Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Matrices of different dimensions in a matrix equation Given two matrices A, B, respectively, and an equation XA=B; how can we obtain X and discuss the system?
\begin{pmatrix}
1 & a\\
a & 1 \\
\end{pmatrix} \begin{pmatrix}
a & 1 \\
a^2 & 0 \\
1 & 1 \\
\end{pmatrix}
I am not familiar with this structure and I can't even picture what the dimensions of X are. Any general matrix I write down looks uncomplete.
| $X$ must be a $3 \times 2$ matrix; we may write
$X = \begin{pmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \\ x_{31} & x_{32} \end{pmatrix}; \tag 1$
we may thus write out the system
$XA = B \tag 2$
as
$\begin{pmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \\ x_{31} & x_{32} \end{pmatrix} \begin{pmatrix} 1 & a \\ a & 1 \end{pmatrix} = \begin{pmatrix} a & 1 \\ a^2 & 0 \\ 1 & 1 \end{pmatrix}; \tag 3$
here
$A = \begin{pmatrix} 1 & a \\ a & 1 \end{pmatrix}, \tag 4$
and
$B = \begin{pmatrix} a & 1 \\ a^2 & 0 \\ 1 & 1 \end{pmatrix}; \tag 5$
we may write out the system (3) as a set of $6$ linear equations in two variables each:
$x_{11} + ax_{12} = a, \tag 6$
$ax_{11} + x_{12} = 1; \tag 7$
$x_{21} + ax_{22} = a^2, \tag 8$
$ax_{21} + x_{22} = 0, \tag 9$
$x_{31} + ax_{32} = 1, \tag{10}$
$ax_{31} + x_{32} = 1; \tag{11}$
the system of equations (6)-(11) is in fact easy to solve, one pair $(x_{i1}, x_{i2})$ at a time; for example, we multiply (6) by $a$:
$ax_{11} + a^2x_{12} = a^2, \tag{12}$
and subtract (7) from (12):
$(a^2 - 1) x_{12} = a^2 - 1, \tag{13}$
whence
$x_{12} = 1, \tag{14}$
provided of course that
$a \ne \pm 1; \tag{15}$
having $x_{12}$ we find via (6) that
$x_{11} = a(1 - x_{12}) = 0; \tag{16}$
the remaining equations (8)-(11) may be similarly solved.
The system (2), (3) may also be addressed in a somewhat more matrix-oriented manner by setting
$Y = \begin{pmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{pmatrix}, \; \mathbf y = (x_{31}, x_{32}), \; C = \begin{pmatrix} a & 1 \\ a^2 & 0 \end{pmatrix}; \mathbf d = (1, 1); \tag {17}$
then (3) may be written in the from
$\begin{pmatrix} Y \\ \mathbf y \end{pmatrix} A = \begin{pmatrix} C \\ \mathbf d \end{pmatrix}, \tag{18}$
leading to
$YA = C, \; \mathbf y A = \mathbf d, \tag{19}$
whence
$Y = CA^{-1}, \; \mathbf y = \mathbf d A^{-1}; \tag{20}$
where it is easily seen that
$A^{-1} = \dfrac{1}{1 - a^2} \begin{pmatrix} 1 & -a \\ -a & 1 \end{pmatrix}. \tag{21}$
We assume $a = \pm 1$ in deriving (17)-(21); in the event that $a = \pm 1$ we see that (8)-(9) become
$x_{21} \pm x_{22} = 1, \; \pm x_{21} + x_{22} = 0, \tag{22}$
which lead to the contradicion
$\pm 1 = 0; \tag{23}$
thus we see there is no solution to (2) when $a = \pm 1$.
| {
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"url": "https://math.stackexchange.com/questions/3054304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a,b\in R$ are distinct numbers satisfying $|a-1|+|b-1|=|a|+|b|=|a+1|+|b+1|$,then find the minimum value of $|a-b|$ If $a,b\in R$ are distinct numbers satisfying $|a-1|+|b-1|=|a|+|b|=|a+1|+|b+1|$,then find the minimum value of $|a-b|$
I squared both sides of $|a-1|+|b-1|=|a|+|b|$ and $|a|+|b|=|a+1|+|b+1|$ and equated them but the equation obtained is messy and i am stuck.Please help.
| Consider the $16$ cases:
$\hspace{3cm}$
$$\begin{align}1) \ &a,b\ge 1 \Rightarrow \\
& \ \ \ \ \ a-1+b-1=a+b=a+1+b+1 \Rightarrow \emptyset;\\
2) \ &0\le a\le 1, b\ge 1 \Rightarrow \\
& \ \ \ \ \ -a+1+b-1=a+b=a+1+b+1 \Rightarrow \emptyset;\\
3) \ &-1\le a\le 0, b\ge 1 \Rightarrow \\
& \ \ \ \ \ -a+1+b-1=-a+b=a+1+b+1 \Rightarrow \\
& \ \ \ \ \ a=-1 \Rightarrow b=1 \Rightarrow (a,b)=(-1,1) \Rightarrow \\
& \ \ \ \ \ |a-b|=2 \ \text{(min)};\\
4, 7, 8) \ &(a,b)=(-1,1);\\
5, 6, 11, 12,15,16) \ &\emptyset;\\
9,10,13,14) \ &(a,b)=(1,-1). \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3054750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
Is the converse of this implication true? If we have,
\begin{align}
|x-2|\leq {1} \iff & -1\leq{x-2}\leq1\\ \iff & 4\leq {x+3}\leq6 \\ \iff & 5\leq {x+4}\leq7.\end{align}
Then In particular,
$ |x-2|\leq {1} \implies 4\leq {|x+3|}\leq6$ and $5\leq {|x+4|}\leq7.$
Why do we have the modulus of $x+3$ and $x+4$? Is it because, the estimates are positive since the bounds are positive, so it is equivalent to stating them with modulus? And is the converse below true?
$4\leq {|x+3|}\leq6$ and $5\leq {|x+4|}\leq7 \implies |x-2|\leq {1}$.
This is a question in my textbook. I feel like the implication can be reversed.
|
Why do we have the modulus of $x+3$ and $x+4$? Is it because, the estimates are positive since the bounds are positive, so it is equivalent to stating them with modulus? And is the converse below true?
Note that:
$$\begin{align}4\le x+3\le 6 &\Rightarrow 4\le |x+3|\le 6, \ \ \text{but} \\
4\le |x+3|\le 6 &\not\Rightarrow 4\le x+3\le 6, \end{align}$$
because $4\le x+3\le 6$ has a solution $x\in [1,3]$:
whereas $4\le|x+3|\le 6$ has a solution $x\in [-9,-7]\cup [1,3]$:
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Ways to prove that $\int_0^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x = 0$. I have managed to solve it in one way, but I became very interested in this failed attempt.
$$
\int_0^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x
= \int_0^\infty \frac{\ln(x)}{(1+x^2)^2} {\rm d}x
- \int_0^\infty \frac{\ln(x)}{(1+x^2)^3} {\rm d}x
$$
We only have to show that those two on the right are equal. And numerical evaluations seem to suggest that they both are in fact $-\frac{\pi}{4}$ but I don't know how to break these down.
I am currently really interested in proving this
$$ \int_0^\infty \frac{\ln(x)}{(1+x^2)^2} {\rm d}x = \int_0^\infty \frac{\ln(x)}{(1+x^2)^3} {\rm d}x = -\frac{\pi}{4} $$
Anyway, here's my trivial solution using $u = \frac1x$:
$$
\begin{align}
\int_0^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x & = \int_0^1 \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x + \int_1^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x \\
& = \int_\infty^1 \frac{\frac{1}{u^2} \ln(\frac1u)}{(1+\frac{1}{u^2})^3} \frac{-1}{u^2} {\rm d}u
+ \int_1^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x \\
& = -\int_1^\infty \frac{\ln(u)}{u(u+\frac{1}{u})^3} {\rm d}u
+ \int_1^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x \\
& = - \int_1^\infty \frac{u^2 \ln(u)}{(1+u^2)^3} {\rm d}u
+ \int_1^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x \\
& = 0
\end{align}
$$
I'm sure there are many more interesting methods for cracking this integral, since it's so closely related to the popular $\int_0^\infty \frac{\ln(x)}{1+x^2} {\rm d}x = 0$. Please share them if you do come up with any.
| $$\int_0^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x=
\int_0^1 \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x+
\int_1^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x
$$
Then change in first integral $x=\frac{1}{t}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Two sides of a triangle are $\sqrt{3}+1$ and $\sqrt{3}-1$ and the included angle is $60^{\circ}$. Find other angles
Two sides of a triangle are $\sqrt{3}+1$ and $\sqrt{3}-1$ and the included angle is $60^{\circ}$. Then find the other angles
My Attempt
Let $a=\sqrt{3}+1$, $b=\sqrt{3}-1$ and $C=60$
$$
c^2=a^2+b^2-2.a.b\cos C\\=(\sqrt{3}+1)^2
+(\sqrt{3}-1)^2-2(\sqrt{3}+1)(\sqrt{3}-1).\frac{1}{2}
=8-2=6\\
\implies c=\sqrt{6}=\sqrt{2}\sqrt{3}\\
\frac{a}{\sin A}=\frac{c}{\sin C}\implies\sin A=\frac{\sqrt{3}+1}{\sqrt{2}\sqrt{3}}\frac{\sqrt{3}}{2}=\frac{\sqrt{3}+1}{2\sqrt{2}}\\
A=75^\circ\quad\&\quad B=45^\circ
$$
But the solution given in my reference is $105^\circ$ and $15^\circ$, what is going wrong with my attempt ?
| By law of cosines we obtain:
$$\cos\alpha=\frac{(\sqrt3+1)^2+(\sqrt6)^2-(\sqrt3-1)^2}{2(\sqrt3+1)\sqrt6}=\frac{\sqrt3+1}{2\sqrt2},$$ which gives $\alpha=15^{\circ}$ and from here $\beta=105^{\circ}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3056699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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A Series for $\pi$
Question: Can we show that $$\sum_{n=0}^\infty(-1)^{n+1}\frac{(2n-3)!!}{(2n+3)!!}=\frac{\pi}{8} $$ ?
According to wolfram alpha this result is true.
Just amateur curiosity, not sure of a starting point to show if this true.
| Considering the simplicity of Micah's answer, I am ashamed to provide this complex one.
Let
$$S_p(x)=\sum_{n=0}^p(-1)^{n+1}\frac{(2n-3)!!}{(2n+3)!!}x^p$$ It write
$$S_p(x)=\frac{\left(8 p^3+36 p^2+46 p+15\right) \,
_2F_1\left(-\frac{1}{2},1;\frac{5}{2};-x\right)+3 (-x)^{p+1} \,
_2F_1\left(1,p+\frac{1}{2};p+\frac{7}{2};-x\right)}{3 \left(8 p^3+36 p^2+46
p+15\right)}$$ Using
$$\,
_2F_1\left(-\frac{1}{2},1;\frac{5}{2};-1\right)=\frac{3 \pi }{8}$$ then
$$S_p(1)=\frac{\pi \left(8 p^3+36 p^2+46 p+15\right)-8 (-1)^p \,
_2F_1\left(1,p+\frac{1}{2};p+\frac{7}{2};-1\right)}{8 \left(8 p^3+36 p^2+46
p+15\right)}$$ that it to say
$$S_p(1)=\frac \pi 8-\frac{(-1)^p }{8 p^3+36
p^2+46 p+15}\, _2F_1\left(1,p+\frac{1}{2};p+\frac{7}{2};-1\right)$$ and $\, _2F_1\left(1,p+\frac{1}{2};p+\frac{7}{2};-1\right)$ looks like an hyperbolic function going asymptotically to $\frac 12$.
Then $S_\infty(1)=\frac{ \pi }{8}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3057278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Calculate $\lim\limits_{ x\to + \infty}x\cdot \sin(\sqrt{x^{2}+3}-\sqrt{x^{2}+2})$ I know that $$\lim\limits_{ x\to + \infty}x\cdot \sin(\sqrt{x^{2}+3}-\sqrt{x^{2}+2})\\=\lim\limits_{ x\to + \infty}x\cdot \sin\left(\frac{1}{\sqrt{x^{2}+3}+\sqrt{x^{2}+2}}\right).$$ If $x \rightarrow + \infty$, then $\sin\left(\frac{1}{\sqrt{x^{2}+3}+\sqrt{x^{2}+2}}\right)\rightarrow \sin0 $. However I have also $x$ before $\sin x$ and I don't know how to calculate it.
| \begin{align}
\lim_{x \to \infty} x \cdot \left( \sin \left( \sqrt{x^2+3}-\sqrt{x^2+2}\right)\right)&=\lim_{x \to \infty} x \cdot \sin \left( \frac{1}{\sqrt{x^2+3}+\sqrt{x^2+2}}\right)\\
&=\lim_{x \to \infty} \frac{x}{\sqrt{x^2+3}+\sqrt{x^2+2}}\\
&=\lim_{x \to \infty} \frac{1}{\sqrt{1+\frac{3}{x^2}}+\sqrt{1+\frac{2}{x^2}}}\\
&= \frac12
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3057710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate $\sum_{r=2}^{\infty} \frac{2-r}{r(r+1)(r+2)}$
Evaluate
$$\sum_{r=2}^{\infty} \frac{2-r}{r(r+1)(r+2)}$$
So in a previous part of the question I calculated that
$$\sum_{r=1}^{n} \frac{2-r}{r(r+1)(r+2)} = \sum_{r=1}^{n}\left( \frac{1}{r}-\frac{3}{r+1}+\frac{2}{r+2}\right)=\frac{n}{(n+1)(n+2)}$$
So my question is then why does $$\sum_{r=2}^{\infty} \frac{2-r}{r(r+1)(r+2)} = -\frac{1}{6}$$
As I though that it would be $\frac{1}{2}$.
| Note that \begin{align}\sum_{r=2}^\infty\left( \frac{1}{r}-\frac{3}{r+1}+\frac{2}{r+2}\right)&=\sum_{r=2}^\infty\left(\frac1r-\frac1{r+1}\right)-2\sum_{r=2}^\infty\left(\frac1{r+1}-\frac1{r+2}\right)\\&=\left(\frac12-\frac13+\frac13-\frac14+\cdots\right)-2\left(\frac13+\frac14-\frac14+\frac15-\cdots\right)\\&=\frac12-\frac23=-\frac16\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Determining the missing digits of $15! \equiv 1\square0767436\square000$ without actually calculating the factorial
$$15! \equiv 1\cdot 2\cdot 3\cdot\,\cdots\,\cdot 15 \equiv 1\square0767436\square000$$
Using a calculator, I know that the answer is $3$ and $8$, but I know that the answer can be calculated by hand.
How to calculate the missing digits? I know that large factorials can be estimated using Stirling's approximation:
$$15! \approx \sqrt{2\pi\cdot 15}
\cdot \left(\frac{15}{e}\right)^{15}$$
which is not feasible to calculate by hand.
The resulting number must be divisible by 9 which means that the digit sum must add up to 9 and is also divisible by 11 which means that the alternating digit sum must be divisible by 11:
$1+ d_0 + 0 + 7 +6 +7 +4 +3+6+d_1+0+0+0 \mod \phantom{1}9 \equiv \,34 + d_0 + d_1 \mod \phantom{1}9 \equiv 0 $
$-1+ d_0 - 0 + 7 -6 +7 -4 +3-6+d_1-0+0-0 \mod 11 \equiv d_0 + d_1 \mod 11 \equiv 0 $
The digits $3$ and $8$, or $7$ and $4$, fulfill both of the requirements.
| Using the divisibility rule for 7 the answer boils down to 3 and 8:
$-368+674+307+1 \mod 7 \equiv 0$
| {
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"timestamp": "2023-03-29T00:00:00",
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Least Squares solution for a symmetric singular matrix I want to solve this system by Least Squares method:$$\begin{pmatrix}1 & 2 & 3\\\ 2 & 3 & 4 \\\ 3 & 4 & 5 \end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} =\begin{pmatrix}1\\5\\-2\end{pmatrix} $$ This symmetric matrix is singular with one eigenvalue $\lambda1 = 0$, so $\ A^t\cdot A$ is also singular and for this reason I cannot use the normal equation: $\hat x = (A^t\cdot A)^{-1}\cdot A^t\cdot b $.
So I performed Gauss-Jordan to the extended matrix to come with $$\begin{pmatrix}1 & 2 & 3\\\ 0 & 1 & 2 \\\ 0 & 0 & 0 \end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix} =\begin{pmatrix}1\\3\\-1\end{pmatrix} $$
Finally I solved the $\ 2x2$ system: $$\begin{pmatrix}1 & 2\\\ 0 & 1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} =\begin{pmatrix}1\\3\end{pmatrix} $$ taking into account that the best $\ \hat b\ $ is $\begin{pmatrix}1\\3\\0\end{pmatrix}$
The solution is then $\ \hat x = \begin{pmatrix}-5\\3\\0\end{pmatrix}$
Is this approach correct ?
EDIT
Based on the book 'Lianear Algebra and its applications' from David Lay, I also include the Least Squares method he proposes: $(A^tA)\hat x=A^t b $
$$A^t b =\begin{pmatrix}5\\9\\13\end{pmatrix}, A^tA = \begin{pmatrix}14 & 20 & 26 \\
20 & 29 & 38 \\
26 & 38 & 50\end{pmatrix}$$
The reduced echelon from the augmented is:
$$ \begin{pmatrix}14 & 20 & 26 & 5 \\
20 & 29 & 38 & 9 \\
26 & 38 & 50 & 13 \end{pmatrix} \sim \begin{pmatrix}1 & 0 & -1 & -\frac{35}{6} \\
0 & 1 & 2 & \frac{13}{3} \\
0 & 0 & 0 & 0
\end{pmatrix} \Rightarrow \hat x = \begin{pmatrix}-\frac{35}{6} \\ \frac{13}{3} \\ 0 \end{pmatrix}$$ for the independent variable case that $z=\alpha , \alpha=0 $
| Note that $$\begin{pmatrix}3&4&5\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=-2$$and $$\Big[2\cdot\begin{pmatrix}2&3&4\end{pmatrix}-\begin{pmatrix}1&2&3\end{pmatrix}\Big]\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}3&4&5\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=2\times 5-1=9$$since this leads to $-2=9$, therefore the solution space is infeasible.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Let $A= \begin{pmatrix} 8&2 \\ -8&-2 \end{pmatrix}$. Find the entry in the first row and second column of $A^{2014}$ I have tried diagonalizing the matrix and obtained:
$A=PDP^{-1}$.
Where:
$P=\begin{pmatrix} 1&1 \\ -4&-1 \end{pmatrix}$
$D=\begin{pmatrix} 0&0 \\ 0&6 \end{pmatrix}$
$P^{-1}=\frac{1}{3}\begin{pmatrix} -1&-1 \\ 4&1 \end{pmatrix}$.
So that $A^{2014}=PD^{2014}P^{-1}$
But i just want to know whether there is an alternate method.
| Hint Show that
$$A^2=6A$$
Note: You can show the stronger statement:
$$\begin{pmatrix} 8&2 \\ -8&-2 \end{pmatrix}\begin{pmatrix} a&b \\ -a&-b \end{pmatrix}=6\begin{pmatrix} a&b \\ -a&-b \end{pmatrix}$$
but this is overkill.
| {
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"timestamp": "2023-03-29T00:00:00",
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Struggling to simplify $w^{3/2}\sqrt{32} - w^{3/2}\sqrt{50}$ to $-w\sqrt{2w}$ I'm asked to simplify $w^{3/2}\sqrt{32} - w^{3/2}\sqrt{50}$ and am provided with the solution: $-w\sqrt{2w}$
I arrived at $9\sqrt{2}$ but I think I'm confused in understanding communitive rule here.
Here is my working:
$w^{3/2}\sqrt{32} - w^{3/2}\sqrt{50}$ = $\sqrt{w^3}\sqrt{32}$ - $\sqrt{w^3}\sqrt{50}$ # is this correct approach? I made the radical exponent a radical
Then:
$\sqrt{32}$ = $\sqrt{4}$ * $\sqrt{4}$ * $\sqrt{2}$ = $2 * 2 * \sqrt{2}$ = $4\sqrt{2}$
$\sqrt{50}$ = $\sqrt{2}$ * $\sqrt{25}$ = $5\sqrt{2}$
So:
$\sqrt{w^3}$$4\sqrt{2}$ - $\sqrt{w^3}5\sqrt{2}$ # should the expressions on either side of the minus sign be considered a single factor? i.e. could I also write as ($\sqrt{w^3}$$4\sqrt{2}$) - ($\sqrt{w^3}5\sqrt{2}$) )?
Then I'm less sure about where to go next. Since I have a positive $\sqrt{w^3}$ and a negative $\sqrt{w^3}$ I cancelled those out and was thus left with $9\sqrt{2}$.
More generally I was not sure of how to approach this and could not fin a justification for taking the path that I did.
How can I arrive at $-w\sqrt{2w}$ per the text book's solution?
| You are on the right track to simplify $ w^\frac{3}{2} \sqrt{32} - w^\frac{3}{2} \sqrt{50} $ to as far as
$$\mathrm{(1)} \qquad \sqrt{w^3} 4 \sqrt{2} - \sqrt{w^3} 5 \sqrt{2} $$
I would make expression $ (1) $ neater and rewrite as
$$\mathrm{(2)} \qquad 4 \sqrt{w^3} \sqrt{2} - 5 \sqrt{w^3} \sqrt{2} $$
These two terms are alike, and combining the two yields
$$\mathrm{(3)} \qquad -\sqrt{w^3} \sqrt{2} $$
According to one of the properties of radicals, $ \sqrt{a} \sqrt{b} = \sqrt{ab} $. Using that property and rearranging,
$$\mathrm{(4)} \qquad -\sqrt{2w^3} $$
Now $ w^3 $ can be expressed as a product involving a perfect square (i.e. $ w^2 $) and a non-perfect square ($ w $). Expression $ (4) $ becomes
$$\mathrm{(5)} \qquad -\sqrt{2w w^2} $$
Simplify to get the desired result
$$ -w \sqrt{2w} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is there a value of $t$ above which these inequalities aren't jointly satisfiable Let $\{a_{1}, ...., a_{8}\} \in [0, 1]$ be such that $\sum \limits_{0 < i \leq 8} a_{i} = 1$ and let $t \in (\frac{1}{2}, 1]$. Is there a value of $t$ above which the following inequalities aren't jointly satisfiable? I know that they are all satisfiable for any value of $t \leq \frac{2}{3}$.
(1) $\frac{a_{1} + a_{2}}{a_{1} + a_{2} + a_{3} + a_{4}} \geq t$
(2) $\frac{a_{1} + a_{5}}{a_{1} + a_{2} + a_{5} + a_{6}} \geq t$
(3) $\frac{a_{4} + a_{8}}{a_{2} + a_{4} + a_{6} + a_{8}} \geq t$
(4) $\frac{a_{7} + a_{8}}{a_{3} + a_{4} + a_{7} + a_{8}} \geq t$
(5) $\frac{a_{1} + a_{3}}{a_{1} + a_{2} + a_{3} + a_{4}} \leq 1 - t$
(6) $\frac{a_{6} + a_{8}}{a_{2} + a_{4} + a_{6} + a_{8}} \leq 1 - t$
| Let $a_1=a_4=\frac{1}{2}$ and everything else equal $0$ (all fractions make sense for these choices since $a_1$ and $a_4$ appear in all denominators). Choose $t=1$. From $(1)$ we get $\frac{1}{2}\geq 1$ and from $(2)$ we get $1\geq 1$, so they are not jointly satisfiable.
(Forgive me if I understood the problem wrong).
| {
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Is there a neat way to find the sign of a real function with radicals like this? This simple function:
$$f(x) = \sqrt{x+2} - 2\sqrt{x+1} + \sqrt{x}$$
is negative for $x>0$ (just checked out with a graphic calculator).
But how to "prove" this algebraically?
The function comes out from the problem of finding which of these numbers $\sqrt{12} - \sqrt{11}$ and $\sqrt{11} - \sqrt{10}$ is bigger.
[BTW, given that $f(x)$ is negative, it should be that $\sqrt{11} - \sqrt{10}$ is bigger than $\sqrt{12} - \sqrt{11}$ !]
Thanks!
| I will prove $f(x)<0$
So i need to prove $\sqrt{x+2}-2\sqrt{x+1}+\sqrt{x}<0\left(x>0\right)$
Or $\sqrt{x+2}+\sqrt{x}<2\sqrt{x+1}$
By square both sides we have: $\sqrt{x^2+2x}<x+1$
Or $x^2+2x<x^2+2x+1\Leftrightarrow0<1$
So $f(x)<0$ it means $f(x)$ is negative for $x>0$
| {
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Find all positive integers a,b,c,d such that $2019^a = b^3+c^3+d^3-5$ Working on a problem...
Find all positive integers a,b,c,d such that $2019^a = b^3+c^3+d^3-5$
Using $(a+b+c)^3$ yields far too many terms, and I cannot come up with any solutions other than guessing and checking.
| HINT: If $a \ge 2$, then $2019^a \equiv 0 \pmod{9}$, while $b^3 \equiv -1, 0, \ \text{or} \ 1 \pmod{9}$ and similarly for $c^3$ and $d^3$.
Do you see a contradiction? After ruling out $a \ge 2$, you only need to solve the $a = 1$ case, i.e. $b^3+c^3+d^3 = 2024$. Since $b,c,d$ are positive integers, none of them can be larger than $\sqrt[3]{2024}$, so there aren't too many possibilities.
| {
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An analogue to Euler's sum of powers conjecture What if instead of considering $$a_1^3+a_2^3+a_3^3=b_1^3 \\
a_1^4+a_2^4+a_3^4+a_4^4=b_1^4 \\
\vdots $$
we instead considered
$$a_1^3+a_2^3+a_3^3=b_1^3+b_2^3 \\
a_1^4+a_2^4+a_3^4+a_4^4=b_1^4+b_2^4+b_3^4 \\
a_1^5+a_2^5+a_3^5+a_4^5+a_5^5=b_1^5+b_2^5+b_3^5+b_4^5 \\
\vdots $$ My gut feeling is that this will have infinitely many solutions.
Note: $a_i,b_i$ are all positive integers.
| Here's a possible strategy to show that there are infinitely many solutions for the equation. For example, let's consider the first one. Suppose that there exists $x, y, z\in \mathbb{Z}$ with $z\neq \pm 1$ and satisfies $x^{3} + y^{3} = z^{3} + 2$. Then this gives one parametrization of the solution as
$$
(z^{m})^{3} + (z^{m})^{3} + (z^{m+1})^{3} = z^{3m}(z^{3} + 2) = z^{3m}(x^{3} + y^{3}) = (xz^{m})^{3} + (yz^{m})^{3}
$$
for $m\geq 0$. Also, we have a solution for the previous equation as $7^{3} + (-5)^{3} = 6^{3} + 2$, so we just proved that the equation $a_{1}^{3} + a_{2}^{3} + a_{3}^{3} = b_{1}^{3} + b_{2}^{3}$ has infinitely many integer solutions.
If you only want positive solutions, it is also possible because we have
$$
1214928^{3} + 3480205^{3} = 3528875^{3} + 2.
$$
Similarly, it is enough to show that there exists an integer $1\leq t \leq k-1$ such that the equation
$$
x_{1}^{k} + x_{2}^{k} + \cdots + x_{k-1}^{k} = ty^{k} + (k-t) = t(y^{k} - 1) + k
$$
has an integer (or positive integer) solution with $y\neq \pm 1$, which will give a parametrization
$$
(y^{m})^{k} + \cdots + (y^{m})^{k} + (y^{m+1})^{k} + \cdots + (y^{m+1})^{k} = (x_{1}y^{m})^{k} + \cdots + (x_{k-1}y^{m})^{k}
$$
where there are $(k-t)$ many $y^{m}$'s and $t$ many $y^{m+1}$'s on LHS.
| {
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Prove whether $xyz =1$ implies that $yzx=1$ or $yxz=1$. Let $x,y,z$ be elements of a group $G$ and $xyz=1$. I am trying to prove whether this implies $yzx=1$ or $yxz=1$.
My proof goes as follows: Let $x^{-1}$ denote the inverse of $x$, then $xx^{-1} = 1 = x(yz)$. By the cancellation property of groups, $x^{-1}= yz$. This implies $(yz)x = x^{-1}x = 1$. Therefore $xyz = 1 \implies yzx = 1$.
By applying the same argument to $y(zx) = 1$ one can prove that $xyz = 1 \implies yxz =1$.
However while trying to my proof online I stumbled upon the following counterexample for the proposition $xyz = 1 \implies yxz =1$. If we take $G$ to be the group of $2\times 2$ matrices and let $x = \left( \begin{array} { c c } { 1 } & { 2 } \\ { 0 } & { 2 } \end{array} \right)$, $y = \left( \begin{array} { l l } { 0 } & { 1 } \\ { 2 } & { 1 } \end{array} \right)$ and $z = \left( \begin{array} { c c } { - 1 / 2 } & { 3 / 4 } \\ { 1 } & { - 1 } \end{array} \right)$. Then $x y z = \left( \begin{array} { c c } { 1 } & { 0 } \\ { 0 } & { 1 } \end{array} \right) = 1$ but $y x z = \left( \begin{array} { c c } { 2 } & { - 2 } \\ { 5 } & { - 9 / 2 } \end{array} \right) \neq 1$.
I don't understand where my proof went wrong.
| You just mixed up some letters. $xyz=1$ implies $yzx=1$ and $zxy=1,$ not $yxz=1.$
| {
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Why do we divide by $ \sqrt{A^2 + B^2} $ to convert a function from the standard form to the normal form? To convert an equation in the standard form, $ Ax + By + C = 0 $ to the normal form $ \cos \omega \ x + \sin \omega \ y = p $, we first divide by $ \sqrt{A^2 + B^2} $. Why?
For example, to convert the equation $ \sqrt 3 \ x + y - 8 = 0 $, we divide by $ \sqrt{A^2 + B^2} = 2 $ to get $ \frac{\sqrt 3}{2} \ x + \frac 12 \ y = 4 $ and then solve for $ \cos \omega = \frac{\sqrt 3}{2} $ and $ \sin \omega = \frac 12 $to find the value of $ \omega $.
I understand that we can't possibly convert this directly, since there is no $ \omega $ for which $ \cos \omega = \sqrt 3 $--but why do we specifically divide by $ \sqrt{A^2 + B^2}?$ Why not any other number to bring the values of $ \cos \omega $ and $ \sin \omega $ between 0 and 1 and then solve it?
| Because the most important property of $\sin$ and $\cos$ must be preserved ($\sin^2x+\cos^2 x=1$).
| {
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If $z=\cos\theta + i\sin \theta$ prove $\frac{z^2-1}{z^2+1}=i\tan\theta$
If $z=\cos\theta + i\sin \theta$ prove $$\frac{z^2-1}{z^2+1}=i\tan\theta$$
Here is my workings, I'm not sure if I've made a mistake or I'm just not spotting what to do next. Any help would be appreciated.
$$\frac{(\cos\theta + i\sin \theta)^2-1}{(\cos\theta + i\sin \theta)^2+1}$$
$$\frac{(\cos^2\theta + 2i\sin \theta \cos\theta - \sin^2\theta)-1}{(\cos^2\theta + 2i\sin \theta \cos\theta - \sin^2\theta)+1}$$
$$\frac{(\cos^2\theta - \sin^2\theta)+( 2i\sin \theta \cos\theta) -1}{(\cos^2\theta - \sin^2\theta)+( 2i\sin \theta \cos\theta)+1}$$
$$\frac{\cos2\theta + i\sin 2\theta -1}{\cos2\theta + i\sin 2\theta +1}$$
I understand how I can do it with using $z=e^{i \theta}$, however I want to solve it using double angle identities.
| Now where you've left off,
use $\cos2y=2\cos^2y-1=1-2\sin^2y\ \ \ \ (1)$
to find $$\frac{\cos2\theta + i\sin 2\theta -1}{\cos2\theta + i\sin 2\theta +1}=\dfrac{-2\sin^2\theta+ i\sin 2\theta}{2\cos^2\theta+ i\sin 2\theta}=\dfrac{2i\sin\theta(\cos\theta+i\sin\theta)}{2\cos\theta(\cos\theta+i\sin\theta)}=?$$
Similarly using Using De Moivre's theorem . $(1+i)^{100}$, $$z^n=(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta $$
use $(1)$ to establish $$\dfrac{z^n-1}{z^n+1}=2i\tan\dfrac{n\theta}2$$
| {
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Solve recursion $a_n=a_{n-1}-6\cdot3^{n-1}$ for $n>0, a_0=0$ $a_n=a_{n-1}-6\cdot3^{n-1}$ for $n>0, a_0=0$
So I calculate first terms
$a_0=0$
$a_1=-6$
$a_2=-24$
$a_3=-78$
I don't see any relation so
$a_n=a_{n-1}-6\cdot3^{n-1}$
$a_{n-1}=a_{n-2}-6\cdot 3^{n-2}$
. . .
$a_2=a_1-6\cdot3^{1}$
$a_1=a_0-6\cdot 3^{0}$
Not sure what to do next, Wolfram solves it in this way:
$a_n=-3\cdot(3^{n}-1)$
How do I get to this point?
| Note that for all $n$ we have $$a_{n+1}-a_n = -6\cdot 3^n $$
so we have also: $$a_n-a_{n-1}=-6\cdot3^{n-1}$$
thus $$a_{n+1}-a_n = 3(-6\cdot3^{n-1}) = 3(a_n-a_{n-1})$$
{or divide this two equations: $${a_{n+1}-a_n \over a_n-a_{n-1}}= {-6\cdot 3^n\over -6\cdot3^{n-1}} = 3$$}
so you have to solve linear recurrence:
$$ a_{n+1}-4a_n+3a_{n-1}=0$$
CAn you do that?
| {
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Correctness of the induction step Prove by induction that for all $ n \geq 2$ that the following:
$\sum_{i = 1}^n \frac{i}{i+1} < \frac{n^2}{n+1}$
Looking at the $n+1$ step, is it safe to assume the following ?
$\sum_{i = 1}^{n+1} \frac{i}{i+1} < \frac{(n+1)^2}{n+2} \iff \frac{n+1}{n+2} \leq \frac{(n+1)^2}{n+2} - \frac{n^2}{n+1}$
Assuming the inequality is true for $n$ we can derive that the inequality for $n+1$ stays true if the growth rate of the left side is lesser or equal to the growth rate of the right side.
We then solve the inequality on the right:
$(n+1)^2 \leq (n+1)^3 -n^2(n+2)$
$n^2+2n+1 \leq n^3+3n^2+3n+1 -n^3-2n^2$
$0 \leq n$
Since the questions assumes $ n \geq 2$ q.e.d.
| No, it is not safe to assume that. Since this is supposed to be an induction proof, what you should assume is that $\sum_{i=1}^n\frac i{i+1}<\frac{n^2}{n+1}$. Then$$\sum_{i=1}^{n+1}\frac i{i+1}=\left(\sum_{i=1}^n\frac i{i+1}\right)+\frac{n+1}{n+2}<\frac{n^2}{n+1}+\frac{n+1}{n+2}.$$So, all that remains to be proved is that$$\frac{n^2}{n+1}+\frac{n+1}{n+2}\leqslant\frac{(n+1)^2}{n+2}.$$
| {
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Alternating versions of Convergent Series summing to half Using $\zeta(2)$ as an example.
$\sum_{n=0}^\infty \frac{1}{n^2}=1+\frac{1}{2^2}+\frac{1}{3^2}+...=\frac{\pi^2}{6} $
When alternating the values gave -$\frac{\zeta(2)}{2}$
$\sum_{n=0}^\infty \frac{(-1)^n}{n^2}=-1+\frac{1}{2^2}-\frac{1}{3^2}+...=-\frac{\pi^2}{12} $
and changing the negative gave $\frac{\zeta(2)}{2}$
$\sum_{n=0}^\infty \frac{(-1)^n}{n^2}=1-\frac{1}{2^2}+\frac{1}{3^2}-...=\frac{\pi^2}{12} $
This seems to only work on convergent series, since something like the Harmonic Series continues to infinity while the Alternating Harmonic Series converges to ln(2)
So why do the alternating versions produce these values so similar to the non alternating version?
| The alternating series you present is $\eta(2)$, the Dirichlet eta function. More specifically,
$$\eta(s) = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}.$$
It turns out that this series, unlike the zeta series, converges for all $s \gt 0$. We can produce a very special relationship to the zeta function $\zeta(s)$ with some algebraic cleverness:
\begin{align} \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} &= 1 - 2^{-s} + 3^{-s} - 4^{-s} + 5^{-s} - 6^{-s} + 7^{-s} - 8^{-s} + \cdots \\
&= \left( 1 + 3^{-s} + 5^{-s} + 7^{-s} + \cdots \right) - \left( 2^{-s} + 4^{-s} + 6^{-s} + 8^{-s} + \cdots \right) \\
&= \left( 1 + 2^{-s} + 3^{-s} + 4^{-s} + 5^{-s} + 6^{-s} + 7^{-s} + \cdots \right) - 2 \left( 2^{-s} + 4^{-s} + 6^{-s} + 8^{-s} + \cdots \right) \\
&= \left( 1 + 2^{-s} + 3^{-s} + 4^{-s} + 5^{-s} + 6^{-s} + 7^{-s} + \cdots \right) - 2 \left( 2^{-s} 1 + 2^{-s} 2^{-s} + 2^{-s} 3^{-s} + 2^{-s} 4^{-s} + \cdots \right) \\
&= \zeta(s) - 2^{1 - s} \left(1 + 2^{-s} + 3^{-s} + 4^{-s} + \cdots \right) \\
&= \zeta(s) - 2^{1 - s} \zeta(s) \\
&= \left( 1 - 2^{1 - s} \right) \zeta(s).
\end{align}
So now, by our new formula $\eta(s) = \left( 1 - 2^{1 - s} \right) \zeta(s)$, we can produce the value of the eta function given the corresponding zeta function, which will differ only by a rational factor.
So if $\zeta(2) = \frac{\pi^2}6$, then $\eta(2) = \left( 1 - 2^{-1} \right) \zeta(2) = \frac12 \frac{\pi^2}6 = \frac{\pi^2}{12},$ as desired.
This formula can also be manipulated into the form $\zeta(s) = \frac{\eta(s)}{1 - 2^{1 - s}}$, which allows us to find a value for zeta in $0 \lt s \lt 1$.
The reason the series for $\eta(1)$ converges (conditionally) is at $s = 1$, that the factor $1 - 2^{1 - s}$ has a simple (i.e. degree one) zero, while $\zeta(s)$ has a simple pole. These two evenly matched factors are battling for control, and their fight lands them at this crossroad. (It's a bit more complicated to show that it converges to $\ln 2$, but it can be done.)
| {
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Naive question about continuous random variable: solution verification The probability density function of the random variable $X$ is given by
$$f(x) =
\begin{cases}
a + bx^{2} & 0 < x < 1 \\
0 & \text{otherwise}
\end{cases}$$
If $\textbf{E}(X) = 3/5$, determine the values of $a$ and $b$
MY SOLUTION
Since $f_{X}$ is a probability density function, we must have
\begin{align*}
\int_{0}^{1}(a + bx^{2})\mathrm{d}x = a + \frac{b}{3} = 1
\end{align*}
On the other hand, according to the definition of expected value, we get
\begin{align*}
\int_{0}^{1}(ax+ bx^{3})\mathrm{d}x = \frac{a}{2} + \frac{b}{4} = \frac{3}{5}
\end{align*}
Solving the linear system of equations obtained, it results that $(a,b) = (3/5,6/5)$.
| Everything's correct; solution is short, but totally understandable!
| {
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Solving equations involving square roots I am a student and I often encounter these type of equations:
$$\sqrt{x^2 + (y-2)^2} + \sqrt{x^2 + (y+2)^2} = 6$$
I usually solve these by taking one term ($\sqrt{x^2 + (y-2)^2}$ for example) to the right hand side but this seems to take more time. Please suggest me some methods which can help me solve these types of problem quickly.
Thanks
| Denote:
$$\begin{cases}x^2 + (y-2)^2=t^2 \\ x^2+(y+2)^2=t^2+8y\end{cases}.$$
Then:
$$\sqrt{x^2 + (y-2)^2} + \sqrt{x^2 + (y+2)^2} = 6 \Rightarrow \\
t+\sqrt{t^2+8y}=6 \Rightarrow \\
t^2+8y=36-12t+t^2 \Rightarrow \\
t=\frac{9-2y}{3}.$$
Plug it into the first equation:
$$x^2+(y-2)^2=\left(\frac{9-2y}{3}\right)^2 \Rightarrow \\
9x^2+9y^2-36y+36=81-36y+4y^2 \Rightarrow \\
9x^2+5y^2=45 \Rightarrow \\
\frac{x^2}{5}+\frac{y^2}{9}=1.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate the constant term of $P(x-1)$ if the remainder of the division of $P(x)$ by $x-3$ is $18$ and $P(x+1) = (x^2 -4)Q(x)+3ax+6$
Assume that $$P(x+1) = (x^2 -4)Q(x)+3ax+6$$
and that the remainder of the division of polynomial $P(x)$ by $x-3$ is $18$. Evaluate the constant term of polynomial $P(x-1)$.
All I could see so far is that the polynomial $P(x)$ should be quadratic because $Q(x)$ is multiplied by quadratic term, which is $x^2$.
| Let $F(x)=P(x-1)$. Of course, the constant term of $F(x)$ is $F(0)=P(-1)$
We write $$P(x)=g(x)\times (x-3)+18$$ Then, of course, $P(3)=18$
Using the equation $$P(x+1) = (x^2 -4)Q(x)+3ax+6$$ We set $$x=-2\implies P(-1)=-6a+6$$
and $$x=2\implies P(3)=6a+6$$ Since we already know that $P(3)=18$ we deduce that $a=2$ hence $$P(-1)=-12+6=-6$$
| {
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Perimeter of orthic triangle If $DEF$ is the orthic triangle of $\triangle ABC$, then prove that
$$\frac{\text{Perimeter of }\triangle DEF}{\text{Perimeter of }\triangle ABC} = \frac{r}{R} $$
where $r$ and $R$ are the inradius and the circumradius of $\triangle ABC$.
My attempt is very simple , I put the side length of orthic triangle in terms of $\cos$ and the side length of $\triangle ABC$ but I can't get the required answer.
| $$\sin{2A}+\sin{2B}+\sin2C=2\sin{(A+B)}\cos{(A-B)}+2\sin{C}\cos{C}$$$$=2\sin{C}(\cos{(A-B)}-\cos{(A+B)})=4\sin{A}\sin{B}\sin{C}$$
$$\text{Now, }\frac{EF}{\sin A}=\frac{AE}{\sin C}=\frac{c\cos A}{\sin C}\implies EF=2R\sin A\cos A=R\sin{2A}\text{ .}$$
$$\therefore \text{Perimeter of }\triangle DEF=4R\sin{A}\sin{B}\sin{C}=\frac{2\Delta}{R}=\frac{r\cdot 2s}{R}$$$$\implies\frac{\text{Perimeter of }\triangle DEF}{\text{Perimeter of }\triangle ABC} = \frac{r}{R}$$
$\blacksquare$
Explanation: $\Delta=\frac{1}{2}ab\sin C=\frac{1}{2}(2R\sin A)(2R \sin B)\sin C=2R^2\sin{A}\sin{B}\sin{C}$
| {
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Why these $2$ methods give the exact same answer for sum of squares? Consider $n = 8$, the sum of squares from $1$ through $8$ is:
$1 \times 1 + 2 \times 2 + 3
\times 3 + 4 \times 4 + 5 \times 5 + 6 \times 6 + 7 \times 7 + 8 \times 8 = 204$.
Also, equal to
$1 \times 8 + 3 \times 7 + 5 \times 6 + 7 \times 5 + 9 \times 4 + 11 \times 3 + 13 \times 2 + 15 \times 1 = 204$.
The second one logic is that I start with $1$, then I increment by $2$ each time and subtract $1$ from the second one, until I reach $1$.
For $n = 2$.
$1 \times 1 + 2 \times 2 = 4 = 1 \times 2 + 3 \times 1$.
For $n = 3$.
$1 \times 1 + 2 \times 2 + 3 \times 3 = 1 \times 3 + 3 \times 2 + 5 \times 1$
The question is, why is it supposed to be the equal the one above? I tried it with a lot of values for $n$?
| We have $$\sum_{k=1}^j (2k-1)\cdot (j+1-k)=(j+1)\sum_{k=1}^j (2k-1)-2\sum_{k=1}^j k^2+\sum_{k=1}^j k$$ $$=(j+1)j^2-\frac{j(j+1)(2j+1)}{3}+\frac{j(j+1)}{2}$$ $$=\frac{j(j+1)(2j+1)}{6}=\sum_{k=1}^j k^2 $$ showing that the equality is no coincidence , but holds for every positive integer $j$.
| {
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} |
Finding the greatest value of $a^2b^3c^2$ if $a+b+c=3$ and all numbers are positive Find the greatest value of $a^2b^3c^2$ if $a+b+c=3$ and all numbers are positive.
Here is my attempt using $\text{AM-GM inequality}$:
$$AM=\frac{a+b+c+a+b+c+b}{7}$$
$$GM=\sqrt[7]{a^2b^3c^2}$$
We have to find the maximum value of the expression under the radical of the $GM$.
$GM$ will be maximum if all terms are equal. Hence,
$$a=b=c=1$$
So the maximum value of the expression should be $1$.
However, this is wrong (at least according to the problem book where I found this question).
I can't figure out what I did wrong. Can you help me?
Source: Resonance DLPD Algebra for JEE Mains and Advanced. Exercise 1, Part II, D-4.
Motivation: I am trying to practice mathematics problems for the JEE Mains and Advanced.
| By AM-GM $$3=2\cdot\frac{a}{2}+3\cdot\frac{b}{3}+2\cdot\frac{c}{2}\geq7\sqrt[7]{\left(\frac{a}{2}\right)^2\left(\frac{b}{3}\right)^3\left(\frac{c}{2}\right)^2}.$$
The equality occurs for $\frac{a}{2}=\frac{b}{3}=\frac{c}{2}$ and $a+b+c=3.$
Can you end it now?
| {
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"url": "https://math.stackexchange.com/questions/3099777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If positive $a$, $b$, $c$, $d$ satisfy ${1\over a+1}+{1\over b+1}+{1\over c+1}+{1\over d+1}=1$, then $abcd\geq 81$
Let $a,b,c,d>0$ satisfying
$${1\over a+1}+{1\over b+1}+{1\over c+1}+{1\over d+1}=1$$
Prove that $abcd\geq 81$
I've tried to apply arithmetic geometric mean inequality or Cauchy-Schwartz inequality.
$${{1\over a+1}+{1\over b+1}+{1\over c+1}+{1\over d+1}\over 4}\geq \left({1\over(a+1)(b+1)(c+1)(d+1)}\right)^{1\over4}$$
or
$$(a+1+b+1+c+1+d+1)\left({1\over a+1}+{1\over b+1}+{1\over c+1}+{1\over d+1}\right)\geq 4^2$$
But this wasn't enough to prove $abcd\geq 81$.
Does anyone have ideas??
| Because by AM-GM $$\frac{a}{a+1}=\frac{1}{b+1}+\frac{1}{c+1}+\frac{1}{d+1}\geq\frac{3}{\sqrt[3]{(b+1)(c+1)(d+1)}}.$$
Id est,
$$\prod_{cyc}\frac{a}{a+1}\geq\prod_{cyc}\frac{3}{\sqrt[3]{(b+1)(c+1)(d+1)}}$$ or
$$\frac{abcd}{\prod\limits_{cyc}(a+1)}\geq\frac{81}{\prod\limits_{cyc}(a+1)}$$ or
$$abcd\geq81.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3102211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove by induction that $\sum_{i=1}^n \frac{1}{\sqrt i} > \sqrt n$ for all integers $n \ge 2$ Having trouble with the concept of proving an inequality.
Q: Prove by induction that $\sum_{i=1}^n \frac{1}{\sqrt i} > \sqrt n$ for all integers $n \ge 2$
Here is what I have so far:
Basis ($n=2$)
$\sum_{i=1}^n \frac{1}{\sqrt i} = \frac{1 + \sqrt 2}{\sqrt 2} > \frac{1 + 1}{\sqrt 2} = \sqrt 2$
Inductive Step
(IH): Let $k \ge 2$ be an integer and suppose that $\sum_{i=1}^k \frac{1}{\sqrt i} > \sqrt k$
We want to prove that $\sum_{i=1}^{k+1} \frac{1}{\sqrt i} > \sqrt {k+1}$
So,
$$\sum_{i=1}^{k+1} \frac{1}{\sqrt i} = \left( \sum_{i=1}^k \frac{1}{\sqrt i} \right) + \frac {1}{\sqrt {k+1}}$$
$$\sum_{i=1}^{k+1} \frac{1}{\sqrt i} > \sqrt k + \frac {1}{\sqrt {k+1}}$$
$$\sum_{i=1}^{k+1} \frac{1}{\sqrt i} > \frac{\sqrt{k+1} \sqrt k}{\sqrt{k+1}} + \frac{1}{\sqrt{k+1}}$$
$$\sum_{i=1}^{k+1} \frac{1}{\sqrt i} > \frac{\sqrt{k+1} \sqrt k + 1}{\sqrt{k+1}}$$
And now I do not understand where to go from here to get my proof. I don't know what I'm supposed to compare here?
| To make the proof work,
you need to show that
$\sqrt{n}+\dfrac1{\sqrt{n+1}}
\gt \sqrt{n+1}
$.
This is the same as
$\begin{array}\\
\dfrac1{\sqrt{n+1}}
&\gt \sqrt{n+1}-\sqrt{n}\\
&=(\sqrt{n+1}-\sqrt{n})\dfrac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}\\
&=\dfrac{1}{\sqrt{n+1}+\sqrt{n}}\\
\end{array}
$
which is obvious.
You can get a more precise result
by noting that
$\dfrac1{\sqrt{n+1}}
=\dfrac{1}{\sqrt{n+1}+\sqrt{n}}
$
implies that
$\dfrac{1}{2\sqrt{n+1}}
\lt \sqrt{n+1}-\sqrt{n}
\lt \dfrac{1}{2\sqrt{n}}
$.
Summing the left side,
$\sum_{n=1}^m \dfrac{1}{2\sqrt{n+1}}
\lt \sum_{n=1}^m (\sqrt{n+1}-\sqrt{n})
=\sqrt{m+1}-1
\lt \sqrt{m+1}
$
so that
$\sum_{n=2}^{m+1} \dfrac{1}{2\sqrt{n}}
\lt \sqrt{m+1}
$
or
$\sum_{n=2}^{m} \dfrac{1}{\sqrt{n}}
\lt 2\sqrt{m}
$
so
$\sum_{n=1}^{m} \dfrac{1}{\sqrt{n}}
\lt 2\sqrt{m}+1
$.
Summing the right side,
$\sum_{n=1}^{m-1} \dfrac{1}{2\sqrt{n}}
\gt \sum_{n=1}^{m-1} (\sqrt{n+1}-\sqrt{n})
=\sqrt{m}-1
$
so that
$\sum_{n=1}^{m-1} \dfrac{1}{\sqrt{n}}
\gt 2\sqrt{m}-2
$
so
$\sum_{n=1}^{m} \dfrac{1}{\sqrt{n}}
\gt 2\sqrt{m}-1+\frac1{\sqrt{m}}
\gt 2\sqrt{m}-1
$.
Therefore
$\sum_{n=1}^{m} \dfrac{1}{\sqrt{n}}$
is between
$2\sqrt{m}-1
$
and
$2\sqrt{m}+1
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3103173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\lim_{n\to\infty} \sum_{i=1}^{n} \left[\frac{4}{n} - \frac{i^2}{n^3}\right]$ $$\lim_{n\to\infty} \sum_{i=1}^{n} \left[\frac{4}{n} - \frac{i^2}{n^3}\right]$$
If I take out $\frac{1}{n}$, it looks like a Riemann sum:
$$\lim_{n\to\infty} \sum_{i=1}^{n} \left[\left(\frac{1}{n}\right)\left(4 - \frac{i^2}{n^2}\right)\right]$$
How should I proceed with solving this limit? The given answer is $\frac{11}{3}$.
| A different, possibly more basic (depending on how you define "basic") solution is to use the sum for the squares of the first $n$ integers (as suggested by RRL in the question comments) being
$$\sum_{i=1}^n i^2 = \cfrac{n\left(n+1\right)\left(2n + 1\right)}{6} = \cfrac{2n^3 + 3n^2 + n}{6} \tag{1}\label{eq1}$$
Thus,
\begin{align}
\lim_{n \to \infty} \sum_{i=1}^n \left[\cfrac{4}{n} - \cfrac{i^2}{n^3}\right] & = \lim_{n \to \infty} \left[\cfrac{4n}{n} - \cfrac{2n^3 + 3n^2 + n}{6n^3} \right] \\
& = \lim_{n \to \infty} \left[4 - \cfrac{1}{3} - \cfrac{1}{2n} - \cfrac{1}{6n^2} \right] \\
& = \cfrac{11}{3} \tag{2}\label{eq2}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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PMF of throwing a die 4 times We throw a fair 6-sided die independently four times and let $X$ denote the minimal value rolled.
*
*What is the probability that $X \ge 4$?
*Compute the PMF of $X$.
*Determine the mean and variance of $X$.
My attempt:
*
*$(1/2)^4$ because that means each of the $4$ rolls, you either get a $4, 5$ or $6$.
*for $X=1: (1/6)(6/6)(6/6)(6/6)$
for $X=2: (1/6)(5/6)(5/6)(5/6)$
for $X=3: (1/6)(4/6)(4/6)(4/6)$
for $X=4: (1/6)(3/6)(3/6)(3/6)$
for $X=5: (1/6)(2/6)(2/6)(2/6)$
for $X=6: (1/6)(1/6)(1/6)(1/6)$
*I can calculate this once I know I did the PMF correctly
Did I do (1) and (2) correctly?
| We throw a fair 6-sided die independently four times and let $X$ denote the minimal value rolled.
What is the probability that $X \ge 4$?
For each roll, the value must be greater than three. This event has probability $\frac{3}{6}=\frac{1}{2}$. As this must occur on each of $4$ rolls, we have:
$$P(X\ge4)=\left(\frac{1}{2}\right)^4=\frac{1}{16}$$
Compute the PMF of $X$.
$$P(X=1)=P(X>0)-P(X>1)=1-\left(\frac{5}{6}\right)^4=1-\frac{625}{1296}=\frac{671}{1296}$$
$$P(X=2)=P(X>1)-P(X>2)=\left(\frac{5}{6}\right)^4-\left(\frac{4}{6}\right)^4=\frac{625}{1296}-\frac{256}{1296}=\frac{369}{1296}$$
$$P(X=3)=P(X>2)-P(X>3)=\left(\frac{4}{6}\right)^4-\left(\frac{3}{6}\right)^4=\frac{256}{1296}-\frac{81}{1296}=\frac{175}{1296}$$
$$P(X=4)=P(X>3)-P(X>4)=\left(\frac{3}{6}\right)^4-\left(\frac{2}{6}\right)^4=\frac{81}{1296}-\frac{16}{1296}=\frac{65}{1296}$$
$$P(X=5)=P(X>4)-P(X>5)=\left(\frac{2}{6}\right)^4-\left(\frac{1}{6}\right)^4=\frac{16}{1296}-\frac{1}{1296}=\frac{15}{1296}$$
$$P(X=6)=P(X>5)-P(X>6)=\left(\frac{1}{6}\right)^4-0=\frac{1}{1296}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3104217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Evaluating $\int_{-5 }^{-2} f(x)\;dx + \int_{1/6}^{1/3}f(x)\,dx+\int_{6/5}^{3/2}f(x)\;dx$, where $f(x)=\left(\frac{x^2-x}{x^3-3x+1}\right)^2$ How to find the value of
$$\int _ { - 5 } ^ { - 2 } \left( \frac { x ^ { 2 } - x } { x ^ { 3 } - 3 x + 1 } \right) ^ { 2 }\,dx + \int _ { \frac 16 } ^ { \frac 13 } \left( \frac { x ^ { 2 } - x } { x ^ { 3 } - 3 x + 1 } \right) ^ { 2 }\,dx+\int _ { \frac 65 } ^ {\frac 32 } \left( \frac { x ^ { 2 } - x} { x ^ { 3 } - 3 x + 1 } \right) ^ { 2 }\,dx$$
My attempt:
I tried to substitute in second and third integral,
such that powers of integration become the same. But I could not find such a substitution.
tried taking out powers of x from denominator
such that differential coefficient gets formed in the numerator
.
| $$I= \int _ { - 5 } ^ { - 2 } \left( \frac { x ^ { 2 } - x } { x ^ { 3 } - 3 x + 1 } \right) ^ { 2 }\,dx + \int _ { \frac 16 } ^ { \frac 13 } \left( \frac { x ^ { 2 } - x } { x ^ { 3 } - 3 x + 1 } \right) ^ { 2 }\,dx+\int _ { \frac 65 } ^ {\frac 32 } \left( \frac { x ^ { 2 } - x} { x ^ { 3 } - 3 x + 1 } \right) ^ { 2 }\,dx$$
$\displaystyle I_{1} = \int^{-2}_{-5}\bigg(\frac{x^2-x}{x^3-3x+1}\bigg)^2dx,$ put $\displaystyle x=1-\frac{1}{u}\rightarrow u=\frac{1}{1-x}$ and $\displaystyle dx = \frac{1}{u^2}du$
$$\displaystyle I_{1}=\int^{\frac{1}{3}}_{\frac{1}{6}}\bigg(\frac{x^2-x}{x^3-3x+1}\bigg)^2\frac{1}{x^2}dx$$
$\displaystyle I_{3}=\int^{\frac{3}{2}}_{\frac{6}{5}}\bigg(\frac{x^2-x}{x^3-3x+1}\bigg)^2dx,$ put $\displaystyle x=\frac{1}{1-v}\rightarrow v=1-\frac{1}{x}$ and $\displaystyle dx = \frac{1}{(1-v)^2}dv$
$$I_{3} =\int^{\frac{1}{3}}_{\frac{1}{6}}\bigg(\frac{x^2-x}{x^3-3x+1}\bigg)^2\frac{1}{(1-x)^2}dx$$
$$I = \int^{\frac{1}{3}}_{\frac{1}{3}}\bigg(\frac{x^2-x}{x^3-3x+1}\bigg)^2\bigg[\frac{1}{x^2}+1+\frac{1}{(1-x)^2}\bigg]dx$$
$$I = \int^{\frac{1}{3}}_{\frac{1}{6}}\bigg(\frac{x^2-x}{x^3-3x+1}\bigg)^2\bigg[\frac{x^{4}-2x^{3}+3x^{2}-2x+1}{x^2(1-x)^2}\bigg]dx$$
$$I = \int^{\frac{1}{3}}_{\frac{1}{6}}\bigg(\frac{x-x^{2}}{x^{3}-3x+1}\bigg)'dx = \frac{x-x^{2}}{x^{3}-3x+1}\bigg|^{\frac{1}{3}}_{\frac{1}{6}}=$$
| {
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"source": "stackexchange",
"question_score": "1",
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How to evaluate $\lim_{n->\infty}\frac{\sqrt[3]{n+1}-\sqrt[3]{n+\cos{}\frac{3}{n}}}{\sqrt[6]{n^2+\sin{\frac{2}{n}}}-\sqrt[3]{n}}?$ I tried to get rid off cube root as written below but still can not get throught the next steps. What should be the right step to take after the steps below? Did I start as I should or do I have to take completely different approach?
$\lim_{n->\infty}\dfrac{\sqrt[3]{n+1}-\sqrt[3]{n+\cos{}\dfrac{3}{n}}}{\sqrt[6]{n^2+\sin{\dfrac{2}{n}}}-\sqrt[3]{n}}=$
$=\lim_{n->\infty}\dfrac{\sqrt[3]{n+1}-\sqrt[3]{n+\cos{}\dfrac{3}{n}}}{\sqrt[6]{n^2+\sin{\dfrac{2}{n}}}-\sqrt[3]{n}}\dfrac{(\sqrt[3]{n+1})^2+\sqrt[3]{n+1}\sqrt[3]{n+\cos{}\dfrac{3}{n}}+(\sqrt[3]{n+\cos{}\dfrac{3}{n}})^2}{(\sqrt[3]{n+1})^2+\sqrt[3]{n+1}\sqrt[3]{n+\cos{}\dfrac{3}{n}}+(\sqrt[3]{n+\cos{}\dfrac{3}{n}})^2}$
$=\lim_{n->\infty}\dfrac{1-\cos\dfrac{3}{n}}{(\sqrt[6]{n^2+\sin{\dfrac{2}{n}}}-\sqrt[3]{n})((\sqrt[3]{n+1})^2+\sqrt[3]{n+1}\sqrt[3]{n+\cos{}\dfrac{3}{n}}+(\sqrt[3]{n+\cos{}\dfrac{3}{n}})^2)}$
| To complete your approach, let
\begin{align*}
a &= \frac{\sqrt[3]{n + 1}}{\sqrt[3]{n}} = \sqrt[3]{1 + \frac{1}{n}} \\
b &= \frac{\sqrt[3]{n + \cos \frac{3}{n}}}{\sqrt[3]{n}} = \sqrt[3]{1 + \frac{1}{n}\cos\frac{3}{n}} \\
c &= \frac{\sqrt[6]{n^2 + \sin \frac{2}{n}}}{\sqrt[3]{n}} = \sqrt[6]{1 + \frac{1}{n^2}\sin\frac{2}{n}} \\
d &= \frac{\sqrt[3]{n}}{\sqrt[3]{n}} = 1
\end{align*}
Then, your expression becomes,
$$\frac{a - b}{c - d} = \frac{(a^3 - b^3)(c^5 + c^4d + c^3d^2 + c^2 d^3 + cd^4 + d^5)}{(c^6 - d^6)(a^2 + ab + b^2)}.$$
Note that $a, b, c, d \to 1$ as $n \to \infty$, so
$$\lim \frac{a - b}{c - d} = \frac{6}{3} \lim \frac{a^3 - b^3}{c^6 - d^6},$$
assuming the limit on the right exists. We have,
$$\frac{a^3 - b^3}{c^6 - d^6} = \frac{\frac{1}{n}\left(1 - \cos \frac{3}{n}\right)}{\frac{1}{n^2} \sin\frac{2}{n}} = \frac{\frac{1}{n}\left(1 - \cos^2 \frac{3}{n}\right)}{\frac{1}{n^2} \sin\frac{2}{n}\left(1 + \cos \frac{3}{n}\right)} = \frac{\frac{\sin^2 \frac{3}{n}}{\frac{9}{n^2}} \cdot 9}{\frac{\sin \frac{2}{n}}{\frac{2}{n}} \cdot 2 \cdot(1 + \cos \frac{3}{n})} \to \frac{9}{4}.$$
Hence, our complete limit is
$$\frac{6}{3} \cdot \frac{9}{4} = \frac{9}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3105716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find positive $K$ such that $\int_0^\infty\left(\frac{1}{\sqrt{2x^2+1}}-\frac{K}{x+1}\right)dx$ converges
Find positive $K$ such that
$$\int_0^\infty\left(\frac{1}{\sqrt{2x^2+1}}-\frac{K}{x+1}\right)dx$$ converges
I used the fact that $\frac{1}{\sqrt{2x^2+1}}-\frac{K}{x+1}>\frac{-K+1/\sqrt2}{1+x}$ for $x>1$ to prove it diverges for $K < 1/\sqrt2$.
And the fact that $\frac{1}{\sqrt{2x^2+1}}-\frac{K}{x+1}<\frac{1-K}{x+1}$ to prove it diverges for $K>1$.
However I am not sure how to prove it converges (or not) in $[\frac1{\sqrt2},1]$.
I think it would not converge for any $K$ becase the denomiantor is sorta linear and this would never be identically $0$ for any $K$.
| Hint
I suggest to have a look at what happens at the bounds.
For $x$ close to $0$, the Taylor expansion of the integrand is
$$\frac{1}{\sqrt{2x^2+1}}-\frac{K}{x+1}=(1-K)+K x-(K+1) x^2+O\left(x^3\right)$$
Now, for infinitely large values of $x$,
$$\frac{1}{\sqrt{2x^2+1}}-\frac{K}{x+1}=\frac{\frac{1}{\sqrt{2}}-K}{x}+\frac{K}{x^2}+O\left(\frac{1}{x^3}\right)$$
Does this tell you something ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3107219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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If $\tan{\frac{x}{2}}=\csc x - \sin x$, then find the value of $\tan^2{\frac{x}{2}}$.
If $\tan{\frac{x}{2}}=\csc x - \sin x$, then find the value of $\tan^2{\frac{x}{2}}$.
HINT: The answer is $-2\pm \sqrt5$.
What I have tried so far:
$$\tan{\frac{x}{2}} = \frac{1}{\sin x}-\sin x$$
$$\tan{\frac{x}{2}} = \frac{1-\sin^2 x}{\sin x}$$
$$\tan{\frac{x}{2}} = \frac{\cos^2 x}{\sin x}$$
I don't know how to solve this problem. Pls help. Thank you :)
| $$\cos^2x=\tan\dfrac x2\sin x=\dfrac{\sin\dfrac x2}{\cos\dfrac x2}\cdot2\sin\dfrac x2\cos\dfrac x2=2\sin^2\dfrac x2=1-\cos x$$
$$\implies\cos x=\dfrac{-1\pm\sqrt5}2$$
As for real $x,\cos x\ge-1,\cos x\ne\dfrac{-1-\sqrt5}2<-1$
Using Weierstrass Substitution $$\dfrac{1-\tan^2\dfrac x2}{1+\tan^2\dfrac x2}=\cos x=\dfrac{-1+\sqrt5}2$$
Now apply Componendo et Dividendo
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Sum of square of binomial coeffcient with positive and negative terms Finding $\displaystyle \binom{2n}{1}^2-2\binom{2n}{2}^2+3\binom{2n}{3}^2-\cdots \cdots -2n\binom{2n}{2n}^2.$
What I've tried:
$$(1-x)^{2n}=\binom{2n}{0}-\binom{2n}{1}x+\binom{2n}{2}x^2+\cdots \cdots +\binom{2n}{2n}x^{2n}$$
$$-2n(1-x)^{2n-1}=-\binom{2n}{1}+2\binom{2n}{2}x-3\binom{2n}{3}x^2+\cdots +n\binom{2n}{2n}x^{2n-1}$$
Sum notation:
$$\sum_{k=0}^{2n} (-1)^{k-1}k\binom{2n}{k}^2$$
| The solution uses the series representation of Legendre polynomials:
$P_n(x)=\frac{1}{2^n}\sum\limits_{k=0}^n \binom{n}{k}^2(x-1)^{n-k}(x+1)^k\tag1$
$\frac{x+1}{x-1}=-1$ at $x=0$ is valid. Extend the original sum (S) in the following way:
$S\frac{(x-1)^{2n}}{2^{2n}}=\frac{1}{2^{2n}}\sum\limits_{k=0}^{2n}\binom{2n}{k}^2 k(\frac{x+1}{x-1})^{k-1}(x-1)^{2n}\tag2$
We can realize that
$k(\frac{x+1}{x-1})^{k-1}=(-\frac{(x-1)^2}{2})\frac {d}{dx}(\frac{x+1}{x-1})^{k}\tag3$
Put it back to eqution (2) and replace the order of sum and derivation we get:
$S(x)\frac{(x-1)^{2n}}{2^{2n}}=-\frac{(x-1)^2}{2}\frac {d}{dx}\Big(\frac{1}{2^{2n}}\sum\limits_{k=0}^{2n}\binom{2n}{k}^2 (\frac{x+1}{x-1})^{k}(x-1)^{2n}\Big)\tag4$
Let's compare the summa part of (4) and $P_n(x)$ we get:
$S(x)\frac{(x-1)^{2n}}{2^{2n}}=-\frac{(x-1)^2}{2}\dfrac {dP_{2n}(x)}{dx}\tag5$
Finally
$S(x)=\frac{2^{2n-1}}{(x-1)^{2n-2}}\dfrac {dP_{2n}(x)}{dx}\tag6$
Applying the recursion relation of the Legendre polynomials we have at $x=0$:
$S(0)= 2n2^{2n-1}P_{2n-2}(0)\tag7$
| {
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"url": "https://math.stackexchange.com/questions/3113707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find the limit without L'Hôpital, fraction with cube root? How can i find this limit without using L'Hôpital (or anything using derivatives for that matter)?
$$\lim_{x \to -8} \frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}} $$
I have tried various substitutions and multiplying with conjugates etc, to no avail. I am probably doing something wrong, but I would really appreciate a good solution so I can follow each step and understand how to solve problems like this.
| Thank you for the pointers, I had to work with it a bit before realising all the steps, but this is what I came up with, posting it here so it can help others.
Substituting $ \sqrt[3]{x} $ with $ t $.
Since $ x \to -8 $, $ t \to \sqrt[3]{-8} = -2 $ the limit is now rewritten as
\begin{align*}
\lim_{t \to -2} \frac{\sqrt{1-t^3}-3}{2 + t}.
\end{align*}
The fraction is multiplied with the conjugate for the numerator
\begin{align*}
\lim_{t \to -2} -\frac{t^3 + 8}{(2+t)(\sqrt{1-t} + 3)}.
\end{align*}
$ 2+t $ is factored out of the numerator and cancelled:
\begin{align*}
\lim_{t \to -2} - \frac{t^2 - 2t + 4}{\sqrt{1-t^3} + 3}.
\end{align*}
It is now possible to plug in the value for $ t $:
\begin{align*}
-\frac{(-2)^2 - 2(-2) + 4}{\sqrt{1-(-2)^3} + 3} = - \frac{12}{6} = -2.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3113867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Show an identity involving a sum of arctangents of algebraic expressions
Show that
$$
2\tan^{-1}\frac{\sqrt{x^2+a^2} - x + b}{\sqrt{a^2-b^2}} + \tan^{-1}\frac{x\sqrt{a^2-b^2}}{b\sqrt{x^2+a^2} + a^2} + \tan^{-1}\frac{\sqrt{a^2-b^2}}{b} = n\pi .
$$
I tried using $$ x= a \tan \theta ,\; b= a \sin\phi,$$
but then calculations are not working out; that is,
I am not able to further simplify.
| Since the roots of $\tan x$ are precisely the integer multiples $n \pi$ of $\pi$, applying $\tan$ to be sides gives that the equation is equivalent to
$$\phantom{(ast)} \qquad \tan[2 \arctan A + \arctan B + \arctan C] = 0 \qquad (\ast)$$
for
$$
A := \frac{\sqrt{x^2 + a^2} - x - b}{\sqrt{a^2 - b^2}} , \qquad
B := \frac{x \sqrt{a^2 - b^2}}{b \sqrt{x^2 + a^2} + a^2} , \qquad
C := \frac{\sqrt{a^2 - b^2}}{b} .
$$
Iterating the tangent sum identity gives a formula for the tangent of the sum of four angles:
$$\tan(\alpha_1 + \alpha_2 + \alpha_3 + \alpha_4)
= \frac{\sum_i \tan \alpha_i - \sum_{i < j < k} \tan \alpha_i \tan \alpha_j \tan \alpha_k}
{1 - \sum_{i < j} \alpha_i \alpha_j + \alpha_1 \alpha_2 \alpha_3 \alpha_4} .$$
So, we can evaluate the left-hand side of $(\ast)$ by setting $\tan \alpha_1 = \tan \alpha_2 = A, \tan \alpha_3 = B, \tan \alpha_4 = C$. Since we only want to check that the sum vanishes, it's show that the numerator is zero, that is, that
$$2 A + B + C - A^2 B - A^2 C - 2 ABC = 0 .$$ Simplifying is a little tedious but straightforward.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Evaluating $\lim_{(x,y)\to(0,0)}\frac{x^2+y^2}{\sin^2y+\ln(1+x^2)}$
$$\lim_{(x,y)\to(0,0)}\frac{x^2+y^2}{\sin^2y+\ln(1+x^2)}$$
If I use a specific path I know I can use Cauchy Theorem to get a number, but how do I prove this for all paths? Thank you!
| What you have to use is the facts, for small $x,y$,
$$ \frac{|y|}{\sqrt2}\le|\sin y|\le |y|, \frac12{x^2}\le\ln (1+x^2)\le x^2$$
and
$$ \lim_{x\to0}\frac{x-\ln(1+x)}{x}=0,\frac{x-\sin x}{x}=0. $$
So
\begin{eqnarray*}
&&\bigg|\frac{x^2+y^2}{\sin^2y+\ln(1+x^2)}-1\bigg|\\
&=&\frac{x^2-\ln(1+x^2)+y^2-\sin^2y}{\sin^2y+\ln(1+x^2)}\\
&\le&2\frac{x^2-\ln(1+x^2)+y^2-\sin^2y}{x^2+y^2}\\
&=&2\frac{x^2-\ln(1+x^2)}{x^2+y^2}+2\frac{y^2-\sin^2y}{x^2+y^2}\\
&\le&2\frac{x^2-\ln(1+x^2)}{x^2}+2\frac{y^2-\sin^2y}{y^2}
&\to$0
\end{eqnarray*}
as $(x,y)\to(0,0)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Evaluate $\int \frac{dt}{(t^2-1)^2}$ I have this problem of evaluating the indefinite integral of a rational function
$$\int \frac{dt}{(t^2-1)^2}$$
and I'm a bit unsure about how to proceed. I could use partial fractions:
$$\frac{1}{(t+1)(t-1)(t+1)(t-1)}$$
$$\frac{1}{(t+1)^2(t-1)^2}$$
$$\frac{A}{t+1} + \frac{B}{t-1} + \frac{C}{t+1} + \frac{D}{(t+1)^2}$$
and go from there... but that seems long and complicated. Is there a trick I'm missing?
| You may find this helpful. Define
$$I_n=\int\frac{dx}{(ax^2+b)^{n+1}}$$
then integrate by parts with $dv=dx$:
$$I_n=\frac{x}{(ax^2+b)^{n+1}}+2(n+1)\int\frac{ax^2}{(ax^2+b)^{n+2}}dx$$
$$I_n=\frac{x}{(ax^2+b)^{n+1}}+2(n+1)\int\frac{ax^2+b}{(ax^2+b)^{n+2}}dx-2b(n+1)\int\frac{dx}{(ax^2+b)^{n+2}}$$
$$I_n=\frac{x}{(ax^2+b)^{n+1}}+2(n+1)I_{n}-2b(n+1)I_{n+1}$$
We then solve for $I_{n+1}$:
$$I_{n+1}=\frac{x}{2b(n+1)(ax^2+b)^{n+1}}+\frac{2n+1}{2b(n+1)}I_n$$
Then replace $n+1$ with $n$ to get
$$I_n=\frac{x}{2bn(ax^2+b)^n}+\frac{2n-1}{2bn}I_{n-1}$$
Which is a recurrence with the base case $$I_0=\frac1{\sqrt{ab}}\arctan\left[x\sqrt{\frac{a}b}\right]$$
Your integral is given by $n=1,\, a=1,\, b=-1$. For the base case, we have
$$I_0=\frac1{\sqrt{-1}}\arctan x\sqrt{-1}=-i\arctan ix=\operatorname{arctanh}x$$
Thus
$$I_1=\frac{x}{2(x^2-1)}+\operatorname{arctanh}x+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3119184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
} |
Finding all rational solutions to $y^2=x^3-432$ In the book "Elliptic curves, number theory and cryptography" by Washington in section 2.5.2 it says that it is somewhat not trivial to show that the only rational solutions to the curve $y^2=x^3-432$ are $(12,\pm36)$ and $\infty$. I am curious as to how one would show this.
| The following is a reduction to $4(x')^3 = (y')^2 + 27$ and $(x'')^3 = 3(y'')^2+16$ (these might be marginally easier?)
Edit: This is for integer solutions (sorry, I misread the question).
Write the equation as $x^3 = y^2 + 432 = (y - \sqrt{-423})(y+\sqrt{-432}) = (y - 12\sqrt{-3})(y + 12\sqrt{-3})$.
The ring of integers of $K = \mathbb{Q}(\sqrt{-3})$ is $\mathcal{O}_K = \mathbb{Z}\big[\frac{1+\sqrt{-3}}{2}\big]$ and it is a UFD.
A prime $\pi$ in $\mathcal{O}_K$ that divides both $y+12\sqrt{-3}$ and $y-12\sqrt{-3}$ divides their difference $24\sqrt{-3} = 2^3 \sqrt{-3}^3$.
So, up to units $\pi = 2$ or $\pi = \sqrt{-3}$ or $\pi$ does not exist.
If $\pi=2$ then $y$ is even, hence $x$ is even, hence $y^2$ is divisible by 8, hence $y$ is divisible by $4$, hence $x^3$ is divisible by $16$, hence $x$ is divisible by $4$. Substitutions $x=4x', y=4y'$ yield $4(x')^3 = (y')^2 + 27$. This has solutions $x'=3$ and $y'=\pm 9$; hence $x = 12$ and $y = \pm 36$.
If $\pi=\sqrt{-3}$ then $y$ is threeven, hence $y^2$ is divisible by $9$, hence $x$ is threeven, hence $y^2$ is divisible by $27$, hence $y$ is divisible by $9$. Substitutions $x=3x'', y=9y''$ yield $(x'')^3 = 3(y'')^2+16$. This has solutions $x''= 4$ and $y'' = \pm 4$; hence $x = 12$ and $y = \pm 36$ again.
If $\pi$ does not exist then the factors $y\pm 12\sqrt{-3} = (y\mp 12) \pm 24\big(\frac{1+\sqrt{-3}}{2}\big)$ are coprime and hence they are both cubes, say $$\begin{align*}(y-12) + 24\bigg(\frac{1+\sqrt{-3}}{2}\bigg) &= \bigg(a + b\frac{1+\sqrt{-3}}{2}\bigg)^3 \\ &= a^3 + 3ab^2\bigg(\frac{1+\sqrt{-3}}{2}\bigg)^2 + 3a^2b\bigg(\frac{1+\sqrt{-3}}{2}\bigg) + b^3\bigg(\frac{1+\sqrt{-3}}{2}\bigg)^3\\ &= a^3 + 3ab^2\bigg(\frac{1+\sqrt{-3}}{2} - 1\bigg) + 3a^2b\bigg(\frac{1+\sqrt{-3}}{2}\bigg) - b^3\\ &= (a^3 - 3ab^2 -b^3) + (3ab^2+3a^2b)\bigg(\frac{1+\sqrt{-3}}{2}\bigg)\end{align*}$$
This implies $3ab(a+b) = 24$, i.e. $ab(a+b) = 8$. Hence both $a$ and $b$ are $\pm$ powers of $2$, say $a = 2^n, b = 2^m$ with $n \leq m$ w.l.o.g. (the equation is symmetric, and $\pm$ won't affect the argument). Then $a+b = 2^n + 2^m = 2^n(1+2^{m-n})$ is also a power of $2$, so $m-n=0$, so $8 = 2^n \cdot 2^n \cdot 2 \cdot 2^n = 2^{3n + 1}$ is a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3119391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Dice Rolls Between Three Players The setup of the game involves three players and a fair, six-sided die. The rules are: If a player rolls a six, they win the game, if they roll an odd number, they pass the die to the player on their right, and if they roll an even number other than six, they pass the die to the player on their left.
The question: What is the probability that the first person to roll the die wins?
Attempt: Broke this up into different cases, suspecting a geometric distribution. Probability that person who rolled first wins on 1st roll is 1/6, probability of winning on the 2nd roll is 0, probability of winning on the 3rd roll is 1/18, and so on. However, I couldn't establish any clear geometric pattern, is this approach a dead end?
| Let $X$ be the event in which player $1$ wins the game, and $T$ denote the person whose turn it is to throw the die. We then find:
$$P(X | T = 1) = \frac{1}{6} + \frac{2}{6} P(X | T = 2) + \frac{3}{6} P(X | T = 3)$$
We know that:
$$P(X | T = 2) = \frac{3}{6} P(X | T = 1) + \frac{2}{6} P(X | T = 3)$$
$$P(X | T = 3) = \frac{2}{6} P(X | T = 1) + \frac{3}{6} P(X | T = 2)$$
$$= \frac{2}{6} P(X | T = 1) + \frac{3}{6} \left(\frac{3}{6} P(X | T = 1) + \frac{2}{6} P(X | T = 3)\right)$$
$$\iff P(X | T = 3) = \frac{21}{30} P(X | T = 1)$$
As such, it follows that:
$$P(X | T = 1) = \frac{1}{6} + \frac{2}{6} \left(\frac{3}{6} P(X | T = 1) + \frac{2}{6} P(X | T = 3)\right) + \frac{3}{6} P(X | T = 3)$$
$$= \frac{1}{6} + \frac{2}{6} \left(\frac{3}{6} P(X | T = 1) + \frac{2}{6} \frac{21}{30} P(X | T = 1)\right) + \frac{3}{6} \frac{21}{30} P(X | T = 1)$$
$$\iff P(X | T = 1) = \frac{30}{73} \approx 0.4110$$
This result can be confirmed using the following Python code:
from random import randint
s = [0, 0, 0]
n = 1000000
for _ in range(n):
p = 0
while True:
t = randint(1, 6)
if t == 6:
s[p % 3] += 1.0 / n
break
elif t == 2 or t == 4:
p += 2
else:
p += 1
print(s)
An example run returned:
[0.411106, 0.300798, 0.288096]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3121048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving that product of general matrices has small spectral radius In a Jacobi type of iteration for finding solution to a linear system $Ax=b$, one writes
$$x_i^{(k+1)} = Gx_i^{(k)}+c,$$
where $x_i$ is the $i$-th component of vector $x$ and $G=D^{-1}N$, $c = D^{-1}b$.
Here we take $A=D+N$, with $D$ the diagonal part of $A$ and $N$ the rest of $A$, i.e. $N = A-D$, so that
$$x_i^{(k+1)} = -D^{-1}Nx_i^{(k)}+D^{-1}b$$
There is a theorem which says that if the spectral radius $\rho(A)<1$ then this scheme converges, so I need to show that $D^{-1}N$ has spectral radius less than $1$.
In my case, $A=\begin{bmatrix}
-6&1 & 1 & 1 &0 & 0 & 0&\dots& 0 \\
1 & 1 & -6 & 1 & 1 & 1 &0&\dots & 0 \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots &\vdots&\ddots&0\\
\vdots & \vdots & \vdots & \vdots & \dots & \dots & \dots&-6 & 1\\
0 & \dots & 0 & 0 & 0 & 1 & 1 & 1 & -6
\end{bmatrix}$, so
$D=diag\{-6,\dots, -6\}$, $\implies D^{-1}=diag\{-1/6, \dots, -1/6\}$,
and $N=\begin{bmatrix}
0&1 & 1 & 1 &0 & 0 & 0&\dots& 0 \\
1 & 1 & 0 & 1 & 1 & 1 &0&\dots & 0 \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots &\vdots&\ddots&0\\
\vdots & \vdots & \vdots & \vdots & \dots & \dots & \dots&0 & 1\\
0 & \dots & 0 & 0 & 0 & 1 & 1 & 1 & 0
\end{bmatrix}$.
One can evaluate the product $D^{-1}N$ to be
$\begin{bmatrix}
0&-1/6 & -1/6 & -1/6 &-1/6 & -1/6 & -1/6&\dots& -1/6 \\
-1/6 & 0 & -1/6 & -1/6 & -1/6 & -1/6 &-1/6&\dots & -1/6 \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots &\vdots&\ddots&-1/6\\
\vdots & \vdots & \vdots & \vdots & \dots & \dots & \dots&0 & -1/6\\
-1/6 & \dots & -1/6 & -1/6 & -1/6 & -1/6 & -1/6 & -1/6 & 0
\end{bmatrix}$
It can be shown that the spectral radius of this product is $1/2$ (if I'm not mistaken). But is there a more elegant way to show that $\rho(D^{-1}N)<1$? I don't really like the direct way of showing this.
| The spectral radius of the last shown matrix for $D^{-1}N$ is $(n-1)/6$, where $n\times n$ is the size of the matrix. Thus the spectral radius is larger than $1$ for $n\ge8$.
However, from the previous parts of the question, $D^{-1}N=-\frac{1}{6}J$ where $$J=\begin{pmatrix}0&1&1&1&0&\cdots&0\\1&0&1&1&1&\cdots&0\\&\ddots&0&0&0&\cdots&1\end{pmatrix}$$ i.e., there is a band of 1s on first three off-diagonals on both sides.
For a symmetric matrix, the spectral radius is equal to the largest value of the numerical range, $\{x^*Ax:\|x\|_2=1\}$.
The extent of the numerical range of $J$ is given by maximizing the following expression for $x=(a_1,\ldots,a_n)$ unit, $$\begin{pmatrix}a_1,&\ldots,&a_n\end{pmatrix}\begin{pmatrix}0&1&\cdots&0\cr 1&0&\cdots&1\cr&\ddots\cr0&1&\cdots&0\end{pmatrix}\begin{pmatrix}a_1\cr\vdots\cr a_n\end{pmatrix}=\sum_{|i-j|\le3,i\ne j}a_ia_j$$ Now
$$\begin{align*}\sum_{|i-j|\le3,i\ne j}a_ia_j
&=2(\sum_{i=1}^{n-1}a_ia_{i+1}+\sum_{i=1}^{n-2}a_ia_{i+2}+\sum_{i=1}^{n-3}a_ia_{i+3})\\
&\le \sum_{i=1}^{n-1}a_i^2+\sum_{i=2}^na_i^2+\sum_{i=1}^{n-2}a_i^2+\sum_{i=3}^na_i^2+\sum_{i=1}^{n-3}a_i^2+\sum_{i=4}^na_i^2\\
&<6\sum_{i=1}^na_i^2=6\end{align*}$$
That is, the spectral radius of this $D^{-1}N$ is less than 1, as required.
| {
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"url": "https://math.stackexchange.com/questions/3121119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
show this inequality $\sum\frac{x}{2+xy}\ge\frac{1}{2}$
Let $x,y,z\ge 0$ such that $$x+y^2+z^3=1.$$
Show that
$$\dfrac{x}{2+xy}+\dfrac{y}{2+yz}+\dfrac{z}{2+zx}\ge\dfrac{1}{2}$$
I try do
$$\sum_{cyc}\dfrac{x}{2+xy}=\sum_{cyc}\dfrac{x^2}{2x+x^2y}\ge\dfrac{(x+y+z)^2}{(2x+2y+2z)+(x^2y+y^2z+z^2x)}$$
it have to prove
$$2(x+y+z)^2\ge (2x+2y+2z)+(x^2y+y^2z+z^2x)$$
it seem this is hold,But I can't prove it
| Now, you can use the Contradiction method.
Indeed, since $$xy+xz+yz\geq x^2y+y^2z+z^2x,$$ it's enough to prove that
$$\sum_{cyc}(2x^2+3xy)\geq2(x+y+z).$$
Let $\sum\limits_{cyc}(2x^2+3xy)<2(x+y+z),$ $x=ka$, $y=kb$ and $z=kc$ such that $k>0$ and
$$\sum_{cyc}(2a^2+3ab)=2(a+b+c).$$
Thus, $$k\sum_{cyc}(2a^2+3ab)<2(a+b+c)=\sum_{cyc}(2a^2+3ab),$$ which gives $0<k<1$ and
$$1=x+y^2+z^3=ka+k^2b^2+k^3c^3<a+b^2+c^3,$$ which is a contradiction because we'll prove now that
$$a+b^2+c^3\leq1.$$
Indeed, since $$\frac{\sum\limits_{cyc}(2a^2+3ab)}{2(a+b+c)}=1,$$ we need to prove that
$$a\left(\frac{\sum\limits_{cyc}(2a^2+3ab)}{2(a+b+c)}\right)^2+b^2\left(\frac{\sum\limits_{cyc}(2a^2+3ab)}{2(a+b+c)}\right)+c^3\leq\left(\frac{\sum\limits_{cyc}(2a^2+3ab)}{2(a+b+c)}\right)^3$$ or
$$\left(\sum_{cyc}(2a^2+3ab)\right)^2\geq$$
$$\geq8c^3(a+b+c)^3+4b^2(a+b+c)^2\sum_{cyc}(2a^2+3ab)+2a(a+b+c)\left(\sum_{cyc}(2a^2+3ab)\right)^2$$ or
$$4(a^5b+a^5c+2b^5c+c^5a+3c^5b)+12a^4b^2+20a^4c^2+6b^4a^2+38b^4c^2+22c^4a^2+46c^4b^2+$$
$$+13a^3b^3+33a^3c^3+63b^3c^3+abc(36a^3+36b^3+64c^3+79a^2b+107a^2c+77b^2a+127b^2c+129c^2a+151c^2b+178abc)\geq0,$$ which is obvious.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do I find the distance from a point to a plane? I am trying to find the distance from point $(8, 0, -6)$ and plane $x+y+z = 6$. I tried solving it but I am still getting it wrong. Can anyone help me on this? Any help I would very much appreciate.
The following is my work:
$$d = \sqrt{(x-8)^2 + (y-0)^2 + (z+6)^2}$$
since $x+y+z = 6$, $z = 6-x-y$, so
\begin{align*}
d &= \sqrt{(x-8)^2 + (y-0)^2 + (-x-y+12)^2} \\
d^2 &= (x-8)^2 + (y-0)^2 + (-x-y)^2
\end{align*}
Find partial derivative $f_x$ and $f_y$ and critical points
\begin{align*}
f_x &= 2(x-8) + 2(-x-y+12) \\
&= 24-2y \quad (\text{set }= 0) \\
&= \text{critical point }y = 4 \\
f_y &= 2y + 2(-x-y+12) \\
&= 24 - 2x \quad (\text{set }= 0) \\
&= \text{critical point }x = 12 \\
\end{align*}
Plug in $x = 12$ and $y = 4$ to original equation
$$d = \sqrt{(x - 8)^2 + (4)^2 + (-12-4+12)^2} = \sqrt{48}$$
| It's $$\frac{|8+0-6-6|}{\sqrt{1^2+1^2+1^2}}=\frac{4}{\sqrt3}.$$
I used the following formula.
The distance from the point $(x_1,y_1,z_1)$ to the plane $ax+by+cz+d=0$ it's
$$\frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}.$$
Also, we can end your way.
Indeed, by C-S
$$\sqrt{(x-8)^2+y^2+(z+6)^2}=\frac{1}{\sqrt3}\sqrt{(1^2+1^2+1^2)((x-8)^2+y^2+(z+6)^2)}\geq$$
$$\geq\frac{1}{\sqrt3}\sqrt{(x-8+y+z+6)^2}=\frac{4}{\sqrt3}$$ and we got the distance again.
| {
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"source": "stackexchange",
"question_score": "2",
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Different condition numbers of $\begin{pmatrix} a & b \\ b & c \end{pmatrix}$
Let $a,b,c \in \mathbb{R}$ and $A := \begin{pmatrix} a & b \\ b & c \end{pmatrix}$ and $\det(A) \neq 0$.
Find the condition number with respect to the 1-, 2- and $\infty$-norm and discuss for which relationships among $a,b,c$ the corresponding linear systems are well conditioned.
What I've done so far
We have $A^{-1} = \frac{1}{ac - b^2}\begin{pmatrix} c & -b \\ - b & a\end{pmatrix}$ and, therefore,
\begin{align}
\kappa_1(A)
& = \| A \|_1 \| A^{-1} \|_1
= \max\{a + b, b + c\} \cdot\max\left\{\frac{c - b}{ac - b^2}, \frac{a - b}{ac - b^2} \right\} \\
& = \big( b + \max(a, c)\big) \cdot \frac{1}{ac - b^2}\big( \max(a, c) - b\big) \\
& = \frac{b^2 - (\max(a, c))^2}{b^2 - ac}
\end{align}
Note: The above steps only work for $\det(A) > 0$ and $a,c > b$ and are the same for the $\| \cdot \|_1$-norm, since the matrix is symmetric.
Now, I'm not sure to find out for which values of $a,b,c$ this would be small (other than that for $a = c$, it's 1 and for $a = b = c$ and $b = a \ge c$ it's undefined) and especially not how to find this out which out plotting it with a computer.
While plotting I noticed that when |b| gets large, the term is very close to one for a sufficiently large plane around zero (i.e for $|b| = 60$ the term is very similar to 1 for all $(x,y) \in [-25,25]^2$).
For $a = -c$, the term becomes $\frac{b^2 - c^2}{b^2 + c^2} = \frac{2 b^2}{b^2 + c^2} - 1$, because $\max(-c,c) = |c|$.
Also, you can use
\begin{align}\tag{$\star$} \label{eq:star}
\max(x,y) = \frac{x + y + |x - y|}{2}
\end{align} to obtain
$$
\kappa_1(A)
= \frac{1}{2}\big( 2b + a + c + |a - c|\big)
\left( \frac{a - b}{ac - b^2} + \left| \frac{2c - a - b}{ac - b^2} \right| \right)
$$
Similar problems arise for the next norm:
We have $\kappa_2 = \sqrt{\rho(A^T A) \rho((A A^T)^{-1})}$ and $A^T A = A A^T = A^2$.
Now, we have
\begin{gather}
A^2 = \begin{pmatrix} a^2 + b^2 & b(a + c) \\ b(a + c) & b^2 + c^2 \end{pmatrix} \\
(A A^T)^{-1} = (A^{2})^{-1}
= \frac{1}{(b^2 - ac)^2} \begin{pmatrix} b^2 + c^2 & -b(a + c) \\ -b(a + c) & a^2 + b^2 \end{pmatrix},
\end{gather}
where $\rho(B)$ is the largest eigenvalue of $B$.
Now we obtain the eigenvalues by finding roots of the [characteristic polynomial][1]:
\begin{gather*}
\det(A^2 - t \cdot I_2)
= (a^2 + b^2 - t)(b^2 + c^2 - t) - b^2(a + c)
\overset{!}{=} 0 \\
\implies t_{1,2} = \frac{a^2 + 2b^2 + c^2 \pm(a + c) \sqrt{(a - c)^2 + 4b^2}}{2}
\end{gather*}
and, [similarly][2]
$$
\rho(A^{-2})
= \max\left\{ \frac{a^2 + 2b^2 + c^2 + \pm (a + c) \sqrt{(a - c)^2 + 4b^2}}{2(b^2 - ac)^2} \right\}
$$
So,
\begin{align}
\kappa_2(A)
& = \sqrt{\max\left\{\frac{a^2 + 2b^2 + c^2 \pm(a + c) \sqrt{(a - c)^2 + 4b^2}}{2} \right\} \cdot \max\left\{ \frac{a^2 + 2b^2 + c^2 \pm (a + c) \sqrt{(a - c)^2 + 4b^2}}{2(b^2 - ac)^2} \right\} } \\
& = \frac{1}{| b^2 - ac |} \max\left\{ \frac{a^2 + 2b^2 + c^2 \pm (a + c) \sqrt{(a - c)^2 + 4b^2}}{2} \right\} \\
& \overset{\eqref{eq:star}}{=} \frac{a^2 + 2b^2 + c^2 + |a + c|\sqrt{(a - c)^2 + 4b^2}}{2| b^2 - ac |}
\end{align}
How do I simplify this result?
| This is not the complete analysis that the instructor requested, but a simplification which can be used to verify certain aspects of the solution.
It is important to note that a matrix is either well-conditioned or it is ill-conditioned. In particular, it does not matter which norm we use to compute the condition number. This follows from the fact that all induced matrix norms are equivalent.
The 2-norm condition number of the matrix $A$ can be computed from the eigenvalues of the matrix $A^TA$. Specifically, we have $$\kappa_2(A) =\sqrt{\frac{\lambda_{\max}(A^TA)}{\lambda_{\min}(A^TA)}}.$$ We have already seen that $$A^TA = \begin{bmatrix} a^2+b^2 & b(a+c) \\ b(a+c) & a^2 + c^2 \end{bmatrix}.$$ Now by Gershgorin's circle theorem and fact that $A^TA$ is symmetric positive semidefinite, the eigenvalues of $A^TA$ can be found in the interval $$I = [a^2+b^2 - |b(a+c)|,a^2+b^2 + |b(a+c)|] \cap [0, \infty).$$ We observe that $0$ can be an eigenvalue only when $$a^2 + b^2 - |b(a+c)| \leq 0.$$ If $a^2 + b^2 - |b(a+c)| > 0$, then the condition number is bounded by
$$ \kappa_2(A) = \sqrt{\frac{a^2 + b^2 + |b(a+c)|}{a^2 + b^2 - |b(a+c)|}} = \sqrt{\frac{ 1 + \frac{|b(a+c)|}{a^2+b^2}}{1 - \frac{|b(a+c)|}{a^2+b^2}}} $$
We conclude that the matrix $A$ is necessarily well-conditioned when
$$ |b(a+c)| \ll a^2+b^2$$ and that it can be ill-conditioned when $$|b(a+c)| \geq a^2+b^2$$
A few sanity check are in order. If $c=-a \not = 0$, then our matrix is orthogonal and perfectly conditioned. If $a=b=c$, then our matrix is singular.
It is possible to say with certainty that the matrix is ill-conditioned when $$|b| < \min\{|a|,|c|\} \ll \max\{|a|,|c|\}.$$ In this situation, Gershgorin's circles for $A$ are disjoint and it follows that one eigenvalue of $A$ is substantially smaller (in absolute value) than the other. This in turn implies that $A$ is ill-conditioned.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3125523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Functional equation to determine $f(-1)$
There's a function $f$ satisfying:
$$f\left(\frac{1}{1−x}\right)+2\cdot f\left(\frac{x−1}{x}\right)=3x$$
Find the value of $f(−1)$.
I have no idea how to solve this one. If anyone could help me to understand it, I would be grateful.
| set $x = -1$ into to the formula of $f$:
$$f\left(\frac{1}{1−x}\right)+2\cdot f\left(\frac{x−1}{x}\right)=3x $$
you'll have for $x = -1$:
$$f\left(\frac{1}{1−(-1)}\right)+2\cdot f\left(\frac{(-1)−1}{(-1)}\right)=-3$$
$$ = f\left(\frac{1}{2}\right)+2\cdot f\left(2\right)=-3$$
Notice that the denominator of is different then $0$
and for $x = \frac{1}{2}$:
$$f\left(2\right)+2\cdot f\left(-1\right)=1.5 $$
for $x=2$:
$$f(-1)+2\cdot f\left(\frac{1}{2} \right)= 6 $$
Then:
$f(-1) = 6 - 2f\left(\frac{1}{2} \right) = \frac{-f(2)}{2} + \frac{3}{4}$ ,
as $ \ -2f\left(\frac{1}{2} \right) = 4f(2) + 6 $
Therefore $f(2) = -2.5 \ \Longrightarrow \ f(-1) = \frac{-2.5}{-2} + \frac{3}{4} = 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3126149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Displaying $ |z+4|=3|z|$ in an Argand diagram Identify in an Argand diagram the points corresponding to:
$$|z+4|=3|z|$$
I am unsure how to approach this question and would appreciate any help anyone can provide with the answer and the intuition behind the method. Thanks.
| Hint: By substituting $z=x+iy$ we get
\begin{align}
|x+4+yi|=3|x+yi|
& \implies \sqrt{(x+4)^2+y^2}=3\sqrt{x^2+y^2} \\
& \implies (x + 4)^2 + y^2 = 9(x^2 + y^2) \\
& \implies (x + 4)^2 - 9x^2 = 8y^2 \\
& \implies y = \pm \sqrt{\frac{(x + 4)^2 - 9x^2}{8}}
=\pm \frac{\sqrt{-8x^2 + 16x + 16}}{2 \sqrt{2}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3126508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find the integral $\int \frac{(\ln(x))^2}{x^3} \, dx$ $$\int \frac{(\ln(x))^2}{x^3} \, dx $$
Starting off with Integration by Parts
$$
\begin{align}
u = \ln(x)^2 &~~~ dv = x^{-3} \\\\
du = 2\ln(x)dx &~~~ v = \frac{x^{-2}}{-2}
\end{align}
$$
$$
\begin{align}
\int \frac{(\ln(x))^2}{x^3} &= (\ln(x))^2 \left( \frac{-x^{-2}}{2} \right) -\frac{2}{2} \int \frac{\ln(x)}{x^2}
\end{align}
$$
Integration by parts again...
$$
\begin{align}
u = \ln(x) &\hspace{10mm} dv = \frac{1}{x^2} \\\\
du = \frac{1}{x}dx &\hspace{10mm} v = -\frac{1}{x} \\\\
-\frac{2}{2} \int \frac{\ln(x)}{x^2} &= -\ln(x) \left(\frac{1}{x} \right) + \int \frac{1}{x^2}dx \\\\
\end{align}
$$
Integration by parts several times $\int \frac{1}{x^2}dx \rightarrow \int \frac{1}{x} \rightarrow \ln(x) + C$
Combining everything together, I get
$$
\int \frac{(\ln(x))^2}{x^3} = (\ln(x))^2 \left( \frac{-x^{-2}}{2} \right) -\ln(x) \left(\frac{1}{x} \right) - \ln(x) + C\\\\
= -\frac{(\ln(x))^2}{2x^2} - \frac{\ln(x)}{x} - \ln(x) + C
$$
Is there a shorter way I could have done this? I think I got the right answer, but I am not really sure either. Checking my answer via differentiate doesn't seem like a feasible test strategy and even without time constraints still seems too complex for my level right now (but then maybe this is why I need the practice)
| $$\begin{align}
\int \frac{(\ln(x))^2}{x^3}dx & = \bigg((\ln(x))^2\times \frac{x^{-2}}{-2}\bigg)-\int\frac{2\ln(x)}{x}\times \frac{x^{-2}}{-2}dx \\
& =\bigg((\ln(x))^2\times \frac{x^{-2}}{-2}\bigg)+\int \ln(x)\times x^{-3}dx \\
& =\bigg((\ln(x))^2\times \frac{x^{-2}}{-2}\bigg)+\bigg(\ln(x)\times \frac{x^{-2}}{-2}\bigg)-\int\frac{1}{x}\times\frac{x^{-2}}{-2}dx \\
& =\bigg((\ln(x))^2\times \frac{x^{-2}}{-2}\bigg)+\bigg(\ln(x)\times \frac{x^{-2}}{-2}\bigg)+\frac{1}{2}\bigg(\frac{x^{-2}}{-2}\bigg)+C \\
& =\bigg(\frac{x^{-2}}{-2}\bigg)\bigg((\ln(x))^2+\ln(x)+\frac{1}{2}\bigg)+C
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3128043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
3-variable symmetric inequality Given $a,b,c>0$ satisfying $a^2+b^2+c^2=3$. Prove that
$$2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)+3(a+b+c)\geq 15.$$
I've tried to use the inequality $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\geq \dfrac{9}{a+b+c}$ as well as AM-HM and the condition $a+b+c \leq 3$ but it still doesn't work.
| It's obviously true by $uvw$, but the Tangent Line method also gives a simple solution:
$$\sum_{cyc}\left(\frac{2}{a}+3a-5\right)=\sum_{cyc}\frac{(a-1)(3a-2)}{a}=$$
$$=\sum_{cyc}\left(\frac{(a-1)(3a-2)}{a}-\frac{a^2-1}{2}\right)=\sum_{cyc}\frac{(a-1)^2(4-a)}{2a}\geq0.$$
The last inequality is true because $3=a^2+b^2+c^2>a^2,$ which gives $0<a<\sqrt3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3130362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What is the integral of $\frac{2x+1}{x^2+1}$ I know the integral can be found by writing it as $$\frac{1}{x^2+1} +\frac{2x}{x^2+1}$$ This gives us an answer of $$\ln(x^2+4) +\frac{1}{2} \tan^{-1} \frac{x}{2}$$ However, when I let $x=2\tan u$, I get
$$\frac{1}{2} \tan^{-1} \frac{x}{2} + 2\ln\left|\text{sec(tan}^{-1} \frac{x}{2})\right|$$ This can be simplified and will give the same results for large values of $x$ but not smaller values. What am I doing wrong. Sorry for the format, I am still trying to figure out Latex.
| You can use a trigonometric substitution if you want, although separating the integrand is probably easier. Since $\tan^2 u+1 = \sec^2 u$, though, try $x = \tan u$. Then $dx = \sec^2u\,du$, and the integral becomes
$$\int \frac{2x+1}{x^2+1}\,dx = \int\frac{2\tan u+1}{\sec^2 u}\cdot \sec^2 u\,du
= \int (2\tan u+1)\,du
= u - \ln|\cos u|+C.
$$
Resubstituting gives
$$\tan^{-1}x - \ln|\cos(\tan^{-1}x)|+C = \tan^{-1}x + \ln\sqrt{x^2+1}+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3130853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Determine all positive integer solutions for $\frac 1x + \frac 1y + \frac 1z + \frac 1{xy} + \frac1{yz} + \frac 1{xz} + \frac 1{xyz} = 1$ I need to determine all positive integer solutions for the equation:
$$\frac 1x + \frac 1y + \frac 1z + \frac 1{xy} + \frac1{yz} + \frac 1{xz} + \frac 1{xyz} = 1.$$
This is how I have tried to do it:
Mulitiplied both sides by $xyz$ to get
$$yz+xz+xy+z+x+y+1=xyz.$$
Factor it:
\begin{align}
x(y+z+1-yz)+yz+y+z &= -1 \\
x(y(1-z)+z+1)+y(1+z)+z &= -1
\end{align}
If $z=0$, we get
\begin{align}
x(y+1)+y=-1 &\iff xy+x+y=-1 \\
&\iff (x+1)(y+1)=0,
\end{align}
which gives us $x=y=-1$.
Is this all positive integer solutions? Or have I missed something?
EDIT: I am stupid.
New attempt.
If $z=1$, I get $2x+2y=-2 \iff x+y=-1$.
Then there is no solutions of positive integers for both $x$ and $y$ at the same time.
If I try for $z=2$, I get
\begin{align}
x(y(1-2)+2+1)+y(1+2)+2=-1 &\iff x(3-y)+3y+2=-1 \\
&\iff 3x+3y+3-xy=0
\end{align}
and I won't get a solution where all the variables are positive integers.
| Let $x\geq y\geq z$.
Thus, $$1\leq\frac{1}{z^3}+\frac{3}{z^2}+\frac{3}{z}$$ or
$$z^3\leq1+3z+3z^2$$ and we got some values of $z$:
$$1\leq z\leq3.$$
Now, for all value of $z$ we can make the similar thing with $y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3131265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
$\frac{a}{a^a+1}+\frac{b}{b^b+1}+\frac{c}{c^c+1}\leq \frac{3}{2}$ with $abc=1$
Let $a,b,c>0$ such that $abc=1$ then we have :
$$\frac{a}{a^a+1}+\frac{b}{b^b+1}+\frac{c}{c^c+1}\leq \frac{3}{2}$$
My try :
The original inequality is equivalent to :
$$a(b^b+1)(c^c+1)+b(a^a+1)(c^c+1)+c(a^a+1)(b^b+1)\leq \frac{3}{2}(a^a+1)(c^c+1)(b^b+1)$$
Or :
$$(a-\frac{3}{2}(a^a+1))(b^b+1)(c^c+1)+(b-\frac{3}{2}(b^b+1))(a^a+1)(c^c+1)+(c-\frac{3}{2}(c^c+1))(a^a+1)(b^b+1)\leq0$$
The function :
$$f(x)=x-\frac{3}{2}(x^{x}+1)$$
Is concave so we can apply Jensen's inequality but it's very ugly !
So have you an alternative proof ?
Thanks in advance for your time .
| We can use also the following way.
Let $a=\sqrt{\frac{x}{y}}$ and $b=\sqrt{\frac{y}{z}}$, where $x$, $y$ and $z$ are positives.
Thus, $c=\sqrt{\frac{z}{x}}$ and by AM-GM and C-S we obtain:
$$\sum_{cyc}\frac{a}{a^a+1}\leq\sum_{cyc}\frac{a}{\frac{a^2+1}{2}+1}\leq\sum_{cyc}\frac{a}{2\sqrt{\frac{a^2+1}{2}\cdot1}}=\frac{1}{\sqrt2}\sum_{cyc}\sqrt{\frac{x}{x+y}}\leq$$
$$\leq\frac{1}{\sqrt2}\sqrt{\sum_{cyc}\frac{x}{(x+y)(x+z)}\sum_{cyc}(x+z)}=\sqrt{\frac{2(xy+xz+yz)(x+y+z)}{(x+y)(x+z)(y+z)}}.$$
Id est, it's enough to prove that
$$\frac{2(xy+xz+yz)(x+y+z)}{(x+y)(x+z)(y+z)}\leq\frac{9}{4}$$ or
$$\sum_{cyc}z(x-y)^2\geq0.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3131403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Function $f$ such that $f(x+a) = \frac12 + \sqrt {f(x)-(f(x))^2}$ is periodic.
Let $f$ be a real valued function defined for all real numbers $x$ such that for some positive constant $a$ the equation $f(x+a) = \frac12 + \sqrt {f(x)-(f(x))^2}$ holds for all $x$. Prove the function $f$ is periodic.
I have tried by replacing $x$ several times but couldn't find the period.
| Note that $f(x)-f^2(x)\ge 0$ and thus $0\leq f(x)\leq 1$. Also $f(x)-\frac{1}{2}\ \ge 0$ which imply $\frac{1}{2}\le f(x)\le 1 $.
Now we compute $f(x+2a).$
$$f(x+2a)=\frac{1}{2}+\sqrt{f(x+a)-f^2(x+a)}\,.$$
$$f(x+a)-f^2(x+a)=f(x+a)-\frac{1}{4}-\sqrt{f(x)-f^2(x)}- f(x)+f^2(x)=$$
$$=f^2(x)-f(x)+\frac{1}{4}.$$
$$f(x+a)-f^2(x+a) =\left(\frac{1}{2}-f(x)\right)^2.$$ and then:
$$f(x+2a)=\frac{1}{2}+|\frac{1}{2}-f(x)|=$$
$$=\frac{1}{2}-\frac{1}{2}+f(x)$$ since $f(x)-\frac{1}{2}\ge 0.$
$$\text {Therefore }\;f(x+2a)=f(x).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3133226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Monotony of the operator exponential $e^A$ I have read somewhere that if $A$ and $B$ are two self-adjoint operators such that $A\geq B$, then we need not have $e^A\geq e^B$. I have tried to find counterexamples using $2\times2$ matrices as well as some online calculators, but without success! I also know that we should avoid commuting $A$ and $B$ for a possible counterexample?
My query is a simple counterexample and whether there is a difference between the finite and the infinite dimensional cases?
Thanks!
Math.
| I am assuming that you define the relation $\geq$ by $\left( X\geq Y\right)
\Longleftrightarrow\left( X-Y\text{ is nonnegative semidefinite}\right) $.
Let $C=\left(
\begin{array}
[c]{cc}
1 & 1\\
1 & 1
\end{array}
\right) $ and $B=\left(
\begin{array}
[c]{cc}
0 & 0\\
0 & 2
\end{array}
\right) $. Set $A=B+C$. Then, $A\geq B$, since $A-B=C$ is nonnegative
semidefinite. (Also, $A$, $B$ and $C$ are nonnegative semidefinite.) But
\begin{equation}
e^{A}-e^{B}=\left(
\begin{array}
[c]{cc}
e^{2-\sqrt{2}}\left( \dfrac{1}{4}\sqrt{2}+\dfrac{1}{2}\right) -e^{\sqrt
{2}+2}\left( \dfrac{1}{4}\sqrt{2}-\dfrac{1}{2}\right) -1 & -e^{\sqrt{2}
+2}\left( \sqrt{2}+1\right) \left( \dfrac{1}{4}\sqrt{2}-\dfrac{1}
{2}\right) -e^{2-\sqrt{2}}\left( \sqrt{2}-1\right) \left( \dfrac{1}
{4}\sqrt{2}+\dfrac{1}{2}\right) \\
\dfrac{1}{4}\sqrt{2}e^{\sqrt{2}+2}-\dfrac{1}{4}\sqrt{2}e^{2-\sqrt{2}} &
\dfrac{1}{4}\sqrt{2}e^{\sqrt{2}+2}\left( \sqrt{2}+1\right) -e^{2}+\dfrac
{1}{4}\sqrt{2}e^{2-\sqrt{2}}\left( \sqrt{2}-1\right)
\end{array}
\right)
\end{equation}
has determinant
\begin{align}
-\dfrac{1}{4}\left( e^{\sqrt{2}}-1\right) \dfrac{e^{2}\left( \sqrt
{2}-2\right) -e^{4}\left( \sqrt{2}+2\right) +e^{2}e^{\sqrt{2}}\left(
\sqrt{2}+2\right) -e^{4}e^{\sqrt{2}}\left( \sqrt{2}-2\right) }{e^{\sqrt{2}
}}\approx-8.436 < 0
\end{align}
and thus fails to be nonnegative semidefinite; thus, $e^{A}\geq e^{B}$ does
not hold.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3133899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Conditional Probability - A die is tossed 5 times. A die is tossed 5 times. Given that the die falls six at least once, what's the probability it falls six at least twice?
Let $A$ be the event the die falls six at least once.
$P(A) = 1 - (\frac{5}{6})^{5}$.
Let $B$ be the event the die falls six at least twice.
We want to find $$P(B | A) = \frac{P(B \cap A)}{P(A)}$$
How do I find $P(B\cap A)$?
EDIT:
$$\begin{align} P(B) & = P(A \cap B). \\ & =1 -\left [ \binom{5}{0} \left (\frac{1} {6} \right)^{0}\left (\frac{5}{6} \right)^{5} + \binom{5}{1}\left (\frac{1}{6} \right )^{1} \left (\frac{5}{6} \right)^{4} \right ]. \\ & = 1 - 2\times \bigg(\frac{5}{6}\bigg)^{5}. \\ & = 1 - .803755 \cdots \\ & = 0.196244856 \cdots \end{align}$$ is approx 19%.
| Let $X$ be a random variable "number of sixes on 5 throws" then $X \sim Bin(5,1/6)$
Then the question reduces to:
$P(X \geq 2 | X \geq 1) = \frac {P ( X \geq 2 \cap X \geq 1)}{P(X \geq 1)}$
As you said the intersection $X \geq 2 \cap X \geq 1$ is $X \geq 2$
So $P(X \geq 2 | X \geq 1) = \frac {P ( X \geq 2 )}{P(X \geq 1)}$
$P ( X \geq 2 ) = 1 - P(X=1) - P(X=0)$
and
$P ( X \geq 1 ) = 1 - P(X=0)$
$P(X=1) = 5 \frac{1}{6} (\frac{5}{6})^4 = (\frac{5}{6})^5$
$P(X=0) = (\frac{5}{6})^5$
and
$P(X \geq 2 | X \geq 1) = \frac {1-2 (\frac{5}{6})^5}{1-(\frac{5}{6})^5}$
$P(X \geq 2 | X \geq 1) = 0.328101 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3134946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
maximum value of $\sum ab-2abc$ If $a+b+c=1$ and $a,b,c\in(0,1)$, then what is the maximum value of $(ab+bc+ca-2abc)$?
What I've tried:
$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\geq 4(ab+bc+ca)$
$(a-b)^2=a^2+b^2-2ab\geq 0$
$a^2+b^2\geq 2ab,b^2+c^2\geq 2bc,c^2+a^2\geq 2ca$
$ab+bc+ca\leq\frac14$
How do I solve it help me please.
| Fixing the value of $a$, we want to find the maximum of
$$
S=ab+bc+ca-2abc=(1-2a)bc+a(1-a).
$$ For $a>\frac 12$, we want to choose $b,c$ with $b+c=1-a$ that makes $bc$ as small as possible. The optimum can be achieved as one of $b$ and $c$ goes arbitrarily close to $0$, but this contradicts $b,c>0$ (But we need to check this case to make sure that $S$ indeed achieves it maximum in the interior). On the other hand, for $a\le \frac12$, we want to choose $b,c$ that makes $bc$ as large as possible. Given that $b+c=1-a$, the maximum value of $bc$ is given by AM-GM; $2\sqrt{bc}\le b+c=1-a\implies bc\le \frac{(1-a)^2}4$ with the equality attained when $b=c=\frac{1-a}2$. Inserting this, we have
$$
S=\frac{(1-2a)(1-a)^2}4+a(1-a)=\frac{(1-a)(2a^2+a+1)}{4}.
$$ By differentiating $S$ with respect to $a$, we have
$$
S'=\frac{a(1-3a)}{2}.
$$ Since $S'>0$ on $(0,\frac13)$ and $S'<0$ on $(\frac13,1)$, we know that $a=\frac 13$, $b=c=\frac{1-a}2=\frac 13$ is optimal. This gives $S\le \frac13-\frac2{27}=\frac7{27}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3136239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Antiderivative of $\frac{(x+2)^2}{\sqrt{1-\frac{1}{x^2}}}$ I am struggling to find the general antiderivative of the following function $f$:
$$f(x)=\frac{(x+2)^2}{\sqrt{1-\frac{1}{x^2}}}$$
I am trying to find the improper integral from $1$ to $2$. I’m aware that the function is undefined at $1$ and I know the method to carry out. I am just unsure on finding the general antiderivative of $f$.
I tried substitution of $1/x$ but I was unable to proceed further.
Can anyone help – I’ve spent a few days on this problem and don’t seem to be getting anywhere. Thanks.
| $$\frac{(x+2)^2}{\sqrt{1-\dfrac1{x^2}}}=\frac{(x+4)(x^2-1)+5x+4}{\sqrt{x^2-1}}=x\sqrt{x^2-1}+4\sqrt{x^2-1}+\frac{5x}{\sqrt{x^2-1}}+\frac4{\sqrt{x^2-1}}.$$
Among these four terms, three have a quasi-immediate antiderivative,
$$I=\frac13(x^2-1)^{3/2}+4\int\sqrt{x^2-1}\,dx+5\sqrt{x^2-1}+4\text{ arcosh}(x).$$
Now by parts,
$$J=\int\sqrt{x^2-1}\,dx=x\sqrt{x^2-1}-\int\frac{x^2}{\sqrt{x^2-1}}dx=x\sqrt{x^2-1}-\int\frac{x^2-1+1}{\sqrt{x^2-1}}dx
\\=x\sqrt{x^2-1}-\text{arcosh}(x)-J.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3141266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $\lim\limits_{n\to\infty}\frac{a_1+a_2+...+a_n}{1+\frac{1}{\sqrt2}+...+\frac{1}{\sqrt{n}}}$ with $a_1=1$ and $a_{n+1}=\frac{1+a_n}{\sqrt{n+1}}$
Let $(a_n)_{n\ge1}, a_1=1, a_{n+1}=\frac{1+a_n}{\sqrt{n+1}}$.
Find $$\lim_{n\to\infty} \frac{a_1+a_2+\cdots+a_n}{1+\frac{1}{\sqrt2}+\cdots+\frac{1}{\sqrt{n}}}$$
These is my try:
I intercalated the limit like that
$$L=\lim_{n\to\infty} \frac{a_1+a_2+\cdots+a_n}{\sqrt{n+1}}\frac{\sqrt{n+1}}{1+\frac{1}{\sqrt2}+\cdots+\frac{1}{\sqrt{n}}}$$.
The second term of the limit tends to 2.
The first one, after Cesaro-Stols, become:
$$\lim_{n\to\infty}a_{n+1}(\sqrt{n+1}+\sqrt{n+2})$$
I tried to intercalate the term $a_n$ between 2 terms in function of n, just like $a_n<\frac{1}{\sqrt{n}}$ or something like that to use the sandwich theorem. Any ideas of this kind of terms? Or other ideas for the problem?
| Stolz–Cesàro is a way to go, but applied to
$S_n=\sum\limits_{k=1}^n a_n$ and $T_n=\sum\limits_{k=1}^n \frac{1}{\sqrt{k}}$, where $T_n$ is strictly monotone and divergent sequence ($T_n >\sqrt{n}$). Then
$$\lim\limits_{n\rightarrow\infty}\frac{S_{n+1}-S_n}{T_{n+1}-T_n}=
\lim\limits_{n\rightarrow\infty}\frac{a_{n+1}}{\frac{1}{\sqrt{n+1}}}=
\lim\limits_{n\rightarrow\infty} \left(1+a_n\right)=\\
\lim\limits_{n\rightarrow\infty} \left(1+\frac{1+a_{n-1}}{\sqrt{n}}\right)=
\lim\limits_{n\rightarrow\infty} \left(1+\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n(n-1)}}+\frac{a_{n-2}}{\sqrt{n(n-1)}}\right)=\\
\lim\limits_{n\rightarrow\infty} \left(1+\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n(n-1)}}+\frac{1}{\sqrt{n(n-1)(n-2)}}+...+\frac{a_1}{\sqrt{n!}}\right)=\\
1+\lim\limits_{n\rightarrow\infty} \left(\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{n-1}}+\frac{1}{\sqrt{(n-1)(n-2)}}+...+\frac{1}{\sqrt{(n-1)!}}\right)\right)$$
Now, for
$$\lim\limits_{n\rightarrow\infty} \left(\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{n-1}}+\frac{1}{\sqrt{(n-1)(n-2)}}+...+\frac{1}{\sqrt{(n-1)!}}\right)\right) \tag{1}$$
we have
$$0<\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{n-1}}+\frac{1}{\sqrt{(n-1)(n-2)}}+\frac{1}{\sqrt{(n-1)(n-2)(n-3)}}+...+\frac{1}{\sqrt{(n-1)!}}\right)<
\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{n-1}}+\frac{1}{\sqrt{(n-1)(n-2)}}+\frac{1}{\sqrt{(n-1)(n-2)}}+...+\frac{1}{\sqrt{(n-1)(n-2)}}\right)
=\\\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{n-1}}+\frac{n-3}{\sqrt{(n-1)(n-2)}}\right)\rightarrow 0$$
Finally, $(1)$ has $0$ as the limit, $\frac{S_{n+1}-S_n}{T_{n+1}-T_n}$ has $1$ as the limit. The original sequence has $1$ as the limit as well.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3141479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Proving $\int_0^\infty \log\left (1-2\frac{\cos 2\theta}{x^2}+\frac{1}{x^4} \right)dx =2\pi \sin \theta$
Prove $$\int_0^\infty \log \left(1-2\frac{\cos 2\theta}{x^2}+\frac{1}{x^4} \right)dx =2\pi \sin \theta$$where $\theta\in[0,\pi]$.
I've met another similar problem,
$$ \int_0^{2\pi} \log(1-2r\cos \theta +r^2) d\theta=2\pi \log^+(r^2) $$
I am curious whether there is any relationship between them.
And I got stuck on the proposition in the title. I found that
$$1-2\frac{\cos 2\theta}{x^2}+\frac{1}{x^4} =\left(\frac{1}{x}-e^{i\theta}\right)\left(\frac{1}{x}+e^{i\theta}\right)\left(\frac{1}{x}-e^{-i\theta}\right)\left(\frac{1}{x}+e^{-i\theta}\right)$$
But I couldn't move on.
Any hints? Thanks in advance.
| We start off by some $x\rightarrow \frac{1}{x}$ substitutions while derivating under the integral sign: $$I(\theta)=\int_0^\infty \log \left(1-2\frac{\cos 2\theta}{x^2}+\frac{1}{x^4} \right)dx\overset{x\rightarrow \frac{1}{x}}=\int_0^\infty \frac{\ln(1- 2\cos(2\theta) x^2 +x^4)}{x^2}dx$$
$$I'(\theta)=4\int_0^\infty \frac{\sin(2\theta)}{x^4-2\cos(2\theta)x^2+1}dx\overset{x\rightarrow \frac{1}{x}}=4\int_0^\infty \frac{\sin(2\theta)x^2}{x^4-2\cos(2\theta)x^2+1}dx$$
Now summing up the two integrals from above gives us:
$$\Rightarrow 2I'(\theta)=4\int_0^\infty \frac{\sin(2\theta)(1+x^2)}{x^4-2\cos(2\theta)x^2+1}dx=4\int_0^\infty \frac{\sin(2\theta)\left(\frac{1}{x^2}+1\right)}{x^2+\frac{1}{x^2}-2\cos(2\theta)}dx$$
$$\Rightarrow I'(\theta)=2\int_0^\infty \frac{\sin(2\theta)\left(x-\frac{1}{x}\right)'}{\left(x-\frac{1}{x}\right)^2 +2(1-\cos(2\theta))}dx\overset{\large x- \frac{1}{x}=t}=2\int_{-\infty}^\infty \frac{\sin(2\theta)}{t^2 +4\sin^2 (\theta)}dt$$
$$=2 \frac{\sin(2\theta)}{2\sin(\theta)}\arctan\left(\frac{t}{2\sin(\theta)}\right)\bigg|_{-\infty}^\infty=2\cos(\theta) \cdot \pi$$
$$\Rightarrow I(\theta) = 2\pi \int \cos(\theta) d\theta =2\pi \sin \theta +C$$
But $I(0)=0$ (see J.G. answer), thus:$$I(0)=0+C\Rightarrow C=0 \Rightarrow \boxed{I(\theta)=2\pi\sin(\theta)}$$
| {
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"url": "https://math.stackexchange.com/questions/3143523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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How obtain $\sqrt{x^2+x-2} - 3 \ln(\sqrt{x-1} + \sqrt{x+2}) + C$ when integrating $\sqrt{\dfrac{x-1}{x+2}}, x > 1$ The book I'm using states two different answers, one being
$\sqrt{x^2+x-2} - 3 \ln(\sqrt{x-1} + \sqrt{x+2}) + C$
I have tried two different approaches.
Approach one was to rewrite the integral like this:
$\int \sqrt{\dfrac{x-1}{x+2}} dx = \int \dfrac{x-1}{\sqrt{x^2+x-2}} dx = \dfrac{1}{2}\int \dfrac{2x + 1 - 3}{\sqrt{x^2+x-2}} dx = \\ \dfrac{1}{2}\int \dfrac{2x + 1}{\sqrt{x^2+x-2}} dx - \dfrac{3}{2}\int \dfrac{1}{\sqrt{x^2+x-2}} dx = \\
\dfrac{1}{2}[2\sqrt{x^2+x-2} +C_1] - \dfrac{3}{2} [\ln |x + \frac{1}{2} + \sqrt{x^2+x-2}| + C_2] = \\
\sqrt{x^2+x-2} - \dfrac{3}{2} \ln |x + \frac{1}{2} + \sqrt{x^2+x-2}|+ C$
Where in the last step I used the fact that $x^2+x-2 = (x+\frac{1}{2})^2 - \frac{9}{4}$ together with the rule that $\int \dfrac{1}{\sqrt{x^2 + a}} dx = \ln|x + \sqrt{x^2+a}|$
The result is correct according to the book, but not the form I want to achieve.
Apprach two was to make the substitution $y = \sqrt{\dfrac{x-1}{x+2}} \implies dx = \dfrac{6y}{(y^2-1)^2} dy$
rewriting as
$\int \dfrac{6y^2}{(y^2-1)^2} dy = 6 \int \dfrac{y^2 - 1 + 1}{(y^2-1)^2} dy = 6\int \dfrac{1}{y^2-1} dy + 6 \int \dfrac{1}{(y^2-1)^2} dy$
And using partial fraction decomposition on both integrals. This gave me the answer $\sqrt{x^2+x-2} + \dfrac{3}{2} \ln |x + \frac{1}{2} - \sqrt{x^2+x-2}| + C$
Should I try a different approach or am I maybe missing some step in the approaches I've already tried?
| I would Substitute $$t=\sqrt{\frac{x-1}{x+2}}$$ then $$x=\frac{2t^2+1}{1-t^2}$$ then $$dx=6\,{\frac {t}{ \left( t-1 \right) ^{2} \left( t+1 \right) ^{2}}}dt$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the formula for the sum: $1+3x^2+5x^4+7x^6+...+(2n+1)x^{2n}$ The formula is supposed to be valid for $x \neq \pm 1$.
Here is how I did it:
$$\bar{S}_n = \frac{d}{dx}\left( x+x^3+x^5+...+x^{2n+1} \right)$$
The term in the brackets is the geometric sequence. So call the sum up to term $x^{2n+1}$ as $S_n$, then:
$$S_n = \frac{x(1-x^{2n})}{1-x^2}$$
and so:
$$\bar{S}_n = \frac{d}{dx} \left( S_n \right)$$
which gives me:
$$\frac{2x^2 (1-x^{2n})}{(1-x^2)^2}+\frac{1}{1-x^2}\left( 1-x^{2n}(1+2n) \right)$$
apparently, that's wrong. Embarrassing I cannot solve such a simple problem.
| Without calculus:
$$S_3:=1+3x^2+5x^4+7x^6
\\=2(1+x^2+x^4+x^6)-1+x^2(1+3x^2+5x^4)
\\=2\frac{1-x^8}{1-x^2}-1+x^2(S_3-7x^6)$$
and more generally
$$S_n=2\frac{1-x^{2n+2}}{1-x^2}-1+x^2(S_n-(2n+1)x^{2n}).$$
So
$$(1-x^2)S_n=2\frac{1-x^{2n+2}}{1-x^2}-1-(2n+1)x^{2n+2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\left|\begin{smallmatrix}1&1&1\\a&b&c\\a^3&b^3&c^3\end{smallmatrix}\right|=(b-a)(c-b)(c-a)(a+b+c)$
Prove that$$\begin{vmatrix}1&1&1\\a&b&c\\a^3&b^3&c^3\end{vmatrix}=(b-a)(c-b)(c-a)(a+b+c)$$
My attempt:
$$\begin{align}\begin{vmatrix}1&1&1\\a&b&c\\a^3&b^3&c^3\end{vmatrix}&=\begin{vmatrix}0&1&0\\a-b&b&c-b\\a^3-b^3&b^3&c^3-b^3\end{vmatrix}\\&=\begin{vmatrix}c-b&a-b\\c^3-b^3&a^3-b^3\end{vmatrix}\\&=(c-b)(a-b)\begin{vmatrix}1&1\\c^2+cb+b^2&a^2+ab+b^2\end{vmatrix}\\&=(c-b)(a-b)[(a^2+ab)-(c^2+cb)]\\\end{align}$$
Where did I go wrong?
| 1) By inspection : The determinant $= 0$ for $a=b$; $a=c$, and $b=c$ (Why?).
Hence $\pm (c-b)$, $\pm(a-b)$, and $\pm (a-c)$ are factors.
You got $(c-b)(a-b)[a^2+ab -(c^2+cb)]$.
Hence $\pm (a-c)$ is a factor of $[a^2+ab -(c^2+cb)]$.
$a^2-c^2+b(a-c)=$
$ (a-c)(a+c)+b(a-c)=$
$(a-c)(a+b+c).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3150198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Triple Pythagorean with $a^2+b^2=c^4$ It is well known that there exist integer solutions to the equation $a^2+b^2=c^2$.
For example, an explicit formula for integer values of $a$ , $b$ , and $c$ is
\begin{align}a&=2mn \\ b&=m^2-n^2 \\ c&=m^2+n^2.\end{align}
I want to find solutions to the equation $a^2+b^2=c^4$. However, I got $a,b,c$ in a general form that substitutes $x$ in integer and get $a,b,c$ in the integer. What should I do first? I'm very stuck!
| Whenever you have $a^2+b^2=c^2$ you also get $(ca)^2+(cb)^2=c^4$ -- does that count for you?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151735",
"timestamp": "2023-03-29T00:00:00",
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$p=a^2+ab+41b^2$ iff $-163$ is a quadratic residue Prove that a prime $p$ can be written as $p=a^2+ab+41b^2$ iff $-163$ is a quadratic residue modulo $p$.
What I have in mind is something like this: look at $\mathbb{Q}[\sqrt{-163}]$ which has the ring of integers $\mathbb{Z} \left [ \dfrac{1+\sqrt{-163}}{2} \right ]$. Now $p$ is of that form iff $p$ is the norm of an element in the ring of integers. It is well known that the ring of intergers in this case is a PID. However, I cannot tie the fact that -163 has to be a quadratic residue to all of this (in one direction, clearly if $p=a^2+ab+41b^2$ then $4p=(2a+b)^2+163b^2$ so $-163$ has to be a quadratic residue). Could somebody help me complete a solution along these lines?
| I love the form "$a^2 + ab + 41b^2$" because it helps confuse humans. The fact that $a$ and $b$ are integers is too small a benefit to make this confusion worthwhile.
Setting $$\theta = \frac{1}{2} + \frac{\sqrt{-163}}{2}$$ we see that $(a + b \theta)(c + d \theta) = p$ gives us two solutions to $a^2 + ab + 41b^2 = p$. For example, $\theta (1 - \theta) = 41$ tells us that $0^2 + 0 \times 41 + 41 + 41 \times 1^2 = 41$ and $1^2 + (1 \times -1) + 41 \times (-1)^2 = 41$. However, if we rewrite $a + b \theta$ as $$\frac{m}{2} + \frac{n \sqrt{-163}}{2},$$ the relationship between a number in this domain and its conjugate becomes much easier to see. In fact, $$\left(\frac{m}{2} - \frac{n \sqrt{-163}}{2}\right) \left(\frac{m}{2} + \frac{n \sqrt{-163}}{2}\right) = N,$$ which in this example is the prime $41$ if we set $m = n = 1$: $$\left(\frac{1}{2} - \frac{\sqrt{-163}}{2}\right) \left(\frac{1}{2} + \frac{\sqrt{-163}}{2}\right) = 41.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $a\sqrt{b^3+1}+b\sqrt{c^3+1}+c\sqrt{a^3+1} \leq 5$ Let $a,b,c$ be nonnegative real numbers such that $a+b+c=3$. Prove that
$$a\sqrt{b^3+1}+b\sqrt{c^3+1}+c\sqrt{a^3+1} \leq 5$$
I found a point at which the equality is attended, say $a=0,b=1,c=2$. But I have no idea how to prove it. I tried to use the AM-GM inequality but then I obtained the more difficult one. Please help me. Thank you very much.
| $\text{WLOG b=mid(a,b,c)} $
By AM-GM and Rearrangement we have:
$$\text{L.H.S}=\sum _{cyc}a\sqrt{b^3+1}=\sum _{cyc}a\sqrt{\left(b+1\right)\left(b^2-b+1\right)}$$
$$\le \sum _{cyc}a\cdot \frac{b+1+b^2-b+1}{2}=\sum_{cyc}\frac{2a+ab^2}{2}=\frac{ab^2+bc^2+ca^2}{2}+3$$
$$\le \frac{b\left(a^2+ac+c^2\right)}{2}+3\le \frac{b\left(a+c\right)^2}{2}+3$$
$$=\frac{2b\left(a+c\right)^2}{4}+3\le \frac{\left(\frac{2\left(a+b+c\right)}{3}\right)^3}{4}+3=5=\text{R.H.S}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int^{2}_{0}\frac{\tan^{-1}(x)}{1+4x}\mathrm dx$
Evaluate $\displaystyle \int^{2}_{0}\frac{\tan^{-1}(x)}{1+4x}\mathrm dx$
My effort:
\begin{align*}
I(a)&=\int^{2}_{0}\frac{\tan^{-1}(ax)}{1+4x}\mathrm dx\\
I'(a) &= \int^{2}_{0}\frac{x}{(1+4x)(1+a^2x^2)}\mathrm dx\\
I'(a) &= \frac{1}{4}\int^{2}_{0}\frac{(1+4x)-1}{(1+4x)(1+a^2x^2)}dx\\
I'(a) &= \frac{1}{4a}\tan^{-1}(2)-\frac{1}{4}\int^{2}_{0}\frac{1}{(1+4x)(1+a^2x^2)}dx
\end{align*}
Then how to proceed? Thank you.
| Firstly, do the integration by parts and and decompose a denominator
$$
\begin{align*}
& \int_0^2 \frac{\tan^{-1} (a x)}{1+4x} dx \\ = &\ \dfrac{\log 3 \tan^{-1} (2a)}{2}-\frac{a}{4}\int_0^2 \frac{\log(1+4x)}{1+a^2x^2} dx\\ = &\ \dfrac{\log 3 \tan^{-1} (2a)}{2}-\frac{a}{8}\int_0^2 \log(1+4x)(\frac{1}{1+iax}+\frac{1}{1-iax}) dx
\end{align*}$$
The obtained integrals can be rewrote through the dilogarithm $\text{Li}_2$ functions. The result have the following form
$$
\begin{align*}
I(a)&= \dfrac{\log 3 \tan^{-1} (2a)}{2}+\frac{i}{8}\left(\text{Li}_2\left(\frac{a}{a-4 i}\right)-\text{Li}_2\left(\frac{9 a}{a-4 i}\right)-\text{Li}_2\left(\frac{a}{a+4 i}\right) \\ + \text{Li}_2\left(\frac{9 a}{a+4 i}\right)-\log (9) \log \left(\frac{8 a+4 i}{-a+4 i}\right)+ \log (9) \log \left(\frac{-8 a+4 i}{a+4 i}\right)\right)
\end{align*}
$$
I think this answer can be simplified, but I do not want to do it.
| {
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"timestamp": "2023-03-29T00:00:00",
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minimum of $a^2+4b^2+c^2$ given $2a+b+3c=20$
If $a,b,c\in\mathbb{R}$ and $2a+b+3c=20.$ Then minimum value of $a^2+4b^2+c^2$ is
what i try
Cauchy schwarz inequality
$$(a^2+(2b)^2+c^2)(2^2+\frac{1}{2^2}+3^2)\geq (2a+b+3c)^2$$
How do i solve it without Cauchy schwarz inequality Help me please
| I think this solution is more simple.
From $$2a+b+3c=20\Leftrightarrow b=20-2a-3c$$
Then $$A=a^2+4\left(20-2a-3c\right)^2+c^2$$
$$=17a^2+48ac-320a+37c^2-480c+1600$$
$$=\frac{1}{17}(17a+24c-160)^2+\frac{1}{17}\left(53\left(c-\frac{240}{53}\right)^2+\frac{27200}{53}\right )$$
$$\ge \frac{1}{17}\cdot \frac{27200}{53}=\frac{1600}{53}$$
The equality occurs when $(a;b;c)=\left(\frac{160}{53};\frac{20}{53};\frac{240}{53}\right)$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\int^{\infty}_{0}\frac{\cos x}{(x^2+1)^2}dx$
Finding value of $\displaystyle \int^{\infty}_{0}\frac{x\sin x}{(x^2+1)^3}dx$
Let $$ I = \frac{1}{2}\int^{\infty}_{0}\sin x\frac{2x}{(x^2+1)^3}dx$$
Integrating by parts
$$ \Rightarrow I = -\frac{\sin x}{(x^2+1)^2}\bigg|^{\infty}_{0}+\int^{\infty}_{0}\frac{\cos x}{(x^2+1)^2}dx=\int^{\infty}_{0}\frac{\cos x}{(x^2+1)^2}dx$$
How do i solve it?
| $$I=\int_0^\infty \frac{x \sin x}{(1+x^2)^3}dx=-\frac14 \underbrace{\frac{\sin x}{(1+x^2)^2}\bigg|_0^\infty}_{=0} +\frac14 \int_0^\infty \frac{\cos x}{(1+x^2)^2}dx $$
Now we have that $I=-\frac14 J'(1)$ because we can take: $$J(a)=\int_0^\infty \frac{\cos x}{a+x^2}dx \Rightarrow J'(a)=-\int_0^\infty \frac{\cos x}{(a+x^2)^2}dx$$
Thus all we need to do is to find $J(a)$ then take a derivate and set $a=1$.
But one can directly find for example here that: $$J(a)=\frac{\pi}{2 \sqrt a }e^{-\sqrt a}\Rightarrow J'(a)=\frac{\pi}{2}\left(-\frac12 a^{-3/2}e^{-\sqrt a} -\frac{1}{\sqrt a} e^{-\sqrt a} \frac{1}{2\sqrt a}\right)$$
$$\Rightarrow I=-\frac14J'(1)=\frac{\pi}{8}\left(\frac12 e^{-1}+\frac12 e^{-1}\right)=\frac{\pi}{8e}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3167180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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Prove that $\sum\limits_{n=1}(-1)^n\frac{x^2+n}{n^2}$ uniformly convergent on $[a,b]$ I used the cauchy criteria, for $p>q$
$$\sup\limits_{x\in [a,b]}\left | S_p(x) -S_q(x)\right |=
\sup\limits_{x}\left | \sum\limits_{k=q+1}^{p}(-1)^n\frac{x^2+n}{n^2} \right |\leq \\
\sup\limits_{x}\sum\limits_{k=q+1}^{p}\left | \frac{x^2+n}{n^2} \right |\leq
\sup\limits_{x}\frac{x^2+q+1}{(q+1)^2}(p-q)=\\
\frac{b^2+q+1}{(q+1)^2}(p-q)$$
Is what I am doing right till now ?
| Since the series is not absolutely convergent, using the bound,
$$\left|\sum_{n=p+1}^q (-1)^n\frac{x^2+n}{n^2}\right| \leqslant \sum_{n=p+1}^q\frac{x^2+n}{n^2},$$
you will not succeed in proving that partial sums of $\sum_{n \geqslant 1}(-1)^n\frac{x^2+n}{n^2}$ satisfy the Cauchy criterion uniformly because the RHS is the difference of partial sums for a divergent series. Note that $\frac{x^2+n}{n^2} = \mathcal{O}\left(\frac{1}{n} \right)$.
For the bounding series to satisfy the Cauchy criterion, we must have for any $\epsilon > 0$,
$$\sup_{x \in [a,b]}\sum_{n=q+1}^{p} \frac{x^2+n}{n^2} < \epsilon $$
for all sufficiently large $q$ and $p > q$.
However,
$$\sup_{x \in [a,b]}\sum_{n=q+1}^{p} \frac{x^2+n}{n^2} \geqslant \sup_{x \in [a,b]}(p-q) \frac{x^2 +q }{p^2} = (p-q) \frac{\max(a^2,b^2) +q }{p^2} $$
Taking $p = 2q$, we have
$$\sup_{x \in [a,b]}\sum_{n=q+1}^{p} \frac{x^2+n}{n^2} \geqslant q \frac{\max(a^2,b^2) +q }{4q^2} = \frac{1 + \max(a^2,b^2)/q}{4},$$
and since the RHS converges to $1/4$ as $q \to \infty$, the Cauchy criterion is violated.
A correct approach
Uniform convergence can be established using Dirichlet's test. The hypotheses are met since the partial sums $\sum_{n=1}^N(-1)^n$are uniformly bounded and it can be shown that as $n \to \infty$, we have
$$\frac{x^2 + n}{n^2 } \downarrow 0,$$
where the convergence is monotonic and uniform for $x \in [a,b]$.
The uniform convergence is easily established since,
$$\frac{\min(a^2,b^2)+n}{n^2} \leqslant \frac{x^2+n}{n} \leqslant \frac{\max(a^2,b^2) +n}{n^2}$$
See if you can finish by showing that $\frac{x^2 + n}{n^2}$ is decreasing with respect to $n$.
Summation by parts is another approach, but that in essence repeats the steps in a proof of Dirichlet's test for uniform convergence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3168087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Permutations - restriction of having number be greater than and even (no repetition of numbers) Using $6$ numbers - $3,4,5,6,7,9$; a $6$-digit number is to be formed.
What is the working to find out how many of these $6$ digit numbers are even and greater than $500,000$?
My attempt:
When the first digit can be either $5,7,9$ and the last digit can be either $4,6$ - the possible amount of numbers is $3 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \cdot 2$. (I'm getting this part correct according to the mark scheme). When the first digit can be either $5,6,7,9$ and the last digit can only be $4$, the possible ammount of numbers is $4 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \cdot 1$ (I'm getting this part wrong according to the mark scheme). The answer I am getting is $240$. The current answer is $168$.
| Observe that the leading digit must be at least $5$. Consider cases, depending on whether the leading digit is even or odd:
*
*If the leading digit is odd, the units digit may be selected from $4$ or $6$. Choose the leading digit from among $5$, $7$, or $9$, choose the units digit, then arrange the remaining digits between them.
*If the leading digit is $6$, the units digit must be $4$. Arrange the remaining digits between them.
In case 1, the number can be selected in $3 \cdot 2 \cdot 4!$ ways. In case 2, it can be selected in $4!$ ways. Hence, there are $$3 \cdot 2 \cdot 4! + 4! = (3 \cdot 2 + 1)4! = 7 \cdot 4! = 168$$ admissible arrangements.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3168194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Problem involving triangle. Find $x$ in the figure
I need to find $x$ in the triangle above.
I tried to do basic things, like sum of a triangle's internal angles $= 180^\circ$ but I only found $2$ equations for $3$ variables
Any help is appreciated
| In triangle $ABC$ you have the following angles $\angle A=120^\circ$ (top corner), $\angle B=20^\circ$ (left corner) and $\angle C=40^\circ$ (right corner). Denote the central point with $D$ and introduce lengths $AD=a, BD=b,CD=c$.
By law of sines applied to triangles $ABD,ACD,BCD$:
$$a\sin20^\circ=b\sin10^\circ\tag{1}$$
$$a\sin100^\circ=c\sin(40^\circ-x)\tag{2}$$
$$b\sin10^\circ=c\sin x\tag{3}$$
From (1) and (2):
$$b=\frac{a\sin20^\circ}{\sin10^\circ}$$
$$c=\frac{a\sin100^\circ}{\sin(40^\circ-x)}$$
Replace that into (3):
$$\frac{a\sin20^\circ}{\sin10^\circ}\sin10^\circ=\frac{a\sin100^\circ}{\sin(40^\circ-x)}\sin x$$
$$\sin20^\circ \sin(40^\circ-x)=\sin100^\circ\sin x$$
$$\sin20^\circ \sin(40^\circ-x)=\cos10^\circ\sin x$$
$$2\sin10^\circ \cos10^\circ \sin(40^\circ-x)=\cos10^\circ\sin x$$
$$2\sin10^\circ \sin(40^\circ-x)=\sin x$$
Sometimes you have to make things more complicated before your are able to jump over the last hurdle: multiply the right side with $1=2\sin30^\circ$.
$$2\sin10^\circ \sin(40^\circ-x)=2\sin x\sin30^\circ$$
$$\cos(-30^\circ+x)-\cos(50^\circ-x)=\cos(x-30^\circ)-\cos(x+30^\circ)$$
$$\cos(50^\circ-x)=\cos(x+30^\circ)$$
For obviously acute angle $x$
$$50^\circ-x=x+30^\circ$$
$$x=10^\circ$$
No calculator needed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3176625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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} |
Finding the value of c for a two variable function to allow continuity. The function f:$\mathbb{R^2}$$\to$$\mathbb{R}$ given by
$$f(x,y)=\begin{cases}\frac{xy^2}{x^2+y^2} & \text{if } (x,y) \neq (0,0) \\
c & \text{if }(x, y) = 0\end{cases}.$$
I am trying to calculate the limit (by any means) to determine the value of constant $c$ such that $f$ is continuous at $(0,0)$.
I have obtained the limit for
$$f(x,y)=\frac{xy^2}{x^2+y^2} \quad \text{if } (x,y) \neq (0,0)$$ through using polar coordinates $L=0$.
I do not know how to find $c$ from here.
| Observe that $$\left | \frac {xy^2} {x^2+y^2} \right | \leq \sqrt {x^2+y^2} < \varepsilon$$ whenever $\sqrt {x^2 + y^2} < \delta,$ where $\delta = \varepsilon.$
So $$\lim\limits_{(x,y) \rightarrow (0,0)} \frac {xy^2} {x^2 + y^2} = 0.$$
For the continuity of $f$ at $(0,0)$ we should have $c=f(0,0) = \lim\limits_{(x,y) \rightarrow (0,0)} \frac {xy^2} {x^2 + y^2} = 0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3177810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does this system of equations have attractors and periodic solutions? I want to solve the following exercise:
Determine the critical points of the system
\begin{align}
\dot{x} &= x^2- y^3\\
\dot{y} &= 2x(x^2 - y)
\end{align}
Are there attractors in this system? Determine a first integral. Do periodic solutions exist?
What I've tried so far:
I think that I've found the critical points of this system. If I inspect
\begin{align}
x^2 - y^3 = 0\\
2x(x^2 - y) = 0
\end{align}
I find that $x = 0$ or $x = \pm \sqrt{y}$.
Case $x = 0$:
$-y^3 = 0$ so that $y = 0$. Hence the first critical point that I find is $(0,0)$.
Case $x = \pm\sqrt{y}$:
$y - y^3 = 0$ so $y=1$ or $y = -1$. If $y = -1$ then $x = \pm i$. If $y = 1$ then $x = \pm1$. Hence I find the critical points $(1, 1), (i, -1), (-1,1),$ and $(-i,-1)$. I think that I've found a first integral by looking at the equation
$$
\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} = \dfrac{dy}{dx} = \dfrac{2x^3 - 2xy}{x^2 - y^3}
$$
Integrating this final equation gives
$$
\int \dfrac{2x^3 - 2xy}{x^2 - y^3}dx = y(y^2 - 1)\log(y^3 - x^2) + x^2 + C
$$
Hence a first integral of this system is
$$
F(x,y) = y(y^2 - 1)\log(y^3 - x^2) + x^2
$$
I tried plotting this function to find out if the system of equations has periodic solutions but I don't get any wiser looking at the plots.
I have the following definitions of an attractor:
A critical point $x = a$ of the equation $\dot{x} = f(x)$ in $\mathbb{R}^n$ is called a positive attractor if there exists a neighborhood $\Omega_a\subset \mathbb{R}^n$ of $x = a$ such that $x(t_0)\in\Omega_a$ implies $\lim_{t\to\infty} x(t) = a$. If a critical point $x = a$ has this property for $t \to - \infty$ then $x = a$ is called a negative attractor.
I think that in order to determine whether either of the critical points is an attractor I need to know what the solution $x(t)$ looks like. I'm not sure how to solve this system though and if I enter the equations in wolfram then I don't get a solution either (I might be missing something). So therefore I tried to linearize the equations. I have:
$$
f(x, y) = \begin{pmatrix}x^2 - y^3\\2x(x^2 - y)\end{pmatrix}, \dfrac{\partial f(x,y)}{\partial(x,y)} = \begin{pmatrix}2x & -3y^2\\6x^2& -2x\end{pmatrix}, \dfrac{\partial^2f(x,y)}{\partial (x,y)^2} = \begin{pmatrix}2 & -6y\\12x & 0\end{pmatrix}, \\\dfrac{\partial^3 f(x,y)}{\partial (x,y)^3} = \begin{pmatrix}0 & -6\\12 & 0\end{pmatrix}
$$
Hence, using the Taylor expansion around $(x,y) = (0,0)$ I get
\begin{align}
\dot{x} &= x^2 - y^3 + \ldots\\
\dot{y} &= 2x^3 + \ldots
\end{align}
but I'm not really sure how I can use this to learn anything new..
Question: How can I learn more about this system so that I will be able to say more about the existence of attractors and periodic solutions? Is my reasoning thus far correct?
| $$\begin{align}
\dot{x} &= x^2- y^3\\
\dot{y} &= 2x(x^2 - y)
\end{align}$$
$$\frac{dy}{dx}=\frac{2x(x^2 - y)}{x^2- y^3}$$
Let $X=x^2$
$$\frac{dy}{dX}=\frac{X - y}{X- y^3}$$
$$\frac{dX}{dy}=\frac{X- y^3}{X-y}$$
Let $X=y+u$
$$\frac{dX}{dy}=1+\frac{du}{dy}=\frac{y+u- y^3}{u}$$
$$\frac{du}{dy}=\frac{y- y^3}{u}$$
$$2udu=2(y-y^3)dy$$
$$u^2=y^2-\frac12 y^4+c$$
$$(X-y)^2=(x^2-y)^2=y^2-\frac12 y^4+c$$
$$x^4-2x^2y=-\frac12 y^4+c$$
The equation of the trajectory is :
$$\boxed{y^4-4x^2y+2x^4=2c}$$
The analytical study of this equation shows that $c\geq -\frac12$ and that the trajectory is a closed curve.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3179667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Are there any other methods to apply to solving simultaneous equations? We are asked to solve for $x$ and $y$ in the following pair of simultaneous equations:
$$\begin{align}3x+2y&=36 \tag1\\ 5x+4y&=64\tag2\end{align}$$
I can multiply $(1)$ by $2$, yielding $6x + 4y = 72$, and subtracting $(2)$ from this new equation eliminates $4y$ to solve strictly for $x$; i.e. $6x - 5x = 72 - 64 \Rightarrow x = 8$. Substituting $x=8$ into $(2)$ reveals that $y=6$.
I could also subtract $(1)$ from $(2)$ and divide by $2$, yielding $x+y=14$. Let $$\begin{align}3x+3y - y &= 36 \tag{1a}\\ 5x + 5y - y &= 64\tag{1b}\end{align}$$ then expand brackets, and it follows that $42 - y = 36$ and $70 - y = 64$, thus revealing $y=6$ and so $x = 14 - 6 = 8$.
I can even use matrices!
$(1)$ and $(2)$ could be written in matrix form:
$$\begin{align}\begin{bmatrix} 3 &2 \\ 5 &4\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}&=\begin{bmatrix}36 \\ 64\end{bmatrix}\tag3 \\ \begin{bmatrix} x \\ y\end{bmatrix} &= {\begin{bmatrix} 3 &2 \\ 5 &4\end{bmatrix}}^{-1}\begin{bmatrix}36 \\ 64\end{bmatrix} \\ &= \frac{1}{2}\begin{bmatrix}4 &-2 \\ -5 &3\end{bmatrix}\begin{bmatrix}36 \\ 64\end{bmatrix} \\ &=\frac12\begin{bmatrix} 16 \\ 12\end{bmatrix} \\ &= \begin{bmatrix} 8 \\ 6\end{bmatrix} \\ \\ \therefore x&=8 \\ \therefore y&= 6\end{align}$$
Question
Are there any other methods to solve for both $x$ and $y$?
| Consider the three vectors $\textbf{A}=(3,2)$, $\textbf{B}=(5,4)$ and $\textbf{X}=(x,y)$. Your system could be written as $$\textbf{A}\cdot\textbf{X}=a\\\textbf{B}\cdot\textbf{X}=b$$ where $a=36$, $b=64$ and $\textbf{A}_{\perp}=(-2,3)$ is orthogonal to $\textbf{A}$. The first equation gives us $\textbf{X}=\dfrac{a\textbf{A}}{\textbf{A}^2}+\lambda\textbf{A}_{\perp}$. Now to find $\lambda$ we use the second equation, we get $\lambda=\dfrac{b}{\textbf{A}_{\perp}\cdot\textbf{B}}-\dfrac{a\textbf{A}\cdot\textbf{B}}{\textbf{A}^2\times\textbf{A}_{\perp}\cdot\textbf{B}}$. Et voilà :
$$\textbf{X}=\dfrac{a\textbf{A}}{\textbf{A}^2}+\dfrac{\textbf{A}_{\perp}}{\textbf{A}_{\perp}\cdot\textbf{B}}\left(b-\dfrac{a\textbf{A}\cdot\textbf{B}}{\textbf{A}^2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3180580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 14,
"answer_id": 4
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Hot to simplify $(1-\frac{1} {N} )(1-\frac{2} {N} )...(1-\frac{r-1} {N} )$ I'm trying to figure out the algebraic simplification from step $6$ to $7$ here
Specifically the left side:
$(1-\frac{1} {N} )(1-\frac{2} {N} )...(1-\frac{r-1} {N} )$
To that:
$1-(\frac{1}{N}+\frac{2}{N} +... +\frac{r-1}{N}) +...$
I tried to multiply out but can't see how they got it and what are the terms after the last three dots.
Thanks.
| In the article the birthmate problem is compared with birthday problem. In step $6$ it is obtained:
$$(1-\frac{1} {N} )(1-\frac{2} {N} )...(1-\frac{r-1} {N} )=\left(1-\frac1N\right)^n$$
Next it is stated:
If we multiply out the left-hand side of Equation $(6)$ to terms of order $1/N$, and
expand the right-hand side to two terms we get:
$$1-(\frac{1}{N}+\frac{2}{N} +... +\frac{r-1}{N}) +\cdots=1-\frac1N+\cdots$$
Indeed, for the left side, consider simpler cases to see the pattern (up to the term $1/N$):
$$\begin{align}(1-\frac{1} {N} )(1-\frac{2} {N} )&=1-\frac1N-\frac2N+\frac{1\cdot 2}{N^2};\\
(1-\frac{1} {N} )(1-\frac{2} {N} )(1-\frac{3} {N} )&=\left(1-\frac1N-\frac2N+\frac{1\cdot 2}{N^2}\right)(1-\frac{3} {N} )=\\
&=1-\frac1N-\frac2N-\frac3N+\frac{1\cdot 2}{N^2}+\frac{1\cdot 3}{N^2}+\frac{2\cdot 3}{N^2}-\frac{1\cdot 2\cdot 3}{N^3};\end{align}$$
For the right side, the binomial theorem is used (up to two terms):
$$\left(1-\frac1N\right)^n=1-\frac{n}{1}\cdot \frac1N+\frac{n(n-1)}{2}\cdot \frac1{N^2}-\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3181215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving that $a,b$ are even integers I'm trying to prove the following theorem:
Let $a,b\in\mathbb{Z}$ . Then $$a^{2}+b^{2}\equiv0\pmod 4 \iff a \;\text{and}\; b\;\text{are even}$$
I always struggle to prove some number is odd or even. How to prove it?
I thought of using the $(a+b)^2=a^2+2ab+b^2$ formula but not sure how.
| If both $a$ and $b$ are odd numbers then $a = 2k_1+1, b=2 k_2+1$ and you have that
$$
a^2+b^2 = (2k_1+1)^2+(2 k_2+1)^2 = 4k_1^2+4 k_2^2 +4k_1+4k_2+2 \equiv 2 (mod\,\, 4)
$$
If one is odd (for instance $a$) and the other is even, then $a =2 k_1+1, b=2k_2$ and
$$
a^2+b^2 = (2k_1+1)^2+(2k_2)^2 = 4k_1^2+4k_2^2+4k_1+1\equiv 1 (mod\,\,4)
$$
If both $a$ and $b$ are even, $a=2k_ 1, b=2k_2$ and
$$
a^2+b^2 = (2k_1)^2+(2k_2)^2=4k_1^2+ 4k_2^2 \equiv 0(mod\,\,4)
$$
So you see that $a^2+b^2$ can only be congruent with 0,1 or 2 (mod 4) and is congruent with 0 if and only if both $a$ and $b$ are even numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3182101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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To find the sum: $\frac {1}{n!} \sum \binom {n}{2+3r} x^{1+r}$
Sum the series:
$$
\frac {x}{2!(n-2)!}+\frac {x^2}{5!(n-5)!}+\frac {x^3}{8!(n-8)!}+....+\frac {x^{\frac {n}{3}}}{(n-1)!},
$$
$n$ being a multiple of $3$.(Math. Tripos, 1899)
My attempt
We may write for an integer $m$, $n=3m$ so that:
$$
S(x)=\frac {x}{2!(n-2)!}+\frac {x^2}{5!(n-5)!}+\frac {x^3}{8!(n-8)!}+....+\frac {x^{\frac {n}{3}}}{(n-1)!}
$$
$$
=\frac {x}{2!(3m-2)!}+\frac {x^2}{5!(3m-5)!}+\frac {x^3}{8!(3m-8)!}+....+\frac {x^m}{(3m-1)!}
$$
$$
=\frac {1}{(3m)!}\left\{ \binom {3m}{2}x+\binom {3m}{5}x^2+\binom {3m}{8}x^3+....+\binom {3m}{3m-1}x^m \right\}
$$
Here it is plain to me that it is some binomial expansion in which every $2$ out of $3$ terms cancel out. The first guess in such a situation is a series consisting the cube roots of unity as well as $\sqrt[3]x.$
$$
(\omega_3+\sqrt[3]x)^{3m}=1+\binom {3m}{1}\omega_3^2\sqrt[3]x+\binom {3m}{2}\omega_3\sqrt[3]{x^2}+....+x^m
$$
$$
S(x)=\frac {\omega_3^2 \sqrt[3]x}{(3m)!}\left\{ \binom {3m}{2}\omega_3\sqrt[3]{x^2}+\binom {3m}{5}\omega_3x^2\sqrt[3]{x^2}+....+\binom {3m}{3m-1}\omega_3x^{m-1}\sqrt[3]{x^2}\right\}
$$
Now if,
$$
S(x)=\frac {\omega_3^2 \sqrt[3]x}{(3m)!} (\omega_3+\sqrt[3]x)^{3m},
$$
then we are left with two conditions:
$$
\binom {3m}{1}+\binom {3m}{4}x+\binom {3m}{7}x^2+....=0,
$$
$$
1+\binom {3m}{3}x+\binom {3m}{6}x^2+....=0
$$
It seems that the whole line of logic is circular. Any help would be greatly appreciated.
| Hint
Consider
$$S=a(1+y)^n+b(1+wy)^n+(1+w^2y)^n$$ where $w$ is a complex cube (as we need every third term) root of unity
We need the coefficients of $y^0,y^1$ to be zero
$a+b+1=0$
$an+bnw+w^2n=0$
Solve for $a,b$
Now find $S$ and compare with given expression
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the remainder when $p(x)$ is divided by $x^2-a^2$ if $p(x)$ leaves remainders $a, -a$ when divided by $x+a, x-a$
Let $a\neq0$ and $p(x)$ be a polynomial of degree greater than $2$. If $p(x)$ leaves remainders $a$ and $-a$ when divided respectively by $x+a$ and $x-a$. Find the remainder when $p(x)$ is divided by $x^2-a^2$
$$
p(x)=q(x).(x+a)+r_1=q(x).(x+a)+a\quad\big[r_1=p(-a)=a\big]\\
p(x)=s(x).(x-a)+r_2=s(x).(x-a)-a\quad\big[r_2=p(a)=-a\big]\\
p(x)=t(x).(x^2-a^2)+r=t(x).(x^2-a^2)+Ax+B\\
p(a)=aA+B=-a\\
p(-a)=-aA+B=a\\
B=0,\;A=-1\implies r=Ax+B=-x
$$
I was wondering Is there another way to solve this problem ?
| One possible approach would be to use partial fractions:
$$\frac{1}{x^2-a^2} = \frac{1}{2a} \left( \frac{1}{x-a} - \frac{1}{x+a}\right) .$$
On the other hand, by the given information,
$$\frac{p(x)}{x-a} = q_1(x) - \frac{a}{x-a}$$
and
$$\frac{p(x)}{x+a} = q_2(x) + \frac{a}{x+a}$$
for some quotient polynomials $q_1, q_2$.
Therefore,
$$ \frac{p(x)}{x^2-a^2} = \frac{1}{2a} \left( \frac{p(x)}{x-a} - \frac{p(x)}{x+a} \right) = \frac{1}{2a} \left( q_1(x) - q_2(x) - \frac{a}{x-a} - \frac{a}{x+a} \right) = \\
q(x) - \frac{x}{x^2 - a^2}$$
where $q(x) = \frac{1}{2a} (q_1(x) - q_2(x))$ is a polynomial. It follows that the remainder of dividing $p(x)$ by $x^2 - a^2$ is $-x$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Show With Induction that $2\cdot2^{0}+3\cdot2^{1}+4\cdot2^{2}+...(n+1)\cdot2^{n-1}=n\cdot2^{n}$ Show with induction that
$2\cdot2^{0}+3\cdot2^{1}+4\cdot2^{2}+5\cdot2^{3}+6\cdot2^{4}+...(n+1)\cdot2^{n-1}=n\cdot2^{n}$
My solution:
Base case 1: n = 1
LHS = $(1+1)\cdot2^{1-1} = 2$
RHS = $1\cdot2^{1}= 2$
Case 2: n = p
When
$LHS_{P}$ =
$RHS_{P}$
$2\cdot2^{0}+3\cdot2^{1}+4\cdot2^{2}+5\cdot2^{3}+6\cdot2^{4}+...(p+1)\cdot2^{p-1}=p\cdot2^{p}$
Case 3: n = p + 1
$LHS_{P+1}$ = $LHS_{P}$ + (p+2)$\cdot2^{p}$
$RHS_{P+1}$ = (p+1)$\cdot2^{p+1}$
So i need to to prove that:
$RHS_{P+1}$ = $RHS_{P}$ + (p+2)$\cdot2^{p}$
$RHS_{P+1}$ = $p\cdot2^{p}$ + (p+2)$\cdot2^{p}$
Am I thinking right here?
$RHS_{P+1}$ =
$(p+1)2^{p+1}=$
$(p+1)\cdot2^{p}\cdot2$ = ?
(Can I get this equal to $p\cdot2^{p}$ + (p+2)$\cdot2^{p}$ ? )
| Hint:
$$\underbrace{2 \cdot 2^0 + 3 \cdot 2^1 + \cdots (p-2) \cdot 2^{p-1} }_{ \text{use the induction hypothesis in this part} }+ (p-1) \cdot 2^{p}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3185481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Very indeterminate form: $\lim_{x \to \infty} \left(\sqrt{x^2+2x+3} -\sqrt{x^2+3}\right)^x \longrightarrow (\infty-\infty)^{\infty}$ Here is problem:
$$\lim_{x \to \infty} \left(\sqrt{x^2+2x+3} -\sqrt{x^2+3}\right)^x$$
The solution I presented in the picture below was made by a Mathematics Teacher
I tried to solve this Limit without using derivative (L'hospital) and Big O notation. Although I get the answer, I don't know if the technique I'm using definitely correct.
And here is my method:
$$\begin{align*}\lim_{x \to \infty} \left(\sqrt{x^2+2x+3} -\sqrt{x^2+3}\right)^x&=\lim_{x \to \infty} \left(\frac {2x}{\sqrt{x^2+2x+3} +\sqrt{x^2+3}}\right)^x\\&=\lim_{x \to \infty}\frac{1}{ \left(\frac {\sqrt{x^2+2x+3} +\sqrt{x^2+3}}{2x}\right)^x}\end{align*}$$
Then, I define a new function here
$$y(x)=\sqrt{x^2+2x+3} +\sqrt{x^2+3}-2x-1$$
We have
$$\begin{align*}
\lim _{x\to\infty} y(x)&=\lim_{x \to \infty}\sqrt{x^2+2x+3} +\sqrt{x^2+3}-2x-1\\
&=\lim_{x \to \infty}(\sqrt{x^2+2x+3}-(x+1))+(\sqrt{x^2+3}-x)\\
&=\lim_{x \to \infty}\frac{2}{\sqrt{x^2+2x+3}+x+1}+ \lim_{x \to \infty}\frac{3}{\sqrt{x^2+3}+x}\\
&=0.
\end{align*}$$
This implies that
$$\lim_{x \to \infty}\frac{2x}{y(x)+1}=\infty $$
Therefore,
$$\begin{align*}
\lim_{x \to \infty}\frac{1}{ \left(\frac {\sqrt{x^2+2x+3} +\sqrt{x^2+3}}{2x}\right)^x}&=\lim_{x \to\infty} \frac{1}{ \left(\frac{y(x)+2x+1}{2x} \right)^x}\\
&=\lim_{x \to\infty} \frac{1}{ \left(1+\frac{y(x)+1}{2x} \right)^x}\\
&=\lim_{x \to \infty}\frac{1}{\left( \left( 1+\frac{1}{\frac{2x}{y(x)+1}}\right)^{\frac{2x}{y(x)+1}}\right)^{\frac{y(x)+1}{2}}}\\
&
\end{align*}$$
Here, we define two functions:
$$f(x)=\left( 1+\frac{1}{\frac{2x}{y(x)+1}}\right)^{\frac{2x}{y(x)+1}},\quad
g(x)=\frac{y(x)+1}{2}.
$$
We deduce that,
$$
\lim_{x\to\infty} f(x)=e>0,\quad \lim_{x\to\infty} g(x)=\frac 12>0.
$$
Thus, the limit $\lim_{x\to\infty} f(x)^{g(x)} $ exists and is finite.
Finally we get,
$$\begin{align*}
\lim_{x \to \infty}\frac{1}{\left( \left( 1+\frac{1}{\frac{2x}{y(x)+1}}\right)^{\frac{2x}{y(x)+1}}\right)^{\frac{y(x)+1}{2}}}
&=\frac{1}{\lim_{x \to \infty}\left( \left( \left( 1+\frac{1}{\frac{2x}{y(x)+1}}\right)^{\frac{2x}{y(x)+1}}\right)^{\frac{y(x)+1}{2}}\right)}\\
&=\frac{1}{\left(\lim_{x\to\infty} \left( 1+\frac{1}{\frac{2x}{y(x)+1}} \right)^{\frac{2x}{y(x)+1}}\right)^{ \lim_{x\to\infty} \frac{y(x)+1}{2}}}\\
&=\frac {1}{e^{\frac12}}=\frac{\sqrt e}{e}.\\&&
\end{align*}$$
Is the method I use correct?
I have received criticisms against my work. What can I do to make the method I use, rigorous? What are the points I missed in the method?
Thank you!
| Your approach is correct but its presentation / application is more complicated than needed here.
Here is how you can use the same approach with much less effort. You have already observed that the base $$F(x) =\sqrt {x^2+2x+3}-\sqrt{x^2+3}$$ tends to $1$ as $x\to\infty $. Now the expression under limit can be written as $$\{F(x) \} ^x=\{\{1+(F(x)-1)\}^{1/(F(x)-1)}\}^{x(F(x)-1)}$$ The inner expression tends to $e$ and the exponent $x(F(x) - 1)\to -1/2$ so that the desired limit is $e^{-1/2}$.
Another part of your approach is that it involves the tricky use of subtracting $2x+1$ from $y(x) $. For those who are experienced in the art of calculus this step is obvious via the approximation $$\sqrt{x^2+2ax+b}\approx x+a$$ but it may appear a bit mysterious for a novice. It is best to either explain this part or remove it altogether as I have done it in my answer.
Also note that your approach uses the following limits / rules (it is not necessary to point them out explicitly unless demanded by some strict examiner) :
*
*$\lim_{x\to\infty} \left(1+\frac{1}{x}\right)^x=e$
*If $\lim_{x\to\infty} f(x) =a>0$ and $\lim_{x\to\infty} g(x) =b$ then $\{f(x) \} ^{g(x)} \to a^b$ as $x\to\infty $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3185610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 6,
"answer_id": 2
} |
Show that there is always one integer $t$ with a least prime factor $> 5$ where $x < t \le x+6$ Let $p_k$ be the $k$th prime.
Let $f_2(x) = \lfloor x\rfloor - \left\lfloor\dfrac{x}{2}\right\rfloor$
For $k > 1$, let:
$f_{p_k}(x) = f_{p_{k-1}}(\lfloor x\rfloor) - f_{p_{k-1}}\left(\left\lfloor\dfrac{x}{p_k}\right\rfloor\right)$
Let $x \ge 1$ be an integer.
Show that $f_5(x+6) - f_5(x) > 0$
I find this problem easy to solve but challenging to answer concisely.
Please let me know if I made a mistake or if my answer could be more concise.
$f_5(x+6) - f_5(x) = \sum\limits_{i|30}\left(\left\lfloor\dfrac{x+6}{i}\right\rfloor-\left\lfloor\dfrac{x}{i}\right\rfloor\right)\mu(i)$
where $\mu(i)$ is the möbius function.
Let $r(x,d)$ be the remainder of $x$ when divided by $d$.
Clearly:
*
*$r(x+6,2) = r(x,2)$
*$r(x+6,3) = r(x,3)$
*$r(x+6,6) = r(x,6)$
So that:
$f_5(x+6) - f_5(x) = (x+6 - x) - \left(\dfrac{x+6 - x}{2}\right) - \left(\dfrac{x+6 - x}{3}\right)- \left(\dfrac{x+6 - x - r(x+6,5) + r(x,5)}{5}\right)+ \left(\dfrac{x+6 - x}{6}\right)+ \left(\dfrac{x+6 - x - r(x+6,10) + r(x,10)}{10}\right) + \left(\dfrac{x+6 - x - r(x+6,15) + r(x,15)}{15}\right) - \left(\dfrac{x+6 - x - r(x+6,30) + r(x,30)}{30}\right)$
$= 2 - \left(\dfrac{6 - r(x+6,5) + r(x,5)}{5}\right)+ \left(\dfrac{6 - r(x+6,10) + r(x,10)}{10}\right) + \left(\dfrac{6 - r(x+6,15) + r(x,15)}{15}\right) - \left(\dfrac{6 - r(x+6,30) + r(x,30)}{30}\right)$
It follows that if $x \not\equiv 4 \pmod 5$:
*
*$\dfrac{6 - r(x+6,5) + r(x,5)}{5} = 1$
If $x \equiv 4 \pmod 5$:
*
*$\dfrac{6 - r(x+6,5) + r(x,5)}{5} - \dfrac{6 - r(x+6,10) + r(x,10)}{10} = 1$
if $r(x,30) < 24$:
*
*$\dfrac{6 - r(x+6,30) + r(x,30)}{30} = 0$
if $r(x,30) \ge 24$:
*
*$\dfrac{6 - r(x+6,15) + r(x,15)}{15} - \dfrac{6 - r(x+6,30) + r(x,30)}{30} = 0$
So that:
$f_5(x+6) - f_5(x) \ge 2 - 1 + 0 = 1$
Edit 1:
I forgot to mention that $x$ is an integer. I have updated the question.
Edit 2:
Updated the recurrence relation in order to make it clearer. One commenter said that the relation wasn't clear.
Edit 3:
Apologies I have no idea why I wrote "kth integer" instead of $k$th prime. Thanks for your patience. Now fixed.
| You can actually solve this using (almost) just very elementary set theory.
Let $X_5$ be the set of integers in $\{x+1,x+2,\ldots, x+6 \}$ that are a multiple of 5. Let $X_3$ be the set of integers in $\{x+1,x_2,\ldots, x+6 \}$ that are a multiple of 3, and let $X_2$ be the set of integers $\{x+1,x+2,x+3,\ldots, x+6 \}$ that are a multiple of 2.
Then $|X_5|$ is either 1 or 2, and if $|X_5|$ is 2 then $X|_5 \cap X_2|$ is at least 1.
Then $|X_3|$ is 2 and $|X_2|$ is 3. And $|X_2 \cap X_3|$ is 1
Case 1: $|X_5| = 2$. Then if $|X_3 \cap X_5 \cap X_2|$ is 1,
then $|X_2 \cup X_3 \cup X_5| = |X_2| + |X_3| + |X_5| - |X_2 \cap X_3| - |X_2 \cap X_5| -|X_3 \cap X_5| + |X_2+X_3+x_5|$ $=3+2+2 -1 -1 -1 + 1 < 6$ so there is an element in $\{x+1,\ldots, x+6 \}$ that is not a multiple of 2, 3, or 5.
If $|X_3 \cap X_5 \cap X_2|$ is 0,
then $|X_2 \cup X_3 \cup X_5| = |X_2| + |X_3| + |X_5| - |X_2 \cap X_3| - |X_2 \cap X_5| + |X_2+X_3+x_5|$ $\geq$ $3+2+2 -1 -1 < 6$ so there is an element in
$\{x+1,\ldots, x+6 \}$ that is not a multiple of 2, 3, or 5.
Can you work through Case 2: $|X_5| =1$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3186750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the fifth expression of Taylor's for $\sin (\tan x)$ around $x=0$
Find the fifth expression of Taylor's for $\sin (\tan x)$ around $x=0$
My try:
$$\sin x=x+r_{1}(x), r_{1}(x)=o(x)$$ $$\tan x=x+\frac{x^3}{3}+\frac{2}{15}x^5+r_{2}(x), r_{2}(x)=o(x^5)$$ So: $$\sin \tan x=(x+\frac{x^3}{3}+\frac{2}{15}x^5)+r_{3}(x+\frac{x^3}{3}+\frac{2}{15}x^5+r_{2}(x)))=x+\frac{x^3}{3}+\frac{2}{15}x^5+r_{4}(x), r_{4}(x)=o(x)$$
That is why fifth expression of Taylor's is $\frac{2}{15}x^5$ for me. However Mathematica say that: $$\sin \tan x=x+\frac{x^3}{6}-\frac{x^5}{40}+o(x^5)$$ Where have I a mistake?
| Hint:
Try starting $$\sin (\tan x)=\tan x - \frac{(\tan x)^3}{3!} + \frac{(\tan x)^5}{5!} +o((\tan x)^5)$$
And you must consider all terms that have $x^5$ in them, for example, $(\tan x)^3= (x+\frac{x^3}{3} +\frac{2}{15}x^5)^3$ have $(\frac{x^5}{3}+\frac{2x^5}{3})$
So, if you gather all the $x^5$ terms respectively, you'll find out that the coefficient of $x^5$ in the series is $$\frac{2}{15} -\frac{1}{6}(\frac{1}{3}+\frac{2}{3})+\frac{1}{5!}= \frac{-3}{120}=\frac{-1}{40}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3187201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit for $e$ and $\frac{1}{e}$ My question concerns the derivation of this: $$e^r = \lim_{n \rightarrow \infty} \left(1 + \frac{r}{n}\right)^n \ \ ...(1).$$
One of the definitions of $e$ is as follows:
$$e = \lim_{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^n.$$
Then, textbooks usually derive equation (1) in the following manner:
\begin{align}
\lim_{n \rightarrow \infty} \left(1 + \frac{r}{n}\right)^n &= \lim_{u \rightarrow \infty} \left(1 + \frac{1}{u}\right)^{ru} \ \ \text{where} \ u = \frac{n}{r}\\
&= \lim_{u \rightarrow \infty} \left(\left(1 + \frac{1}{u}\right)^u\right)^r \\
&= \left(\lim_{u \rightarrow \infty} \left(1 + \frac{1}{u}\right)^u\right)^r \\
&= e^r.
\end{align}
This argument is fine if $r > 0$ since $u \rightarrow \infty$ as $n \rightarrow \infty$, but when $r < 0$, $u \rightarrow - \infty$ as $n \rightarrow \infty$.
How can I extend the proof for (1) where $r$ is any real number?
When $r = 0$, $\lim_{n \rightarrow \infty} (1+0/n)^n = 1$ (Although, I should be careful about evaluating limits that look like $``1^{\infty}"$.)
Here's my attempt so far for the case where $r < 0$:
\begin{align}
\lim_{n \rightarrow \infty} \left(1 + \frac{r}{n}\right)^n &= \lim_{u \rightarrow \infty} \left(1 - \frac{1}{u}\right)^{-ru} \ \text{where} \ u = -\frac{n}{r} \\
&= \left(\lim_{u \rightarrow \infty} \left(1 - \frac{1}{u}\right)^u\right)^{-r}.
\end{align}
My question boils down to how to show the following limit from the definition above for $e$ $$\lim_{n \rightarrow \infty} \left(1 - \frac{1}{n}\right)^n = \frac{1}{e}.$$
Thanks.
| Consider proving the reciprocal, i.e.
$$
\lim \left(1 - \frac 1n\right)^{-n} = \mathrm e.
$$
The expression inside could be rewritten as
$$
\left(1 - \frac 1n \right)^{-n} = \left(\frac {n-1}n\right)^{-n} = \left(\frac n{n-1}\right)^n = \left(1 + \frac 1{n-1}\right)^n = \left(1 + \frac 1{n-1}\right)^{n-1} \cdot \left(1 + \frac 1{n-1}\right).
$$
Now use the definition of $\mathrm e$:
$$
\lim_n \left(1 + \frac 1{n-1}\right)^{n-1} = \mathrm e.
$$
Since
$$
\lim_n \left(1 + \frac 1{n-1}\right) = 1,
$$
we conclude that
$$
\lim_n \left(1 - \frac 1n \right)^{-n} = \lim_n \left(1 + \frac 1{n-1}\right)^{n-1} \cdot \lim_n \left(1 + \frac 1{n-1}\right) = \mathrm e \cdot 1 = \mathrm e.
$$
by the law of arithmetic operations of limits.
Therefore the original limit is $1/\mathrm e$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3187576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
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find the maximum value of $xy + yz +zx$ find the maximum value of $$xy + yz +zx$$given that $x+2y+z=4$
my attempt :
$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx) $
or $2S=2(xy+zx+zy)=(x+y+z)^2 -x^2-y^2-z^2=(4-y)^2-x^2-y^2-z^2$
$2S=-x^2-z^2-8y+16=-x^2-z^2+4x+4z$
from the the above we can say due to symmetry maximum value occurs at $x=z$
hence $S=-x^2+4x$ whose maximum is 4
is this right or/and is there a better way ??
| With a Lagrangian multiplier $\lambda$ in a Lagrangian $L:=xy+yz+zx+\lambda (4-x-2y-z)$, $0=\partial_x L=y+z-\lambda$ etc. gives $\lambda=y+z=\frac{z+x}{2}=x+y$. Comparing the first and last of these expressions for $\lambda$ gives $x=z$, so $y+z=z$ and $y=0$. Then $x=\frac{4-2\times 0}{2}=2$, so $z=2,\,xy+yz+zx=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3192014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find Taylor series for $\sin x$ at $\frac{\pi}{6}$ and estimate $\sin 38^\circ$ I want to see if I understand Taylor series correctly. Mind checking this work?
(a) So $f(x) = \sin{x}$ and $a = \frac{\pi}{6}$, $n = 4$ (I want to find $T_4(x)$ and I want to find the accuracy of the approximation when $x$ lies in the interval $[0,\frac{\pi}{3}]$.
So I write:
$$f'(x) = \cos{x},\quad
f''(x) = -\sin{x},\quad
f''(x) = -\cos{x},$$
$$
f^{(4)}(x) = \sin{x},\quad
f^{(5)}(x) = \cos{x}$$
and thus
$$f'(\frac{\pi}{6}) = \frac{\sqrt{3}}{2},\quad
f''(\frac{\pi}{6}) = \frac{-1}{2},\quad
f'''(\frac{\pi}{6}) = \frac{-\sqrt{3}}{2}$$
$$f^{(4)}(\frac{\pi}{6}) = \frac{1}{2},\quad
f^{(5)}(\frac{\pi}{6}) = \frac{\sqrt{3}}{2},$$
which gives
$$T_4(x) = \frac{1}{2} + \frac{\sqrt{3}}{2}(x-\frac{\pi}{6}) + \frac{\frac{-1}{2}(x-\frac{\pi}{6})^2}{2!} + \frac{\frac{-\sqrt{3}}{2}(x-\frac{\pi}{6})^3}{3!} + \frac{\frac{1}{2}(x-\frac{\pi}{6})^4}{4!}.$$
Question 1: Is the Taylor polynomial of the 4th degree above correct?
(b) Now I try to find the error in approximating $\sin\frac{\pi}{6}$ by the first 4 terms using Taylor's Inequality in my textbook:
First of all, $f^{(5)}(x) = \cos{x}$. For $x$ with $0 \leq x \leq \frac{\pi}{3}$, $\cos x$ is largest at $0$ so $f^{(5)}(x) \leq 1$ and $M = 1$.
So $$R_5(x) = \frac{1}{5!}(x-a)^5$$
I'm stuck here.
Question 2: Can I say that $|x-a| < d$ where d = $\frac{\pi}{6}$?
(c) To estimate $\sin 38^{\circ}$, I write:
$$\sin 38^{\circ} = \sin(\frac{38}{180}\pi) = \sin(\frac{19}{90}\pi)$$
Question 3: Do I just plugin that value to the $T_4$ expression?
Question 4: More generally, what does it mean to be a Taylor series about $\frac{\pi}{6}$?
| May be faster.
Let $x=y+\frac \pi 6$ making
$$\sin(x)=\sin \left(y+\frac{\pi }{6}\right)=\frac 12 \big( \sqrt{3} \sin (y)+\cos (y)\big)$$ for which the derivatives at $y=0$ are very simple.
As a result
$$\frac 12 \big( \sqrt{3} \sin (y)+\cos (y)\big)= \sum_{n=0}^\infty \frac{\sqrt{3} \sin \left(\frac{\pi n}{2}\right)+\cos \left(\frac{\pi
n}{2}\right)}{2 (n!)}\, y^n$$
Replace $y$ by $x-\frac{\pi }{6}$.
| {
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"url": "https://math.stackexchange.com/questions/3193017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Different ways of computing probability
There are twelve unique watches and five men.
Each of the five men are asked to choose a watch for themselves.
What is the probability that at-least two of them choose the same watch.
I could think of two different approaches of solving above problem.
1st approach:
$$ \text{answer} = 1 - \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 }{ 12^5} \approx 0.618055$$
2nd approach: Consider $x$ such that
$$\begin{aligned}
x = &\hphantom{{}+{}} \text{no of ways exactly}~\textbf{two}~\text{men choose same watch} \\
&{}+ \text{no of ways exactly}~\textbf{three}~\text{men choose same watch} \\
&{}+\text{no of ways exactly}~\textbf{four}~\text{men choose same watch} \\
&{}+ \text{no of ways exactly}~\textbf{five}~\text{men choose same watch} \\
= &\hphantom{{}+{}} {5\choose 2} \cdot 12 \cdot {11\choose 3} \cdot 3! \\
&+ {5\choose 3} \cdot 12 \cdot {11\choose 2} \cdot 2! \\
&+ {5\choose 4} \cdot 12 \cdot {11\choose 1} \cdot 1! \\
&+ {5\choose 5} \cdot 12 \cdot {11\choose 0} \cdot 0! \\
={} &118800 + 13200 + 660 + 12 = 132672
\end{aligned}$$
Then $\text{answer} = x\, / \,12^5 = 132672\, / \, 12^5 \approx 0.53317$
The two approaches have different answers.
I believe the first approach has no issues, which means i am counting short in the second approach. But not able to figure out how?
| As DJohnM pointed out in the comments, it is possible that there is more than one watch that at least two men pick. This can happen in two ways: three men pick one model and two other men pick a different model or two men pick one model, two other men pick a second model, and one man chooses a third model.
Three men choose one model and two other men choose a second model: There are $\binom{5}{3}$ ways that three of the men can select the same model and twelve models for them to choose. There are eleven ways for the other two men to pick the same model. Hence, there are
$$\binom{5}{3}\binom{12}{1}\binom{2}{2}\binom{11}{1} = 1320$$
such distributions.
Two men choose one model, two other men choose a second model, and the fifth man picks a third model: There are $\binom{12}{2}$ ways to select which two models are each picked by two men, $\binom{5}{2}$ ways for two of the men to pick the more expensive of those two models (or some other feature that distinguishes them if they are the same price), $\binom{3}{2}$ ways for two of the other three men to pick the other of those models, and ten ways for the remaining man to pick one of the remaining models. Hence, there are
$$\binom{12}{2}\binom{5}{2}\binom{3}{2}\binom{1}{1}\binom{10}{1} = 19800$$
such distributions.
Adding these to the total gives
$$118800 + 13200 + 660 + 12 + 1320 + 19800 = 153792$$
Dividing this amount by $12^5$ yields $\approx 0.6180555556$, which agrees with the answer obtained from your simpler first approach.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving a trigonometric system of equations related to addition formulas. I have the following trigonometric system of equations.
$
\begin{align*}
&& \sin(x)\cos(y) &= \frac{1+\sqrt{3}}{4} \\
&& \cos(x)\sin(y) &= \frac{-1+\sqrt{3}}{4} \\
\end{align*}
$
I am trying to find a quick way to solve this system for $x$ and $y$ with $0<x,y<\frac{\pi}{2}$. I have managed to find one set of solutions ($x=\frac{\pi}{4}$ and $y=\frac{\pi}{12}$) but I am unable to find the other pair which I can clearly see after graphing the system.
I was able to deduce, by making use of the addition formula for $sin$, that:
$\begin{align*}
&&\sin(x + y) = \frac{\sqrt{3}}{2} \implies x + y = \frac{\pi}{3}
\end{align*}$
Using this fact, I found my pair of solutions by forming a quadratic in $\tan(x)$ after playing around with addition formula for $\tan$.
I am certain there is an easier way to solve the problem that I am missing. Any tips for solving this system would be greatly appreciated, thanks.
| Hint:
$$\begin{align*}
\sin(x + y) &= \frac{\sqrt{3}}{2} &\implies x + y = \left\{\frac{\pi}{3},\frac{2\pi}{3}\right\}+2\pi n\\
\sin(x - y) &= \frac{1}{2} &\implies x - y = \left\{\frac{\pi}{6},\frac{5\pi}{6}\right\}+2\pi n\\
\end{align*}
$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3195283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
a tricky trig questions from Step If $\theta + \phi + \psi = \pi/2$, show that $\sin^2 \theta + \sin^2 \phi + \sin^2 \psi + 2 \sin \theta \sin \phi \sin \psi = 1$.
By taking $\theta = \phi =\pi/5$ in this equation, or otherwise, show that $\sin(\pi/10)$ satisfies the equation
$$8x^3 + 8x^2 − 1 = 0$$
I got stuck in the first part. I want to prove this by making connection
with $\sin(\theta + \phi + \psi)=1$,but I failed.
| If $\theta+\varphi+\psi=\pi/2$, then $$1=\sin^2\theta+\cos^2\theta=\sin^2\theta+\sin^2(\varphi+\psi)$$
But, since $\sin(\varphi+\psi)=\sin\varphi\cos\psi+\sin\psi\cos\varphi$, we have $$\sin^2(\varphi+\psi)=\sin^2\varphi\cos^2\psi+\sin^2\psi\cos^2\varphi+2\sin\varphi\cos\psi\sin\psi\cos\varphi.
$$
Since $\sin^2\varphi\cos^2\psi=\sin^2\varphi(1-\sin^2\psi)$ and $\sin^2\psi\cos^2\varphi=\sin^2\psi(1-\sin^2\varphi)$. Replacing it on the previous equation, we get
\begin{align}\sin^2(\varphi+\psi)&=\sin^2\varphi+\sin^2\psi-2\sin^2\psi\sin^2\varphi+2\sin\varphi\cos\psi\sin\psi\cos\varphi.\\
&=\sin^2\varphi+\sin^2\psi+2\sin\varphi\sin\psi(\cos\psi\cos\varphi-\sin\psi\sin\varphi)\\
&=\sin^2\varphi+\sin^2\psi+2\sin\varphi\sin\psi(\cos(\varphi+\psi))\\
&=\sin^2\varphi+\sin^2\psi+2\sin\varphi\sin\psi\sin\theta
\end{align}
Replacing this on the very first equation, we get what we wanted,
$$ 1=\sin^2\theta+\sin^2\varphi+\sin^2\psi+2\sin\varphi\sin\psi\sin\theta$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3195445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve for x: $x^3-\lfloor x\rfloor=5$
Solve for x:$$x^3-\lfloor x\rfloor=5$$
My Attempt:
$x^3-5=\lfloor x\rfloor$
Now, $x-1<\lfloor x\rfloor\leq x$
$x-1<x^3-5\leq x$
Not able to proceed from here
| Since $$ x^3-x-6< x^3-x-5\leq 0$$ we have $$(x-2)(x^2+2x+3)<0$$
Since $x^2+2x+3 = (x+1)^2+2 $ we have $ \boxed{x<2}$.
On the other side we have $$0<x^3-x-4<x(x^2-1)$$
so $\boxed{x\in (-1,0)\cup (1,\infty)}$. Both together we get $$\boxed{x\in (-1,0)\cup (1,2)}$$
Now if $x\in (-1,0)$ then $\lfloor x\rfloor = -1$ so $x^3 = 4$ so no solution.
and if $x\in (1,2)$ then $\lfloor x\rfloor=1$ so $x^3= 6$ so $x=\sqrt[3]{6}$ is a solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3195877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
if $ A \in R^{n \times n}$ , $A > 0$ and $ b \in R^n$ then the function $\frac{1}{2}\langle Ax,x\rangle - \langle b,x\rangle$ is convex in $R^n$ Show by direct estimates that if $ A \in R^{n \times n}$ , $A > 0$ and $ b \in R^n$ then the function
$$\frac{1}{2}\langle Ax,x\rangle - \langle b,x\rangle$$ with $x$ is convex on $R^n$.
My approach: A function $g : i \rightarrow R$ is said to be convex if $g(tx + (1-t)y) \le tg(x) + (1-t)g(y)$, $\forall x,y$ in $i$, and $0 \le t \le 1$
Hence, $g(x) = \frac{1}{2}\langle Ax,x\rangle - \langle b,x\rangle$ with $x$, $\Rightarrow g(tx + (1-t)y) = \frac{1}{2}\langle A[(tx+(1-t)y)], tx + (1-t)y\rangle - \langle b, tx + (1-t)y\rangle$.
From here I got stuck expanding the term, any help is highly appreciated.
| Since you're not comfortable showing $f(x) = x^2$ is convex, let's begin with that. Naturally, you can prove this by showing that $f''(x) \ge 0$ for all $x$, but we want to prove it by definition.
Suppose $x, y \in \Bbb{R}$ and $\lambda \in [0, 1]$. Then
\begin{align*}
&\lambda f(x) + (1 - \lambda) f(y) - f(\lambda x + (1 - \lambda)y) \\
= \; &\lambda x^2 + (1 - \lambda)y^2 - (\lambda x + (1 - \lambda)y)^2 \\
= \; &\lambda x^2 + (1 - \lambda)y^2 - \lambda^2 x^2 - 2\lambda x(1 - \lambda) y - (1 - \lambda)^2 y^2 \\
= \; &\lambda(1 - \lambda)x^2 + \lambda(1 - \lambda)y^2 - 2\lambda(1 - \lambda)xy \\
= \; &\lambda(1 - \lambda)(x^2 - 2xy + y^2) \\
= \; &\lambda(1 - \lambda)(x - y)^2 \ge 0.
\end{align*}
Thus,
$$\lambda f(x) + (1 - \lambda) f(y) \ge f(\lambda x + (1 - \lambda)y),$$
and $f$ is convex.
Now, suppose $x, y \in \Bbb{R}^n$ and $\lambda \in [0, 1]$. By linearity of $\sqrt{A}$, triangle inequality, and positive scalar homogeneity of the norm respectively,
\begin{align*}
\|\sqrt{A}(\lambda x + (1 - \lambda)y)\| &= \|\lambda \sqrt{A} x + (1 - \lambda) \sqrt{A} y\| \\
&\le \|\lambda \sqrt{A} x\| + \|(1 - \lambda) \sqrt{A} y\| \\
&= \lambda\|\sqrt{A} x\| + (1 - \lambda) \|\sqrt{A} y\|.
\end{align*}
Thus the map $x \mapsto \| \sqrt{A} x\|$ is also convex. Using this and the previous inequality,
\begin{align*}
&\lambda\|\sqrt{A} x\|^2 + (1 - \lambda)\|\sqrt{A} y\|^2 - \|\sqrt{A}(\lambda x + (1 - \lambda)y)\|^2 \\
\ge \; &\lambda\|\sqrt{A} x\|^2 + (1 - \lambda)\|\sqrt{A} y\|^2 - (\lambda\|\sqrt{A} x\| + (1 - \lambda)\|\sqrt{A} y\|)^2 \\
= \; &\lambda(1 - \lambda)(\|\sqrt{A} x\| - \|\sqrt{A} y\|)^2 \ge 0.
\end{align*}
Therefore, the map $x \mapsto \|\sqrt{A} x\|^2$ is convex, as required. Use the fact that
$$\|\sqrt{A} x\|^2 = \langle \sqrt{A} x, \sqrt{A} x\rangle = \langle\sqrt{A} \sqrt{A} x, x\rangle = \langle A x, x\rangle,$$
as $\sqrt{A}$ is positive and hence Hermitian.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3198198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Value of $|z_{1}|^2+|z_{2}|^2+|z_{3}|^2$ is
If $|z_{1}+z_{2}|=|z_{1}|-|z_{2}|=2$ and $|2z_{2}+2i(z_{3}-z_{2})|=|2iz_{3}+(1-2i)z_{2}|=10$
Where $z_{1}=3+4i.$ Then value of $|z_{1}|^2+|z_{2}|^2+|z_{3}|^2$ is
Try: Let $z_{2}=a+ib$ and $z_{3}=c+id$. Then we have $(a+c)^2+(b+d)^2=4.$ and $\sqrt{a^2+b^2}-\sqrt{c^2+d^2}=2$
I have seems that there is an easy way other then substituting $z_{1}=a+ib$ and $z_{2}=c+id.$
Could some help me to solve it , Thanks
| We have $$|z_{1}|=\sqrt{3^2+4^2}=5,\qquad\qquad |z_{2}|=|z_{1}|-2=5-2=3\tag1$$
Now, we see that
$$|2z_{2}+2i(z_{3}-z_{2})|=|2iz_{3}+(1-2i)z_{2}|=10$$
is equivalent to
$$|\alpha+\beta|=|\alpha-\beta|\tag2$$
and
$$|\alpha+\beta|=10\tag3$$
where
$$\alpha=2i(z_3-z_2)+\frac 32z_2, \qquad\beta=\frac 12z_2$$
From $(2)$, we have
$$(\alpha+\beta)(\bar\alpha+\bar\beta)=(\alpha-\beta)(\bar\alpha-\bar\beta)\iff \alpha\bar\beta+\overline{\alpha\bar\beta}=0\iff \alpha\bar\beta=ki$$
where $k\in\mathbb R$.
So, from $(3)$, we get
$$\begin{align}|\alpha+\beta|=10&\iff (\alpha+\beta)(\bar\alpha+\bar\beta)=100
\\\\&\iff \left(\frac{ki}{\bar\beta}+\beta\right)\left(\frac{-ki}{\beta}+\bar\beta\right)=100
\\\\&\iff \frac{k^2}{|\beta|^2}+|\beta|^2=100
\\\\&\iff \frac{4k^2}{9}+\frac 94=100
\\\\&\iff k=\pm\frac 34\sqrt{391}\tag4\end{align}$$
Also, we have
$$\begin{align}\alpha\bar\beta=ki& \iff\left(2i(z_3-z_2)+\frac 32z_2\right)\left(\frac 12\overline{z_2}\right)=ki
\\\\&\iff i(z_3\overline{z_2}-|z_2|^2)+\frac 34|z_2|^2=ki
\\\\&\iff z_3=\frac{9+k+\frac{27}{4}i}{\overline{z_2}}\tag5
\end{align}$$
It follows from $(1)(4)(5)$ that
$$|z_1|^2+|z_2|^2+|z_3|^2=5^2+3^2+\frac{(9+k)^2+(\frac{27}{4})^2}{|\overline{z_2}|^2}=\color{red}{\frac{145\pm 3\sqrt{391}}{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3203077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Show that $5^n\mid2^{5^{n-1}}+3^{5^{n-1}}$ for $n \gt 0$. Show that $5^n\mid 2^{5^{n-1}}+3^{5^{n-1}}$ for $n \gt 0$.
I tried using Euler's Theorem: $2^{4\cdot5^{n-1}} \equiv 1 \mod{5^n}$ and $3^{4\cdot5^{n-1}} \equiv 1 \mod{5^n}$ but I can't find a way to use it.
I also tried showing that the sum of the remainders of $2^{5^{n-1}}+3^{5^{n-1}}$ is a multiple of $5^n$ with induction but got stuck again.
How do I prove this kind of statement?
| Simply induct on $n$. Base cases are clear. Suppose, for $k=n-1$, $5^k \mid 2^{5^{k-1}}+3^{5^{k-1}}$. The goal is to prove now the claim for $k=n$. Let $a=2^{5^{k-1}}$ and $b=3^{5^{k-1}}$. What we want to prove is, $5^n\mid a^5+b^5$, provided $5^{n-1}\mid a+b$. Now, using $a^5+b^5=(a+b)(a^4-a^3 b +a^2 b^2 -ab^3+b^4)$, and the fact that, $a\equiv -b\pmod{5}$, it follows that, $a^4-a^3 b +a^2 b^2 -ab^3+b^4\equiv 5b^4\equiv 0\pmod{5}$. Since $5^{n-1}\mid a+b$, their product is therefore divisible by $5^n$, as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3203339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Completion for this proof that $\sum_{k=0}^n {n \choose k} \cos\left((k+1)x\right) = 2^n \cos^n\frac x 2 \cos\frac{(n+2)x}{2}$ There's this exercise in a textbook:
$$\sum_{k=0}^n {n \choose k} \cos\left((k+1)x\right) = 2^n \cos^n\frac x 2 \cos\frac{(n+2)x}{2}$$
I've been missing from some classes at school and I copied the notes from one of my classmates, but this proof given by my teacher seems incomplete to me, because it only considers certain cases for x. However, I ploted both sides of the equality in the title and it works for every x. So how can this proof be completed?
Proof:
$$ S_1 = \sum_{k=0}^n {n \choose k} \cos\left((k+1)x\right) \\
S_2 = \sum_{k=0}^n {n \choose k} \sin\left((k+1)x\right) \\
S_1 + iS_2 = \sum_{k=0}^n {n \choose k}\cos\left((k+1)x\right) + i\sin\left((k+1)x\right) \\
z = \cos x + i\sin x \\
z^k = \cos kx + i\sin kx \hspace{1in}\text{(Moivre)} \\
\begin{align*} S_1 + iS_2 & = \sum_{k=0}^n {n \choose k}z^{k+1} \\ &= z\sum_{k=0}^n{n \choose k}z^k \\ &= z(1+z)^n \hspace{1in} \text{(Newton)}\end{align*}$$
$$|z| = 1, \varphi_1 = arg(z) = x \\
\begin{align*}
|1+z| &= \sqrt{(1+\cos x)^2 + \sin^2 x} \\ &= \sqrt{1 + 2\cos x + \cos^2 x + \sin^2 x} \\ &= \sqrt{2 + 2\cos x} \\ &= \sqrt{2(1+\cos x)} \\ &= \sqrt{2\cdot 2\cos^2 \frac x 2} \\ &= 2\left|\cos\frac x 2\right| \\ &= \color{red}{2\cos\frac x 2} \end{align*}$$
$$ 1+z = |1+z|\left(\cos \varphi_2 + i\sin\varphi_2\right) \\
\begin{align*}
\color{red}{\varphi_2 } &\color{red}{= \arctan\frac{\sin x}{1 + \cos x}} \\
&=\arctan\frac{2\sin\frac x 2\cos\frac x 2}{2\cos^2\frac x 2} \\ &= \arctan\left(\tan\frac x 2\right) \\
&= \color{red}{\frac x 2}\end{align*}$$
$$ z(1+z)^n = \left(\cos x + i\sin x\right)\left[2\cos\frac x 2\left(\cos\frac x 2 + i\sin\frac x 2\right)\right]^n = 2^n\cos^n\frac x 2 \left[\cos\left(x + \frac{nx}{2}\right) + i\sin\left(x + \frac{nx}{2}\right)\right] = S_1 + iS_2 \\
S_1 = {\rm Re}(S_1 + iS_2) = 2^n\cos^n\frac x 2 \cos \frac {(n+2)x}{2}$$
In red, I put the parts where the teacher only considered certain cases for x.
| $$
\begin{aligned}
S=& \sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right) \cos((k+1) x)+i \sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right) \sin (k+1) x \\
=& \sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right) e^{(k+1) x i}=e^{x i} \sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right)\left(e^{x i}\right)^{k}
\end{aligned}
$$
By Binomial Theorem, $$
\begin{aligned}
S &=e^{x i}\left(1+e^{x i}\right)^{n} \\
&=e^{x i}\left(e^{\frac{x i}{2}}\left(e^{\frac{x i}{2}}+e^{-\frac{x}{2} i}\right) \right) ^{n}\\
&=e^{x i} \cdot e^{\frac{n x}{2} i}\left(2 \cos \frac{x}{2}\right)^{n} \\
&=2^{n} \cos ^{n} \frac{x}{2} e^{\frac{(n+2)x}{2} i}
\end{aligned}
$$
Comparing the real parts on both sides yields
$$\sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right) \cos((k+1) x)= 2^{n} \cos ^{n} \frac{x}{2} \cos \frac{(n+2)x}{2} $$
As a bonus, $$\sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right) \sin((k+1) x)= 2^{n} \cos ^{n} \frac{x}{2} \sin \frac{(n+2)x}{2} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3205309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Poles of MIMO transfer function matrix. I am looking for the poles of the following transfer function matrix:
$$G_a(s)=\begin{bmatrix} \frac{1}{(s+1)(s+2)} & \frac{-1}{(s+1)(s+2)} \\ \frac{s^2+s-4}{(s+1)(s+2)} & \frac{2s^2-s-8}{(s+1)(s+2)} \\ \frac{s-2}{s+1} & \frac{2(s-2)}{s+1} \end{bmatrix}$$
There are, as I understand, a total of six $1 \times 1$ principle minors and two $2 \times 2$ principle minors.
The multiplicity of $s =-1$ in all $1 \times 1$ minors is $1$. The multiplicity of $s = -1$ in both $2 \times 2$ minors is $4$. So the multiplicity of $s = -1$ equals $4 - 1 =3$. Is this correct?
And what about $s+2$ because this term isn't present in all $1 \times 1$ principle minors?
I only found one file explaining this and it only covered really basic examples and not a $3 \times 2$ transfer funciton matrix.
| The poles can be found in the least common denominator of all principle minors. The minors of order $1$ are:
\begin{align}
M1_{11}=\frac{1}{(s+1)(s+2)}, \ M1_{12}=\frac{-1}{(s+1)(s+2)}, \ M1_{21}=\frac{s^2+s-4}{(s+1)(s+2)},\\
M1_{22}=\frac{2s^2-s-8}{(s+1)(s+2)}, \ M1_{31}=\frac{s-2}{s+1}, \ M1_{32}=\frac{2(s-2)}{s+1}
\end{align}
And the minors of order $2$ are:
\begin{align}
M2_{1}=\frac{3s(s-2)}{(s+1)^2(s+2)}, \ M2_{2}=M2_{3}=\frac{3(s-2)}{(s+1)^2(s+2)}
\end{align}
The least common denominator is: $(s+1)^2(s+2)$. The poles are thus:
\begin{align}
s=-1 \quad \text{(2x)} \\
s=-2
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3208143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Find extreme values for $f(x,y,z)=xyz$ given the constraints $g_{1}(x,y,z)=x+y+z-5$ and $g_{2}(x,y,z)=xy+yz+zx-8$ using Lagrange multipliers method. I want to calculate extreme values for $f(x,y,z)=xyz$ given the constraints $g_{1}(x,y,z)=x+y+z-5$ and $g_{2}(x,y,z)=xy+yz+zx-8$ using Lagrange multipliers method. Im skeptical about my solution for this problem which goes as follow. First I got:
$$\nabla f= (yz)i+(xz)j+(xy)k,$$
$$\nabla g_{1}=i+j+k,$$
and
$$\nabla g_{2}=(y+z)i+(x+z)j+(y+x)k.$$
So from having the equality $\nabla f = \lambda_{1} \nabla g_{1} + \lambda_{2} \nabla g_{2}$ I got the following equation system:
\begin{align*}
yz&= \lambda_{1} + \lambda_{2}(y+z) \\
xz&=\lambda_{1} + \lambda_{2}(x+z)\\
xy&=\lambda_{1} + \lambda_{2}(y+z)\\
\end{align*}
But Im really stucked finding the extrem values from the last equations system. So far, I realize that if I sum up the three equations from the system and the way $g_{2}(x,y,z)$ is defined I obtained:
$$xy+yz+zx=8=3(\lambda_{1})+3(\lambda_{2})(x+y+z).$$
But from the way $g_{1}(x,y,z)$ is defined I got that last equation is
$$8=3(\lambda_{1})+3(\lambda_{2})(5).$$
So I found by trial and error that $\lambda_{1}=1$ and $\lambda_{2}=\frac{1}{3}$, im not sure if there is more possible values for $\lambda_{1}$ and $\lambda_{2}$ satysfing the last equation. Anyways, from here I have been trying to find the values for $x,y$ and $z$ substituting the values I obtained for $\lambda_{1}$ and $\lambda_{2}$ in the equation from the original system of equations. For example, from the first equation I got
$$yz=1+\frac{1}{3}(y+z)$$
But finding $x,y$ and $z$ that way is hard. I´ve been thinking also that from some equation before I have that:
$$8=3\lambda_{1}+3\lambda_{2}(x+y+z)=3+(x+y+z)$$.
The problem of finding $x,y$ and $z$ this way is that I got a lot of points satisfying this last equation. Just to list a few:
$$(5,0,0),(0,5,0),(0,0,5),(1,1,3),...$$. Basically, all the points $(x,y,z) \in \mathbb{R}^{3}$ which satisfy $g_{1}(x,y,z)=0$ but I can tell there is something wrong from the fact a lot of these points dont satisfy the equations system obtained from $\nabla f = \lambda_{1} \nabla g_{1} + \lambda_{2} \nabla g_{2}$.
| Here $f(x,y,z) = xyz$, $g(x, y, z) = x+y+z-5$, $h(x, y, z) = xy+yz+zx-8$
So $L(x, y, z, \lambda, \mu) = f(x, y, z) + \lambda g(x, y, z) + \mu h(x, y, z)$
$L_x = yz + \lambda +\mu(y+z) = 0$ --(1)
$L_y = xz + \lambda +\mu(x+z) = 0$ --(2)
$L_z = xy + \lambda +\mu(x+y) = 0$ --(3)
$L_{\lambda} = x+y+z = 5$ --(4)
$L_{\mu} = xy+yz+zx = 8$ --(5)
Multiply (1),(2),(3) by $x,y,z$ respectively, we get
$xyz +\lambda x +\mu(xy+xz) = 0$ --(6)
$xyz +\lambda y +\mu(xy+yz) = 0$ --(7)
$xyz +\lambda z +\mu(xz+yz) = 0$ --(8)
from (6) - (7) we get $(x-y)(\lambda+\mu z) = 0$ --(9)
from (7) - (8) we get $(y-z)(\lambda+\mu x) = 0$ --(10)
from (8) - (6) we get $(z-x)(\lambda+\mu y) = 0$ --(11)
Now if we take $x=y$ from (9), we see that it satisfies (10) and (11) also where $\lambda = -\mu x = -\mu y$, we are $\textbf{not}$ assuming here $z=x=y$
Then putting the vaue of x in (4) we get $2y+z = 5$, putting these values in (5) we get
$y^2 + 2y(5-2y) = 8$ or $3y^2 - 10y + 8 = 0$
Solving for y we get $y = 2,4/3$, so one solution is $2,2,1$ another solution is $4/3,4/3,7/3$
The solution $4/3,4/3,7/3$ gives the maximum value 4.148148,
solution $2,2,1$ gives the minimum value 4.
I am actually amazed to see such a narrow band!.
| {
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"url": "https://math.stackexchange.com/questions/3213915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the rank of the following matrix I am thinking on the following problem and have some results but I need a little help. Can some one give me a little hint?
Suppose $a_{ij}= \cos (i +j )$ in the matrix $A$ find $\operatorname{rank}(A)$.
Here is my attempt:
We know that:
$${\cos(\theta)} = {{e^{i \theta} + e^ {-i \theta} }\over{2}}$$
so we can write :
$$
A={1 \over 2} \begin{pmatrix}
e^{2i} + e^ {-2i} & e^{3i} + e^ {-3i} & \cdots & e^{i(2+n-1)} + e^ {-i(2+n-1)} \\
e^{3i} + e^ {-3i} & e^{4i} + e^ {-4i} & \cdots & e^{i(3+n-1)} + e^ {-i(3+n-1)} \\
\vdots & \vdots & \ddots & \vdots \\
e^{i(2+n-1)} + e^ {-i(2+n-1)} & e^{i(3+n-1)}+ e^ {-i(3+n-1)} & \cdots & e^{i(2+2n-2)}+ e^ {-i(2+2n-2)} \\
\end{pmatrix}
$$
Now we try to show that $\det(A)=0$ , which show us $\operatorname{rank}(A) < n$. First we show that for odd $n$ , suppose $n=2k+1$, we use middle column of matrix , to change first and last column of the matrix to compute determinant, the middle column will be $(k+1)^\mathrm{th}$ column so suppose :
$$
C_{k+1}={1 \over 2} \begin{pmatrix}
e^{i(2+k)} + e^ {-i(2+k)} \\
e^{i(2+k+1)} + e^ {-i(2+k+1)} \\
\vdots \\
e^{i(2+k+n-1)} + e^ {-i(2+k+n-1)} \\
\end{pmatrix}
$$
Now if we compute $C_{k+1} \times (-e^{-k})+C_1$ we have :
$$
{1 \over 2} \begin{pmatrix}
e^ {-2i} - e^{-i(2+2k)} \\
e^ {-3i} - e^{-i(2+2k+1)} \\
\vdots \\
e^ {-i(2+n-1)} - e^{-i(2+2k+n-1)} \\
\end{pmatrix}
$$
and now if we compute $C_{k+1} \times (-e^k) +C_n$ we have :
$$
{1 \over 2} \begin{pmatrix}
e^{-i(2+2k)} - e^{-2i} \\
e^{-i(2+2k+1)} - e^{-3i} \\
\vdots \\
e^{-i(2+2k+n-1)} - e^{-i(2+n-1)} \\
\end{pmatrix}
$$
So it is clear if we add $C_{k+1} \times (-e^k) +C_n$ to the $C_1$, $A$ will be $0$ which mean $\det(A)=0$.
For the case $n$ even, I was able to cease a column with $n-1$ zeroes but I could not show that $\det(A)=0$, but I used mathematica and verified for $n \in \{ 3,4,5 \}$ determinant is $0$, also at first I thought $\operatorname{rank}(A)=n$ but now I know it is not and get a little bit confused, can some one give me a little hint?
Thanks.
| Let
$$
A=\begin{pmatrix}
\cos(2) & \cos(3) & \cdots & \cos(2+n-1) \\
\cos(3) & \cos(4) & \cdots & \cos(3+n-1) \\
\vdots & \vdots & \ddots & \vdots \\
\cos(2+n-1) & \cos(2+n) & \cdots & \cos(2+n-1+n-1) \\
\end{pmatrix}
$$
Suppose $V_i$ is the $i$th row of the $A$ , it is clear:
$$
V_i=\begin{pmatrix}
\cos(2+i-1) & \cos(2+i-1+1) & \cdots & \cos(2+i-1+n-1) \\
\end{pmatrix}
$$
Using $\cos(i+j)=\cos(i)\cos(j)-\sin(i)\sin(j)$ and $\cos(i)-\cos(j)=-2\sin({(i-j) \over 2})\sin({(i+j) \over 2})$ we show that :
$$
V_i-{{V_i-V_{i-2}} \over 2}-V_{i-1}\cos(1)=0
$$
With computin ${{V_i-V_{i-2}} \over 2}$ we have :
$$
{{V_i-V_{i-2}} \over 2}=\begin{pmatrix}
-\sin(1)\sin(2+i-2) & -\sin(1)\sin(2+i-1) & \cdots & -\sin(1)\sin(2+i-2+n-1) \\
\end{pmatrix}
$$
No using the identities we can write :
$$
V_i=\begin{pmatrix}
\cos(1)\cos(2+i-2)-\sin(1)\sin(2+i-2) & \cos(1)\cos(2+i-1)-\sin(1)\sin(2+i-1) & \cdots & \cos(1)\cos(2+i-2+n-1)-\sin(1)\sin(2+i-2+n-1) \\
\end{pmatrix}
$$
Thus $V_i-{{V_i-V_{i-2}} \over 2}$ equals to :
$$
V_i=\begin{pmatrix}
\cos(1)\cos(2+i-2) & \cos(1)\cos(2+i-1) & \cdots & \cos(1)\cos(2+i-2+n-1) \\
\end{pmatrix}
$$
Now it is easy to see that :
$$
V_i-{{V_i-V_{i-2}} \over 2}-V_{i-1}\cos(1)=0
$$
It is show that all the $V_i, 3\le i\le n$ genrated by first two row thus $rank(A)=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3214949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Double sum factorial manipulation $$\sum_{B = 0}^{n-1} \sum_{A = 0}^{n-B-1} \frac{(n-1)!}{B!(n-B-1)!} \frac{(n-B-1)!}{A!(n-B-A-1)!} \frac{A}{A+B+1}$$
This is driving me nuts! Is there anyway to reduce
$$\sum_{B = 0}^{n-1} \sum_{A = 0}^{n-B-1} \frac{(n-1)!}{A!B!(n-A-B-1)!} \frac{A}{A+B+1}$$
beyond
$$\sum_{B = 0}^{n-1} \sum_{A = 0}^{n-B-1} \frac{(n-1)!}{(A-1)!B!(n-A-B-1)!} \frac{1}{A+B+1}$$
I can't figure out how to deal with that $\frac{1}{A+B+1}$ term in any way that brings it inside the $\frac{1}{n-A-B-1}$ term. Is this not possible?
|
We obtain
\begin{align*}
\color{blue}{\sum_{B=0}^{n-1}}&\color{blue}{\sum_{A=0}^{n-B-1}\binom{n-1}{B}\binom{n-1-B}{A}\frac{A}{A+B+1}}\\
&=\sum_{B=0}^{n-1}\binom{n-1}{B}\sum_{A=1}^{n-1-B}\binom{n-1-B}{A}A\int_{0}^{1}z^{A+B}\,dz\tag{1}\\
&=\int_{0}^{1}\sum_{B=0}^{n-1}\binom{n-1}{B}\sum_{A=1}^{n-1-B}\binom{n-2-B}{A-1}(n-1-B)z^{A+B}\,dz\tag{2}\\
&=\int_{0}^{1}\sum_{B=0}^{n-1}\binom{n-1}{B}(n-1-B)\sum_{A=0}^{n-2-B}\binom{n-2-B}{A}z^{A+B+1}\,dz\tag{3}\\
&=\int_{0}^{1}\sum_{B=0}^{n-1}\binom{n-1}{B}(n-1-B)z^{B+1}(1+z)^{n-2-B}\,dz\tag{4}\\
&=\int_{0}^{1}\sum_{B=0}^{n-1}\binom{n-2}{B}(n-1)z^{B+1}(1+z)^{n-2-B}\,dz\tag{5}\\
&=(n-1)\int_{0}^{1}z\sum_{B=0}^{n-2}\binom{n-2}{B}z^B(1+z)^{n-2-B}\,dz\\
&=(n-1)\int_{0}^{1}z(1+2z)^{n-2}\,dz\tag{6}\\
&=\frac{n-1}{4}\int_{1}^{3}(u-1)u^{n-2}\,du\tag{7}\\
&=\frac{n-1}{4}\int_{1}^{3}\left(u^{n-1}-u^{n-2}\right)\,du\\
&=\frac{n-1}{4}\left.\left(\frac{1}{n}u^n-\frac{1}{n-1}u^{n-1}\right)\right|_{1}^{3}\\
&=\frac{n-1}{4}\left(\frac{3^n-1}{n}-\frac{3^{n-1}-1}{n-1}\right)\\
&\,\,\color{blue}{=\frac{3^{n-1}}{2}-\frac{3^n-1}{4n}}
\end{align*}
Comment:
*
*In (1) we use the integral to get rid of the denominator. We also start the inner sum with $A=1$.
*In (2) we use the binomial identity $\binom{p}{q}=\binom{p-1}{q-1}\frac{p}{q}$.
*In (3) we shift the index of the inner sum to start with $A=0$.
*In (4) we apply the binomial theorem.
*In (5) we use the binomial identity $\binom{p}{q}=\binom{p-1}{p-q-1}\frac{p}{p-q}$.
*In (6) we use again the binomial theorem.
*In (7) we substitute $1+2z=u, 2dz=du$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3217254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to simplify $\frac{1+\frac{1}{2^p} + \frac{1}{3^p} + \frac{1}{4^p} + \cdots}{1 - \frac{1}{2^p} + \frac{1}{3^p} - \frac{1}{4^p} + \cdots}$ Let $p$ is real number which satisfies$\quad p > 1$
How can I simplify the fraction
$$\frac{1+\frac{1}{2^p} + \frac{1}{3^p} + \frac{1}{4^p} + \cdots}{1 - \frac{1}{2^p} + \frac{1}{3^p} - \frac{1}{4^p} + \cdots}$$
Numerator is $\zeta(p)$ but I don't know the closed form of denominator.
Is there any idea to simplify this fraction?
| You might want to try
\begin{align*}
\frac{1+\frac1{2^p}+\frac1{3^p}+\frac1{4^p}+\cdots}{1-\frac1{2^p}+\frac1{3^p}-\frac1{4^p}+\cdots}&=\frac{1-\frac1{2^p}+\frac1{3^p}-\frac1{4^p}+\cdots}{1-\frac1{2^p}+\frac1{3^p}-\frac1{4^p}+\cdots}+2\cdot\frac{\frac1{2^p}+\frac1{4^p}+\frac1{6^p}\cdots}{1-\frac1{2^p}+\frac1{3^p}-\frac1{4^p}+\cdots}\\
&=1+2\cdot\frac{\frac1{2^p}+\frac1{4^p}+\frac1{6^p}\cdots}{1-\frac1{2^p}+\frac1{3^p}-\frac1{4^p}+\cdots}\\
&=\ldots
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3219584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
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