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Find the linear transformation of a matrix knowing 4 linear transformations I'm stuck with an exercise where they give me 4 linear transformations $T:M_{2\times2}\to\Bbb R$ for 4 matrices 2x2 and then they ask me for the linear transformation of a fifth matrix.
$$ T
\begin{pmatrix}
1 &0\\
0 & 0\\
\end{pmatrix}
=3,
T\begin{pmatrix}
0 &1\\
1 & 0\\
\end{pmatrix}
=-1,
T
\begin{pmatrix}
1 &0\\
1 & 0\\
\end{pmatrix}
=0,
T
\begin{pmatrix}
0 &0\\
0 & 1\\
\end{pmatrix}
=0,
T
\begin{pmatrix}
a &b\\
c & d\\
\end{pmatrix}
=?$$
and I have worked with this situation for $T:\Bbb R^n \to\Bbb R^m$ but never with matrices so I have the doubt that how is supposed to find 4 constants that will multiply de 4 matrices I know to compose the fifth matrix to finally can find the linear transformation of this last one. (I have tried adding up the 4 matrices as follows)
$$
\begin{pmatrix}
1 &0\\
0 & 0\\
\end{pmatrix}
+
\begin{pmatrix}
0 &1\\
1 & 0\\
\end{pmatrix}
+
\begin{pmatrix}
1 &0\\
1 & 0\\
\end{pmatrix}
+
\begin{pmatrix}
0 &0\\
0 & 1\\
\end{pmatrix}=
\begin{pmatrix}
a &b\\
c & d\\
\end{pmatrix}
$$
\begin{cases}
2=a & \\
1=b & \\
2=c\\
1=d\\
\end{cases}
and then use the coefficients
$$
2\begin{pmatrix}
1 &0\\
0 & 0\\
\end{pmatrix}
+
1\begin{pmatrix}
0 &1\\
1 & 0\\
\end{pmatrix}
+
2\begin{pmatrix}
1 &0\\
1 & 0\\
\end{pmatrix}
+
1\begin{pmatrix}
0 &0\\
0 & 1\\
\end{pmatrix}=
\begin{pmatrix}
a &b\\
c & d\\
\end{pmatrix}
$$
and I just want to know if I continue in this way or I have committed any mistakes, thanks for your time <3
|
What is given can be translated as follows. Denote, in a standard way,
$$E_1=\begin{pmatrix}1&0\\0&0 \end{pmatrix},\quad E_2=\begin{pmatrix}0&1\\0&0 \end{pmatrix}, \quad E_3=\begin{pmatrix}0&0\\1&0 \end{pmatrix},\quad E_4=\begin{pmatrix}0&0\\0&1\end{pmatrix}.$$
We have $\:M=\begin{pmatrix}a&b\\c&d \end{pmatrix}=aE_1+bE_2+cE_3+dE_4$, so
$$T(M)=aT(E_1)+bT(E_2)+cT(E_3)+dT(E_4).$$
Now, we are given that $T(E_1)=3,\enspace T(E_2)+T(E_3)=-1,\enspace T(E_1)+T(E_3)=0, \enspace T(E_4)=0$. Can you proceed?
|
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"timestamp": "2023-03-29T00:00:00",
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|
Laurent series centered at $z_0=1$ with $1<|z-1|<\infty$ There is a function $f(z)=\dfrac{1}{z(z-1)}$, and it can be expanded as $\dfrac{1}{z-1}=\dfrac{1}{z}\cdot \dfrac{1}{1-\frac{1}{z}}$. Furthermore, $\dfrac{1}{1-\frac{1}{z}}=1+\dfrac{1}{z}+\dfrac{1}{z^2}...$ I substituted $w=z-1$ and expanded to get my Laurent series: $\dfrac{1}{(z-1)^2}+\dfrac{1}{(z-1)^3}+\cdots$. Did I approach this correctly?
|
It is not clear for what you are using $w = z-1$ and the link between $f(z)$ and the remainder of your argument is unclear. However, you have nearly got the right answer.
Instead, you could justify the result by writing,
\begin{align}
f(z) &= \frac{1}{z(z-1)} \\
&= \frac{1}{(z-1)^2(1+\frac{1}{z-1})}
\end{align}
and when $\lvert z-1 \rvert > 1$ the second parentheses in the denominator can be expanded as a convergent infinite series, so that,
\begin{align}
f(z) &= \frac{1}{(z-1)^2} \cdot \Big(1 - \frac{1}{z-1}+\frac{1}{(z-1)^2}-\cdots\Big) \\
&=\frac{1}{(z-1)^2} - \frac{1}{(z-1)^3}+\frac{1}{(z-1)^4} -\cdots
\end{align}
which is nearly the same as your result, derived directly from the original expression. Notice the alternating sign in the sum.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove by induction that $1*3+2*3^2+3*3^3 + \cdots + n*3^n = \dfrac{3}{4}(3^n(2n-1)+1)$ I'm trying to prove this using induction
$1*3+2*3^2+3*3^3 + \cdots + n*3^n = \dfrac{3}{4}(3^n(2n-1)+1)$
So far I have:
*
*Base case: true
*Induction step:
$\dfrac{3}{4}(3^n(2n-1)+1)+(n+1)*3^{n+1}=\dfrac{3}{4}(3^{n+1}(2(n+1)-1)+1)$
here is where I get stuck on simplification of both left and right side so that they are equal
$\dfrac{3}{4}(3^n(2n-1)+1)+3^{n+1}+3^{n+1}n=\dfrac{3}{4}(3^{n+1}(2(n+1)-1)+1)$
Thanks for your help!
|
In the induction step you want to prove the increment from $n$ to $n+1$ on both sides are the same, namely,
$$(n+1)3^{n+1} = \dfrac{3}{4}(3^{n+1}(2n+1)-3^n(2n-1))\tag1$$
By looking at the increment you got rid of the annoying constant $1$ from the RHS. Then you notice immediately that you can divide $3^{n+1}$ from both sides, and $(1)$ becomes
$$n+1 = \dfrac{1}{4}(3(2n+1)-(2n-1))=\frac 14 (6n+3-2n+1)=\frac 14 (4n+4)=n+1$$
then you are done.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove $\frac{n}{\sum_{k=1}^n{\frac{1}{\frac{1}{k}+a_k}}}-\frac{n}{\sum_{k=1}^n{\frac{1}{a_k}}}\geqslant \frac{2}{n+1}$ Let $a_k>0,k=1,2,\cdots, n$. Prove that
$$
\frac{n}{\sum_\limits{k=1}^n{\frac{1}{\frac{1}{k}+a_k}}}-\frac{n}{\sum_\limits{k=1}^n{\frac{1}{a_k}}}\geqslant \frac{2}{n+1}
$$
My attempt: multiply both sides by the denominators, and it is equivalent to prove:
$$
n\sum_{k=1}^n{\frac{1}{a_k}}-n\sum_{k=1}^n{\frac{1}{\frac{1}{k}+a_k}}\geqslant \frac{2}{n+1}\left( \sum_{k=1}^n{\frac{1}{a_k}} \right) \left( \sum_{k=1}^n{\frac{1}{\frac{1}{k}+a_k}} \right) \\
\sum_{k=1}^n{\frac{1}{a_k}}-\sum_{k=1}^n{\frac{1}{\frac{1}{k}+a_k}}\geqslant \frac{2}{n\left( n+1 \right)}\left( \sum_{k=1}^n{\frac{1}{a_k}} \right) \left( \sum_{k=1}^n{\frac{1}{\frac{1}{k}+a_k}} \right)\\ \sum_{k=1}^n{\left( \frac{1}{a_k}-\frac{1}{\frac{1}{k}+a_k} \right)}\geqslant \frac{2}{n\left( n+1 \right)}\sum_{k=1}^n{\frac{1}{a_k}}\sum_{p=1}^n{\frac{1}{\frac{1}{p}+a_p}}
\\ \sum_{k=1}^n{\left( \frac{\frac{1}{k}}{a_k\left( \frac{1}{k}+a_k \right)} \right)}\geqslant \frac{2}{n\left( n+1 \right)}\sum_{k=1}^n{\sum_{p=1}^n{\frac{1}{a_k\left( \frac{1}{p}+a_p \right)}}}
$$
But how to proceed further?
|
I know the question has already been answered, and the answer is quite nice, but I was thinking along a different line:
$$\sum\limits_{k=1}^{n} \frac{\frac{1}{k}}{a_k\left(\frac{1}{k}+a_k\right)} \geq \frac{2}{n(n+1)}\left(\sum\limits_{k=1}^{n}\frac{1}{a_k}\right)\left(\sum\limits_{k=1}^{n}\frac{1}{\frac{1}{k}+a_k}\right)$$
Since $\displaystyle \sum\limits_{k=1}^{n}k = \dfrac{n(n+1)}{2}$, we have
$$\left(\sum\limits_{k=1}^{n} k \right)\left(\sum\limits_{k=1}^{n} \frac{\frac{1}{k}}{a_k\left(\frac{1}{k}+a_k\right)}\right) \geq \left(\sum\limits_{k=1}^{n}\frac{1}{a_k}\right)\left(\sum\limits_{k=1}^{n}\frac{1}{\frac{1}{k}+a_k}\right)$$
$$\left(1+2+3+ \cdot \cdot \cdot +n\right)\cdot \left(\frac{1}{a_1(1+a_1)}+\frac{\frac{1}{2}}{a_2(\frac{1}{2}+a_2)}+\cdot\cdot\cdot + \frac{\frac{1}{n}}{a_n(\frac{1}{n}+ a_n)} \right) \geq \\ \left(\frac{1}{a_1}+\frac{1}{a_2}+\cdot\cdot\cdot+\frac{1}{a_n}\right)\cdot\left(\frac{1}{1+a_1}+\frac{1}{\frac{1}{2}+a_2}+\cdot\cdot\cdot+\frac{1}{\frac{1}{n}+a_n}\right)$$
These multiplications can be expanded into sums which can be organized into rows and columns, like an $n \times n$ matrix, say '$B\geq C$'
$$
\frac{1}{a_1(1+a_1)}+\frac{\frac{1}{2}}{a_2(\frac{1}{2}+a_2)}+\cdot\cdot\cdot + \frac{\frac{1}{n}}{a_n(\frac{1}{n}+ a_n)}\\
\frac{2}{a_1(1+a_1)}+\frac{1}{a_2(\frac{1}{2}+a_2)}+\cdot\cdot\cdot + \frac{\frac{1}{n}}{a_n(\frac{2}{n}+ a_n)}
\\
\vdots\\
\frac{n}{a_1(1+a_1)}+\frac{\frac{n}{2}}{a_2(\frac{1}{2}+a_2)}+\cdot\cdot\cdot + \frac{1}{a_n(\frac{1}{n}+ a_n)}\\
\geq \\
\frac{1}{a_1(1+a_1)}+\frac{1}{a_1(\frac{1}{2}+a_2)}+\cdot\cdot\cdot + \frac{1}{a_1(\frac{1}{n}+ a_n)}\\
\frac{1}{a_2(1+a_1)}+\frac{1}{a_2(\frac{1}{2}+a_2)}+\cdot\cdot\cdot + \frac{1}{a_2(\frac{1}{n}+ a_n)}\\
\vdots\\
\frac{1}{a_n(1+a_1)}+\frac{1}{a_n(\frac{1}{2}+a_2)}+\cdot\cdot\cdot + \frac{1}{a_n(\frac{1}{n}+ a_n)}
$$
Now note that the diagonals are equal, so we don't need to worry about them.
However, take for example the opposite entries of the diagonals,
First the entries in the first 'matrix' $B$ are of the form $b_{ij}=\dfrac{\frac{i}{j}}{a_j\left(\frac{1}{j}+a_j\right)}$ while the entries in the second 'matrix' $C$ are of the form $c_{ij}=\dfrac{1}{a_i\left(\frac{1}{j}+a_j\right)}$
then $c_{ij}+c_{ji} \leq b_{ij} + b_{ji}$:
$$\frac{1}{a_i\left(\frac{1}{j}+a_j\right)}+\frac{1}{a_j\left(\frac{1}{i}+a_i\right)} \leq \frac{\frac{i}{j}}{a_j\left(\frac{1}{j}+a_j\right)}+\frac{\frac{j}{i}}{a_i\left(\frac{1}{i}+a_i\right)}\\
a_j\left(\frac{1}{i}+a_i\right)+a_i\left(\frac{1}{j}+a_j\right) \leq \frac{i}{j}a_i\left(\frac{1}{i}+a_i\right)+\frac{j}{i}a_j\left(\frac{1}{j}+a_j\right)\\
\frac{1}{i}a_j+a_i a_j+\frac{1}{j}a_i + a_i a_j \leq \frac{1}{j}a_i+\frac{i}{j}a^{2}_i+\frac{1}{i}a_j+\frac{j}{i}a^{2}_j\\
2ija_i a_j \leq i^2 a^{2}_i+ j^2 a^{2}_j$$
and
$$0 \leq i^2 a^{2}_i - 2ija_i a_j + j^2 a^{2}_j = (ia_i-ja_j)^2$$
the inequalities hold because $a_k>0$ for $k=1,2,...,n$
Therefore the sum of the upper and lower triangular parts of $B$ is $\geq$ to the sum of the of the upper and lower triangular parts of $C$.
Sorry for the extra long answer.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $c^2=a^2+b^2-2ab\cos\left(C-\frac{KS}{3}\right)$ holds on a smooth surface.
Problem: Given a infinitely small geodesic triangle $\triangle ABC$ on a smooth surface, denote the corresponding edges as $a,b,c.$ Prove: the area of $\triangle ABC$ (denote as $S$) and the Gauss curvature on $C$ ( denote as $K$) satisfies
$$c^2=a^2+b^2-2ab\cos\left(C-\frac{KS}{3}\right).$$
Since the triangle is infinitesimal, therefore it is naturally to use the first fundamental form and the Gauss-Bonnet theorem. However, I got stuck at how to handle the $\frac{KS}{3}$ in the equation. Since the coefficient is $\frac{1}{3}$, if using Gauss-Bonnet theorem, we would include new variables $\angle A,\angle B$. Could anyone help me out? Thanks a lot in advance!
|
Comment @ robjohn : Can we check for both signs of const $K?$
So for constant $K$ approximation we have $ \cos \tilde C$
$$= \cos (C+ \dfrac{KS}{3})= \cos (C\pm \dfrac{A+B+C-\pi}{3}) $$
K>0
For an equilateral spherical triangle $( a=b=c,A=B=C=\pi/2) $
$$\text{RHS}=\cos 2 \pi/3= -\dfrac12$$
$$ c^2= a^2+b^2+ab = 3 a^2 \to c= {\sqrt 3 a}, \text{ a contradiction }$$
K<0
For an equilateral hyperbolic spiky triangle $(A=B=C=\pi/6,a=b=c),$
$$ = \cos ( \dfrac{4C+A+B-\pi}{3}) = \cos (\dfrac{2C-A-B+\pi}{3}), $$
$$\text{RHS}=\cos \pi/3= \dfrac12$$
$$ c^2= a^2+b^2-ab = a^2 ,\; c=a.$$
Please comment what I am missing here in the K>0 approximation.
|
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|
Polar to cartesian form of $r=\cos(4θ)$?
Consider an equation in polar coordinates, $r = \cos(4θ)$. Find the equation of the curve in the first quadrant in Cartesian coordinates.
This is for an assignment and this is what help I have received so far from user170231-
$$\begin{align}
r(\theta)&=\cos(4\theta)\\[1ex]
&=\cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta\\[1ex]
&=\frac{r^4\left(\cos^4\theta-6\cos^2\theta\sin^2\theta+\sin^4\theta\right)}{r^4}\\[1ex]
&=\frac{(r\cos \theta)^4-6(r\cos \theta)^2(r\sin \theta)^2+(r\sin \theta)^4}{\left(r^2\right)^2}
\end{align}$$
He told me to see if I could solve it now, and here's what I could come up with:
$$\sqrt{x^2+y^2} = \frac{x^4-6x^2y^2+y^4}{(x^2+y^2)^2}$$
Is this step correct? If so, am I done with the transformation of $r=\cos(4θ)$ to cartesian for the first quadrant only? If not, where should I go from here? Any help is appreciated.
|
Welcome to MSE! Here you can see how $r=cos(4\theta)$ is negative and positive and the importance of solving for both $\pm \sqrt{x^2+y^2}$
|
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|
Is there any way to calculate $\binom{10}{1}-\binom{10}{3}+\binom{10}{5}-\binom{10}{7}+\binom{10}{9}$ faster? During solving a problem I get this expression:
$$\binom{10}{1}-\binom{10}{3}+\binom{10}{5}-\binom{10}{7}+\binom{10}{9}$$
To calculate it normally, I do this way:
$$\binom{10}{1}-\binom{10}{3}+\binom{10}{5}-\binom{10}{7}+\binom{10}{9}=2\binom{10}{1}-2\binom{10}{3}+\binom{10}{5}$$
And here we have to evaluate $\binom{10}{3}$ and $\binom{10}{5}$. after all I got $32$ as the result correctly.
My question: Is there any other way to calculate this expression easier (faster)? I suspect there is because the answer is $32=2^5$ and I guess maybe there is other way to calculate it for example by combinations or other ways.
|
$$(1+i)^{10} = {10\choose 0} + {10\choose 1}i -{10\choose 2}-{10\choose 3}i +{10\choose 4}+ ... -{10\choose 10} $$
$$(1-i)^{10} = {10\choose 0} - {10\choose 1}i -{10\choose 2}+{10\choose 3}i +{10\choose 4}+...-{10\choose 10}$$
So $$...={(1+i)^{10}-(1-i)^{10}\over 2i} = {1\over 2i}((2i)^5-(-2i)^5) = 32$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Range of error in truncated division $\lfloor\frac{a\pm 1}{b\pm 1}\rfloor$ With $a,b\in\mathbb{N}^+$, $b > 1$ and $2\lfloor\log_2(b+\Delta_b)\rfloor\ge\lfloor\log_2(a+\Delta_a)\rfloor$ where $\Delta_n\in\{-1,0,1\}$ is my guess that
$$x + \Delta_x = \left\lfloor\frac{a+\Delta_a}{b+\Delta_b}\right\rfloor$$
correct? If the difference would be restricted to the numerator it would be
$$\left\lfloor \frac{a - 1}{b} \right\rfloor = \left\lfloor \frac{a}{b} + \frac{b - 1}{b} \right\rfloor -1$$
and
$$\left\lfloor \frac{a + 1}{b} \right\rfloor = \left\lfloor \frac{a}{b} + \frac{1}{b} \right\rfloor$$
respectively if I remember it right but the denominator is involved, too. So, how do I (dis)prove my assumption?
Edited for clarity:
$\Delta_n$ is meant to be a variable, so with all possible variations of the division the outcome is one of $x + 0$, $x - 1$, or $x + 1$ where $x + 0$ is the outcome of $\lfloor (a + 0 / (b + 0) \rfloor$. My assumption is that for all of the variations
$\lfloor(a+1)/(b+1)\rfloor$,
$\lfloor(a+1)/(b+0)\rfloor$,
$\lfloor(a+1)/(b-1)\rfloor$,
$\lfloor(a+0)/(b+1)\rfloor$,
$\lfloor(a+0)/(b-1)\rfloor$,
$\lfloor(a-1)/(b+1)\rfloor$,
$\lfloor(a-1)/(b+0)\rfloor$,
$\lfloor(a-1)/(b-1)\rfloor$, the result $x'$ is at most one unit away from $x = \lfloor (a + 0 / (b + 0) \rfloor$
|
The largest and smallest numbers that can be expressed as $\frac{a+\Delta_1}{b+\Delta_2}$ with $\Delta_1,\Delta_2\in\{-1,0,1\}$ are $\frac{a+1}{b-1}$ and $\frac{a-1}{b+1}$, respectively. We can calculate that
$$\frac{a+1}{b-1}-\frac{a}{b}=\frac{a+b}{b(b-1)}$$
and
$$\frac{a}{b}-\frac{a-1}{b+1}=\frac{a+b}{b(b+1)}.$$
If these differences are both at most $1$, that is, if $a\leq b^2-2b$, then the floors of any of these and $\frac{a}{b}$ can't differ by more than $1$; if $\lfloor y\rfloor\geq\lfloor x\rfloor+2$, then
$$y\geq \lfloor y\rfloor\geq \lfloor x\rfloor+2>x+1.$$
In general, it's a bad idea to use the same letter to refer to different variables (even if they vary among the same set); I've used $\Delta_1$ and $\Delta_2$ to clarify what's happening here.
|
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|
How can I find the reduction formula for $I_n=\int^1_0{x^6 (1+x^3)^{n}}dx$ I need to find the reduction formula for $I_n=\int^1_0{x^6 (1+x^3)^{n}}dx$ so that I may find the relationship between $I_4$ and $I_3$, I have done the following:
Setting up for Integration by parts
$$u=(1+x^3)^n \implies u^{'}=3nx^2(1+x^3)^{n-1}$$ $$v^{'}=x^6 \implies v=\frac{1}{7}x^7$$
Applying Integration by parts
$$I_n=[\frac{1}{7}x^7(1+x^3)^n]^1_0 - \frac{3n}{7}\int^1_0{x^9(1+x^3)^{n-1}dx}$$
$$=\frac{2^n}{7}-\frac{3n}{7}\int^1_0{x^9(1+x^3)^{n-1}dx}$$
However I have no idea how to bring the integral of $\frac{3n}{7}\int^1_0{x^9(1+x^3)^{n-1}dx}$ into the form of $\int^1_0{x^6(1+x^3)^{n-1}dx}$ so that I may substitute it as $I_{n-1}$.
How can I do it, so that I may find a relationship between $I_4$ and $I_3$ ?
|
For completeness, by $$x^{6}\left(1+x^{3}\right)^{n}-x^{6}\left(1+x^{3}\right)^{n-1}=x^{6}\left(1+x^{3}\right)^{n-1}\left(1+x^{3}-1\right)= x^{9}\left(1+x^{3}\right)^{n-1}, $$ we have $$
\begin{aligned}
I_{n} &=\frac{2^{n}}{7}-\frac{3 n}{7}\left(I_{n}-I_{n-1}\right) \\
7I_{n} &=2^{n}-3 n I_{n}+3 n I_{n-1} \\
I_{n} &=\frac{1}{7+3 n}\left(2^{n}+3 n I_{n-1}\right)
\end{aligned}
$$
|
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|
Find $\lim_{m \to \infty} m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) \right]$ where $N$ is Poisson random variable with mean $m$ Let $N$ be a Poisson random variable with the mean parameter $m$.
We are interested in finding the following limit
\begin{align}
\lim_{m \to \infty} m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) \right].
\end{align}
Things that I tried:
First, I found an upper bound by using Jensen's inequality
\begin{align}
\lim_{m \to \infty} m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) \right]\le \lim_{m \to \infty} m \log\left( \frac{ E \left[ N \right]+\frac{1}{2}}{m+1} \right)= \lim_{m \to \infty} m \log\left( \frac{ m+\frac{1}{2}}{m+1} \right)=-\frac{1}{2}.
\end{align}
For the lower bound I tried to use the CLT argument from a related question in here:
\begin{align}
m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) \right]= m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) | N \ge \frac{m}{k} \right] P[ N \ge \frac{m}{k} ]+ m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) | N < \frac{m}{k} \right] P[ N < \frac{m}{k} ]
\end{align}
for some $k>0$. Via the CLT it can be shown that $P[ N < \frac{m}{k} ] \to 0$ for all $0<k <1$, and we have that
\begin{align}
\lim_{m\to \infty}m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) \right]= \lim_{m\to \infty}m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) | N \ge \frac{m}{k} \right] .
\end{align}
Now if we use the lower bound at this point, we get
\begin{align*}
\lim_{m\to \infty}m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) | N \ge \frac{m}{k} \right] \ge \lim_{m\to \infty}m \log\left( \frac{ \frac{m}{k}+\frac{1}{2}}{m+1} \right) =-\infty,
\end{align*}
for all $k \in (0,1)$.
|
We will show that $lim_m m E\ln \frac{N+ \frac12}{m+1} = -1$.
We have $$m E\ln \frac{N+ \frac12}{m+1} = Em\ln \frac{N+ \frac12}{m+\frac12} + Em\ln \frac{m+ \frac12}{m+1} = Em\ln \frac{N+ \frac12}{m+\frac12} + m \ln (1 - \frac{0.5}{m+1})$$
and thus
$$lim_m m E\ln \frac{N+ \frac12}{m+1} = \lim_m E\ln \frac{N+ \frac12}{m+\frac12} -\frac12.$$
It's sufficient to show that $\lim_m E\ln \frac{N+ \frac12}{m+\frac12} \to -\frac{1}2$.
Condsider $m \ge 100$, $1+Y_i \sim Pois(1)$, $\xi_m \sim Pois(m - [m])$. Thus $\sum_{i=1}^{[m]} (Y_i + 1) + \xi_m = N$ in distribution.
Fix $x$. $$P(\frac{\xi_m}{\sqrt{m}} \ge x) \le \frac{E \frac{\xi_m}{\sqrt{m}} }{x} \to 0, m \to \infty$$
thus $ \frac{\xi_m}{\sqrt{m}} \to 0$ in probability.
It follows from CLT that $\frac{ \sum_{i=1}^{[m]} Y_i }{\sqrt{[m]}} \to N(0,1)$. Hence
$$\frac{N-m}{\sqrt{m}} = \frac{ \sum_{i=1}^{[m]} Y_i }{\sqrt{[m]}}\frac{\sqrt{[m]}}{\sqrt{m}} + \frac{\xi_m}{\sqrt{m}} \to N(0,1).$$
Put $$\eta_m = \frac{m}{m+\frac12}\cdot \frac{N-m}{\sqrt{m}} = \frac{\sqrt{m}(N-m)}{m+\frac12}$$
Then $\eta_m \to N(0,1)$ and $\eta_m \ge \frac{m}{m+\frac12}\frac{-m}{\sqrt{m}} \ge -\sqrt{m} + \frac{1}{3\sqrt{m}}$.
Further,
$$\lim_m E\ln \frac{N+ \frac12}{m+\frac12} = E m \ln(1 + \frac{\eta_m}{\sqrt{m}}).$$
Put $g(x) = \ln(1+x) - (x - \frac{x^2}2)$. Then
$$ E m \ln(1 + \frac{\eta_m}{\sqrt{m}}) = E m( \frac{\eta_m}{\sqrt{m}} - \frac12 ( \frac{\eta_m}{\sqrt{m}})^2 + g( \frac{\eta_m}{\sqrt{m}}) ) = - \frac12 + Emg( \frac{\eta_m}{\sqrt{m}}) $$
and it's sufficient to show that $Emg( \frac{\eta_m}{\sqrt{m}}) \to 0$.
Put $A = [-\sqrt{m}+ \frac{1}{3\sqrt{m}}, - m^{\frac{3}{8}}], B= [-m^{\frac{3}{8}}, m^{\frac{3}{8}}], C = [m^{\frac{3}{8}}, \infty)$. Then
$$|Emg( \frac{\eta_m}{\sqrt{m}})| \le Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in A} + Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in B} + Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in C} \le $$
$$\le m \max_{x \in A} |g(\frac{x}{\sqrt{m}})|P(\eta_m \in A) + m \max_{x \in B} |g(\frac{x}{\sqrt{m}})|P(\eta_m \in B) + Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in C}$$
As $m\ge 100$, we know that $\max_{x \in A} |g(\frac{x}{\sqrt{m}})| = |\ln(\frac{1}{3m})| \le 2 \ln m $.
Further, by Taylor formula
$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} \cdot \frac{1}{(1+\theta_x x)^3}$, $0 \le \theta \le 1$. Thus
$g(x) = \frac{x^3}{3} \cdot \frac{1}{(1+\theta_x x)^3}$
and $\max_{x \in B} |g(\frac{x}{\sqrt{m}})| = O(\frac{1}{ m^{\frac98 } })$.
We have $|g(x)| \le 2x^2$ for $x > 0$.
Thus $Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in C} \le E m 2 (\frac{\eta_m}{\sqrt{m}})^2 I_C = 2 E \eta_m^2 I_C$.
Further there will be heuristic arguments.
$2 E \eta_m^2 I_C \approx 2 E N^2(0,1) I_{(N(0,1) > m^{\frac38})} \to 0$ hence $Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in C} \to 0$.
$P(\eta_m \in A) \approx P(N(0,1) \in A) \le P(N(0,1) \le -m^{\frac38})$ and thus
$Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in A} \le m \max_{x \in A} |g(\frac{x}{\sqrt{m}})| P(N(0,1) \le -m^{\frac38}) \to 0$.
It's sufficient to show that$Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in B} \to 0$.
But $Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in B} \le m \max_{x \in B} |g(\frac{x}{\sqrt{m}})| = m O(\frac{1}{ m^{\frac98 } }) = o(1)$, q.e.d.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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|
Evaluate the double integral $\int_{0}^{16} \int_{\sqrt[4]{x}}^{2} \sqrt{2 y^{5}+1} d y d x$ $\int_{0}^{16} \int_{\sqrt[4]{x}}^{2} \sqrt{2 y^{5}+1} d y d x$
I considered changing the order of integration but that I do not think it does anything as it just leads me to evaluating dy still.
Another possibility I considered is converting the integral to polar coordinates but again the square root bogs me in solving this integral. Any help or hint is appreciated. Thank you.
|
I think changing the order of the integration is best:
$$\sqrt[4]{x}\leq y \leq 2$$
$$0 \leq x \leq 16$$
is the same as
$$0\leq x \leq y^4$$
$$0 \leq y \leq 2$$
from which the integral becomes $\int_0^2\int_0^{y^4}\sqrt{2y^5 +1}dxdy=\int_0^2 {y^4}\cdot \sqrt{2y^5 +1}dy$
which is easy to solve given the substitution mentioned in the comments.
let $u=2y^5 +1$, then $du= 10y^4 dy$ and $dy=\frac{du}{10y^4}$
$$\int_0^2 {y^4}\cdot \sqrt{2y^5 +1}dy = \int_1^{65} {y^4}\cdot \sqrt{u}\frac{du}{10y^4}= \frac{1}{10}\int_1^{65} \sqrt{u} du = \frac{1}{10}\cdot \frac{2\cdot u^{\frac{3}{2}}}{3}\Bigg|_1^{65}$$ $$= \frac{1}{15} \cdot \left(65^{\frac{3}{2}}-1\right)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
$a^2 + b^2 + c^2 + 6\ge 3(a + b + c), abc = 1$ I have not solved inequalities in a while, so I am a little rusty. Could you help me with this inequality I have found?
$$a^2 + b^2 + c^2 + 6 \ge 3(a + b + c),$$ where $a, b, c > 0$ and $abc = 1$
My initial idea was $a ^ 2 + 2 \ge 2\sqrt 2a$ and the inequalities with $b $ and $c$, then adding these 3, we get, $a ^ 2 + b ^ 2 + c ^ 2 + 6 \ge 2\sqrt 2(a + b+ c)$, but then we get to $2\sqrt2 > 3$, which is false.
Edited: I found some variants of the original problem.
Problem 1: Let $a, b, c > 0$. Prove that $a^2 + b^2 + c^2 + 6 + (abc - 1) \ge 3(a+b+c)$.
Problem 2: Let $a, b, c$ be reals with $abc \le 1$. Prove that $a^2 + b^2 + c^2 + 6 \ge 3(a+b+c)$.
|
Let $t = \sqrt{ab}$ and
$$f(a,b,c) = a^2+b^2+c^2 + 6 - 3(a+b+c).$$
Suppose $c = \min\{a,\,b,\,c\},$ then $c \leqslant 1,$ so
$$\sqrt{a}+\sqrt{b} \geqslant 2\sqrt{t} \geqslant 2.$$
We have
$$f(a,b,c) - f(t,t,c) = a^2+b^2-2t^2 - 3(a+b-2t) = \left[\left(\sqrt{a}+\sqrt{b}\right)^2-3\right]\left(\sqrt{a} - \sqrt{b}\right)^2 \geqslant 0.$$
Therefore $f(a,b,c) \geqslant f(t,t,c),$ and
$$f(t,t,c) = f\left(t,t,\frac{1}{t^2}\right) = \frac{(2t^4-2t^3+2t+1)(t-1)^2}{t^4} \geqslant 0.$$
The proof is completed.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Trying to show that $r = \frac{1}{C}\left( \frac{1}{1 + e\cos{(\theta + \omega)} } \right)$ is an ellipse I am trying to prove that planets move in ellipses,
I watched this video: https://www.youtube.com/watch?v=DurLVHPc1Iw
and read this: https://arxiv.org/pdf/1009.1738.pdf .
But both sources end up with this as an equation for the paths that planets follow :
$$
r = \frac{1}{C}\left( \frac{1}{1 + e\cos{(\theta + \omega)} } \right),
$$
where $C=G\frac{m_{1}+m_{2}}{h^{2}}$ which is just a constant, as is $\omega$, and $e$ is the eccentricity, so this is an ellipse with the focus at the origin as long as $e<1$ and when I put the equation in graphing software it gives me ellipse.
But I don't understand why it is an ellipse? Can anyone show me how it is an ellipse? Either from the normal polar coordinates of a ellipse centred at the origin and shifting it (I couldn't get the algebra to work out when I added ae to the x coordinate) or just by explaining it conceptionally?
|
We have
$$
\begin{split}
r&= \dfrac{1}{C} \left(\dfrac{1}{1+e \cos(\theta+\omega)}\right) \\
&= \dfrac{1}{C+Ce \cos(\theta+\omega)} \\
&= \dfrac{1}{C+Ce \cos(\theta+\omega)} \cdot \dfrac{1/C}{1/C} \\
&= \dfrac{1/C}{1+ e\cos(\theta+\omega)}
\end{split}
$$
Write $P= 1/C$. Then we have
$$
\begin{split}
r= \dfrac{P}{1+e \cos(\theta+\omega)} \\
r+er \cos(\theta+\omega)= P \\
r= P - er \cos(\theta+\omega)
\end{split}
$$
Now let $x= r \cos \theta$. Then noting that in our case, $r^2= x^2+y^2$, we have
$$
\begin{split}
r&= P - er \cos(\theta+\omega) \\
r&= P - ex \\
r^2&= P^2 - 2Pex + e^2 x^2 \\
x^2+y^2&= P^2 - 2Pex + e^2 x^2 \\
x^2 - e^2x^2 +2Pex + y^2&= P^2 \\
(1-e^2)x^2 +2Pex + y^2&= P^2 \\
x^2 + \dfrac{2Pex}{1-e^2} + \dfrac{y^2}{1-e^2}&= \dfrac{P^2}{1-e^2}
\end{split}
$$
Now it is just a routine complete the square in $x$ and we have
$$
\left(x+ \dfrac{Pe}{1-e^2} \right)^2 + \dfrac{y^2}{1-e^2}= \dfrac{P^2}{(1-e^2)^2}
$$
Of course, this is the same as
$$
\dfrac{(x+Pe/(1-e^2))^2}{P^2/(1-e^2)^2} + \dfrac{y^2}{P^2/(1-e^2)} = 1
$$
So that we have an ellipse 'centered' at the point $(-Pe/(1-e^2), 0)$, semimajor axis length $P/(1-e^2)$, semiminor axis $P/\sqrt{1-e^2}$, and with foci at distance
$$
\sqrt{A^2 - B^2}= \sqrt{\dfrac{P^2}{(1-e^2)^2} - \dfrac{P^2}{1-e^2}}= \dfrac{P|e|}{1-e^2}
$$
|
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|
determinant of matrix in polynomial Suppose $X=\sqrt{k^2 - k_1^2}, Y= \sqrt{k^2 - k_2^2},Z=\sqrt{k^2 - k_3^2}, S=\sqrt{k^2 - k_4^2}, T= \sqrt{k^2 - k_5^2}$. \
And there is $5 \times 5 $ matrix as given below.
\begin{equation}
\begin{vmatrix}
a_{11} & a_{12} & a_{13} & a_{14}*S & a_{15}*T\\
a_{21}*X & a_{22}*Y & a_{23}*Z & a_{24} & a_{25}\\
a_{31}*X & a_{32}*Y & a_{33}*Z & 0 & 0\\
a_{41} & a_{42} & a_{43} & a_{44}*S & a_{45}*T\\
a_{51}*X & a_{52}*Y & a_{53}*Z & a_{54}& a_{55}\\
\end{vmatrix} = 0,
\end{equation}
I have to find the determinant in terms of k with inter powers. By usual exapnsion method (squaring the terms), finding the expression for k is getting complex. Is there any other approach to find the expression for above determinant. And that expression should not contain fraction powers of k, means expression should have $X^2, Y^2,Z^2,S^2,T^2$ instead of $X, Y,Z,S,T$.
|
Here is a special case of your matrix
$$
\begin{vmatrix}
1 & 0 & 0 & 0 & 0\\
0& 0 & 0 & 1 & 0\\
0 & 0 & Z & 0 & 0\\
0 & 1 & 0 & 0 & 0\\
0 & 0 & 0 & 0& 1\\
\end{vmatrix}.
$$
It evaluates to $Z=\sqrt{k^2 - k_3^2}$. (Or is it $-Z$? No matter.) You can't achieve what you are asking for.
|
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|
Prove that integral converges How can I prove that this integral converges?
$$\int\limits_1^{\infty}\frac{\ln{x}}{(x-1)\sqrt{x^{2}-1}}\ dx$$
I tried to show that function $0 < \frac{\ln{x}}{(x-1)\sqrt{x^{2}-1}}$ is decreasing and is continious for $x$ in $(1,\infty)$. Furthermore the above integral converges when series $\sum_{1}^{\infty}\frac{\ln{x}}{(x-1)\sqrt{x^{2}-1}}$ converges.
Are there better methods?
|
Observe that for $x\geq 2$ we have $\frac{x}{2}\geq 1$ and $\frac{x^2}{2}\geq 2\geq 1$ . Hence,
\begin{align}
(x-1)\cdot \sqrt{x^2-1}&\geq (x-\frac{x}{2})\cdot \sqrt{x^2-\frac{x^2}{2}}\\
&=\frac{x}{2}\cdot \sqrt{\frac{x^2}{2}}\\
&=\frac{x^2}{2\sqrt{2}}
\end{align}
Hence for every $x\geq 2$
$$\frac{1}{(x-1)\cdot \sqrt{x^2-1}}\leq \frac{2\sqrt{2}}{x^2}$$
This shows that
\begin{align}
\int_{2}^{\infty}\frac{\ln x}{(x-1)\cdot \sqrt{x^2-1}}\,dx&\leq \int_{2}^\infty \frac{2\sqrt{2}\ln x}{x^2}\,dx\\
&=2\sqrt{2}\int_{2}^\infty \biggl(\frac{-1}{x}\biggr)'\cdot \ln x\,dx\\
&=2\sqrt{2}\biggl(-\frac{1}{x}\ln x \biggr|_{2}^{\infty}\biggr)+2\sqrt{2}\int_{2}^\infty\frac{1}{x^2}\,dx\\
&=\sqrt{2}\ln 2+2\sqrt{2}\biggl(-\frac{1}{x}\biggr|_{2}^\infty\biggr)\\
&=\sqrt{2}\ln 2+\sqrt{2}
\end{align}
Now, to treat the case where $x>1$ and $x$ being close to $1$, observe that
$$(x-1)\leq (x-1)\cdot \sqrt{x^2-1}\leq x\cdot (x-1)$$
for all $x>1$. Hence, for $x>1$
$$\frac{\ln x}{x\cdot (x-1)}\leq \frac{\ln x}{(x-1)\cdot \sqrt{x^2-1}}\leq \frac{\ln x}{x-1}$$
Using DLH to evaluate the limits of the right and left hand side of the above inequality we obtain
$$\lim_{x\to 1^+}\frac{\ln x}{(x-1)\cdot \sqrt{x^2-1}}=1$$
This shows also that
$$\int_{1}^2 \frac{\ln x}{(x-1)\cdot \sqrt{x^2-1}}\,dx<\infty$$
|
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|
Elementary proof (involving no series expansion) that $\tan x \approx x+\frac{x^3}{3}$ for small $x$ For small values of $x$, the following widely used approximations follow immediately by Taylor expansion:
*
*$\sin x \approx x$
*$\cos x \approx 1-\frac{x^2}{2}$
*$\tan x \approx x +\frac{x^3}{3}$
I am looking for a justification of these approximations without the use of series expansions.
By purely geometric considerations, it is easy to see that for small values of $x$, we have
$$ \sin x < x < \tan x.$$
Division by $\cos x$ and an application of the squeeze lemma yield
$$\frac{\sin x}{x}\xrightarrow{x\to0}1$$
and hence the approximation (1.). Using the identity $\cos x = 1-2 \sin^2 \frac{x}{2}$, the approximation (2.) also follows.
Can one justify the approximation (3.) by a similarly elementary argument without using Taylor expansion?
I tried around using the angle addition theorems, but I did not really get anywhere, mainly because I could not make the factor $\frac13$ appear anywhere.
|
Look at the trigonometric circle and approximate the chord $s$ with the angle $\theta$.
Here $x=\cos \theta$ and $y=\sin \theta$.
From the big triangle, you have $x^2+y^2=1$ which is obvious, and from the small circle, you have $s^2 = y^2 + (1-x)^2$. But with approximation $s \approx \theta$ the above is
$$\theta^2 \approx y^2 + (1-x)^2 \tag{1}$$
Now sub $y=\sqrt{1-x^2}$ to get the well-known approximation
$$ \theta^2 \approx 1-x^2 + (1-x)^2 = 2 (1-x) $$
or
$$ x = \cos \theta \approx 1 - \tfrac{1}{2} \theta^2 \tag{2}$$
Now use (2) in (1) to get $\theta^2 \approx y^2 + \tfrac{1}{4} \theta^4$, but since we are keeping this a second order approximation $\theta^4 \approx 0$ and
$$ y = \sin \theta \approx \theta \tag{3} $$
Finally, $\tan \theta = \tfrac{y}{x} \approx \tfrac{\theta}{1-\tfrac{1}{2} \theta^2}$. But use the property $\tfrac{1}{1-z} = 1 + \frac{1}{1-z} z \approx 1 + z$ with $z=\tfrac{1}{2} \theta^2$ to get the final approximation.
$$ \tfrac{y}{x} = \tan \theta \approx \theta \left( 1 + \tfrac{1}{2} \theta^2 \right) \tag{4a} $$
But note that the last one is based on two approximations.
Another approach with $\tan$ would be to approximate once $\tan \theta = \tfrac{y}{x} = \theta \frac{\sqrt{1-\tfrac{1}{2}z}}{1-z}$ with $z=\tfrac{1}{2}\theta^2$ and the approximation $ \tan \theta \approx \theta \left( 1 + \tfrac{3}{4} z \right)$
$$ \tfrac{y}{x} = \tan \theta \approx \theta \left( 1 + \tfrac{3}{8} \theta^2 \right) \tag{4b} $$
For example with $\theta=0.6$ you have $\tan(0.6) = 0.68413681$, $0.6 \left( 1 + \tfrac{1}{2} 0.6^2 \right)=0.708$ an 3.5% error, and $0.6 \left( 1 + \tfrac{3}{8} 0.6^2 \right)=0.681$ an 0.45% error.
So (4b) is at least an order of magnitude better at approximating $\tan \theta$ than (4a). A graphical comparison of the error $f(x)/\tan x-1$ is shown below:
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $(1+x+x^2)^n = C_0 + C_1x + C_2 x^2...$, then find $C_0C_1 -C_1C_2 + C_2C_3...$ $$(1+x+x^2)^n = C_0 + C_1 x + C_2 x^2...C_n x^n$$
And
$$(1-\frac 1x +\frac{1}{x^2})^n = C_0 -\frac{C_1}{x}+\frac{C_2}{x^2}...$$
So if we multiply, the two equations, the given series will be obtained by the coefficient of $x^{-1}$ on the LHS.
The LHS$$-(\frac{x^4+x^2 +1}{x^2})^n$$
The given answer is 0, but I don’t understand how $x^{-1} $ can have coefficient 0. If it is, how do I prove it?
|
$$
(1+x+x^2)(1-\frac 1x +\frac{1}{x^2}) = x^2 + 1 + \frac{1}{x^2}
$$
so that
$$
(1+x+x^2)^n(1-\frac 1x +\frac{1}{x^2})^n = (x^2 + 1 + \frac{1}{x^2})^n
$$
contains only even powers of $x$. In particular, the coefficient of $x^{-1}$ is zero.
|
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|
The maximum value of $\left| {\operatorname{Arg}\left( {\frac{1}{{1 - z}}} \right)} \right|$ for $|z|=1$,$z\ne1$ The maximum value of $\left| {\operatorname{Arg}\left( {\frac{1}{{1 - z}}} \right)} \right|$ for $|z|=1$,$z\ne1$=_____
My approach is as follow
Already this question is solved Maximum value of argument but I would like to slve by my approach which is illustrated below
$z = \cos \theta + i\sin \theta ;\theta \notin \left\{ 0 \right\}$ because $z\ne1$
$\left| {\operatorname{Arg}\left( {\frac{1}{{1 - \cos \theta - i\sin \theta }}} \right)} \right|$
$\left| {\operatorname{Arg}\left( {\frac{1}{{2{{\sin }^2}\frac{\theta }{2} - 2i\sin \frac{\theta }{2}\cos \frac{\theta }{2}}}} \right)} \right|$
$\left| {\operatorname{Arg}\left( {\frac{1}{{2\sin \frac{\theta }{2}\left( {\sin \frac{\theta }{2} - i\cos \frac{\theta }{2}} \right)}}} \right)} \right| \Rightarrow \left| {\operatorname{Arg}\left( {\frac{{\left( {\sin \frac{\theta }{2} + i\cos \frac{\theta }{2}} \right)}}{{2\sin \frac{\theta }{2}}}} \right)} \right|$
$\left| {\operatorname{Arg}\left( {\frac{{\left( {\cos \left( {\frac{\pi }{2} - \frac{\theta }{2}} \right) + i\sin \left( {\frac{\pi }{2} - \frac{\theta }{2}} \right)} \right)}}{{2\sin \frac{\theta }{2}}}} \right)} \right|$
How do we get the required answer which if $\frac{\pi}{2}$
|
Hint:
Arg$\left(\dfrac12+i\cdot\dfrac{\cot\dfrac\theta2}2\right)=$Arg$\left(\cot\dfrac\theta2\right)=$Arg$\left(\tan\left(\dfrac{\pi-\theta}2\right)\right)$
Use this
|
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"timestamp": "2023-03-29T00:00:00",
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|
How does $\cos\arcsin(\frac{3}{5})\cos\arctan(\frac{7}{24})-\sin\arcsin(\frac{3}{5})\sin\arctan\left(\frac{7}{24}\right)$ simplify to $\frac{3}{5}$? The question is to prove $\arcsin\left(\frac{3}{5}\right)+\arctan\left(\frac{7}{24}\right)=\arccos\left(\frac{3}{5}\right)$ which can be easily done by taking cos of both side and drawing triangles. However, the worked solutions does a simplification from LHS to RHS instead, namely $$\cos\arcsin\left(\frac{3}{5}\right)\cos\arctan\left(\frac{7}{24}\right)-\sin\arcsin\left(\frac{3}{5}\right)\sin\arctan\left(\frac{7}{24}\right)=\frac{3}{5}$$which I don't understand. Can someone please explain?
|
The provided solution first employs the angle addition identity $$\cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta,$$ with the choices $$\alpha = \arcsin \frac{3}{5}, \quad \beta = \arctan \frac{7}{24}.$$ That is to say, $$\sin \alpha = \frac{3}{5}, \quad \tan \beta = \frac{7}{24}.$$ In order to proceed with the evaluation, we now recall $$\sin^2 \theta + \cos^2 \theta = 1, \\ \tan^2 \theta + 1 = \sec^2 \theta.$$ So $$\frac{9}{25} = \sin^2 \alpha = 1 - \cos^2 \alpha,$$ hence $$\cos \alpha = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}.$$ Similarly, $$\frac{49}{24^2} = \tan^2 \beta = \sec^2 \beta - 1,$$ or $$\cos \beta = \frac{24}{25}.$$ It follows that $\sin \beta = \cos \beta \tan \beta = \frac{7}{25}$ and $$\cos(\alpha + \beta) = \frac{4}{5} \frac{24}{25} - \frac{3}{5} \frac{7}{25} = \frac{3}{5}.$$ This solution assumes that $-\frac{\pi}{2} \le \arcsin x \le \frac{\pi}{2}$ and $-\frac{\pi}{2} < \arctan x < \frac{\pi}{2}$.
|
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|
Prove the roots of $z^7+7z^4+4z+1=0$ lie inside a circle of radius $2$. Problem: Prove that all the roots of the equation $z^7+7z^4+4z+1=0$ lie inside the circle of radius $2$ centered at the origin.
This question came in the roots of unity section of a summer course. My attempts have been to factor and hope something nice pops out, which ended up in the following: $$z(z^3+2)^2+3z^4+1=0$$ $$\frac{z^8-1}{z-1}+3z^4=(z^4+z)(z^2+z-3).$$ They all look horrendous. Other thoughts would be to use some inequality to prove $|z|<2$ (am-gm, cauchy, etc. seem to fail me). Not entirely sure what to do from here.
|
Assume that $z$ is a root of the equation with $|z| \ge 2$. Then
$$
1 = \left|\frac{7z^4+4z+1}{z^7} \right|
\le \left|\frac{7}{z^3}\right| + \left|\frac{4}{z^6}\right| +\left| \frac{1}{z^7}
\right| \\
\le \frac{7}{8} + \frac{4}{64} + \frac{1}{128} = \frac{121}{128} < 1
$$
gives a contradiction.
The idea is to show that if $|z| \ge 2$ then the modulus of $z^7$ is larger than the modulus of $7z^4+4z+1$, so that the sum of these terms can not be zero.
Remark: According to WolframAlpha, the equation $z^7 + 7z^4+4z+1 = 0$ has one solution $z \approx -1.86283$, so that the estimate $|z| < 2$ for all solutions is not bad.
|
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|
Prove $H$ is a normal subgroup of $K$ $G$ is a set.
$$G=\left\{\begin{vmatrix}1 & 0 \\ 0 & 1\end{vmatrix}, \begin{vmatrix}0 & 1 \\-1 & 0\end{vmatrix}, \begin{vmatrix}-1 & 0 \\ 0 & -1\end{vmatrix}, \begin{vmatrix}0 & -1 \\ 1 & 0\end{vmatrix}, \begin{vmatrix}i & 0 \\ 0 & i\end{vmatrix}, \begin{vmatrix}0 & i \\ -i & 0\end{vmatrix}, \begin{vmatrix}-i & 0 \\ 0 & i\end{vmatrix}, \begin{vmatrix}0 & -i \\ i & 0\end{vmatrix}\right\}$$ and $(G, \times)$ is a group.
If
$$H=\left\{\begin{vmatrix}1 & 0 \\ 0 & 1\end{vmatrix}, \begin{vmatrix}-1 & 0 \\ 0 & -1\end{vmatrix}\right\}$$ and
$$K=\left\{\begin{vmatrix}1 & 0 \\ 0 & 1\end{vmatrix}, \begin{vmatrix}-1 & 0 \\ 0 & -1\end{vmatrix}, \begin{vmatrix}i & 0 \\ 0 & i\end{vmatrix}, \begin{vmatrix}-i & 0 \\ 0 & -i\end{vmatrix}\right\},$$
then $H$ and $K$ is a subgroup of $G$, $H$ is a normal subgroup of $K$, and $K$ is a normal subgroup of $G$.
My questions are :
a. How do I prove that $H$ is a normal subgroup of $K$?
I knew that to prove if $H$ is a normal subgroup I have to prove that each left coset of $H$ in K is a right coset of $H$ in $K.$ But I don't know how apply that to the matrices.
b. What is every element of factor group $G/H?$
|
a. $H=\{I,-I\}$, so, for all $k\in K$, $kH=\{kI,-kI\}=\{k,-k\}$,
and also $Hk=\{Ik,-Ik\}=\{k,-k\}$,
so indeed each left coset of $H$ in $K$ is a right coset of $H$ in $K$, so $H\lhd K$.
b. One element of $G/H$, for example,
is $\left\{I\begin{bmatrix}0 & i \\ -i & 0\end{bmatrix}, -I\begin{bmatrix}0 & i \\ -i & 0\end{bmatrix}\right\}$ =$ \left\{\begin{bmatrix}0 & i \\ -i & 0\end{bmatrix}, \begin{bmatrix}0 & -i \\ i & 0\end{bmatrix}\right\}.$
Now can you find the others?
|
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|
Evaluate $360 (\frac{1}{15}+\frac{1}{105}+\frac{1}{315}+⋯).$ Evaluate $360$ ($\frac{1}{15}$+$\frac{1}{105}$+$\frac{1}{315}$+$\cdots$).
My Work :-
Well we can clearly see that $15 = 1*3*5$ ; $105 = 3*5*7$ ; $315 = 5*7*9$. So I basically know the pattern, but I want to know how to apply that like maybe we can break $\frac{1}{15}$ into $x*\frac{1}{1}\pm y*\frac{1}{3}\pm z*\frac{1}{5} $ or something like that so that we can cancel some terms in the whole sum.
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$T_n = \dfrac{1}{(2n-1)(2n+1)(2n+3)}$
Let $V_n= \dfrac{1}{(2n-1)(2n+1)}=(2n+3)T_n$
$V_{n} - V_{n+1}=\dfrac{1}{(2n-1)(2n+1)}-\dfrac{1}{(2n+1)(2n+3)}=\dfrac{2}{(2n-1)(2n+1)(2n+3)}=2T_n$
Let $\sum_{n=1}^{\infty}T_n=S$
Therefore,
$\sum_{n=1}^{\infty}(V_n - V_{n+1})=(V_1-V_{\infty})=2S$
$V_1=\frac{1}{3},V_{\infty}=0$
Therefore, $S=\frac{1}{6}$ and $360S = 60$
A telescopic series!
|
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|
How to evaluate $\sum_{k=1}^{n} \frac{f^3_k}{3(f^2_k-f_k)+1}, ~~f_k=k/n$ Very often the summations at pre-degree level are done by using differencing-telescoping and by some symmetry properties. The following summation is fabricated keeping one thing in mind, I may tell it later. $$\sum_{k=1}^{n} \frac{f^3_k}{3(f^2_k-f_k)+1},~~ f_k=k/n.$$
The question is: How will you evaluate this sum?
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There a many ways to do it, I presume.
$$S_n=\sum_{k=1}^n \frac{k^3}{3 k^2 n-3 n^2k+n^3}=\sum_{k=1}^n \frac{k^3}{3(k-a)(k-b) }$$ where
$$a=\frac{n}{6} \left(3-i \sqrt{3}\right)\qquad \text{and} \qquad b=\frac{n}{6} \left(3+i \sqrt{3}\right)$$
$$S_n=\sum_{k=1}^n \Bigg[\frac{a^3}{3 (a-b) (k-a)}-\frac{b^3}{3 (a-b) (k-b)}+\frac{a+b}{3}+\frac{k}{3}\Bigg]$$
$$S_n=\frac{2 a^3 \left(H_{n-a}-H_{-a}\right)+n (a-b) (2 a+2 b+n+1)+2 b^3
\left(H_{-b}-H_{n-b}\right)}{6 (a-b)}$$
$$S_n=\frac{1}{18} \left(9 n+9+\frac{4 \pi n \sin (\pi n)}{\cos (\pi n)-\cosh
\left(\frac{\pi n}{\sqrt{3}}\right)}\right)=\frac{n+1}2$$
|
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|
Positive coefficients after the substitution? Let $n>1$ be an integer and $\Phi_n(x)$ be the $n$ th cyclotomic polynomial.
If we subsititute $x$ by $x+1$ , do we always get a polynomial with positive coefficients ? If yes, how can this be proven ? I also want to prove that no
coeffcient upto the $\varphi(n)$ th is missing.
Examples :
*
*$\Phi(6)=x^2-x+1$ transforms into $x^2+x+1$
*$\Phi(24)=x^8-x^4+1$ transforms into $x^8+8x^7+28x^6+56x^5+69x^4+52x^3+22x^2+4x+1$
Upto $n=3\ 200$ , this is the case. Some special cases can be easily proven, for example that $n$ is a power of $2$ or that $n$ is a prime. But I have no idea for a general proof.
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Deal with the case $n=1,$ $ n = 2$ separately.
Notice that $ \Phi_1(x+1) = (x+1) - 1 = x$, which has a non-positive constant term.
This will be the only coefficient of $ \Phi_n (x+1)$ which is non-positive.
For $ n > 2$,
Let $A_n$ denote the set of integers $ 0 < k < n$ such that $ \gcd( n,k) = 1$.
Let $A_n^*$ denote the set of integers $ 0 < k < n/2$ such that $ \gcd( n,k) = 1$.
$$\Phi_n(x) = \prod_{j \in A_n } (x - e^{2k\pi/n}) = \prod_{j \in A_n^*} \left( x^2 - 2 \cos \frac{ 2\pi j}{n} \cdot x + 1 \right) .$$
Let $ f_m (x) = x^2 - 2mx + 1 $.
Since $f_m(x+1) = x^2 + 2(1-m)x + 2 (1-m)$, so the coefficients of $f_m(x+1)$ are positive for $ m < 1$.
Thus, for $n > 2$, since $ \cos \frac{2 \pi j } { n} < 1$ for $ j \in A_n^*$
$$\Phi_n(x+1) = \prod_{j \in A_n^*} f_{\cos \frac{2\pi j }{n}}(x+1) ,$$
where the latter is a product of polynomials with positive coefficients, hence $\Phi_n(x+1)$ has positive coefficients for up to the degree $ \phi(n)$.
|
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|
For which values of integer $k$, does the equation $x^2+y^2+z^2=kxyz$ have positive integer solutions $(x, y, z)$ For which values of integer $k$, does the equation $x^2+y^2+z^2=kxyz$ have positive integer solutions $(x, y, z)$
I immediately thought of saying that from symmetry we have that $x\le y \le z$.
Also, $y^2+z^2 \equiv 0 mod x$, $x^2+z^2\equiv 0mody$ and $x^2+y^2\equiv 0modz$.
Moreover through trial and error I worked out that the solutions for $k$ must be $k=1$ or $k=3$ but I have not managed to prove it. I attempted to use inequalities, but that didn't work out either. Could you please explain to me how to solve this question?
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this is called a CW answer; recommend beginning with
Equation with Vieta Jumping: $(x+y+z)^2=nxyz$.
Here are a number of posts about
Vieta Jumping/ Hurwitz/Markov Grundlösung
Let $x$ and $y$ be positive integers such that $xy \mid x^2+y^2+1$.
Diophantine quartic equation in four variables
Is it true that $f(x,y)=\frac{x^2+y^2}{xy-t}$ has only finitely many distinct positive integer values with $x$, $y$ positive integers? !!!!!!!!! +++++++
http://math.stackexchange.com/questions/1411049/if-a-b-are-positive-integers-and-ab-1-mid-a%C2%B2-b%C2%B2-then-prove-that-q
Proving that all terms in sequence are positive integers (Recursive)
Diophantine equation $(x+y)(x+y+1) - kxy = 0$
Find the postive integers such $xy+x+y\mid x^2+y^2-2$
Find the integer values of c
Find all solutions to the diophantine equation $(x+2)(y+2)(z+2)=(x+y+z+2)^2$
Characterize the integers $a,b$ satisfying: $ab-1|a^2+b^2$
Equation with Vieta Jumping: $(x+y+z)^2=nxyz$. Follows Hurwitz very closely!! (x+y+z)^2 = nxyz
Showing that $m^2-n^2+1$ is a square LEMMA +
Showing that $m^2-n^2+1$ is a square LEMMA -
How prove infinitely many postive integers triples $(x,y,z)$ such $(x+y+z)^2+2(x+y+z)=5(xy+yz+zx)$
Find all possible value of c
What are the solutions of the equation $3np+3n+2=n^2+p^2$, with n and p positive integers?
Prove there is no $x, y \in \mathbb Z^+ \text{ satisfying } \frac{x}{y} +\frac{y+1}{x}=4$
Integer points on a hyperbola two spiral arms
$m+n+p-1=2\sqrt{mnp}$ in positive integers
https://artofproblemsolving.com/community/c6h1726220_if_fraction_integer_then_equal_to_5 Poland 1991 training
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|
Evaluating $\lim_{x\to\infty}(\sqrt{x^2+1} - x)(x+1) $ The question is $$\lim_{x\to\infty}(\sqrt{x^2+1} - x)(x+1) $$ I know the answer is $\frac{1}{2}$ and I found it using this equality :
$$(\sqrt{x^2+1} - x)(x+1) = \frac{x+1}{\sqrt{x^2+1} + x}$$
But is there any other way to solve this? Any hints would be appreciated.
|
Just for fun, try letting $x=\tan\theta=\sin\theta/\cos\theta$ with $\theta\to\pi/2^-$, and use
$$\sqrt{\tan^2\theta+1}=\sqrt{\sec^2\theta}=\sec\theta={1\over\cos\theta}$$
so that
$$\begin{align}
(\sqrt{x^2+1}-x)(x+1)
&=\left({1\over\cos\theta}-{\sin\theta\over\cos\theta}\right)\left({\sin\theta\over\cos\theta}+1\right)\\
&={(1-\sin\theta)(\sin\theta+\cos\theta)\over\cos^2\theta}\\
&={(1-\sin\theta)(\sin\theta+\cos\theta)\over1-\sin^2\theta}\\
&={\sin\theta+\cos\theta\over1+\sin\theta}
\end{align}$$
We get
$$\lim_{x\to\infty}(\sqrt{x^2+1}-x)(x+1)=\lim_{\theta\to\pi/2^-}{\sin\theta+\cos\theta\over1+\sin\theta}={1+0\over1+1}={1\over2}$$
|
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|
Real numbers $a$ and $b$ satisfy $a^3+b^3-6ab=-11$. Prove that $-\frac{7}{3}
Real numbers $a$ and $b$ satisfy $a^3+b^3-6ab=-11$. Prove that $$-\frac{7}{3}<a+b<-2$$
I have shown that $a+b<-2$. My approach: $-3=8-11=a^3+b^3-6ab+2^3=\frac{1}{2}(a+b+2)((a-b)^2+(a-2)^2+(b-2)^2)$. From this we must have that $a+b<-2$.
Please give some idea/hint for the other part.
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Showing that $a+b > - \frac73$:
$$(a-2)^2+(b-2)^2+(a-b)^2\geq(a-2)^2+(b-2)^2 \geq\frac{(a+b)^2}{2}-4(a+b)+8$$
Now using the previous result $a + b < -2$ we get:
$$\frac{(a+b)^2}{2}-4(a+b)+8> 18$$
From OP's equation in the question, this gives
$$a+b=-\frac{6}{(a-2)^2+(b-2)^2+(a-b)^2}-2>-\frac{6}{18} -2 = -\frac{7}{3}. \qquad \Box$$
|
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|
Integral of $1/x^2$ without power rule I was wondering if it was possible to evaluate the following integral without using the power rule for negative exponents
\begin{equation*}
\int \frac{1}{x^2} \; dx
\end{equation*}
When using integration by parts, you end up with the same integral in the rhs so it seems out of luck
\begin{equation*}
\int \frac{1}{x^2} \; dx = \frac{\ln x}{x} + \int \frac{\ln x}{x^2} \; dx
\end{equation*}
This question is inspired by this blackpenredpen's video
Using integration by parts with $u = \frac{1}{x^2}$ and $v = x$ is NOT accepted. If you write
\begin{equation*}
\int \frac{1}{x^2} \; dx = \frac{1}{x} + 2 \int \frac{1}{x^2} \; dx
\end{equation*}
you are still implicitly using the power rule to compute the derivative of $\frac{1}{x^2}$ so it is not correct per the rules.
The same rules apply for $u$-substitution which implicitly use the power rule. See the following with $u = \frac{1}{x}$ such that
\begin{equation*}
\int \frac{1}{\left(\frac{1}{x}\right)^2} \left(\frac{1}{x}\right)' \; dx = \int \frac{1}{u^2} \; du
\end{equation*}
and here the power rule is also considered used to compute the derivative of $\frac{1}{x}$ although it can be subject to discussion (geometric proof, limit definition of derivative, etc...)
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Let $x=\cos \theta +i\sin \theta$. By De Moivre's Theorem, $$\frac{1}{x^2}=x^{-2}=\cos (-2\theta)+i\sin (-2\theta)=\cos (2\theta)-i\sin(2\theta)$$
Also $$dx=(-\sin \theta+i\cos \theta) d\theta$$
The integral becomes $$\int( \cos 2\theta-i\sin 2\theta)(-\sin \theta+i\cos\theta)d\theta\\
=\int (-\sin \theta\cos 2\theta+i\cos \theta\cos 2\theta+i\sin\theta\sin 2\theta+\cos \theta\sin2\theta)d\theta\\
=\int -\frac{1}{2}(\sin 3\theta-\sin \theta)+i\frac{1}{2}(\cos 3\theta+\cos \theta)+i\frac{1}{2}(\cos\theta-\cos 3\theta)+\frac{1}{2}(\sin 3\theta+\sin\theta)d\theta \\ =\int \sin \theta +i\cos \theta d\theta=-\cos \theta +i\sin\theta +C\\
=-(\cos \theta-i\sin \theta)+C=-(\cos (-\theta)+i\sin (-\theta))+C=-\frac{1}{x}+C$$
You can find $\int x^a\:dx,a\in \mathbb Z$ in a similar manner.
|
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|
Determine if there's a linear map such that... Today in the class we were solving a following problem:
Determine if there's a linear map $\phi : R^3 \rightarrow R^3$ such that $\phi(2,1,0)=(3,2,1), \phi(4,2,1) = (4,1,0), \phi(2,2,1) = (2,3,3), \phi(2,2,2) = (0,0,1)$. It was then solved by creating an augmented matrix with the inputs for $\phi$ written in rows in the left hand side of the matrix and outputs of $\phi$ in the right hand side, also in rows. Then the matrix has been row reduced until an identity appeared in the left hand side... it seemed to me like some black magic. I have no idea why or how it works. This whole method doesn't make any sense to me. Is there any better way of doing it or some explanation why something like this works?
|
$$A=\left(
\begin{array}{ccc}
a & b & c \\
d & e & f \\
g & h & i \\
\end{array}
\right)$$
$$A\cdot (2,1,0)=(3,2,1),A\cdot (4,2,1)=(4,1,0),A\cdot (2,2,1)=(2,3,3),A\cdot (2,2,2)=(0,0,1)$$
$$\begin{cases}
2 a + b=3\\
2 d + e=2\\
2 g + h=1\\
4 a + 2 b + c=4\\
4 d + 2 e + f=2\\
4 g + 2 h + i=1\\
2 a + 2 b + c=2\\
2 d + 2 e + f=3\\
2 g + 2 h + i=3\\
\end{cases}
$$
which gives
$$a= 1,b= 1,c= -2,d= -1,e= 4,f= -3,g= -\frac{3}{2},h= 4,i= -2$$
These results satisfy the fourth set of conditions
$$\begin{cases}
2 a + 2 b + 2 c=0\\
2 d + 2 e + 2 f=0\\
2 g + 2 h + 2 i=1\\
\end{cases}
$$
Therefore the matrix is
$$
A=\left(
\begin{array}{ccc}
1 & 1 & -2 \\
-1 & 4 & -3 \\
-\frac{3}{2} & 4 & -2 \\
\end{array}
\right)
$$
|
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|
Solving the integral of $\frac {dx} {1+x^3}$
Compute integral $$\int_0^{\infty} \frac{dx}{1+x^3}$$.
I used the formula from complex variables by fisher (2.6 formula (9)) that $$\int_0^{\infty} \frac{dx}{1+x^a}=\frac {\pi} {a\sin(\pi/a)}$$ This means $$\int_0^{\infty} \frac{dx}{1+x^3}=\frac {\pi} {3\sin(\pi/3)}$$ Is that simply the answer or do I need to expand it to find all concrete values?
Additionally, from this answer here, the improper integral gives a different result:
$$-\frac{1}{6} \ln |x^2-x+1|+\frac{1}{\sqrt{3}} \arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{1}{3}\ln|x+1|$$
is this result equal to the one I got above?
|
Next, simplify
$$
F(x)=-\frac{1}{6}\ln|x^2-x+1|+\frac{1}{\sqrt{3}}\arctan{\frac{2x-1}{\sqrt{3}}}+\frac{1}{3}\ln|x+1|
$$
$$
=\frac{1}{\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{1}{3}\ln|x+1|-\frac{1}{3}\ln\sqrt{|x^2-x+1|}
$$
$$
=\frac{1}{\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{1}{3}\ln\left(\frac{|x+1|}{\sqrt{|x^2-x+1|}}\right).
$$
Then
$$\int_0^\infty \frac{dx}{1+x^3}=\lim_{X\rightarrow\infty}F(X)-F(0).$$
Compute the limit, and you are done.
|
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|
Nested equilateral triangles Let triangle $ABC$ is an equilateral triangle. Triangle $DEF$ is also an equilateral triangle and it is inscribed in triangle $ABC \left(D\in BC,E\in AC,F\in AB\right)$. Find $\cos\measuredangle DEC$ if $AB:DF=8:5$.
Firstly, I would be very grateful if someone can explain to me how I am supposed to draw the diagram. Obviously I have made it by sight.
Let $\measuredangle DEC=\alpha$. We can note that $\triangle AEF \cong \triangle BFD \cong CDE$. This is something we can always use in such configuration. So $$AE=BF=CD, $$ $$AF=BD=CE.$$ Let $AB=BC=AC=8x$ and $DF=DE=EF=5x$. If we denote $CD=y,$ then $CE=AC-AE=AC-CD=8x-y$. Cosine rule on $CED$ gives $$25x^2=(8x-y)^2+y^2-2y\cos60^\circ(8x-y)$$ which is a homogenous equation. I got that $\dfrac{y}{x}=4\pm\sqrt{3}.$ Now using the sine rule on $CED$ $$\dfrac{CD}{DE}=\dfrac{\sin\alpha}{\sin60^\circ}\Rightarrow \sin\alpha=\dfrac{\sqrt{3}}{10}\cdot\dfrac{y}{x}=\dfrac{4\sqrt3\pm3}{10}.$$ Now we can use the trig identity $\sin^2x+\cos^2x=1$ but it doesn't seem very rational. Can you give me a hint? I was able to find $\sin\measuredangle DEC$ in acceptable way, but I can't find $\cos\measuredangle DEC$...
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WLOG, $AB = 8, DF = 5$. Say $\angle DEC = \theta$
If $CD = x$ then $CE = 8 - x$
Applying sine law in $\triangle CDE$,
$\displaystyle \frac{\sin 60^0}{5} = \frac{\sin \theta}{x} = \frac{\sin(60^0+\theta)}{8-x}$
From first two, we have $\sin \theta = \frac{x \sqrt3}{10}$
From first and third, $\sin 60^0 \cos \theta + \cos 60^0 \sin\theta = \frac{4\sqrt3}{5} - \frac{x \sqrt3}{10}$
$\implies \cos\theta = \frac{8}{5} - \frac{3x}{10}$
Applying $sin^2\theta + \cos^2\theta = 1$, we get
$x^2 - 8x+\frac{39}{3} = 0 \implies x = 4 \pm \sqrt3$
So, $\cos\theta = \frac{4 \mp 3\sqrt3}{10}$ (please note $\pm$ for $x$ and corresponding $\mp$ for $\cos \theta$)
|
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|
Transformation of the integrand of an elliptic integral to the general form of an elliptic curve First, some background.
An ellipse is defined by the following equation:
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\implies y=b\sqrt{1-\frac{x^2}{a^2}}.$
To get the arclength of the ellipse, we use calculus to obtain:
$4\int_{0}^{a}\sqrt{1+\frac{b^2x^2}{a^2(a^2-x^2)}}dx.$
We introduce a change of variables $x=at$ to get:
$4a\int_{0}^{1}\sqrt{1+\frac{a^2b^2t^2}{a^2(a^2-a^2t^2)}}dt=4a\int_{0}^{1}\sqrt{\frac{1-t^2(1-b^2/a^2)}{1-t^2}}dt.$
Let $e^2=1-\frac{b^2}{a^2}$ be the square of the eccentricity of the ellipse, then the integral becomes:
$\int_{0}^{1}\sqrt{\frac{1-e^2t^2}{1-t^2}}dt.$
This is an elliptic integral.
Let $u=u(t)=\sqrt{\frac{1-e^2t^2}{1-t^2}}$ be the integrand, then:
$u^2(1-t^2)=1-e^2t^2.$
I believe that this equation defines an elliptic curve of the form:
$y^2=x^3+ax+b$
Such that:
$4a^3+27b^2\neq0.$
How does one transform the equation in terms of $u$ into an elliptic curve?
|
$y^2=(1+e^2t^2)/(1+t^2)$
If $2\ne 0$,
$y=z/(1+t^2)$ gives $z^2=(1+e^2t^2)(1+t^2)=e^2(t^2+a)^2+b$
$z=w+ie(t^2+a)$ gives $w^2 +2wie(t^2+a)=b$
$t=u/w$ gives $w^2 +2ieu^2/w+2wiea=b$
or $$w^3+2w^2iea-bw=-2ieu^2$$
If $e\ne 0,3\ne 0$ it can be transformed into $U^2= W^3+AW+B$
|
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|
Calculate $\displaystyle \lim_{x \to 3} \frac{\sqrt{19-x} - 2\sqrt[4]{13+x}}{\sqrt[3]{11-x} - x + 1}$ Calculate:
$$\displaystyle \lim_{x \to 3} \frac{\sqrt{19-x} - 2\sqrt[4]{13+x}}{\sqrt[3]{11-x} - x + 1}$$
The problem with that case is that the roots are in different powers so multiplication in nominator and denominator by conjugate is not an option (at least I think it's not).
|
The conjugation/rationalization approach works, but is somewhat tedious. The key is to introduce the right terms in order to force the numerator and denominator into a differences of integer powers:
$$\begin{align}
\frac{\sqrt{19-x}-2\sqrt[4]{13+x}}{\sqrt[3]{11-x}-x+1}&=\frac{\left((19-x)^2\right)^{\frac14}-\left(2^4(13+x)\right)^{\frac14}}{(11-x)^{\frac13}-\left((x-1)^3\right)^{\frac13}}\\[1ex]
&=\frac{a^{\frac14}-b^{\frac14}}{c^{\frac13}-d^{\frac13}}\\[1ex]
&=\frac{a^{\frac14}-b^{\frac14}}{c^{\frac13}-d^{\frac13}}\times\frac{a^{\frac34}+a^{\frac12}b^{\frac14}+a^{\frac14}b^{\frac12}+b^{\frac34}}{a^{\frac34}+a^{\frac12}b^{\frac14}+a^{\frac14}b^{\frac12}+b^{\frac34}}\times\frac{c^{\frac23}+c^{\frac13}d^{\frac13}+d^{\frac23}}{c^{\frac23}+c^{\frac13}d^{\frac13}+d^{\frac23}}\\[1ex]
&=\frac{\left(a^{\frac14}\right)^4-\left(b^\frac14\right)^4}{\left(c^{\frac13}\right)^3-\left(d^\frac13\right)^3}\times\frac{c^{\frac23}+c^{\frac13}d^{\frac13}+d^{\frac23}}{a^{\frac34}+a^{\frac12}b^{\frac14}+a^{\frac14}b^{\frac12}+b^{\frac34}}\\[1ex]
&=\frac{a-b}{c-d}\times\frac{c^{\frac23}+c^{\frac13}d^{\frac13}+d^{\frac23}}{a^{\frac34}+a^{\frac12}b^{\frac14}+a^{\frac14}b^{\frac12}+b^{\frac34}}
\end{align}$$
(where I hope the replacements of $a,b,c,d$ are obvious)
Upon simplification (with $x\neq3)$, we have
$$\frac{a-b}{c-d}=\frac{153-54x+x^2}{12-4x+3x^2-x^3}=\frac{(51-x)(3-x)}{(3-x)(4+x^2)}=\frac{51-x}{4+x^2}$$
and the remaining fraction of rational powers is continuous at $x=3$. Then the limit is
$$\lim_{x\to3}\frac{\sqrt{19-x}-2\sqrt[4]{13+x}}{\sqrt[3]{11-x}-x+1}=\frac{48}{13}\times\frac{12}{256}=\boxed{\frac9{52}}$$
|
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|
Prove that for any complex numbers: $a,b,c$ , $|a|=|b|=|c|=1$ and $abc \neq -1$ number $\frac{a+b+c+ab+ac+bc}{1+abc}$ is a real number.
Prove that for any complex numbers: $a,b,c$ , $|a|=|b|=|c|=1$ and $abc \neq -1$ number $$x = \frac{a+b+c+ab+ac+bc}{1+abc}$$ is a real number.
I wanted to calculate $2 \cdot Im(x) = x- \overline x$ and show that it's equal to zero:
$$\frac{a+b+c+ab+ac+bc}{1+abc} - \frac{\overline{a}+\overline{b}+\overline{c}+\overline{ab}+\overline{ac}+\overline{bc}}{1+\overline{abc}} =$$ $$=a+b+c+ab+ac+bc-\overline{a}-\overline{b}-\overline{c}-\overline{ab}-\overline{ac}-\overline{bc} = $$ $$=2(Im(a) + Im(b)+Im(c)+Im(ab)+Im(ac)+Im(bc))=$$$$=2(Im(a+b+c+ab+bc+ac))$$But I have no idea what to do next or what am I missing.
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Consider $(a+b+c+ab+ac+bc)(1+\overline {abc})$.
When you multiply out remember that $a\overline {a}=1$.
|
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|
Link between thetrahedral numbers and combinatorics The triangular numbers are $1, 3=2+1, 6=3+2+1$ and the $n$-th triangular number is
$$\binom{n+1}{2}=\frac{n(n+1)}{2}=n+(n-1)+\ldots+2+1.$$
There is a neat explation that the n-te triangular number is $\binom{n+1}{2}$: Consider $n+1$ people. Then there are $\binom{n+1}{2}$ pairs. But you could also count the pairs differently (see here): You can match the first person with $n$ different persons, the second person can be paired with $n-1$ different persons, the third with $n-2$, $\ldots$. Hence $\binom{n+1}{2}=n+(n-1)+(n-2)+\ldots$.
Now the tetrahedral numbers $1, 4=1+3, 10=1+3+6,\ldots$ can be written as $\binom{n+2}{3}$. Is there also an combinatorical explaination as for triangular numbers? I just found this explanation but I can't see clearly why $\binom{n+2}{3}$ should be the same as $1+3+6+\ldots+\frac{n(n+1)}{2}$.
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I found an analogous explaintion using people.
Let's assume we want to calculate $\binom{5}{3}$, i.e. the number of subsets of order 3 with 5 elements / people {A,B,C,D,E}.
*
*Subgroups of size 3 with A are all possible pairs using {B,C,D,E} and
A itself, i.e. $\binom{4}{2}$.
*Subsets of size 3 with B (and without A, because we counted all
subsets with A already) are all possible pairs using {C,D,E} and B
itself, i.e. $\binom{3}{2}$.
*Subsets of size 2 with C (and without A and B, because we counted
these subsets already) are all possible pairs using {D,E} and C
itself, i.e. $\binom{2}{2}$.
Hence all possible subsets are $\binom{4}{2}+\binom{3}{2}+\binom{2}{2}=\binom{5}{3}$ and in general
$$\binom{n+2}{3}=\binom{2}{2}+\binom{3}{2}+\binom{4}{2}+\ldots +\binom{n+1}{2}.$$
|
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|
Find in closed form.$\int_0^{\frac{\pi}{2}}\left(\frac{\ln(1+\cos(x))}{1+\sin^2(x)}\right)\cos(x)dx$ Find in closed form .
$\displaystyle\int_0^{\frac{\pi}{2}}\left(\frac{\ln(1+\cos(x))}{1+\sin^2(x)}\right)\cos(x)dx$
My work
We put $t=\cos(x)$ we find $\displaystyle\int_0^{\frac{\pi}{2}}\left(\frac{\ln(1+\cos(x))}{1+\sin^2(x)}\right)\cos(x)dx=\frac{1}{2}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\beta(\frac{1}{2},\frac{2-n}{2})+\frac{3}{4}\eta(1)-\frac{\pi^2}{24}-\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{2n^2+3n}$
But :
$\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\beta(\frac{1}{2},\frac{2-n}{2})=?$ and $\sum_{n=1}^{\infty}\frac{1}{2n^2+3n}$
Where $\eta(s)$is eta function
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I did not check your calculations.
$$\frac{1}{2n^2+3n}=\frac{1}{3 n}-\frac{2}{3 (2 n+3)}$$
$$S_p=\sum_{n=1}^{p}\frac{1}{2n^2+3n}=\frac{1}{3}H_p-\frac{1}{3} \left(\psi \left(p+\frac{5}{2}\right)-\psi
\left(\frac{5}{2}\right)\right)$$Using asymptotics
$$S_p=\left(\frac{8}{9}-\frac{2 \log (2)}{3}\right)-\frac{1}{2 p}+O\left(\frac{1}{p^2}\right)$$
On the other side
$$T_p=\sum_{n=1}^{p}(-1)^n\frac{ B\left(\frac{1}{2},\frac{n-2}{2}\right)}{n}$$ is a nice monster in terms of hypergeometric function but its limit is simply
$$\frac{1}{2}-\frac{\pi ^2}{8}$$
|
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|
How to evaluate $\int _0^1\frac{\ln \left(1+\sqrt{x}\right)}{1+x^2}\:dx$ How can I approach:
$$\int _0^1\frac{\ln \left(1+\sqrt{x}\right)}{1+x^2}\:dx$$
I tried the usual differentiation under the integral sign but it didn't work so well.
I also tried to rewrite it the following 2 ways:
$$\int _0^1\frac{\ln \left(1+\sqrt{x}\right)}{1+x^2}\:dx=\int _0^1\frac{\ln \left(\sqrt{x}+x\right)}{1+x^2}\:dx-\frac{1}{2}\underbrace{\int _0^1\frac{\ln \left(x\right)}{1+x^2}\:dx}_{-G}$$
$$\int _0^1\frac{\ln \left(1+\sqrt{x}\right)}{1+x^2}\:dx=\underbrace{\int _0^1\frac{\ln \left(1-x\right)}{1+x^2}\:dx}_{\frac{\pi }{8}\ln \left(2\right)-G}-\int _0^1\frac{\ln \left(1-\sqrt{x}\right)}{1+x^2}\:dx$$
Yet I'm still stuck with those other integrals, any help is appreciated.
Managed to numerically find that:
$$\int _0^1\frac{\ln \left(1+\sqrt{x}\right)}{1+x^2}\:dx=\frac{\pi }{16}\ln \left(2\right)+\frac{\pi }{4}\ln \left(1+\sqrt{2}\right)-\frac{1}{2}G$$
But I still dont know how to approach it.
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I do not think that I should arrive to the result but here are my attempts.
Trying to work the antiderivative
$$\int \frac{\log \left(1+\sqrt{x}\right)}{1+x^2}\,dx=2\int \frac{ t \log (t+1)}{t^4+1}\,dt$$ Using partial fractions
$$\frac{ t }{t^4+1}=\frac{ t }{(t-a)(t-b)(t-c)(t-d)}$$ where $(a,b,c,d)$ are the roots of unity
$$\frac{ t }{t^4+1}=\frac A{t-a}+\frac B{t-b}+\frac C{t-c}+\frac D{t-d}$$
$$A=\frac{a}{(a-b) (a-c) (a-d)} \qquad \qquad B=\frac{b}{(b-a) (b-c) (b-d)}$$
$$C=\frac{c}{(c-a) (c-b) (c-d)} \qquad \qquad D=\frac{d}{(d-a) (d-b) (d-c)}$$
So, four integrals
$$I_k=\int \frac{\log(t+1)}{t-k}\,dt=\text{Li}_2\left(\frac{t+1}{k+1}\right)+\log (t+1) \log
\left(1-\frac{t+1}{k+1}\right)$$
$$J_k=\int_0^1 \frac{\log(t+1)}{t-k}\,dt=\log (2) \log \left(\frac{k-1}{k+1}\right)+\text{Li}_2\left(\frac{2}{k+1}\right)-\text{Li}_2\left(\frac{1}{k+1}\right)$$ Now, starts the nightmare !
Since @Quanto provided a nice answer, I give up.
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|
show this inequality $x^{n+1}+y^{n+1}\ge x^n+y^n$ let $x,y>0$ and $n$ be positive integer.if
$$x^{2n+1}+y^{2n+1}\ge 2$$
show that
$$x^{n+1}+y^{n+1}\ge x^n+y^n$$
maybe use Holder inequality: for example
$$(x^{n+1}+y^{n+1})^n(1+1)\ge (x^n+y^n)^{n+1}$$
so we must prove
$$\dfrac{1}{2}(x^n+y^n)^{n+1}\ge (x^n+y^n)^n$$
or
$$x^n+y^n\ge 2$$
this last inequality seem it is not hold.so How to prove my question.Thanks
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Since at the point $x=y$, equality holds, and multiple derivatives are $0$, it appears likely that a solution with derivatives is needed. Here is one:
Following @River Li's calculations, it suffices to prove, for $x \ge 1$, that (let $n$ be a parameter)
$$
f(x) = \ln (x^{n+1} + 1) - \ln (x^n + 1)- \frac{1}{2n+1}\ln \frac{x^{2n+1} + 1}{2} \ge 0
$$
This holds with equality for $x=1$, so we prove that $f(x)$ is increasing with $x$. The derivative is
$$
f'(x) = \frac1x \Big[ \frac{(n+1)x^{n+1}}{x^{n+1} + 1} - \frac{nx^{n}}{x^{n} + 1}- \frac{x^{2n +1}}{x^{2n +1} + 1} \Big]
$$
Multiplying out gives
$$
f'(x) = \frac{x^{n-1}}{ (x^{n+1} + 1) (x^{n} + 1) (x^{2n +1} + 1)} \Big[x - n + nx - x^{2n +1} + n x^{2n +2} - n x^{2n +1}\Big]
$$
Hence it suffices to prove that the last term in brackets is $\ge 0$, slightly rewritten:
$$
g(x) = x(n+1) - n + x^{2n +1}(-1 -n + n x)\ge 0
$$
Again we have equality for $x=1$. Further, note that $x(n+1) - n \ge 0$ holds always for $x \ge 1$, and $-1 -n + n x \ge 0 $ holds for $x\ge \frac{n+1}{n}$, so we only need to consider the range $x \in [1 \; \frac{n+1}{n}]$. In that range, we need to prove
$$
\frac{x(n+1) - n }{1 + n - n x} \ge x^{2n +1}
$$
Taking logarithms gives
$$
h(x) = \ln (x(n+1) - n ) - \ln (1 + n - n x) - (2n + 1 ) \ln x \ge 0
$$
Since we have equality for $x=1$, we use the above method again and prove that $h(x)$ is increasing. Taking derivatives gives that we need to prove
$$
h'(x) = \frac{n+1}{x(n+1) - n }+ \frac{n}{1 + n - n x} - (2n + 1 ) \frac1x \ge 0
$$
or, multiplying out,
$$
h'(x) = \frac{1}{x(x(n+1) - n) (1 + n - n x)} \Bigg[ n(2n^2 + 3n + 1)(x - 1)^2 \Bigg] \ge 0
$$
But this is clearly true for $x \in [1 \; \frac{n+1}{n}]$, which proves the claim. $\qquad \Box$
|
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Mistake proving that $3 = 0$ I was watching this video in which they proved that $3 = 0$ and the objective is to find the mistake in the proof. They did the following:
*
*Let $x$ be a solution for the equation: $x^2 + x + 1 = 0$ $(1)$.
*Because $x \neq 0$ we can devide both sides by $x$: $x + 1 + \frac{1}{x} = 0$ $(2)$
*From eq. $(1)$ we conclude that $x +1 = -x^2$ and if we substitude that in eq. $(2)$ we get: $-x^2 + \frac{1}{x} = 0$ $(3)$
*Simplifying eq $(3)$ we get: $x^2 = \frac{1}{x} \iff x^3 = 1$, so we get that $x = 1$ is a solution.
*If we substitude the solution $x = 1$ back into eq. $(1)$ we get $1^2 + 1 + 1 = 0 \iff 3 = 0$
I'll leave the answer and the rest of my question as a spoiler if you first want to try this yourself:
He explains that the mistake is when we substitute $x + 1=-x^2$ into eq. $(2)$, because when we do so we are actually adding the solution $x = 1$. The thing is that in the video he only says that and that got my wondering, why is that so?
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You made a mistake When you said that if $$x^3=1$$
then $$x=1$$
In fact,
$$x^3=1\iff x^3-1=0\iff (x-1)(x^2+x+1)=0$$
$$\implies x=1 \text{ xor } x^2+x+1=0$$
|
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|
How to express the series of $[\tan^{-1}(x)][\tanh^{-1}(x)]$ as $x^2+\left(1-\frac{1}{3}+\frac{1}{5}\right)\frac{x^6}{3}...$ How to express the series of $[\tan^{-1}(x)][\tanh^{-1}(x)]$ as
$x^2+\left(1-\dfrac{1}{3}+\dfrac{1}{5}\right)\dfrac{x^6}{3}+\left(1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}\right)\dfrac{x^{10}}{5}+\left(1-\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{13}\right)\dfrac{x^{14}}{7}+...$
I have tried to use Cauchy product or long multiplication, but the result, although of course, are equivalent, is not as elegantly expressed as this expression. This is taken from Ferrar "A textbook of convergence" page 123.
For example, for $x^6$, I have $\dfrac{2}{5}-\dfrac{1}{9}$, while for $x^{10}$, I have $\dfrac{2}{9}+\dfrac{1}{25}-\dfrac{2}{21}$. I don't know how to make it pop out the partial Leibnitz series for the coefficients.
|
Since,
for $|x| < 1$,
$\arctan(x)
=\sum_{n=0}^{\infty} \dfrac{(-1)^nx^{2n+1}}{2n+1}
$,
so
$\arctan(\sqrt{x})
=\sum_{n=0}^{\infty} \dfrac{(-1)^nx^{n+\frac12}}{2n+1}
=\sqrt{x}\sum_{n=0}^{\infty} \dfrac{(-1)^nx^n}{2n+1}
$.
Similarly,
for $|x| < 1$,
$arctanh(x)
=\sum_{n=0}^{\infty} \dfrac{x^{2n+1}}{2n+1}
$,
so
$arctanh(\sqrt{x})
=\sum_{n=0}^{\infty} \dfrac{x^{n+\frac12}}{2n+1}
=\sqrt{x}\sum_{n=0}^{\infty} \dfrac{x^n}{2n+1}
$.
Therefore
for $|x| < 1$,
$\arctan(\sqrt{x})arctanh(\sqrt{x})
=x\sum_{n=0}^{\infty}
x^n\sum_{k=0}^n \dfrac{(-1)^k}{2k+1}\dfrac1{2(n-k)+1}
$
so
$\begin{array}\\
\arctan(x)arctanh(x)
&=x^2\sum_{n=0}^{\infty}
x^{2n}\sum_{k=0}^n \dfrac{(-1)^k}{2k+1}\dfrac1{2(n-k)+1}\\
&=\sum_{n=0}^{\infty}
x^{2n+2}\sum_{k=0}^n \dfrac{(-1)^k}{2k+1}\dfrac1{2(n-k)+1}\\
&=\sum_{n=1}^{\infty}
x^{2n}\sum_{k=0}^{n-1} \dfrac{(-1)^k}{2k+1}\dfrac1{2(n-1-k)+1}\\
&=\sum_{n=1}^{\infty}
x^{2n}c_n\\
\end{array}
$
The coefficients are thus
$c_n
=\sum_{k=0}^{n-1} \dfrac{(-1)^k}{2k+1}\dfrac1{2(n-1-k)+1}
$.
According to Wolfy,
the first few terms are
$x^2 + \dfrac{13 x^6}{45} + \dfrac{263 x^{10}}{1575} +
\dfrac{36979x^{14}}{315315}+O(x^{18})
$.
It looks like
$c_{2n} = 0$
and
$c_{2n+1} \ne 0$.
The first should be
straightforward to prove
by reversing the order of summation.
A similar technique
should prove that
$c_{2n+1} > 0$.
As to a closed form
for the sum,
I don't know.
|
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Evaluating $\int_{0}^{\infty} \frac{x^8 -1}{x^{10} + 1} dx$
How to evaluate $$\int_{0}^{\infty} \frac{x^8 -1}{x^{10} + 1} dx$$
Attempt:
$$f(x) = \frac{x^8-1}{x^{10}+1}$$
$$I = \int_{0}^{\infty} \frac{x^8 -1}{x^{10} + 1} dx$$
Substituting $t = \frac{1}{x}$,
$$I = \int_{0}^{\infty} \dfrac{\frac{1}{x^8} -1}{\frac{1}{x^{10}} + 1} x^2dx$$
$$I = \int_{0}^{\infty} \frac{x^{4} - x^{12}}{x^{10} + 1} dx$$
Adding the two versions of $I$,
$$2I = \int_{0}^{\infty} \frac{x^8 - x^{12} -1 + x^4}{x^{10} + 1} dx = \int_{0}^{\infty} \frac{(x^8-1)(1 - x^{4}) }{x^{10} + 1} dx$$
However, that does not help much.
Putting it in WolframAlpha gives $I = 0$
I am pretty sure there is an elegant solution which I am not able to find it.
|
with $x = e^u$ and $dx = e^u du,$ I got
$$ \int_{- \infty}^\infty \frac{e^{4u} - e^{-4u}}{e^{5u} + e^{-5u}} du = \int_{- \infty}^\infty \; \frac{ \sinh 4u \; }{ \cosh 5u \;} du$$
which is an odd integrand over symmetric endpoints
|
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|
Show that for every triangle $ABC$ $\frac{1}{h_b}+\frac{1}{h_c}-\frac{1}{h_a}=\frac{1}{r_a}$ Show that for every triangle $ABC$ $$\dfrac{1}{h_b}+\dfrac{1}{h_c}-\dfrac{1}{h_a}=\dfrac{1}{r_a}.$$
Let $S_{ABC}=S$. Then $$S=\dfrac{ah_a}{2}=\dfrac{bh_b}{2}=\dfrac{ch_c}{2}\Rightarrow h_a=\dfrac{2S}{a};h_b=\dfrac{2S}{b};h_c=\dfrac{2S}{c}.$$ On the other hand, $$S=r_a(p-a)\Rightarrow r_a=\dfrac{S}{p-a}.$$ where $r_a$ is the exradii.
So $$\dfrac{1}{h_b}+\dfrac{1}{h_c}-\dfrac{1}{h_a}=\dfrac{b+c-a}{2S}.$$ I can't see that this is equal to $\dfrac{1}{r_a}=\dfrac{p-a}{S}.$ Thank you in advance!
|
If $p$ is the semiperimeter $\frac {a+b+c}2$:
$$\frac1{h_b}+\frac1{h_c}-\frac1{h_a}=\frac {b+c-a}{2\color{red}S} = \frac {a+b+c-2a}{2\color{red}S} = \frac{\frac {a+b+c}2-\frac {2a}2}{\color{red}S} = \frac {p-a}{\color{red}S} = \frac1{r_a}$$
|
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|
Find all $x,y \in \mathbb{N}$ such that $x^2+y^2-8x=9$ This is a math Olympiad problem.
my attempt :
By solving the quadratic equation in $x$, i’ve got this:
$$x=\frac{8 \pm \sqrt{100-4y^2}}{2}$$
And from this it’s easy to see that $100-4y^2$ it’s a perfect square and it’s divisible by $4$.
$$\cases{100-4y^2=m^2 \\ 100-4y^2=4n }$$
Dividing the first equation by the second we get:
$$m^2=4n \iff m=2\sqrt{n} \implies n=a^2$$
By plugging this value to $100-4y^2=4n$:
$$100-4y^2=4a^2 \iff 25=a^2+y^2$$
so :$y \in \{0,5,3,4\}$
And by plugging these values to $x=\frac{8 \pm \sqrt{100-4y^2}}{2}$, we get:(we ignore the cases where $x \notin \mathbb N$)
$$(x,y) \in \{(9,0),(4,5),(8,3),(0,3),(1,4),(7,4)\}$$
I think that my solution is a bit long, so if is there a short solution please post it.
|
From the equation, we have
$$(x-4)^2+y^2 = 25$$
Hence, we can deduce that $(|x-4|,y) = (0,5),(3,4),(4,3)$ or $(5,0)$.
The solution is then
$$(x,y) = (4,5),(7,4),(1,4),(0,3),(8,3),(9,0)$$
|
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|
PDF of a rectangle If I want to find the $c$ of a $PDF$ when it's given:
$f_{X,Y}\left(x,y\right)=c\:\:\:\:\left(The\:area\:in\:blue\right),\:otherwise:\:0$
I try to do that:
$$\int _{\frac{1}{2}}^1\:\int _{-x+\frac{3}{2}}^1\:c\,dy\,dx+\int _0^1\:\int _{-x+\frac{1}{2}}^{-x+1}\:c\,dy\,dx=1$$
But I got $c=\frac{8}{5}$ and the result is $c=2$.
(I know we can calculate it with by calculating the area of the triangles but I am wondering to know why my way is not working)
|
$$
\begin{aligned}
\iint\limits_{\text{square}}f(x, y)dxdy &= \iint\limits_{\text{white triangle}}0dxdy+
\iint\limits_{\text{blue trapezoid}}cdxdy+
\iint\limits_{\text{white trapezoid}}0dxdy+
\iint\limits_{\text{blue triangle}}cdxdy = \\
&= c\left(\iint\limits_{\text{blue trapezoid}}dxdy+
\iint\limits_{\text{blue triangle}}dxdy\right) = \\
&= \left|
\begin{aligned}
&\text{Integral over the blue trapezoid can be represented as a sum of two integrals:}\\
&\qquad\text{integral over a blue parallelogram (left to the }x = \frac{1}{2}\text{) and }\\
&\qquad\text{integral over a blue triangle (right to the }x = \frac{1}{2}\text{)}
\end{aligned}
\right| = \\
&= c\left(\int\limits_0^{\frac{1}{2}}dx\int\limits_{-x + \frac{1}{2}}^{-x + 1}dy +
\int\limits_{\frac{1}{2}}^1dx\int\limits_{0}^{-x + 1}dy +
\int\limits_{\frac{1}{2}}^1dx\int\limits_{-x + \frac{3}{2}}^1dy
\right) = \\
&= c\left(\int\limits_0^\frac{1}{2}\frac{1}{2}dx + \int\limits_\frac{1}{2}^1(-x+1)dx + \int_\frac{1}{2}^1\left(x-\frac{1}{2}\right)dx\right) = \\
&= c\left(\frac{1}{4} - \left.\frac{(-x+1)^2}{2}\right|_\frac{1}{2}^1+
\left.\frac{\left(x-\frac{1}{2}\right)^2}{2}\right|_\frac{1}{2}^1\right) = \\
&= c\left(\frac{1}{4} + \frac{1}{8} + \frac{1}{8}\right) = c\cdot\frac{1}{2} = 1 \Rightarrow c = 2
\end{aligned}
$$
|
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|
sum of floor function simplification Let $n \in \mathbb{N}$.
I'm trying to show $\sum_{i=1}^{\infty} \left(\left \lfloor{\frac{n+2^k-1}{2^i}}\right \rfloor - \left \lfloor{\frac{n-1}{2^i}}\right \rfloor \right) = 2^k-1$.
I know that when $2^i > n-1$, the right floor function will become zero.
I'm not sure how I cancel out the $n$'s.
I also tried solving the sum of each floor function separately, but didn't get anywhere.
Any suggestions will be appreciated!
Thanks.
|
What you're trying to show is not always true. For example, let $n = 4$, so $n - 1 = 3$, and $k = 1$, so $2^k = 2$. This means $n + 2^k - 1 = 5$. However, using that each summation term is non-negative, the first few terms then become
$$\begin{equation}\begin{aligned}
& \sum_{i=1}^{\infty} \left(\left\lfloor{\frac{n+2^k-1}{2^i}}\right\rfloor - \left\lfloor{\frac{n-1}{2^i}}\right\rfloor\right) \\
& = \left(\left\lfloor\frac{5}{2}\right\rfloor - \left\lfloor\frac{3}{2}\right\rfloor\right) + \left(\left\lfloor\frac{5}{4}\right\rfloor - \left\lfloor\frac{3}{4}\right\rfloor\right) + \ldots \\
& = (2 - 1) + (1 - 0) + \ldots \\
& = 1 + 1 + \ldots \\
& = 2
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Note I've used that all of the remaining terms, i.e., the "$\ldots$" part, are $0$. This is larger than your right side of $2^{k} - 1 = 1$.
Note your statement would be correct with an appropriate restriction between $k$ and $n$. In particular, something like $2^k \gt n - 1$ works. This is because, for $i \le k$, each of the summation terms becomes
$$\begin{equation}\begin{aligned}
\left\lfloor{\frac{n+2^k-1}{2^i}}\right\rfloor - \left\lfloor{\frac{n-1}{2^i}}\right\rfloor & = \left(\left\lfloor{\frac{2^k}{2^i} + \frac{n-1}{2^i}}\right\rfloor\right) - \left\lfloor{\frac{n-1}{2^i}}\right\rfloor \\
& = \left(2^{k-i} + \left\lfloor{\frac{n-1}{2^i}}\right\rfloor\right) - \left\lfloor{\frac{n-1}{2^i}}\right\rfloor \\
& = 2^{k-i}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
For $i \gt k$, then $2^i \ge 2(2^k) \gt n + 2^k - 1$ and $2^i \gt n - 1$, so both floor terms become $0$. Thus, the total sum would be that of a geometric series of
$$\sum_{i=1}^{k}(2^{k-i}) = 2^k - 1 \tag{3}\label{eq3A}$$
which then matches your right side.
|
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|
Showing $\sum_{k=1}^{n}k(k+1)(k+2)\cdots(k+r) = \frac{n(n+1)(n+2)\cdots(n+r+1)}{r+2}$ I was solving a question and saw a pattern. Can someone prove it, please?
We know
$$\sum_{k=1}^{n}k = \frac{n(n+1)}{2}$$
$$\sum_{k=1}^{n}k(k+1) = \frac{n(n+1)(n+2)}{3}$$
$$\sum_{k=1}^{n}k(k+1)(k+2) = \frac{n(n+1)(n+2)(n+3)}{4}$$
So we see the pattern... Can anyone give a proof of why:
$$\sum_{k=1}^{n}k(k+1)(k+2)\cdots(k+r) = \frac{n(n+1)(n+2)\cdots(n+r+1)}{r+2}$$
|
You have
\begin{align}
k(k+1)(k+2)\cdots(k+r) &= k(k+1)(k+2)\cdots(k+r)\frac{r+2}{r+2}\\
&= k(k+1)(k+2)\cdots(k+r)\frac{(k+r+1)-(k-1)}{r+2}\\
&= \frac{k(k+1)(k+2)\cdots(k+r)(k+r+1)}{r+2}-\frac{(k-1)k(k+1)(k+2)\cdots(k+r)}{r+2}\\
\end{align}
Make the sum, you deduce easily that
$$\sum_{k=1}^{n}k(k+1)(k+2)\cdots(k+r) = \frac{n(n+1)(n+2)\cdots(n+r+1)}{r+2}$$
|
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|
Integral solutions to the diophantine equation $y^2=x^3+2017$. Case 1: $y^2=0 \mod{3}$
\begin{align*}
& y^2 = 3b \quad (b=3k^2) \implies x^3=3a +2 \\
& 3b = 3a + 2 + 2017 \implies b-a = 673 \\
& k^2 = \frac{a-2}{3} + 225 \\
\end{align*}
Then I just took $a =2$ since 225 is already a perfect square. Thus I got the solution $(x,y) = (2,\pm 45)$. This is the only solution I was able to get.
I tried to do the second case, i.e, $y^2 \equiv 1 \mod 3$ as I did Case 1 but did not have any luck. Here are my proceedings.
Let $y^2 = 3b + 1$ and $x^3 = 3a$ where $a = 9k^3$
Then doing the same thing above we get,
\begin{equation*}
k^3 = \frac{b+3}{9} - 75
\end{equation*}
Now I am stuck, since I cannot find any solutions to the above equation.
|
Comment:
You can try this method:
Take floor of $2017^{\frac 13}=12$
You can check 12 and less to see it gives a solution:
$2017=12^3+17^2=y^2-x^3$
Which gives $y=\pm 17$ and $x=-12$
Among other numbers bellow 12 only 2 gives integer for y:
$2017=45^2-2^3=y^2-x^3$
Which give what you found; $x=2$ and $y=\pm 45$
This method is for small numbers.
|
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|
Finding length of diagonal in parallelogram given two sides and other diagonal A parallelogram has sides of length 11 cm and 13 cm and has one diagonal 20 cm long the length of the other diagonal is what I am supposed to find. Now my answer came out to be 13.2 cm.
Finding the area of the first triangle:
Perimeter = $11 + 13 + 20 = 44$ cm
Semi-Perimeter = $44 ÷ 2 = 22$ cm
Area = $\sqrt{p(p - a)(p - b) (p - c)} = \sqrt{22(22 - 11)(22 - 13)(22 - 20)} = \sqrt{4356} = 66 \text{cm}^2$
Find the area of the parallelogram:
Area of parallelogram = $2 \times \ $area of triangle
Area = $66 * 2 = 132 cm²$
Find the other diagonal:
Let the other diagonal be x
$1/2 (diagonal 1 * diagonal 2) = Area $
$1/2 (20x) = 132 $
$10x = 132 $
$x = 13.2 cm $
However, surprisingly my answer wasn't in the options. Now my friend sent me this attachment from the website topper and claimed that 20 is the correct answer so can you guys pls tell me whether 13.2 cm is the correct answer or 20? Do pls also tell whether what is the mistake done by the other one? Thanks in advance for your help.
|
Obviously, both diagonals cannot be $20$, otherwise you would have a rectangle, and this would imply $11^2 + 13^2 = 20^2$, which is false.
One solution that is accessible to you is to employ Heron's formula as you have done, but extend the computation. You already found that the area of the parallelogram is $66(2) = 132$. However, in order to find the other diagonal, you must solve the equation $$66 = \sqrt{s(s-11)(s-13)(s-c)},$$ where $s = (11+13+c)/2$ and $c$ is the desired diagonal. You already know $c = 20$ satisfies this equation; the goal is to find the other positive root. Some algebra will give $$66 = \frac{1}{4} \sqrt{-2304 + 580c^2 - c^4},$$ after which obtaining the nontrivial positive root should be straightforward.
Alternatively, we can employ trigonometry: we know by the Law of Cosines that $$20^2 = 11^2 + 13^2 - 2(11)(13) \cos C,$$ hence the other diagonal has length $$c^2 = 11^2 + 13^2 - 2(11)(13) \cos (180^\circ - C) = 11^2 + 13^2 + 2(11)(13) \cos C.$$ In fact, this is how the diagonal identity can be shown: for general side lengths $a, b$ with one given diagonal $d$, we find the other diagonal $c$ obeys $$c^2 + d^2 = 2(a^2 + b^2), \tag{1}$$ since we must have the simultaneous solution $$\begin{align} c^2 &= a^2 + b^2 - 2ab \cos C \\ d^2 &= a^2 + b^2 - 2ab \cos (180^\circ - C) \end{align}$$ and since $\cos (180^\circ - C) = -\cos C$, we add the two equations to get $c^2 + d^2 = 2a^2 + 2b^2$, proving Equation $(1)$.
|
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|
How to compute $1 \times 2 \times 3 \times 4 + 3 \times 4 \times 5 \times 6 + ... + 97 \times 98 \times 99 \times 100$ How to compute $S = 1 \times 2 \times 3 \times 4 + 3 \times 4 \times 5 \times 6 + ... + 97 \times 98 \times 99 \times 100$
Was thinking $\frac{S}{24} = {4\choose 4} + {6\choose 4} + {8\choose 4} + ... + {100\choose 4}$ but how do I sum this up?
|
Hint:
The $n$th term $T(n)$ is $$(2n-1)(2n)(2n+1)(2n+2) =16n^4+16n^3-4n^2-4n$$
WLOG $$T(n)=P(n+1)-P(n)$$ where $P(m)=\sum_{r=0}^ua_rm^r$
$$\sum_{r=0}^5a_r((n+1)^r-n^r)$$
$$=a_1+a_2(2n+1)+a_3(3n^3+3n+1)+a_4(4n^3+6n^2+4n+1)a+a_5(5n^4+10n^3+10n^2+5n+1)+\cdots$$
Clearly, $a_r=0$ for $r\ge6$
Now compare the coefficients of $n^4,n^4,n^2,n$ and constants to find $a_j;0\le j\le5$
Comparing the coefficients of $n^4,$ $$5a_5=16$$
|
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|
Evaluate $\int_{0}^{1} \frac{r^3}{ \sqrt {1-r^2}}dr$ Evaluate
$$\int_{0}^{1} \frac{r^3}{ \sqrt {1-r^2}}dr$$
My attempt : I put $t= \sqrt{1-r^2}$ now $dt/dr= \frac{-r}{2\sqrt {1-r^2}}$ $$\implies dr=\frac{2\sqrt {1-r^2}}{r}dt$$
$$\int_{0}^{1} \frac{r^3}{ \sqrt {1-r^2}}\frac{2\sqrt {1-r^2}}{r}dt$$
$$= 2\int_{0}^{1} r^2 dt$$
$t= \sqrt{1-r^2}\implies r^2= t^2-1$
$$2\int_{0}^{1} t^2-1 dt= 2[\frac{t^3}{3} -t]_{0}^{1}==-4/3$$
|
Or just integration by parts:
\begin{eqnarray*}\int_0^1 r^2 \frac{r}{\sqrt{1-r^2}}dr
& = & \left.-\sqrt{1-r^2}\cdot r^2\right|_0^1 + \int_0^1 (2r)\sqrt{1-r^2}\;dr \\
& = & \left.-\frac 23\left(1-r^2\right)^{\frac 32}\right|_0^1 \\
& = & \frac 23
\end{eqnarray*}
|
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|
$\lim_{n\to\infty} \frac{3^n}{(1+3)(1+3^2)(1+3^3)\ldots (1+3^n)}$ It is visible that the result is 0, but I can't calculate it.
$\lim_{n\to\infty} \frac{3^n}{(1+3)(1+3^2)(1+3^3)\ldots (1+3^n)}$
It occurred to me to express it as a product and take a ratio test, but I'm not sure
|
$$
f(n)=\frac{3^n}{(1+3)(1+3^2)\cdots(1+3^n)}
=\frac{\frac{3^n}{3^1 3^2 \cdots 3^n}}{\frac{1+3^1}{3^1}\frac{1+3^2}{3^2}\cdots\frac{1+3^n}{3^n}}\\
=\frac{\frac{3^n}{3^{\frac{n(n+1)}2}}}{(1+3^{-1})(1+3^{-2})\cdots(1+3^{-n})}
=\frac{2\cdot 3^{-\frac{n(n-1)}{2}}}{\left(-1;\frac{1}{3}\right)_{n+1}}
$$
in terms of a $q$-Pochhammer symbol in the denominator.
For large $n$ this denominator approaches a constant,
$$
\lim_{n\to\infty}\left(-1;\frac{1}{3}\right)_{n+1}
= \left(-1;\frac{1}{3}\right)_{\infty}
\approx 3.129868037134023\ldots,
$$
in Mathematica,
QPochhammer[-1, 1/3] // N
and therefore $\lim_{n\to\infty}f(n)=0$.
|
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|
Symmetry in the quadratic form matrix The following is an excerpt from Greene's Econometric Analysis (7th Edition). Therein, the author states the matrix in quadratic forms must be symmetric. I would like to know why. What if, for instance, $a_{12} \neq a_{21}$?
|
$$\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=ax^2+(b+c)xy+dy^2$$ can be symmetrized as
$$\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}a&\frac{b+c}2\\\frac{b+c}2&d\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=ax^2+2\frac{b+c}2xy+dy^2.$$
So a quadratic form has a symmetric matrix WLOG.
|
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|
Prove that the real root of $X^3-X^2+2X-1$ is the square of the real root of $X^3+X^2-1$. Context: I'm studying the family of functions of the form:
$$y=a \ln(x) + b \ln(1-x)$$
for $0 \leq x \leq 1$ and $(a,b) \in \mathbb{N} \times \mathbb{N}$ where $\mathbb{N} = \{0, 1, 2,\dots\}$.
When "all" the functions of the family are drawn on the same picture, one gets (click to enlarge):
What I'm calling "pillars" make their appearance. They are located on precise abscissas $x_{c/d}$ which are algebraic numbers, roots of $X^c-(1-X)^d$ polynomials for some positive rational $c/d$ (in irreducible form). To each pillar is associated a vertical wavelength (clearly visible on the picture) which is equal to $-\frac{\ln(x_{c/d})}{d}=-\frac{\ln(1-x_{c/d})}{c}$. I compared the wavelength of the $(c/d=5)$ pillar (for which $x_5$ is the multiplicative inverse of $\rho$, the plastic number) with the wavelength of the $(c/d=3/2)$ pillar (marked green), and found numerically that they were equal. So, the asked question is to prove that, in any of its forms:
*
*$$ -\frac{\ln(x_{3/2})}{2} = -\frac{\ln(x_{5})}{1} $$
*$$ x_{3/2} = x_{5}^2 $$
*the real root of $X^3-(1-X)^2$ is the square of the real root of $X^5-(1-X)^1$
*the real root of $X^3-X^2+2X-1$ is the square of the real root of $X^3+X^2-1$
|
We will prove the last statement.
Let $\alpha$ be a root of $p_1:= X^3 + X^2-1$, i.e., $\alpha^3 +\alpha^2-1 = 0$. We will prove that $\beta := \alpha^2$ is a root of $p_2:= X^3-X^2+2X-1$.
We have the following identities:
$$\beta = \alpha^2,$$
$$\beta^2 = \alpha^4 = \alpha\cdot \alpha^3 = \alpha (-\alpha^2 +1) = -\alpha^3 + \alpha,$$
$$\beta^3 = \alpha^6 = (\alpha^3)^2 = (-\alpha^2+1)^2 = \alpha^4 -2\alpha^2 +1 = -\alpha^3 -2\alpha^2 + \alpha +1.$$
Then,
$$\beta^3 - \beta^2 +2\beta -1 = (-\alpha^3 -2\alpha^2 + \alpha +1) - (-\alpha^3 + \alpha) +2(\alpha^2) -1 = 0.$$
Thus $\beta$ is a root of $p_2$. Note that we did not require $\alpha$ to be real, this means that the roots of $p_2$ are precisely the squares of the roots of $p_1$. Particularly, the square of the real root of $p_1$ is a (the) real root of $p_2$.
|
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|
Determine the generator for the cyclic group formed by the solutions of $x^9 = 1$.
Find the solutions to $x^9 = 1$ and determine the generator for the cyclic group formed by the solutions.
The equation can be factored as $(x^3 - 1)(x^6 + x^3 + 1) = 0$ and the solutions are $$\begin{align*}
x &= 1\\
x &= -\sqrt[\leftroot{1}\uproot{3}9]{-1} \\
x &= (-1)^{2/9} \\
x &= -\sqrt[\leftroot{1}\uproot{3}3]{-1} \\
x &= (-1)^{4/9} \\
x &= -(-1)^{5/9} \\
x &= (-1)^{2/3} \\
x &= -(-1)^{7/9} \\
x &= (-1)^{8/9}
\end{align*}$$ but I'm unsure how to determine the generator. How are these found generally and what would be my options to start determining it?
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The multiplicative group of $9$-th roots is isomorphic to the additive group $\mathbf Z/9\mathbf Z$, under the isomorphism:
$$\mathbf Z/9\mathbf Z\longrightarrow U_9,\quad k\bmod9\longmapsto\mathrm e^{\tfrac{2ik\pi}9}.$$
Under this isomorphism, generators of $\mathbf Z/9\mathbf Z$ map onto generators of $U_9$, i.e. primitive $9$-th roots of unity. Therefore it is enough to know the generators of $\mathbf Z/9\mathbf Z$, which is standard.
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"timestamp": "2023-03-29T00:00:00",
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|
What is the meaning of $A =\prod_{jI am trying to parse this proof in Hungerford's book:
When he gets into computing this:
$$A =\prod_{\substack{j<k\\ j,k \neq c,d}}^{}(i_j-i_k)$$
What does this means?
*
*Does it means that $j<k$ and $j\neq c$ and $k\neq d$; or
*Does it means that $j<k$ and $j\neq c$ and $k\neq d$ and $h\neq d$ and $k\neq c$?
I computed the indices I'd get in both cases in a simple example where $c=3$, $d=5$ and $j=9$ and it for the first case, I'd get the following indices:
$$\begin{array}{cccccccc}
\begin{array}{cc}
1 & 2 \\
\end{array}
& \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\
\begin{array}{cc}
1 & 3 \\
\end{array}
&
\begin{array}{cc}
2 & 3 \\
\end{array}
& \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\
\begin{array}{cc}
1 & 4 \\
\end{array}
&
\begin{array}{cc}
2 & 4 \\
\end{array}
&
\begin{array}{c}
\text{$\bullet$ $\bullet$} \\
\end{array}
& \text{} & \text{} & \text{} & \text{} & \text{} \\
\begin{array}{c}
\text{$\bullet$ $\bullet$} \\
\end{array}
&
\begin{array}{c}
\text{$\bullet$ $\bullet$} \\
\end{array}
&
\begin{array}{c}
\text{$\bullet$ $\bullet$} \\
\end{array}
&
\begin{array}{c}
\text{$\bullet$ $\bullet$} \\
\end{array}
& \text{} & \text{} & \text{} & \text{} \\
\begin{array}{cc}
1 & 6 \\
\end{array}
&
\begin{array}{cc}
2 & 6 \\
\end{array}
&
\begin{array}{c}
\text{$\bullet$ $\bullet$} \\
\end{array}
&
\begin{array}{cc}
4 & 6 \\
\end{array}
&
\begin{array}{cc}
5 & 6 \\
\end{array}
& \text{} & \text{} & \text{} \\
\begin{array}{cc}
1 & 7 \\
\end{array}
&
\begin{array}{cc}
2 & 7 \\
\end{array}
&
\begin{array}{c}
\text{$\bullet$ $\bullet$} \\
\end{array}
&
\begin{array}{cc}
4 & 7 \\
\end{array}
&
\begin{array}{cc}
5 & 7 \\
\end{array}
&
\begin{array}{cc}
6 & 7 \\
\end{array}
& \text{} & \text{} \\
\begin{array}{cc}
1 & 8 \\
\end{array}
&
\begin{array}{cc}
2 & 8 \\
\end{array}
&
\begin{array}{c}
\text{$\bullet$ $\bullet$} \\
\end{array}
&
\begin{array}{cc}
4 & 8 \\
\end{array}
&
\begin{array}{cc}
5 & 8 \\
\end{array}
&
\begin{array}{cc}
6 & 8 \\
\end{array}
&
\begin{array}{cc}
7 & 8 \\
\end{array}
& \text{} \\
\begin{array}{cc}
1 & 9 \\
\end{array}
&
\begin{array}{cc}
2 & 9 \\
\end{array}
&
\begin{array}{c}
\text{$\bullet$ $\bullet$} \\
\end{array}
&
\begin{array}{cc}
4 & 9 \\
\end{array}
&
\begin{array}{cc}
5 & 9 \\
\end{array}
&
\begin{array}{cc}
6 & 9 \\
\end{array}
&
\begin{array}{cc}
7 & 9 \\
\end{array}
&
\begin{array}{cc}
8 & 9 \\
\end{array}
\\
\end{array}$$
For the second case, I'd get the following indices:
$$\begin{array}{cccccccc}
\begin{array}{cc}
1 & 2 \\
\end{array}
& \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\
\begin{array}{c}
\diamond \;\diamond \\
\end{array}
&
\begin{array}{c}
\diamond \;\diamond \\
\end{array}
& \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\
\begin{array}{cc}
1 & 4 \\
\end{array}
&
\begin{array}{cc}
2 & 4 \\
\end{array}
&
\begin{array}{c}
\text{$\bullet$ $\bullet$} \\
\end{array}
& \text{} & \text{} & \text{} & \text{} & \text{} \\
\begin{array}{c}
\text{$\bullet$ $\bullet$} \\
\end{array}
&
\begin{array}{c}
\text{$\bullet$ $\bullet$} \\
\end{array}
&
\begin{array}{c}
\text{$\bullet$ $\bullet$} \\
\end{array}
&
\begin{array}{c}
\text{$\bullet$ $\bullet$} \\
\end{array}
& \text{} & \text{} & \text{} & \text{} \\
\begin{array}{cc}
1 & 6 \\
\end{array}
&
\begin{array}{cc}
2 & 6 \\
\end{array}
&
\begin{array}{c}
\text{$\bullet$ $\bullet$} \\
\end{array}
&
\begin{array}{cc}
4 & 6 \\
\end{array}
&
\begin{array}{c}
\diamond \;\diamond \\
\end{array}
& \text{} & \text{} & \text{} \\
\begin{array}{cc}
1 & 7 \\
\end{array}
&
\begin{array}{cc}
2 & 7 \\
\end{array}
&
\begin{array}{c}
\text{$\bullet$ $\bullet$} \\
\end{array}
&
\begin{array}{cc}
4 & 7 \\
\end{array}
&
\begin{array}{c}
\diamond \;\diamond \\
\end{array}
&
\begin{array}{cc}
6 & 7 \\
\end{array}
& \text{} & \text{} \\
\begin{array}{cc}
1 & 8 \\
\end{array}
&
\begin{array}{cc}
2 & 8 \\
\end{array}
&
\begin{array}{c}
\text{$\bullet$ $\bullet$} \\
\end{array}
&
\begin{array}{cc}
4 & 8 \\
\end{array}
&
\begin{array}{c}
\diamond \;\diamond \\
\end{array}
&
\begin{array}{cc}
6 & 8 \\
\end{array}
&
\begin{array}{cc}
7 & 8 \\
\end{array}
& \text{} \\
\begin{array}{cc}
1 & 9 \\
\end{array}
&
\begin{array}{cc}
2 & 9 \\
\end{array}
&
\begin{array}{c}
\text{$\bullet$ $\bullet$} \\
\end{array}
&
\begin{array}{cc}
4 & 9 \\
\end{array}
&
\begin{array}{c}
\diamond \;\diamond \\
\end{array}
&
\begin{array}{cc}
6 & 9 \\
\end{array}
&
\begin{array}{cc}
7 & 9 \\
\end{array}
&
\begin{array}{cc}
8 & 9 \\
\end{array}
\\
\end{array}$$
Where $\bullet \;\bullet$ and $\diamond \;\diamond$ are the deleted pairs of indices. My guess is that it means the second case because it seems only this way would allow us to have $\sigma(A)=A$, since no $(i_j - i_k)$ would be changed. Is my reasoning correct?
|
I feel like Hungerford may have painted himself into a corner with that notation, so let me offer you a different way of doing exactly what the proof is doing.
For a fixed $n$, consider commuting variables $x_1,\ldots,x_n$ and form the Vandermonde matrix $V=V(x_1,\ldots,x_n)$ with $V(x)_{ij} = x_i^{j-1}$. It is a pleasant exercise to verify that $
\det V$ is equal to $\prod_{i<j}(x_i-x_j)$ and, in particular, it evaluates to a non-zero value whenever we pick distinct values for the variables, as Hungerford stated. Let us call this polynomial $\Delta$.
The symmetric group $S_n$ acts on the variables $x_i$ by its action on the indices, so in particular it acts on each polynomial in the variables $x_1,\ldots,x_n$ and hence on $\Delta $.
The first observation we make is that for a transposition the determinant changes by $-1$, since in the way we have written it, it is clear two rows are swapped. This means that if $\sigma$ is written as a product of $N$ transpositions, then $\sigma \Delta = (-1)^N\Delta$. Assume that $\sigma$ is also written as a product of $M$ traspositions.
To conclude that $N=M$ modulo $2$, it suffices we evaluate $\Delta$ at a the point say $p=(1,2,\ldots,n)$ as suggested in your post, in which case $\Delta(p)\neq 0$ and we obtain that $(-1)^N = (-1)^M$.
|
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|
Algebraic simplification of $(a-b)(2b-a+2)(2b+1) - (b+1)(2a-2b-1)(2b-a)$ When trying to replicate a proof about Catalan numbers, I came across the following simplification $$(a-b)(2b-a+2)(2b+1) - (b+1)(2a-2b-1)(2b-a)= a^2+a$$
This is confirmed by writing out all the terms (and by WolframAlpha). My question - is there a clever substitution or factorisation that makes this obvious / less tedious? (Or can anyone provide a more succinct explanation of the answer linked above?)
Thanks
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Original problem : $$\color{green}{(a-b)}\color{red}{(2b-a+2)}\color{green}{(2b+1)} - (b+1)\color{blue}{(2a-2b-1)}(2b-a)$$
*
*) Notice how
$\color{green}{(a-b)(2b+1)}=-2b^2+2ab-b+a=b(-2b+2a-1)+a=b\color{blue}{(2a-2b-1)}+a$
$$\Rightarrow\color{blue}{(2a-2b-1)}=\dfrac{\color{green}{(a-b)(2b+1)}-a}{b}$$
*) We find the same for the second part of the equation:
$(b+1)(2b-a)=2b^2-ab+2b-a=b\color{red}{(2b-a+2)}-a$
$$\Rightarrow\color{red}{(2b-a+2)}=\dfrac{(b+1)(2b-a)+a}{b}$$
Plugging in we get:
$$(a-b)\color{red}{\dfrac{(b+1)(2b-a)+a}{b}}(2b+1) - (b+1)(2a-2b-1)(2b-a)=$$
$$\dfrac{(a-b)}{b}\cdot[(b+1)(2b-a)+a]\cdot(2b+1) - (b+1)(2a-2b-1)(2b-a)=$$
$$\dfrac{(a-b)}{b}\cdot(2b+1)\cdot[(b+1)(2b-a)+a] - (b+1)(2a-2b-1)(2b-a)=$$
$$\dfrac{(a-b)}{b}\cdot(2b+1)\cdot a + \dfrac{(a-b)}{b}(2b+1)(b+1)(2b-a) - (b+1)(2a-2b-1)(2b-a)=$$
$$\dfrac{(a-b)}{b}(2b+1)a + (b+1)(2b-a)\left(\dfrac{(a-b)}{b}(2b+1)+(2a-2b-1)\right)$$
$$\dfrac{b\color{blue}{(2a-2b-1)}+a}{b}a + (b+1)(2b-a)\left(\dfrac{(a-b)}{b}(2b+1)+(2a-2b-1)\right)$$
$$\dfrac{b\color{blue}{(2a-2b-1)}+a}{b}a + (b+1)(2b-a)\left(\dfrac{b\color{blue}{(2a-2b-1)}+a}{b}+\dfrac{\color{green}{(a-b)(2b+1)}-a}{b}\right)$$
$$\dfrac{b\color{blue}{(2a-2b-1)}+a}{b}a + (b+1)(2b-a)\left(\dfrac{b\color{blue}{(2a-2b-1)}+\color{green}{(a-b)(2b+1)}}{b}\right)$$
$$\dfrac{b\color{blue}{(2a-2b-1)}+a}{b}a + (b+1)(2b-a)\left(\dfrac{(2ab-2b^2-b)+\color{green}{-2b^2+2ab-b+a}}{b}\right)$$
$$\dfrac{b\color{blue}{(2a-2b-1)}+a}{b}a + (b+1)(2b-a)\left(\dfrac{4ab-4b^2-2b+a}{b}\right)$$
$$\dfrac{b\color{blue}{(2a-2b-1)}+a}{b}a + \dfrac{b\color{red}{(2b-a+2)}-a}{b}\left({4ab-4b^2-2b+a}\right)$$
$$\color{blue}{(2a-2b-1)}a+\dfrac{a^2}{b} + \left(\color{red}{(2b-a+2)}-\dfrac{a}{b}\right)\left({4ab-4b^2-2b+a}\right)$$
There is definitely something in this 'equation', but it definitely won't save you either time or paper.
|
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|
Finding a polynomial $f(x)$ that when divided by $x+3$ yields quotient $2x^2-x+7$ and remainder $10$ I'm struggling to grasp this particular question:
When a polynomial $f(x)$ is divided by $x+3$, the quotient is $2x^2-x+7$ and the remainder is $10$. What is $f(x)$?
This is what I did:
$$\begin{align}
f(x) &= (x+3)(2x^2-x+7) \\
&= 2x^3-x^2+7x+6x^2-3x+21 \\
&= 2x^3+5x^2+4x+21
\end{align}$$
Making sure it is correct:
Then, I realized that the remainder has to be 10 not 0...I have no idea how to do that
|
In general, when you divide a function $p(x)$ by $q(x)$ and you get $f(x)$ as the quotient with a remainder of $r(x)$, you can alyways represent the answer as: $\frac{r(x)}{q(x)} + f(x)$
So you can rewrite your question algebraically as $\frac{10}{x+3} + 2x^2-x+7$ and then can simplify as follows:
$\frac{10}{x+3} + 2x^2-x+7 = \frac{10 + (x+3)(2x^2-x+7)}{x+3} = \frac{10 + 2x^3 -x^2+7x+6x^2-3x+21}{x+3} = \frac{2x^3 +5x^2 + 4x + 31}{x+3}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove inequality $\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right) \tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)\le\frac13 $ How to determine the range of the function
$$f(x)=\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right)
\tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)
$$
It it is straightforward to verify that $f(x)$ is even and
$$f(0)= \frac13, \>\>\>\>\> \lim_{x\to\pm \infty} f(x) \to -\infty$$
which implies $f(x) \in (-\infty,\frac13]$, i.e.
$$\tan \left( \frac\pi2 \frac{(1+x)^2}{3+x^2}\right)
\tan \left( \frac\pi2 \frac{(1-x)^2}{3+x^2}\right)\le \frac13
$$
and is visually confirmed below
However, it is not obvious algebraically that $f(x)$ monotonically decreases away from $x=0$. The standard derivative tests are not viable due to their rather complicated functional forms.
So, the question is how to prove the inequality $f(x) \le \frac13$ with rigor.
Note that it is equivalent to proving
$$\cot \left(\frac{\pi(1+x)}{3+x^2}\right)
\cot \left(\frac{\pi(1-x)}{3+x^2}\right)\le \frac13
$$
|
Note that
$$f(x)=\tan\left(\frac{\pi}{2}\frac{(1+x)^2}{3+x^2}\right)\tan\left(\frac{\pi}{2}\frac{(1-x)^2}{3+x^2}\right)=\frac{2\cos(2\pi/(3+x^2))}{\cos(2\pi x/(3+x^2))-\cos(2\pi/(3+x^2)} +1,$$
which is slightly easier to take the derivative:
$$f'(x)=\frac{8 \pi x \cos(\frac{2 \pi x}{3 + x^2}) \sin(\frac{2 \pi}{3 + x^2}) - 4 \pi (x^2-3) \cos(\frac{2 \pi}{3 + x^2}) \sin(\frac{2 \pi x}{3 + x^2})}{(3 + x^2)^2 \left(\cos(\frac{2 \pi}{3 + x^2}) - \cos(\frac{2 \pi x}{3 + x^2})\right)^2}$$
Obviously the denominator is positive, so we need to show that for $x \geq 0$:
$$8 \pi x \cos\left(\frac{2 \pi x}{3 + x^2}\right) \sin\left(\frac{2 \pi}{3 + x^2}\right) - 4 \pi (x^2-3) \cos\left(\frac{2 \pi}{3 + x^2}\right) \sin\left(\frac{2 \pi x}{3 + x^2}\right) \leq 0.$$
This can be further simplified to
$$-2 \pi \left((x-1)(x+3) \sin\left(\frac{2 \pi(x-1)}{3 + x^2}\right) + (x+1)(x-3) \sin\left(\frac{2 \pi(x+1)}{3 + x^2}\right)\right)\leq 0\tag{1}$$
ant thus we need to show that for $x\geq 0$
$$(x-1)(x+3) \sin\left(\frac{2 \pi(x-1)}{3 + x^2}\right) \geq -(x+1)(x-3) \sin\left(\frac{2 \pi(x+1)}{3 + x^2}\right)$$
It seems we have to split it into intervals (probably two $x\in(0,1)$ and $x\in(1,\infty)$), and investigate the inequalities on the intervals, but I am stuck here at the moment.
UPDATE:
We see that (1) is an odd function, so it sufficient to investigate only one part, e.g. we need to show that for $x\geq 0$,
$$g(x)=(x-1)(x+3)\sin\left(\frac{2\pi(x-1)}{3+x^2}\right)\geq 0$$
We see that for $x\geq 0$, the equation has exacly 2 roots, $x=0$ and $x=1$ and therefore we can look at the sign in the subintervals. For example,plugging $x=0.5$, $g(0.5)=1.44022$ and $g(2)=3.909$. Thus the function $g(x)\geq 0$, and therefore $f(x)$ is non-increasing for $x\geq0$ (and by the oddness of (1), non-decreasing for $x\leq 0$).
If we look at the extrema $x=0$, and $x=1$, we see that $f(0)=\frac{1}{3}$ and $\lim_{x\to 1}f(x)=0$, therefore $f(0)=\frac{1}{3}$ is global maximum.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Complex Quadratic, what did I do wrong? $$z^2-2(1-2i)z-8i=0$$
Here is my working:
$$\begin{align}
x_{1,2} &= \frac{2-4i \pm \sqrt{(4i-2)^2 + 4\cdot 8i}}{2} \\
&=\frac{2-4i \pm \sqrt{-16+4-16i + 32i}}{2} \\
&= \frac{2-4i \pm \sqrt{-12+16i}}{2} \\
&= \frac{2-4i \pm \sqrt{4(-3+4i)}}{2} \\
&= \frac{2-4i \pm 2\sqrt{(4i-3)}}{2} \\
&= 1-2i \pm \sqrt{4i-3}
\end{align}$$
Let $a+bi = \sqrt{4i-3}$
$$a^2-b^2+abi=4i-3$$
$$a^2-b^2=-3, ab=4$$
$$b=\frac4a$$
$$a^2-\left(\frac{4}{a}\right)^2=-3$$
$$a^4+3a^2-16=0$$
$$t^2+3t-16=0$$
$$t_{1,2}=\frac{-3\pm\sqrt{9+4\cdot 16}}{2}$$
Answers should be $z=2, z=-4i,$ I don't know what I am doing wrong, I can get it right when substituting $z=a+bi$ but I don't when using this method.
|
The mistake is when you square $a+bi$, you should get $a^2-b^2+\color{red}2abi$ and not $a^2-b^2+abi$.
$$a^4+3a^2-4=0$$
$$(a^2+4)(a^2-1)=0$$
$$a=\pm 1, b = \pm 2$$
Hence,
$$x_{1,2}=1-2i\pm(1+2i)$$
which gives you the desired solution of $2$ and $-4i$.
|
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|
What is the expression $\sqrt{8\cdot32\cdot(-3)^2}$ equal to? What is the expression $\sqrt{8\cdot32\cdot(-3)^2}$ equal to?
Sorry for the basic question. I am a little confused when solving such problems. They are very easy, I know, but still... Which is the easiest algorithm and how would you solve it? We are supposed to use the rule that $$\sqrt{a^2b}=a\sqrt{b},a\ge0, b\ge0,$$ right? What would you do to solve the problem? I was thinking about $$8=4\times2\\32=8\times4\\(-3)^2=3^2$$ but are we supposed to continue this to $$8=4\times2=2\times2\times2=2^3\\32=8\times4=2^3\times2^2=2^5?$$ How can we still use $\sqrt{a^2b}=a\sqrt{b}$ then? Thank you in advance!
|
$\sqrt{8\cdot32\cdot(-3)^2} = \sqrt{8\cdot32\cdot9} = 3 \cdot \sqrt{8\cdot32} = 3 \cdot \sqrt{16\cdot16} = 3 \cdot 16 = 48$
|
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|
How should I finish this by completing to square? I am trying to calculate the integral:
$$\int_{-\infty}^{+\infty}\exp\left[-\frac{a^2(k-k_0)^2}{2} + ikx-i\frac{hk^2}{2m}t\right]dk$$
I thought on maybe mark $k-k_0=u$ but I am not see how it gonna help me
P.S the integral is only with respect to $k$. $k_0$ is constant also like the rest of paramteters
$i$ is the imagenery unit
|
$$
-\frac{a^2(k-k_0)^2}{2} + ikx-i\frac{hk^2}{2m}t
= -\frac12 a^2 k^2 + a^2 k_0 k - \frac12 a^2 k_0^2 + ikx - i\frac12 \frac1m ht k^2 \\
= -\frac12 (a^2+i\frac{ht}{m}) k^2 + (a^2 k_0 + ix) k - \frac12 a^2 k_0^2 \\
= -\frac12 (a^2+i\frac{ht}{m}) \left( k^2 - 2 \frac{a^2 k_0 + ix}{a^2 + i\frac{ht}{m}} k + \frac{a^2 k_0^2}{a^2+i\frac{ht}{m}} \right) \\
= -\frac12 (a^2+i\frac{ht}{m}) \left( \left[ k - \frac{a^2 k_0 + ix}{a^2 + i\frac{ht}{m}} \right]^2 + \left[ \frac{a^2 k_0^2}{a^2+i\frac{ht}{m}} - \left(\frac{a^2 k_0 + ix}{a^2 + i\frac{ht}{m}}\right)^2 \right] \right) \\
$$
|
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Prove : $\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\sqrt{\frac{9}{4}+\frac{3}{2}\frac{(a-b)^2(a+b+c)}{(a+b)(b+c)(c+a)}}$ It's an inequality based on two found on the website MSE (see the reference):
Let $a,b,c>0$ then we have:
$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\sqrt{\frac{9}{4}+\frac{3}{2}\frac{(a-b)^2(a+b+c)}{(a+b)(b+c)(c+a)}}$$
Lemma 1 :
$a,b,c>0$ then we have :
$$\sum_{cyc}\frac{a}{b+c}\geq P(a,b,c)=\sqrt{\frac{9}{4}+\frac{9}{4}\frac{(c^4+a^4+b^4-c^2a^2-b^2a^2-c^2b^2)}{((a+b+c)\frac{3}{4}+\frac{3}{4}(abc)^{\frac{1}{3}})(a+b)(b+c)(c+a)}+\frac{(c^2+a^2+b^2-ca-ba-cb)(a+b+c)}{(a+b)(b+c)(c+a)}}$$
Proof of lemma 1 :
First we remark that the inequality is homogenous and we can try the substitution $3u=a+b+c$, $3v^2=ab+bc+ca$ and $w^3=abc$ and apply the uvw's method .
We have :
$$a^4+b^4+c^4=(9u^2-6v^2)^2-2(9v^4-6uw^3)$$
$$a^2b^2+b^2c^2+c^2a^2=9v^4-6uw^3$$
$$a^2+b^2+c^2=9u^2-6v^2$$
And :
$\left(\frac{((3u)((3u)^2-4(3v^2))+5w^3)}{3u3v^2-w^3}+2\right)^2\geq \frac{9}{4}+\frac{(9/4)((9u^2-6v^2)^2-2(9v^4-6uw^3)-(9v^4-6uw^3))+(2.25u+0.75w)(9u^2-9v^2)(3u)}{(2.25u+0.75w)(3u3v^2-w^3)}$
it's enough to find an extreme value of our expression for the extreme value of $w^3$ wich happens for an equality case of two variables .
Since the last inequality is homogeneous, we can assume that $b=c=1$.
$$\frac{2}{a+1}+\frac{a}{2}\geq\sqrt{\frac{9}{4}+\frac{9}{4}\frac{(a^4+1-2a^2)}{((a+2)\frac{3}{4}+\frac{3}{4}(a)^{\frac{1}{3}})(a+1)^2(2)}+\frac{(a^2+1-2a)(a+2)}{(a+1)^2(2)}}$$
Now it seems to be clear : we get a polynomial with a root equal to one . See the factorization by Wolfram alpha .
End of the proof of the lemma 1
Remains to show that $ a\geq b \geq c>0$:
$$P(a,b,c)\geq\sqrt{\frac{9}{4}+\frac{3}{2}\frac{(a-b)^2(a+b+c)}{(a+b)(b+c)(c+a)}}$$
Wich is not hard I think .
Question :
How to show it ?
Reference :
M. A. Rozenberg, “uvw–Method in Proving Inequalities”, Math. Ed., 2011, no. 3-4(59-60), 6–14
If $x,y,z>0$, prove that: $\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge \sqrt{2}\sqrt{2-\frac{7xyz}{(x+y)(y+z)(x+z)}}$
Stronger than Nesbitt inequality
|
The given inequality is homogenious by $\;(a,b,c)\;$ and symmetric by $\;(a,b).\;$
Let WLOG
$$a+b+c = 6,\quad a+b=s,\quad ab=p,\tag1$$
then it suffices to prove the inequality in the form of
$$\dfrac a{6-a}+\dfrac b{6-b}+\dfrac{6-a-b}{a+b} \ge \sqrt{\dfrac94
+9\,\dfrac{(a-b)^2}{(a+b)(6-a)(6-b)}},\tag2$$
with the both positive sides, wherein the square root function monotonically increases in $\;[0,\infty).\;$
Therefore, one may to square both of the inequality sides and eliminate the positive denominators, with the represntations
\begin{align}
&\bigg(\big(a(6-b)+b(6-a)\big)(a+b)+(6-a-b)(6-a)(6-b)\bigg)^2\\[4pt]
&\ge \dfrac94\bigg(\big((6-a)(6-b)(a+b)\big)^2-4(a-b)^2(6-a)(6-b)(a+b)\bigg),
\tag3 \end{align}
or
$$\dfrac49\big(6s^2-2ps + 6(6-s)^2+p(6-s)\big)^2$$
$$\ge (36s-6s^2+ps)^2-4(s^2-4p)(36s-6s^2+ps),$$
or
$$f(s,p)\ge 0,\tag4$$
where
$$s^2\ge4p,\quad 6\ge s,\tag5$$
$$f(s,p)=\big(144-48s+8s^2+4p-2ps\big)^2 - (36s-8s^2+8p+ps)^2+4(s^2-4p)^2$$
$$=\big(144-84s+16s^2-4p-3ps\big)\big(144-12s+12p-ps\big)+4(s^2-4p)^2$$
$$\ge\dfrac34\big(192-112s+20s^2-s^3\big)(12-s)(12+p)$$
$$=\dfrac34(s-4)^2(12-s)^2(12+p)\ge0.$$
Proved!
|
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Show that the $p$-series $\sum_{n=1}^\infty \frac{1}{n^p}$ is convergent iff $p \gt 1$. Show that the $p$-series $\sum_{n=1}^\infty \frac{1}{n^p}$ is convergent iff $p \gt 1$.
Attempt:
For the right direction:
Let $\sum_{n=1}^\infty \frac{1}{n^p}$ is convergent. Then, $\lim\limits_{n\to \infty} \frac{1}{n^p} = 0$.
In particular, $\lim\limits_{n\to \infty} \frac{1}{n^p} = 0$ for $p \gt 0$. But, for $0 \lt p \le 1$, the series $\sum_{n=1}^\infty \frac{1}{n^p}$ is divergent. Hence, we must have $p \gt 1$. $\Box$
Proof that series $\sum_{n=1}^\infty \frac{1}{n^p}$ is divergent:
We know that $n^p \le n$ for all positive integers $n$ and $0 \lt p \le 1$.
Then,
\begin{equation*}
\frac{1}{n} \le \frac{1}{n^p}.
\end{equation*}
Now, since the partial sums of the harmonic series are not bounded, this
inequality shows that the partial sums of the $p-$series are not bounded when $0 \lt p \le 1$. Hence, the $p-$series diverges for theses values of $p. \Box$
For the left direction:
Let $p \gt 1$. Let $f(x) = \frac{1}{x^p}$ for all $x \in [1.\infty)$. Then, $f$ is a positive, continous, and decreasing sequence. Hence, we can apply the Integral Test here. Notice that
\begin{align*}
\int_1^\infty \frac{1}{x^p} &= \lim\limits_{b \to \infty} \int_1^b \frac{1}{x^p} \\
&= \lim\limits_{b\to \infty} \left[\frac{1}{1-p}x^{1-p}\right]_1^b \\
&= \lim\limits_{b \to \infty} \left( \frac{1}{1-p}b^{1-p} - \frac{1}{1-p} \right) \\
&= \frac{1}{p-1} \lim\limits_{b \to \infty} \left(1 - \frac{1}{b^{p-1}} \right) \\
&= \frac{1}{p-1}.
\end{align*}
Thus,
\begin{equation*}
\int_1^\infty \frac{1}{x^p}dx
\end{equation*}
is convergent. Consequently, $\int_1^\infty \frac{1}{n^p}dn$ is convergent. By the Integral Test,
we have that
\begin{equation*}
\sum_{n=1}^\infty \frac{1}{n^p}
\end{equation*}
is convergent. $\Box$
Am I correct, especially for the right direction ?
|
Here's something you may use:
Consider $p \leq 0$ first. In this case, $\frac{1}{n} = n^{-1} \geq 0$. The limit of this implies that it doesn't converge to $0$ as $n \rightarrow \infty$ and so $\sum_{n = 1}^{\infty} \frac{1}{n^p}$ does not converge.
Now assume $p > 0$. Then $\frac{1}{n^p}$ is a decreasing sequence of positive real numbers. By applying the Dyadic criterion, we have:
$\sum_{n = 1}^{\infty} \frac{1}{n^p}$ converges iff $\sum_{n = 0}^{\infty} 2^n \frac{1}{(2^n)^p}$ converges.
And so $\sum_{n = 0}^{\infty} (2^{1-p})^n$.
Let $q = 2^{1 - p}$ then the sum converges iff $q = 2^{1 - p} < 1$, which is iff $p > 1$.
|
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|
Solve one system of equations Solve the system of equations
$$ ax + by + cz = 0, $$
$$ bcx + cay + abz = 0, $$
$$ xyz + abc (a^3x + b^3y + c^3z) = 0 $$
I tried solving this using cross multiplication method but got stuck at one point :
$$x/ab^2-ac^2 = y/bc^2-ba^2 = z/ca^2-cb^2 = k (say) $$
I substituted the values in the third equation, and after simplifying ended up being here :
$$ k(abc)[a^2c^2(b^2 + a^2c^2 - b^2c^2 - a^2)] $$
|
Here is a systematic method that brings back your issue to solving a linear system.
I assume $abc \ne 0$. Setting
$$X=x/a, Y=y/b, Z=z/c \ \ \ \text{and} \ \ \ A=a^2,B=b^2,C=c^2,$$
then replacing $x,y,z$ by $aX,bY,cZ$ resp., and inverting equ. 1, with equ. 2, your system can be written under the form:
$$\begin{pmatrix}1&1&1\\A&B&C\\A^2&B^2&C^2\end{pmatrix}\begin{pmatrix}X\\Y\\Z\end{pmatrix}=\begin{pmatrix}0\\0\\-XYZ\end{pmatrix}$$
Otherwise said:
$$\begin{pmatrix}1&1&1\\A&B&C\\A^2&B^2&C^2\end{pmatrix}\begin{pmatrix}1/(YZ)\\1/(XZ)\\1/(XY)\end{pmatrix}=\begin{pmatrix}0\\0\\-1\end{pmatrix}$$
It remains to solve a simple linear system (with a so-called "Vandermonde matrix", always invertible, because its determinant is $(B-A)(C-A)(C-B) \ne 0$) then obtain $X,Y,Z$ out of the result.
|
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Tennis match probability - is my logic incorrect? A tennis match consists of sets, until one player wins three sets. Player $A$ has a two-thirds chance of winning. Player $B$ has a one-third chance of winning. If they play a match, what is the probability of $A$ winning?
My attempt:
The minimum games for $A$ victory is $3$, and the maximum is $5$ (both win $2$, then next round is decider)
I have probability = win in three + win in four + win in five
=$\left(\frac{2}{3}\right)^{3}+ {4\choose 1}\ \left(\frac{1}{3}\right)\left(\frac{2}{3}\right)^{3}+{5\choose2}\left(\frac{1}{3}\right)^{2}\left(\frac{2}{3}\right)^{3}$
Which is more than $1$!
I don't understand why this is wrong though
|
Your working: $\displaystyle \small \left(\frac{2}{3}\right)^{3}+ {4\choose 1}\ \cdot \frac{1}{3} \cdot \left(\frac{2}{3}\right)^{3}+{5\choose2}\left(\frac{1}{3}\right)^{2}\left(\frac{2}{3}\right)^{3}$
Please see the second term. You are allowing any $1$ of the $4$ sets to be won by player B. But if there are $4$ sets, player A cannot win first three sets. Same logic for the next term.
So it should be,
$\displaystyle \small \left(\frac{2}{3}\right)^{3}+ {3\choose 1}\ \cdot \frac{1}{3} \cdot \left(\frac{2}{3}\right)^{3}+{4\choose2}\left(\frac{1}{3}\right)^{2}\left(\frac{2}{3}\right)^{3}$
|
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|
Prove that $\frac{10^n-1}{9n}$ is integer when $n=3^k$ I have no idea how to go about proving this. The furthest I've gotten is to say that the sequence equals 1/1, 11/2, 111/3, etc.
So that means that $\frac{10^a-1}{9}$ must be divisible by 3.
$\frac{10^a-1}{9} = 3b \implies 10^a-1 = 27b \implies 27b+1 = 10^a$.
When is that last statement true?
|
The statement is equivalent to proving $$10^{3^k}-1\equiv_{3^{k+2}}0$$
Notice that $\phi(3^{k+2})=3^{k+2}-3^{k+1}=2\cdot 3^{k+1}$, so that for a unit $a$ we have $$a^{2\cdot 3^{k+1}}\equiv_{3^{k+2}}1$$ We also have
Claim: $$10^{3^k}\equiv_{3^{k+2}}10^{3^{k+1}}$$
Proof: $$10^{3^k}\equiv_{3^{k+2}}10^{3^{k+1}-2\cdot 3^{k}}\equiv_{3^{k+2}}10^{3^{k+1}}$$
Let $t=10^{3^k}$, then the above claim may be written as $$t^3\equiv_{3^{k+2}}t\iff t(t+1)(t-1)\equiv_{3^{k+2}}0$$ Both $t,t+1$ are units since $t,t+1\not\equiv_{3}0$, therefore we must have $t-1\equiv_{3^{k+2}}0$.
|
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|
Limit of a continued product $\prod_{r=3}^n \frac{(r^3+3r)^2}{r^6-64} $ as $n \to \infty$ Evaluate $\lim_{n\to \infty }\prod_{r=3}^n \frac{(r^3+3r)^2}{r^6-64} $
I don't have any ideas on how to approach the problem. Hints or solutions are appreciated.
|
If you factor the general term, you get:
$$\frac{r^2 \left(r^2+3\right)^2}{(r-2) (r+2)
\left(r^2-2 r+4\right) \left(r^2+2 r+4\right)}$$
Notice that $r^2-2r + 4 = (r-1)^2 + 3,\ r^2+2r + 4 = (r+1)^2+3,$
So the product telescopes: and almost everything cancels. (the final answer is $\frac{72}7.)$
|
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Are there random variables $X,Y$ such that $X,Y,X+Y \sim N(0,1)$? Let's suppose we have two, not necessarily independent, random variables $X,Y \sim N(0,1)$.
Then, is it possible that $X+Y \sim N(0,1)$?
What I have managed to show is that because $Var(X+Y) = Var(X) + Var(Y) + 2Cov(X,Y)$ it would be necessary that $Cov(X,Y) = -1/2$.
But I'm still looking for an example or else a proof that it is not possible.
|
The pdf of the sum of two absolutely continuous random variables is simply given by the convolution of the pdf's of the variables. In our case, $f_X(x) = f_Y(x) = \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}$, so $f_{X + Y}(x) = \int_{- \infty}^\infty f_X(t)f_Y(x - t) dt = \frac{1}{4} \int_{- \infty}^\infty \exp(- \frac{t^2 + (x - t)^2}{2}) dt = \frac{1}{2 \pi} \int_{- \infty}^\infty \exp(- \frac{(t \sqrt{2})^2 - 2 t \sqrt{2} \frac{x}{\sqrt{2}} + \frac{x^2}{2} + \frac{ x^2}{2}}{2}) dt = \frac{1}{2 \pi} e^{-\frac{ x^2}{4}} \int_{- \infty}^\infty \exp(- \frac{(t \sqrt{2} - \frac{x}{\sqrt{2}})^2}{2}) dt = \frac{1}{2 \sqrt{2} \pi} e^{- \frac{x^2}{4}} \int_{- \infty}^\infty e^{- \frac{u^2}{2}} du = \frac{1}{2 \sqrt{2} \pi} e^{- \frac{x^2}{4}} \sqrt{2 \pi} = \frac{1}{2 \sqrt{\pi}} e^{- \frac{x^2}{4}}.$ In particular, the sum of two normals of mean $0$ and variance $1$ is not a normal of mean $0$ and variance $1$. I hope this helps. :)
|
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How many five-digit numbers can be formed using digits $0, 1, 2, 6, 7, 8,9$ which are divisible by $3$ and $5$? How many five-digit numbers can be formed using digits $0, 1, 2, 6, 7, 8$ and $9$ which are divisible by $3$ and $5$, without any of the digits repeating?
For the number to be divisible by $5$ it must end with $0$ (or $5$, but we don't have it in the given problem). So we have $V_6^4$ numbers which are divisible by $5$ but how can we exclude those which aren't divisible by $3$?
$120$
|
$0$ is last digit. Note that there are only 2 numbers from the set whose mod 3 is 1.
Case 1: Pick $0$ $x$ s.t. $x \mod 3 = 1$: Not possible
Case 2: Pick exactly 1 digit x s.t. $x \mod 3 = 1$. We must pick exactly 1 digit x s.t. $x \mod 3 = 2$. Remaining are filled with $x \mod 3 = 0$.
Sum = $2 * 2 * (4!) = 96$
case 3: Pick 2 digits x s.t. $x \mod 3 = 1$: We must pick exactly 2 digits x s.t. $x \mod 3 = 2$.
Sum = $4! = 24$
Ans = $96 + 24 = 120$
|
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A biquadratic equation to have at least two real roots Which of the equations have at least two real roots?
\begin{aligned}
x^4-5x^2-36 & = 0 & (1) \\
x^4-13x^2+36 & = 0 & (2) \\
4x^4-10x^2+25 & = 0 & (3)
\end{aligned}
I wasn't able to notice something clever, so I solved each of the equations. The first one has $2$ real roots, the second one $4$ real roots and the last one does not have real roots. I am pretty sure that the idea behind the problem wasn't solving each of the equations. What can we note to help us solve it faster?
|
$$ x^4 -13 x^2 + 36 = (x^2 + 6)^2 - 25 x^2 = (x^2 + 5x+6) (x^2 - 5x+6) $$
and both quadratic factors have real roots(positive discriminants).
$$ 4 x^4 -10 x^2 + 25 = (2x^2 + 5)^2 - 30 x^2 = (2x^2 + x \sqrt{30}+5) (2x^2 - x \sqrt{30}+5) $$
and both quadratic factors have negative discriminants. OR
$$ 4 x^4 -10 x^2 + 25 = (2x^2 - 5)^2 +10 x^2 $$
is always positive
|
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Find $\mathbb{E}(X)$ and Var$(X)$ from the c.d.f. Suppose a child plays outside in the yard. On their own, they come back inside at a random time uniformly distributed on the interval [0,1] (Take the units to be hours.) However, if the child is not back in 50 minutes, their mother brings them in. Let X be the time when they come back in.
*
*What is the cumulative distribution function F of X?
*Find E[X].
*Find Var[X].
My Attempt
*
*The c.d.f. is pretty straight forward from the problem.
\begin{equation}
F(s)=
\begin{cases}
0, & s<0\\
s, & 0\leq s<5/6\\
1, & s\geq 5/6
\end{cases}
\end{equation}
*Differentiating the c.d.f. gives us the p.d.f. $f(s)=1$ for $0\leq s <5/6$.
Thus, our expectation is
$$\mathbb{E}(X)=\int^{5/6}_0xdx=0.3472$$
*To find the variance, we also need $\mathbb{E}(X^2)$
$$\mathbb{E}[X^2]=\int^{5/6}_0x^2dx=0.1929$$
So the variance is
$$\text{Var}(X)=\mathbb{E}[X^2]-\mathbb{E}[X]^2=0.1929-0.3472^2=0.0723$$
Can I get verification on my answers? I am a little skeptical on the expectation because I was told that the expectation for a uniform distribution, Unif[a, b], can be calculated as $\mathbb{E}(X)=\frac{a+b}{2}=\frac{5/6+0}{2}=0.4167$, which is different from what I got using the integral method.
|
Another way to solve this is by ysing the heaviside step function $H(x)$ and the dirac delta function $\delta(x)$:
\begin{align*}
F(s)=
\begin{cases}
0, & s<0\\
s, & 0\leq s<5/6\\
\frac{5}{6} + \frac{1}{6}\cdot H(x-\frac{5}{6}), & s\geq 5/6
\end{cases}
\end{align*}
We can then write the pdf in terms of the dirac delta function, which is\begin{align*}f(s)=
\begin{cases}
0, & s<0\\
1, & 0< s<5/6\\
\frac{1}{6}\delta(x-\frac{5}{6}), & s\geq 5/6\\
\end{cases}\end{align*}
The expectation is better computed as an integral over
\begin{align*}
\mathbb E [X] &= \int\limits_{x=-\infty}^{0} x\cdot \mathbb P[X=x] + \int\limits_{x=0}^{\frac{5}{6}} x\cdot \mathbb P[X=x]+ \int\limits_{x=\frac{5}{6}}^{\infty} x\cdot \mathbb P[X=x]\\
&= \int\limits_{x=-\infty}^{0} x\cdot f[X=x]dx + \int\limits_{x=0}^{\frac{5}{6}} x\cdot f[X=x]dx+ [X=\frac{5}{6}]\cdot\frac{1}{6} \int\limits_{x=\frac{5}{6}}^{\infty}\delta(x-\frac{5}{6})\\
&= \int\limits_{x=-\infty}^{0} x\cdot 0dx + \int\limits_{x=0}^{\frac{5}{6}} x\cdot 1 dx+ \frac{5}{6}\cdot \frac{1}{6}\int\limits_{x=-\infty}^{\infty}\delta(x-\frac{5}{6})\\
&= 0 + \frac{25}{36\cdot 2}+ \frac{5}{6}\cdot \frac{1}{6}\cdot 1\\
&= \frac{35}{72}
\end{align*}
Similarly, one can solve $\mathbb E [X^2]$
\begin{align*}
\mathbb E [X^2] &= \int\limits_{x=-\infty}^{0} x^2\cdot \mathbb P[X=x] + \int\limits_{x=0}^{\frac{5}{6}} x^2\cdot \mathbb P[X=x]+ \int\limits_{x=\frac{5}{6}}^{\infty} x^2\cdot \mathbb P[X=x]\\
&= 0 + \int\limits_{x=0}^{\frac{5}{6}} x^2\cdot f[X=x]dx+ [X=\frac{5}{6}]^2\cdot\frac{1}{6} \int\limits_{x=\frac{5}{6}}^{\infty}\delta(x-\frac{5}{6})\\
&= \int\limits_{x=0}^{\frac{5}{6}} x^2dx+ \frac{5^2}{6^2}\cdot \frac{1}{6}\int\limits_{x=-\infty}^{\infty}\delta(x-\frac{5}{6})\\
&= 0 + \frac{125}{216\cdot 3}+ \frac{25}{36}\cdot \frac{1}{6}\cdot 1\\
&= \frac{200}{648}=\frac{25}{81}
\end{align*}
|
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Help understanding the solution of $m^2 = n^3 -4$. The material I'm studying has the solution for $m^2 = n^3 + 4$ but I can't quite understand it. It starts like this:
First suppose, for a contradiction, that $m$ is odd. Then $n^3 = m^2 − 4 = (m + 2)(m − 2)$. Any common factors of $m + 2$ and $m − 2$ divide $(m + 2) − (m − 2) = 4$ and so, $\gcd(m + 2, m − 2) = $1$.
It follows that $m + 2$ and $m − 2$ are both cubes.
a) How did they get to the conclusion $\gcd(m+2, m-2)=1$? I mean, at this stage of the proof, I don't know.
$m\neq 10$ so I could think if $m=10$ then gcd(12,8) $=4\neq 1$.
b) Why does it follow that $m+2$ and $m-2$ are both cubes, and not that their multiplication is a cube?
Let's carry on the proof.
[...] No two odd cubes differ by 4, giving a contradiction.
Since $m$ is even there exists $m'$ such that $m = 2m'$. We thus have that $4m'^2 = n^3 + 4$ and $n$ is divisible by $2$ so we write $n = 2n'$. Now, $m'^2 = 2n'^3 + 1$ and so $m'$ is odd so let $m' = 2r + 1$. Substituting this into the previous equation tells us that $2r^2 + 2r = n'^3$.
No problems with the above, let's carry on
Finally,$r^2 + r = r (r + 1)$ and since $r$ and $r + 1$ are coprime we must have that one of $r$ and $r + 1$ is a cube and the other is four times a cube.
c) Why are we considering $r^2+r$ when we have $2r^2+2r=n'^3$? Yes, they have a common factor of $2$, but we don't know this about $n'$ do we? we can't cancel them.
d) ``since $r$ and $r + 1$ are coprime we must have that one of $r$ and $r + 1$ is a cube and the other is four times a cube'' ????
Any help with the above is really appreciated.
|
a) You do know $m \neq 10$ because you have assumed $m$ odd. From "any common factors of $m+2$ and $m-2$ divide $(m+2)-(m-2)=4$," we can conclude that the only possible common factors of $m+2$ and $m-2$ are $1$, $2$, or $4$. But if $m$ is odd, then so are $m+2$ and $m-2$. So $2$ and $4$ cannot be divisors. This leaves $1$ as the greatest common divisor.
b) If $a$ and $b$ are not themselves cubes, yet $ab=c^3$ is a cube, then $a$ and $b$ must share a common factor greater than $1$. An example (not a proof) would be $3\cdot 9 = 27=2^3$, and $3$ and $9$ share a common factor of $3$. To see why this holds, consider the prime factorization of $c$ and note that each prime must appear with an exponent that is a multiple of $3$. If $a$ and $b$ are not cubes, then they each have some prime divisor that appears with an exponent not a multiple of $3$. Then you can show $a$ and $b$ must share a prime divisor.
c) Nothing has been canceled, just factoring $n'^3$ as $2r(r+1)$.
d) See the answer to b). The reasoning is similar. We know that $2$ divides $n'^3$, so $8$ must divide $n'^3$. Only one of $r$ and $r+1$ is even, so $4$ must divide one of them. So if we divide $2r(r+1)=n'^3$ by $8$, we get $\frac{r(r+1)}{4}=n''^3$ where $n'=2n''$. This is a product of two coprime integers (either $\frac{r}{4}$ and $r+1$ or $r$ and $\frac{r+1}{4}$) equal to a cube, hence both factors are cubes. Thus one of $r$ and $r+1$ is a cube and the other is $4$ times a cube.
|
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|
How can one simplify $\sum_{n=1}^8\frac{\sin10n^\circ}{\cos5^\circ\cos10^\circ\cos20^\circ}$? I have trouble solving the following question:
Consider the following expression:
$$\sum_{n=1}^8\frac{\sin10n^\circ}{\cos5^\circ\cos10^\circ\cos20^\circ}$$
The value of the above expression can be fully simplified into the form $a\sqrt b$. What is $a+b$?
I know that $$\sin\alpha+\sin\beta=2\sin\left(\frac{\alpha+\beta}2\right)\cos\left(\frac{\alpha-\beta}2\right)$$
but I'm not too sure what pairs of $\sin$ I should apply it on, or if this is even the right approach to this problem. Could someone give me hints for solving this problem(Am I on the right track)?
|
$\textbf{Hint:}$
Use the formula that, $\sum_{r=0}^{n-1}\sin(a+rh) = \frac{\sin(\frac{nh}{2})\sin(a+(n-1)\frac{h}{2})}{\sin(\frac{h}{2})}$
so, Let $$P = \sum_{n=1}^{8}\frac{\sin(10n^\circ)}{\cos(5^\circ)\cos(10^\circ)\cos(20^\circ)}$$
$$=\frac{1}{\cos(5^\circ)\cos(10^\circ)\cos(20^\circ)}\sum_{n=1}^{8}\sin(10n^\circ)$$
$$=\frac{1}{\cos(5^\circ)\cos(10^\circ)\cos(20^\circ)}\cdot\frac{\sin(\frac{8\cdot10}{2})\sin(10+7\cdot\frac{10}{2})}{\sin(\frac{10}{2})}$$
$$=\frac{\sin(40^\circ)\sin(45^\circ)}{\cos(5^\circ)\cos(10^\circ)\cos(20^\circ)\sin(5^\circ)}$$
$$=2\cdot\frac{\sin(40^\circ)\sin(45^\circ)}{2\cos(5^\circ)\sin(5^\circ)\cos(10^\circ)\cos(20^\circ)}$$
$$=4\cdot\frac{\sin(40^\circ)\sin(45^\circ)}{2\sin(10^\circ)\cos(10^\circ)\cos(20^\circ)}$$
$$=8\cdot\frac{\sin(40^\circ)\sin(45^\circ)}{2\sin(20^\circ)\cos(20^\circ)}$$
$$=8\frac{\sin(40^\circ)\sin(45^\circ)}{\sin(40^\circ)}$$
$$P=8\sin(45^\circ)$$
So, $P=4\cdot\sqrt{2}$,
therefore $a = 4$ & $b=2$, so $a+b=6$
I hope this was helpful, :)
|
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|
If $x^{x^{x+1}}=\sqrt{2}$, then evaluate $x^{x^{p}}$, where $p = 2x^{x+1}+x+1$ I can't figure out how to give a proper form to this expression to use the root of two.
If
$$x^{x^{x+1}}=\sqrt{2}$$
find the value of $W$ if
$$W=x^{x^{p}} \quad\text{where}\; p = 2x^{x+1}+x+1$$
EDIT: This is an algebraic manipulation problem with the exponents. There was an error in the previous version (see the Edit History) that I have corrected.
|
Assuming $x^{x^{x+1}} = \sqrt 2$,
\begin{align*}
x^{x^{2x^{x+1} + x + 1}} &= x^{x^{2x^{x+1}} \cdot x^{x+1}} \\
&= x^{\left(x^{x^{x+1}}\right)^{2} \cdot x^{x+1}} \\
&= x^{\left(\sqrt{2}\right)^{2} \cdot x^{x+1}} \\
&= x^{2 \cdot x^{x+1}} \\
&= \left(x^{x^{x+1}}\right)^{2} \\
&= \left(\sqrt{2}\right)^{2} \\
&= 2
\end{align*}
|
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|
Power of little o - asymptotic - series Consider the following, when $n \to + \infty$:
$$\log\left(1 + \frac{-1}{q}n^{\frac{1-q}{q}} +o(n^{-1}) \right)= R_1(n^{\frac{-1}{q}}) + o(n^{-1}).$$
And I am trying to determine the polynomial $R_1$. From a lemma, I know that this polynomial satisfies $R_1(0)=0$ and is of degree at most $q$, where $q$ is an integer $\geq 3$.
My problem is with the little-o notations, I don't see how to work with them.
I used the Taylor expansion of $\log(1+x)$ and have the following expression for the left-hand side :
$$\sum_{j=1}^{\infty} (-1)^{j+1} \frac{1}{j} \left(\frac{-1}{q} n^{\frac{1-q}{q}} + o(n^{-1})\right)^j $$
Can anybody help ? Thanks a lot inadvance !
|
If you expand the logarithm using $\log(1+w)=w+\mathcal{O}(w^2)$, you find
\begin{align*}
& - \frac{1}{q}\frac{1}{{n^{1 - 1/q} }} + o\!\left( {\frac{1}{n}} \right) + \mathcal{O}(1)\left( { - \frac{1}{q}\frac{1}{{n^{1 - 1/q} }} + o\!\left( {\frac{1}{n}} \right)} \right)^2
\\ &
= - \frac{1}{q}\frac{1}{{n^{1 - 1/q} }} + o\!\left( {\frac{1}{n}} \right) + \mathcal{O}(1)\left( {\frac{1}{{q^2 }}\frac{1}{{n^{2 - 2/q} }} + o\!\left( {\frac{1}{{n^{2 - 1/q} }}} \right) + o\!\left( {\frac{1}{{n^2 }}} \right)} \right)
\\ &
= - \frac{1}{q}\frac{1}{{n^{1 - 1/q} }} + o\!\left( {\frac{1}{n}} \right) + \mathcal{O}(1)o\!\left( {\frac{1}{n}} \right) = - \frac{1}{q}\frac{1}{{n^{1 - 1/q} }} + o\!\left( {\frac{1}{n}} \right).
\end{align*}
since, as $q\geq 3$,
$$
\frac{1}{{n^{2 - 2/q} }} \le \frac{1}{{n^{2 - 2/3} }} = \frac{1}{{n^{1 + 1/3} }} = o\!\left( {\frac{1}{n}} \right)
$$
and
$$
\frac{1}{{n^{2 - 1/q} }} \le \frac{1}{{n^{2 - 1/3} }} = \frac{1}{{n^{1 + 2/3} }} = o\!\left( {\frac{1}{n}} \right).
$$
|
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|
Calculate integral $\int\limits_{0}^{2\pi}\frac{dx}{\left ( 1+n^2\sin^2 x \right )^2}$ I recently saw the integral problem
$$\int\limits_{0}^{2\pi}\frac{dx}{\left ( 1+n^2\sin^2 x \right )^2}$$
and tried to solve it. Below is what I did.
Interesting to look at other easier solutions.
$$\int\limits_{0}^{2\pi}\frac{dx}{\left ( 1+n^2\sin^2 x \right )^2}=4\int_{0}^{\pi /2}\frac{dx}{\left ( 1+n^2\sin^2 x \right )^2}\\\overset{t=\operatorname{tg} x}{=}\int\limits_{0}^{\infty }\frac{1+t^2}{\left ( 1+\left ( 1+n^2 \right )t^2 \right )^2}dt\\
\overset{t=\frac{y}{\sqrt{1+n^2}}}{=}\frac{4}{\left ( 1+n^2 \right )\sqrt{1+n^2}}\int\limits_{0}^{\infty }\frac{1+n^2+y^2}{\left ( 1+y^2 \right )^2}dy\\ \overset{y=\operatorname{tg} \theta }{=}\frac{4}{\left ( 1+n^2 \right )\sqrt{1+n^2}}\int_{0}^{\pi /2}\left ( 1+n^2\cos^2 \theta \right )d\theta \\
=\frac{\pi \left ( 2+n^2 \right )}{\left ( 1+n^2 \right )\sqrt{1+n^2}}$$
|
Just for the fun !
When there is a square in denominator, we never know ! Trying
$$\int\frac{dx}{\left ( 1+n^2\sin^2 (x) \right )^2}=\frac{P(x)}{ 1+n^2\sin^2 (x)} $$ Differentiate both sides and remove the denominators
$$(1+n^2 \sin ^2(x))P'(x)-n^2 \sin(2x) P(x)=1$$ which is not very difficult to integrate. So, by the end
$$\frac{P(x)}{ 1+n^2\sin^2 (x)}=\frac{n^2+2 }{2
\left(n^2+1\right)^{3/2}}\tan ^{-1}\left(\sqrt{n^2+1} \tan (x)\right)-\frac{n^2 \sin (2 x)}{2 \left(n^2+1\right) \left(n^2 \cos (2 x)-n^2-2\right)} + C $$ Integrating between $0$ and $\frac \pi 2$ and multiplying by $4$, the result
$$\int\limits_{0}^{2\pi}\frac{dx}{\left ( 1+n^2\sin^2 (x) \right )^2}=\frac{ \left(n^2+2\right)}{\left(n^2+1\right)^{3/2}}\, \pi$$
|
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|
Misprint(s) In "Combinatorial Identities" by Riordan? On page 38 in the exercises I was able to confirm that
$$a_{kj}(x)=(x+1)\dots(x+k-1)a_{k-1,j}(x)+x^{k-1}a_{k-1,j-1}(x+1).$$
For $k\ge{}j$ the author defines $b_{kj}(x)$ via
$$a_{kj}(x)=(x+j+1)^{k-j-1}(x+j+2)^{k-j-2}\dots(x+k-1)b_{kj}(x),a_{kk}(x)=b_{kk}(x),$$
where $a_{kk}(x)=x^{k-1}(x+1)^{k-2}\cdots(x+k-2)$.
In both its 1968 and 1979 edition on pages 38 and 39, I am confident that the author intended
$$b_{kj}(x)=(x+1)\dots(x+j)b_{k-1,j}(x)+x^{k-1}b_{k-1,j-1}(x+1)$$
and not
$$b_{kj}(x)=(x+1)\dots(x+j)b_{k-j,j}+x^{k-1}b_{k-1,j-1}(x+1).$$
The absence of the argument $x$ in the second line makes me suspect two typographical errors.
Furthermore, I am confused by the sum that defines $b_j(x,y)$. A few lines prior to its definition it is stated that $k\ge{j}$ and then its stated that
$$b_j(x,y)=\sum_{k=j}b_{kj}(x)y^{k-j}$$.
The author concludes $b_1(x,y)=[1-(x+1)y]^{-1}(1-xy)^{-1}$ which I suspect arises from
$$[1-(x+1)y](1-xy)b_1(x,y)=1.$$
I am hopeful that somebody has deciphered the possible errors or has some clues how to proceed. Thank you!
Sincerely,
John Majewicz, Ph.D.
|
First question: Your assumption is correct. There are two typos as indicated by you.
We have the recurrence relation
\begin{align*}
a_{k,0}(x)&=(x+1)\cdots(x+k-1)a_{k-1,0}(x)\\
&=(x+1)^{k-1}(x+1)^{k-2}\cdots(x+k-1)\\
a_{k,j}(x)&=(x+1)\cdots(x+k-1)a_{k-1,j}(x)\\
&\qquad+x^{k-1}a_{k-1,j-1}(x+1)\qquad\qquad\qquad\qquad 0<j<k\tag{1}\\
a_{k,k}(x)&=x^{k-1}a_{k-1,k-1}(x+1)\\
&=x^{k-1}(x+1)^{k-2}\cdots(x+k-2)\\
a_{1,1}(x)&=a_{0,0}(x)=1
\end{align*}
According to the relation
\begin{align*}
\color{blue}{a_{k,j}(x)}&\color{blue}{=(x+j+1)^{k-j-1}(x+j+2)^{k-j-2}}\\
&\color{blue}{\qquad\cdot\ldots\cdot(x+k-1)b_{k,j}(x)\qquad\qquad\qquad k\geq j}\tag{2}\\
\color{blue}{a_{k,k}(x)}&\color{blue}{=b_{k,k}(x)}
\end{align*}
we derive by putting (2) into (1)
\begin{align*}
&(x+j+1)^{k-j-1}(x+j+2)^{k-j-2}\cdots(x+k-1)b_{k,j}(x)\\
&\qquad=(x+1)\cdots(x+k-1)(x+j+1)^{k-j-2}(x+j+2)^{k-j-3}\\
&\qquad\qquad\cdot\ldots\cdot(x+k-2)b_{k-1,j}(x)\\
&\qquad\qquad+x^{k-1}(x+j+1)^{k-j-1}(x+j+2)^{k-j-2}\cdots(x+k-1)b_{k-1,j-1}(x+1)\\
&\qquad=(x+1)\cdots(x+j)(x+j+1)^{k-j-1}(x+j+2)^{k-j-2}\cdots(x+k-1)b_{k-1,j}(x)\\
&\qquad\qquad+x^{k-1}(x+j+1)^{k-j-1}(x+j+2)^{k-j-2}\cdots(x+k-1)b_{k-1,j-1}(x)\\
\end{align*}
from which after division by $(x+j+1)^{k-j-1}(x+j+2)^{k-j-2}\cdots(x+k-1)$
\begin{align*}
b_{k,j}(x)&=(x+1)\cdots(x+j)\color{blue}{b_{k-1,j}(x)}\\
&\qquad+x^{k-1}b_{k-1,j-1}(x+1)\tag{3}
\end{align*}
follows.
Second question:
From
\begin{align*}
b_j(x,y)\sum_{k\geq j}b_{k,j}(x)y^{k-j}
\end{align*}
we want to derive $b_1(x,y)$. Since
\begin{align*}
b_1(x,y)=\sum_{k\geq 1}b_{k,1}(x)y^{k-1}\tag{4}
\end{align*}
we start with calculating $b_{k,1}(x)$. We obtain from (3) with $b_{1,1}(x)=1$
\begin{align*}
\color{blue}{b_{k,1}(x)}&=(x+1)b_{k-1,1}(x)+x^{k-1}b_{k-1,0}(x+1)\\
&=(x+1)^2b_{k-2,1}(x)+(x+1)x^{k-2}+x^{k-1}\\
&=(x+1)^3b_{k-3,1}(x)+(x+1)^2x^{k-3}+\cdots+x^{k-1}\\
&\vdots\\
&=(x+1)^{k-1}+(x+1)^{k-2}x+\cdots+x^{k-1}\\
&=\sum_{j=0}^{k-1}(x+1)^{k-1-j}x^j\\
&=(x+1)^{k-1}\sum_{j=0}^{k-1}\left(\frac{x}{x+1}\right)^j\\
&=(x+1)^{k-1}\cdot\frac{1-\left(\frac{x}{x+1}\right)^k}{1-\frac{x}{x+1}}\\
&=(x+1)^k\left(1-\left(\frac{x}{x+1}\right)^k\right)\\
&\,\,\color{blue}{=(x+1)^k-x^k}\tag{5}
\end{align*}
Putting (5) into (4) we obtain
\begin{align*}
\color{blue}{b_1(x,y)}&=\sum_{k\geq 1}b_{k,1}(x)y^{k-1}\\
&=\sum_{k\geq 1}(x+1)^ky^{k-1}-\sum_{k\geq 1}x^ky^{k-1}\\
&=(x+1)\sum_{k\geq 0}\left((x+1)y\right)^k-x\sum_{k\geq 0}(xy)^k\\
&=\frac{x+1}{1-(x+1)y}-\frac{x}{1-xy}\\
&\,\,\color{blue}{=\frac{1}{(1-(x+1)y)(1-xy)}}
\end{align*}
according to the claim.
|
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Matrix Multiplication - Express a row as a linear combination $$ Let \ A \ = \begin{bmatrix} 1 & 2 \\ 4 & 5 \\ 3 & 6 \\ \end{bmatrix} and
\ let \ B = \begin{bmatrix} 0 & 1 & -3 \\ -3 & 1 & 4 \end{bmatrix} $$
Express the third row of AB as a linear combination of the rows of B
$$ AB \ = \ \begin{bmatrix} -6 & 3 & 5 \\ -15 & 9 & 8 \\ -18 & 9 & 15 \end{bmatrix} $$
3rd row of AB would be
$$ \begin{bmatrix} 0 \ (-18) & 1 \ (9) & -3 \ (15) \\ -3 \ (-18) & 1 \ (9) & 4 \ (15) \end{bmatrix} $$
ANS: So the third row represented as a linear combination of the rows of B is given by:
$$ -18 \ \begin{bmatrix} \ \ \ 0 \\ -3 \end{bmatrix} \ \
+ \ \ 9 \ \begin{bmatrix} \ \ 1 \ \\ \ \ 1 \ \end{bmatrix} \ \
+ \ \ 15 \ \begin{bmatrix} -3 \\ \ \ 4 \end{bmatrix} \ \ $$
Is my answer correct?
If not any suggestion or help would be appreciated.
Thanks for your time and cooperation from now.
|
\begin{align} \textrm{third row of } AB
&= \begin{bmatrix}
\begin{bmatrix} 3 & 6 \end{bmatrix} \begin{bmatrix} 0 \\ -3 \end{bmatrix} &
\begin{bmatrix} 3 & 6 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} &
\begin{bmatrix} 3 & 6 \end{bmatrix} \begin{bmatrix} -3 \\ 4 \end{bmatrix}
\end{bmatrix} \\
&= \begin{bmatrix}
3 \cdot \color{red}0 + 6 \cdot \color{blue}{(-3)} &
3 \cdot \color{red}1 + 6 \cdot \color{blue}1 &
3 \cdot \color{red}{(-3)} + 6 \cdot \color{blue}4
\end{bmatrix} \\
&= 3 \begin{bmatrix} \color{red}0 & \color{red}1 & \color{red}{-3} \end{bmatrix} + 6 \begin{bmatrix} \color{blue}{-3} & \color{blue}1 & \color{blue}4 \end{bmatrix} \\
&= 3(\textrm{first row of } B) + 6(\textrm{second row of } B).
\end{align}
|
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Taylor Expansion of the first four non-zero terms $$f(x) = \sqrt{1+x}\sin(4x)$$
Okay, so at this point, I am able to find the answer to these kinds of problems when the outside term is x to the power of a whole number, but this is two terms added together then to the 1/2 power.
The sin expansion looks like this:
$$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$$
and the sin expansion of sin(4x), looks like this: (I'm pretty sure)
$$\sin(4x) = 4x - \frac{4x^3}{3!} + \frac{4x^5}{5!} - \frac{4x^7}{7!} + \cdots$$
but, what does this look like:
$$\sqrt{1+x}\sin(4x) = \cdots$$
|
Near $0$, you have$$\sin(4x)=4x-\frac{4^3}{3!}x^3+\cdots$$and$$\sqrt{1+x}=(1+x)^{1/2}=1+\frac x2-\frac{x^2}8+\frac{x^3}{16}-\frac{5x^4}{128}+\cdots,$$and therefore$$\sqrt{1+x}\sin(4x)=4x+2x^2-\frac{67x^3}6-\frac{61x^4}{12}+\cdots$$The RHS of this equality is what one gets after ignoring the terms of the product$$\left(4x-\frac{4^3}{3!}x^3\right)\times\left(1+\frac x2-\frac{x^2}8+\frac{x^3}{16}-\frac{5x^4}{128}\right)$$whose degree is greater than $4$.
|
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|
Find the rank of $\begin{pmatrix} 1 &1 &0 &1 \\ 1 &2 &2 &1 \\ 3 &4 &2 &3 \end{pmatrix}$ Found this exercise in Serge Lang's Introduction to Linear Algebra:
Find the rank of the matrix $$\begin{pmatrix} 1 &1 &0 &1 \\ 1 &2 &2 &1 \\ 3 &4 &2 &3 \end{pmatrix}$$
So my process to solve it is as follows. First, I set a system
$$ x \begin{pmatrix} 1 \\ 1 \\ 0 \\ 1 \end{pmatrix} + y \begin{pmatrix} 1 \\ 2 \\ 2 \\ 1 \end{pmatrix} + z \begin{pmatrix} 3 \\ 4 \\ 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix} $$
$$\begin{cases}
x + y + 3z = 0 \\
x + 2y + 4z = 0 \\
2y + 2z = 0 \\
x + y +3z = 0
\end{cases}$$
Immediately we see that we can ignore the last equation. Then we can subtract the first one from the second one so we get
$$\begin{cases}
x + y +3z = 0 \\
y + z = 0 \\
2y + 2z = 0
\end{cases}$$
The third one is $2$ times the second one so we can remove it. Finally we have
$$\begin{cases}
x + y +3z = 0 \\
y + z = 0 \\
\end{cases}$$
Since this is a system of two equations in three unknowns, it has a non-trivial solution and thus, both are linearly dependent. Therefore, the rank is $1$
I must have made some mistake since the answers at the back of the book state that the solution is $2$ but I don't see where I'm wrong
|
The first two rows are independent (by inspection: they're not multiples of each other).
Then $r_3=2r_1+r_2$, as you can see.
So the row rank is $2$. So the rank is $2$.
|
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|
Prove that the product of four consecutive natural numbers are not the square of an integer
Prove that the product of four consecutive natural numbers are not the square of an integer
Would appreciate any thoughts and feedback on my suggested proof, which is as follows:
Let $f(n) = n(n+1)(n+2)(n+3) $.
Multiplying out the expression and refactoring it in a slightly different way gives
$$f(n) = n^4 + 6n^3+11n^2+6n \\= n^4 + 6n^3 + 9n^2 + 2n^2 + 6n = (n^2 + 3n)^2 + 2n(n+3). \tag{1}\label{1} $$
We want to show that the only possible way for $ f(n) $ to be the square of an integer is if $ f(n) = (n^2 + 3n +1 ) ^2. $ We show this by proving that $ (n^2+3n)^2 < f(n) < (n^2+3n+2)^2 $. The left-hand side follows immediately from $(1)$, since $ 2n(n+3) > 0 $ for all $ n \geq 1 $, and the right-hand side can be verified by multiplying out both sides:
$$
\begin{align}
(n^2+3n)^2 + 2n(n+3) &< (n^2+3n+2)^2 \\ \iff
n^4 + 6n^3 + 11n^2 + 6n &< n^4 + 9n^2 + 4 + 6n^3+4n^2+12n \\ \iff
0 &< 2n^2 + 6n + 4
\end{align}
$$
which is true for all $n \geq 1 $. Now we note that $n^2+3n = n(n+3)$ is even since one of the factors $n$ or $n+3$ is even for all $n$. It follows that $ n^2+3n+1$ must be odd, and so $ (n^2+3n+1)^2 $ must be odd. But $ f(n) $ must be even, since either $n$ and $(n+2)$ is even, or $(n+1)$ and $(n+3)$ is even and an even number multiplied by an odd number is an even number. So $f(n) \neq (n^2 + 3n +1)$ and therefore $f(n)$ cannot be the square of an integer for all $n \geq 1 $.
|
Your solution is perfect and detailed.
Another alternative shorter approach based on observation is if we can consider $n(n+2)=k$ and then redefine the function as:
$$\begin{align*}f(k) &= k(k+2)\\ f(k) &= (k+1)^2-1\end{align*}$$
Now, we need to prove that $f(k)$ is not the square of an integer.
Well, there exists no two squares of a nonzero integer that differ by 1.
Hope that helps!!
|
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|
Find infimum and closure of $A$ Let $$
\mathbf{A}=\left\{\frac{\mathbf{3 m}+\mathbf{2 n}}{5\mathbf{m}+7 \mathbf{n}}: \mathbf{m}, \mathbf{n} \geq \mathbf{3}\right\}
$$
Now find :
*
*$inf(A)$
*$int(A)$ , interior points
*$cl(A)$ , closure of $A$
With euclidean metric on $\mathbb{R}$
We have $int(A) =\emptyset $ because $A$ is countable. also if $m$ be fixed then
$$ lim _{n \to \infty} \frac{\mathbf{3 m}+\mathbf{2 n}}{5\mathbf{m}+7 \mathbf{n}}= \frac{2}{7}$$
and if $n$ be fixed we have :
$$ lim _{m \to \infty} \frac{\mathbf{3 m}+\mathbf{2 n}}{5\mathbf{m}+7 \mathbf{n}}= \frac{3}{5}$$
Also for $m=n $ we have :
$$ lim _{m \to \infty} \frac{\mathbf{5 m}}{12\mathbf{m}}= \frac{5}{12}$$
|
$$\dfrac{3m + 2n}{5m + 7n} = k$$
$$(3m + 2n) = k(5m + 7n)$$
$$(7k -2)n + (5k-3)m = 0$$
$$-\dfrac{7k-2}{5k-3}= \dfrac{m}{n} > 0$$
so $\dfrac{7k-2}{5k-3} < 0$ and $\dfrac{2}{7} \leq k \leq \dfrac{3}{5}$. You can prove that $A$ is dense in $[2/7, 3/5]$, so $inf(A) = \dfrac{2}{7}$, $int(A) =\{\}$ and $cl(A) = [2/7, 3/5]$.
|
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|
Scalar "exchange" in determinants Consider the following:
Find
$$
\begin{vmatrix}
x^2+1 & xy & xz \\
xy & y^2+1 & yz \\
xz & yz & z^2+1 \\
\end{vmatrix}
$$
Multiplying $R_1,R_2,R_3$ with $x,y,z$ and dividing $C_1,C_2,C_3$ by $x,y,z$ respectively, we get
$$
\begin{vmatrix}
x^2+1 & x^2 & x^2 \\
y^2 & y^2+1 & z^2 \\
z^2 & z^2 & z^2+1 \\
\end{vmatrix}
$$
I have seen this technique of distributing scalars and taking them out being used extensively to simplify determinants quickly.
Is this a known "trick" in linear algebra? What is the intuition behind this? Are there some specific patterns of determinants where this technique proves helpful?
|
*
*Multiplying the $k$th row by $c$ is accomplished by multiplying on the left by the matrix $M(k,c)$ given by
$$ M(k,c)_{ij}=\begin{cases} c, & i=j=k,\\ 1, &i=j\neq k, \\ 0 &\text{otherwise} \end{cases}$$
*Multiplying the $k$th column by $c$ is done by multiplying by $M(k,c)$ on the right.
*Determinants of products are products of determinants: $|AB|=|A| |B|$, more generally $|A_1\dots A_n| = |A_1|\dots |A_n|$.
*Let $$ A=\begin{bmatrix}
x^2+1 & xy & xz \\
xy & y^2+1 & yz \\
xz & yz & z^2+1 \\
\end{bmatrix} , B=\begin{bmatrix}
x^2+1 & x^2 & x^2 \\
y^2 & y^2+1 & z^2 \\
z^2 & z^2 & z^2+1 \\
\end{bmatrix}$$
Then points 1 and 2 let us translate the given algorithm into $$B=M(1,x)M(2,y)M(3,z)AM(1,\frac1x)M(2,\frac1y)M(3,\frac1z)$$
since $|M(1,x)| |M(1,1/x)|=1$, the result follows from point 3.
|
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|
Prove by induction that $3^{n-2} \geq n^5$ for $n\geq20$ I am new to induction proofs and wanted to know if my reasoning behind the following proof is correct. Here are my steps :
*
*$n=18 \Rightarrow 3^{18} \geq 20^5$
*Given that $3^{n-2} \geq n^5$ is true for n, show $3^{n-1} \geq (n+1)^5$
$3^{n-1} = 3\times(3)^{n-2} \geq 3n^5$
If $3n^5 \geq (n+1)^5$, then $3\times (3)^{n-2} \geq 3n^5 \geq (n+1)^5 \Leftrightarrow 3^{n-1} \geq (n+1)^5$
I show this by induction for all $n \geq 20$.
*
*$n=18 \Rightarrow 3(20)^{5} \geq 21^5$
*$3(n+1)^{5} = 3n^5+15n^4+30n^3+30n^2+15n+3 \geq (n+1)^5+15n^4+30n^3+30n^2+15n+3$
$3n^5+15n^4+30n^3+30n^2+15n+3 \geq (n+2)^5-10n^4-40n^2-60n-28$
$(3n^5+25n^4+70n^3+30n^2+75n+31) = (n+2)^5 + (2n^5+15n^4+30n^3+10n^2+10n+1)\geq (n+2)^5$
Since $(2n^5+15n^4+30n^3+10n^2+10n+1) > 0$ for $n\geq20$, it follows that $(n+2)^5+(2n^5+...+1) \geq (n+2)^5$.
Therefore, $3\times (3)^{n-2} \geq 3n^5 \geq (n+1)^5 \Leftrightarrow 3^{n-1} \geq (n+1)^5$
|
It seems like the limit is less ambitious than it could be. Let's aim for $n=15$.
Here is a $n=15$ base case to show $15^5 < 3^{13} $:
$$\begin{align}
15^5 &= 3^5\cdot 5^5\\
&<3^{5}\cdot 27^2\cdot 5\\
&=3^{11}\cdot 5\\
&<3^{13}\\
\end{align}$$
Then for $n\ge 15$,
$$\begin{align}
(n+1)^5 &= n^5 + 5n^4 + 10n^3+10n^2+5n+1 \\
&< n^5 + 5n^4 + 10n^3+10n^2+6n \\
&< n^5 + 5n^4 + 10n^3+11n^2 \\
&< n^5 + 5n^4 + 11n^3 \\
&< n^5 + 6n^4 \\
\color{#0B2}{\small(\text{so far only needing }n>11)}\qquad\qquad &< 2n^5 \\
\color{#0B2}{\small\text{inductive hypothesis}}\qquad \qquad &< 2\cdot 3^{n-2} \\
&< 3^{n-1}
\end{align}$$
as required.
|
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|
How to evaluate $\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4}$? Is it possible to evaluate the sum:
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4}$$
I expect it may be related to $\zeta^{\prime} (2)$:
$$\zeta^{\prime} (2) = - \sum_{k=2}^{\infty} \frac{\ln(k)}{k^2}$$
Is there an identity that works for my series, involving the natural logarithm, that is similar to the identity that:
$$\sum_{n=0}^{\infty} \frac{1}{(n+a)(n+b)} = \frac{\psi(a) - \psi(b)}{a-b}$$
Also potentially related, the Lüroth analogue of Khintchine’s constant can be defined as the following:
$$\sum_{n=1}^{\infty} \frac{\ln (n)}{n(n+1)}$$
as mentioned here.
After some work, the following can be shown:
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \frac{5\ln(2) + 4\ln(3)}{16} + \frac{1}{2} \sum_{k=3}^{\infty} \frac{1}{k} \text{tanh}^{-1} \left( \frac{2}{k} \right)$$
and furthermore:
$$\sum_{k=3}^{\infty} \frac{1}{k} \text{tanh}^{-1} \left( \frac{2}{k} \right) = \int_{0}^{2} \left( \frac{\left(1-\pi x \cot(\pi x) \right)}{2x^2} + \frac{1}{x^2 - 1} + \frac{1}{x^2 -4} \right) \, dx$$
EDIT
I have derived yet another form for my sum of interest, however, I found this one interesting as it seems like it could potentially be solvable?
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \int_{0}^{\infty} \left( \frac{\psi^{(0)} (s+3) + \gamma}{(s+2)(s-2)} - \frac{25}{16 (s-2)(s+1)} \right) \, ds$$
From this, it is possible to obtain the following:
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \frac{\pi \gamma}{4} i + \frac{25}{48} (\ln (2) - i \pi) - \frac{1}{8} + \frac{1}{16} i \pi + \frac{1}{4} \int_{0}^{i \pi} \psi^{(0)} \left( \frac{4}{1+ e^{u}} \right) \, du$$
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = \frac{\pi \gamma}{4}i+\frac{25}{48} (\ln (2)-i \pi )+\frac{7 i \pi }{48}-\frac{1}{8}-\frac{\ln (2)}{3} -2 \int_0^{\infty } \frac{t \ln (\Gamma (1-i t))}{\left(t^2+4\right)^2} \, dt$$
$$\sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} = -\frac{1}{8}-\frac{i \pi }{4}+\frac{i \gamma \pi }{4}-\frac{\ln (2)}{16} - 2 \int_{0}^{\infty} \frac{t \ln (\Gamma (-i t)) }{(4+t^2)^2} \, dt$$
$$\implies \sum_{k=3}^{\infty} \frac{\ln (k)}{k^2 - 4} =\frac{25}{48} \ln (2) -\frac{1}{8} + \int_{1}^{\infty} \frac{\ln (v-1) \text{li} (v^2)}{v^5} \, dv$$
Where $\text{li}$ is the logarithmic integral function.
$$\sum_{k=3}^{\infty} \frac{\ln(k)}{k^2-4} = \frac{3 \ln (2)}{16} - \frac{\pi^2+1}{8} - \frac{\pi}{2} \int_{0}^{\infty} \sin(4\pi x) (\psi (x) - \ln (x)) \, dx$$
|
This is a partial answer and not very rigorous (I don't know how to make it rigorous) but i think this way might be able to produce some results.
$$ \sum_{k=3}^{\infty} \frac{\ln(k)}{k^2-4} - \sum_{k=3}^{\infty} \frac{\ln(k)}{k^2} = 4\sum_{k=3}^{\infty} \frac{\ln(k)}{k^2(k^2-4)} $$
Now Consider the function
$$f(a) = \sum_{k=3}^{\infty} \frac{1}{k^a(k^2-4)}$$
such that :
$$\lim_{a \to 2}\, \frac{d}{da}\, f(a) = -\sum_{k=3}^{\infty} \frac{\ln(k)}{k^2(k^2-4)} $$
Then using
$$ \int_0^1 x^{k-1} \, dx = \frac{1}{k} $$
One can show that if $a$ is an EVEN INTEGER greater than $0$
$$ \sum_{k=3}^{\infty} \frac{1}{k^a(k^2-4)} = \left.\left(\frac{1}{2^{a+2}x}+\frac{(4^{\frac{a}{2}+1}-3)x}{3*2^{a+2}}+\frac{(2a+1)x^2}{4*2^{a+2}} +\frac{\ln(1-x)}{2^{a+2}x^2}-\frac{x^2\ln(1-x)}{2^{a+2}}-\sum_{n=1}^{\frac{a}{2}}\frac{\operatorname{Li}_{(a-2n+2)}(x)}{2^{2n}} \right)\right\vert_{x=0}^{x=1} $$
Hopefully someone is able to produce some results from this , cheers.
Edit #1
$$ \sum_{k=3}^{\infty} \frac{1}{k^a(k^2-4)} = \frac{1}{3} +\frac{1}{2^{a+3}} + \frac{1}{2^{a+4}}+\frac{a}{2^{a+3}} - \sum_{n=1}^{\frac{a}{2}}\frac{\zeta{(a-2n+2)}}{2^{2n}} $$
Edit #2 (27/05/2021)
One last remark before i stop trying the problem. Once Again , not rigorous.
Consider the following identity for EVEN INTEGERS and $ x\ge 2$ which i will not prove but you can find in a similar manner here in result #20 :
https://mathworld.wolfram.com/InfiniteProduct.html
$$\prod_{n=3}^{\infty} \left(1-\frac{4}{n^x}\right) = \frac{1}{6\, \pi^{x/2} \left(1-\frac{1}{2^{x-2}}\right) i^{x/2-1}} \prod_{n=1}^{x/2} \sin\left(\pi \,2^{2/x} \,(-1)^{2n/x}\right) $$
Taking Natural Logs of both sides and differentiating with respect to x and some switching around we can obtain a neat little identity :
$$ -\frac{i \pi}{4} - \frac{4 \ln{(2)}}{2^x-4} - \frac{\ln(\pi)}{2} = 4 \sum_{n=3}^{\infty}\frac{\ln(n)}{n^x-4} - \frac{d}{dx} \sum_{n=1}^{x/2} \ln\left(\sin\left(\pi \,2^{2/x} \,(-1)^{2n/x}\right)\right) $$
Note : One can encounter similar sums of these forms the one with the $**\ln(\sin)**$ inside and these arise when dealing with Catalan constant ,$\zeta(3)$ and PolyGamma functions as far as i know, we simply don't understand their behavior yet. Providing further evidence that at least right now , the sum you asked for may not have a closed form.
|
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|
Divisibility of an expression by 11 How to prove that $5^{5n+1}+4^{5n+2}+3^{5n}$ where $n\in \mathbb{N}$ is divisible by $11$ using mathematical induction?
I have tried and got to this $$5 \cdot 25 \cdot 25 \cdot 5^{5k+1}+4\cdot 16 \cdot 16 \cdot 4^{5k+2}+3\cdot 9 \cdot 9 \cdot 3^{5k}$$
and I even skipped the first step.
Thanks in advance.
(I am sorry if the tags are not relevant to this question)
|
Let $f(n)=5^{5n+1}+4^{5n+2}+3^{5n}$
$$f(m+1)-3^5\cdot f(m)=5^{5(m+1)+1}+4^{5(m+1)+2}-3^5(5^{5m+1}+4^{5m+2})$$
$$=5^{5m+1}(5^5-3^5)+4^{5m+2}(4^5-3^5)$$ which is divisible by $11$
as $3^5=9(22+5)\equiv1\pmod{11}, 5^5=5^2\cdot5^3\equiv3\cdot4\equiv1\pmod{11}$
Similarly, $4^5\equiv1\pmod{11}$
$$\implies11|f(m)\iff11\mid f(m+1)$$
|
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|
Evaluate $\sum_{r=0}^n 2^{n-r} \binom{n+r}{r}$
Evaluate: $$\sum_{r=0}^n 2^{n-r} \binom{n+r}{r}$$
This looks like an unusual hockey stick sum. Here are my attempts:
Method 1:
The sum is equivalent to
$$S=\sum_{r=0}^n 2^{n-r} \binom{n+r}{n}=\sum_{r=0}^n 2^{r} \binom{2n-r}{n-r}$$
and I could evaluate neither of these.
Method 2:
$$S=\text{coefficient of $x^n$ in }:$$
$$2^n(1+x)^n\Bigg( 1+\frac{1+x}{2}+\left(\frac{1+x}{2}\right)^2+\cdots+\left(\frac{1+x}{2}\right)^n \Bigg)$$
$$=(1+x)^n\left(\frac{(1+x)^{n+1}-2^{n+1}}{(x-1)}\right)$$
It looks like a heavy task to collect all the $x^n$ coefficients from this expression.
Im out of ideas. Any hint is appreciated.
From Putnam 2020, Q A2
|
Let
$$
S_n:=\sum_{k=0}^n\frac{\binom{n+k}{k}}{2^{n+k}}
$$
Obviously $S_0=1$. Assume that equality
$$S_{n-1}=1\tag1$$
is valid for some $n$. Then it is valid for $n+1$ as well:
$$
\begin{align}
S_n
&=\sum_{k=0}^n\frac{\binom{n+k}{k}}{2^{n+k}}\\
&=\sum_{k=0}^n\frac{\binom{n+k-1}{k-1}+\binom{n-1+k}{k}}{2^{n+k}}\\
&=\frac12\sum_{k=0}^{n-1}\frac{\binom{n+k}{k}}{2^{n+k}}+\frac12\sum_{k=0}^n\frac{\binom{n-1+k}{k}}{2^{n-1+k}}\\
&=\frac12S_n-\frac{\binom{2n}{n}}{2^{2n+1}}+\frac12S_{n-1}+\frac{\binom{2n-1}{n}}{2^{2n}}\\
&=\frac12S_n+\frac12S_{n-1}\implies S_n=S_{n-1}\stackrel{I.H.}=1.
\end{align}
$$
Thus, by induction the equality $(1)$ is valid for all integer $n\ge0$.
Accordingly your sum is $2^{2n}S=4^n$.
|
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|
How to integrate $\int_0^{\infty}\frac{1}{(1+x)(1+x^2)}$ What is the method of integrating the following:
$$\int_0^{\infty}\frac{1}{(1+x)(1+x^2)}$$
I tried doing it via using partial fractions and deduce that:
$$\frac{\ln(x+1)}{2} - \frac{\ln(x^2+1)}{4} + \frac{\arctan(x)}{2} + C$$
However I am unsure how to use the limits given in the question.
I suspect there is a way to do this using contour integration and this is probably the most straightforward method but I cannot see how to do it. Any help would be appreciated.
|
HINT:
Note that we have
$$\begin{align}
\frac12\log(x+1)-\frac14\log(x^2+1)&=\frac14 \log\left(\frac{x^2+2x+1}{x^2+1}\right)\\\\
&=\frac14 \log\left(1+\frac{2x}{x^2+1}\right)
\end{align}$$
And we have the estimates for $x>0$
$$0\le \log\left(1+\frac{2x}{x^2+1}\right)\le \frac{2x}{x^2+1}$$
Can you finish now?
ALTERNATIVE APPROACH:
Let $I$ be given by the integral $I=\int_0^\infty \frac{1}{(x+1)(x^2+1)}\,dx$. We write the integral as the sum of integrals, the first from $0$ to $1$ and the second from $1$ to $\infty$. The we enforce the substitution $x\mapsto 1/x$ in the second integral. Proceeding we have
$$\begin{align}
I&=\int_0^1 \frac{1}{(x+1)(x^2+1)}\,dx+\color{blue}{\int_1^\infty \frac{1}{(x+1)(x^2+1)}\,dx}\\\\
&\overbrace{=}^{\color{blue}{x\mapsto 1/x}}\int_0^1 \frac{1}{(x+1)(x^2+1)}\,dx+\color{blue}{\int_1^0 \frac1{(1+1/x)(1+1/x^2)}\,\left(-\frac1{x^2}\right)\,dx}\\\\
&=\int_0^1 \frac{1}{(x+1)(x^2+1)}\,dx+\color{blue}{\int_0^1 \frac{x}{(x+1)(x^2+1)}\,dx}\\\\
&=\int_0^1 \frac1{1+x^2}\,dx\\\\
&=\frac\pi4
\end{align}$$
And we are done!
|
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|
Convergence of $ I=\int_{0}^{+\infty} \frac{1}{\sqrt{t}} \cdot \sin \left(t+\frac{1}{t}\right) \, dt $ What have i done?
$I$ is improper at $0$ and $+\infty$. The function inside is continuous on $(0 ;+\infty)$ and then integrable on any closed interval contained in $(0 ;+\infty).$
$$
I=\int_0^1 \frac{1}{\sqrt{t}} \sin \left(t+\frac{1}{t}\right) \cdot d t+\int_1^{+\infty} \frac{1}{\sqrt{t}} \sin \left(t+\frac{1}{t}\right)
\, dt
$$
$$
\begin{array}{l}
\text { As }\left|\frac{\sin \left(t+\frac{1}{t}\right)}{\sqrt{t}} \right| \leqslant \frac{1}{\sqrt{t}} \text { and }\left(\int_0^1 \frac{1}{\sqrt{t}} \cdot d t\right) \text { converges } , \\
\text { the comparison test states that } \int_0^1 \frac{\sin \left(t+\frac{1}{t}\right)}{\sqrt{t}} \,dt \text { is } \\
\text { absolutely convergent. Thus } I_1 = \int_0^1 \frac{\sin \left(t+\frac{1}{t}\right)}{\sqrt{t}} \, d t \text { converges. }
\end{array}
$$
$$
\begin{array}{l}
I_2=\int_1^{+\infty} \frac{1}{\sqrt{t}} \cdot \sin \left(t+\frac{1}{t} \right) \, d t \\
I_2=\int_1^{+\infty}\left(\frac{1}{\sqrt{t}} \sin (t) \cos \left(\frac{1}{t}\right)+\frac{1}{\sqrt{t}} \sin \left(\frac{1}{t}\right) \cos (t)\right) \,dt \\
\text { The taylor expansion of } \cos \left(\frac{1}{t}\right) \text { and } \sin \left(\frac{1}{t}\right) \\
\text { at first order about } +\infty \text { gives: } \\
\cos \left(\frac{1}{t}\right)=1-\frac{1}{2 \cdot t^{2}} \\
\sin \left(\frac{1}{t}\right)=\frac{1}{t} \\
\text { Thus, }
\end{array}
$$
$$
I_2=\int_1^{+\infty}\left(\frac{1}{\sqrt{t}} \cdot \sin (t) \cdot\left(1-\frac{1}{2 t^{2}}\right)+\frac{1}{\sqrt{t}} \cos (t) \cdot \frac{1}{t}\right) \cdot d t
$$
$$
I_2=\int_1^{+\infty}\left(\frac{\sin (t)}{\sqrt{t}}-\frac{\sin (t)}{2 t^{5 / 2}}+\frac{\cos (t)}{t^{3 / 2}}\right) \cdot d t
$$
For the same previous reasons (when studying $ I_1$ convergence. The convergence of $
\int_{1}^{+\infty} \frac{\sin (t)}{t^{1 / 2}} \cdot d t
$ can be proved using Abel criterion) ), $ I_2$ converges, as sum of convergent integrals.
My questions: Is my method correct? Is there any other nice way to study the convergence of $I$? I stopped the taylor expansion of $\cos(1/x)$ and $\sin(1/x)$ at the first order; but will it always work (stopping it at the first order)?can i have a counterexample?
Thanks in advance!
|
We can go ahead and actually compute the value of this integral by recognizing
$$I = \int_0^\infty\frac{2\,dt}{2\sqrt{t}}\sin\left((\sqrt{t})^2+\frac{1}{(\sqrt{t})^2}\right)$$
which suggests using the substitution $s = \sqrt{t}$
$$I = \int_0^\infty 2\sin\left(s^2+ \frac{1}{s^2}\right) = \int_{-\infty}^\infty \sin\left(\left[s-\frac{1}{s}\right]^2+2\right)$$
Then we can use the following theorem
$$\operatorname{p.v.}\int_{-\infty}^\infty f\left(x-\frac{1}{x}\right)\,dx = \operatorname{p.v.}\int_{-\infty}^\infty f(x)\,dx$$
to compute the previous integral
$$I = \operatorname{p.v.}\int_{-\infty}^\infty \sin(x^2+2)\:dx = \sin(2)\int_{-\infty}^\infty \cos(x^2)\:dx + \cos(2)\int_{-\infty}^\infty \sin(x^2)\:dx$$
$$I = \sqrt{\frac{\pi}{2}}\left[\sin(2)+\cos(2)\right]$$
using the known value of $\int_{\Bbb{R}}\sin(ax^2)\:dx = \int_{\Bbb{R}}\cos(ax^2)\:dx = \sqrt{\frac{\pi}{2a}}$
|
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|
Finding generating function and closed formula Find generating function and closed formula $1,0,1,0,1,0,1,0,1,...$
Solution Attempt)
$$\begin{align}G(x) &= 1 + 0x + 1x^2 + 0x^3 + 1x^4 + 0x^5 + 1 x^6 + 0 x^7 + 1 x^8 + \dots\\
&= 1 + x^2 + x^4 + x^6 + x^8 + \dots\end{align}$$
$$G(x) = \sum_{k = 0}^{∞} x^{2k} = \sum_{k = 0}^{∞} (x^{2})^k = \frac{1}{1-x^2}$$
|
Following Ethan remark, we have
$$u_{2n}=1=|\sin((2n+1)\frac{\pi}{2})|$$
and
$$u_{2n+1}=0=|\sin((2n+2)\frac{\pi}{2})|$$
So, we can take
$$f(x)=|\sin((x+1)\frac{\pi}{2})|$$
|
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|
Limit of $\left(y+\frac23\right)\mathrm{ln}\left(\frac{\sqrt{1+y}+1}{\sqrt{1+y}-1} \right) - 2 \sqrt{1+y}$ We consider the limit for large $y$ of the following expression :
$$\left(y+\frac23\right)\mathrm{ln}\left(\dfrac{\sqrt{1+y}+1}{\sqrt{1+y}-1} \right) - 2 \sqrt{1+y}.$$
Many references state that the large $y$ behaviour of this last expression is $y^{-3/2}$. I have been trying to show this in any way possible but I'm only hitting dead ends. Wolframalpha gives a Puiseux series of that expression which makes this behaviour explicit, but I do not understand how to get to this series either. Could you please help me on this?
|
Considering $$A=\left(y+\frac23\right)\log\left(\dfrac{\sqrt{1+y}+1}{\sqrt{1+y}-1} \right) - 2 \sqrt{1+y}$$ first simplify
$$A=\left(y+\frac{2}{3}\right) \log \left(\frac{y+2 \sqrt{y+1}+2}{y}\right)-2
\sqrt{y+1}$$ For large values of $y$, you could start using $y=\frac 1x$ to make
$$A=\frac{(2 x+3) \log \left(2 x+2 \sqrt{x (x+1)}+1\right)-6 \sqrt{x (x+1)}}{3 x}$$ Now, Taylor series
$$\sqrt{x (x+1)}=x^{1/2}+\frac{x^{3/2}}{2}-\frac{x^{5/2}}{8}+\frac{x^{7/2}}{16}+O\left(x^{9/2}\right)$$
$$ \log \left(2 x+2 \sqrt{x (x+1)}+1\right)=2 x^{1/2}-\frac{x^{3/2}}{3}+\frac{3 x^{5/2}}{20}-\frac{5 x^{7/2}}{56}+O\left(x^{9/2}\right)$$
$$A=\frac{8 x^{3/2}}{45}-\frac{4 x^{5/2}}{35}+O\left(x^{7/2}\right)$$ Back to $y=\frac 1x$ gives the result.
As I used to say : we are always closer to $0$ than to $\infty$
|
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|
Solving for integers $m>8, n > 0$ for $2^m - 3^n = 13$ Let $m,n$ be integers such that:
$$2^m - 3^n = 13$$
$m > 8$ since $2^8 - 3^5 = 13$.
I am trying to either find a solution or prove that no solution exists.
I tried to use an argument similar to this one for $2^m - 3^n = 5$ where $m > 5$.
$3^n \equiv -13 \pmod {512}$ if and only if $n \equiv{69} \pmod {128}$
Is there a straight forward way to complete the argument? Is there a better way to answer the question for $2^m - 3^n = 13$ with $m > 8$?
Edit:
Adding context for question:
I have been trying to understand why it is so difficult to establish a lower bound to $2^m - 3^n$ when $2^m > 3^n$. This came out of thinking about the Collatz Conjecture.
It occurs to me that for $m \ge 3$, $2^m - 3^n$ is congruent to either $5$ or $7$ modulo $8$ (since $-3^{2i} \equiv 7 \pmod 8$ and $-3^{2i+1} \equiv 5 \pmod 8$)
For $2^m - 3^n \equiv 7 \pmod {12}$, $m$ and $n$ are even, so the lower bound is at least $3^{n/2}$ since:
$$2^m - 3^n = (2^{m/2} - 3^{n/2})(2^{m/2} + 3^{m/2}) > 0$$
and
$$2^{m/2} - 3^{m/2} \ge 1$$
So, to reach a lower bound, I need to better understand the implications of $2^m - 3^n \equiv 5 \pmod 8$ and $2^m - 3^n \equiv 7 \pmod 8$ when $2^m - 3^n \not\equiv 7 \pmod {12}$.
I am also working to better understand this blog post by Terence Tao.
Edit 2:
I think that I can complete the argument using the congruence classes of $257$.
Here's my thinking:
Since $n \equiv 69 \pmod {128}$, $3^n \equiv 224$ or $33 \pmod {257}$
Then $2^m \equiv 3^n + 13 \pmod {257}$ which means $2^m \equiv 237$ or $46 \pmod {257}$
But there is no such solution of $2^m$ since for each $2^m$, there exists an integer $i$ such that $2^m \equiv \pm 2^{i} \pmod {257}$ and there is no such $i$ where $\pm 2^{i} \equiv {237} \pmod {257}$ or $\pm 2^{i} \equiv {46} \pmod {257}$
|
we have $$ 256(2^x - 1) = 243(3^y - 1) $$
and we assume $x,y \geq 1$
since $2^x \equiv 1 \pmod {243}$ we calculate that $162 | x.$
Next, $2^{162} - 1$ is divisible by the prime $262657.$
since $3^y \equiv 1 \pmod {262657}$ we calculate that $14592 | y.$ In particular, $256|y.$
Well, $3^{256} - 1$ is divisible by 1024. So that
$ 256(2^x - 1) $ is divisible by 1024, which is a contradiction of $x \geq 1$
similar:
https://math.stackexchange.com/users/292972/gyumin-roh
Exponential Diophantine equation $7^y + 2 = 3^x$ Gyumin Roh answer from 2015
Elementary solution of exponential Diophantine equation $2^x - 3^y = 7$.
Elementary solution of exponential Diophantine equation $2^x - 3^y = 7$. ME! 41, 31, 241, 17
Finding solutions to the diophantine equation $7^a=3^b+100$ 343 - 243 =
100
http://math.stackexchange.com/questions/2100780/is-2m-1-ever-a-power-of-3-for-m-3/2100847#2100847
The diophantine equation $5\times 2^{x-4}=3^y-1$
Equation in integers $7^x-3^y=4$
Solve in $\mathbb N^{2}$ the following equation : $5^{2x}-3\cdot2^{2y}+5^{x}2^{y-1}-2^{y-1}-2\cdot5^{x}+1=0$
Solve Diophantine equation: $2^x=5^y+3$ for non-negative integers $x,y$. 128 - 125 = 3
Diophantine equation power of 7 and 2
Find natural numbers a,b such that $|3^a-2^b|=1$ did +-1
Finding all natural $x$, $y$, $z$ satisfying $7^x+1=3^y+5^z$
|
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|
Complete the Matrix The eigenvectors of A (corresponding to the eigenvalue 2) = \begin{bmatrix}
1 & 2 \\
-1 & 4 \\
\end{bmatrix}
can be solved by solving the matrix equation:
Here is my working:
My answers were $4$ and $-1/2$ which were apparently wrong, though I am wondering if I made a typo when I put in $-1/2$ (I am unable to check further).
|
I don't know where the matrix $\begin{pmatrix} 4 & -2 \\ 1 & -1/2 \end{pmatrix}$ comes from.
The way I have seen it done is that $\begin{pmatrix} x \\ y \end{pmatrix}$ is a $\lambda$-eigenvector if
$\begin{pmatrix} 1-\lambda & 2 \\ -1 & 4-\lambda \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}= \begin{pmatrix} 0 \\ 0 \end{pmatrix}$.
So when $\lambda = 2$ we want:
$\begin{pmatrix} -1 & 2 \\ -1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$.
Of course $\begin{pmatrix} -1 & 2 \\ -1 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} -x+2y \\ -x+2y \end{pmatrix}$, so the vector works when $x=2y$.
So we want vectors $\begin{pmatrix} 2y \\ y \end{pmatrix}$ with $y\neq 0$.
|
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|
(rectified) proof by induction - Fibonacci Sequence Define the sequence $ (a_n) $ by induction, putting $ a_1 = a_2 = 1 $ and $ a_ {n + 2} = a_ {n + 1} + a_n $, $ \forall n \in \mathbb{N} $ , thus obtaining the Fibonacci sequence $ (1,1,2,3,5,8,13,...)$.
Write $ x_n = \frac{a_n}{a_ {n + 1}} $ and prove that $ \lim x_n = c $, where $ c $ is the only positive number such that $ \frac{1}{c + 1} = c $ (ie $ c $ is the root of the equation $ c^{2} +c-1 = 0$).
Could someone help me continue my induction prove?
$$(a_n) = \begin{cases}
1 & \mbox{for}\quad a_1\\
1 & \mbox{for}\quad a_2 \hspace{1cm} (\forall n \in \mathbb{N})\\
a_{n+1}-a_{n+2} & \mbox{for}\quad a_n
\end{cases}$$
Suppose $ \lim x_n = c $, then $ \lim x_{n + 1} = c $, since $ x_ {n + 1} $ is a subsequence of $ x_n $.
We can write
\begin{align*}
a_{n + 2} = a_{n + 1} + a_{n} \Rightarrow \frac{a_{n + 2}}{a_{n + 1}} &= \frac{a_{n}+a_{n + 1}}{a_{n + 1}}\\
&= 1 +\frac{a_{n}}{a_{n + 1}}\\
\end{align*}
We know that
\begin{align*}
x_n=\frac{a_n}{a_{n + 1}}\
\end{align*}
follow that
$$x_{n+1}=\frac{a_{n+1}}{a_{n+2}}$$
therefore
$$ \frac{1}{x_{n + 1}}=x_n+1\Rightarrow x_{n+1}=\frac{1}{x_n+1}.$$
We have
$$\lim x_n=\lim x_{n+1}=c$$
Hence
\begin{align*}
\lim x_n=c &\Rightarrow \frac{1}{c+1}=c\\
&\Rightarrow 1=c^2+c \\
&\Rightarrow 0=c^2+c-1.
\end{align*}
Thus
\begin{align*}
c&=\frac{-1+\sqrt{1^2-4(1)(-1)}}{2\cdot(1)}\\
&= \frac{-1+\sqrt{5}}{2}
\end{align*}
Consequently
$$x_n \longrightarrow c=\frac{-1+\sqrt{5}}{2}$$
|
There are several mistakes/typos in your proof, and I suggest you go over your proof much more carefully. Once you have reached the equation
$$
\frac{1}{x_{n+1}} = 1+x_n
$$
you can simply apply the limit as $n \to \infty$ from both sides as the limits being finite.
|
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|
Simplifying $\left({\sqrt{x} + \frac{1}{\sqrt{x}}}\right)^2 - \left({\sqrt{x} - \frac{1}{\sqrt{x}}}\right)^2 $ Hi can someone help me please simplify the following showing the working out step by step?
$$
\left({\sqrt{x} + \frac{1}{\sqrt{x}}}\right)^2 - \left({\sqrt{x} - \frac{1}{\sqrt{x}}}\right)^2
$$
I can't get the answer matching the text book but I'd also like to get an idea of the most idiomatic way to solve it in terms of steps.
What I attempt to do based on what I've learned so far is to:
*
*try and simplify the contents of the parens using conjugate and then LCM
*then square
*then handle the subtraction.
But my answer ends up incorrect.
So even steps just to simplify say the left hand term (without the squaring step) would be helpful.
My simplifying the left hand looks like:
Use Conjugate:
$$
\left(\sqrt{x} + \frac{(1)(\sqrt{x})}{(\sqrt{x})(\sqrt{x})}\right)^2
$$
$$
\left(\sqrt{x} + \frac{(\sqrt{x})}{x}\right)^2
$$
Use LCM:
$$
\left(\frac{(\sqrt{x})(x)}{x} + \frac{(\sqrt{x})}{x}\right)^2
$$
$$
\left(\frac{(x)(\sqrt{x}) + \sqrt{x}}{x}\right)^2
$$
Then I'm not sure next best step.
|
Using $a^2-b^2 = (a+b)(a-b)$ we get
$$\left(\sqrt{x} + \frac{1}{\sqrt{x}} + \sqrt{x} - \frac{1}{\sqrt{x}}\right)\left(\sqrt{x}+\frac{1}{\sqrt{x}} - \sqrt{x}+\frac{1}{\sqrt{x}}\right) = \left(2\sqrt{x}\right)\left(\frac{2}{\sqrt{x}}\right) = 4$$
Hence, we get our answer as $4$.
Hope it helps.
|
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|
Prove that $\left(\frac{2N}{2N+1}\right)^{\frac{2N+1}{2N+2}}\left(\frac{2}{1}\right)^{\frac{1}{2N+2}} < 1$ for all $N \geq 1$ I would like to prove that for all $N\geq 1$ we have,
$$\mathcal{P}(N) = \left(\frac{2N}{2N+1}\right)^{\frac{2N+1}{2N+2}}\left(\frac{2}{1}\right)^{\frac{1}{2N+2}} < 1.$$
Some basic simulations and worked out examples convince me that this inequality indeed holds true. I have tried to solve this problem by induction. Clearly, for $N=1$ we have,
$$\mathcal{P}(1) = \left(\frac{2}{3}\right)^{\tfrac{3}{4}}\cdot\left(\frac{2}{1}\right)^{\tfrac{1}{4}} \approx 0.8774 < 1.$$
Now assume the inequality holds for $N$, then for $N+1$ we have,
\begin{align}
\mathcal{P}(N+1) &=\left(\frac{2N+2}{2N+3}\right)^{\tfrac{2N+3}{2N+4}}\cdot\left(\frac{2}{1}\right)^{\tfrac{1}{2N+4}} \\
&= \left(\left(\frac{2N}{2N+1}\right)\left(\frac{2N+1}{2N}\cdot\frac{2N+2}{2N+3}\right)\right)^{\left(\frac{2N+1}{2N+2}\right)\left(\frac{2N+2}{2N+1}\cdot\frac{2N+3}{2N+4}\right)}\cdot\left(\frac{2}{1}\right)^{\frac{1}{2N+2}\frac{2N+2}{2N+4}}\\[1em]
&= \left(\frac{2N+2}{2N+3}\right)^{\left(\frac{2N+1}{2N+2}\right)\left(1 + \frac{1}{2(N+1/2)(N+2)}\right)}\cdot\left(\frac{2}{1}\right)^{\frac{1}{2N+2}\left(1 - \frac{1}{N+2}\right)}\left(\frac{2N+1}{2N}\cdot\frac{2N+2}{2N+3}\right)^{\frac{2N+3}{2N+4}}\\
&= \small\left(\frac{2N+2}{2N+3}\right)^{\left(\frac{2N+1}{2N+2}\right)}\cdot\left(\frac{2}{1}\right)^{\frac{1}{2N+2}}\cdot \left(\frac{2N+2}{2N+3}\right)^{\frac{1}{2(N+1/2)(N+2)}}\left(\frac{1}{2}\right)^{\frac{1}{N+2}} \left(\frac{2N+1}{2N}\cdot\frac{2N+2}{2N+3}\right)^{\frac{2N+3}{2N+4}}\\
&= \mathcal{P}(N) \cdot \left(\frac{2N+2}{2N+3}\right)^{\frac{1}{2(N+1/2)(N+2)}}\cdot\left(\frac{1}{2}\right)^{\frac{1}{N+2}} \cdot\left(\frac{2N+1}{2N}\cdot\frac{2N+2}{2N+3}\right)^{\frac{2N+3}{2N+4}}
\end{align}
Now from here we know that the first three terms are all smaller than 1 ($\mathcal{P}(N) < 1$ by induction hypothesis). However the last term is larger than one. For the proof by induction to work out, we need that this last term cancels against,
$$\left(\frac{2N+2}{2N+3}\right)^{\frac{1}{2(N+1/2)(N+2)}}\cdot\left(\frac{1}{2}\right)^{\frac{1}{N+2}}.$$ But I do not see how it does. Any help is greatly appreciated.
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If you could show that it is an increasing function of real positive $N$ and its limit is $1$ then that would be enough.
Showing the limit is $1$ as $N \to \infty$ looks easy enough: each component heads towards $1$
while the derivative seems to be $\mathcal P'(x)=\frac{{{\left( \frac{x}{2x+1}\right) }^{\frac{2x+1}{2x+2}}}\left( x\mathrm{log}\left( \frac{x}{2x+1}\right) +x+1\right) }{x{{\left( x+1\right) }^{2}}}$ which seems to be positive for positive $x$, so $\mathcal P(x)$ must be strictly less than $1$ for all positive real $x$ and thus $\mathcal P(N)$ must be strictly less than $1$ for all positive integer $N$
|
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|
Prove that $\sum_{n=1}^\infty \frac{2}{2n}-\frac{2}{2n+1}=2-2\ln(2)$ The question is simple, to prove that $\sum_{n=1}^\infty{\frac{2}{2n}-\frac{2}{2n+1}}=2-2\ln(2)$.
I did it by wirtting down some terms of the series, rearranging as follows:
$$\sum_{n=1}^\infty \frac{2}{2n}-\frac{2}{2n+1}$$
$$2\sum_{n=1}^\infty \frac{1}{2n}-\frac{1}{2n+1}$$
$$2\left[\frac 12 - \frac 13 + \frac 14 - \frac 15+ \frac 16- \frac 17+ \frac 18- \frac 19+ \ldots\right]$$
$$2\left[1-\left(1-\frac 12 + \frac 13 -\frac 14 +\frac 15-\frac 16+\frac 17-\frac 18+\frac 19-\ldots\right)\right]$$
and noticing that the series inside the round parenthesis corresponds to $$1-\frac 12 + \frac 13 -\frac 14 +\frac 15-\frac 16+\frac 17-\frac 18+\frac 19-\ldots=\ln(2)$$ follows that:
$$2[1-\ln(2)]=2-2\ln(2)$$
So what I'm wanting to clarify is if is there any other way to improve this answer, a more "elegant" way to come up to the infinite sum of the series.
|
Well, you asked for an elegant way, which depends on your perspective for elegance. I came up with a solution using integrals.
Simplifying the sum,
$$ S =2\sum_{n=1}^{\infty} \frac{1}{2n}-\frac{1}{2n+1} = \sum_{n=1}^{\infty} \dfrac{1}{n(2n+1)} $$
Also,
$$ \int_0^1 x^{2n}\ \mathrm{d} x = \dfrac{1}{2n+1} $$
Substituting this result into our sum,
$$ S = \sum_{n=1}^{\infty} \int_0^1 \frac{x^{2n}}{n}\ \mathrm{d} x $$
Now, for $ |x| \lt 1 $,
$$\ln(1-x)= -\sum_{n=1}^{\infty} \dfrac{x^n}{n} $$
Replacing $ x $ with $ x^2 $
$$ \ln(1-x^2)= -\sum_{n=1}^{\infty} \dfrac{x^{2n}}{n} $$
So, our sum will be transformed to
$$ \begin{align*} S &= -\int_0^1 \ln(1-x^2)\ \mathrm{d} x \\ &= - \int_0^1 \ln(1-x)\ \mathrm{d} x -\int_0^1 \ln(1+x)\ \mathrm{d} x \end{align*} $$
Using the susbtitution $ 1-x \mapsto x $ on the first integral and $1+x \mapsto x $ on the second integral
$$ \begin{align*} S &= -\int_0^1 \ln{x}\ \mathrm{d} x - \int_1^2 \ln(x)\ \mathrm{d} x \\ &= -\int_0^2 \ln{x}\ \mathrm{d} x \\ &= -\left. \biggl( x \ln{x}-x \biggr) \right|_0^2 \quad , \text{via IBP} \\ &= 2-2\ln{2} \end{align*} $$
|
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|
Solution Verification: Find Extrema points of the function $f(x,y)=2^{3x+8y}$ that are on $x^2+y^2=1$.
$f(x,y)=2^{3x+8y}$
$x^2+y^2=1$
Polar Coordinates approach:
Let $x=cos(t), y=sin(t), 0\le t \le 2\pi$ (We can see that this solved the circle equation).
Substituting them into my function:
$f = 2^{3\cos(t)+8\sin(t)} \Longrightarrow f=e^{(3\cos(t)+8\sin(t))ln(2)}$
Taking derivative to find critical points:
$f' = \ln(2)(-3\sin(t)+8\cos(t))e^{(3\cos(t)+8\sin(t))ln(2)}=0$
Now $e^x>0$, so I need to find when
$-3\sin(t)+8\cos(t)=0 \Longrightarrow 3 \sin(t)=8\cos(t) \Longrightarrow \tan(t) = \frac{8}{3} \Longrightarrow t=\arctan(\frac{8}{3}), \arctan(\frac{8}{3}) + \pi $
So I get two Points:
$(\cos(\arctan(\frac{8}{3})) ,\sin(\arctan(\frac{8}{3})))$
$(\cos(\arctan(\frac{8}{3}) + \pi), \sin(\arctan(\frac{8}{3}))$
and now I need to find $f_{xx}, f_{yy}, f{xy}$ in order to know which point is minimum and which is maximum:
$f_{x}=8^x3ln(2)$
$f_y = 256^y 8ln(2)$
$f_{xx} = 9\ln^2(2) 8^x$
$f_{yy} = 64ln^2(2) 256^y$
$f_{xy} = 0$
$f_{yx} = 0$.
for first point: $f_{xx} = 8.97$, $f_{yy} = 5530.131$. So $f_{xx}*f_{yy} - 0 > 0$ , and $f_{xx} > 0 $, which means it is a local minimum point.
for second point: $f_{xx} = 2.03$ , $f_{yy} = 0.17$. and same as above, it is a minimum point.
Would appreciate any help in approving this solution or finding mistakes in it, and if there's any more efficient approaches, thanks in advance.
EDIT: Approach from Parcly Taxel answer:
Checking maximum and minimum for $3x+8y$:
$\phi (x,y,\lambda) = 3x+8y + \lambda(x^2+y^2-1)$
$\phi_x = 3 + 2\lambda x = 0 \Longrightarrow x= \frac{-3}{2\lambda}$
$\phi_y = 8 + 2 \lambda y = 0 \Longrightarrow y= \frac{-4}{\lambda}$
$\phi_{\lambda} = x^2 +y^2 -1 = 0 \Longrightarrow \frac{9}{4\lambda^2} + \frac{16}{\lambda^2}-1=0 \Longrightarrow 9+64=4\lambda^2 \Longrightarrow \lambda^2 = \frac{73}{4} \Longrightarrow \lambda = \pm \frac{\sqrt{73}}{2}$
So: $x= \pm 0.3511 $
$y = \pm 0.9363$
$(0.3511, 0.9363) , (-0.3511, -0.9364)$
|
You could have made your work much simpler by noticing that $2^z$ is monotonically increasing – maximising/minimising $2^{3x+8y}$ is equivalent to maximising/minimising $3x+8y$. Then Lagrange multipliers (or any other reasonable method, like plotting the objective function and constrant and eyeballing) will work just fine.
|
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|
Prob. 2.10 in Bourne & Kendall: These four points lie on a circle Here is Prob. 2.10 in the book Vector Analysis and Cartesian Tensors by D. E. Bourne and P. C. Kendall, 3rd edition:
Show that the four points with position vectors
$$
\mathbf{r}_1, \mathbf{r}_2, \frac{r_2}{r_1} \mathbf{r}_1, \frac{r_1}{r_2} \mathbf{r}_2,
$$
where $\mathbf{r}_1 \neq \mathbf{0}$ and $\mathbf{r}_2 \neq \mathbf{0}$, lie on a circle.
My Attempt:
Relative to a rectangular cartesian coordinate system $Oxyz$, let $\mathbf{r}_1 = \left( a_1, b_1, c_1 \right)$ and $\mathbf{r}_2 = \left( a_2, b_2, c_2 \right)$, where
$$
a_1^2 + b_1^2 + c_1^2 > 0 \qquad \mbox{ and } \qquad a_2^2 + b_2^2 + c_2^2 > 0,
$$
and as $\mathbf{r}_1, \mathbf{r}_2, \frac{r_2}{r_1} \mathbf{r}_1, \frac{r_1}{r_2} \mathbf{r}_2$ constitute four points on the circle, so we must also have
$$
a_1^2 + b_1^2 + c_1^2 \neq a_2^2 + b_2^2 + c_2^2.
$$
As a circle is a plane figure, so we can also assume without any loss of generality that both $\mathbf{r}_1$ and $\mathbf{r}_2$ lie in the $xy$-plane so that $c_1 = c_2 = 0$. Therefore we obtain
$$
a_1^2+ b_1^2 > 0 \qquad \mbox{ and } \qquad a_2^2 + b_2^2 > 0,
$$
and also
$$
a_1^2+ b_1^2 \neq a_2^2 + b_2^2.
$$
Let $x^2 + y^2 + 2gx + 2fy + c = 0$, $z = 0$ be the required circle. Then substituting the coordinates of the heads of the four vectors we obtain
$$
\begin{align}
2a_1 g + 2 b_1 f + c &= - \left( a_1^2+ b_1^2 \right), \\
2a_2 g + 2 b_2 f + c &= - \left( a_2^2+ b_2^2 \right), \\
2 \sqrt{ a_2^2+ b_2^2} a_1 g + 2 \sqrt{ a_2^2+ b_2^2} b_1 f + c &= a_2^2 + b_2^2, \\
2 \sqrt{ a_1^2 + b_1^2 } a_2 g + 2 \sqrt{ a_1^2 + b_1^2} b_2 f + c &= a_1^2 + b_1^2.
\end{align}
$$
Is what I have done so far correct? If so, then what next? How to proceed from here? Or, is there a more direct solution?
|
You can just observe that the origin has the same power with respect to the two pairs of points collinear with it:
$$
|\mathbf{r}_1|\cdot \left|\frac{r_2}{r_1} \mathbf{r}_1\right|
=r_1r_2=
|\mathbf{r}_2|\cdot \left|\frac{r_1}{r_2} \mathbf{r}_2\right|.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4154969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
If $|f(x)-f(y)| \leq A |x-y|^\alpha$, then $|\hat{f}(n) | \leq C(1+|n|)^{-\alpha}$ Let $f$ be a $2\pi$ periodic continuous function and let
\begin{align}
\hat{f}(n) = \frac{1}{2\pi} \int_0^{2\pi} e^{-inx} f(x) dx
\end{align}
I want to show if
\begin{align}
|f(x)-f(y)| \leq A |x-y|^\alpha, \quad0< \alpha \leq 1
\end{align}
then $|\hat{f}(n) | \leq C(1+|n|)^{-\alpha}$ for some $C>0$.
What I know is following
\begin{align}
|f(x) - f(y)| \leq A |x-y|^\alpha \quad \Rightarrow \quad |\frac{f(x)}{x}| \leq A |x|^{\alpha-1}.
\end{align}
And from MVT, $\exists c \in [0,x]$ such that $|f'(c)| \leq A |x|^{\alpha-1}$. And from the range of $\alpha$, if $\alpha=1$, $|f'(c)| \leq 1$, and for other case $0<\alpha<1$, $f'(c)=0$.
The hints from the textbook say $\hat{f}(n) = -\frac{1}{2\pi} \int_{0}^{2\pi} e^{-inx} f(x-\frac{\pi}{n}) dx$)
But I have
\begin{align}
\hat{f}(n) &= \frac{1}{2\pi} \int_0^{2\pi} e^{-inx} f(x) dx \\
& = \frac{1}{2\pi} \int_{-\frac{\pi}{n}}^{2\pi - \frac{\pi}{n}} e^{-in(x-\frac{\pi}{n})} f(x-\frac{\pi}{n}) dx \\
&
= - \frac{1}{2\pi} \int_{-\frac{\pi}{n}}^{2\pi - \frac{\pi}{n}} e^{-inx} f(x-\frac{\pi}{n})dx
\end{align}
which is slightly different.
Assuming hint is correct, even so I have no idea of showing $|\hat{f}(n) | \leq C(1+|n|)^{-\alpha}$
|
Comment from @Medo
\begin{align}
|\hat{f}(n) - \hat{f}(1)| &= \frac{1}{2\pi} \left| \int_0^{2 \pi} (e^{-inx} f(x) - e^{-ix} f(x) ) dx \right| \\
&
= \frac{1}{2\pi} \left| \int_0^{2\pi} \left( e^{-inx} f(x) - f(x+\frac{1}{n} ) - f(x+\frac{1}{n}) - e^{-ix} f(x) \right) dx \right| \\
& \leq \frac{1}{2\pi} \int_0^{2 \pi} \left( |f(x) - f(x+\frac{1}{n}) | + |f(x+\frac{1}{n}) - f(x)| \right) dx
\end{align}
Then in the last step using Holder continuity condition, I can see $|\hat{f}(n) - \hat{f}(1)| \leq C (1+ |n|^{-\alpha} )\leq C (1+|n|)^{-\alpha}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4156413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
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