can center an open ball $V$ of radius $y$ such that $V\subseteq C$ and $V\cap B(\mathbf{0},1)=\varnothing$ belongs to the line connecting the vector $\bf 0$ and the vector $\bf 1$, all of whose coordinates are $1.$ Thus $x=t\mathbf{1}$ for some real $0<t<1$ and also $|x|>1$ (because $x\notin B(\mathbf{0},1)$ and also $x$ must be an interior point of the complement of this ball). But then I don't know what to do. I would like to be as rigorous as possible.
Thank you for any help.
| Consider a point $p ∈ [-1, 1]^n \setminus B(\mathbf{0}, 1)$. Observe that $B(p, r) ⊆ [-1, 1]^n \setminus B(\mathbf{0}, 1)$ if and only if $r$ is not greater than the distance from $p$ to any of the faces of the cube and also not greater than the distance from $p$ to $B(\mathbf{0}, 1)$. But we can calculate those!
For $n = 2$ and $p = (x, y)$ the distances from the faces are $1 - x$, $1 - y$, $x - (-1) = 1 + x$, and $y - (-1) = 1 + y$. The distance from the ball is $\sqrt{x^2 + y^2} - 1$.
To find the supremum it is enough to maximize the minimum of the distances. By the symmetry of the problem, we may suppose $0 ≤ x ≤ y ≤ 1$, and hence consider only minimum of $1 - y$ and $\sqrt{x^2 + y^2} - 1$. We consider two cases, depending on which of the quantities is lesser. This is decided by condition $1 - y ≤ \sqrt{x^2 + y^2} - 1$, which is in our situation equivalent to $2 \sqrt{1 - y} ≤ x$.
If $2 \sqrt{1 - y} ≤ x$, then $r = 1 - y$, which we are maximizing. Equivalently, we are minimizing $y$ under condition $0 ≤ 2 \sqrt{1 - y} ≤ x ≤ y ≤ 1$. Note that as $y$ decreases from $1$ to $0$, the quantity $2 \sqrt{1 - y}$ increaces from $0$ to $2$. Hence, when miniminzing $y$ under the condition we obtain $x = y = 2\sqrt{1 - y}$.
If $2 \sqrt{1 - y} ≥ x$, then $r = \sqrt{x^2 + y^2} - 1$, which we are maximizing. $r$ inscreases as $x$ increases, so we put $x = \min(y,\, 2\sqrt{1 - y})$ and we maximize $\sqrt{\min(y,\, 2\sqrt{1 - y})^2 + y^2} - 1$ for $0 ≤ y ≤ 1$. We again consider two cases: if $y ≤ 2 \sqrt{1 - y}$, then we are maximizing $\sqrt{y^2 + y^2} - 1 = \sqrt{2} y - 1$ or equivalently $y$ under conditions $0 ≤ y ≤ 2\sqrt{1 - y},\, 1$. By the same observation on behavior of the two quantities we obtain again $y = 2 \sqrt{1 - y}$. Otherwise if $y ≥ 2 \sqrt{1 - y}$, we maximize $\sqrt{4 (1 - y) + y^2} - 1 = 1 - y$ and therefore minimize $y$ under $0 ≤ 2 \sqrt{1 - y} ≤ y ≤ 1$, which we have already done.
In all cases we get $x = y = 2 \sqrt{1 - y}$ with solution $x = y = 2 \sqrt{2} - 2$ and $r = 1 - y = \sqrt{x^2 + y^2} - 1 = 3 - 2 \sqrt{2} = (\sqrt{2} - 1)/(\sqrt{2} + 1)$, which is the desired result.
The computation can be hopefully generalized to arbitrary $n$. We can again suppose $0 ≤ x_1 ≤ x_2 ≤ … ≤ x_n ≤ 1$ and maximize $\min(1 - x_n,\, \sqrt{∑_{i = 1}^n x_i^2} - 1)$.
Also note that $(\sqrt{n} - 1)/(\sqrt{n} + 1) \to 1$ as $n \to ∞$. Therefore, at higher dimensions the smaller balls have almost the same radius as $B(\mathbf{0}, 1)$, and so are intersecting. They are thouching exactly when $n = 9$. More precisely, every two of the $2^9$ balls with centers sharing all but one coordinate are touching.
I gave this problem to a colleague of mine and he found the result for $n = 2$ a different way: he considered a copy of the square rotated by $45$ degrees and then inscribed the small circle we a looking for to one of the resulting triangles. Then by using simple trigonometry we get $r = \tan^2(π/8)$. As a byproduct we obtain $\tan^2(π/8) = (\sqrt{2} - 1)/(\sqrt{2} + 1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1954214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to find the determinant of a 5x5 matrix How do I find the determinant of this?
$$\begin{bmatrix}
0& 6& −2& −1& 5\\
0& 0& 0& −9& −7\\
0& 15& 35& 0& 0\\
0 &−1 &−11& −2& 1\\
−2 &−2& 3& 0& −2\end{bmatrix}$$
I tried doing row reductions but every time I get a $0$ and I gain a number. I'm not really sure how to do this by cofactors
| By using a Laplace expansion along the first column the problem immediately boils down to computing $R=-2\cdot\det(M)$ with
$$ \det M=\det\begin{pmatrix}6&-2&-1& 5 \\ 0 & 0 & -9 & -7 \\ 15 & 35 & 0 & 0 \\ -1&-11&-2&1\end{pmatrix}=-5\cdot\det\begin{pmatrix}6&-2&1& 5 \\ 0 & 0 & 9 & -7 \\ 3 & 7 & 0 & 0 \\ -1&-11&2&1\end{pmatrix}$$
hence
$$ R = 10\left[-9\det\begin{pmatrix}6&-2& 5 \\ 3 & 7 & 0 \\ -1&-11&1\end{pmatrix}-7\det\begin{pmatrix}6&-2&1 \\ 3 & 7 & 0 \\ -1&-11&2\end{pmatrix}\right]$$
$$ R = 10\left[-9\det\begin{pmatrix}11&53& 0 \\ 3 & 7 & 0 \\ -1&-11&1\end{pmatrix}-7\det\begin{pmatrix}6&-2&1 \\ 3 & 7 & 0 \\ -13&-7&0\end{pmatrix}\right]$$
$$ R = 10\left[-9\cdot(11\cdot 7-53\cdot 3)-7\cdot\left(-7\cdot 3+7\cdot 13\right)\right]=\color{red}{2480}.$$
| {
"language": "en",
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"source": "stackexchange",
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Find $\Pr(Y\le 2X)$ for two iid random variables of a given pdf Consider two independent random variables $X$ and $Y$.
Let both PDFs are defined by $$f_X(x)=\begin{cases}\displaystyle\frac{4-2x}{3}&\text{if }x\in(0,1),\\0&\text{otherwise.}\end{cases}$$
Find $\Pr(Y \le 2X)$.
I solved this the following process.
*
*I've found
$$
\Pr(Y \le 2X | X=c) = \begin{cases}
0 & \text{if }c\le0\\
-\frac{4}{3}(c^2-2c) & \text{if } 0\lt c \lt \frac12\\
1 & \text{if }c\ge\frac12
\end{cases}
$$
*I've solved
\begin{align}
\Pr(Y \le 2X) &= \int_{-\infty}^{\infty} \Pr(Y\le 2X|X=x) f_X(x)~dx\\
&= \int_{0}^{1} \Pr(Y\le 2X|X=x) \cdot \frac{4-2x}{3} ~dx\\
&= \int_{0}^{\frac12} -\frac{4}{3}(x^2-2x) \cdot \frac{4-2x}{3} ~dx + \int_{\frac12}^1 1 \cdot \frac{4-2x}{3} ~dx\\
&= \frac{121}{216}
\end{align}
Is there anything wrong? Also, is there the better solution? Thank you for reading my question.
| Yes, indeed, that's the right integral. But not the right value.
$$\begin{align}\mathsf P(Y\leq 2X) ~=~& \int_0^{1/2}\frac{4-2x}3\int_{0}^{2x}\frac{4-2y}3\,\mathsf d y\,\mathsf d x +\int_{1/2}^1\frac{4-2x}3\int_0^1 \frac{4-2y}3\,\mathsf d y\,\mathsf d x \\[1ex] ~=~& \int_0^{1/2}\frac{4-2x}3\frac{4(2x-x^2)}3\,\mathsf d x +\int_{1/2}^1\frac{4-2x}3\mathsf d x \\[1ex] ~=~& \int_0^{1/2}\frac{32x-32x^2+8x^3}9\,\mathsf d x +\int_{1/2}^1\frac{4-2x}3\mathsf d x \\[1ex]=~& \frac 89\left[\frac{x^4}4-\frac{4x^3}3+2x^2\right]_{x=0}^{x=1/2}+\frac 13\left[4x-x^2\right]_{x=1/2}^{x=1} \\[2ex]=~& \dfrac{157}{261} \end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $x^3 \equiv 0,1,8 \mod{9}$ (Critique my solution) My solution is based on the idea that it is sufficient to consider the cubes less than 9. Ie $0^3 \equiv 0, 1^3\equiv 1, 2^3 \equiv 8$. Since these are the only cubes in the residue class$\mod{9}$ these are the only possibilities for $x^3 \mod{9}$
Is this a sufficient motivation?
New solution:
By computation, $0^3 \equiv 3^3 \equiv 6^3 \equiv 0 \mod{9}$ and $ 1^3 \equiv 4^3 \equiv 7^3 \equiv 1 \mod{9}$ and $ 2^3 \equiv 5^3 \equiv 8^3 \equiv 8 \mod{9}$. For any $x$, we can can calculate $x^3\mod{9}$ by reducing $x \equiv y \mod{9}$ where $0 \leq y \leq 8$ , and so $x^3 \equiv y^3 \mod 9$ where $y^3$ is one of our cases above, ie either 0,1 or 8.
| According to that reasoning, since $1^3\equiv1,2^3\equiv8,3^3\equiv0\pmod{27}$, we have that $x^3\equiv0,1,8\pmod{27}$ for any $x$. But $4^3\equiv10\pmod{27}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to prove statement about sum We have such sum: $$\sum_{n=1}^{\infty}\frac{(-1)^{(n+1)}}{n^p}, 0<p<1$$
How to prove that sum is between 1/2 and 1?
| Let $ \quad f(x) = 1/x^{p} \quad \colon \{ x \ge 1 \,, \space 0 \lt p \lt 1 \} \space\Rightarrow\space 1/x \lt 1/x^{p} \lt 1 $
$ \Rightarrow \quad f^{\prime}(x) = -p/x^{p+1} \lt 0 \space\Rightarrow\space f^{\prime\prime}(x) = p(p+1)/x^{p+2} \gt 0 \space\Rightarrow\space f(x) \space $ is convex
$$
\begin{align}
& \\
& \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{p}} = 1 + \sum_{n=2}^{\infty} \frac{(-1)^{n+1}}{n^{p}} = 1 + \sum_{n=1}^{\infty} \frac{(-1)^{n+2}}{(n+1)^{p}} = 1 - \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(n+1)^{p}} \Rightarrow \\
& \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(n+1)^{p}} = 1 - \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{p}} \Rightarrow \small \sum_{n=1}^{\infty} \left[ \frac{1}{(2n)^{p}} - \frac{1}{(2n+1)^{p}} \right] = 1 - \sum_{n=1}^{\infty} \left[ \frac{1}{(2n-1)^{p}} - \frac{1}{(2n)^{p}} \right] \\
\end{align}
$$
$$
\begin{align}
& \\
& \because \space 2n \lt 2n+1 \Rightarrow \frac{1}{2n} \gt \frac{1}{2n+1} \Rightarrow \frac{1}{(2n)^{p}} \gt \frac{1}{(2n+1)^{p}} \Rightarrow \frac{1}{(2n)^{p}} - \frac{1}{(2n+1)^{p}} \gt 0 \Rightarrow \\
& \sum_{n=1}^{\infty} \left[ \frac{1}{(2n)^{p}} - \frac{1}{(2n+1)^{p}} \right] = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{(n+1)^{p}} = 1 - \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{p}} \gt 0 \Rightarrow \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{p}} \color{red}{\lt 1} \\
\end{align}
$$
$$
\begin{align}
& \\
& \because \space f(x) \space \text{convex} \space\Rightarrow\space f(x) \space \text{midpoint convex} \space\Rightarrow\space \small f \left( \frac{x_{1} + x_{2}}{2} \right) \le \frac{f(x_{1}) + f(x_{2})}{2} \\
& \text{Let} \space \small x_{1} = 2n - 1 \,, \space x_{2} = 2n + 1 \normalsize \Rightarrow \small f \left( \frac{2n - 1 + 2n + 1}{2} \right) = f(2n) \le \frac{f(2n - 1) + f(2n + 1)}{2} \normalsize \Rightarrow \\
& \frac{1}{(2n-1)^{p}} + \frac{1}{(2n+1)^{p}} \ge \frac{2}{(2n)^{p}} \quad\Rightarrow\quad \frac{1}{(2n-1)^{p}} - \frac{1}{(2n)^{p}} \ge \frac{1}{(2n)^{p}} - \frac{1}{(2n+1)^{p}} \space\space\Rightarrow \\
& \sum_{n=1}^{\infty} \left[ \small \frac{1}{(2n-1)^{p}} - \frac{1}{(2n)^{p}} \normalsize \right] \ge \sum_{n=1}^{\infty} \left[ \small \frac{1}{(2n)^{p}} - \frac{1}{(2n+1)^{p}} \normalsize \right] \Rightarrow \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{p}} \ge 1 - \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{p}} \Rightarrow \\
& 2 \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{p}} \ge 1 \Rightarrow \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{p}} \color{red}{\ge 1/2} \\
\end{align}
$$
| {
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"question_score": "2",
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If $p(x)$ is a cubical polynomial with $p(1)=3,p(0)=2,p(-1)=4$, then what is $\int_{-1}^{1} p(x)dx$? Q.If $p(x)$ is a cubical polynomial with $p(1)=3,p(0)=2,p(-1)=4$,Then $\int_{-1}^{1} p(x)dx$=__?
My attempt:
Let $p(x)$ be $ax^3+bx^2+cx+d$
$p(0)=d=2$
$p(1)=a+b+c+d=3$
$p(-1)=-a+b-c+d=4$
From them,we get $b=3.5$,$d=2$ and $a+c=-0.5$
I could not progress any further.
| It could be helpful to do the integration step as well:
\begin{align}
\int_{-1}^1p(x)dx&=\int_{-1}^1\left(ax^3+bx^2+cx+d\right)\,\text{d}x\\
&=\color{blue}{\frac{a}{4}x^4\big|_{x=-1}^{x=1}}+\color{red}{\frac b3x^3 \big|_{x=-1}^{x=1}}+\color{green}{\frac c2 x^2\big|_{x=-1}^{x=1}}+\color{black}{dx\big|_{x=-1}^{x=1}}\\
&=\color{blue}{0}+\color{red}{\frac b3 \times 2} + \color{green}{0} + \color{black}{d \times 2}
\end{align}
Note that with $d=2$ you have $a+b+c=1$ and $-a+b-c=2$. If you sum the latter euqations on both sides you obtain $2b=3$ or $b=\frac32$. Hence your integral becomes $$\frac{\frac32}{3}\times 2+2 \times 2 = 5.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1962163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Probability that a number is divisible by 11 The digits $1, 2, \cdots, 9$ are written in random order to form a nine digit number. Then, the probability that the number is divisible by $11$ is $\ldots$
I know the condition for divisibility by $11$ but I couldn't guess how to apply it here.
Please help me in this regard. Thanks.
| A number is divisible by 11 if and only if the sum of the odd-position digits and the even-position digits differ by a multiple of 11. Now, the sum of the digits from 1 to 9 is odd, so the difference in any such number must be either 11 or 33. The only subsets of $\{1,2,3,4,5,6,7,8,9\}$ where the difference between the subset and its complement is $33$ have six or more elements; this is impossible, since there are 5 odd-position digits and 4 even-position digits. Thus the two sets of digits differ by 11, and then the larger sum is $28$ and the smaller is $17$.
So we want collections of 4 or 5 integers from $\{1,2,3,4,5,6,7,8,9\}$ whose sum is $28$. There are eleven such (it's pretty straightforward to simply enumerate these by hand):
\begin{align*}
&\{4, 7, 8, 9\}, \{5, 6, 8, 9\}, \{1, 3, 7, 8, 9\}, \{1, 4, 6, 8, 9\}, \\
&\{1, 5, 6, 7, 9\}, \{2, 3, 6, 8, 9\}, \{2, 4, 5, 8, 9\}, \{2, 4, 6, 7, 9\}, \\
&\{2, 5, 6, 7, 8\}, \{3, 4, 5, 7, 9\}, \{3, 4, 6, 7, 8\}.
\end{align*}
Each of these yields $24\cdot 120$ possible numbers (arrange the 5-set any way you want, and the 4-set any way you want), for a total of $11\cdot 24\cdot 120 = 31680$ possibilities. (Note that this corresponds roughly to the numeric estimate of $0.08$ given in the comments).
| {
"language": "en",
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"source": "stackexchange",
"question_score": "22",
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How to show $\tanh^{-1}(3i)=i\tan^{-1}(3)$? I want to find all the complex values associated with $\tanh^{-1}(3i)$. On WolframAlpha, it computes $\tanh^{-1}(3i)=i\tan^{-1}(3)$. I know that $\tanh^{-1}(z)=\frac{1}{2} \log{\frac{1+z}{1-z}}$ and so
$$\begin{align}
\tanh^{-1}(3i)&=\frac{1}{2} \log{\frac{1+3i}{1-3i}}\\
&=\frac{1}{2}\log \frac{-8+6i}{10}\\
&=\frac{1}{2}\log \frac{-4+3i}{5}\\
&=\frac{1}{2}\log \frac{-4+3i}{5}\\
&=\frac{1}{2}\Big[\ln(5)+\arg(-4+3i)-[\ln(5)+\arg(5)]\Big]\\
&=\frac{1}{2}\Big[\arg(-4+3i)+2\pi n\Big]\\
\end{align}$$
Where do I go from here to simplify it to $i\tan^{-1}(3)$?
| put $\arg(-4+3i)=2\theta$
$\displaystyle\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}=-\frac{3}{4} $
$\displaystyle\tan\theta=3$
$\displaystyle\theta=i\tan^{-1}3$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Equation which has to be solved with logarithms I need some help how to solve these equations for $x$. I think I have to use logarithms but still not sure how to do it and would be really grateful if someone could explain me.
$x^2 \cdot 2^{x + 1} +2 ^{\lvert x - 3\rvert + 2}
= x^2 \cdot 2^{\lvert x - 3\rvert + 4} + 2^{x - 1}$
$(x^2 - 7x + 5)^{x^2-2x-15} = 1$
| 1) Making $2^{x-1}=a$ and $2^{|x-3|+2}=b$ you have $$4ax^2+b=4bx^2+a\iff(4x^2-1)(a-b)=0$$ This gives $x=\frac 12$ and $x\ge3$
2) You have two independent possibilities $$x^2-7x+5=1\\x^2-2x-15=0$$
| {
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Supremum and infimum of asinx + bcosx I am trying to draw conclusions about the supremum and infimum of $F=\{a\sin x+b\cos x: x \in \mathbb{R}\}$ where $a,b \in \mathbb{R}$ are parameters. I am able to rewrite the function as
$$\sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}sinx+\frac{b}{\sqrt{a^2+b^2}}cosx\right).$$
Notice that $\left(\frac{a}{\sqrt{a^2+b^2}}\right)^2 + \left(\frac{b}{\sqrt{a^2+b^2}}\right)^2=1$, so $\frac{a}{\sqrt{a^2+b^2}}$ and $\frac{b}{\sqrt{a^2+b^2}}$ can be rewritten as $\cos\theta$ and $\sin\theta$, respectively. Then I continue $\sqrt{a^2+b^2}\left(\cos\theta \sin x + \sin\theta \cos x\right)$ which is $\sqrt{a^2+b^2}\sin(x+\theta)$. I conclude that $F=\left[-\sqrt{a^2+b^2};\sqrt{a^2+b^2}\right]$, and so:
1) $F$ is bounded both from above and below.
2) $\text{sup } F=\sqrt{a^2+b^2}, \text{inf }F=-\sqrt{a^2+b^2}$.
3) $\text{sup }F \in F, \text{inf }F \in F$.
Is this work correct?
| Checking with calculus, we have the extreme point of
\begin{align}
f(x) = a \cos x + b\sin x \ \ \Rightarrow \ \ f'(x) = -a\sin x + b\cos x.
\end{align}
Solving for the critical points
\begin{align}
b\cos x - a\sin x = 0
\end{align}
leads to the cases
Case 1: $a\neq 0$
Then it follows
\begin{align}
\tan x = \frac{\sin x}{\cos x}=\frac{ kb/\sqrt{a^2+b^2}}{ ka/\sqrt{a^2+b^2}}=\frac{b}{a}
\end{align}
where $|k|=1$ which means either
\begin{align}
f(x) = \frac{a^2k}{\sqrt{a^2+b^2}}+\frac{b^2k}{\sqrt{a^2+b^2}}=\sqrt{a^2+b^2}
\end{align}
or
\begin{align}
f(x) = \frac{a^2k}{\sqrt{a^2+b^2}}+\frac{b^2k}{\sqrt{a^2+b^2}}=-\sqrt{a^2+b^2}
\end{align}
Case 2 $a = 0$ We immediately have the maximum is $|b|$ and min is $-|b|$.
| {
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combinatorics: sum of product of integer compositions I am trying to solve a problem from Stanley's book, it says:
Fix $k,n \in \mathbb{P}$. Show that:
\begin{align}
\sum a_1 a_2 \cdots a_k = \binom{n+k-1}{2k-1}
\end{align}
where the sum ranges over all compositions $(a_1 , a_2 , \ldots , a_k)$ of $n$ into $k$ parts.
I am trying to reason like this:
we need to find the coefficient $c_n = \sum a_1 a_2 \cdots a_k$ from this generating function --
\begin{align}
\sum_n c_n x^n &= \sum_n \sum a_1 a_2 \cdots a_k x^n \\
&= \sum_n \sum a_1 a_2 \cdots a_k x^{a_1 + a_2 + \cdots + a_k}\\
&= \sum_n \sum a_1x^{a_1} a_2x^{a_2} \cdots a_kx^{a_k}
\end{align}
after that, I have no clue, how do I solve this ?
moreover, what is the range in the inner sum ?
If we consider Mark Riedel's answer, and assume $n=4$, $k=2$; then the sum will be
\begin{align}
\sum (z + 2z^2)^2 = z^2 + 4z^3 + 4z^4
\end{align}
On the other hand the compositions will be $(1,3), (2,2), (3,1)$, therefore the above sum will be counted as:
\begin{align}
(1.3)z^{1+3} + (2.2)z^{2+2} + (3.1)z^{3+1} &= 1z^1.3z^3 + 2z^2.2z^2 + 3z^3.1z^1\\
&= 3z^4 + 4z^4 + 3z^4 = 10z^4
\end{align}
what's going on? what am I missing?
| This is a supplement to @MarkoRiedel's answer which should clarify OPs question. At first we look at the compositions of $n=4$ which consists of two terms ($k=2$) and look at the corresponding terms of the generating function.
\begin{array}{crl}
1+3\qquad&\qquad x^1\cdot3x^3=&3x^4\\
2+2\qquad&\qquad2 x^2\cdot 2 x^2=&4x^4\\
1+3\qquad&\qquad3 x^3\cdot x^1=&3x^4\\
\end{array}
When looking at the three compositions above, we see the first summand is either $1$ or $2$ while the second summand is either $3$ or $2$. In general the first summand is a positive integer encoded as generating function
\begin{align*}
(x^1+2x^2+3x^3+\cdots)=\sum_{n=1}^\infty n x^n=x\frac{d}{dx}\left(\frac{1}{1-x}\right)=\frac{x}{(1-x)^2}
\end{align*}
The same holds for all the other summands of a composition. Since we want to multiply the summands and $k=2$ we have to consider
\begin{align*}
(x^1+2x^2+3x^3+\cdots)^2=\frac{x^2}{(1-x)^4}\tag{1}
\end{align*}
We use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$.
The number of compositions of $n=4$ is according to (1)
\begin{align*}
[x^4](x^1+2x^2+3x^3+\cdots)^2&=[x^4](x^1+2x^2+3x^3)^2\tag{2}\\
&=([x^3]+2[x^2]+3[x^1])(x^1+2x^2+3x^3)\tag{3}\\
&=3+2\cdot 2+3\\
&=\color{blue}{10}
\end{align*}
Comment:
*
*In (2) we see it is sufficient to consider terms with exponents up to $3$, since higher exponents do not contribute to $x^4$.
*In (3) we use the linearity of the coefficient of operator and apply the rule $$[x^p]x^qA(x)=[x^{p-q}]A(x)$$
On the other hand we obtain with $n=4$ and $k=2$
\begin{align*}
\binom{n+k-1}{2k-1}=\binom{5}{3}=\color{blue}{10}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1976347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Difficult limit problem involving sine and tangent I encountered the following problem:
$$\lim_{x\to 0} \left(\frac 1{\sin^2 x} + \frac 1{\tan^2x} -\frac 2{x^2} \right)$$
I have tried to separate it into two limits (one with sine and the other with tangent) and applied L'Hôpital's rule, but even third derivative doesn't work.
I also tried to simplify the expression a bit:
$$\frac 1{\sin^2 x} + \frac 1{\tan^2 x} = \frac{1+\cos^2 x}{\sin^2 x} = \frac{ 1}{1-\cos x} + \frac 1{1+\cos x} -1$$
But I cannot make it work either. I would like answers with series expansion. Thanks in advance.
| Since $\sin x$ and $\tan x$ are “almost equal” to $x$, separating into a sum of two limits seems like an interesting approach. The first limit is
$$
\lim_{x\to0}\left(\frac{1}{\sin^2x}-\frac{1}{x^2}\right)=
\lim_{x\to0}\frac{x^2-\sin^2x}{x^2\sin^2x}
$$
Finding a suitable Taylor expansion of the numerator is easy:
$$
x^2-\sin^2x=
(x-\sin x)(x+\sin x)=
\Bigl(\frac{x^3}{6}+o(x^3)\Bigr)\bigl(2x+o(x)\bigr)=\frac{1}{3}x^4+o(x^4)
$$
For the second limit you don't even need to remember the Taylor expansion of the tangent (but it's easier if you do):
$$
\lim_{x\to0}\left(\frac{1}{\tan^2x}-\frac{1}{x^2}\right)=
\lim_{x\to0}\frac{x^2\cos^2x-\sin^2x}{x^2\sin^2x}
$$
Now
$$
(x\cos x-\sin x)(x\cos x+\sin x)=
\Bigl(x-x\frac{x^2}{2}-x+\frac{x^3}{6}+o(x^3)\Bigr)
\bigl(2x+o(x)\bigr)=-\frac{2}{3}x^4+o(x^4)
$$
Hence your limit is
$$
\frac{1}{3}-\frac{2}{3}=-\frac{1}{3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1976711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Solution to System of Linear Differential Equations with Variable Coefficients I'm stuck trying to solve the system $\frac{dx(t)}{dt} = A(t)x(t)$ with $A(t) =\frac{1}{2}\begin{pmatrix}
ln(t+1) & ln(t-1)\\
ln(t-1) & ln(t+1)
\end{pmatrix}$.
I think I could try computing $e^{\int A(z) dz}$ and that should give me a fundamental matrix for the solutions? (I failed miserably at that though.)
| We shall consider our initial time at $t_0 = 1$ for this post.
Since
\begin{align}
A(t) = \frac{1}{2}
\begin{pmatrix}
\ln (t+1) & \ln (t-1)\\
\ln (t-1) & \ln (t+1)
\end{pmatrix}
\end{align}
then it follows
\begin{align}
\int^t_{1} A(s)\ ds = \frac{1}{2}
\begin{pmatrix}
(t+1)\ln (t+1) +(t+1) & (t-1)\ln(t-1) + (t-1)\\
(t-1)\ln(t-1) + (t-1) & (t-1)\ln(t+1) + (t+1)
\end{pmatrix}+I
\end{align}
where $I$ is the identity matrix. Now, let us rewrite the above solution as
\begin{align}
\int^t_1 A(s)\ ds = M(t) + I
\end{align}
which means we need to find the exponetial matrix
\begin{align}
\exp\left( M(t)+I \right) = \exp [M(t)] \exp(I)=\exp[M(t)].
\end{align}
Before we compute anything, we shall look at the eigen-decomposition of $M$. Observe since
\begin{align}
M(t) = \begin{pmatrix}
a(t) & b(t)\\
b(t) & a(t)
\end{pmatrix}
\end{align}
which means the characteristic polynomial is given by
\begin{align}
p(z) = (z-a)^2-b^2 = z^2-2az+(a^2-b^2).
\end{align}
Hence the roots of $M(t)$ for any fixed time is given by $a(t)-b(t)$ and $a(t)+b(t)$. Thus, it follows
\begin{align}
M(t) =
\frac{1}{2}
\begin{pmatrix}
1 & 1\\
1 & -1
\end{pmatrix}
\begin{pmatrix}
a(t)+b(t) & 0\\
0 & a(t)-b(t)
\end{pmatrix}
\begin{pmatrix}
1 & 1\\
1 & -1
\end{pmatrix}
\end{align}
which means
\begin{align}
\exp[M(t)] = \frac{1}{2}
\begin{pmatrix}
1 & 1\\
1 & -1
\end{pmatrix}
\begin{pmatrix}
e^{a(t)+b(t)} & 0\\
0 & e^{a(t)-b(t)}
\end{pmatrix}
\begin{pmatrix}
1 & 1\\
1 & -1
\end{pmatrix}.
\end{align}
The rest of the computation is straightforward.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1979528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The value of following integral $$\frac {x^2\cos (x)}{1+e^x} dx$$ from $[-\pi,\pi] $ ? Now I converted to $$R\frac{e^{ln (x^2)+ix}}{1+e^x}dx $$ wherw R is real part now it isnt an even or odd so those manipulations are useless. $1+e^x=u $ thus $e^xdx=du $ but from here I can go nowhere.
| Following @Sangchul Lee's comment (although the idea is pretty straight forward to someone who is aware of basic calculus technics) we have:
\begin{align*}
\int_{-\pi}^{\pi} \frac{x^2 \cos x}{e^x+1} \, {\rm d}x &\overset{u=-x}{=\! =\! =\! =\!} \int_{-\pi}^{\pi} \frac{x^2 \cos x}{e^{-x} +1} \, {\rm d}x \\
&= \int_{-\pi}^{\pi} \frac{x^2 \cos x e^x}{1+e^x} \, {\rm d}x\\
&= \frac{1}{2}\left ( \int_{-\pi}^{\pi} \frac{x^2 \cos x}{e^x+1} \, {\rm d}x + \int_{-\pi}^{\pi} \frac{x^2 \cos x e^x}{e^x+1} \right ) \, {\rm d}x\\
&= \frac{1}{2} \int_{-\pi}^{\pi} \frac{x^2 \cos x \left ( e^x+1 \right )}{e^x+1} \, {\rm d}x \\
&= \frac{1}{2} \int_{-\pi}^{\pi} x^2 \cos x \, {\rm d}x \\
&= -2 \pi
\end{align*}
The last integral is dealt with integration by parts twice and easy quite easy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1979669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the coefficient of $x^{80}$ in the power series expansion $\dfrac{x^2(1+x^2+x^5)}{(1-x)^2(1+x)}$ Find the coefficient of $x^{80}$ in the power series expansion $$\dfrac{x^2(1+x^2+x^5)}{(1-x)^2(1+x)}.$$
I don't know how to find coefficients in power series, solutions are greatly appreciated!
| $$\frac{x^2(1+x^2+x^5)}{(1-x)^2(1+x)}=x^2(1+x^2+x^5)\frac{1}{4}\left(\frac{2}{1-x^2}+\frac{1}{(1-x)^2}\right)$$
then use the following
$${\frac {1}{1-x^2}}=\sum _{n=0}^{\infty }x^{2n}$$
$${\frac {1}{(1-x)^{2}}}=\sum _{n=1}^{\infty }nx^{n-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1985027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Show that $|z| = 1$ if and only if $\bar{z} = \frac{1}{z}$. Maybe a very stupid question but I am stuck. Show that $|z| = 1$ if and only if $\bar{z} = \frac{1}{z}$.
Is it enough to simply multiply, i.e. $z\bar{z} = \frac{1\times z}{z} = 1$? Showhow I feel this is not correct. I know that if $z = \pm 1$ or $z \pm i$ then $|z| = 1$. Am I supposed to draw the circle $|z| = 1$? But what does $\frac{1}{z}$ represent?
If someone could give me a hint.
| $z = a + bi$
for $\frac{1}{z} = \bar{z}$
$\frac{1} {a + bi} = a - bi$
$\frac{1} {(a + bi)} \frac{a - bi}{a - bi} = a - bi$
$\frac{a - bi}{a^2 + b^2} = a - bi$
cancel and rearrange
$(a^2 + b^2) = 1$
recalling the formula for |z|
$|z| = \sqrt{a^2 + b^2} = \sqrt 1 = 1$
the other way
if |z| = 1 then $a^2 + b^2 = 1$
$z = a + bi$
again
$\frac{1}{z} = \frac{a - bi} {a^2 + b^2} = \frac{a - bi}{|z|^2}$
$= \frac{a - bi}{1^2} = a - bi = \bar z$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1985829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
What is the curved asymptote of $\frac{x+1}{\sqrt{3x-2}}$? Please excuse any non-technical language I use...
I'm confused by the process for finding the curved asymptote (as $x$ gets really large) of the graph $y=\frac{x+1}{\sqrt{3x-2}}$.
I first looked at the (naive?) approach of separating the fraction up into two parts:
\begin{align} y &= \frac{x+1}{\sqrt{3x-2}} \\
&= \frac{x}{\sqrt{3x-2}} + \frac{1}{\sqrt{3x-2}} \\
\text{as } x \rightarrow \infty, \quad y &\rightarrow \frac{x}{\sqrt{3x-2}}
\end{align}
Seemed ok, rationale being that the 1/sqrt part goes to zero compared to the ~ sqrt() part.
Then after a bit of searching I came to this video presented here: https://www.youtube.com/watch?v=8aWFSNbl2Ho and tried a similar method as presented...
\begin{align}
y &= \frac{x+1}{\sqrt{3x-2}} \\
&= \frac{x+1}{\sqrt{3x-2}} \times \frac{\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{x}}} \\
&= \frac{\sqrt{x}+\frac{1}{\sqrt{x}}}{\sqrt{3-\frac{2}{x}}} \\ \\
\text{as } x \rightarrow \infty, \quad y &\rightarrow \sqrt{\frac{x}{3}}
\end{align}
Which looks close(ish).
Wolfram Alpha however has different ideas: http://www.wolframalpha.com/input/?i=y%3D(x%2B1)%2Fsqrt(3x-2)+asymptotes
It gives the relation $x=3y^2-\frac{8}{3}$ which rearranges to a similar curve to my attempt:
$$y=\frac{1}{3}\sqrt{3x-8}$$
Am I on the right track with any of my attempts? I feel like I'm closer with the 2nd attempt but I'm not allowed to disappear both the $1/\sqrt{2}$ and the $2/x$ parts like that.
| Using equivalents, when $x$ is large $$x+1\sim x \qquad , \qquad \sqrt{3x-2}\sim \sqrt{3x}\qquad \implies y \sim \frac{x}{\sqrt{3x}}=\frac{\sqrt x}{\sqrt{3}}$$
Letting $x=\frac 1t$, you also could write $$y=\frac{1+\frac 1t}{\sqrt{\frac 3t -2}}=\frac 1{\sqrt t}\frac{1+t}{\sqrt{ 3 -2t}}$$ Now, use Taylor expansion around $t=0$ (or the generalized binomial theorem) $$\sqrt{ 3 -2t}=\sqrt{3}-\frac{t}{\sqrt{3}}-\frac{t^2}{6 \sqrt{3}}+O\left(t^3\right)$$ $$\frac{1}{\sqrt{ 3 -2t}}=\frac{1}{\sqrt{3}}+\frac{t}{3 \sqrt{3}}+\frac{t^2}{6 \sqrt{3}}+O\left(t^3\right)$$ $$\frac{1+t}{\sqrt{ 3 -2t}}=\frac{1}{\sqrt{3}}+\frac{4 t}{3 \sqrt{3}}+O\left(t^2\right)$$ $$y=\frac{1}{\sqrt{3} \sqrt{t}}+\frac{4 \sqrt{t}}{3 \sqrt{3}}+O\left(t^{3/2}\right)$$ Now, replace $t$ by $\frac 1x$ to get $$y=\sqrt{\frac x 3}+\frac 4{2\sqrt{3x}}+\cdots$$ which shows the curve asymptote and how it is approached.
Edit
Concerning the different approximations, let us consider $$\Delta_1=\frac{x+1}{\sqrt{3x-2}}-\sqrt{\frac x 3}\qquad , \qquad \Delta_2=\frac{x+1}{\sqrt{3x-2}}-\frac{1}{3}\sqrt{3x-8}$$ and use Taylor expansion of the ratio. You should get $$\frac{\Delta_1}{\Delta_2}=\frac{1}{2}-\frac{7}{96 x}+O\left(\frac{1}{x^{3/2}}\right)$$ that is to say that you are twice better.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1987344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Inverse of function $f : \mathbb{N}\times\mathbb{N}\to\mathbb{N}$ Matthew Szudzik's mapping function is another approach on mapping $\mathbb{N}\times\mathbb{N}$ to $\mathbb{N}$, however, I have trouble finding the inverse of it. The function is
$$f(a, b)=\left\{\begin{array}{ll}a^2+a+b&\mbox{, if }a\geq b\\a+b^2&\mbox{, else}\end{array}\right.$$
He states that the inverse of it is
$$f^{-1}(c)=\left\{\begin{array}{ll}(c-\lfloor\sqrt{c}\rfloor^2,\lfloor\sqrt{c}\rfloor)&\mbox{, if } c - \lfloor\sqrt{c}\rfloor^2<\lfloor\sqrt{c}\rfloor\\(\lfloor\sqrt{c}\rfloor, c-\lfloor\sqrt{c}\rfloor^2-\lfloor\sqrt{c}\rfloor)&\mbox{, else}\end{array}\right.$$
But, how did he find it?
| $f(a,b) = a^2 + a + b=c; a\ge b$
$f(a,b) = a+b^2=c; a< b$.
So make wild guesses. If $a' = -b'$ we have $f(a',b) = a'^2 = c$ so $g(c) = (\sqrt{c}, -\sqrt{c})$ will work if $c$ is a perfect square. But obviously $c$ needn't be a perfect square. But we can be close.
Suppose $n^2 \le c < (n+1)^2$ or in other words $n \le \sqrt{c}< n+1$. Then $c = n^2 + n + (c-n^2 - n)$. If $c-n^2 -n \le n$ we'd have $f(n,(c-n^2 -n)) = c$. $n^2 < c < (n+1)^2\implies 0< c - n^2<(n+1)^2 -n^2 = (n+1 +n)(n+1 -n) = 2n+1 \implies 0\le c-n^2 < 2n + 1\implies c-n^2-n < n + 1\implies c-n^2 -n \le n$.
So $g(c) = (\lfloor \sqrt{c} \rfloor, c - \lfloor \sqrt{c} \rfloor^2 - \lfloor \sqrt{c} \rfloor)$ will do. But this requires $ c - \lfloor \sqrt{c} \rfloor^2 - \lfloor \sqrt{c} \rfloor)\in \mathbb N$. i.e $ c - \lfloor \sqrt{c} \rfloor^2 > \lfloor \sqrt{c} \rfloor) $
Other wise we must solve for $f(a,b) = a+b^2=c; a< b$ first.
If $n^2 \le c < (n+1)^2$ then $f(c-n^2, n) = c$ if $c-n^2 < n$. As $n^2 \le c < n^2 + 2n+1$ we know $0\le c-n^2 < 2n+1$.
So $h(c) = (c - \lfloor \sqrt{c} \rfloor^2 , \lfloor \sqrt{c} \rfloor)$ will do if $c - \lfloor \sqrt{c} \rfloor^2>0$ which only fails if $n^2 \le c < n^2 + 1$ . If this is the case then $c - \lfloor \sqrt{c} \rfloor^2 < 1 \le \lfloor \sqrt{c} \rfloor$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1990235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find all Pairs of Integers $(x,y)$ , $x \gt y \ge 2$ such that $x^y=y^{x-y}$ Find all Pairs of Integers $(x,y)$ , $x \gt y \ge 2$ such that $x^y=y^{x-y}$
My Try:
if $y=2$ then $$x^2=2^{x-2}$$ and Taking log on both sides we get
$$2 \log x=(x-2) \log 2$$ i.e.,
$$2 \log_2 x=x-2$$ if $x=2^k$ then we get
$$2k=2^k-2$$ or
$$k+1=2^{k-1}$$
By Trying different values we get $k=3$ is the only solution.
so one pair is $(x,y)=(8,2)$
Similarly if $y=3$ we get
$$k+1=3^{k-1}$$ which is satisfied by $k=2$ so another pair is
$(x,y)=(9,3)$
Is this approach correct?
| For
$x^y=y^{x-y}
$
I find that
$x=8, y=2$
and
$x=9, y=3$
are the only solutions.
Must have $x > y$.
$x^y
=y^{x-y}
$
so
$(xy)^y = y^x
$.
Let
$x = ry$ where
$r > 1$.
$(ry^2)^y = y^{ry}$
so
$ry^2 = y^r$
or
$r = y^{r-2}$
or
$y = r^{1/(r-2)}$.
If $x = 2y$
then
$(2y)^y = y^y$
which has no solution.
If
$1 < r < 2$,
then
$r-2 < 0$
so
$y = r^{1/(r-2)}
< 1$,
so no solution.
Therefore $r > 2$.
$r^{1/(r-2)} = 2$
for $r = 4$.
Since
$r^{1/(r-2)} $
is decreasing for
$r \ge 4$,
must have
$2 < r \le 4$.
If $r=4$
then $y=2$
and $x = ry = 8$.
$(xy)^y = (16)^2 = 256$
and
$y^x = 2^8 = 256$
so this is a solution!
If
$r^{1/(r-2)} = 3$,
then
$r=3$.
Then
$y = 3$
and $x = ry = 9$.
$(xy)^y = 27^3 = 3^9$
and
$y^x = 3^9$,
so this is a solution!
Let
$d = (x, y)$
so
$x = ad, y = bd$
with
$4b >a > 2b, (a, b) = 1$.
From
$(xy)^y = y^x
$,
$(abd^2)^{bd} = (bd)^{ad}
$
or
$(abd^2)^{b} = (bd)^{a}
$
or
$a^bb^bd^{2b} = b^ad^a$
or
$a^b = b^{a-b}d^{a-2b}$.
Must have
$b=1$
since
$(a, b) = 1$.
This becomes
$a = d^{a-2}$
or
$d = a^{1/(a-2)}$.
As before,
the only integer solutions
to $d = a^{1/(a-2)}$
are
$a=4, d=2$
and
$a=3, d=3$
which we have found already.
Therefore
$x=8, y=2$
and
$x=9, y=3$
are the only solutions.
Note on the
derivative of
$r^{1/(r-2)}$.
$\begin{array}\\
(r^{1/(r-2)})'
&= r^{1/(r-2)} (\frac1{r(r-2)}-\frac{\log(r)}{(r-2)^2}\\
&= r^{1/(r-2)} \frac{(r-2)-r\log(r)}{r(r-2)^2}\\
& < 0
\qquad\text{for } r > 2\\
\end{array}
$
so
$r^{1/(r-2)} < 2$
for
$r > 4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1992626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How To Write A Formula For The First N Terms of a Sequence? So I have the sequence $$\{5, 15, 45,\cdots\}$$ and I figured out that the formula to find a particular term is $$S_n = 5 \times 3^{n-1}$$ but how do I use this to find the sum of the first $n$ terms?
|
Geometric Summation: Given the sequence$$a,ar,ar^2,ar^3,\ldots,ar^{n-1}\tag1$$
We have the sum as$$S=a\left(\frac {1-r^n}{1-r}\right)\tag2$$
Where $S$ is the total sum and $a$ is the first term of the sequence.
Proof: Multiplying $S$ by $r$, we obtain another equation$$Sr=ar+ar^2+ar^3+\ldots+ar^n\tag3$$
And subtracting $Sr$ from $S$, we have$$S-Sr=a-ar^n=a\left(1-r^n\right)\implies \boxed{S=a\frac {1-r^n}{1-r}}\tag4$$
For example, if you wanted to find the sum of the first two terms $5+15$, we see that $a=5$, $r=3$ and $n=2$. Thus, the sum is$$S=5\cdot\frac {1-3^2}{1-3}=20$$
Trying out the first $5$ sums, we have $5+15+45+135+405$. Therefore,$$S=5\cdot\frac {1-3^5}{1-3}=605$$
And verifying, we see that indeed, the sum is $605$.$$5+15+45+135+405=605$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1993493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $f(x)= \lim_ {n \to \infty} \sum_{r=1}^{n} \frac{n}{n^2+x^2r^2}$, find the required value We have
$$f(x)= \lim_ {n \to \infty} \sum_{r=1}^{n} \frac{n}{n^2+x^2r^2}$$, then find the value of $\sum_{k=1}^{3} k f(k)$.
Could someone give me slight hint as how to find $f(x)$ here.
| $$f(x)= \lim_ {n \to \infty} \sum_{r=1}^{n} \frac{n}{n^2+x^2r^2}$$
$$= \lim_ {n \to \infty} \frac{1}{n}\sum_{r=1}^{n} \frac{n^2}{n^2+x^2r^2}$$
$$= \lim_ {n \to \infty} \frac{1}{n}\sum_{r=1}^{n} \frac{1}{1+x^2\cdot \frac{r^2}{n^2}}$$
$$= \lim_ {h \to 0} \text{h} \sum_{r=1}^{n} \frac{1}{1+x^2(rh)^2}$$
Then using Riemann summation, we get
$$f(x)=\int_0^1 \frac{1}{1+x^2y^2}dy$$
$$=\frac{1}{x}\cdot \int_0^x \frac{d(xy)}{1+(xy)^2}$$
$$=\frac{\arctan x}{x}$$
So $$\boxed{f(x)=\frac{\arctan x}{x}}$$
Hence, $$\sum_{k=1}^{3} k f(k)=1f(1)+ 2f(2)+ 3f(3)=\arctan 1+\arctan 2+\arctan 3= \pi$$
Hope this helps you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1999572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Prove that, for all non-negative real numbers $x,y,z$ that satisfy $x+y+z=1, x^2y+y^2z+z^2x≤4/27$ Prove that, for all non-negative real numbers $x, y, z$ that satisfy $x + y + z = 1$,
$$x^2 y + y^2 z + z^2 x \leq \frac {4}{27}
$$
I'm having trouble with this question. I suspect it may have a fairly simple proof using the AM-GM inequality and certain substitutions, however, I have been unable to complete such a proof.
| Let $\{x,y,z\}=\{a,b,c\}$, where $a\geq b\geq c$.
Hence, by Rearrangement and AM-GM we obtain:
$$x^2y+y^2z+z^2x=x\cdot xy+y\cdot yz+z\cdot zx\leq a\cdot ab+b\cdot ac+c\cdot bc=$$
$$=b(a^2+ac+c^2)\leq b(a+c)^2=4b\left(\frac{a+c}{2}\right)^2\leq4\left(\frac{b+\frac{a+c}{2}+\frac{a+c}{2}}{3}\right)^3=\frac{4}{27}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
An Identity for Pell-numbers The Pell-numbers are defined recursively by:
$P_0 = 0,
P_1 = 1$ and $
P_{n+2} = 2P_{n+1} + P_n$
I am stuck trying to prove the identity:
$P_{2n+1}^2 - 1 = 4P_n P_{n+1}(P_{n+1}^2 - P_n^2)$
A proof would be great, otherwise a list of known Pell-identies would also help
| You know $P_{n+2} = 2 P_{n+1} + P_{n}$ $\Rightarrow$ $P_{n+2}-(1+\sqrt{2})P_{n+1}= (1-\sqrt{2})(P_{n+1}-(1+\sqrt{2})P_{n})$. In other word, we have
\begin{align*}
\frac{P_{n+2}-(1+\sqrt{2})P_{n+1}}{P_{n+1}-(1+\sqrt{2})P_{n}} = (1-\sqrt{2}).
\end{align*}
Since $P_0 = 0, P_1 = 2, P_2 = 2$, we deduce that
\begin{align*}
P_{n} = \frac{(1+\sqrt{2})^n - (1-\sqrt{2})^n }{2\sqrt{2}}.
\end{align*}
You refer to Pell_number.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate a determinant with pattern A friend of mine asked me to calculate the value of a perticular determinant, but after spending quite some time on it I'm unable to find a solution to it. The given determinant is:
$$
\begin{vmatrix}
0 & 1 & \cdots &n-2&n-1 \\
n-1 & 0 & \cdots &n-3&n-2 \\
\vdots & \vdots & \ddots & \vdots &\vdots\\
1 & 2 & \cdots & n-1 &0 \\
\end{vmatrix}
$$
Also my friend is supposed to solve this determinant by more or less elementary methods, like elementary row operations, transpose... and no eigenvalues, nor adjoints.
First of all it seems like the value of the determinant is:
$$
\frac{1}{2n}\begin{vmatrix}
0 & n & \cdots &n&n \\
n & 0 & \cdots &n&n \\
\vdots & \vdots & \ddots & \vdots &\vdots\\
n & n & \cdots & n &0 \\
\end{vmatrix}
$$
Anyway I tried adding all the rows to the last one and by taking $\frac{n(n-1)}{2}$ in front to have a row of 1s, but it didn't helped me much.
Then I tried to use the fact that:
$$
\begin{vmatrix}
A & C \\
0 & B \\
\end{vmatrix} = \det(A) \cdot \det(B)
$$, where $A,B$ are square matrices. By subtracting the $(\frac{n}{2} + i)$-th column from the $i$-th ($i=0,1,2...$) and then adding the $i$-th row to the $(\frac{n}{2} + i)$-th row, one can obtain the wanted form, but again nothing more. Also this method pretty much fails when $n$ is odd.
My best attepmt was to notice that $A_{ij} + A_{ji} = n$ and using $\det(A^{T}) = \det(A)$, but unfortunatelly
$$\begin{vmatrix}
0 & n & \cdots &n&n \\
n & 0 & \cdots &n&n \\
\vdots & \vdots & \ddots & \vdots &\vdots\\
n & n & \cdots & n &0 \\
\end{vmatrix} \not = \begin{vmatrix}
0 & 1 & \cdots &n-2&n-1 \\
n-1 & 0 & \cdots &n-3&n-2 \\
\vdots & \vdots & \ddots & \vdots &\vdots\\
1 & 2 & \cdots & n-1 &0 \\
\end{vmatrix} + \left( \begin{vmatrix}
0 & 1 & \cdots &n-2&n-1 \\
n-1 & 0 & \cdots &n-3&n-2 \\
\vdots & \vdots & \ddots & \vdots &\vdots\\
1 & 2 & \cdots & n-1 &0 \\
\end{vmatrix}\right)^T$$
I tried to decompose the LHS by using the $n-$linearity of the determinants, by then one gets $2^n$ determinants on the RHS similar to the first one, but things are messy.
| I take it this is a Toeplitz matrix (diagonals are constant).
Subtract row $n-1$ from row $n$, then row $n-2$ from row $n-1$, ... row $1$ from row $2$. Then subtract column $1$ from each other column.
Then add $1/n$ times the second, third, ..., last columns to the first column. The result should be upper triangular.
| {
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} |
Show that $(1-\beta)(p+\beta)=1$
If $$(7+4\sqrt{3})^n = p+\beta,$$ where $n$ and $p$ are positive integers and $\beta$ is a proper fraction, then show that $$(1-\beta)(p+\beta)=1.$$
I cant even understand how to express the term in a positive number and a proper fraction. I would appreciate any hint.
| Fun question! The key realization is that
$$
(7 + 4\sqrt{3})^n + (7 - 4\sqrt{3})^n \in \mathbb{Z}
$$
(do you see why?) and moreover, that
$$
0 < (7 - 4\sqrt{3}) < 1,
$$
so that
$$
0 < (7 - 4\sqrt{3})^n < 1,
$$
for all natural numbers $n$.
It follows from here that
\begin{align*}
p &= (7 + 4\sqrt{3})^n + (7 - 4\sqrt{3})^n - 1 \\
\beta &= 1 - (7 - 4\sqrt{3})^n.
\end{align*}
Now to finish, we see directly from the above that
\begin{align*}
p + \beta &= (7 + 4\sqrt{3})^n \\
1 - \beta &= (7 - 4\sqrt{3})^n.
\end{align*}
Multiply them together and see what you get.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2002928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solve for $x$ $2\log_4(x+1) \le 1+\log_4x$ $2\log_4(x+1) \le 1+\log_4x$
I did:
$$2\log_4(x+1) \le 1+\log_4x \Leftrightarrow \log_4(x^2+1) \le 1+\log_4(x) \Leftrightarrow \log_4(\frac{x^2+1}{x}) \le 1 \Leftrightarrow 4 \ge \frac{x^2+1}{x} \Leftrightarrow 4x \ge n^2 +1 \Leftrightarrow 0\ge x^2 -4x +1$$
Using the quadratic formula this becomes $x \le 2 + \sqrt{3}$ $\lor$ $x\le 2 - \sqrt{3}$
But my book says that the solution is $[1,1] = [1]$
What did I do wrong?
| write your inequation like that:
$$\log_4(x+1)^2\le \log_4 4+\log_4 x$$ thus $$\log_4 (x+1)^2\le \log_4 4x$$ this is equivalent to $$(x+1)^2\le 4x$$ can you proceed?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2006674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Evaluate $\lim_\limits{x \to +\infty}\left(x\ln (1+x)-x\ln x + \arctan\frac{1}{2x}\right)^{x^2\arctan x}$ $$\lim_{x \to +\infty}\left(x\ln (1+x)-x\ln x + \arctan\frac{1}{2x}\right)^{x^2\arctan x}$$
My attempt
\begin{align*}
&=\exp \lim_\limits{x \to +\infty} x^2\arctan x \cdot\ln\left[x\ln (1+x)-x\ln x + \arctan\frac{1}{2x}\right]\\
&=\exp \lim_{x \to +\infty} \frac{\pi}{2}x^2 \ln\left[\frac{\ln x}{\frac{1}{x}} + \frac{\ln \left(1+\frac{1}{x}\right)}{\frac{1}{x}}-\frac{\ln x}{\frac{1}{x}} + \arctan\frac{1}{2x}\right]\\
&=\exp \lim_{x \to +\infty} \frac{\pi}{2}x^2\ln\left[1 + \arctan \frac{1}{2x}\right]
\end{align*}
Here I don't know what to do. One thing I did was to multiply and divide by $\arctan \frac{1}{2x}$ in order to get rid of the logarithm. But I ended up with
$$\frac{\pi}{2}x^2\arctan \frac{1}{2x}$$
$\arctan \frac{1}{2x} \to \pi/2$ as $x \to +\infty$ but that doesn't lead to the right answer (I'd get a limit of $+\infty$).
| Let $f(x)$ be your function.
we have
$x\ln(1+\frac{1}{x})+\arctan(\frac{1}{2x})=$
$=x(\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3})+(\frac{1}{2x}-\frac{1}{24x^3})+\frac{1}{x^3}\epsilon(x)$
$=1+\frac{1}{3x^2}(1+\epsilon(x))$
thus
$\ln(f(x))\sim \frac{1}{3}\arctan(x)\;\; (x\to+\infty)$
and
$$\lim_{x\to+\infty}f(x)=e^{\frac{\pi}{6}}.$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
interesting scalene triangle trigonometry problem the problem is to find the minimum value of cos(2α)+cos(2β)+cos(2γ) in a scalene triangle. So what should the value of α,β, and γ be?
I already transformed the expression using γ=180∘−(α+β) to cos(2α)+cos(2β)+cos(2α+2β)
so how can I finish the problem if this is a good way?
Thank you
| Given $A+B+C = \pi$ and $$\cos 2A+\cos 2 B+\cos 2C = 2\cos(A+B)\cos(A-B)+\cos 2C$$
So $$\cos 2A+\cos 2B+\cos 2C = -2\cos C\cdot \cos(A-B)+2\cos^2 C-1$$
$$ = -2\cos C\left[\cos(A-B)-\cos(C)\right]-1=-2\cos C\left[\cos(A-B)+\cos(A-B)\right]$$
$$ = -4\cos A \cos B\cos C-1$$
So $$\cos 2A+\cos 2B+\cos 2C = -1-4\cos A\cos B\cos C\geq -1+\frac{4}{8}=-\frac{3}{2}$$
Above we have used $$\cos A\cos B\cos C\leq \frac{1}{8}$$
and equality hold when $\displaystyle A=B=C=\frac{\pi}{3}$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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} |
Manipulation of conditions of roots of a quadratic equation. How do I write $a^5+b^5$ in terms of $a+b$ and $ab$. Also is there any general way of writing $a^n+b^n$ in terms of $a+b$ and $ab$?
| This is what I found when I first came across this question. Now let me see if I can find the steps to get Lozenges' expression. First of all, Dr. Sonhard's factorization is correct:
$$a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4).$$
Were Zlatan's sign doubt correct, we would get, I suppose, either $ab$ or $a-b$ as the first factor, but with $ab$ we get a $a^5b$ term, which is not on the LHS, and with $a-b$ the last term would be $-b^5$, whereas we have $b^5$. In fact, a general way to factoize $a^n+b^n$ with $n$ odd is:
$$a^n+b^n=(a+b)\cdot\sum_{k=0}^{n-1}(-1)^{n-k-1}a^kb^{n-k-1}.$$
Let us verify this:
\begin{align*}
(a+b)\cdot\sum_{k=0}^{n-1}(-1)^{n-k-1}a^kb^{n-k-1}={}&\sum_{k=0}^{n-1}(-1)^{n-k-1}a^{k+1}b^{n-k-1}+\sum_{k=0}^{n-1}(-1)^{n-k-1}a^kb^{n-k}={} \\
{}={}&a^n+\sum_{k=0}^{n-2}(-1)^{n-k-1}a^{k+1}b^{n-k-1}+b^n+\sum_{k=1}^{n-1}(-1)^{n-k-1}a^kb^{n-k}={} \\
{}\underset{\substack{|\\\ell=k+1}}{=}{\hspace{-.35cm}}&a^n+b^n+\sum_{\ell=1}^{n-1}(-1)^{n-\ell}a^\ell b^{n-\ell}+\sum_{k=1}^{n-1}(-1)^{n-k-1}a^kb^{n-k}.
\end{align*}
Notice how the two sums cancel out, since the terms are all identical except for the sign. Note also that if $n$ were even then we would get $-b^n$ instead of $b^n$.
Lozenges says that:
$$a^5+b^5=5 a^2 b^2 (a+b)-5 a b (a+b)^3+(a+b)^5.$$
tatan commented suggesting to use:
\begin{align*}
a^4+b^4={}&(a^2)^2+(b^2)^2 \\
-a^3b-ab^3={}&-ab(a^2+b^2).
\end{align*}
With that, we rewrite:
$$a^5+b^5=(a+b)((a^2)^2+(b^2)^2-ab(a^2+b^2)+a^2b^2).$$
But this does not seem to lead me anywhere, since a sum of squares is something I do not know how to factorize. So let us go back to $a^5+b^5$ and add and subtract $(a+b)^5$, which is pretty natural since we have a sum of fifth powers:
$$a^5+b^5=a^5+b^5-(a+b)^5+(a+b)^5=(a+b)^5-5a^4b-10a^3b^2-10a^2b^3-5ab^4.$$
It seems now pretty natural to collect a $-5ab$:
$$a^5+b^5=(a+b)^5-5ab(a^3+2a^2b+2ab^2+b^3).$$
That looks a lot like a cube there, right? There are 2's instead of 3's, that's all. So we add and subtract $-6ab(a^2b+ab^2)$:
$$a^5+b^5=(a+b)^5-5ab(a^3+2a^2b+2ab^2+b^3)-5ab(a^2b+ab^2)+5ab(a^2b+ab^2)=(a+b)^5-5ab(a^3+3a^2b+3ab^2+b^3)+5ab\cdot ab(a+b).$$
It looked pretty natural to collect an $ab$ in that last term, right? Oh, but this is just Lozenges' expression! So we are done.
Update
Picking up from where I dropped tatan's suggestion:
$$a^5+b^5=(a+b)((a^2)^2+(b^2)^2-ab(a^2+b^2)+a^2b^2).$$
Apply Lozenges' comment to this post:
\begin{align*}
a^5+b^5={}&(a+b)((a^2+b^2)^2-2a^2b^2-ab(a+b)^2+ab\cdot2ab+a^2b^2)={} \\
{}={}&(a+b)((a^2+b^2)^2-ab(a+b)^2+a^2b^2)={} \\
{}={}&(a+b)(((a+b)^2-2ab)^2-ab(a+b)^2+a^2b^2)={} \\
{}={}&(a+b)((a+b)^4-4ab(a+b)^2+4a^2b^2-ab(a+b)^2+a^2b^2)={} \\
{}={}&(a+b)((a+b)^4-5ab(a+b)^2+5a^2b^2)={} \\
{}={}&(a+b)^5-5ab(a+b)^3+5a^2b^2(a+b),
\end{align*}
which is Lozenges' expression, IIRC. So that is another way to get it, and probably what tatan had in mind when writing his comment.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How does one solve a bivariate normal density function? If the exponent of $e$ of a bivariate normal density is
$$\frac{-1}{54} *(x^2+4y^2+2xy+2x+8y+4) \\\text{find } \sigma_{1},\sigma_{2} \text{ and } p \text{ given that } \mu_{1} =0 \text{ and } \mu_{2}=-1. $$
One must use this definition to solve.
A pair of random variables $X$ and $Y$ have a bivariate normal distribution and they are referred to as jointly normally distribed random variables if and if their joint probability density is given by
If $X$ and $Y$ have a bivariate normal distribution normal distribution the conditional density of $Y$ given $X =x$ is a normal distribution with the mean
$\mu_{Y|x} = \mu_{2} + \rho \cdot \frac{\sigma_2}{\sigma_1}\cdot (x-\mu_{1})$
and the variance
$\sigma^{2}_{Y|x} = \sigma^{2}_{2}(1-\rho^2)$
and the conditonal density of $X$ given $Y=y$ is a normal distribution with the mean
$\mu_{X|y} = \mu_{1} + \rho\cdot \frac{\sigma_{1}}{\sigma_{2}}\cdot (y-\mu_{2})$
and the variance
$\sigma^{2}_{X|y} = \sigma^{2}_{1}(1-\rho^2)$
Does one re arrange the formula so that one can just plug in what $\mu_{1} \text{ and } \mu_{2}$ are? Does one substitute like this?
$$f(x,y) = \frac{-1}{54} *(x^2+4y^2+2xy+2x+8y+4) \\ = \frac{1}{-54} \left[ (x-\mu_{1})^2 +2(x-\mu_{1})(y-\mu_{2})+4(y-\mu_{2})^2 \right]$$
EDIT : Thanks to the valiant work of Anatoly I understand one needs to compare coefficients in order to derive $\rho,\sigma_{1},\sigma_{2}$
$$\color{lime}{(1)}: \frac{1}{2(1-p^2)\sigma^2_{1}} = \frac{1}{54}$$
However after this one I do not know what to do does anyone know how to do the rest?
| If we substitute $\mu_1=0$ and $\mu_2=-1$ in the exponent of $e $ of the joint probability density formula, we get
$$\displaystyle -\frac 1{2(1-\rho^2)}\left[\left(\frac{x}{\sigma_1}\right)^2
-2\rho \left(\frac{x}{\sigma_1}\right) \left(\frac{y+1}{\sigma_2}\right) +\left(\frac{y+1}{\sigma_2}\right)^2\right]$$
Expanding it, we have
$$-\frac 1{2(1-\rho^2)} \left[\frac{1}{\sigma_1^2}x^2 +\frac{1}{\sigma_2^2} y^2
- \frac{ 2\rho }{\sigma_1 \sigma_2} xy - \frac{ 2\rho }{\sigma_1 \sigma_2} x + \frac{2y}{\sigma_2^2} + \frac{1}{\sigma_2^2} \right]$$
which can be written, if we move a given factor $k^2$ into the brackets, as
$$-\frac {1}{2k^2(1-\rho^2)} \left[\frac{k^2}{\sigma_1^2}x^2 +\frac{k^2}{\sigma_2^2} y^2
- \frac{ 2k^2 \rho }{\sigma_1 \sigma_2} xy - \frac{ 2 k^2\rho }{\sigma_1 \sigma_2} x + \frac{2k^2y}{\sigma_2^2} + \frac{k^2}{\sigma_2^2} \right]$$
Comparing this expression with
$$-\frac {1}{54} \left[x^2 +4 y^2+2 xy +2 x + 8y + 4 \right]$$
which is provided in the OP we directly get $\sigma_1=k \,\,$, $\sigma_2=k/2\,\,$ (note that both variances are by definition positive), and $\rho=-1/2\,\,$. So we have
$$-\frac {1}{2k^2 \left(1-\left (\frac {1}{2} \right)^2 \right)} =-\frac {1}{54} $$
and then $k=\pm 6\,\,$. This leads to $\sigma_1^2=36\,\,$ and $\sigma_2^2=9\,\,$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2011353",
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"source": "stackexchange",
"question_score": "3",
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If the matrices $A^3 = 0$, $B^3=0$ and $AB=BA$ then show this: The question: If $A$ and $B$ are square matrices of the same type such that $A^3=0$, $B^3=0$ and $AB=BA$. Show that
$$\left(I+A+\frac{A^2}{2!}\right)\left(I+B+\frac{B^2}{2!}\right)=I+(A+B)+\frac{(A+B)^2}{2!}+\frac{(A+B)^3}{3!}+\frac{(A+B)^4}{4!}$$
This is how I started.
$$\left(I+A+\frac{A^2}{2!}\right)\left(I+B+\frac{B^2}{2!}\right)=I+(A+B)+\frac{A^2+B^2}{2!}+AB+\frac{AB^2}{2!}+\frac{A^2B}{2!}+\frac{A^2B^2}{2!2!}$$
I tried to get
$$\frac{A^2+B^2}{2!}+AB+\frac{AB^2}{2!}+\frac{A^2B}{2!}+\frac{A^2B^2}{2!2!}$$
to be equal to $$\frac{(A+B)^2}{2!}+\frac{(A+B)^3}{3!}+\frac{(A+B)^4}{4!}$$ But I'm not getting there.
| I would do it without brute force:
*
*Since $A^3=0$ we have
$$
e^A=I+A+\frac{A^2}{2!}.
$$
Similar for $B$.
*The LHS of your expression is then $e^Ae^B$. For commuting matrices it is true that
$$
e^Ae^B=e^{A+B}.
$$
*Finally notice that if $A^3=B^3=0$ and $AB=BA$ then $(A+B)^5=0$ (e.g. do the binomial expansion and see that you get at least $A^3$ or $B^3$ in all terms). Thus, the RHS is precisely $e^{A+B}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2012593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Disproving a ring Homomorphism Problem My question is:
$\varphi=\left\{
\begin{array}{c l}
T &\mbox{$\longrightarrow R$} \\
\begin{pmatrix}
a & 0\\
b & c
\end{pmatrix} & \mbox{$\longrightarrow$ $\begin{pmatrix}
a & b\\
0 & c
\end{pmatrix}$}
\end{array}\right.$
Show that this function is not a ring homomorphism
I get that the definition of $f$ a ring homomorphism is when satisfies
$f(a + b) = f(a) + f(b)$ for all $a$ and $b$ in $\mathbb{R}$
and $f(ab) = f(a) f(b)$ for all $a$ and $b$ in $\mathbb{R}$
But I can't see how to prove the original question. Any help will be appreciated.
| I will call $f$ to your $\varphi$. Look that if you have:
$$
A= \begin{pmatrix}
1 & 0\\
1 & 0
\end{pmatrix},
B= \begin{pmatrix}
0 & 0\\
1 & 1
\end{pmatrix}
$$
Then,
$$\begin{pmatrix}
1 & 0\\
1 & 0
\end{pmatrix}*\begin{pmatrix}
0 & 0\\
1 & 1
\end{pmatrix}= \begin{pmatrix}
0 & 0\\
0 & 0
\end{pmatrix}$$
So
$$f(A*B)=\begin{pmatrix}
0 & 0\\
0 & 0
\end{pmatrix}$$
But:
$$f(A)*f(B)=\begin{pmatrix}
1 & 1\\
0 & 0
\end{pmatrix}*\begin{pmatrix}
0 & 1\\
0 & 1
\end{pmatrix}= \begin{pmatrix}
0 & 2\\
0 & 0
\end{pmatrix}$$
Making a contradiction to supposing $f$ is an homomorphism.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2013078",
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"source": "stackexchange",
"question_score": "1",
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How does one prove that this function is injective? As the title suggests, I am trying to prove that the following function is injective but at the moment I am stuck.
$$f(x)=-x^5-16x -1 $$
$$-x_1^5 - 16x_1 -1 = -x_2^5 - 16x_2 -1$$
$$x_1^5 + 16x_1= x_2^5 + 16x_2$$
What should be the next step?
Could you also tell me if the following identities are correct?
$$sin\frac{1}{1+x_1^2}=sin\frac{1}{1+x_2^2}$$
$$\frac{1}{1+x_1^2}=\frac{1}{1+x_2^2}$$
| I suppose that your function is defined on $\mathbb{R}$. Assume that $f(x)=f(y)$. This means that $-x^{5}-16x-1=-y^{5}-16y-1$, so $x^5+16x=y^5+16y$. From this we get $$(x-y)(x^4+x^3y+x^2y^2+xy^3+y^4+16)=0.$$
Then, $x=y$ or $x^4+x^3y+x^2y^2+xy^3+y^4+16=0$. If $x=y$ we're done. Otherwise, $x^4+x^3y+x^2y^2+xy^3+y^4+16=0$. But if $x,y>0$ thus the LHS is positive, and therefore cannot be equal to $0$. So at least one of $x$ and $y$ must be negative. Suppose that it's $y$ and write $y=-z$ with $z>0$. We have $$x^4-x^3z+x^2z^2-xz^3+z^4+16=0.$$
But $x^4-x^3z+x^2z^2-xz^3+z^4+16=(x-z)^2(x^2+xz+z^2)+x^2z^2+16>0$. So it must be $x=y$, and hence $f$ is injective.
| {
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"timestamp": "2023-03-29T00:00:00",
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In a $\triangle ABC,\angle B=60^{0}\;,$ Then range of $\sin A\sin C$
In a $\triangle ABC,\angle B=60^{0}\;,$ Then range of $\sin A\sin C$
$\bf{My\; Attempt:}$
Using Sin formula:
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$
So $\displaystyle \frac{a}{\sin A} = \frac{b}{\sin 60^0}\Rightarrow \sin A = \frac{a}{b}\cdot \frac{\sqrt{3}}{2}$ and $\displaystyle \frac{c}{\sin C} = \frac{b}{\sin 60^0}\Rightarrow \sin C = \frac{c}{b}\cdot \frac{\sqrt{3}}{2}$
So $$\sin A\sin C = \frac{3}{4}\cdot \frac{ac}{b^2}$$
Now using Cosine formula:
$$\cos B = \cos 60^0 = \frac{1}{2}= \frac{a^2+c^2-b^2}{2ac}\Rightarrow b^2=a^2+c^2-ac$$
So $$\sin A\sin C = \frac{3}{4}\bigg[\frac{ac}{a^2+c^2-ac}\bigg] = \frac{3}{4}\bigg[\frac{1}{\frac{a}{c}+\frac{c}{a}-1}\bigg]\leq \frac{3}{4}$$
Using $\bf{A.M\geq G.M},$ We get $$\frac{a}{c}+\frac{c}{a}\geq 2\Rightarrow \frac{a}{c}+\frac{c}{a}-1\geq 1$$
$\bf{ADDED::}$ Using Jensen Inequality:: For $f(x) = \ln(\sin x)\;,$
$$\ln(\sin A)+\ln(\sin C)\leq 2\cdot \ln \sin \left(\frac{A+C}{2}\right) = \ln \cos 30^0 = 2\cdot \ln \frac{\sqrt{3}}{2} = \ln \frac{3}{4}$$
But I do not understand how to calculate the lower bound for $\sin A\sin C$.
Thanks in advance!
| It is sufficient to consider the function
$$f(x)=\sin (x)\sin(\frac{2\pi}{3}-x)$$ restreint to the domain $D=\{x|\space 0\lt x\lt\dfrac{2\pi}{3}\}$.
It is easy to find $f$ is positive and has a maximun at the point $x=\dfrac{\pi}{3}\in D$; furthermore the infimum of $f(x)$ is equal to $0$ taken to the neighborhood of both $0$ and $\dfrac{2\pi}{3}$.
Thus the range of $f$ is the semi-open interval $(0,\space\frac34]$.
| {
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Show that the set of numbers whose Euler function is less than a certain value has an upper bound Let $S = \{n \in \mathbb{Z} \mid \varphi(n) \le 12 \}$ be the set of number whose Euler function is less or equal 12. Prove that $S$ has an upper bound.
In the example $S= \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 18, 20, 21, 22, 24, 26, 28, 30, 36, 42\}$ and I'd like to show formally that no $k > 42 \in S$.
I've demostrated some particular cases:
*
*if $k > 13$ is prime, then $\varphi(k) = k - 1 > 12$, thus $k \notin S$.
*if $k^\alpha$ is a power of a prime $k > 13$, then $\varphi(k^\alpha) = \varphi(k^{\alpha-1})\cdot(k-1) > (k-1) > 12$, thus $k^\alpha \notin S$.
For the general case, I've thought about factorization. So $k = p_1^{\alpha_1} \cdot \ldots \cdot p_n^{\alpha_n}$, where each $p_i$ must be in $\{2, 3, 5, 7, 11, 13\}$.
I can't however conclude my proof.
| The Fundamental theorem of arithmetic allows to write $k = p_1^{\alpha_1} \cdot \ldots \cdot p_n^{\alpha_n}$.
As $\varphi(p) = p-1$, $p$ must be less than 13. So the only primes allowed are $\{2,3,5,7,11,13\}$. Consider every possibile combination:
*
*$p_1 = 13$ gives $13$ and $13 \cdot 2 = 26$
*$p_1 = 11$ gives $11$ and $11 \cdot 2 = 22$
*$p_1 = 7$ gives $7$, $7 \cdot 2 = 14$, $7 \cdot 2^2 = 28$, $7 \cdot 3 = 21$, $7 \cdot 2 \cdot 3 = 42$
*$p_1 = 5$ gives $5$, $5 \cdot 2 = 10$, $5 \cdot 2^2 = 20$, $5 \cdot 3 = 15$, $5 \cdot 2 \cdot 3 = 30$
*$p_1 = 3$ gives $3$, $3^2 = 9$, $3 \cdot 2 = 6$, $3 \cdot 2^2 = 12$, $3 \cdot 2^3 = 24$, $3^2 \cdot 2 = 18$, $3^2 \cdot 2^2 = 36$
*$p_1 = 2$ gives $2$, $2^2 = 4$, $2^3=8$, $2^4 = 16$
By definition we also have $\varphi(1) = 1$.
Every combination has been explored, so the list provided above is complete.
| {
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Let $p$ be a prime of the form $3k+2$ that divides $a^2+ab+b^2$. Prove that $a,b$ are both divisible by $p$. [Solution verification] Problem:Let $p$ be a prime of the form $3k+2$ that divides $a^2+ab+b^2$ for some $a,b\in \mathbb{Z}$. Prove that $a,b$ are both divisible by $p$.
My Attempt: $a^2+ab+b^2\equiv 0 \pmod p\Rightarrow a^3\equiv b^3\pmod p\Rightarrow a^{3k}\equiv b^{3k}\pmod p.$
Next, observe that due to FLT we have $a^{3k+1}\equiv b^{3k+1}\pmod p.$ Now if $p\not|a$ and $p\not|b$, then $\gcd(a,p)=\gcd(b,p)=1.$ Therefore we can use can conclude that $$a^{\gcd(3k,3k+1)}\equiv b^{\gcd(3k,3k+1)}\pmod p\Rightarrow a\equiv b\pmod p.$$ Therefore $$a^2+ab+b^2\equiv 0 \pmod p\Rightarrow 3b^2\equiv 0\pmod p \text{ and } 3a^2\equiv 0\pmod p.$$ Which implies that $p|3$ which is a contradiction. Hence Proved.
I would like to know whether this proof is correct or not. I am unsure about the use of $\gcd$ in the exponent. Moreover, I acknowledge that this question has been asked before, but I've not seen any answer using this fact explicitly. The fact being: Let $\gcd(a,m)=\gcd(b,m)=1$, then if $a^{x}\equiv b^x\pmod m$ and $a^y\equiv b^y\pmod m\Rightarrow a^{\gcd(x,y)}\equiv b^{\gcd(x,y)}\pmod m.$
| Here's a correct different solution. If $p=3k+2$ is an odd prime, $p\mid a^2+ab+b^2$, where $a,b\in\mathbb Z$, then $$a^2+ab+b^2\equiv 0\pmod{p}$$
$$\stackrel{\cdot 4}\iff (a+2b)^2\equiv -3a^2\pmod{p}$$
$$\iff (2a+b)^2\equiv -3b^2\pmod{p}$$
For contradiction, let either $p\nmid a$ or $p\nmid b$. Then either
$$\left((a+2b)a^{-1}\right)^2\equiv -3\pmod{p}$$
or $$\left((2a+b)b^{-1}\right)^2\equiv -3\pmod{p}$$
In both cases $-3$ is a quadratic residue mod the odd prime $p=3k+2$, which contradicts Quadratic Reciprocity.
If $p=2$ and $2\mid a^2+ab+b^2$ for some $a,b\in\mathbb Z$, then either $a\equiv 0\pmod{2}$ or $a\equiv 1\pmod{2}$; in the first case $2\mid b^2$, i.e. $b\equiv 0\pmod{2}$, and in the second case $$a^2+ab+b^2\equiv 1+b+b^2\equiv 1+b+b$$
$$\equiv 1+2b\equiv 1\not\equiv 0\pmod{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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if a,b,c,d are integers such that 5 divides ($a^4+b^4+c^4+d^4$) then 5 also divides (a+b+c+d) i want to show that
if a,b,c,d are integers such that 5 divides ($a^4+b^4+c^4+d^4$) then 5 also divides (a+b+c+d) but im not quite sure how to aproach this problem.
| Mod $5$, we have that the only squares are $0,1,4$. The squares of these are $0,1$. So, we have each of $a^4,b^4,c^4,d^4\in\lbrace 0,1\rbrace $, and their sum is $0$ (becaue $5$ divides them). Because of this, each of $a^4,b^4,c^4,d^4\equiv 0\pmod{5}$. Because of this, we have that $a,b,c,d\equiv 0\pmod{5}$, so their sum is divisible by $5$.
An elaboration on "the only squares" argument:
The only numbers mod $5$ are $0,1,2,3,4$.
Lets see how these look like when we square them, and then square them again (take their $4$th power).
\begin{array}{|c|c|c|c|}
\hline
\text{Number}& \text{Square} & \text{Square mod 5} & \text{4th Power} & \text{4th Power mod 5} \\ \hline
0 &0 &0 &0 & 0\\ \hline
1 &1 &1 &1 & 1\\ \hline
2 &4 &4 &16 & 1\\ \hline
3 & 9 & 4 & 16 & 1 \\\hline
4 & 16 & 1 & 1 & 1\\\hline
\end{array}
Look at this last column. The only possibilities if we take the $4$th power of a number $\pmod{5}$ are $0$ or $1$.
So, each of $a^4,b^4,c^4,d^4$ must be either $0$ or $1$ mod $5$, so if we know they all add up to $0$ (mod $5$), then we know that none of them can be $1$. The have to all be $0$.
So, we have that $a^4 \equiv b^4\equiv c^4\equiv d^4\equiv 0\pmod{5}$.
So, we know that $a\equiv b\equiv c\equiv d\equiv 0\pmod{5}$ (If we start at the right hand side of the table and move to the left, we see that if something to the $4$th power is $0$ mod $5$, the original thing had to be $0$ mod $5$). So, we get that $a+b+c+d\equiv 0+0+0+0\equiv 0\pmod{5}$, so it has to be divisible by $5$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How can I derive what is $1\cdot 2\cdot 3\cdot 4 + 2\cdot 3\cdot 4\cdot 5+ 3\cdot 4\cdot 5\cdot 6+\cdots + (n-3)(n-2)(n-1)(n)$ ?? I'd like to evaluate the series $$1\cdot 2\cdot 3\cdot 4 + 2\cdot 3\cdot 4\cdot 5+ 3\cdot 4\cdot 5\cdot 6+\cdots + (n-3)(n-2)(n-1)(n)$$
Since I am a high school student, I only know how to prove such formula's (By principal of mathematical induction). I don't know how to find result of such series.
Please help. I shall be thankful if you guys can provide me general solution (Since I have been told that there exist a general solution by my friend who gave me this question).
| As shown in this post,
$$ \sum_{k=1}^n x^k = x \sum_{k=1}^{n} \binom{n}{k} (x-1)^{k-1}$$
Differentiate five times:
$$ \sum_{k=1}^n k (k-1)(k-2)(k-3) x^{k-4} = \frac{d^5}{dx^5} \left[ x \sum_{k=1}^{n} \binom{n}{k} (x-1)^{k-1} \right] \tag{1}$$
For left side,
$$ \sum_{k=1}^n k (k-1)(k-2)(k-3)(k-4) x^{k-4}$$
We can start this sum:
$$ \sum_{k=5}^n (k-1)(k-2)(k-3) x^{k-4}$$
And, I can sub $k-5 =u$, this turns into:
$$ \sum_{u=0}^{n-5} (u+4)(u+3)(u+2)(u+1)x^u$$
And putting this back in(1):
$$ \sum_{u=0}^{n-5} (u+4)(u+3)(u+2)(u+1)x^u= \frac{d^5}{dx^5} \left[ x \sum_{k=1}^{n} \binom{n}{k} (x-1)^{k-1} \right] \tag{1}$$
Now shift $ n \to n +5$, this leads to:
$$ \sum_{u=0}^{n} (u+4)(u+3)(u+2)(u+1)x^u= \frac{d^5}{dx^5} \left[ x \sum_{k=1}^{n+5} \binom{n+5}{k} (x-1)^{k-1} \right] \tag{1}$$
Evaluate RHS using leibniz product rule and take limit on both side as $x \to 1$, you get the formula
| {
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"timestamp": "2023-03-29T00:00:00",
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If $ \sin \alpha + \sin \beta = a $ and $ \cos \alpha + \cos \beta = b $ , then show that $\sin(\alpha + \beta) = \frac {2ab } { a^2 + b^2} $ I've been able to do this, but I had to calculate $ \cos (\alpha + \beta) $ first. Is there a way to do this WITHOUT calculating $\cos(\alpha+\beta)$ first ?
Here's how I did it by calculating $\cos(\alpha+\beta)$ first
$ a^2 + b^2 = \sin ^2 \alpha + \sin ^2 \beta + 2 \sin \alpha \sin \beta + \cos ^2 \alpha + \cos ^2 \beta + 2 \cos \alpha \cos \beta $
$a^2 + b^2 = (\sin^2\alpha + \cos^2\alpha) + (\sin ^2 \beta + \cos^2 \beta) + 2(\cos\alpha\cos\beta + \sin\alpha\sin\beta)$
$a^2 + b^2 = 2 (1 + \cos(\alpha-\beta))$
$ \frac{a^2 + b^2}{2} = (1 + \cos(\alpha - \beta))$
$ b^2 - a^2 = (\cos ^2\alpha - \sin^2\alpha) + (\cos^2 \beta - \sin^2\beta) + 2\cos\alpha\cos\beta - 2\sin\alpha\sin\beta$
$b^2 - a^2 = (\cos^2\alpha - (1 - cos^2\alpha)) +(1-\sin^2\beta) - \sin^2\beta)) + 2(\cos\alpha\cos\beta - \sin\alpha\sin\beta) $
$b^2 - a^2 = 2 (\cos^2\alpha - \sin^2\beta + \cos(\alpha+\beta))$
$b^2 - a^2 = 2(\cos(\alpha+\beta)\cos(\alpha-\beta)+\cos(\alpha+\beta))$
$\frac{b^2 - a^2}{2} = \cos(\alpha+\beta)\{\cos(\alpha-\beta) + 1 \}$
$\frac{b^2 - a^2}{2} = \cos(\alpha+\beta)\{\frac{b^2+a^2}{2}\}$
$\cos(\alpha+\beta) = \frac {a^2 + b^2 } {a^2 - b^2}$
Then I just calculated $\sin(\alpha + \beta)$ by $1 - \cos^2(\alpha+\beta)$
| How about:
$$\begin{array}{}
a&=\sin\alpha+\sin\beta&=2\sin\frac 12(\alpha+\beta)\cos\frac 12(\alpha -\beta) \\
b&=\cos\alpha+\cos\beta&=2\cos\frac 12(\alpha+\beta)\cos\frac 12(\alpha -\beta) \\
ab&=4\sin\frac 12(\alpha+\beta)\cos\frac 12(\alpha+\beta)\cos^2\frac 12(\alpha -\beta)&=2\sin(\alpha+\beta)\cos^2\frac 12(\alpha -\beta) \\
a^2+b^2&=4\big(\sin^2\frac 12(\alpha+\beta)+\cos^2\frac 12(\alpha+\beta)\big)\cos^2\frac 12(\alpha -\beta)&=4\cos^2\frac 12(\alpha -\beta) \\
\frac{2ab}{a^2+b^2}&=\sin(\alpha+\beta)
\end{array}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Real values of $a$ for $x^4+(a-1)x^3+x^2+(a-1)x+1=0$ to have at least two negative roots
Problem Statement:-
Find all the values of $a$ for which the equation
$$x^4+(a-1)x^3+x^2+(a-1)x+1=0$$
possess at least two negative roots.
I know that there has been a post regarding this same problem here, but I have a different issue with the problem than that posted in the OP in the above post.
So, I don't have an issue of finding a solution and coincidentally I had the same approach as the one posted by RicardoCruz.
But still if you have a solution than you are more than welcome to post one, well come on who doesn't want some good reputation change ;P.
My Attempt at a solution:-
For those who wanna skip to my main issue may start to read from the next quote.
We can see very clearly that $x=0$ isn't a root of the equation $$x^4 +(a-1)x^3 +x^2 +(a-1)x+1=0$$
hence it is safe to divide the equation throughout by $x^2$ for arriving at the condition which the problem wants.
So, we get
$$\left(x^2+\dfrac{1}{x^2}+2\right)+(a-1)(x+\dfrac{1}{x})-1=0\\
\implies \left(x+\dfrac{1}{x}\right)^2+(a-1)(x+\dfrac{1}{x})-1=0$$.
Now, let $z=x+\dfrac{1}{x}$ then the equation becomes
$$z^2+(a-1)z-1=0\\
\implies z=\dfrac{-(a-1)\pm\sqrt{\left(a-1\right)^2+4}}{2}$$
Now, its necessary to note that $\sqrt{\left(a-1\right)^2+4}\gt(a-1)$, hence for the roots to be negative $$\left(x+\dfrac{1}{x}\right)=\dfrac{-(a-1)-\sqrt{\left(a-1\right)^2+4}}{2}$$
As, we know that $\text{A.M. $\ge$ G.M.}$ for non negative real numbers, and as we are dealing with negative roots of the equation, so we will be applying $\text{A.M. $\ge$ G.M.}$ to $-x,-\dfrac{1}{x}$, we get
$$\dfrac{-\left(x+\dfrac{1}{x}\right)}{2}\ge1\implies x+\dfrac{1}{x}\le-2\\
\implies \left(x+\dfrac{1}{x}\right)=\dfrac{-(a-1)-\sqrt{\left(a-1\right)^2+4}}{2}\le -2$$
But as the roots are supposed to be distinct hence, we have to exclude the case of equality, because it occurs when $x=\dfrac{1}{x}=-1$, which results in $x+\dfrac{1}{x}=-2\implies (x-1)^2=0$, hence $x$ has a repeated root. So, we get
$$\dfrac{-(a-1)-\sqrt{\left(a-1\right)^2+4}}{2}\lt -2\implies \sqrt{\left(a-1\right)^2+4}\gt 5-a$$
Now, this is where I am having problem, that's right solving the "inequality". I always had thought of this before but never tried to test it before, so bear with it and follow me on to a trip to solve inequalities in a weird way.
I did a case study for the inequality, which is as follows:-
Case-1:-
If $(5-a)\gt 0\implies a\lt 5$, then there is no problem in squaring the inequality as both sides are positive.
$$\left(a-1\right)^2+4\gt a^2+25-10a\implies a\gt \dfrac{5}{2}$$
$$\therefore a\in\left(\dfrac{5}{2},5\right)$$
Case-2:-
If $5-a\lt0$, then we might be faced with a problem cause, there might be a case where $|5-a|\gt\sqrt{\left(a-1\right)^2+4}$, which would make the inequality $\sqrt{\left(a-1\right)^2+4}\gt 5-a$ on squaring change the sign of inequality to be in accordance with the condition $|5-a|\gt\sqrt{\left(a-1\right)^2+4}$.
So, on solving the inequality we get,
$$\sqrt{\left(a-1\right)^2+4}\gt 5-a\implies (a-1)^2+4\lt(5-a)^2$$
Now, why did I change the sign of inequality was due to the fact that I was finding the interval where(as per the condition I stated before) $|5-a|\gt\sqrt{\left(a-1\right)^2+4}$. So, just like $5\gt-6$ but on squaring the inequality we get $25\lt36$. So that's what I did.
This results in $a\lt\dfrac{5}{2}\cup a\gt5\implies a\in\emptyset$
So, on combining both the cases we get $a\in\left(\dfrac{5}{2},5\right)$
But from RobertCruz's solution and the books answer it seems that $ a\in\left(\dfrac{5}{2},\infty\right)$ is the correct solution, so what am I doing wrong here.
| Just posting an alternative solution, as the comments already helped you to debug your solution:
For quartic polynomials, there is a closed-form analytical solution for the zeroes. However horrible that formula may seem, you don't need it here. You only need the discriminant.
A polynomial $ax^4+bx^3+cx^2+dx+e=0$ has two real solutions, when the discriminant is negative:
$$\Delta=256 a^3 e^3 - 192 a^2 b d e^2 - 128 a^2 c^2 e^2 + 144 a^2 c d^2 e$$
$$ - 27 a^2 d^4 + 144 a b^2 c e^2 - 6 a b^2 d^2 e - 80 a b c^2 d e + 18 a b c d^3 $$
$$+ 16 a c^4 e - 4 a c^3 d^2 - 27 b^4 e^2 + 18 b^3 c d e - 4 b^3 d^3 - 4 b^2 c^3 e + b^2 c^2 d^2<0$$
In your case, the polynomial can be written also as
$$\underbrace{x^4+x^2+1}_{>0} + (a-1)\underbrace{(x+x^3)}_{>0 \text{ if } x>0}=0$$
for which any potential roots are negative if $a-1>0$ and positive otherwise.
Therefore, having checking for $a-1>0$ and negative discriminant ensures two negative real roots. You get a third degree polynomial in $u=(a-1)^2$.
$$144+8u-23u^2-4u^3<0$$
It turns out this factors:
$$(4u-9)(4+u)^2>0$$
$$|u|>9/4$$
$$a-1>3/2$$
Also, you can quickly prove that your polynomial cannot have more than two real roots. Either by checking other tests on the link above (checking the value for $P$ and $D$) or by writing
$$x=-\frac{1}{1-a}\frac{1+x^2+x^4}{1+x^2}=-\frac{1}{a-1}\left(1+\frac{x^4}{1+x^2}\right)$$
RHS has no inflection points, so a straight line can intersect it twice at most (definition of convexity).
| {
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Show that $1^3 + 2^3 + ... + n^3 = (n(n+1)/2)^2$ by induction I have some with proving by induction. I cannot find a solution for the inductive step: $1^3 + 2^3 + ... + n^3 = (n(n+1)/2)^2$
I already did the induction steps:
Basis: P(1) = $1^3 = (1(1+1)/2)^2$ (This is true)
Inductive step: Assume $P(k) = ((k)(k+1)/2)^2$
To be proven: $((k)(k+1)/2)^2 + (k+1)^3 = ((k+1)(k+2)/2)^2$
My problem is that I do not know how I can put the $ + (k+1)^3$ inside $((k)(k+1)/2)^2$.
Simplifying the left and right part of the statement does not help:
Simplifying the left side: $((k)(k+1)/2)^2 + (k+1)^3 = ((k^2+k)/2)^2 + (k+1)^3 $
Simplifying the right side: $((k+1)(k+2)/2)^2 = ((k^2+3k+2)/2)^2$
So i am left with: $((k^2+k)/2)^2 + (k+1)^3 = ((k^2+3k+2)/2)^2$
That is the same as: $1/4 (k^2+k)^2 + (k+1)^3 = 1/4((k^2+3k+2))^2$
Going further with the left side:$1/4 (k^2+k)^2 + (k+1)^3 = (1/4)(k^4 + 2k^3 + k^2) + k^3+3 k^2+3 k+1$
Going further with the right side: $1/4((k^2+3k+2))^2 = 1/4 (k^4+6 k^3+13 k^2+12 k+4)$
Now I am stuck with: $(1/4)(k^4 + 2k^3 + k^2) + k^3+3 k^2+3 k+1 = 1/4 (k^4+6 k^3+13 k^2+12 k+4)$
Now I am kind of left with garbage. Am I missing something? What do I do wrong? Where can I find a good resources to learn how to solve this issue?
| First, show that this is true for $n=1$:
$\sum\limits_{k=1}^{1}k^3=(1(1+1)/2)^2$
Second, assume that this is true for $n$:
$\sum\limits_{k=1}^{n}k^3=(n(n+1)/2)^2$
Third, prove that this is true for $n+1$:
$\sum\limits_{k=1}^{n+1}k^3=$
$\color\red{\sum\limits_{k=1}^{n}k^3}+(n+1)^3=$
$\color\red{(n(n+1)/2)^2}+(n+1)^3=$
$((n+1)(n+2)/2)^2$
Please note that the assumption is used only in the part marked red.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2024028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
How can $\frac{x^3-4x^2+4x}{x^2-4}$ be both $0$ and "undefined" when $x = 2$? Suppose I have a function defined as $$F(x)= \frac{x^3-4x^2+4x}{x^2-4}$$
Now I want to find the value of $F(2)$. I can do it in 2 ways:
*
*Put $x=2$ and solve the function. It will give:
$$F(2)=\frac{0}{0}$$ which is not defined.
*Solve $F(x)$ first and then put $x=2$.
$$F(x)= \frac{x(x-2)^2}{(x-2)(x+2)}=\frac{x(x-2)}{x+2}$$
It will give $${F(2)=\frac{0}{4}}$$ which is zero.
How can zero equal not defined?
| What you have found is a simplification for $F(x)$, provided $x\neq 2, x\neq -2$. The denominator $(x^2 - 4)$ of the original function makes it undefined at $x = 2, \;x=-2:$ $(2^2-4) = (-2)^2 - 4 = 0$
So your simplification
$$F(x)= \frac{x(x-2)^2}{(x-2)(x+2)}=\frac{x(x-2)}{x+2}$$ is valid, $\forall x \in \mathbb R \setminus\{-2, 2\}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2027810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
How can simultaneous sinusoidal equations be solved? I have come across a set of simultaneous equations which I can't figure out how to solve. I have 3 equations and only two unknowns, but they are angular quantities and feature in the equations as sinusoidal functions of the angular quantities.
The system of equations is:
$$ \begin{Bmatrix}\cos\psi\sin\theta+\cos\theta\sin\phi\sin\psi \\ \sin\psi\sin\theta-\cos\psi\cos\theta\sin\phi \\ \cos\phi\cos\theta \end{Bmatrix} = \begin{Bmatrix} x \\ y \\ z \end{Bmatrix} $$
Where: x, y, z and psi are known and phi and theta are unknown.
Is it possible to rearrange these equations to solve for theta and phi using the other terms?
| Note that
$$
\begin{array}{l}
\left( {\begin{array}{*{20}c}
x \\
y \\
z \\
\end{array}} \right) = \left( {\begin{array}{*{20}c}
{\cos \psi \sin \theta + \cos \theta \sin \phi \sin \psi } \\
{\sin \psi \sin \theta - \cos \psi \cos \theta \sin \phi } \\
{\cos \phi \cos \theta } \\
\end{array}} \right) = \\
= \left( {\begin{array}{*{20}c}
{\sin \theta \cos \psi + \cos \theta \sin \phi \sin \psi } \\
{\sin \theta \sin \psi - \cos \theta \sin \phi \cos \psi } \\
{\cos \theta \cos \phi } \\
\end{array}} \right) = \\
= \sin \theta \left( {\begin{array}{*{20}c}
{\cos \psi } \\
{\sin \psi } \\
0 \\
\end{array}} \right) + \cos \theta \left( {\begin{array}{*{20}c}
{\sin \phi \sin \psi } \\
{ - \sin \phi \cos \psi } \\
{\cos \phi } \\
\end{array}} \right) = \\
= \sin \theta \left( {\begin{array}{*{20}c}
{\cos \psi } \\
{\sin \psi } \\
0 \\
\end{array}} \right) + \cos \theta \left( {\begin{array}{*{20}c}
{\sin \phi \cos \left( {\psi + \pi /2} \right)} \\
{\sin \phi \sin \left( {\psi + \pi /2} \right)} \\
{\cos \phi } \\
\end{array}} \right) = \\
= \sin \theta \;{\bf u} + \cos \theta \;{\bf v} \\
\end{array}
$$
where $\bf u$ and $\bf v$ are given in spherical coordinates, and can be seen to be
unitary vectors and orthogonal to each other.
Their combination with $sin \theta$ and $cos \theta$ is therefore a circle in their plane.
For the system to have solutions, the vector $\bf p=(x,y,z)$ shall be on the unitary sphere.
That given, there will be multiple solutions for the three angles, corresponding to the
infinite main circles passing through $\bf p$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2029025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Advice on proof in linear algebra. I just wrote my first proof in linear algebra so I'd love some advice on the things that go well and what could be improved upon. It's a proof by induction.
Theorem:
Let $A_n$ be a $n\times n$ matrix of the form:
$\begin{pmatrix}
2 & 1 & 0 & 0 && & \cdots & 0\\
1 & 2 & 1 & 0 && & \cdots & 0\\
0 & 1 & 2 & 1 && & \cdots & 0\\
0 & 0 & 1 & 2 && & \cdots & 0\\
\vdots& & & & \ddots & && \vdots\\
0&\cdots&&&&2&1&0\\
0&\cdots&&&&1&2&1\\
0&\cdots&&&&0&1&2\\
\end{pmatrix}$
Then $det(A_n)=3(n-1)$.
Proof:
We'll give a proof by mathematical induction.
Let $n=2$:
$\begin{vmatrix}
2&1\\
1&2\\
\end{vmatrix}=4-1=3$.
Let $n=3$:
$\begin{vmatrix}
2&1&0\\
1&2&1\\
0&1&2\\
\end{vmatrix}=2\begin{vmatrix}
2&1\\
1&2\\
\end{vmatrix}-\begin{vmatrix}
1&0\\
1&2\\
\end{vmatrix}=8-2=6$
Let $n=4$:
$\begin{vmatrix}
2&1&0&0\\
1&2&1&0\\
0&1&2&1\\
0&0&1&2\\
\end{vmatrix}=2\begin{vmatrix}
2&1&0\\
1&2&1\\
0&1&2\\
\end{vmatrix}-\begin{vmatrix}
1&0&0\\
1&2&1\\
0&1&2\\
\end{vmatrix}=2\begin{vmatrix}
2&1&0\\
1&2&1\\
0&1&2\\
\end{vmatrix}-\begin{vmatrix}
2&1\\
1&2\\
\end{vmatrix}=12-3=9$
Notice how due to the repetitive nature of our matrix, we'll candevise a recursive formula for our determinant: $|A_n|=2|A_{n-1}|-|A_{n-2}|$.
If we assume: $|A_n|=3(n-1)$, $|A_{n-1}|=3(n-2)$, $|A_{n-2}|=3(n-3)$, then:
$|A_n|=6(n-2)-3(n-3)=3(n-1)$
Since we checked $n=2, 3$; we conclude that $|A_n|=3(n-1)$ for all $n \in \mathbb{N} | n>1$
| Let $d_n = \det A_n$. Note that $d_1 = 2, d_2 = 3$.
For the induction step, look at the (clunky) picture below:
Note that $d_n = 2 d_{n-1} -d_{n-2}$.
Now show that $n \mapsto 3(n-1)$ satisfies the equation with the same
initial conditions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2029176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solving inequality involving floor function We have this inequality (over real numbers) :
$$x^2-2x\le \frac{\sqrt{1-\lfloor x\rfloor^2}}{\lfloor x \rfloor + \lfloor -x \rfloor}$$
How we can solve it using both of algebraic and geometric methods ?
| $$\begin{cases}x= \lfloor x\rfloor+\{x\}\\-x= \lfloor -x\rfloor+1-\{x\}\end{cases}\Rightarrow\lfloor x\rfloor+\lfloor -x\rfloor=-1$$
$$\sqrt{1-\lfloor x\rfloor^2}=\begin{cases}0 \text{ if } -1\le x\lt0\\1\text{ if } 0\le x\lt1\\0 \text{ if } 1\le x\lt2\end{cases}$$ Hence
$$\frac{\sqrt{1-\lfloor x\rfloor^2}}{\lfloor x\rfloor+\lfloor -x\rfloor}=\begin{cases}0 \text{ if } -1\le x\lt0\\-1\text{ if } 0\le x\lt1\\0 \text{ if } 1\le x\lt2\end{cases}$$
It follows that if $f(x)=x^2-2x$ then since $$f((-1,0))=(0,3)\\f([0,1))=[-1,0]\\f([1,2))=[-1,0)$$ the solution is $$1\le x\lt 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2032279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Differential equation with integral that is hard to compute Task and my attempt to solve it
$$xy'=\sqrt{2x^2+y^2}-y$$
I have tried substitution like $y=z·x$ but finally got very difficult integral
$$ \int \frac{dz}{\sqrt{2+z^2}-2z}=\int\frac{dx}{x}$$
at the left side.
Maybe i have to do something different to avoid difficult computation?
Please explain me how to solve it.
| *
*Multiply by $\frac{\sqrt{2+z^2}+2z}{\sqrt{2+z^2}+2z}$:
$$
\int\frac{dz}{\sqrt{2+z^2}-2z}=\int\frac{\sqrt{2+z^2}+2z}{(2+z^2)-4z^2}\,dz=\int\frac{\sqrt{2+z^2}}{2-3z^2}\,dz+\int\frac{2z}{2-3z^2}\,dz.
$$
The second integral is simple.
*In the first integral do the change of variables $t=z+\sqrt{z^2+2}$:
$$
\int\frac{\sqrt{2+z^2}}{2-3z^2}\,dz=\int\frac{(t^2+2)^2}{t(-3t^4+20t^2-12)}\,dt.
$$
*Another change of variables $u=t^2$ gives
$$
\int\frac{(t^2+2)^2}{t(-3t^4+20t^2-12)}\,dt=\int\frac{\frac12(u+2)^2}{u(-3u^2+20u-12)}\,du=\frac12\int\frac{(u+2)^2}{u(u-6)(2-3u)}\,du
$$
which can be calculated by partial fractions.
However, I wouldn't wish finding $y(x)$ explicitly after all that.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2032716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve system of equation using matrices (4 variables) The Question:
Solve using matrices.
$$2w-2x-2y+2z=10\\w+x+y+z=-5\\3w+x-y+4z=-2\\w+3x-2y+2z=-6$$
My work:
$$
\begin{bmatrix}
2&-2&-2&2&10\\
1&1&1&1&-5\\
3&1&-1&4&-2\\
1&3&-2&2&-6\\
\end{bmatrix}
\rightarrow
\begin{bmatrix}
1&-1&-1&1&5\\
0&2&2&0&-10\\
0&4&-1&1&-11\\
0&-2&3&-1&1\\
\end{bmatrix}
\rightarrow\\
\begin{bmatrix}
1&-1&-1&1&5\\
0&2&2&0&-10\\
0&0&-5&0&9\\
0&0&5&-1&-9\\
\end{bmatrix}
\rightarrow
\begin{bmatrix}
1&-1&-1&1&5\\
0&2&2&0&-10\\
0&0&-5&0&9\\
0&0&0&-1&0\\
\end{bmatrix}\\[6ex]
z=0, \;y=\frac{-9}5,\; x=\frac{34}5,\; w=10
$$
The correct answer is $z=-1, \;y=-2,\; x=-3,\; w=1$. What did I do wrong?
| In the second matrix of your work, entry $a_{3,3}$ should be $2$, not $-1$, and entry $a_{3,5}$ should be $-17$, not $-11$. Also, the whole fourth row seems wrong. It appears that the four row operations performed in that step should have been:
\begin{align}
&1.\text{ Replace R1 with $\frac12\times$ R1.} \\
&2.\text{ Replace R2 with R2 - R1.} \\
&3.\text{ Replace R3 with R3 - 3$\times$R1.} \\
&4.\text{ Replace R4 with R4 - R1.}
\end{align}
It appears that you did the first two of those correctly, but not the third and fourth ones.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2036797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to show that $f(x,y)=x^4+y^4-3xy$ is coercive? How to show that $f(x,y)=x^4+y^4-3xy$ is coercive ?
This is my attempt :
$$f(x,y)=x^4+y^4-3xy$$
$$f(x,y)=x^4+y^4\left(1-\frac{3xy}{x^4+y^4}\right)$$
As $||(x,y)|| \to \infty $ , $\frac{3xy}{x^4+y^4} \to 0$
So $||(x,y)|| \to \infty $ , $f(x,y)=x^4+y^4-3xy \to \infty$.
Is this a valid method ?
I think it is not a rigorous proof. can anyone help me with a good proof ?
| Your answer looks right.
Here's another way. Observe
\begin{align}
xy \le \frac{x^2+y^2}{2} \ \ \text{ and }\ \ x^4+y^4\geq \frac{1}{2}(x^2+y^2)^2
\end{align}
which means
\begin{align}
x^4+y^4-3xy \geq&\ x^4+y^4-\frac{3}{2}(x^2+y^2) \\
\geq&\ \frac{1}{2}(x^2+y^2)^2-\frac{3}{2}(x^2+y^2)\\
=&\ \frac{1}{2}\left(x^2+y^2-\frac{3}{2}\right)^2-\frac{9}{8}.
\end{align}
Hence it follows as $x^2+y^2\rightarrow \infty$, we see that $x^4+y^4-3xy\rightarrow \infty$.
Note: Using the above inequalities you could show
\begin{align}
1-\frac{3xy}{x^4+y^4} \geq 1- \frac{3(x^2+y^2)}{2(x^4+y^4)} \geq 1-\frac{3}{(x^2+y^2)}
\end{align}
which means for $x^2+y^2$ sufficiently big we have that
\begin{align}
1-\frac{3}{(x^2+y^2)} \ge \frac{1}{2}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2041278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
find $g$ such that $g\circ f=h$ Let
$$f(x)=\dfrac{2x+3}{x-1} \quad \mbox{and}\quad h(x)=\dfrac{6x^2+8x+11}{(x-1)^2} $$
find polynom $g$ such that $g\circ f=h$
Indeed,
$$g\circ f=h \iff \forall x\in \mathbb{R}\quad g\circ f(x)=h(x) $$
\begin{align}
g\circ f(x)&=h(x)\\
g\left(f(x)\right)&=h(x)\\
g\left(\dfrac{2x+3}{x-1}\right)&=\dfrac{6x^2+8x+11}{(x-1)^2}
\end{align}
Let $y=\dfrac{2x+3}{x-1}$
\begin{align}
y&=\dfrac{2x+3}{x-1}\\
x&=\dfrac{3+y}{y-2}
\end{align}
\begin{align}
g\left(y\right)&=\dfrac{6\left(\dfrac{3+y}{y-2}\right)^2+8\left(\dfrac{3+y}{y-2}\right)+11}{\left(\dfrac{3+y}{y-2}-1\right)^2}\\
&=\dfrac{6\left(\dfrac{3+y}{y-2}\right)^2+8\left(\dfrac{3+y}{y-2}\right)+11}{\left(\dfrac{3+y}{y-2}-1\right)^2} \\
&=\dfrac{6\left(3+y\right)^2+8\left(3+y\right)+11\left(y-2 \right)}{25} \\
&=\dfrac{6y^2+55y+56}{25} \\
\end{align}
Then $g(x)=\dfrac{6x^2+55x+56}{25}$ such that $g\circ f=h$
AM i right ?
| It is wrong. After you have written the expression of $g(y)$ we should have
$$ g(y) = \frac{6(3+y)^{2} + 8(3+y)(y-2) + 11(y-2)^{2}}{25} $$
instead of your expression. On simplifying we get $g(y) = y^{2}+2$, i.e,
$$ g(x) = x^{2} + 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2041532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve for $x$ : $x^4 = 5(x-1)(x^2 - x +1)$
Solve for $x$ : $x^4 = 5(x-1)(x^2 - x +1)$
I am having difficulty factoring the whole equation so that I can equate each factor to $0$ one by one and get the roots accordingly.
| Divide both sides by $x^4$ and change variable to $y = \frac{x-1}{x^2}$, we have
$$\begin{align}1 = 5y(1-y)
\iff & 4y(y-1) = -\frac45 \iff
(2y-1)^2 = \frac15\\
\implies & y = \frac12\left(1 \pm \frac{1}{\sqrt{5}}\right) = \frac{\sqrt{5}\pm 1}{2\sqrt{5}} = \frac{2}{5 \mp \sqrt{5}}
\end{align}
$$
Substitute $y$ back by $\frac{x-1}{x^2}$, we get
$$
x^2 - \frac{5\mp\sqrt{5}}{2} (x - 1) = 0
\iff \left(x - \frac{5\mp\sqrt{5}}{4}\right)^2 = \frac{5\mp\sqrt{5}}{2} -
\left(\frac{5\mp\sqrt{5}}{4}\right)^2
= -\frac{5 \pm \sqrt{5}}{8}
$$
This leads to
$\displaystyle\;x = \frac{5 - \epsilon \sqrt{5} \pm i \sqrt{2(5 + \epsilon \sqrt{5})}}{4}$ where $\epsilon = \pm 1$.
If one insists on factoring the equation first, one can treat $x^2$ and $(x-1)$ as two seperate units and rearrange them:
$$\begin{align}
& x^4 = 5(x-1)(x^2 - x + 1)\\
\iff & x^4 - 5x^2(x-1) + 5(x-1)^2 = 0\\
\iff & \left(x^2 - \frac52(x-1)\right)^2 = \left(\frac{25}{4} - 5\right)(x-1)^2 = \frac{5}{4}(x-1)^2\\
\iff & \left(x^2 - \frac{5 + \sqrt{5}}{2}(x-1)\right)\left(x^2 - \frac{5 - \sqrt{5}}{2}(x-1)\right) = 0
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2043887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Computation of eigenvalues/vectors of a $9\times 9$ matrix I have a symmetric matrix (with real coefficients) and I need to compute its eigenvalues and eigenvectors. My matrix depends on 3 parameters $(\nu_1,\nu_2,\nu_3)$ that are not independent (in fact we have $\nu_1^2+\nu_2^2+\nu_3^2=1$). If I am using Maple to compute the eigenvalues/vectors, I get an ugly answer since I don't know how to use the fact that I know that $\nu_1^2+\nu_2^2+\nu_3^2=1$ and consequently, there is certainly a lot of simplification that Maple doesn't do.
Is there a software that I can use to do this kind of computation?
For example, how could I if the matrix is
$$
B := \begin{pmatrix}
0 & \nu_1 & \nu_2 & \nu_3 & 0 & 0 & \nu_3 & \nu_2 & \nu_1 \\
\nu_1 & 0 & 0 & 0 & 0 & 0 &0 & 0 & \nu_2 \\
\nu_2 & 0 & 0 & 0 & 0 & 0 &0 & 0 & \nu_3 \\
\nu_3 & 0 & 0 & 0 & 0 & 0 &0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 &0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 &0 & 0 & 0 \\
\nu_3 & 0 & 0 & 0 & 0 & 0 &0 & 0 & 0 \\
\nu_2 & 0 & 0 & 0 & 0 & 0 &0 & 0 & 0 \\
\nu_1 & \nu_2 & \nu_3 & 0 & 0 & 0 &0 & 0 & 0 \\
\end{pmatrix}
$$
| One can do this with most of the CASs, by taking the condition $a^2+b^2+c^2-1:=0$ as another polynomial equation for computing a Gröbner basis. It depends on your specific matrix, whether or not the complexity of the system of polynomial equations is still manageable.
Edit: the characteristic polynomial of $B$ is given by
$$
f(t)=t^9 - 2t^7(a^2 + b^2 + c^2) - 2t^6ab(a + c) + t^5(a^2c^2 - 2ab^2c +
b^4 + b^2c^2 + c^4).
$$
Using $a^2+b^2+c^2=1$ we obtain
$$
f(t):=t^5 \cdot(t^4 - 2t^2 + 2tb( - ac + b^2 + c^2 - 1) - 2ab^2c + b^4 + c^2).
$$
So we have $\lambda=0$ eigenvalue with multiplicity $5$, and the other four eigenvalues the zeroes of the polynomial in brackets.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2046526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\int_{ - \infty }^\infty {\frac{{{e^{7\pi x}}}}{{{{\left( {{e^{3\pi x}} + {e^{ - 3\pi x}}} \right)}^3}\left( {1 + {x^2}} \right)}}dx}$
How to evaluate this integral?
$$\int_{ - \infty }^\infty {\frac{{{e^{7\pi x}}}}{{{{\left( {{e^{3\pi x}} + {e^{ - 3\pi x}}} \right)}^3}\left( {1 + {x^2}} \right)}}dx}$$
Maybe we can start by $$\int_0^\infty {\frac{{dx}}{{({x^2} + 1)\cosh ax}}} = \frac{1}{2}\left[ {{\psi _0}\left( {\frac{a}{{2\pi }} + \frac{3}{4}} \right) - {\psi _0}\left( {\frac{a}{{2\pi }} + \frac{1}{4}} \right)} \right]$$ in this. Then take the derivative with respect to $a$, but I'm failed to solve it!
| $$I=\frac{1}{8} \int_{-\infty}^{\infty}
\frac{\cosh(7\pi z)}{(1+z^2)\cosh^3(3\pi z)} \text{d}z$$
Now we're considering the function
$$f(z)=\frac{\cosh(7\pi z)\psi^{(0)}\left ( 1-iz\right ) }{\cosh^3(3\pi z)}$$
From the residue theorem and calculate the residues at $$z=\frac{i}{6},\frac{i}{2},\frac{5i}{6}$$
We finally obtain
$$
\int_{-\infty}^{\infty} \frac{e^{7\pi z}}
{(e^{3\pi z}+e^{-3\pi z})^3(1+z^2)} \text{d}z
=\frac{3375\pi^3-12000\sqrt{3}\pi^2+25760\pi+26784\sqrt{3} }{27000\pi^2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2051160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Determine the vectors of components For the polynomial vector space $\mathbb{R}[x]$ of degree $\leq 3$ we have the following three bases:
$$B_1 = \{1 - X^2 + X^3, X - X^2, 1 - X + X^2, 1 - X\} , \\
B_2 = \{1 - X^3, 1 - X^2, 1 - X, 1 + X^2 - X^3\}, \\
B_3 = \{1, X, X^2, X^3\}$$
How can we determine the following vectors of components $\mathbb{R}^4$ ?
$\Theta_{B_1}(b)$ for all $b \in B_1$
and
$\Theta_{B_3}(b)$ for all $b \in B_1$
Could you give me hint?
Do we use the transformation matrix? If yes, how?
$$$$
EDIT:
I have seen the following notes :
$\Theta_{B_1}(b\in B_1)=i$-th comlumn of idenity, since it shown always at itself, and $\Theta_{B_3}(b\in B_1)=i$-th column of $B_1$.
Why does this hold?
| Hint:
$$
B_1=
\begin{bmatrix}
1&X&X^2&X^3
\end{bmatrix}
\begin{bmatrix}
1&0&1&1\\
0&1&-1&-1\\
-1&-1&1&0\\
1&0&0&0
\end{bmatrix}
$$
$$
B_2=
\begin{bmatrix}
1&X&X^2&X^3
\end{bmatrix}
\begin{bmatrix}
1&1&1&1\\
0&0&-1&0\\
0&-1&0&1\\
-1&0&0&-1
\end{bmatrix}
$$
$$
B_3=
\begin{bmatrix}
1&X&X^2&X^3
\end{bmatrix}
\begin{bmatrix}
1&0&0&0\\
0&1&0&0\\
0&0&1&0\\
0&0&0&1
\end{bmatrix}
$$
Note that the columns of the matrices above are the matrices for the various bases with respect to $B_3$.
For example, consider the first vector of $B_1$, $1-X^2+X^3$
$$
\Theta_{B_3}\!\left(1-X^2+X^3\right)=\begin{bmatrix}1\\0\\-1\\1\end{bmatrix}
$$
since
$$
1-X^2+X^3=
\overbrace{\begin{bmatrix}
1&X&X^2&X^3
\end{bmatrix}}^{B_3}
\begin{bmatrix}
1\\0\\-1\\1
\end{bmatrix}
$$
This is how the matrix for $B_1$ was created.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2051986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The Most general solution satisfying equations $\tan x=-1$ and $\cos x=1/\sqrt{2}$ The most general value of $x$ satisfying the equations $\tan x=-1$ and $\cos x=1/\sqrt{2}$, is found to be $x=2n\pi+\frac{7\pi}{4}$.
My approach:
$$
\frac{\sin x}{\cos x}=-1\implies \sqrt{2}\sin x=-1\implies\sin x=\frac{-1}{\sqrt{2}}=\sin (\frac{\pi}{4}+\frac{\pi}{2})=\sin\frac{3\pi}{4}\\\implies x=n\pi+(-1)^n\frac{3\pi}{4}
$$
If I consider the cosine function
$$
\cos x=\frac{\sin x}{\tan x}=-\sin x=-\sin\frac{3\pi}{4}=\cos(\frac{3\pi}{4}+\frac{\pi}{2})=\cos\frac{5\pi}{4} \implies x=2n\pi+\frac{5\pi}{4}
$$
Is their anything wrong with my approach ?How do I compare different forms of general solutions without inputting for $n$ ?
| Here is what you got wrong:
$$\frac{\sin x}{\cos x}=-1\implies \sqrt{2}\sin x=-1\implies\sin x=\frac{-1}{\sqrt{2}}$$
So you got $\sin x<0$ and $\cos x>0$. Then you are in the $4$th quadrant.
$$\sin x=\frac{-1}{\sqrt{2}}=\sin \left(\frac{7\pi}{4}\right) \Rightarrow x=\frac{7\pi}{4}+2n\pi$$
P.S.: You got wrong because $\frac{5\pi}{4}$ is in $3$th quadrant and $\frac{3\pi}{4}$ is in $2$th quadrant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2052250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
General solution to $(\sqrt{3}-1)\cos x+(\sqrt{3}+1)\sin x=2$ $(\sqrt{3}-1)\cos x+(\sqrt{3}+1)\sin x=2$ is said to have a general solution of $x=2n\pi\pm\frac{\pi}{4}+\frac{\pi}{12}$.
My Approach:
Considering the equation as
$$
a\cos x+b\sin x=\sqrt{a^2+b^2}\Big(\frac{a}{\sqrt{a^2+b^2}}\cos x+\frac{b}{\sqrt{a^2+b^2}}\sin x\Big)=\sqrt{a^2+b^2}\big(\sin y.\cos x+\cos y.\sin x\big)=\sqrt{a^2+b^2}.\sin(y+x)=2
$$
$\frac{a}{\sqrt{a^2+b^2}}=\sin y$ and $\frac{b}{\sqrt{a^2+b^2}}=\cos y$.
$$
{\sqrt{a^2+b^2}}=\sqrt{8}=2\sqrt{2}\\\tan y=a/b=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{\frac{\sqrt{3}}{2}.\frac{1}{\sqrt{2}}-\frac{1}{2}.\frac{1}{\sqrt{2}}}{\frac{\sqrt{3}}{2}.\frac{1}{\sqrt{2}}+\frac{1}{2}.\frac{1}{\sqrt{2}}}=\frac{\sin(\pi/3-\pi/4)}{\sin(\pi/3+\pi/4)}=\frac{\sin(\pi/3-\pi/4)}{\cos(\pi/3-\pi/4)}=\tan(\pi/3-\pi/4)\implies y=\pi/3-\pi/4=\pi/12
$$
Substituting for $y$,
$$
2\sqrt{2}.\sin(\frac{\pi}{12}+x)=2\implies \sin(\frac{\pi}{12}+x)=\frac{1}{\sqrt{2}}=\sin{\frac{\pi}{4}}\\\implies \frac{\pi}{12}+x=n\pi+(-1)^n\frac{\pi}{4}\implies x=n\pi+(-1)^n\frac{\pi}{4}-\frac{\pi}{12}
$$
What's going wrong with the approach ?
| converting the given equation in $$\tan$$ we get
$${\frac { \left( 1+\sqrt {3} \right) \left( 1- \left( \tan \left( x/2
\right) \right) ^{2} \right) }{1+ \left( \tan \left( x/2 \right)
\right) ^{2}}}+2\,{\frac { \left( \sqrt {3}-1 \right) \tan \left( x/2
\right) }{1+ \left( \tan \left( x/2 \right) \right) ^{2}}}-2=0
$$
simplifying and factorizing we obtain
$$-1/3\,{\frac { \left( \sqrt {3}+3 \right) \left( \tan \left( x/2
\right) +2-\sqrt {3} \right) \left( 3\,\tan \left( x/2 \right) -
\sqrt {3} \right) }{1+ \left( \tan \left( x/2 \right) \right) ^{2}}}
=0$$
thus you have to solve
$$- \left( \sqrt {3}+3 \right) \left( -\tan \left( x/2 \right) -2+
\sqrt {3} \right) \left( -3\,\tan \left( x/2 \right) +\sqrt {3}
\right)
=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2053720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Find all integer solutions $(x,y)$ such that $2x^2 + y^2 = 3x^2y$
Find all integer solutions $(x,y)$ such that $2x^2 + y^2 = 3x^2y$.
We can rearrange the given equation to $$y^2 = x^2(3y-2)\tag1$$ Thus $3y-2$ must be a perfect square and so $3y-2 = k^2$.
How can we continue?
| You have that $3y-2$ divides $y^2$.
Notice that $(3y-2,y^2)| (3y-2,y)^2$.
On the other hand, if $d$ divides $3y-2$ and $y$ we have $d|2$. So $(3y-2,y^2)| 4$.
Therefore $3y-2|4$
Therefore $y=0,1$ or $2$.
If $y=0$ we have $x=0$.
If $y=1$ we have $x=\pm 1$
If $y=2$ we have $x=\pm 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2056506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
How to factorize $a^2-b^2-a+b+(a+b-1)^2$? The answer is $(a+b-1)(2a-1)$ but I have no idea how to get this answer.
| $$(a+b-1)^2=a^2+b^2+1-2a-2b+2ab$$
$$\implies a^2-b^2-a+b+(a+b-1)^2=2a(a+b)-\{2a+(a+b)\}+1$$
$$ =2a\{a+b-1\}-\{a+b-1\}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2063343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
How to integrate $\int_a^b (x-a)(x-b)\,dx=-\frac{1}{6}(b-a)^3$ in a faster way? $\displaystyle \int_a^b (x-a)(x-b)\,dx=-\frac{1}{6}(b-a)^3$
$\displaystyle \int_a^{(a+b)/2} (x-a)(x-\frac{a+b}{2})(x-b)\, dx=\frac{1}{64}(b-a)^4$
Instead of expanding the integrand, or doing integration by part, is there any faster way to compute this kind of integral?
| You can do this with substitution and Cavalieri's formula:
$$
\int_0^1 u^n \,du = \frac{1}{n+1}
$$
For the first one, let $u= \frac{x-a}{b-a}$. Then $x = (b-a)u +a$, which means $x-a = (b-a)u$ and $x-b = (b-a)(u-1)$. Also $dx=(b-a)\,du$. So
\begin{align*}
\int_a^b (x-a)(x-b)\,dx
&= (b-a)^3 \int_0^1 u (u-1)\,du
= (b-a)^3 \int_0^1 (u^2 - u) \\
&= (b-a)^3 \left(\frac{1}{3}-\frac{1}{2}\right)
= -\frac{1}{6}(b-a)^3
\end{align*}
For the second one, let $u= \frac{2}{a-b}\left(x-\frac{a+b}{2}\right)$. Then:
\begin{align*}
x-a &= \frac{a-b}{2}(u-1) \\
x-b &= \frac{a-b}{2}(u+1) \\
x - \frac{a+b}{2} &= \frac{a-b}{2}u \\
dx &= \frac{a-b}{2}\,du
\end{align*}
So
\begin{align*}
\int_a^{(a+b)/2} (x-a)\left(x-\frac{a+b}{2}\right)(x-b)\,dx
&= \left(\frac{a-b}{2}\right)^4\int_1^0(u-1)u(u+1)\,du
= -\frac{(b-a)^4}{16}\int_0^1 (u^3-u)\,du \\
&= -\frac{(b-a)^4}{16}\left(\frac{1}{4}-\frac{1}{2}\right)
= -\frac{1}{64}(b-a)^4
\end{align*}
This has less integration, but it took some time to get the substitution right.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2065639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 4
} |
How to see there exists const $C$ such that $\frac{d^{n-m}}{dx^{n-m}}(1-x^2)^n=C(1-x^2)^m\frac{d^{n+m}}{dx^{n+m}}(1-x^2)^n$ This comes up in relation to Legendre functions. The claim is made that for $n =0,1,2,3,\cdots$ and $m=0,1,2,3,\cdots,n$, there is a constant $C_{n,m}$ such that
$$
\frac{d^{n-m}}{dx^{n-m}}(1-x^2)^n=C_{n,m}(1-x^2)^{m}\frac{d^{n+m}}{dx^{n+m}}(1-x^2)^n
$$
A typical suggested proof is writing out all of the polynomial coefficients. Does anyone have a clever way to see this must be true? I've tried many things with differentiation, but I can't find any clean way to see this.
Example: Consider $n=3,m=1$ where $(1-x^2)^n=1-3x^2+3x^4-x^6$, and
\begin{align}
\frac{d^{3-1}}{dx^{3-1}}(1-x^2)^3&=-6+36x^2-30x^4 \\
(1-x^2)^1\frac{d^4}{dx^4}(1-x^2)^3&=(1-x^2)(72-360x^2) \\
&=72-72x^2-360x^2+360x^4 \\
&=72-432x^2+360x^4 \\
&=(-12)(-6+36x^2-30x^4)
\end{align}
| Suppose we seek to determine the constant $Q$ in the equality
$$ Q_{n,m} \left(\frac{d}{dz}\right)^{n-m} (1-z^2)^n
= (1-z^2)^m \left(\frac{d}{dz}\right)^{n+m}
(1-z^2)^n$$
where $n\ge m.$ We will compute the coefficients on $[z^q]$ on the LHS
and the RHS. Writing $1-z^2 = (1+z)(1-z)$ we get for the LHS
$$\sum_{p=0}^{n-m} {n-m\choose p}
{n\choose p} p! (1+z)^{n-p}
\\ \times {n\choose n-m-p} (n-m-p)! (-1)^{n-m-p} (1-z)^{m+p}
\\ = (n-m)! (-1)^{n-m}
\sum_{p=0}^{n-m} {n\choose p} {n\choose n-m-p}
(1+z)^{n-p} (-1)^p (1-z)^{m+p}.$$
Extracting the coefficient we get
$$(n-m)! (-1)^{n-m}
\sum_{p=0}^{n-m} {n\choose p} {n\choose n-m-p} (-1)^p
\\ \times \sum_{k=0}^{n-p} {n-p\choose k}
(-1)^{q-k} {m+p\choose q-k}.$$
We use the same procedure on the RHS and merge in the $(1-z^2)^m$ term
to get
$$(n+m)! (-1)^{n+m}
\sum_{p=0}^{n+m} {n\choose p} {n\choose n+m-p} (-1)^p
\\ \times \sum_{k=0}^{n+m-p} {n+m-p\choose k}
(-1)^{q-k} {p\choose q-k}.$$
Working in parallel with LHS and RHS we treat the inner sum of the LHS
first, putting
$${m+p\choose q-k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{q-k+1}} (1+z)^{m+p}
\; dz$$
to get
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{q+1}} (1+z)^{m+p}
\sum_{k=0}^{n-p} {n-p\choose k} (-1)^{q-k} z^k \; dz
\\ = \frac{(-1)^q}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{q+1}} (1+z)^{m+p} (1-z)^{n-p}
\; dz.$$
Adapt and repeat to obtain for the inner sum of the RHS
$$\frac{(-1)^q}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{q+1}} (1+z)^{p} (1-z)^{n+m-p}
\; dz.$$
Moving on to the two outer sums we introduce
$${n\choose n-m-p} =
\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{n-m-p+1}} (1+w)^n
\; dw$$
to obtain for the LHS
$$\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{n-m+1}} (1+w)^n
\\ \times \frac{(-1)^q}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{q+1}} (1+z)^{m} (1-z)^{n}
\sum_{p=0}^{n-m} {n\choose p} (-1)^p w^p \frac{(1+z)^p}{(1-z)^p}
\; dz\; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{n-m+1}} (1+w)^n
\\ \times \frac{(-1)^q}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{q+1}} (1+z)^{m} (1-z)^{n}
\left(1-w\frac{1+z}{1-z}\right)^n
\; dz\; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{n-m+1}} (1+w)^n
\\ \times \frac{(-1)^q}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{q+1}} (1+z)^{m}
(1-z-w-wz)^n
\; dz\; dw.$$
Repeat for the RHS to get
$$\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{n+m+1}} (1+w)^n
\\ \times \frac{(-1)^q}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{q+1}} (1-z)^{m}
(1-z-w-wz)^n
\; dz\; dw.$$
Extracting coefficients from the first integral (LHS) we write
$$(1-z-w-wz)^n = (2-(1+z)(1+w))^n
\\ = \sum_{k=0}^n {n\choose k} (-1)^k (1+z)^k (1+w)^k 2^{n-k}$$
and the inner integral yields
$$(-1)^q \sum_{k=0}^n {n\choose k} (-1)^k {m+k\choose q}
(1+w)^k 2^{n-k}$$
followed by the outer one which gives
$$(-1)^q \sum_{k=0}^n {n\choose k} (-1)^k {m+k\choose q}
{n+k\choose n-m} 2^{n-k}.$$
For the second integral (RHS) we write
$$(1-z-w-wz)^n = ((1-z)(1+w)-2w)^n
\\ = \sum_{k=0}^n {n\choose k} (1-z)^k (1+w)^k
(-1)^{n-k} 2^{n-k} w^{n-k}$$
and the inner integral yields
$$(-1)^q \sum_{k=0}^n {n\choose k} {m+k\choose q} (-1)^q
(1+w)^k (-1)^{n-k} 2^{n-k} w^{n-k}$$
followed by the outer one which produces
$$\sum_{k=0}^n {n\choose k} {m+k\choose q}
{n+k\choose k+m} (-1)^{n-k} 2^{n-k}.$$
The two sums are equal up to a sign and the RHS for the coefficient on
$[z^q]$ is obtained from the LHS by multiplying by
$$\frac{(n+m)!}{(n-m)!} (-1)^{n-q}.$$
Observe that powers of $z$ that are present in the LHS and the RHS
always have the same parity, the coefficients being zero otherwise
(either all even powers or all odd). Therefore $(-1)^{n-q}$ is in fact
a constant not dependent on $q$, the question is which. The leading
term has degree $2n-(n-m)=n+m=(2n-(n+m))+2m$ on both sides and the
sign on the LHS is $(-1)^n$ and on the RHS it is $(-1)^{n+m}.$ The
conclusion is that the queried factor is given by
$$\bbox[5px,border:2px solid #00A000]{
Q_{n,m} = (-1)^m \frac{(n+m)!}{(n-m)!}.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2066340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Help verify the proof that $\min \left({\lfloor{N/2}\rfloor, \lfloor{(N + 2)/p}\rfloor}\right) = \lfloor{(N + 2)/p}\rfloor$ for $N \ge p$ Well, I think that I have this correct, however my proof seams cumbersome, so can someone check these results or suggest a simplified proof.
Prove that ($N \ge p$)
\begin{equation*}
\min \left({\left\lfloor{\frac{N}{2}}\right\rfloor, \left\lfloor{\frac{N + 2}{p}}\right\rfloor}\right) = \left\lfloor{\frac{N + 2}{p}}\right\rfloor
\end{equation*}
for prime $p \ge 3$ with $N \ge p$ and $N \in \mathbb{Z}^{+}$. To see this, let $N = k\, p + m$ where $k \in \mathbb{Z}^{+}$ and $m \in \left\{{0}\right.$, $1$, $\cdots$, $\left.{p - 1}\right\}$ by the Quotient Remainder Theorem, then show that this equation holds for all values of $k$. Starting with
\begin{equation*}
\left\lfloor{\frac{N + 2}{p}}\right\rfloor = \left\lfloor{\frac{k\, p + m + 2}{p}}\right\rfloor = k + {\delta}_{2}
\end{equation*}
where
\begin{equation*}
{\delta}_{2} =
\left\lfloor{\frac{m + 2}{p}}\right\rfloor =
\begin{cases}
0, & \text{for } m \le p - 3, \\
1, & \text{for } m \in \left\{{p - 2, p - 1}\right\}.
\end{cases}
\end{equation*}
Also
\begin{equation*}
\left\lfloor{\frac{N}{2}}\right\rfloor =
\left\lfloor{\frac{k\, p + m}{2}}\right\rfloor =
\left\lfloor{\frac{k\, p}{2} + \frac{m}{2}}\right\rfloor.
\end{equation*}
Therefore
\begin{equation*}
k + {\delta}_{2} \le \left\lfloor{\frac{k\, p}{2} + \frac{m}{2}}\right\rfloor.
\end{equation*}
If $k$ is even then we have
\begin{equation*}
k + {\delta}_{2} \le \frac{k\, p}{2} + \left\lfloor{\frac{m}{2}}\right\rfloor.
\end{equation*}
Solving for $k$ gives
\begin{equation*}
k \ge \frac{2 \left({{\delta}_{2} - \left\lfloor{m/2}\right\rfloor}\right)}{p - 2}.
\end{equation*}
If ${\delta}_{2} = 0$ or ${\delta}_{2} = 1$ then $k \ge 2$ since $k$ is even and ${\delta}_{2} - \left\lfloor{m/2}\right\rfloor \le 1$. Now if $k$ is odd then we have three cases to consider starting with
\begin{equation*}
k + {\delta}_{2} \le \frac{\left({k + 1}\right) p}{2} + \left\lfloor{\frac{m - p}{2}}\right\rfloor.
\end{equation*}
Solving for $k$ gives
\begin{equation*}
k \ge \frac{2 \left({{\delta}_{2} - p/2 - \left\lfloor{\left({m - p}\right)/2}\right\rfloor}\right)}{p - 2}.
\end{equation*}
With $m = 0$, then ${\delta}_{2} = 0$ and
\begin{equation*}
k \ge \frac{2}{p - 2} \left({- \frac{p}{2} + \left\lceil{\frac{p}{2}}\right\rceil}\right) =
\frac{1}{p - 2} \ge 1
\end{equation*}
since $\left\lceil{p/2}\right\rceil = \left({p - 1}\right)/2 + \left\lceil{1/2}\right\rceil = \left({p + 1}\right)/2$ with $p \ge 3$. For the last two cases $m = p - 2$ or $m = p - 1$, $\delta = 1$, then
\begin{equation*}
k \ge \frac{4 - p}{p - 2} \ge 1
\end{equation*}
for $p \ge 3$. Thus the primary equation holds for $p \ge 3$.
| Show that
\begin{equation*}
\left\lfloor{\frac{N}{2}}\right\rfloor < \left\lfloor{\frac{N + 2}{p}}\right\rfloor
\end{equation*}
is not valid. Let $N = k\, p + m$ where $k \in \mathbb{Z}^{+}$ and $m \in \left\{{0}\right.$, $1$, $\cdots$, $\left.{p - 1}\right\}$ by the Quotient Remainder Theorem. Then show this holds for all values of $k$ and $m$. Starting with the left-hand side
\begin{equation*}
\left\lfloor{\frac{N}{2}}\right\rfloor =
\left\lfloor{\frac{p\, k + m}{2}}\right\rfloor =
\left\lfloor{\frac{2\, k + \left({p - 2}\right) k + m}{2}}\right\rfloor =
k + \left\lfloor{\frac{\left({p - 2}\right) k + m}{2}}\right\rfloor
\end{equation*}
then the right-hand side
\begin{equation*}
\left\lfloor{\frac{N + 2}{p}}\right\rfloor =
\left\lfloor{\frac{p\, k + m + 2}{p}}\right\rfloor =
k + \left\lfloor{\frac{m + 2}{p}}\right\rfloor
\end{equation*}
then
\begin{equation*}
\left\lfloor{\frac{\left({p - 2}\right) k + m}{2}}\right\rfloor < \left\lfloor{\frac{m + 2}{p}}\right\rfloor =
{\delta}_{2} =
\begin{cases}
0, & \text{for } m \le p - a - 1, \\
1, & \text{for } m \in \left\{{p - a, \cdots, p - 2, p - 1}\right\}.
\end{cases}
\end{equation*}
Then
\begin{equation*}
\min \left({\left\lfloor{\frac{N}{2}}\right\rfloor, \left\lfloor{\frac{N + 2}{p}}\right\rfloor}\right) = \left\lfloor{\frac{N + 2}{p}}\right\rfloor
\end{equation*}
for $p \ge 3$. To see this, establish a contradiction. Then
\begin{equation*}
\left\lfloor{\frac{\left({p - 2}\right) k + m}{2}}\right\rfloor < {\delta}_{2}.
\end{equation*}
Now with $m \in \left\{{0}\right.$, $1$, $\cdots$, $\left.{p - 3}\right\}$, then ${\delta}_{2} = 0$ and $\left({p - 2}\right) k + m \ge 1$ with $\left\lfloor{\left[{\left({p - 2}\right) k + m}\right]/2}\right\rfloor \ge 0$ which leads to $0 < 0$ which is a contradiction. Next $m = p - 2$ and ${\delta}_{2} = 1$ then
\begin{equation*}
\left\lfloor{\frac{\left({p - 2}\right) \left({k + 1}\right)}{2}}\right\rfloor < 1
\end{equation*}
with $\left({p - 2}\right) \left({k + 1}\right) \ge 2$ leads to $1 < 1$ which is a contradiction. Next $m = p - 1$ and ${\delta}_{2} = 1$ then
\begin{equation*}
\left\lfloor{\frac{\left({p - 2}\right) k + p - 1}{2}}\right\rfloor < 1
\end{equation*}
with $\left({p - 2}\right) k + p - 1 \ge 3$ leads to $1 < 1$ which is a contradiction. Therefore there are no cases.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2066562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Common proof for $(1+x)(1+x^2)(1+x^4)...(1+x^{2^n})=\dfrac{1-x^{2^{n+1}}}{1-x} $ I'm asking for an alternative (more common?) proof of the following equality, more specifically an alternative proof for the inductive step:
$$(1+x)(1+x^2)(1+x^4)...(1+x^{2^n})=\dfrac{1-x^{2^{n+1}}}{1-x} (x\neq 1)$$
This is how I proved it:
Basecase: substitute $1$ for $n$, everything works out.
Inductive step: assume that $$\prod _{i=1}^{n}(1+x^i)=\dfrac{1-x^{2^{n+1}}}{1-x}$$
then $$\prod _{i=1}^{n+1}(1+x^i)=\dfrac{1-x^{2^{n+1}}}{1-x} (1+x^{n+1})$$
Let $a$, $b$ and $c$ be positive real numbers, then $\dfrac {c}{a}=b\Leftrightarrow ab=c$, thus
$$\dfrac {\left( \dfrac{1-x^{2^{n+2}}}{1-x}\right)}{\left( \dfrac{1-x^{2^{n+1}}}{1-x}\right)}=1-x^{2^{n+1}} \Leftrightarrow \dfrac{1-x^{2^{n+1}}}{1-x} (1+x^{n+1}) = \dfrac{1-x^{2^{n+2}}}{1-x}$$
$$\dfrac {\left( \dfrac{1-x^{2^{n+2}}}{1-x}\right)}{\left( \dfrac{1-x^{2^{n+1}}}{1-x}\right)}=\dfrac {1-x^{2^{n+2}}}{1-x^{2^{n+1}}}$$
Applying polynomial division, we see that indeed $\dfrac {1-x^{2^{n+2}}}{1-x^{2^{n+1}}} = 1+x^{2^{n+1}}$.
Thus $\dfrac{1-x^{2^{n+1}}}{1-x} (1+x^{n+1}) = \dfrac{1-x^{2^{n+2}}}{1-x}$.
However, the exercise was in a chapter on binomial coefficients and pascal's triangle, furthermore we didn't mention polynomial division in class. Which makes me think that there was another solution that I was "supposed" to see. How was I supposed to prove it?
| Expanding $(1+x)(1+x^2)(1+x^4)\cdots(1+x^{2^n})$ gives $1+x+x^2+\cdots +x^{2^{n+1}-1}$ because every integer $k$ with $0 \le k \le 2^{n+1}-1$ has a unique binary representation (as a sum of powers of $2$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2067075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
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