Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove that $\sqrt[3]{18 + \sqrt{325}} + \sqrt[3]{18 - \sqrt{325}} = 3$ without using Cardano's formula. (Hint, what is $(3\pm \sqrt{13})^3$ Prove that $\sqrt[3]{18 + \sqrt{325}} + \sqrt[3]{18 - \sqrt{325}} = 3$ without using Cardano's formula. (Hint, what is $(3\pm \sqrt{13})^3$
I have that $$(3 + \sqrt{13})^3 = 144 + 40 \sqrt{13} $$ and $$(3 - \sqrt{13})^3 = 144 - 40 \sqrt{13} $$
A cursory look into Bombelli's method led me to the following system of equations:
$$\sqrt[3]{18 + \sqrt{325}} = a + b^{1/2}$$
$$\sqrt[3]{18 - \sqrt{325}} = a - b^{1/2}$$
I am unsure how to solve this system of equations without making a mess of the radicals...I know however that the given cube roots on the LHS of the above system are solutions to the cubic $x^3 + 3x = 36 $
| Let $$\sqrt[3]{18 + \sqrt{325}} + \sqrt[3]{18 - \sqrt{325}}=x$$
Thus, $$18 + \sqrt{325}+ 18 - \sqrt{325}+3\left(\sqrt[3]{18 + \sqrt{325}} + \sqrt[3]{18 - \sqrt{325}}\right)\sqrt[3]{18 + \sqrt{325}}\sqrt[3]{18 - \sqrt{325}}=x^3$$ or
$$36+3x\cdot(-1)=x^3$$ or
$$x^3-3x^2+3x^2-9x+12x-36=0$$ or
$$(x-3)(x^2+3x+12)=0.$$
Can you end it now?
Also, by your hint:
$$\sqrt[3]{18 + \sqrt{325}} + \sqrt[3]{18 - \sqrt{325}} =\frac{3}{2}+\frac{1}{2}\sqrt{13}+\frac{3}{2}-\frac{1}{2}\sqrt{13}=3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3509357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solution verification: $\lim_{x\to 0}\frac{e^{2\sin x}\cosh x-1}{\sqrt[3]{x(\cosh x-\cos x)}}$
Find (without L'Hospital):$$\lim_{x\to 0}\frac{e^{2\sin x}\cosh x-1}{\sqrt[3]{x(\cosh x-\cos
x)}}$$
My attempt:
$$\lim_{x\to 0}\cosh x=1$$
$$\lim_{x\to 0}e^{2\sin x}\cosh x-1=\lim_{x\to 0}\frac{e^{2\sin x}-1}{2\sin x}\cdot2\sin x=2\lim_{x\to 0}\sin x$$
$$\lim_{x\to 0}(\cosh x-\cos x)=\lim_{x\to 0}(1-\cos x)=\lim_{x\to 0}\frac{1-\cos x}{x^2}\cdot x^2=\frac{1}{2}\lim_{x\to 0}x^2$$
$$\lim_{x\to 0}\sqrt[3]{x(\cosh x-\cos x)}=\lim_{x\to 0}\sqrt[3]{x\cdot\frac{1}{2}x^2}=\frac{1}{\sqrt[3]{2}}\lim_{x\to 0}x$$
$$2\sqrt[3]{2}\lim_{x\to 0}\frac{\sin x}{x}=2\sqrt[3]{2}$$
Is this legitimate?
| $$L=\lim_{x \rightarrow 0}\frac{e^{2 \sin x} \cosh x-1}{\sqrt[3]{x(\cosh x-\cos x)}}= \lim_{x\rightarrow 0} \frac{(1+2x+O(x^2))(1+x^2/2+O(x^4))-1}{{\sqrt[3]{x^3+O(x^5)}}}$$ $$\implies L= \lim_{x\rightarrow 0} \frac{2x+x^2/2+x^3+O(x^4)}{x}=2.$$
Here we have used $\cosh x \approx 1+x^2/2 +O(x^4), \cos x \approx 1-x^2/2+O(x^4), \sin x \approx x|O(x^3),$ $e^{x}\approx 1+x+O(x^2),$ when $x \rightarrow 0$
| {
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"timestamp": "2023-03-29T00:00:00",
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Number of matrices $A \in \{-1,0,1\}^{3 \times 3}$ such that the trace of $A A^{T}$ is $3$
The number of $3 \times 3$ matrices $A,$ with entries from the set $\{-1,0,1\}$ such that the sum of the diagonal elements of $AA^{T}$ is $3$, is __ .
What I tried:
Let $\displaystyle A=\begin{pmatrix}
a& b & c\\
d& e & f \\
g& h & i
\end{pmatrix}$ and $\displaystyle A^{T}=\begin{pmatrix}
a& d & g \\
b& e & h \\
c & f & i
\end{pmatrix}$
$$AA^{T}=\begin{pmatrix}
\sum a^2 & .. & ..\\
.. & \sum d^2 &.. \\
..& ...& \sum g^2
\end{pmatrix}$$
Given $a^2+b^2+c^2+d^2+e^2+f^2+g^2+h^2+i^2=3$
given $a,b,c,d,e,f,g,h,i\in \{-1,0,1\}$
Out of $ 9 ,$ any three is $1$ and rest all are zero(like $1,1,1,0,0,0,0,0,0$)
$$\binom{9}{3}\cdot \frac{9!}{6!\cdot 3!}=$$
but answer is different. How do I solve it? Help me, please.
| You are very close.
First choose three of the slots out of the $9$ to set the corresponding square, $x^2$ to $1$. Each each of the location, we have two choices.
Hence $$2^3 \cdot \binom{9}{3}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find a basis of a subspace spanned by matrices $G = \langle \begin{bmatrix}1&1\\1&1\\ \end{bmatrix}, \begin{bmatrix}1&1\\1&0\\ \end{bmatrix}, \begin{bmatrix}2&-3\\1&1\\ \end{bmatrix},\begin{bmatrix}4&-1\\3&2\\ \end{bmatrix}$
First I write all the matrices as vectors:
$M_1 = \begin{bmatrix}1\\1\\1\\1 \end{bmatrix}$
$M_2 = \begin{bmatrix}1\\1\\1\\0 \end{bmatrix}$
$M_3 = \begin{bmatrix}2\\-3\\1\\1 \end{bmatrix}$
$M_4 = \begin{bmatrix}4\\-1\\3\\2 \end{bmatrix}$
Then I found the rref:
$$\begin{bmatrix}1 & 1 & 1 & 1\\1 & 1 & 1 & 0\\2 & -3 & 1 & 1\\4 & -1 & 3 & 2 \end{bmatrix} \rightarrow (...) \rightarrow \begin{bmatrix}1 & 0 & \frac{4}{5} & 0\\0 & 1 & \frac{1}{5} & 0\\0 & 0 & 0 & 1\\0 & 0 & 0 & 0 \end{bmatrix}$$
The basis I get is
$$(\begin{bmatrix}1 &0\\4/5&0 \end{bmatrix},\begin{bmatrix}0 &1\\\frac{1}{5}&0 \end{bmatrix}),\begin{bmatrix}0 &0\\0&1 \end{bmatrix}$$
Is this correct? It doesn't even come close to the solution in my book which is
$$(\begin{bmatrix}1 &1\\1&1 \end{bmatrix},\begin{bmatrix}1 &1\\1&0 \end{bmatrix}),\begin{bmatrix}2 &-3\\1&1 \end{bmatrix}$$
| Let's check it out.
$$e_1=\begin{bmatrix}1 & 0 \\ \tfrac{4}{5} & 0 \end{bmatrix} \\ e_2 = \begin{bmatrix}0 & 1 \\ \tfrac{1}{5} & 0 \end{bmatrix} \\ e_3 = \begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}$$
We have:
$$\begin{bmatrix}1 & 1 \\ 1 & 1\end{bmatrix} = e_1+e_2+e_3 \\ \begin{bmatrix}1 & 1 \\ 1 & 0\end{bmatrix} = e_1+e_2 \\ \begin{bmatrix}2 & -3 \\ 1 & 1\end{bmatrix} = 2e_1-3e_2+e_3 \\ \begin{bmatrix}4 & -1 \\ 3 & 2\end{bmatrix} = 4e_1-e_2+2e_3$$
You have three matrices, which is definitely the size of the basis given in the book. Each of the four matrices you were given are in the span of the matrices you found. The three matrices you found are linearly independent. This means the two bases span the same subspace.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Cases where Heaviside cover up method fails for partial fraction expansion. Why can't I use Heaviside coverup method to find partial fraction expansion of:
$$F(s) = \frac{s^3}{(s^2+1)^2}$$
For example:
$$F(s) = \frac{s^3}{(s^2+1)^2} = \frac{r_1}{(s^2+1)^2} + \frac{r_2}{(s^2+1)}$$
$$r_1 = (s-\alpha)^r~ F(s) ~\bigg|_{s=\alpha}$$
$$r_1= s^3 \bigg|_{s=-1} = -1$$
$$r_2= \frac{d}{ds}[s^3] \bigg|_{s=-1} = \bigg[3s^2\bigg]_{s=-1} = 3$$
$$r_2 = \bigg( \frac{d}{ds} \Big[ (s-\alpha)^r ~ F(s)\Big] \bigg)~\bigg|_{s=\alpha}$$
$$F(s) = \frac{s^3}{(s^2+1)^2} = \frac{-1}{(s^2+1)^2} + \frac{3}{(s^2+1)}$$
however, if I check the result:
$$F(s) = \frac{-1}{(s^2+1)^2} + \frac{3(s^2+1)}{(s^2+1)(s^2+1)}$$
$$F(s) = \frac{-1 +3s^2+3}{(s^2+1)^2}$$
$$F(s) = \frac{3s^2+2}{(s^2+1)^2} \ne \frac{s^3}{(s^2+1)^2}$$
Why does heaviside coverup fail for this case?
I look at order of polynomial in numerator, its 3, and the polynomial order for the denominator is 4, so it should work, but it doesn't. Does heaviside coverup method only work for first order polynomial factors?
| $$F(s) = \frac{s^3}{(s^2+1)^2}$$
Is easy to decompose:
$$F(s) = \frac{s^3+s-s}{(s^2+1)^2}$$
$$F(s) = \frac{s(s^2+1)-s}{(s^2+1)^2}$$
$$F(s) = \frac{s}{(s^2+1)}-\frac{s}{(s^2+1)^2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $(G,*)$ be a group with $(a*b)^3=a^3*b^3$ and $(a*b)^5=a^5*b^5\forall a,b∈G$. Show that $(G,*)$ is abelian.
Let $(G, *)$ be a group such that $(a * b)^3 = a^3 * b^3$ and $(a * b)^5 = a^5 * b^5$ for all $a, b ∈ G$. Show that $(G, *)$ is abelian.
My attempt:
I need to show $ab = ba$.
$$(ab)^3 = a^3b^3 \implies a(ba)^2b = a^3b^3 \implies (ba)^2 = a^2b^2$$
$$(ab)^5 = a^5b^5 \implies a(ba)^4b = a^5b^5 \implies (ba)^4 = a^4b^4$$
$$(ba)^4 = ((ba)^2)^2=(a^2b^2)^2 = a^4b^4$$
$$a^2b^2a^2b^2=a^4b^4$$
$$b^2a^2=a^2b^2$$
But what next?
| You also have $(ba)^3=b^3a^3$, as the statement holds for all group elements. Hence, $(ab)^2=b^2a^2$, by left and right cancellation. So, by combining with your last equality, you can get $ab=ba$.
| {
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"timestamp": "2023-03-29T00:00:00",
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When $\frac {a^3-b^3}{a^2-b^2}$ is an integer? If $\frac {a^3-b^3}{a^2-b^2}$ is an integer, then supposing $a-b \ne 0$ we have that also$\frac {a^2+ab+b^2}{a+b}$ is an integer.
For which $a, b\in\mathbb Z$, the fraction $\frac {a^2+ab+b^2}{a+b}$ is an integer?
| A trivial example is for $a,b \in \mathbb{Z}$ such that $a+b=1$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Is there alternative factoring of a quintic equation? In a paper here
the author appears to be able to factor a Bring-Jerrard quintic making
$$P=2mn(m^2-n^2)(m^2+n^2)=2m^5n-2mn^5\\ \implies n^5-m^4n+\frac{P}{2m}=0 \rightarrow x^5+px+q=0$$
become
$$(x^3+bx^2+cx+d)(x^2+ex+f)=0$$
but I haven't been able to follow how he got there. If I could, I would have what I need to find the one or more valid values of $n$ in the equation:
$$n^5-m^4n+\frac{P}{2m}=0$$
given that I will know the values of $P$ and $m$.
Can anyone help me figure out how the 'factored' equation would look in terms of $p,q$?
| It seems that the claim of the paper is not true.
The author finally got the following equation
$$b^4-2b^3\bigg(\frac{q-p+\sqrt{q^2-q}}{q-\sqrt{q^2-q}}\bigg)+q-p+\sqrt{q^2-q}=0$$
Here, let us consider one example. We have
$$x^5-31x+30=(x^3+3x^2+7x+15)(x^2-3x+2)$$
This means that one of the possible values of $b$ is $3$ for $(p,q)=(-31,30)$.
However, the above equation does not have a solution $b=3$ for $(p,q)=(-31,30)$.
Since we have
$$(x^3+bx^2+cx+d)(x^2+ex+f)$$
$$=x^5+(b+e)x^4+(eb+c+f)x^3+(bf+ec+d)x^2+(cf+ed)x+df=0$$
if we compare this with $x^5+px+q$, then we get the following system
$$\begin{cases}b+e=0
\\eb+c+f=0
\\bf+ec+d=0
\\cf+ed=p
\\df=q\end{cases}$$
from which we want to represent $b,c,d,e,f$ by $p,q$.
Now, we have
$$\begin{align}&\begin{cases}b+e=0
\\eb+c+f=0
\\bf+ec+d=0
\\cf+ed=p
\\df=q\end{cases}\\\\&\stackrel{\text{eliminating $e$}}{\implies}
\begin{cases}e=-b
\\(-b)b+c+f=0
\\bf+(-b)c+d=0
\\cf+(-b)d=p
\\df=q\end{cases}
\\\\&\stackrel{\text{eliminating $f$}}{\implies}\begin{cases}e=-b
\\df=q
\\-b^2d+cd+q=0
\\bq-bcd+d^2=0
\\bd^2=cq-pd
\end{cases}
\\\\&\stackrel{\text{eliminating $b$}}{\implies}
\begin{cases}e=-b
\\df=q
\\bd^2=cq-pd
\\c^2dq^2-cqd^2p+d^3p^2-d^2pcq-cd^5-qd^4=0
\\c^2dq^2-cq^3+dpq^2-cd^2pq-d^4q=0
\end{cases}
\\\\&\stackrel{\text{eliminating $c$}}{\implies}
\begin{cases}e=-b
\\df=q
\\bd^2=cq-pd
\\c(-d^2pq-d^5+q^3)=dpq^2-d^3p^2
\\d^{10} + p qd^7 + p^3d^6 - 2 q^3d^5 - p^2 q^2 d^4- p q^4d^2 + q^6=0
\end{cases}\end{align}$$
So, we have to solve the following equation for $d$ :
$$d^{10} + p qd^7 + p^3d^6 - 2 q^3d^5 - p^2 q^2 d^4- p q^4d^2 + q^6=0$$
whose degree is $10$.
In conclusion, if we want to find $b,c,d,e,f$ such that
$$x^5+px+q=(x^3+bx^2+cx+d)(x^2+ex+f)$$
then, in general, we have to solve an equation whose degree is $10$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find such $x$ so that set $\{\sin x, \sin2x, \sin3x\}$ coincides with the set $\{\cos x, \cos2x, \cos3x\}$ I recently started learning set theory so my attempt might be looking silly.
I set the sums of elements to be equal:
$$
\begin{split}
\sin x+\sin 2x+\sin 3x&=\cos x+\cos 2x+\cos 3x\\
\sin 2x+2\sin 2x\cos x&=\cos 2x+2\cos 2x \cos x\\
(\sin 2x-\cos 2x)(2\cos x+1)&=0\\
\end{split}
$$
*
*1)
$\sin2x-\cos2x=0$
$\tan2x=1$
$x=\pi/8+\pi n/2$
*
*2)
$\cos x=-1/2$
$x=\pm\pi/3+\pi k$
Turns out that the solution of 1) is the correct answer to the initial problem, but the solution of 2) isn't.
| Your first equation is necessary, but not sufficient. If $\tan 2x=1$ then $\tan x=\pm\sqrt{2}-1$ and $\tan 3x=\frac{1+\tan x}{1-\tan x}=\frac{\pm\sqrt{2}}{2\mp\sqrt{2}}=\pm\sqrt{2}+1=\frac{1}{\tan x}$, so that works. By contrast, if $\cos x=-\frac12$ then $\cos 2x=2\cos^2x-1=-\frac12$ and $\cos3x=\cos x(2\cos 2x-1)=-1$, so $\sin x=\pm\frac{\sqrt{3}}{2}$ isn't obtained.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove for any matrix $B$, we have $y^{\top}By = y^{\top}Ay$ with $A$ symmetric I need to handle the following question:
Prove for any matrix $B$, we have $y^TBy = y^TAy$ with $A$ symmetric.
Now, we can of course simply pick $B$ to be $A^T$, but that would be pretty trivial.
Does anyone have a tip or a hint how to start? I thought we might consider decomposing the symmetric matrix into two seperate matrices (symmetric and skew symmetric), but I don't know whether that helps
| Note that for any square matrix $B$ we may define
$B_{sym} = \dfrac{B + B^T}{2}, \tag 1$
and
$B_{skew} = \dfrac{B - B^T}{2}; \tag 2$
we have
$B_{sym}^T = \dfrac{B^T + (B^T)^T}{2} = \dfrac{B + B^T}{2} = B_{sym}, \tag 3$
and
$B_{skew}^T = \dfrac{B^T - (B^T)^T}{2} = \dfrac{B^T - B}{2} = -B_{skew}, \tag 4$
and
$B_{sym} + B_{skew} = \dfrac{B + B^T}{2} + \dfrac{B - B^T}{2} = \dfrac{B + B^T + B - B^T}{2} = \dfrac{2B}{2} = B. \tag 5$
and
$B_{sym} - B_{skew} = \dfrac{B + B^T}{2} - \dfrac{B - B^T}{2} = \dfrac{B + B^T - B + B^T}{2} = \dfrac{2B^T}{2} = B^T; \tag 6$
therefore, since $y^TB_{skew}y$ is a scalar quantity
$y^TB_{skew}y = (y^TB_{skew}y)^T = y^TB_{skew}^T(y^T)^T = -y^TB_{skew}y, \tag 7$
which evidently implies
$y^TB_{skew}y = 0 \tag 8$
over any field $\Bbb F$ with
$\text{char}(\Bbb F) \ne 2; \tag 9$
in such cases we thus conclude that
$y^TBy = y^T(B_{sym} + B_{skew})y= y^TB_{sym}y + y^TB_{skew}y=y^TB_{sym}y; \tag{10}$
now by virtue of (3) we may take
$A = B_{sym}. \tag{11}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Does this sequence of function's sum uniformly converge? $Q)$ For the countable set, $\mathbb{Q}$ (rational number's set)
$$(0,1) \cap \mathbb{Q} = \{a_1, a_2, \ldots, a_n ,\ldots\}$$
Define $f_n : [0,1] \to \mathbb{R}$ by $\begin{cases}
1/n, & x=a_n \\
0 & x \neq a_n
\end{cases}$
Determine uniformly converge $\sum_{n=1}^\infty f_n(x)$ on $[0,1]$
My guess this is not uniformly converge
since $a_n \in (0,1)$ into $f_n(x)$
Then, $\sum_{n=1}^{\infty} f_n(a_n)=\sum_{n=1}^{\infty}{ 1 \over {n}} $
Therefore, $\sum_{n=1}^\infty f_n(x)$ does not uniformly converge.
What do you think? Any help would be appreciated.
| It does converge uniformly if $a_1,a_2,a_3,\ldots$ are all distinct.
For example suppose we have
$$
\begin{array}{lcll}
a_1 & = & 1/2 & (\text{denominator} = 2) \\
a_2, a_3 & = & 1/3,\,\, 2/3 & (\text{denominator} = 3) \\
a_4, a_5 & = & 1/4, \, \,3/4 & (\text{denominator} = 4) \\
a_6,a_7,a_8,a_9 & = & 1/5,\,\, 2/5,\,\, 3/5,\,\, 4/5 & (\text{denominator} = 5) \\
a_{10}, a_{11} & = & 1/6,\,\,5/6 & (\text{denominator} = 6) \\
\text{etc.}
\end{array}
$$
Then
$$
\sum_{n=1}^{11} f_n(x) =
\begin{cases}
1 & \text{if } x = 1/2, \\
1/2 & \text{if } x=1/3, \\
1/3 & \text{if } x=2/3, \\
1/4 & \text{if } x = 1/4, \\
1/5 & \text{if } x = 3/4, \\
1/6 & \text{if } x = 1/5, \\
1/7 & \text{if } x = 2/5, \\
1/8 & \text{if } x = 3/5, \\
1/9 & \text{if } x = 4/5, \\
1/10 & \text{if } x = 1/6, \\
1/11 & \text{if } x = 5/6, \\
0 & \text{if } x = \text{anything else.}
\end{cases}
$$
At no point do we have $1+\frac 1 2 + \frac 1 3 + \cdots + \frac 1 {11}.$
This sum of $11$ terms differs from the sum of infinitely many terms by $1/12$ if $x=a_{12}$ and by less than that if $x={}$any other number. The biggest discrepancy between the sum of infinitely many terms and the sum of finitely many is always $1/(n+1)$ when $n$ terms have been added.
| {
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How to evaluate an infinite product of idempotent stochastic matrices? How to find the infinite products of the following matrices?
I have a set of $n\times n$ doubly stochastic matrices, all of them are idempotent and when they act on a vector, what they do is to take the average of two consecutive elements in a vector. That is, each matrix is in the form of $I_k\oplus\pmatrix{\frac12&\frac12\\ \frac12&\frac12}\oplus I_{n-k-2}$. For example, when $n=4$, the set comprises of the following members:
$$
\begin{pmatrix}
\frac12 & \frac12 & 0 & 0 \\
\frac12 & \frac12 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{pmatrix},
\ \ \begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & \frac12 & \frac12 & 0 \\
0 & \frac12 & \frac12 & 0 \\
0 & 0 & 0 & 1 \\
\end{pmatrix}
\ \text{ and }\
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & \frac12 & \frac12 \\
0 & 0 & \frac12 & \frac12 \\
\end{pmatrix}.
$$
When they are drawn randomly to form an infinite sequence, the infinite product converges to:
$$
\frac14
\begin{pmatrix}
1&1&1&1\\ 1&1&1&1\\ 1&1&1&1\\ 1&1&1&1
\end{pmatrix} = \frac{1}{4}J_4.
$$
This is for $4$-by-$4$ matrices. By numerical experiments, I saw that for $n$-by-$n$ matrices, we get the same result, i.e. the product is:
$$\frac{1}{n}J_n.$$
Is there any way to prove that? (Although all the matrices are diagonalizable, they are multiplied in random order infinite times, so it is not straightforward how to prove that the result will converge.)
| it's a actually a very simple problem. As a gut check note that your infinite product is bounded -- the product of $k$ stochastic matrices (n x n) has a Frobenius norm $\leq n$ for all natural numbers $k$.
It's enough to prove that
$\mathbf e_j^T\cdot \text{"infinite product"} = \frac{1}{n}\mathbf 1^T$
WP1, for each standard basis vector. $\mathbf e_j$
There's no need to reinvent the wheel here: your problem is just a standard finite state time homogenous markov chain, in disguise.
I.e. for your specific example (with the obvious generalization the the n x n case) what you have is
$P =
\frac{1}{3}\begin{pmatrix}
\frac12 & \frac12 & 0 & 0 \\
\frac12 & \frac12 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{pmatrix} + \frac{1}{3}\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & \frac12 & \frac12 & 0 \\
0 & \frac12 & \frac12 & 0 \\
0 & 0 & 0 & 1 \\
\end{pmatrix}
+ \frac{1}{3}\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & \frac12 & \frac12 \\
0 & 0 & \frac12 & \frac12 \\
\end{pmatrix} = \begin{pmatrix}
\frac{5}{6} & \frac{1}{6} & 0 & 0 \\
\frac{1}{6} & \frac{2}{3} & \frac{1}{6} & 0 \\
0 & \frac{1}{6} & \frac{2}{3} & \frac{1}{6} \\
0 & 0 & \frac{1}{6} & \frac{5}{6} \\
\end{pmatrix}
$
(I took 'they are drawn randomly' to mean uniform at random but this can be tweaked without much complication so long as each probability is $\in (0,1)$)
In effect you just applied total probability by conditioning on some unrelated event (say a roll of dice). To clarify: each transition matrix is applied with probability $\frac{1}{3}$ based on some event (roll of dice) that is independent of past selection and more to the point, it is independent of the current state we are in -- i.e. your selection mechanism preserves the markov property.
$P$ is doubly stochastic, has a single communicating class and is aperiodict (e.g. because there is at least one positive number on the diagonal).
so
$\mathbf e_j^T\cdot \prod_{k=1}^\infty P = \frac{1}{n}\mathbf 1^T$
by standard results from markov chains, renewal theory or Perron Frobenius theory
which completes the proof.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $\sum_{n=1}^∞ \frac{n^{2}}{n^{8}+n^{4}+1}$ Find $\sum_{n=1}^∞ \frac{n^{2}}{n^{8}+n^{4}+1}$
This series is equivalent to $\frac{1}{2} \sum_{n=1}^∞ \frac{1}{n^{4} - n^{2} +1} - \frac{1}{n^{4} + n^{2} +1}$
Them I'm stuck. Any hint$?$
| You properly wrote $$\sum_{n=1}^\infty \frac{n^{2}}{n^{8}+n^{4}+1}=\frac 12\left(\sum_{n=1}^\infty\frac{1}{n^{4} - n^{2} +1}-\sum_{n=1}^\infty\frac{1}{n^{4} + n^{2} +1} \right)$$
Now, write
$$\frac{1}{n^{4} - n^{2} +1}=\frac{1}{(n^2-a)(n^2-b)}=\frac 1 {a-b}\left(\frac 1 {n^2-a}-\frac 1 {n^2-b}\right)$$
$$\frac{1}{n^{4} + n^{2} +1}=\frac{1}{(n^2-c)(n^2-d)}=\frac 1 {c-d}\left(\frac 1 {n^2-c}-\frac 1 {n^2-d}\right)$$
which means that you face a series of
$$\sum_{n=1}^\infty\frac{1}{n^2-k}=\frac{1-(\pi \sqrt{k}) \cot \left(\pi \sqrt{k}\right)}{2 k}$$
For the first summation, you have $a=\frac{1}{2}-\frac{i \sqrt{3}}{2}$ and $b=\frac{1}{2}+\frac{i \sqrt{3}}{2}$. For the second summation $c=-\frac{1}{2}-\frac{i \sqrt{3}}{2}$ and $d=-\frac{1}{2}+\frac{i \sqrt{3}}{2}$. That is to say
$$\sqrt a=\frac{\sqrt{3}}{2}-\frac{i}{2} \qquad \sqrt b=\frac{\sqrt{3}}{2}+\frac{i}{2}\qquad \sqrt c=\frac{1}{2}-\frac{i \sqrt{3}}{2}\qquad \sqrt d=\frac{1}{2}+\frac{i \sqrt{3}}{2}$$
All of the above makes
$$\sum_{n=1}^\infty\frac{1}{n^{4} - n^{2} +1}=\frac{1}{6} \left(\frac{\pi \left(\sqrt{3} \sin \left(\sqrt{3} \pi \right)+3 \sinh
(\pi )\right)}{\cosh (\pi )-\cos \left(\sqrt{3} \pi \right)}-3\right)$$
$$\sum_{n=1}^\infty\frac{1}{n^{4} + n^{2} +1}=\frac{1}{6} \left(\sqrt{3} \pi \tanh \left(\frac{\sqrt{3} \pi }{2}\right)-3\right)$$
$$\color{blue}{\sum_{n=1}^\infty \frac{n^{2}}{n^{8}+n^{4}+1}=\frac{\pi \sqrt 3}{12} \left(\frac{\sin \left(\sqrt{3} \pi \right)+\sqrt{3} \sinh (\pi )+\tanh \left(\frac{\sqrt{3}
\pi }{2}\right) \left(\cos \left(\sqrt{3} \pi \right)-\cosh (\pi )\right) } { \cosh (\pi )-\cos \left(\sqrt{3} \pi \right)}\right)}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the point, if any, the graph of $f(x) = \sqrt{8x^2+x-3}$ has a horizontal tangent line Section 2.5 #14
Find the point, if any, the graph of $f(x) = \sqrt{8x^2+x-3}$ has a horizontal tangent line.
Okay, so having a horizontal tangent line at a point on the graph means that the slope of that tangent line is zero. The derivative of a function is another function that tells us the slope of the tangent line at any given point on the graph of the original function.
Thus, to find where the graph of $f(x) = \sqrt{8x^2+x-3}$ has a horizontal tangent line, we need to take the derivative, set it equal to zero, and solve for $x$. This will give us the $x$-coordinate of where the graph of $f(x)$ has a horizontal tangent line. To find the corresponding $y$ value, we plug the $x$ value that we found into the original equation.
In this problem, when we plug the $x$ value we find into the original equation, we get an imaginary number, which means that no point on the graph of $f(x)$ has a horizontal tangent line, and thus our answer is DNE, does not exist. Let's go through the motions!!!
$f(x) = \sqrt{8x^2+x-3}=(8x^2+x-3)^{1/2}$
$f'(x) = \frac{d}{dx}(8x^2+x-3)^{1/2}$
Time do the chain rule!!!
$$\begin{align}
f'(x) &= \frac{(8x^2+x-3)^{-1/2}}{2}\frac{d}{dx}(8x^2+x-3)\\
&= \frac{(8x^2+x-3)^{-1/2}}{2}(16x+1)\\
&= \frac{(16x+1)}{2(8x^2+x-3)^{1/2}}
\end{align}$$
Alright, we have our derivative. We want to find horizontal tangent lines, so we set this equal to zero and solve for $x$
$$0 = \frac{(16x+1)}{2(8x^2+x-3)^{1/2}}$$
multiplying both sides of the equation by $2(8x^2+x-3)^{1/2}$ we get
$0 = (16x+1)$
And thus $x = \frac{-1}{16}$
Now, we plug this value into the original equation to get the corresponding $y$ value, because remember, we are looking for a point on the graph where the horizontal line is tangent, so our answer will be in $(x,y)$ format, is it exists, (which in this case, it won't)..
$f(\frac{-1}{16}) = \sqrt{8(\frac{-1}{16})^2+\frac{-1}{16}-3}$
But $8(\frac{-1}{16})^2+\frac{-1}{16}-3<0$, so taking its square root will give us an imaginary number. Thus the answer is DNE
| First off, I assume you are working in $\mathbb{R}$, yes? What is the domain of your original function? Since $$f\left(x\right)=\sqrt{8x^{2}+x-3}\text{, the domain is implied: }8x^{2}+x-3\ge 0.$$Solve this to find the domain. Then, when you find the $x$-value of the point where the tangent is $0$, see if that point is in the domain and if it is, you should end up with a real $y$-value, i.e., $y\in\mathbb{R}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the minimum of $abcd+\sqrt{(a^2+1)(b^2+1)(c^2+1)(d^2+1)}$ when $ab+bc+cd+da+ac+bd=6$ If $a,b,c,d\geq0$ are such that $ab+bc+cd+da+ac+bd=6$, then what is the minimum value of $$f(a,b,c,d)=abcd+\sqrt{(a^2+1)(b^2+1)(c^2+1)(d^2+1)}$$ ?
My attempts: I think the minimum $5$ is obtained at $a=b=c=d=1$. I tried the Hölder inequality: $$f(a,b,c,d)\geq abcd + (1+\sqrt{a b c d})^2,$$ but the last term is $1<5$ for $d=0$.
Also, direct AM-GM on the two terms of $f$ did not work either. What to do?
Remark: This problem is from AoPS. There, it is also asked what happens for $ab+bc+cd+da+ac+bd=7$, which seems even harder.
| For $a=b=c=d=1$ we obtain a value $5$.
We'll prove that it's a minimal value.
Indeed, let $a+b+c+d=4u$, $ab+ac+bc+ad+bd+cd=6v^2,$
$abc+abd+acd+bcd=4w^3$ and $abcd=t^4$.
Thus, by AM-GM $v^4\geq t^4,$ $v^2=1$ and we need to prove that:
$$abcd+\sqrt{a^2b^2c^2d^2+\sum\limits_{cyc}a^2b^2c^2+\frac{1}{4}\sum\limits_{sym}a^2b^2+\sum\limits_{cyc}a^2+1}\geq5$$ or
$$t^4+\sqrt{t^8+(16w^6-12v^2t^4)v^2+(36v^4-32uw^3+2t^4)v^4+(16u^2-12v^2)v^6+v^8}\geq5v^4$$ or
$$t^8-10v^4t^4+16v^2w^6+25v^8-32uv^4w^3+16u^2v^6\geq25v^8-10v^4t^4+t^8$$ or
$$(uv^2-w^3)^2\geq0$$ and we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
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For real numbers $x,y$ and $z$ if $9x^2+16y^2+4z^2=108$ and $6xy+4yz+3zx=54$. Find value of $x^2+y^2+z^2$. Question
For real numbers $x,y$ and $z$ if $9x^2+16y^2+4z^2=108$ and $6xy+4yz+3zx=54$. Find value of $x^2+y^2+z^2$.
My calculations
I found
$$(3x+4y+2z)^2=108+4×54=324$$
I have no clue how to proceed further .
But for check of validity of question I let wolfram solve it . But results left me more puzzled .
Now how can I get to this result .
There is one more picture
Now why I don't get same result $x^2+y^2+z^2$
Feel free to edit , comment and advise . Thank You
| $$0=9x^2+16y^2+4z^2-2(6xy+4yz+3zx)=\frac{1}{2}((3x-4y)^2+(3x-2z)^2+(4y-2z)^2),$$
which gives $$3x=4y=2z.$$
Now, let $3x=t.$
Thus, $$t^2+t^2+t^2=108$$ or $$t^2=36,$$ which gives
$$x^2+y^2+z^2=\frac{t^2}{9}+\frac{t^2}{16}+\frac{t^2}{4}=4+\frac{9}{4}+9=\frac{61}{4}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the minimum value of $x$ s.t. $\sqrt{\left(\frac{x+y}{2}\right)^3}+\sqrt{\left(\frac{x-y}{2}\right)^3}=27$ Let $x,y\in \mathbb{R}$ such that $$\sqrt{\left(\frac{x+y}{2}\right)^3}+\sqrt{\left(\frac{x-y}{2}\right)^3}=27$$. Find the minimize value of $x$.
[Edit by Michael Rozenberg] I tried to use AM-GM to retire the radical but failed:
$$27=\sqrt{\left(\frac{x+y}{2}\right)^3}+\sqrt{\left(\frac{x-y}{2}\right)^3}\geq2\sqrt[4]{\left(\frac{x+y}{2}\right)^3\left(\frac{x-y}{2}\right)^3}=$$
$$=2\sqrt[4]{\frac{(x^2-y^2)^3}{64}}=\sqrt[4]{\frac{(x^2-y^2)^3}{4}}.$$
Help me
| Let $\frac{x-y}{2}=a$ and $\frac{x+y}{2}=b$.
Thus, $a\geq0$, $b\geq0$, $x=a+b$ and $$\sqrt{a^3}+\sqrt{b^3}=27.$$
Now, let $f(x)=\sqrt{x^3}.$
Thus, $f$ is a convex function on $[0,+\infty)$ and by Karamata $$27=\sqrt{a^3}+\sqrt{b^3}\leq\sqrt{(a+b)^3}+\sqrt{0^3},$$ which gives
$$x=a+b\geq9.$$
The equality occurs for $b=0$, which says that $9$ is a minimal value.
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality with x,y,z fractions $\frac{x}{y}+\frac{y}{z+x}+\frac{z}{x}\ge 2$ If $x,y,z>0$, show:
$$\frac{x}{y}+\frac{y}{z+x}+\frac{z}{x}\ge 2$$
I expand and to prove
$$x^3 - 2 x^2 y + x^2 z + x y^2 - x y z + y z^2\ge 0$$
I don't know how to do this.
| This might be pretty cumbersome to prove, if you don't see the trick is to sum $1$ in both sides:
$$\frac{x}{y}+\frac{y}{z+x}+\frac{z}{x} + 1\geq 3$$
And this follows directly from AM-GM:
$$\frac{x}{y}+\frac{y}{z+x}+\left(\frac{z}{x} + 1\right) = \frac{x}{y}+\frac{y}{z+x}+\frac{z+x}{x} \geq 3\sqrt[3]{\frac{x}{y}\cdot \frac{y}{z+x}\cdot \frac{z+x}{x}} = 3$$
with equality when $x=y$ and $z = 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral of $\int{\frac{2x-3}{(3x^2-2x+4)\sqrt{x^2-3x+1}}}dx$ I need to solve the following integral:
$$\int{\frac{2x-3}{(3x^2-2x+4)\sqrt{x^2-3x+1}}}dx$$
First I tried factorizing the denominators but it wasn't helpful. Then I thought about completing the square in $3x^2-2x+4$ but nothing came from there. I don't know if there might be any useful trigonometric substitution I could use or maybe some other substitution. I also tried with $t = \sqrt{x^2-3x+1}$ and $t = \frac{1}{x^2-2x+4}$ but also got me nowhere. Thanks for reading and your help.
| $$\int\dfrac{2x - 3}{\left(3x^2 - 2x + 4\right)\sqrt{x^2 - 3x + 1}}\,\mathrm dx\equiv2\int\dfrac{2x - 3}{\left(3x^2 - 2x + 4\right)\sqrt{\left(2x - 3\right)^2 - 5}}\,\mathrm dx$$
Let $u = 2x - 3\implies\mathrm du = 2\,\mathrm dx$. Therefore,
$$\int\dfrac{2x - 3}{\left(3x^2 - 2x + 4\right)\sqrt{\left(2x - 3\right)^2 - 5}}\,\mathrm dx\equiv2\int\dfrac u{\left(3u^2 + 14u + 31\right)\sqrt{u^2 - 5}}\,\mathrm du$$
Let $u = \sqrt 5\sec(v)\implies \mathrm du = \sqrt5\sec(v)\tan(v)\,\mathrm dv$. Therefore,
$$\begin{align}\int\dfrac u{\left(3u^2 + 14u + 31\right)\sqrt{u^2 - 5}}\,\mathrm du&\equiv\int\dfrac{5\sec^2(v)\tan(v)}{\left(15\sec^2(v)+ 14\sqrt5\sec(v) + 31\right)\sqrt{5\sec^2(v) - 5}}\,\mathrm dv \\ &\stackrel{\sec^2(v) = 1 + \tan^2(v)}=\sqrt5\int\dfrac{\sec^2(v)}{15\sec^2(v) + 14\sqrt5\sec(v) + 31}\,\mathrm dv\end{align}$$
Perform tangent half-angle substitution.
$$\int\dfrac{\sec^2(v)}{15\sec^2(v) + 14\sqrt5\sec(v) + 31}\,\mathrm dv\equiv\int\dfrac{\left(1 + \tan^2\left(\frac v2\right)\right)^2}{\left(1 - \tan^2\left(\frac v2\right)\right)^2\left(\frac{15\left(1 + \tan^2\left(\frac v2\right)\right)^2}{\left(1 - \tan^2\left(\frac v2\right)\right)^2} + \frac{14\sqrt5\left(1 + \tan^2\left(\frac v2\right)\right)}{1 - \tan^2\left(\frac v2\right)} + 31\right)}\,\mathrm dv$$
Let $t = \tan\left(\dfrac v2\right)\implies\mathrm dv = \dfrac2{1 + t^2}\,\mathrm dt$. Therefore,
$$\begin{align}&\int\dfrac{\left(1 + \tan^2\left(\frac v2\right)\right)^2}{\left(1 - \tan^2\left(\frac v2\right)\right)^2\left(\frac{15\left(1 + \tan^2\left(\frac v2\right)\right)^2}{\left(1 - \tan^2\left(\frac v2\right)\right)^2} + \frac{14\sqrt5\left(1 + \tan^2\left(\frac v2\right)\right)}{1 - \tan^2\left(\frac v2\right)} + 31\right)}\,\mathrm dv\\&\equiv-\int\dfrac{1 + t^2}{\left(7\sqrt5 - 23\right)t^4 + 16t^2 - 7\sqrt5 - 23}\,\mathrm dt\end{align}$$
Now, factor the denominator and perform partial fraction decomposition.
| {
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How many of the integers between $1$ and $200$ are odd numbers or divisible by $3$ or divisible by $5$? How many of the integers between $1$ and $200$ are odd numbers or divisible by $3$ or divisible by $5$?
\begin{align*}
A_1 & = \left\lfloor{\frac{200}{3}} \right\rfloor = 66 && \text{(divisible by $3$)}\\
A_2 & = \left\lfloor{\frac{200}{5}} \right\rfloor = 40 && \text{(divisible by $5$)}\\
A_3 & = \left\lfloor{\frac{200}{2}} \right\rfloor = 100 && \text{(odd)}\\
| A_1 \cap A_2 | & = \left\lfloor{\frac{200}{3 \cdot 5}}\right\rfloor = 13\\
| A_1 \cap A_3 | & = \left\lfloor{\frac{200}{3 \cdot 2}}\right\rfloor = 33\\
| A_2 \cap A_3 | & = \left\lfloor{\frac{200}{5 \cdot 2}} \right\rfloor= 20\\
| A_1 \cap A_2 \cap A_3 | & = \left\lfloor{\frac{200}{5 \cdot 2 \cdot 3}}\right\rfloor = 6
\end{align*}
Therefore, by the principle exclusion inclusion theorem
$= 66 + 40 + 100- (13 + 33 + 20) + 6 = 146$
Is this logically right?
| In this simple case you don't really need the Inclusion/Exclusion principle $-$ you can do it 'by hand'.
First, count the even numbers between $1$ and $30$ that are divisible by $3$ or $5$: these are $6,10,12,18,20,24,$ and $30$. So there are $7$ of them. (Or you could use Inclusion/Exclusion on this part only: $\frac{30}{6}+\frac{30}{10}-\frac{30}{30}=7$.)
Next, note that this pattern repeats exactly every $30$ numbers; so between $1$ and $210$, there are $49$ of them. Subtract $204$ and $210$ to get $47$ between $1$ and $200$.
Now add the odd numbers to get $147$.
| {
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Closed-form expression for $F(x,y) = \int_0^1 \frac{\sqrt{t(1-t)}}{(t+x)^2 (t+y)} \mathrm{d}t$? I am considering the following function
$$F(x,y) = \int_0^1 \frac{\sqrt{t(1-t)}}{(t+x)^2 (t+y)} \mathrm{d}t,$$
which is well-defined for any $x > 0$ and $y \geq 0$.
Is there a hope to obtain a closed form formula with respect to $x$ and $y$?
For instance, according to Mathematical, we have that
$$F(x,0) = \int_0^1 \frac{\sqrt{t(1-t)}}{(t+x)^2 t} \mathrm{d}t = \frac{\pi}{x \sqrt{x (x+2)}}.$$
Remark: To give a bit of context, the function $F$ appears when I consider the quadratic optimization problem of the form $\min_{\mathbf{x} \in \mathrm{R}^N} \lVert \mathbf{A} \mathbf{x} - \mathbf{y} \rVert_2^2 + \lambda \lVert \mathbf{x} \rVert_2^2$ and I try to understand the behavior of $\lVert \widehat{\mathbf{x}} - \mathbf{x}_0 \rVert_2^2$ with $\widehat{\mathbf{x}}$ the unique optimizer and $\mathbf{x}_0$ the vector we aim at recovering, with $\mathbf{y} = \mathbf{A} \mathbf{x}_0 + \mathbf{n} \in \mathbb{R}^M$ and $\mathbf{n}$ an i.i.d. Gaussian vector. The values $x$ and $y$ above appear as functions of $\lambda$ and $\gamma = \lim M/N$ when $N\rightarrow \infty$ when the matrix $\mathbf{A}$ is i.i.d. Gaussian and its spectrum behaves according to the Marchenko-Pastur law.
| NoName's approach is more elegant, but the integral can be evaluated using contour integration.
Consider the complex function $$g(z) = \frac{\sqrt{z} \sqrt{z-1}}{(z+x)^{2}(z+y)}$$ where $x$ and $y$ are positive parameters, $y \ne x$ , and $0 < \arg(z), \arg(z-1) < 2 \pi.$
The function is well defined on $\mathbb{C} \setminus [0,1]$ and real-valued on the real axis to the right of $z=1$.
Since $g(z) \sim \mathcal{O} \left(\frac{1}{z^{2}}\right) $ as $|z| \to \infty$, the residue of $g(z)$ at complex infinity is $0$.
If we integrate clockwise around a dog-bone/dumbbell contour, we get $$ \begin{align} \int_{0}^{1} \frac{\sqrt{t} \sqrt{(1-t)e^{i \pi}}}{(t+x)^{2}(t+y)} \, \mathrm dt + \int_{1}^{0}\frac{\sqrt{te^{ 2 \pi i}} \sqrt{(1-t)e^{i \pi}}}{(t+x)^{2}(t+y)} \, \mathrm dt &= 2i \int_{0}^{1} \frac{\sqrt{t} \sqrt{1-t}}{(t+x)^{2}(t+y)} \, \mathrm dt \\ &=2 \pi i \left(\operatorname{Res}[g(z), -x] + \operatorname{Res}[g(z), -y] \right), \end{align}$$
where $$\operatorname{Res}[g(z), -y] = \frac{\sqrt{y e^{ i \pi}} \sqrt{(y+1) e^{ i \pi }}}{(-y+x)^{2}} = -\frac{\sqrt{y(1+y)}}{(x-y)^{2}} $$
and $$ \begin{align} \operatorname{Res}[g(z), -x] &= \lim_{ z \to -x} \frac{\mathrm d}{\mathrm dz} \frac{\sqrt{z}\sqrt{z-1}}{z+y} \\ &= \lim_{ z \to -x} \frac{\left(\frac{\sqrt{z-1}}{2 \sqrt{z}} + \frac{\sqrt{z}}{2\sqrt{z-1}}\right)(z+y) - \sqrt{z} \sqrt{z-1} }{(z+y)^{2}} \\ &= \frac{\left(\frac{\sqrt{(x+1)e^{ i \pi}}}{2 \sqrt{xe^{ i \pi}}} + \frac{\sqrt{xe^{ i \pi}}}{2\sqrt{(x+1)e^{ i \pi}}}\right)(-x+y) - \sqrt{xe^{i \pi}} \sqrt{(x+1)e^{ i \pi}} }{(-x+y)^{2}}\\ &= \frac{\left(\frac{\sqrt{x+1}}{2 \sqrt{x}} + \frac{\sqrt{x}}{2 \sqrt{x+1}} \right)(y-x) + \sqrt{x}\sqrt{x+1}}{(x-y)^{2}} \\ &=\frac{(2x+1)(y-x) + 2x(1+x) }{2 \sqrt{x(1+x)}(x-y)^{2}} \\&= \frac{x+y+2xy}{2 \sqrt{x(1+x)}(x-y)^{2}}. \end{align} $$
Therefore, $$\int_{0}^{1} \frac{\sqrt{t} \sqrt{1-t}}{(t+x)^{2}(t+y)} \, \mathrm dt = \frac{\pi}{2} \frac{x+y+2xy-2 \sqrt{xy(1+x)(1+y)}}{\sqrt{x(1+x)}(x-y)^{2}}. $$
The result should hold for $y=0$, but the case $y=x$ has to be done separately.
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"language": "en",
"url": "https://math.stackexchange.com/questions/3547135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Prove $\sum_{k=1}^r\frac{1}{k+r}=\sum_{k=1}^{2r}\frac{(-1)^{k-1}}{k}$ How to prove that
$$\sum_{k=1}^r\frac{1}{k+r}=\sum_{k=1}^{2r}\frac{(-1)^{k-1}}{k}\tag1$$
We know that both sides are equal to $H_{2r}-H_r$ but I am trying to convert the left side to the right side using only series manipulations and without going through $H_{2r}-H_r$
There is nothing I could try but we know that
$$\sum_{k=1}^r\frac{1}{k+r}=\sum_{k=1+r}^{2r}\frac{1}{k}$$
What next? Thank you.
By the way, its easy to prove it using integration,
$$\sum_{k=1}^r\frac{1}{k+r}=\int_0^1\sum_{k=1}^r x^{k+r-1}\ dx=\int_0^1\frac{x^r-x^{2r}}{1-x}\ dx$$
$$=\int_0^1\frac{1-x^{2n}-(1-x^r)}{1-x}\ dx=H_{2r}-H_r\tag2$$
and we proved here
$$\overline{H}_{2r}=\sum_{k=1}^{2r}\frac{(-1)^{k-1}}{k}=H_{2r}-H_r\tag3$$
Hence by $(2)$ and $(3)$ , $(1)$ is proved
| As has been pointed out by @Greg Martin, one way to get the result you desire using only manipulations of series would be to essentially run the proof of the result I gave here in reverse.
Starting with $\sum_{k = 1}^r \frac{1}{k + r}$ reindexing $k \mapsto k - r$ we have
\begin{align}
\sum_{k = 1}^r \frac{1}{k + r} &= \sum_{k = r + 1}^{2r} \frac{1}{k}\\
&= \sum_{k = 1}^{2r} \frac{1}{k} - \sum_{k = 1}^n \frac{1}{k}\\
&= \left (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{2r - 1} + \frac{1}{2r} \right ) - \left (1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{r} \right )\\
&= 1 + \left (\frac{1}{2} - 1 \right ) + \frac{1}{3} + \left (\frac{1}{4} - \frac{1}{2} \right ) + \frac{1}{5} + \left (\frac{1}{6} - \frac{1}{3} \right ) + \cdots\\
&\qquad \cdots + \frac{1}{2n - 1} + \left (\frac{1}{2n} - \frac{1}{n} \right )\\
&= 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2n - 1} - \frac{1}{2n}\\
&= \sum_{k = 1}^{2r} \frac{(-1)^{k + 1}}{k},
\end{align}
as desired.
Of course this is just $H_{2r}$ and $H_r$ in disguise. So let us try again starting from the other side.
\begin{align}
\sum_{k = 1}^{2r} \frac{(-1)^{k + 1}}{k} &= 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots + \frac{1}{2r - 1} - \frac{1}{2r}\\
&= \left (1 - \frac{1}{2} \right ) + \left (\frac{1}{3} - \frac{1}{4} \right ) + \cdots + \left (\frac{1}{2r - 1} - \frac{1}{2r} \right )\\
&= \left (1 + \frac{1}{2} - 2 \cdot \frac{1}{2} \right ) + \left (\frac{1}{2} + \frac{1}{4} - 2 \cdot \frac{1}{4} \right ) + \cdots\\
& \qquad \cdots + \left (\frac{1}{2r - 1} + \frac{1}{2r} - 2 \cdot \frac{1}{2r} \right )\\
&= \left (1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2r - 1} + \frac{1}{2r} \right ) - 2 \left (\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2r} \right )\\
&= \left (1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{2r - 1} + \frac{1}{2r} \right ) - \left (1 + \frac{1}{2} + \cdots + \frac{1}{r} \right )\\
&= \frac{1}{r + 1} + \frac{1}{r + 2} + \cdots + \frac{1}{2r}\\
&= \sum_{k = 1}^r \frac{1}{k + r},
\end{align}
which is essentially the same thing.
Finally, this identity is known as the Botez-Catalan identity.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Figuring out the formula of a recurrence relation I've got a recurrence relation which I'm stuck on figuring out the formula for:
The relation is: $T(n) = 2 T(n/4) + 1$ where $T(1) = 1$
So for my tests, I did:
$T(4) = 2 T(4/4) + 1 = 2 T(1) + 1 = 2\times1+1 = 3$
$T(16) = 2 T(16/4) + 1 = 2 T(4) + 1 = 2\times (2\times1+1) +1 = 7$
$T(64) = 2 T (64/4) + 1 = 2 T(16) + 1 = 2\times ( 2\times (2\times1+1) +1 )+1 = 15$
So I've noticed that by writing out all the multiplications, the number of $2$'s and number of $1$'s seems to be equal to the power of $4$, i.e. at $T(64)$, equivalent to $4^3$ , there are $3$ lots of $2$'s and $1$'s in the multiplication (not including the $*1$ which doesn't change the answer anyway) so I thought that the formula may then be:
$T(4^n ) = 2n + 1n$
But when I tried it out it didn't seem to work, I've also noticed the actual answers go up by the answer to the previous relation $ + 4* $the power of the previous $n$, e.g. at:
$T(16)$, the answer is $7$, answer to the previous is $3$, power of the previous is $1$ $(4^1 ) 3 + 4x1 = 7$
$T(64)$, the answer is $15$, answer to the previous is $7$, power of the previous is $2$ $(4^2 ) 7 + 4*2 = 15$
| We will develop the recurrence:
$T(n) = 2T(n/4) + 1$
$T(4n) = 2T(n) + 1$
$T(16n) = 2T(4n) + 1$ $\to T(16n) = 2[2t(n)+1] + 1 = 4T(n) + 2 + 1$
$T(64n) = 2T(16n) + 1$ $\to T(64n) = 2[4t(n)+2 + 1] + 1 = 8T(n) + 4 + 2 + 1$
$...$
$T(4^pn) = 2T(4^{p-1}n) + 1$ $\to T(4^pn) = 2^pT(n) + \sum_{i=0}^p 2^i$
$\to T(4^pn) = 2^pT(n) + 2^p - 1$
but it was given that $T(1) = 1$ so:
$\to T(4^p) = 2^p*1 + 2^p - 1$
$\to T(4^p) = 2^{p+1} - 1$ (I)
doing: $z = 4^p$ we have to:
$2p = log_2 z$ $\to p = \frac{1}{2}log_2 z $
replacing in (I) have to:
$T(z) = 2^{\frac{1}{2}log_2 z + 1} - 1$
$T(z) = 2*2^{\frac{1}{2}log_2 z} - 1$
$T(z) = 2\sqrt{z} - 1$
since z is a generic variable, we finally return to:
$T(n) = 2\sqrt{n} - 1$
I hope this may have helped.
| {
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"url": "https://math.stackexchange.com/questions/3551232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
maximum value of $ac+bd$
If $a,b,c,d\in \mathbb{R}$ and $a^2+b^2\leq 2$ and $c^2+d^2\leq 4.$ Then maximum value of $ac+bd$ is
what i try
$(ac+bd)^2+(ad-bc)^2=(a^2+b^2)(c^2+d^2)$
$\underbrace{(ac+bd)^2}_{\max}=\underbrace{(a^2+b^2)(c^2+d^2)}_{\max}-\underbrace{(ad-bc)^2}_{\min}$
$(ac+bd)^2\leq 8\Rightarrow (ac+bd)\leq 2\sqrt{2}$
but answer is $3$
How do i get right answer help me please
| $2 \sqrt2 $ is the correct maximum value.The value $3$ which is larger than $2 \sqrt2 $ is never attained. The bound $2 \sqrt 2$ can also be obtained by Cauchy - Schwarz inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3552345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove if $5 \nmid n$, then $n^2 = 5k \pm 1$ I am having difficulty finishing this proof. At first, the proof is easy enough. Here's what I have thus far:
Because $5 \nmid n$, we know $\exists q \in \mathbb{Z}$ such that $$n = 5q + r$$ where $0 < r < 4$. Note $r \neq 0$ because if $r = 0$, then $5 \mid n$. Also note that $n^2 = 25q^2 + 10qr + r^2$. Then we have four cases: when $r=1$, $r=2$, $r = 3$, and $r = 4$. This is where I run into difficulty. In each of these cases, we can prove that either $n^2 = 5k + 1$ or $n^2 = 5k - 1$ for some integer $k$, but I cannot see how to prove both for each case. Any ideas?
As a side note on how I went to prove each case, I simply plugged $r$ into the formula $n^2 = 25q^2 + 10qr + r^2$. This results in $n^2 = 25q^2 + 10q + 1$.
Continuing, we get $n^2 = 5(5q^2 + 2q) + 1$, and because $5q^2 + 2q$ is still an integer, this is of the form $n^2 = 5k + 1$ for some integer $k$. But I cannot find how to make the 1 a negative to prove both cases.
| As you showed, $n^2=25q^2+10qr+r^2$, where $r=1, 2, 3,$ or $4$.
When $r=1$, $n^2=25q^2+10qr+1=5(5q^2+2qr)+1=5k+1.$
When $r=2$, $n^2=25q^2+10qr+4=5(5q^2+2qr+1)-1=5k-1.$
When $r=3$, $n^2=25q^2+10qr+9=5(5q^2+2qr+2)-1=5k-1.$
When $r=4$, $n^2=25q^2+10qr+16=5(5q^3+2qr+3)+1=5k+1.$
(Note: I did not mean to imply that $k$ in one equation is the same as $k$ in another.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3553233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Roots of the equation $(x^2+3x+4)^2+3(x^2+3x+4)+4=x$ The equation $(x^2+3x+4)^2+3(x^2+3x+4)+4=x$ has
(A) all its solution real but not all positive
(B) only two of its solution real
(C) two of its solution positive and two negative
(D) none of its solution real
My approach is as follow $(x^2+3x+4)^2+3(x^2+3x+4)+4-x=y$
$f(x)=y=x^4+6x^3+20x^2+32x+32$
$f(-x)=y=x^4-6x^3+20x^2-32x+32$
Using Descartes rule no positive roots but the possible ways we can have either 4,2,0 negative roots.
$f'(x)=4x^3+18x^2+40x+32$
$f''(x)=12x^2+36x+40$ which is imaginary
From here I am not able to approach.
| Hint (edited): Let $f(x) = x^2+3x+4$. Then $f(x) > x $ for all reals $x$, hence $f(f(x)) > f(x) > x$ for all reals $x$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $x+y$ and $y+z$ are even, prove $x+z$ is even I'm stuck on a practice problem for an upcoming midterm. Can someone tell me if I'm starting off on the right foot here?
Q: $x,y,z \in \mathbb{Z}$. $x + y$ and $y + z$ are even. Prove that $x + z$ is even.
I thought it would be easier to prove using the contrapositive so I am proving that if $x+z$ is odd, then either $x+y$ or $y+z$ are odd.
By the definition of an odd number, $x+z = 2n + 1, n \in \mathbb{Z}$.
Case 1 : suppose x is even. $x = 2a, a \in\mathbb{Z}$.
Then $x+z = 2(a+b) + 1 = (2a) + (2b + 1)$, where $n = a+b, a,b \in\mathbb{Z}$.
This means $z$ must be odd.
The same logic holds for $x+y$, meaning that $y$ must also be odd.
Therefore, $y = 2c + 1$.
It follows that $y+z = 2b +1 + 2c + 1 = 2d$, where $d = b+c+1$.
Case 2 : suppose $x$ is odd. $x = 2a +1, a \in\mathbb{Z}$.
Then $x+z = 2(a+b) + 1 = (2a +1) + (2b)$, where $n = a+b, a,b\in\mathbb{Z}$.
This means $z$ must be even.
The same logic holds for $x+y$, meaning that $y$ must also be even.
Therefore, $y = 2c$.
It follows that $y+z = 2b + 2c = 2d$, where $d = b+c$.
I just don't think this proves that one of the two expressions is odd. Any help would be much appreciated!
| Suppose $x+y$ and $y+z$ are even. Then $x+y=2a$ and $y+z=2b$ for some $a,b\in\Bbb Z$, so that
$$\begin{align}
x+z&=(x+y)+(y+z)-2y\\
&=2a+2b-2y\\
&=2(a+b-y).
\end{align}$$
But $a+b-y\in \Bbb Z$. Thus $x+z$ is even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3554238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
$z^2 = -1$ Why what i do is wrong? I want to calculate $z^2 = -1$ i think i should get to $z = i$.
$-1 = -1+0i = 1(\cos(2\pi k)+i\sin(2\pi k)), k \in Z$
$z^2 = r^2(\cos(2 \theta) + i\sin(2\theta))$
So i get:
$$r^2(\cos(2 \theta) + i\sin(2\theta)) = 1(\cos(2\pi k)+i\sin(2\pi k))$$
Namely:
$r = 1, 2 \theta = 2 \pi k \Rightarrow \theta = \pi k$
So for $k = 0$ i get $z_0 = 1$ for $k=1: z_1 = -1$
What i do wrong?
| It should be
$-1 = -1+0i = 1(\cos(\pi(2 k+1))+i\sin(2\pi(2 k+1))), k \in Z$
$z^2 = r^2(\cos(2 \theta) + i\sin(2\theta))$
So $r^2(\cos(2 \theta) + i\sin(2\theta)) = 1(\cos(\pi(2 k+1))+i\sin(2\pi (2k+1)))$
Namely, $r = 1, 2 \theta = \pi(2 k+1)\Rightarrow \theta = \pi (k+\frac12)$.
So for $k = 0$ you get $z_0 = i$, and, for $k=1, z_1 = -i$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3555035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Minimum value of $f(a,b) = a^2+ab+b^2-3a-6b+11$
Minimum value of
$f(a,b) = a^2+ab+b^2-3a-6b+11$ for all $a,b\in \mathbb{R}$
what i try
Let $$k=a^2+ab+b^2-3a-6b+11$$
$$k=a^2+(b-3)a+b^2-6b+11$$
$\displaystyle k=\bigg[a^2+(b-3)a+\frac{(b-3)^2}{4}+b^2-6b+11-\frac{(b-3)^2}{4}\bigg]$
$$\displaystyle k = \bigg[\bigg(a+\frac{b-3}{2}\bigg)^2+3b^2-18b+35\bigg]$$
$$k=\bigg[\bigg(a+\frac{b-3}{2}\bigg)^2+3(b-3)^2+8\bigg]\geq 8$$
but answer is $20$
Help me please
| $$a^2+ab+b^2-3a-6b+11=b^2+(a-6)b+a^2-3a+11=$$
$$=\left(b+\frac{a-6}{2}\right)^2+a^2-3a+11-\frac{(a-6)^2}{4}=$$
$$=\left(b+\frac{a-6}{2}\right)^2+\frac{3a^2+8}{4}\geq2.$$
The equality occurs for $a=0$ and $b=3,$ which says that we got a minimal value.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that this sum of products of binomial coefficient can be simplified like this $a , b, c$ are positive integers such that ${a}\geq{b+c}$.
Prove that $$\sum_{i=0}^{b}{\binom{a}{i+c}\binom{b}{i}}=\binom{a+b}{b+c}$$
| $\binom{a}{i+c}$ is the coefficient of $x^{i+c}$ from $(x+1)^{a}$ while $\binom{b}{i}$ is the coefficient of $\frac{1}{x^{i}}$ from $(\frac{1}{x}+1)^{b}$.
$\sum_{i=0}^{b}\binom{a}{i+c}\binom{b}{i}$ is the coefficient of $x^{c}$ from $(x+1)^{a}(\frac{1}{x}+1)^b=\frac{(x+1)^{a+b}}{x^{b}}$ which is $\binom{a+b}{b+c}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3557249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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An indeterminate limit form of infinity/infinity I am trying to solve the limit:
$$\lim_{x\to\infty}x^\frac{5}{3}\left(\left(x+\sin\left(\frac{1}{x}\right)\right)^\frac{1}{3}-x^\frac{1}{3}\right)$$
I was trying to find a way to bring it into a fraction form to apply L'Hospital's rule, and I tried using
$$a-b=\frac{a^3-b^3}{a^2+ab+b^2}$$
But it made it even more complex and after applying L'Hospital's rule I got stuck with all the terms. Is there a smarter way to evaluate it?
| I think using the difference of cubes should work:$$\lim_{x\to\infty}x^\frac{5}{3}\left(\left(x+\sin\left(\frac{1}{x}\right)\right)^\frac{1}{3}-x^\frac{1}{3}\right)=\\
\lim_{x\to\infty}\frac{x^\frac{5}{3}\sin\frac{1}{x}}{\left(x+\sin\left(\frac{1}{x}\right)\right)^\frac{2}{3}+\left(x+\sin\frac{1}{x}\right)^{\frac{1}{3}}x^\frac{1}{3}+x^\frac{2}{3}}=\\
\lim_{x\to\infty}\frac{x^\frac{5}{3}x^{-1}}{\left(x+x^{-1}\right)^\frac{2}{3}+\left(x+x^{-1}\right)^\frac{1}{3}x^\frac{1}{3}+x^\frac{2}{3}}=\\
\lim_{x\to\infty}\frac{x^\frac{2}{3}}{x^\frac{2}{3}+x^\frac{2}{3}+x^\frac{2}{3}}=\frac{1}{3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3559942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove the three lines are concurrent.
Let $O$ be the circumcenter of $\triangle ABC$ with $\angle
A=60^{\circ}$, $P$ be an arbitary point on the circumcircle of
$\triangle BOC$, and $D,E,F$ be the circumcenters of $\triangle
BPC,\triangle CPA, \triangle APB$ respectively. Prove $AD,BE,CF$ are
concurrent.
Some intermediate results:
*
*$AD$ bisects $\angle BAC$,and $OD \perp BC$;
*$O,P$ are isogonal conjugate points of $\triangle DEF$.
| Let us denote by $A',B',C$ respectively the intersections of $AD$, $BE$, $CF$ with the sides $BC$, $CA$, $AB$.
We have also denoted by $a_1,a_2;b_1,b_2,c_1,c_2$ the (lengths of the) angles delimited by lines $AA'D$, $BB'E$, $CC'F$ from $\hat A, \hat B,\hat C$.
Then we have the relations:
$$
\begin{aligned}
a_1+a_2 &=\hat A =60^\circ\ ,\\
b_1+c_2 &=180^\circ-\widehat{BPC}=180^\circ-\widehat{BOC}=180^\circ-2\hat A
% =180^\circ-120^\circ\\ &
=60^\circ\ ,\\
b_2+c_1
&= 180^\circ-a_1-a_2-b_1-c_2 =60^\circ\ .
\end{aligned}
$$
Then working in the circle $(F)=(ABP)$ we have
$$
\widehat{AFB}
=\overset\frown{APB}
=\overset\frown{AP}
+\overset\frown{PB}
=2b_2+2a_1\ .
$$
So $\widehat{FAB}=90^\circ-\frac 12\widehat{AFB}=90^\circ-a_1-b_2$, giving
$\widehat{FAC}=90^\circ+a_2-b_2$.
Similarly,
$\widehat{FBC}=(90^\circ-a_1-b_2)+(b_1+b_2)=90^\circ+b_1-a_1$.
Let us find a formula for the proportion $C'A:C'B$ using (only) the above data.
$$
\begin{aligned}
\frac{C'A}{C'B}
&=
\frac
{\operatorname{Area}AFC}
{\operatorname{Area}BFC}
=
\frac
{AF\cdot AC\cdot \sin \widehat{FAC}}
{BF\cdot BC\cdot \sin \widehat{FBC}}
=
\frac
{AC\cdot \sin (90^\circ+a_2-b_2)}
{BC\cdot \sin (90^\circ+b_1-a_1)}
\ .
\\[3mm]
&\qquad\text{ Similarly:}
\\[3mm]
\frac{B'A}{B'C}
&=
\frac
{\operatorname{Area}AEB}
{\operatorname{Area}CEB}
=
\frac
{AE\cdot AB\cdot \sin \widehat{EAB}}
{CE\cdot CB\cdot \sin \widehat{ECB}}
=\frac
{AB\cdot \sin (90^\circ+a_1-c_1)}
{CB\cdot \sin (90^\circ+c_2-a_2)}
\ .
\end{aligned}
$$
As already observed in the OP, the point $D$ is on the angle bisector of $\hat A$, which gives the third needed proportion $D'A:D'B$. Putting all together, and with omission of signs,
$$
\begin{aligned}
\frac{B'C}{B'A}\cdot
\frac{C'A}{C'B}\cdot
\frac{A'B}{A'C}
&=
\frac
{BC\cdot \sin (90^\circ+c_2-a_2)}
{AB\cdot \sin (90^\circ+a_1-c_1)}
\cdot
\frac
{AC\cdot \sin (90^\circ+a_2-b_2)}
{BC\cdot \sin (90^\circ+b_1-a_1)}
\cdot
\frac
{AB}
{AC}
\\
&=
\frac
{\sin (90^\circ+c_2-a_2)}
{\sin (90^\circ+a_1-c_1)}
\cdot
\frac
{\sin (90^\circ+a_2-b_2)}
{\sin (90^\circ+b_1-a_1)}
\\
&=1\ ,
\end{aligned}
$$
because we can express all angles in terms of $a_1,b_1,c_1$, and then
$$
\begin{aligned}
c_2-a_2=(60^\circ-b_1)-(60^\circ-a_1)=a_1-b_1\ ,\\
a_2-b_2=(60^\circ-a_1)-(60^\circ-c_1)=c_1-a_1\ ,
\end{aligned}
$$
(and since the two values $\sin(90\pm ?)$ coincide,) we obtain the cancellations of sine factors.
Now use the reciprocal of the theorem of Ceva to obtain $AA'$, $BB'$, $CC'$ concurrent.
$\square$
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If $a > b$ and $x > 2$ show $\frac{2^{a+1}-1}{2^a}\frac{x^{b+1}-1}{x^b(x-1)} > \frac{2^{b+1}-1}{2^b}\frac{x^{a+1}-1}{x^a(x-1)}$ Given integers $a > b \geq 0$ and $p>2$, show that the following inequality holds
$$\dfrac{2^{a+1}-1}{2^a}\dfrac{x^{b+1}-1}{x^b(x-1)} > \dfrac{2^{b+1}-1}{2^b}\dfrac{x^{a+1}-1}{x^a(x-1)}$$
This is trivial if $b=0$ as we'll end up with the inequality
$$\dfrac{2^{a+1}-1}{2^a} > \dfrac{x^{a+1}-1}{x^a(x-1)}$$
which can be shown using a variety of methods such as taking the derivative of both sides. I want to show that this inequality holds for all values of $a,b,x$ which I am able to plot out trivially, however I get stuck down the line.
Here is what I've tried so far:
$$\dfrac{2^{a+1}-1}{2^a}\dfrac{x^{b+1}-1}{x^b(x-1)} > \dfrac{2^{b+1}-1}{2^b}\dfrac{x^{a+1}-1}{x^a(x-1)}\equiv\dfrac{2^{a+1}-1}{2^a}\dfrac{x^{b+1}-1}{x^b} > \dfrac{2^{b+1}-1}{2^b}\dfrac{x^{a+1}-1}{x^a}$$
then we cross multiply to find
$$
\dfrac{2^b}{2^{b+1}-1}\dfrac{2^{a+1}-1}{2^a} > \dfrac{x^b}{x^{b+1}-1}\dfrac{x^{a+1}-1}{x^a} \equiv \dfrac{2^{a+b+1}-2^b}{2^{a+b+1}-2^a}>\dfrac{x^{a+b+1}-x^b}{x^{a+b+1}-x^a}
$$
but from here I find myself stuck without a direction forward. Any help would be appreciated!
| $
f(x)=\dfrac{x^{a+b+1}-x^b}{x^{a+b+1}-x^a}=\dfrac{1-\frac{1}{x^{a+1}}}{1-\frac{1}{x^{b+1}}}\\
f'(x)=\dfrac{\frac{a+1}{x^{a+2}}-\frac{b+1}{x^{b+2}}-\frac{a-b}{x^{a+b+3}}}{\left(1-\frac{1}{x^{b+1}}\right)^2}=\dfrac{(a+1)x^{b+1}-(b+1)x^{a+1}-a+b}{x^{a+b+3}\left(1-\frac{1}{x^{b+1}}\right)^2}\\
g(x)=(a+1)x^{b+1}-(b+1)x^{a+1}-a+b\\
g(1)=0\\g'(x)=(a+1)(b+1)(x^b-x^a)\\
x>1\,\Rightarrow\,g'(x)<0\,\Rightarrow\,g(x)<0\,\Rightarrow\,f'(x)<0\,\Rightarrow\,f(2)>f(x)
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3561551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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if $x$ is odd, show that $x^3+x$ has a remainder 2 when divided by 4 I did part of this question but am stuck and don't know how to continue
I let $x= 2k +1$
Also noticed that $x^3+x = x(x^2+1)$
therefore
$4m+2 = 2k+1((2k+1)^2+1)$
I simplified this and ended up with
$4m+2 = 8k^3+12k^2+8k+2$
I don't know how to continue from and prove that $x^3+x$ has remainder 2 when divided by 4
| Consider $f(x) =x^3+x-2$ instead. This can be factorised as $(x-1)(x^2+x+2)$ instead using the remainder theorem ($1^3+1-2 = 0$).
If $x = 1 \pmod 2$, then $x -1 \equiv 0 \pmod 2$ and $x^2+x+2 \equiv 1^2+1+2 \equiv 4 \equiv 0 \pmod 2$. Therefore $x^3+x-2 \equiv 0 \pmod 4$, so $x^3+x \equiv 2 \pmod 4$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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System of logarithmic and exponential equations $\log_{7}(x^2-x+1)=\log_{2}(y^2-1)-\log_{2}(x+1)$ Solve the following system in $\mathbb{R}$.
$$\log_{7}(x^2-x+1)=\log_{2}(y^2-1)-\log_{2}(x+1)$$
$$\log_{7}(y^2-y+1)=\log_{2}(z^2-1)-\log_{2}(y+1)$$
$$\log_{7}(z^2-z+1)=\log_{2}(x^2-1)-\log_{2}(z+1)$$
I considered the functions $f,g:[1,\infty]\longrightarrow \mathbb{R}$
$f(x)=\log_{7}(x^2-x+1)+\log_{2}(x+1) $ and $g(x)=\log_{2}(x^2-1)$ both being increasing on $[1,\infty]$
And so the system is equivalent with:
$f(x)=g(y) \\ f(y)=g(z)\\f(z)=g(x)$
Suppose that $x \geq y$ then $f(x) \geq f(y) \Leftrightarrow $ $g(y) \geq g(z) \Leftrightarrow y \geq z $ then $f(y) \geq f(z) \Leftrightarrow $ $g(z) \geq g(x) \Leftrightarrow z \geq x $ this implies $x \geq y \geq z \geq x$ so $x=y=z$
Therefore the system can be rewritten as
$\log_{7}(x^2-x+1)=\log_{2}(x-1)=a$
So I have a new system
$x^2-x+1=7^a \\ x-1=2^a$
But i don't know how to prove that the only solution of this system is $x=3$ , any ideas?
| Substitute $x=2^a+1$ in the first equation to get:
$$4^a+2^a+1=7^a\implies \left(\frac{4}{7}\right)^a+\left(\frac{2}{7}\right)^a+\left(\frac{1}{7}\right)^a=1$$
Clearly, the left hand side is decreasing, so we can have at most one solution, and that is $a=1$, which gives $x=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3563491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Minimum distance between a surface and a point I created this problem which is inspired by some problems I encountered in this site.
What is the minimum distance between point $(1,1,1)$ and a surface defined by $z-xy=10$?
The answer is $\sqrt{18}$ but I am curious how others solve this kind of problem.
| We want to minimize
$$\sqrt{(x-1)^2+(y-1)^2+(xy+10-1)^2}$$
which can be written as
$$\begin{align}&\sqrt{x^2-2x+1+y^2-2y+1+x^2y^2+18xy+81}
\\\\&=\sqrt{(y^2+1)x^2+(18y-2)x+y^2-2y+83}
\\\\&=\sqrt{(y^2+1)\bigg(x^2+\frac{18y-2}{y^2+1}x\bigg)+y^2-2y+83}
\\\\&=\sqrt{(y^2+1)\bigg(\bigg(x+\frac{9y-1}{y^2+1}\bigg)^2-\bigg(\frac{9y-1}{y^2+1}\bigg)^2\bigg)+y^2-2y+83}
\\\\&=\sqrt{(y^2+1)\bigg(x+\frac{9y-1}{y^2+1}\bigg)^2-\frac{(9y-1)^2}{y^2+1}+y^2-2y+83}
\\\\&=\sqrt{(y^2+1)\bigg(x+\frac{9y-1}{y^2+1}\bigg)^2+\frac{(y^2-y-8)^2}{y^2+1}+18}\end{align}$$
So, the minimum distance is $\color{red}{3\sqrt{2}}$ which is attained when
$$x+\frac{9y-1}{y^2+1}=y^2-y-8=0\quad \text{and}\quad z=xy+10$$
i.e.
$$(x,y)=\bigg(\frac{1\pm\sqrt{33}}{2},\frac{1\mp\sqrt{33}}{2},2\bigg)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3563960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The Theorems of Ceva and Menelaus
In the parallelogram $ABCD$ $M\in AB, AM:MB=2:7$ and $N\in BC, BN:NC=4:5$. If $DM\cap AN=O$, calculate $AO:ON;DO:OM$.
For the first ratio let $BC \cap DM=Q$. By Menelaus' theorem $\dfrac{AO}{ON}\cdot\dfrac{NQ}{QB}\cdot\dfrac{BM}{MA}=1 \Rightarrow \dfrac{AO}{ON}=\dfrac{BQ}{QN}\cdot\dfrac{AM}{BM}$. If we look at the similar triangles $\triangle BMQ \sim AMD \Rightarrow \dfrac{BQ}{AD}=\dfrac{BQ}{BC}=\dfrac{BM}{AM}=\dfrac{7}{2}$ or $BQ=\dfrac{7}{2}BC$. Now, $\dfrac{AO}{ON}=\dfrac{BQ}{QN}\cdot\dfrac{AM}{BM}=\dfrac{\dfrac{7}{2}BC}{BQ+BN}\cdot\dfrac{2}{7}=\dfrac{\dfrac{7}{2}BC}{\dfrac{7}{2}BC+\dfrac{4}{9}BC}\cdot\dfrac{2}{7}=\dfrac{63}{71}\cdot\dfrac{2}{7}=\dfrac{18}{71}$. Thales theorem tells us that
$\dfrac{AO}{ON}=\dfrac{DO}{OQ}=\dfrac{18}{71}$.
How to find $DO:OM$?
|
Let [.] denote areas. Since the triangles ADN and AMN share the same base AN,
$$\frac{DO}{OM} = \frac{[ADN]}{[AMN]}
= \frac{\frac12[ABCD]}{\frac29[ABN]} =
= \frac{\frac12[ABCD]}{\frac29\cdot\frac4{2\cdot9}[ABCD]} = \frac{81}{8}$$
Similarly,
$$\frac{AO}{ON} = \frac{[DAM]}{[DNM]}
= \frac{[DAM]}{[ABCD]-[DAM]-[NBM]-[DNC] } $$
$$=
= \frac{\frac2{2\cdot9}[ABCD]}
{(1-\frac2{2\cdot9}- \frac7{2\cdot9}\cdot\frac49- \frac5{2\cdot9})[ABCD]} = \frac{18}{71}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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System of equations $x^3+y=y^2\ \& \ y^3+z=z^2\ \& \ z^3+x=x^2$
Solve over reals:
$$ \begin{cases} x^3+y=y^2\\ y^3+z=z^2\\ z^3+x=x^2\\ \end{cases} $$
I think the only solution is $x=y=z=0$. If the variables are non-zero multiplying the equations:
$$(x-1)(y-1)(z-1)=x^2y^2z^2 > 0$$
Here, there are two cases: all three variables are greater than $1$ or only one.
If $x,y,z>1$, it's easy to prove there are no solutions because $y^2>x^3$ a.s.o. gives $xyz<1$.
How can I prove there are no solutions when only variable, say $x$, is greater than $1$?
| I will prove by contradiction that there exists no solutions for $x > 1$ and $y,z < 1$. Suppose $(x,y,z)$ is such a solution.
Iteration #1: First, note that:
$$
z^3 = x^2 - x = x(x - 1) > 0 \implies z > 0
$$
Thus, $0 < z < 1$. We see that:
$$
y^3 = z^2 - z = z(z - 1)
$$
For $0 < z < 1$, $-\frac{1}{4} < z(z - 1) < 0$, so $-2^{-\frac{2}{3}} < y < 0$. For this range of values of $y$, since $y(y - 1)$ is decreasing for $y < 0$, we have that:
$$
x^3 = y^2 - y < 2^{-\frac{4}{3}} + 2^{-\frac{2}{3}} = 1.0268108 \implies x < 1.00885823
$$
Iteration #2: Applying this inequality to the third equation again, we have:
$$
z^3 = x(x - 1) < 1.00885823(1.00885823 - 1) = 0.0089367 \implies z < 0.20752
$$
For $0 < z < 0.20752$, we have that:
$$
0.20752(0.20752 - 1) < y^3 < 0 \implies -0.54787 < y < 0
$$
Once again, we have:
$$
x^3 = y(y - 1) < -0.54787 \left(-0.54787 - 1\right) < 1 \implies x < 1
$$
This contradicts that $x > 1$. Therefore, no non-trivial solutions exist.
| {
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"timestamp": "2023-03-29T00:00:00",
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Given $a_1=1$ and $ a_{n+1}=a_n + 1 +\frac{a_n}{a_{n+1}}$ prove $\lim\limits_{n\to\infty }( a_{n+1} - a_n)=2$ Given a sequence $(a_n)_{n\geq 1}$ such that $a_1=1$ and $ a_{n+1}=a_n + 1 +\frac{a_n}{a_{n+1}}$ where $ n\geq1$
prove that $\lim\limits_{n\to\infty }( a_{n+1} - a_n)=2$
I tried to prove that $a_n$ is increasing but it I just can't get around to it. Thank you in advance!
| Let's address first the necessity that $a_n \geq \frac{1}{2}$ for any $n$ so that the sequence is well defined. Suppose that $a_n \geq \frac{1}{2}$, then there are two options for $a_{n+1}$:
$$a_{n+1}=\frac{1}{2}(1+a_n\pm \sqrt{a_n^2+6a_n+1})$$
The positive option is good and it satisfies $a_{n+1}\geq \frac{1}{2}$. Now, let's prove that if we choose $a_{n+1}=\frac{1}{2}(1+a_n- \sqrt{a_n^2+6a_n+1})$, then $a_{n+2}$ won't be defined anymore. We have:
$$a_{n+2}-a_{n+2}(a_{n+1}+1)-a_{n+1}=0$$
and the discriminant is:
$$\Delta = a_{n+1}^2+6a_{n+1}+1=a_{n+1}(a_n+1)+a_n+6a_{n+1}+1=a_{n+1}(7+a_n)+a_n+1$$
$$=\frac{1}{2}\left[(a_n+1)(a_n+9)-(a_n+7)\sqrt{a_n^2+6a_n+1}\right]$$
We are going to prove that this discriminant is negative:
$$(a_n+1)(a_n+9)<(a_n+7)\sqrt{a_n^2+6a_n+1}$$
After squaring and expanding, this simplifies to:
$$a_n^2+8a_n >2$$
which is true since $a_n \geq \frac{1}{2}$. Now, that we have proved we always have to choose the option greater than $\frac{1}{2}$, we can start proving the question.
Since we established we must have $a_n\geq \frac{1}{2}$ for any $n$, from the recurrence formula, $a_n$ is increasing and:
$$a_{n+1}-a_n=1+\frac{a_n}{a_{n+1}} < 2$$
therefore the sequence $a_{n+1}-a_n$ is bounded. Also
$$a_{n+1}^2 = a_{n+1}a_n +a_{n+1}+a_{n} \geq a_{n+1}a_n+2a_{n} = a_n(a_{n+1}+2) > a_na_{n+2}$$
This implies that
$$(a_{n+2}-a_{n+1})-(a_{n+1}-a_n) = \frac{a_{n+1}^2-a_na_{n+2}}{a_{n+1}a_{n+2}} > 0$$
the sequnce $a_{n+1}-a_n$ is bounded and increasing (thus convergent) and $\frac{a_n}{a_{n+1}}$ is also bounded above by $1$ and increasing (it converges to $1$). Therefore:
$$\lim_{n \to \infty} (a_{n+1}-a_{n}) = 1+\lim_{n \to \infty} \frac{a_n}{a_{n+1}} = 2$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Combinatorics: Number of ways to distribute 12 balls in group, with given conditions
The number of ways in which 12 identical balls can be grouped in four marked non empty sets A,B,C,D such that n(A) < n(B) is ?
The answer given is 70.
This is how I solved it:-
Now to distribute 12 identical balls to 4 different group such that no group is empty, we have $a+b+c+d = 12$, $$^{12-1}C_{4-1}$$ ways but we cannot have $a=b$ (as $a$ and $b$ are variables and interchangeable n(A) < n(b) and n(A) > n(b) mean the same), which also means that all $a, b, c,d$ should not be equal (as $a,b,c,d$ are variables and interchangeable) i.e. we need to exclude cases of $a=b=c=d$, but there is only one case that is possible which is $a=b=c=d=3$,
thus total ways $=$ $$^{12-1}C_{4-1}-1$$ but that equals 164 but answer given is 70. What am I doing wrong? How will you solve the question?
| Answer will be $$\dfrac{^{12-1}C_{4-1} - K}{2}$$
Here K are the cases when $a =b$, which can be calculated using $2a + c + d =12$, which is
Coefficient of $x^{12}$ in $(x + x^2 + .... + x^{12})^2(x^2 + x^4 + ...... + x^{12})$
$\implies$ coefficient of $x^{12}$ in $(1 - x)^{-2} (x - x^{13})^2 (x^2 - x^{14}) (1 - x^2)^{-1}$
$\implies$ coefficient of $x^{8}$ in $(1 - x)^{-2} (1 + x^2 + x^4 + ..... + x^{12})$.
Now you can calculate easily.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving $\int_0^1 \sqrt{x \left(\sqrt{-3 x^2+2 x+1}-x+1\right)} \, dx=\frac{7 \pi }{12 \sqrt{6}}$ I was given that $$\int_0^1 \sqrt{x \left(\sqrt{-3 x^2+2 x+1}-x+1\right)} \, dx=\frac{7 \pi }{12 \sqrt{6}}.$$
Which is numerically correct. I tried a few substitutions, none of which lead to anything valuable. I'd like you to help me on the problem. Thanks in advance!
|
Let
$$y(x)= \sqrt{x \left(\sqrt{-3 x^2+2 x+1}-x+1\right)}\implies
[y^2-x(1-x)]^2=x^2(1+2x-3x^2)$$
The integral $I=\int_0^1 y(x)dx$ represents the shaded area of the left graph. Rotate the curve with $x=\frac1{\sqrt2}(u+v)$ and $y=\frac1{\sqrt2}(u-v)$ for a symemtric functional form,
$$7u^4+12u^3v+26u^2v^2+12uv^3+7v^4=\sqrt2(6u^3+10u^2v+10uv^2+6v^3)$$
In polar coordinates $u=r\cos\theta$ and $v=r\sin\theta$, the function reads
$$r(\theta)=\frac{2\sqrt2(\sin\theta+\cos\theta)(3+\sin2\theta)}{7+6\sin2\theta+3\sin^22\theta}$$
which is shown in the right graph. The equivalent area integral is,
$$I = \int_{-\pi/4}^{\pi/4} \frac12r^2(\theta)d\theta
=\int_{0}^{\pi/2} \frac{8\cos^2t(3+\cos2t)^2}{(7+6\cos2t+3\cos^22t)^2}dt $$
where the variable change $t=\frac\pi4-\theta$ is made. Rewrite the integral with $\cos2t =\frac2{\sec^2t}-1$ and then let $u=\tan t$,
$$I = \int_{0}^{\pi/2} \frac{2(1+\sec^2t)^2\sec^2t}{(\sec^4t+3)^2}dt
=\int_{0}^{\infty} \frac{2(u^2+2)^2}{(u^4+2u^2+4)^2}du$$
$$=\frac13\frac{u^3+u}{u^4+2u^2+4}\bigg|_0^\infty+\frac13\int_{0}^{\infty} \frac{u^2+5}{u^4+2u^2+4}du$$
$$=\frac7{12}\int_{0}^{\infty} \frac{2+u^2}{u^4+2u^2+4}du
+\frac1{4}\int_{0}^{\infty} \frac{2-u^2}{u^4+2u^2+4}du
=\frac7{12}\cdot \frac\pi{\sqrt6}+\frac1{4}\cdot 0=\frac{7\pi}{12\sqrt6}$$
| {
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"url": "https://math.stackexchange.com/questions/3576992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Convergence of $\sum_{n=2}^{\infty}\frac{1}{n\ln (n)}$ WITHOUT using the integral test. Yesterday I was tutoring a high school student on the convergence of infinite series. We encounter the following series on his practice assignment:
$$\sum_{n=2}^{\infty}\frac{1}{n\ln (n)}$$
He's at the precalculus level so he hasn't seen the integral test which would be the standard method for determining that this particular series diverges.
His teacher though gave him a hint that the formula for infinite interest compound was to be used. I assumed that we had to use something along the lines of:
$$e^x=\lim_{n\to\infty}\left( 1+\frac{x}{n}\right)^n\implies x=\lim_{n\to\infty} n \ln \left(1+\frac{x}{n}\right)$$
I couldn't do it in the end, it takes me to weird places. Is there really a way of determining this series is divergent without using the integral test?
Edit: The tests that the teacher taught my student and thus the only ones allowed are the following:
nth term test, Comparison/limit comparison test, alternating series test, ratio test.
| Replace $\ln$ by $\log$ base 2, larger than
$$ \frac{1}{2 \cdot 1} + $$
$$ \frac{1}{3 \cdot 2} + \frac{1}{4 \cdot 2} + $$
$$ \frac{1}{5 \cdot 3} + \frac{1}{6 \cdot 3} + \frac{1}{7 \cdot 3} + \frac{1}{8 \cdot 3} + $$
$$ \frac{1}{9 \cdot 4} + \frac{1}{10 \cdot 4} + \frac{1}{11 \cdot 4} + \frac{1}{12 \cdot 4} +\frac{1}{13 \cdot 4} + \frac{1}{14 \cdot 4} + \frac{1}{15 \cdot 4} + \frac{1}{16 \cdot 4} + $$
$$ \frac{1}{17 \cdot 5} + \frac{1}{18 \cdot 5} + \frac{1}{19 \cdot 5} + \frac{1}{20 \cdot 5} +\frac{1}{21 \cdot 5} + \frac{1}{22 \cdot 5} + \frac{1}{23 \cdot 5} + \frac{1}{24 \cdot 5} + \frac{1}{25 \cdot 5} + \frac{1}{26 \cdot 5} + \frac{1}{27 \cdot 5} + \frac{1}{28 \cdot 5} +\frac{1}{29 \cdot 5} + \frac{1}{30 \cdot 5} + \frac{1}{31 \cdot 5} + \frac{1}{32 \cdot 5} +$$
This is bigger than
$$ \frac{1}{2} + \frac{2}{8} + \frac{4}{24} + \frac{8}{64} + \frac{16}{160} + \dots$$
or
$$ \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} + \frac{1}{10} + \dots$$
| {
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Prove $\sum\limits^{n-1}_{i=1} i\ln (i) \leq \frac{n^2}{2}\ln (n)-\frac{n^2}{4}+\frac{1}{4}$ for $n\geq 1$. I'm quite stuck, truly. I have tried proving it directly, using induction, showing that $$n\ln (n) \leq \Big(\frac{(n+1)^2}{2}\ln (n+1)-\frac{(n+1)^2}{4}+\frac{1}{4}\Big)-\Big( \frac{n^2}{2}\ln (n)-\frac{n^2}{4}+\frac{1}{4}\Big)$$
as well as using the fact that $$\sum ^n_{i=1}i\ln (i)\leq \sum ^n_{i=1}i\ln (n)=\frac{n(n-1)}{2}\ln (n
)$$
but haven't been able to achieve much. The inequality does hold though, I've checked using desmos.
| It's not clear to me what the sticking point is. If you have proven that $$n\ln (n) \leq \Big(\frac{(n+1)^2}{2}\ln (n+1)-\frac{(n+1)^2}{4}+\frac{1}{4}\Big)-\Big( \frac{n^2}{2}\ln (n)-\frac{n^2}{4}+\frac{1}{4}\Big)$$ then you are basically done, as this shows that the change in the left side of your desired inequality is less than the change of the right side.
To be explicit, let $f(n):=\sum_{i=1}^{n-1} i \ln (i)$ and let $g(n):= \frac{n^2}{2}\ln (n)-\frac{n^2}{4}-\frac{1}{4}$. Then $f(n+1)-f(n)$ is the left side of the above inequality, and $g(n+1)-g(n)$ is the right side. By induction, if $f(n)\leq g(n)$, then $$f(n+1) = \left(f(n+1)-f(n)\right)+f(n) \leq \left(g(n+1)-g(n)\right) + g(n) = g(n+1)$$ and together with $f(2)=0\leq 2\ln(2)-\frac{5}{4} = g(2)$ you're done. (The reason I chose $n=2$ as a base case rather than $n=1$ is because it makes the following argument work more smoothly. The $n=1$ case is trivial.)
The hard part is showing that the top inequality holds. Here's one way to do so:
$$
\begin{align}
& \Big(\frac{(n+1)^2}{2}\ln (n+1)-\frac{(n+1)^2}{4}+\frac{1}{4}\Big)-\Big( \frac{n^2}{2}\ln (n)-\frac{n^2}{4}+\frac{1}{4}\Big) \\
= & \frac{n^2}{2}\left(\ln (n+1)-\ln(n)\right) + n\ln(n+1) + \frac{1}{2}\ln(n+1) - \frac{n}{2} - \frac{1}{4} \\
= & \frac{n^2}{2}\ln (1+\frac{1}{n}) + n\ln(n+1) + \frac{1}{2}\ln(n+1) - \frac{n}{2} - \frac{1}{4}
\end{align}
$$
Using $\ln(1+\frac{1}{n}) \geq \frac{1}{n}-\frac{1}{2n^2}$ (from the Taylor series expansion of $\ln$) on the first term, $\ln(n+1)\geq\ln(n)$ on the second term, and $\ln(n+1)\geq 1$ (for $n\geq 2$) for the third term, we find that the above is
$$\geq\frac{n}{2}-\frac{1}{4} + n\ln(n)+\frac{1}{2}-\frac{n}{2}-\frac{1}{4} = n\ln(n)$$as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $\frac{\sin^4x}{a}+\frac{\cos^4x}{b}=\frac{1}{a+b},$ then show that $\frac{\sin^6x}{a^2}+\frac{\cos^6x}{b^2}=\frac{1}{(a+b)^2}$
Question: If $\frac{\sin^4x}{a}+\frac{\cos^4x}{b}=\frac{1}{a+b},$ then show that $\frac{\sin^6x}{a^2}+\frac{\cos^6x}{b^2}=\frac{1}{(a+b)^2}$.
My approach: Since $$\frac{\sin^4x}{a}+\frac{\cos^4x}{b}=\frac{1}{a+b} \\ \implies \left(\frac{\sin^4x}{a}+\frac{\cos^4x}{b}\right)^2=\frac{1}{(a+b)^2} \\ \implies \frac{\sin^6x}{a^2}+\frac{\cos^6x}{b^2}-\sin^2x\cos^2x\left(\frac{\sin^2x}{a}-\frac{\cos^2x}{b}\right)^2=\frac{1}{(a+b)^2}.$$
Therefore, if we can prove that $$\frac{\sin^2x}{a}-\frac{\cos^2x}{b}=0,$$ then we are done. But, I am not able to prove the same.
| A cool question: write $\left(1+\frac{b}{a}\right)\sin(x)^4+\left(1+\frac{a}{b}\right)\cos(x)^4=1$ split the parantheses as $$(\sin(x)^2+\cos(x)^2)^2-2\cos(x)^2\sin(x)^2+\frac{a}{b}\cos(x)^4+\frac{b}{a}\sin(x)^4=1,$$ now you get $2\cos(x)^2\sin(x)^2=\frac{a}{b}\cos(x)^4+\frac{b}{a}\sin(x)^4$ which is possible if and only if $\frac{a}{b}\cos(x)^2=\sin(x)^2$ as it is said $\cos(x)^2\left(1+\frac{a}{b}\right)=1$ you have them all in terms of $a$ and $b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
The equation $x^4-2x^3-3x^2+4x-1=0$ has four distinct real roots $x_1,x_2,x_3,x_4$ such that $x_1The equation $x^4-2x^3-3x^2+4x-1=0$ has four distinct real roots $x_1,x_2,x_3,x_4$ such that $x_1<x_2<x_3<x_4$ and product of two roots is unity, then:
$Q-1$: Find $x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_4+x_3\cdot x_4$
$Q-2$: Find $x_2^3+x_4^3$
My attempt is as follows:-
$A-1$ : First I tried to find any trivial root, but was not able to find any. After that I tried following:-
$$x_1\cdot x_2+x_1\cdot x_3+x_1\cdot x_4+x_2\cdot x_3+x_2\cdot x_4+x_3\cdot x_4=-3$$
$$x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_4+x_3\cdot x_4=-3-x_1\cdot x_4-x_2\cdot x_3$$
$$x_1\cdot x_2\cdot x_3\cdot x_4=-1$$
$$x_1\cdot x_4=\dfrac{-1}{x_2\cdot x_3}$$
$$x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_4+x_3\cdot x_4=-3-x_1\cdot x_4-x_2\cdot x_3$$
$$x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_4+x_3\cdot x_4=-3-x_2\cdot x_3+\dfrac{1}{x_2\cdot x_3}$$
But from here I was not able to proceed as I was not able to calculate value of $x_2\cdot x_3$
$A-2$ : $(x_2+x_4)(x_2^2+x_4^2-x_2\cdot x_4)$
Now here I was not getting any idea for how to proceed.
Please help me in this.
| Doing it in OP's way
$f(x)=x^4-2x^3-3x^2+4x-1=0$
let its rootsa be $a,b,c,d$, and let $a+b=u$ and $ab=v.$
Then by Vieta's formulas:
$$a+b+c+d=2~~~(1) \implies c+d=2-u$$
$$abcd=-1 ~~~~~(2) \implies cd=-1/v$$
$$ab+bc+cd+ac+bd+ad=-3~~~(3) \implies v-1/v+(a+b)(c+d)=-3 \implies v-1/v+u(2-u)=-3$$
$$abc+bcd+acd+bcd=-4~~~(4) \implies ab(c+d)+cd(a+b) =-4 \implies v(2-u)-(1/v)u=-4$$By putting $v=1$ in (3) we get $u^2-2u-3=0 \implies u=3,-1$
Next $a+b=3, ab=1; a+b=-1,ab=1$ give $$a,
b=\frac{3\pm \sqrt{5}}{2};~~ a,b=\frac{-1\pm \sqrt{5}}{2}$$
These for are the roots which can ve arranged ascending order as
$$x=\frac{-1-\sqrt{5}}{2},\frac{3-\sqrt{5}}{2},\frac{-1+\sqrt{5}}{2},\frac{3+\sqrt{5}}{2}~~~~(5)$$
Interestingly, (4) when $v=1$ also gives $u=3$, again.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Calculate the limit (verifying my answer). I have to calculate
$$\lim_{x\rightarrow 1^+} \frac{\sin(x^3-1)\cos(\frac{1}{1-x})}{\sqrt{x-1}}$$
Making the substitution $x-1=y$ and doing the math, I get that,
$$=\lim_{y\rightarrow 0^+} \frac{\sin(y(y^2+3y+3))}{y(y^2+3y+3)}\cdot\cos\Big(\dfrac{1}{y}\Big)\cdot\sqrt{y}(y^2+3y+3)$$
Since the first fraction goes to $0$. I have to worry with the $\cos(1/y)$, but I realized that $\cos(x)$ is bounded above and below, then, this kind of function times something that goes to $1$ results in $0$ (I studied this theorem). Since $\sqrt{y}$ goes to $0$. Then, the asked limit is $0$. Is that correct?
| $(x^3-1)=(x-1)(x^2+x+1)$;
$|\dfrac{\sqrt{x-1}(x^2+x+1)\sin (x^3-1)}{(x^3-1)}\cdot$
$\cos (\frac{1}{1-x})|=$
$(\sqrt{x-1}(x^2+x+1))\cdot \dfrac{\sin(x^3-1)}{(x^3-1)}$
$\cdot |\cos (\frac{1}{1-x})|.$
Take the limit $x \rightarrow 1^+$.
Note :
1) Limit of first term $=0$;
2) Use $\lim_{y \rightarrow 0^+}\dfrac{\sin y}{y}=1$;
3)$ |\cos z| \le 1$ for $z$ real.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3589642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Cauchy-Schwarz inequality for $a_1^4 + a_2^4 + \cdots + a_n^4 \geqslant n$
Let $a_1+a_2,...,a_n \in \mathbb{R}.$ Show that if $a_1+a_2+...+a_n=n$, then $$a_1^4+a_2^4+...+a_n^4 \geqslant n.$$
The proposed solution for this was the following:
Using the Cauchy-Schwarz inequality twice we get
$a_1^4+a_2^4+...+a_n^4 \geqslant \frac{(a_1^2+a_2^2+...+a_n^2)^2}{n} \geqslant \frac{\frac{((a_1+a_2+...+a_n)^2)^2}{n}}{n} = \frac{(\frac{n^2}{n})^2}{n} = n$
I can see that we can deduce this straight from the definition, but where on earth does the denominator $n$ come for $a_1^4+a_2^4+...+a_n^4 \geqslant \frac{(a_1^2+a_2^2+...+a_n^2)^2}{n}$.
From Cauchy-Schwarz we can come up with $a_1^4+a_2^4+...+a_n^4 \geqslant (a_1^2+a_2^2+...+a_n^2)^2$, but I don't see where the denominator comes from. Could someone enlighten me?
| The denominator comes from the following:
$$n(a_1^4+...+a_n)^4=(1^2+...+1^2)(a_1^4+_...+a_n^4)\geq(a_1^2+...+a_n^2)^2,$$ which gives
$$a_1^4+...+a_n^4\geq\frac{(a_1^2+...+a_n^2)^2}{n}.$$
Another way: $$\sum_{i=1}^n(a_i^4-1)=\sum_{i=1}^n(a_i-1)(a_i^3+a_i^2+a_i+1)=$$
$$=\sum_{i=1}^n((a_i-1)(a_i^3+a_i^2+a_i+1)-4(a_i-1))=\sum_{i=1}^n(a_i-1)^2(a_i^2+2a_i+3)\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Show that for $n\geqslant1,\ \displaystyle \binom{2n}{n} = \dfrac{1\cdot3\cdot5\cdot\cdots\cdot(2n-1)}{2\cdot4\cdot6\cdot\cdots\cdot2n}2^{2n}$
Show that for $n\geqslant1,$ $$\displaystyle \binom{2n}{n} = \dfrac{1\cdot3\cdot5\cdot\cdots\cdot(2n-1)}{2\cdot4\cdot6\cdot\cdots\cdot2n}2^{2n}.$$
This is an exercise from the Preliminaries of David Burton's Elementary Number Theory.
I started by simplifying the $\text{RHS}$ into something that could enable me to use mathematical induction.
$$
\begin{align*}
\text{RHS} &= \dfrac{\displaystyle\sum_{k=1}^{n}(2k-1)}{\displaystyle\sum_{k=1}^{n}2k}\cdot2^{2n}\\
&=\frac{n(n+1)-n}{n(n+1)}\cdot2^{2n}\\
&=\frac{2^{2n}n}{n+1}.
\end{align*}
$$
To be proved is that
$$\binom{2n}{n}=\frac{2^{2n}n}{n+1}.$$
To get $\binom{2n}{n}$ on the $\text{RHS}$, I did the following.
$$
\begin{align*}
2^{2n}&=(1+1)^{2n}\\
&=\binom{2n}{0}+\binom{2n}{1}+\binom{2n}{2}+\cdots+\binom{2n}{n}+\cdots+\binom{2n}{2n}.
\end{align*}
$$
From $\displaystyle \binom{n}{k}=\binom{n}{n-k}$, it follows that
$$\begin{align*}
2^{2n}=\left[\binom{2n}{0}+\binom{2n}{2n}\right]+\left[\binom{2n}{1}+\binom{2n}{2n-1}\right]+\cdots+\binom{2n}{n},
\end{align*}$$
but I can't proceed any further.
| Let $ n $ be a positive integer greater than $ 1 \cdot $
Separating the odd factors from the even ones in the product $ \prod\limits_{k=1}^{2n}{k} $, gives the following : $$ \prod_{k=1}^{2n}{k}=\prod_{k=1}^{n}{\left(2k\right)}\prod_{k=0}^{n-1}{\left(2k+1\right)} $$
Meaning, we have $ \prod\limits_{k=0}^{n-1}{\left(2k+1\right)}=\frac{\prod\limits_{k=1}^{2n}{k}}{\prod\limits_{k=1}^{n}{\left(2k\right)}}=\frac{\left(2n\right)!}{2^{n}n!}\cdot $
Thus, $$ \frac{\prod\limits_{k=0}^{n-1}{\left(2k+1\right)}}{\prod\limits_{k=1}^{n}{\left(2k\right)}}=\frac{\left(2n\right)!}{2^{2n}\left(n!\right)^{2}}=\frac{1}{2^{2n}}\binom{2n}{n} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3591661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Integrate $\int \frac{16x+16}{(x^2+2x+8)^9}dx$ using u substitution. Section 5.2
Can somebody verify this solution for me?
Integrate $\int \frac{16x+16}{(x^2+2x+8)^9}dx$ using u substitution.
Let $u=x^2+2x+8$. Then $\frac{du}{dx}=2x+2$ and so $\frac{du}{2x+2}=dx$. Thus we have:
$\int \frac{16x+16}{(x^2+2x+8)^9}dx$
$= \int \frac{16x+16}{u^9}\frac{du}{2x+2}$
$= \int \frac{8(2x+2)}{u^9}\frac{du}{2x+2}$
$= \int \frac{8}{u^9}du$
$= 8 \int u^{-9} du$
$= 8 \frac{u^{-8}}{-8}+C$
$= 8 \frac{(x^2+2x+8)^{-8}}{-8}+C$
$= -(x^2+2x+8)^{-8}+C$
| In this case, very conveniently, the numerator is (a multiple of) the derivative of the polynomial being raised to the $-9$- th power.
Thus this function is easy to integrate. Use the power rule and the chain rule.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Expressing logarithms in terms of $a$ and $b$ Given that $\log_{10}{5}=a$ and $\log_{3}{10}=b$, express $\log_{48}{65}$ in terms of $a$ and $b$.
I tried various transformations of $\log_{48}{65}$ and got:
$$\log_{48}{65} =\frac{\log_{10}{65}}{\log_{10}{48}}=\frac{a+\log_{10}{13}}{\frac{1}{b}+4\log_{10}2}=\frac{a+\log_{10}{13}}{\frac{1}{b}+4-4a}$$
I'm stuck here since I'm really not sure what to do with $\log_{10}{13}$. Have I approached this problem in a wrong way, and if so, what is a better way to approach it?
| I'd just put all in terms of $\log_{10} = \log$ (just for familiarity with days of yore):
$\begin{align*}
\log 5
&= a \\
\log 2
&= \log 10 - \log 5 \\
&= 1 - a \\
\log_3 10
&= \frac{\log 10}{\log 3} \\
&= \frac{1}{\log 3} \\
&= b \\
\log_{48} 65
&= \frac{\log 65}{\log 48} \\
&= \frac{\log 5 + \log 13}{\log 3 + \log 16} \\
&= \frac{(a + \log 13)}{1/b + \log 16} \\
&= \frac{b (a + \log 13)}{1 + 4 b \log 2} \\
&= \frac{b (a + \log 13)}{1 + 4 b (1 - a)}
\end{align*}$
Can't get rid of the 13, as a prime it can't be written as a product of powers of 10 and 3.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3595122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Equation $x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$
The value of $x$ satisfying the equation $x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$ is given by
*
*$2 \cos(10°)$
*$2 \cos(20°)$
*$2 \cos(40°)$
*$2 \cos(80°)$
Using $x=2\cos(\theta)$ we get
$2\cos(\theta)=\sqrt{2+\sqrt{2-\sqrt{2+2\cos\theta}}}$
$2\cos(\theta)=\sqrt{2+\sqrt{2-2\cos\left(\frac{\theta}{2}\right)}}$
$2\cos(\theta)=\sqrt{2+2\sin\left(\frac{\theta}{4}\right)}$
$4\cos^2(\theta)={2+2\sin\left(\frac{\theta}{4}\right)}$
After this step I am not able to solve it
| My illustration is outlined below
$\begin{array}{l}
4{\cos ^2}\theta = 2\left( {1 + \sin \frac{\theta }{4}} \right)\\
4{\cos ^2}{10^o} \ne 2\left( {1 + \sin \frac{{{{10}^o}}}{4}} \right)\\
4{\cos ^2}{10^o} \ne 2\left( {1 + \sin {{2.5}^o}} \right)\\
4{\cos ^2}{10^o} \ne 2\left( {1 + \cos {{87.5}^o}} \right)\\
\\
4{\cos ^2}{20^o} \ne 2\left( {1 + \sin \frac{{{{10}^o}}}{4}} \right)\\
4{\cos ^2}{20^o} \ne 2\left( {1 + \sin {5^o}} \right)\\
4{\cos ^2}{20^o} \ne 2\left( {1 + \cos {{85}^o}} \right)\\
\\
4{\cos ^2}{40^o} = 2\left( {1 + \sin \frac{{{{40}^o}}}{4}} \right)\\
4{\cos ^2}{40^o} = 2\left( {1 + \sin {{10}^o}} \right)\\
4{\cos ^2}{40^o} = 2\left( {1 + \cos {{80}^o}} \right)\\
\\
4{\cos ^2}{80^o} \ne 2\left( {1 + \sin \frac{{{{80}^o}}}{4}} \right)\\
4{\cos ^2}{80^o} \ne 2\left( {1 + \sin {{20}^o}} \right)\\
4{\cos ^2}{80^o} \ne 2\left( {1 + \cos {{70}^o}} \right)
\end{array}$\
Hence the option is $0=40^o$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3596196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What's the limit of the string?
Find $$\lim_{n\to\infty} (x_n\sqrt{n})^{\sqrt{n^2-1}},$$ where $ x_{n+1} = \frac{x_n}{\sqrt{1+x_n^2}}$ and $x_1 = 2$.
I showed $x_n \to 0$, $x_n\sqrt{n} \to 1$, but i don't know how to solve limit properly.
| Strating from @dan_fulea nice solution
$$x_n=
\sqrt{\frac {4}{4n-3}}$$we could have nice approximation of
$$y_n=\big(x_n\sqrt{n}\big)^{\sqrt{n^2-1}}=\left(1+\frac 3{4n-3}\right)^{\frac 12\sqrt{n^2-1}}$$ Take logarithms
$$\log(y_n)={\frac 12\sqrt{n^2-1}}\,\log\left(1+\frac 3{4n-3}\right)$$
Now, use Taylor series for large values of $n$
$$\sqrt{n^2-1}=n-\frac{1}{2 n}-\frac{1}{8 n^3}+O\left(\frac{1}{n^5}\right)$$
$$\log\left(1+\frac 3{4n-3}\right)=\frac{3}{4 n}+\frac{9}{32 n^2}+\frac{9}{64 n^3}+O\left(\frac{1}{n^4}\right)$$
$$\log(y_n)=\frac{3}{8}+\frac{9}{64 n}-\frac{15}{128 n^2}+O\left(\frac{1}{n^3}\right)$$
$$y_n=e^{\log(y_n)}=e^{3/8}\left(1+\frac{9}{64 n}-\frac{879}{8192 n^2}+O\left(\frac{1}{n^3}\right) \right)$$
Just for the fun, using your pocket calculator for $n=10$; the exact value is $1.47381$ whle le above approximation gives $1.47389$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3599459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Rotation matrix and its powers A is a rotation matrix in ${\mathbb{R}}^2$ by angle $θ=30°$.
I know the matrix:
$A= \begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2}\\\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}$
I can write $A$ to another form:
$$
A=
\begin{bmatrix}
\frac{\sqrt{3}}{2} \\
-\frac{1}{2} \\
\frac{1}{2} \\
\frac{\sqrt{3}}{2}\\
\end{bmatrix}
$$
The question is how many of $A$'s powers are linearly independent?
So i have $A$ , $A^2$, $A^3$, ... , $A^{12} =$ identity matrix
$A = -A^{7}$
$A^2 = -A^{8}$
.
.
.
$A^6 = - A^{12}$
So I know that the numbers of linearly independent vectors $\leqslant 6$.
As these are 4 dimensional vectors, there are maximum 4 linearly dependent ???
How many of $A$'s powers are linearly independent?
| The characteristic equation for matrix $A$ is $t^2-\sqrt 3t+1=0$. Using Cayley-Hamilton theorem
$A^2-\sqrt 3 A+I_3=0$
$\implies A^2=\sqrt 3 A-I_3$.....(*)
So $A^2$ is linear combination of $A$ and $A^0$. Multiply (*) by $A$ repeatedly to observe that each $A^n, (n\ge 2)$ is linear combination of $A$ and $A^0$. So, only $A$ and $A^0$ are LI.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3599936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Showing $\frac34=\sqrt{1-\frac{1}{2}\sqrt{1-\frac{1}{4}\sqrt{1-\frac{1}{8}\sqrt{\cdots}}}}$ $$\frac34=\sqrt{1-\frac{1}{2}\sqrt{1-\frac{1}{4}\sqrt{1-\frac{1}{8}\sqrt{\cdots}}}}$$
We cannot use the Herschfeld's Convergence Theorem because you can see that it's negative. I have tried some generalization but nothing good. I have looked for some algorithm on the web like Landau's algorithm but we speak about and infinite nested radicals so I was wondering myself if there is not a recursive formula existing.
So if you have some papers or a proof, I will be thankful.
Any helps is greatly appreciated
Thanks a lot.
| Induction at work:
$$\frac{3}{4}=\sqrt{1-\frac{7}{16}}=
\sqrt{1-\color{blue}{\frac{1}{2}}\cdot\color{red}{\frac{7}{8}}}=\\
\sqrt{1-\frac{1}{2}\sqrt{1-\frac{15}{64}}}=
\sqrt{1-\frac{1}{2}\sqrt{1-\color{blue}{\frac{1}{4}}\cdot\color{red}{\frac{15}{16}}}}=\\
\sqrt{1-\frac{1}{2}\sqrt{1-\frac{1}{4}\sqrt{\frac{225}{256}}}}=
\sqrt{1-\frac{1}{2}\sqrt{1-\frac{1}{4}\sqrt{1-\color{blue}{\frac{1}{8}}\cdot\color{red}{\frac{31}{32}}}}}$$
Do you see the pattern?
At every step (in blue/red) we have something like $\color{blue}{\frac{1}{2^{n-2}}}\cdot\color{red}{\frac{2^n-1}{2^n}}$, then
$$\color{blue}{\frac{1}{2^{n-2}}}\cdot\color{red}{\frac{2^n-1}{2^n}}=
\color{blue}{\frac{1}{2^{n-2}}}\sqrt{\frac{2^{2n}-2^{n+1}+1}{2^{2n}}}=\\
\color{blue}{\frac{1}{2^{n-2}}}\sqrt{1-\color{blue}{\frac{1}{2^{n-1}}}\cdot\color{red}{\frac{2^{n+1}-1}{2^{n+1}}}}$$
and the result follows ...
| {
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"source": "stackexchange",
"question_score": "3",
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$(A -3I)(A + 2I) = 0$ implies $A = 3I$ or $A = -2I$? I know that square matrices' multiplication $XY=0$ does not imply $X=0$ or $Y=0$.
However, what if $X = A - 3I $, $Y = A + 2I $, given $I$ the identity matrix.
$$(A - 3I)(A + 2I) = 0$$
If it goes with the same rule of $XY=0$, then $A$ can be one matrix other than $-3I$ and $2I$.
But, we have the restriction that we can find an $A$ such that $A - 3I$ and $A + 2I$ are the multiplication that produce $0$.
Can I still find a matrix $A$ which is not equal to $3I$ or $-2I$?
| To find all solutions in the $2 \times 2$-case, write $A= \begin{pmatrix}a & b\\ c & d\end{pmatrix}$. The condition $$(A-3I)(A+2I) = 0$$
leads to the system
$$\begin{cases}(a-3)(a+2) + bc = 0 \\ (a-3)b + b(d-2) = 0 \\ c(a+2) + c(d-3) = 0 \\ bc + (d-3)(d+2) = 0\end{cases}$$
In particular, $a= 3, b = 0, c = 0, d = -2$ satisfies this, so the matrix
$$A:=\begin{pmatrix} 3 & 0 \\ 0 & -2\end{pmatrix}$$
presents a counterexample.
There are even more solutions: see https://www.wolframalpha.com/input/?i=solve+for+a%2Cb%2Cc%2Cd%3A+%28a-3%29%28a%2B2%29%2Bbc+%3D+0%2C+%28a-3%29b+%2B+b%28d-2%29+%3D+0%2C+c%28a%2B2%29+%2B+c%28d-3%29+%3D+0%2C+bc+%2B+%28d-3%29%28d%2B2%29+%3D+0+
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3602117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Distances from Morley triangle to edges of the original triangle Morley's trisector theorem states that in any triangle, the three points of intersection of the adjacent angle trisectors form an equilateral triangle, as illustrated in the left-hand diagram below. Let's call this equilateral triangle the Morley triangle and its edge length $m$.
I am interested the distances from the Morley triangle to the edges of the original triangle as illustrated in blue in the right-hand diagram.
Empirically it seems that these distances lie between $\frac{\sqrt 3}{2}m$ and $m$, depending on the angles of the original triangle. Is there simple proof?
| Let $\Delta XYZ$ ($X$ is opposite to $A$, $Y$ is opposite to $B$ and $Z$ is opposite to $C$) be the Morley's triangle of $\Delta ABC.$
Thus, $$S_{\Delta ZAB}=\frac{c^2\sin\frac{\alpha}{3}\sin\frac{\beta}{3}}{2\sin\frac{\alpha+\beta}{3}}=\frac{ch_Z}{2},$$
which gives $$h_Z=\frac{c\sin\frac{\alpha}{3}\sin\frac{\beta}{3}}{\sin\frac{\alpha+\beta}{3}}=\frac{2R\sin\gamma\sin\frac{\alpha}{3}\sin\frac{\beta}{3}}{\sin\left(60^{\circ}-\frac{\gamma}{3}\right)}.$$
We need to prove that $$\frac{\sqrt3}{2}m\leq h_Z\leq m.$$
The right inequality.
We need to prove that:
$$\frac{2R\sin\gamma\sin\frac{\alpha}{3}\sin\frac{\beta}{3}}{\sin\left(60^{\circ}-\frac{\gamma}{3}\right)}\leq8R\sin\frac{\alpha}{3}\sin\frac{\beta}{3}\sin\frac{\gamma}{3}$$ or
$$\sin\gamma\leq4\sin\frac{\gamma}{3}\sin\left(60^{\circ}-\frac{\gamma}{3}\right),$$ which is true because
$$4\sin\frac{\gamma}{3}\sin\left(60^{\circ}-\frac{\gamma}{3}\right)-\sin\gamma=8\sin\frac{\gamma}{3}\sin^2\left(15^{\circ}-\frac{\gamma}{6}\right)\cos\left(30^{\circ}+\frac{\gamma}{3}\right)\geq0.$$
The left inequality.
We need to prove that:
$$4\sqrt3R\sin\frac{\alpha}{3}\sin\frac{\beta}{3}\sin\frac{\gamma}{3}\leq\frac{2R\sin\gamma\sin\frac{\alpha}{3}\sin\frac{\beta}{3}}{\sin\left(60^{\circ}-\frac{\gamma}{3}\right)}$$ or
$$2\sqrt3\sin\frac{\gamma}{3}\sin\left(60^{\circ}-\frac{\gamma}{3}\right)\leq\sin\gamma,$$ which is equivalent to
$$\cos\left(30^{\circ}-\frac{\gamma}{6}\right)\sin\frac{\gamma}{3}\sin\frac{\gamma}{6}\sin^2\left(30^{\circ}-\frac{\gamma}{6}\right)\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3604897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove $x^2 + y^2 + z^2\geq xy + xz + yz$ Result: Let $, , ∈ ℝ$. Then we have $^2 + ^2 + ^2 ≥ + + $
Need some help proving this, just a few steps with work.
Was thinking you start with $x^2+y^2+z^2−xy−xz−yz$
then factor?
can anyone show me how to solve this?
$x^2+y^2+z^2≥xy+yz+zx⇔ $
$⇔2(x^2+y^2+z^2)≥2(xy+yz+zx)⇔$
$⇔x^2−2xy+y^2+y^2−2yz+z^2+z^2−2xz+x^2≥0⇔$
$⇔(x−y)^2+(y−z)^2+(z−x)^2≥0$
Should I do anything else? Or is this right?
| Another way.
Let $y=x+u$ and $z=x+v$.
Thus, $$x^2+y^2+z^2-xy-xz-yz=$$
$$=x^2+(x+u)^2+(x+v)^2-x(x+u)-x(x+v)-(x+u)(x+v)=$$
$$=x^2+x^2+2xu+u^2+x^2+2xv+v^2-x^2-xu-x^2-xv-x^2-xv-xu-uv=$$
$$=u^2-uv+v^2=\left(u-\frac{v}{2}\right)^2+\frac{3v^2}{4}\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3607257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Prove $\frac{1+a^2}{1-a^2}+\frac{1+b^2}{1-b^2}+\frac{1+c^2}{1-c^2}\ge \frac{15}{4}$ Let $1>a>0$, $1>b>0$, $1>c>0$ and $a+b+c=1$. Prove that
$$
\frac{1+a^2}{1-a^2}+\frac{1+b^2}{1-b^2}+\frac{1+c^2}{1-c^2}\ge \frac{15}{4}.
$$
I saw the following solution. Let $x=\frac{2}{1-a^2}$, $y=\frac{2}{1-b^2}$, $z=\frac{2}{1-c^2}$, then, using AM-GM inequality, we get
$$
x+y+z-3\ge 3 \sqrt[3]{xyz}-3=\frac{27}{4}-3=\frac{15}{4}\; for \; x=y=z=\frac{9}{4}.
$$
Is it correct?
| Let $f(t):=\tfrac{1+t^2}{1-t^2}$ so $f^{\prime\prime}(t)=\tfrac{4(1+3t^2)}{(1-t^2)^3}>0$. By Jensen's inequality, $f(a)+f(b)+f(c)\ge 3f(\tfrac13)=\frac{15}{4}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Prove that the last digit of $\frac{n(n+1)}{2}$ is never 9 The original question was
If the last digit of $\sum_1^n n^3$ is 1, then the last digit of $\sum_1^n n$ is ______?
The sum of the cubes of natural numbers is equal to the square of the sum of the natural numbers. Since the last digit of the sum of cubes is 1, the last digit of the sum of numbers can be either 1 or 9. Substituting n=1..13 in the formula $\frac{n(n+1)}{2}$ gives numbers which end in 1, but never a number ending in 9. The answer key also mentions the answer to be 1, but not 9.
Please conclusively prove why $\sum_1^n n$ can never end with a 9.
| If $\dfrac{n(n+1)}2=r, 8r+1=(2n+1)^2$
Now for any odd number $2n+1, (2n+1)^2\equiv1,5,9\pmod{10}$
$8r+1\equiv1,5,9\pmod{10}$
$\iff8r\equiv0,4,8\pmod{10}$
$\iff4r\equiv0,2,4\mod5$
$\iff r\equiv0,3,1\pmod5$
$\implies r\equiv0,0+5,3,3+5,1,1+5\pmod{10}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Given $a, b, c>0$, prove $\frac{a^4}{a+b}+\frac{b^4}{b+c}+\frac{c^4}{c+a}\geq \frac{1}{2}(a^{2}c+b^{2}a+c^{2}b)$
Given $a,b,c>0$, prove that $$\frac{a^4}{a+b}+\frac{b^4}{b+c}+\frac{c^4}{c+a}\geq \frac{1}{2}(a^{2}c+b^{2}a+c^{2}b).$$
My attempt:
I have that $$\frac{a^4}{a+b}+\frac{c^2(a+b)}{4}\geq a^{2}c$$
$$\frac{b^4}{b+c}+\frac{a^2(b+c)}{4}\geq b^{2}a$$
$$\frac{c^4}{c+a}+\frac{b^2(c+a)}{4}\geq c^{2}b$$
Adding them up yields $$\sum \frac{a^4}{a+b}\geq\frac{3}{4}\sum a^2c-\frac{1}{4}\sum a^2b.$$
So it boils down to proving that $a^2b+b^2c+c^2a\leq ab^2+bc^2+ca^2$, which isn't correct.
Could you help me with this problem?
| We need to prove that: $$\sum_{cyc}\left(\frac{a^4}{a+b}-\frac{1}{2}ab^2\right)\geq0$$
or
$$\sum_{cyc}\frac{a(2a^3-ab^2-b^3)}{a+b}\geq0$$ or
$$\sum_{cyc}\frac{a(a-b)(2a^2+2ab+b^2)}{a+b}\geq0$$ or
$$\sum_{cyc}\left(\frac{a(a-b)(2a^2+2ab+b^2)}{a+b}-\frac{5}{6}(a^3-b^3)\right)\geq0$$ or
$$\sum_{cyc}\frac{(a-b)^2(7a^2+9ab+5b^2)}{a+b}\geq0$$ and we are done!
Now we see that our inequality is true even for any $a$, $b$ and $c$
such that $a+b>0$, $a+c>0$ and $b+c>0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the value of $a_0a_1a_2\cdots a_n\left(\frac{1}{a_0}+\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}\right)$
Given that the sequence $\left\{a_n\right\}$ satisfies $a_0 \ne 0,1$ and $$a_{n+1}=1-a_n(1-a_n)$$ $$a_1=1-a_0$$
Find the value of $$a_0a_1a_2\cdots a_n\left(\frac{1}{a_0}+\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}\right)$$
We actually get the terms of the sequence as:
$$a_1=1-a_0$$
$$a_2=1-a_1a_0$$
$$a_3=1-a_2a_1a_0$$
$$\vdots$$
$$a_{n+1}=1-a_na_{n-1}a_{n-2}\cdots a_0$$
Any way from here?
| As we get,
$$a_1=1-a_0\\
a_2=1-a_0a_1\\
\begin{align}\\
\therefore \frac{1}{a_0}+\frac{1}{a_1} &=\frac{1}{a_0}+\frac{1}{1-a_0}\\
&=\frac{1-a_0+a_0}{a_0(1-a_0)}\\
&=\frac{1}{a_0(1-a_0)}\\
&=\frac{1}{a_0a_1}\\
\end{align}$$
$$\begin{align}\\
\frac{1}{a_0}+\frac{1}{a_1}+\frac{1}{a_2} &=\frac{1}{a_0a_1}+\frac{1}{a_2}\\
&=\frac{1}{a_0a_1}+\frac{1}{1-a_0a_1}\\
&=\frac{1-a_0a_1+a_0a_1}{a_0a_1(1-a_0a_1)}\\
&=\frac{1}{a_0a_1a_2}\\
\end{align}$$
$$.\\
.\\
.\\
.$$
$$\begin{align}\\
\frac{1}{a_0}+\frac{1}{a_1}+....\frac{1}{a_{n-1}}+\frac{1}{a_n}
&={\begin{aligned}\\
\frac{1}{a_0...a_{n-2}a_{n-1}}+&\frac{1}{1-a_0....a_{n-2}a_{n-1}}\\
\end{aligned}}\\
&=\frac{1-a_0....a_{n-2}a_{n-1}+a_0....a_{n-2}a_{n-1}}{a_0....a_{n-2}a_{n-1}(1-a_0....a_{n-2}a_{n-1})}\\
&=\frac{1}{a_0....a_{n-2}a_{n-1}a_n}\\
\end{align}$$
$$\begin{aligned}\\
\therefore a_0a_1....a_{n-1}a_n\left(\frac{1}{a_0}+\frac{1}{a_1}+....\frac{1}{a_n}\right)
&=1\\
\end{aligned}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Absolute extrema on closed boundary I solved a question and want to check whether I did any mistake or not.
Question. Given $f(x,y)=2y^3 +2x^2y-x^2 -y$, find the absolute min and max over the region $\{x^2+y^2 \le 1\}$.
First, I check the interior points so $f_x = 4xy-2x = 0 \implies x = 0$ or $y = \frac{1}{2}$. $f_y = 6y^2 + 2x^2 -1 = 0$. Then, to solve these simultaneously, I set $x = 0$ in $f_y$ so I get
$$6y^2 -1 = 0 \implies y= \pm \frac{1}{\sqrt{6}}.$$
However, setting $y= \frac{1}{2}$ in $f_y$, I get $\frac{3}{2} + 2x^2 -1 =0 \implies 2x^2 = -\frac{1}{2}$ so that we have no solution. (Is everything okay here ?)
Now, I check the boundary points: $x^2 + y^2 = 1\implies x^2 = 1-y^2$ and plugging this into $f$, we get the following:
$$f(x,y)=2y^3 + 2(1-y^2)y -(1-y^2) -y = y^2 + y -1.$$
To find the extreme points for $f$, $-1\le y \le 1$ and $f' = 2y+1 =0$ must be satisfied so in the end, we get $4$ points:$(0,-1),(\pm \frac{3}{2},-\frac{1}{2}),(0,1)$.
To sum up, there are $6$ critical points in total so $f$ attains its global max $1$ at $(0,1)$ and global min $-\frac{5}{4}$ at the points $(\pm \frac{3}{2},-\frac{1}{2})$.
| On the interior, consider the function
$$
\phi(\varepsilon) = f( \varepsilon x' + (1-\varepsilon)x^*)
$$
where $x^*$ is a local maximizer of $f$ on some convex set $K$. Then a necessary condition is
$$
\phi'(0) = \nabla f(x^*)'(x'-x^*) \le 0, \quad \forall x'\in K
$$
On the interior of the set, like you said, the gradient must vanish since the above condition must hold for all $x'\in K$, so those candidate points are valid.
On the boundary of the feasible set, you basically used the implicit function theorem to derive the standard Lagrangian tangency conditions. The concern is that $x^2 = 1-y^2$ really describes a correspondence and not a function, so substituting and simplifying might be a mistake. A Lagrangian approach would be to set
$$
\mathcal{L} = 2y^3 + 2x^2 y - x^2 - y - \lambda (x^2+y^2-1)
$$
with necessary conditions
$$
4xy - 2x - \lambda 2x = 0
$$
$$
6y + 2x^2 - 1 - \lambda 2y = 0
$$
$$
-(x^2 + y^2 -1) =0.
$$
When $x=0$, the other conditions give $y^2 = 1$ and $y(6-2\lambda)=1$, with candidate solutions $(x^*,y^*,\lambda^*) = (0,1,5/2)$ and $(x^*,y^*,\lambda^*) = (0,-1,7/2)$. If $y=0$, the system cannot be solved, since it requires $x^2=1$ and $2x^2=1$.
When $x \neq 0$ and $y\neq 0$, the KKT system becomes
$$
4y - 2 - \lambda 2 = 0
$$
$$
6y + 2x^2 - 1 - \lambda 2y = 0
$$
$$
-(x^2 + y^2 -1) =0.
$$
The first two equations imply
$$
2y -1 = \dfrac{6y+2x^2-1}{2y}
$$
or
$$
4y^2 - 8y -2x^2 +1= 0.
$$
If $x^2=1-y^2$, this implies $6y^2-8y-1 = 0$, which doesn't seem to characterize the same critical points as in your answer.
Hope that helps! Sorry if there are any mistakes!
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit calculation (L'Hospital?) I need to calculate the following limit:
$$
\lim_{x\to0}{\lim}\frac{x-\arcsin\left(x\right)}{\sin^{3}(x)}
$$
but I have no idea how to do it. I tried to use L'Hospital (as it meets the conditions), but it looks like it's only get worse:
\begin{align}\lim_{x\to0}\frac{f'\left(x\right)}{g'\left(x\right)}&=\lim_{x\to 0}\frac{1-\frac{1}{\sqrt{1-x^{2}}}}{3\sin^{2}\left(x\right)\cos\left(x\right)}=\lim_{x\to 0}\frac{\sqrt{1-x^{2}}-1}{\sqrt{1-x^{2}}\cdot\left(3\sin^{2}\left(x\right)\cos\left(x\right)\right)}\\
&=\lim{x\to 0}\biggl(1-\frac{1}{\sqrt{1-x^{2}}}\biggr)\biggl(\frac{1}{3\sin^{2}\left(x\right)\cos\left(x\right)}\biggr)
\end{align}
Someone told me to apply L'Hospital again (and maybe it could work), but it gets really complicated.
Maybe there is another (more simple) way?
Thanks in advance :)
| You can first rewrite the limit as
$$
\lim_{x\to0}\frac{x-\arcsin x}{x^3}\frac{x^3}{\sin^3x}
$$
Since the last fraction has limit $1$, you can concentrate on
$$
\lim_{x\to0}\frac{x-\arcsin x}{x^3}=
\lim_{x\to0}\frac{1-\dfrac{1}{\sqrt{1-x^2}}}{3x^2}=
\lim_{x\to0}\frac{\sqrt{1-x^2}-1}{x^2}\frac{1}{3\sqrt{1-x^2}}
$$
The second fraction has limit $1/3$, so you can concentrate on
$$
\lim_{x\to0}\frac{\sqrt{1-x^2}-1}{x^2}=\lim_{x\to0}\frac{-\dfrac{x}{\sqrt{1-x^2}}}{2x}=-\frac{1}{2}
$$
by an easy verification.
Hence the global limit is
$$
1\cdot\frac{1}{3}\cdot\left(-\frac{1}{2}\right)=-\frac{1}{6}
$$
If you don't remember the Taylor expansion of the arcsine, it is not difficult to find it up to degree $3$, which is all it's needed here.
Indeed, the derivative is $(1-x^2)^{-1/2}$, so the Taylor expansion up to degree $2$ is
$$
1-\frac{1}{2}(-x^2)+o(x^2)=1+\dfrac{1}{2}x^2+o(x^2)
$$
Hence the Taylor expansion of the arcsine is
$$
c+x+\frac{1}{6}x^3+o(x^3)
$$
and $c=0$ because $\arcsin 0=0$. So your limit is
$$
\lim_{x\to0}\frac{x-x-x^3/6+o(x^3)}{(x+o(x))^3}=-\frac{1}{6}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that the angle function between two points on the unit circle satisfies the triangle inequality With $S^1$ as the unit circle:
$$
S^1 = \{ x \in \mathbb{R}^2 \mid x_1^2 + x_2^2 = 1 \} \\
$$
And with $\arg$ as the well defined argument function that returns the principle value:
$$
\arg(x) : \mathbb{C} \to (-\pi,+\pi]
$$
We can define the angle distance function as:
\begin{align*}
\angle &: S^1 \times S^1 \to [0,\pi] \\
\angle(x,y) &= | \arg((x_1 + i \cdot x_2) / (y_1 + i \cdot y_2)) | \\
\end{align*}
What we want to show is:
\begin{align*}
\angle(x,z) &\le \angle(x,y) + \angle(y,z) \\
| \arg((x_1 + i \cdot x_2) / (z_1 + i \cdot z_2)) | &\le | \arg((x_1 + i \cdot x_2) / (y_1 + i \cdot y_2)) | + | \arg((y_1 + i \cdot y_2) / (z_1 + i \cdot z_2)) | \\
\end{align*}
I'm stuck on what to do next
| To complete this problem with help from zugzug and the others:
We can define the angle distance function as follows with $k$ as the appropriate integer constant to result in a value in $[0,\pi]$:
\begin{align*}
\angle &: S^1 \times S^1 \to [0,\pi] \\
%\angle(x,y) &= \abs{\arg((x_1 + i \cdot x_2) / (y_1 + i \cdot y_2))} \\
\angle(x,y) &= \arg(x_1 + i \cdot x_2) - \arg(y_1 + i \cdot y_2) + 2k\pi \\
\end{align*}
\begin{align*}
\angle(x,z) &= \arg(x_1 + i \cdot x_2) - \arg(z_1 + i \cdot z_2) + 2k\pi \\
\angle(x,y) &= \arg(x_1 + i \cdot x_2) - \arg(y_1 + i \cdot y_2) + 2k\pi \\
\angle(y,z) &= \arg(y_1 + i \cdot y_2) - \arg(z_1 + i \cdot z_2) + 2k\pi \\
\end{align*}
With some algebra:
\begin{align*}
\angle(x,z) &= \angle(x,y) + \angle(y,z) + 2k'\pi \\
\end{align*}
In this last expression, $k'$ can be either $0$ or $-1$ so we can conclude that:
\begin{align*}
\angle(x,z) &\le \angle(x,y) + \angle(y,z) \\
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Quick way to compute/estimate the expected value of the maximum of a number of dices. I am preparing for trading interviews and met this problem that essentially boiled down to computing/estimating (they key is speed) the expected value of the maximum face of two independent dice rolls. I understand that you may compute it through the formula $E[X] = \sum_{n} nP\{X \geq n\}$. However, I think it is not that fast. Is there any trick that people use to quickly get a sense of this problem? Thank you in advance!
| The simplest thing here is to view the problem graphically. For a pair of standard six-sided dice you can visualise the maximum value in a grid as shown below. You can see that the number of values of each number increases by the odd numbers --- $1, 3, 5, 7, ..., 11$. For dice with more sides, this pattern continues, with a longer series of odd numbers.
Using this pattern, we can see that for a pair of $w$-sided dice the expected value of the maximum roll is:
$$\begin{aligned}
\mathbb{E}(M) = \sum_{m=1}^w m \cdot \frac{2m-1}{w^2}
&= 1 \times \frac{1}{w^2} + 2 \times \frac{3}{w^2} + 3 \times \frac{5}{w^2} + \cdots + w \times \frac{2w-1}{w^2}. \\[6pt]
\end{aligned}$$
Computing this expectation requires us to compute the sum of products of natural numbers and the corresponding odd numbers. The exact formula is:
$$\begin{aligned}
\mathbb{E}(M) = \frac{1}{w^2} \sum_{m=1}^w m (2m-1)
&= \frac{1}{w^2} \Bigg[ \frac{w(w+1)(2w+1)}{3} - \frac{w(w+1)}{2} \Bigg] \\[6pt]
&= \frac{1}{w^2} \cdot \frac{2w(w+1)(2w+1)-3w(w+1)}{6} \\[6pt]
&= \frac{1}{w^2} \cdot \frac{w(w+1)(4w-1)}{6} \\[6pt]
&= \frac{(w+1)(4w-1)}{6w}. \\[6pt]
&= \frac{4w + 3w-1}{6w}. \\[6pt]
\end{aligned}$$
For a six-sided dice we have $w=6$ so that:
$$\mathbb{E}(M)
= \frac{4 \cdot 6^2 + 3 \cdot 6 - 1}{6 \cdot 6}
= \frac{144 + 18 - 1}{36}
= \frac{161}{36} = 4.47 \bar{2}.$$
If you are looking for a quick-and-dirty approximation, you could approximate this sum by an integral:
$$\begin{aligned}
\mathbb{E}(M)
&\approx \int \limits_0^w m \cdot \frac{2m-1}{w^2} dm \\[6pt]
&= \frac{1}{w^2} \int \limits_0^w (2m^2-m) dm \\[6pt]
&= \frac{1}{w^2} \bigg[ \frac{2}{3} m^3 - \frac{1}{2} m^2 \bigg]_{m=0}^{m=w} \\[6pt]
&= \frac{2}{3} w - \frac{1}{2} \\[6pt]
&= \frac{4w^2-3w}{6w}. \\[6pt]
\end{aligned}$$
This is quite close to the exact formula; only the last term in the formula is missing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3616884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to bring elliptical equation $2x^2+2y^2+3xy-x-y=0$ into canonical form I have this ellipse: $$2x^2+2y^2+3xy-x-y=0$$
Canonical form is $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
How can I bring my ellipse to that canonical form? It seems like I need some substitution.
| I tried to look for a good way to substitute something but failed miserably ...
So instead I tried the following :
Firstly I found the centre of the ellipse by solving the equations :
$$\frac{\delta f}{\delta x} = 0 \ and \ \frac{\delta f}{\delta y} = 0$$
So the centre came out to be ($\frac{1}{7}$,$\frac{1}{7}$)
The major and minor axes will pass through this point so let the slope of the line with acute angle to the x-axis be m. (m>0)
One of the axes : $(y-\frac{1}{7})=m(x-\frac{1}{7})$
Subsequently the other axis is : $(y-\frac{1}{7})=-\frac{1}{m}(x-\frac{1}{7})$
Rearrange these equations a bit
Now the canonical form of ellipse can be rewritten as :
$$\frac{(distance \ from \ an \ axis )^2}{a^2} +\frac{(distance \ from \ other \ axis )^2}{b^2} = 1$$
The distance of a point(x,y) from a line $y=mx +c$ is given by $\frac{|y-mx-c|}{\sqrt{m^2+1}}$
So write the equations of assumed axes as $y=mx+c$ and fill them in the above equation
$$\frac{(y-mx-\frac{1}{7}+\frac{m}{7})^2}{(1+m^2)a^2}+\frac{(my+x+\frac{1}{7}-\frac{m}{7})^2}{(1+m^2)b^2}=1$$
Now simply compare coefficients of obtained equation and given equation. The coefficients of x and y are equal in given equation, so let us just do that in our equation.
$$\frac{m^2}{(1+m^2)a^2} + \frac{1}{(1+m^2)b^2} = \frac{1}{(1+m^2)a^2} + \frac{m^2}{(1+m^2)b^2}$$
$$\frac{m^2 -1}{(1+m^2)a^2} = \frac{m^2 -1}{(1+m^2)b^2}$$
So either $m=\pm 1$ or $a=\pm b$
We can eaisilly discard the possibility of $a=\pm b$ as the would mean it is a circle, but in the given equation coefficient of $xy$ is 3 (which must be $0$ for it to be a circle)
Also notice how for m=1 , the equation will have non-zero constant term; so only possible value is 1
Now plug in the value of m as -1 and again compare coefficients
Compare coefficients of $xy$:
$$-\frac{1}{b^2}+\frac{1}{a^2} = 3..........(1)$$
Compare coefficients of $x^2$:
$$\frac{1}{a^2}+\frac{1}{b^2}=4..........(2)$$
solve (1) and (2) to get $\frac{1}{a^2}$ and $\frac{1}{b^2}$;plug in the obtained values in the equation to get the required equation in canonical form.
$$\frac{7(y+x-\frac{2}{7})^2}{4} +\frac{(x-y+\frac{2}{7})^2}{4} = 1$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Let $n > 1$ be a natural number and $p$ be a prime such that $p\mid n^3-1$ and $n\mid p-1$. Prove that $4p-3$ is a perfect square.
Question: Let $n > 1$ be a natural number and $p$ be a prime such that $p\mid n^3-1$ and $n\mid p-1$. Prove that $4p-3$ is a perfect square.
My solution: Given that $p\mid(n^3-1)$, thus we have $p\mid(n-1)$ or $p\mid(n^2+n+1)$. We also have $n\mid(p-1)$. Let us make cases.
Case 1: $p\mid(n-1)$ and $n\mid(p-1)$. Since $p-1\ge 1$ and $n>1$. Thus we can conclude that $p\mid(n-1)\implies p\le n-1$ and $n\mid(p-1)\implies n\le p-1$. Thus we have $n\le p-1\le n-2\implies n\le n-2\implies 0\le-2$, which is a clear contradiction. Hence $p\not\mid(n-1)$.
Case 2: $p\mid(n^2+n+1)$ and $n\mid(p-1)$. Thus, $\exists k\in\mathbb{Z},$ such that $pk=n^2+n+1$ and $\exists q'\in\mathbb{Z}$, such that $p=nq'+1$. Now $n^2+n+1\equiv 1\pmod n$ and $p\equiv 1\pmod n\implies pk\equiv k\pmod n\implies k\equiv pk=n^2+n+1\equiv 1\pmod n\implies k\equiv 1\pmod n.$
Thus $\exists q\in\mathbb{Z}$, such that $k=nq+1$. Hence we have $$p=nq'+1 \text{ and } k=nq+1\\ \implies pk=n^2qq'+n(q+q')+1=n^2+n+1\\\implies n(1-qq')=q+q'-1.$$
Now $p=nq'+1\implies nq'=p-1$ and $p-1\ge 1$. Thus $nq'\ge 1$. Now since $n>1$, hence $q'\ge 1$. This again implies that $nq'>1\implies p>2\implies p\ge 3.$
Now $pk=n^2+n+1>3\implies pk>3.$ Now $p\ge 3\implies k\ge 1.$ Now again we have $k=nq+1$. Since, $k\ge 1\implies nq+1\ge 1\implies nq\ge 0\implies q\ge 0$ $(\because n>1).$
Thus $q\ge 0,q'\ge 1\implies qq'\ge 0.$ Also $q+q'\ge 1\implies q+q'-1\ge 0.$ This implies that $n(1-qq')=q+q'-1\ge 0\implies n(1-qq')\ge 0\implies 1-qq'\ge 0 \implies qq'\le 1.$
Therefore, we simultaneously have $qq'\ge 0$ and $qq'\le 1$, which implies that $qq'=0$ or $qq'=1$.
Let us make cases.
Case 1: $qq'=1\implies q=1$ and $q'=1$. Thus $q+q'-1=1$ and $n(1-qq')=0\implies n(1-qq')\neq q+q'-1$, which is a contradiction to the fact that $n(1-qq')=q+q'-1$. Hence $qq'\neq 1$.
Case 2: $qq'=0\implies q=0$. Thus $n=q'-1\implies q'=n+1$, which in turn implies that $p=n(n+1)+1.$ Thus $4p-3=4n(n+1)+1=4n^2+4n+1=(2n+1)^2$.
Hence after analyzing all the cases we can conclude that $4p-3$ is a perfect square and is equal to $(2n+1)^2$, and we are done.
P.S.: I have edited the post and have attached the correct solution to the question.
| As @user760870 said: from $pk>1$ and $p\geq 3$ it is not necessary true $k>1$.
*
*Case 2. If we multiply both divisibility relations we get: $$pn\mid pn^2+pn+p-n^2-n-1\implies pn\mid p-n^2-n-1$$
If $p\geq n^2+n+1$ then $p= n^2+n+1$ sp $$4p-3 = 4n^2+4n+1 = (2n+1)^2$$
If $p\leq n^2+n+1$ then $$pn\leq n^2+n+1-p\implies p\leq {n^2+n+1\over n+1}<n+1$$
So $p\leq n$ and $n\leq p-1$. A contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $x,y,z \in \mathbb{R}$, $x^2+4y^2+16z^2=48, xy+4yz+2zx=24$, then find $x^2+y^2+z^2$.
If $x,y,z \in \mathbb{R}$ are such that $x^2+4y^2+16z^2=48$ and $xy+4yz+2zx=24$, then find $x^2+y^2+z^2$.
I can find the answer if I find the value of $x+y+z$ and $xy+yz+zx$. But I don't know how to do that. I found that $$(x+2y+4z)^2=144 \implies x+2y+4z=±12$$
But I can't progress after this.
| Now, by C-S $$144=3\cdot48=(1+1+1)(x^2+4y^2+16z^2)\geq(x+2y+4z)^2=144,$$
where the equality occurs for $$(1,1,1)||(x,2y,4z),$$
which gives $$(x,y,z)=(4,2,1)$$ or
$$(x,y,z)=(-4,-2,-1)$$ and $$x^2+y^2+z^2=21.$$
| {
"language": "en",
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"source": "stackexchange",
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Find locus of $S$ denoting set of complex numbers $\frac{z+1}{z-3}$, where $z$ varies over set of $|z|=1$.
Question: Let $S$ denote the set of all complex numbers of the form $\frac{z+1}{z-3}$, where $z$ varies over the set of all complex numbers with $|z|=1.$ Find the locus of the points in set $S$.
My approach:
Let $z=x+iy$, with $x,y\in\mathbb{R}$ such that $|z|=1\implies |z|^2=1$. This implies that we must have $x^2+y^2=1$. Now $z+1=(x+1)+iy$ and $z-3=(x-3)+iy$.
Thus
\begin{align*}
\frac{z+1}{z-3}&=\frac{(z+1)(\overline{z-3})}{|z-3|^2}\\
&=\frac{(z+1)(\overline{z}-3)}{|z-3|^2}\\
&=\frac{x^2-2x-3+y^2-4iy}{(x-3)^2+y^2}\\
&=\frac{-2x-2-4iy}{10-6x}\\
&=\frac{-x-1-2iy}{5-3x}.
\end{align*}
Thus we have
$$
\Re\left(\frac{z+1}{z-3}\right)=\frac{x+1}{3x-5}
$$
and
$$
\Im\left(\frac{z+1}{z-3}\right)=\frac{2y}{3x-5},$$ and our task is to find a relationship between these two given that $|z|=1$.
For our ease let us have $\alpha=\frac{z+1}{z-3}\,\, \forall z$ satisfying $|z|=1$.
Now we obtain two useful information using the triangle inequality. We have
$$
|z+1|\le |z|+1=2
$$
and
$$
|z-3|\le |z|+3=4.
$$
From here we can conclude that $$|z+1|^2=(x+1)^2+y^2=2+2x\le 4 \iff 1+x\le 2\text{ and } |z-3|^2=5-3x\le 16.$$
Thus we have $$0\le 1+x\le 2 \text{ and } 0\le 5-3x\le 16\implies -1\le x\le 1.$$
Observe that even from $x^2+y^2=1$, we can directly conclude that $-1\le x,y\le 1$.
But this doesn't help much to find a relationship between $\Re(\alpha)$ and $\Im(\alpha)$.
How to proceed?
| You should have proceeded with your approach. You have a calculation mistake which must be fixed first. The real part of $(z+1)/(z-3)$ should be $(x+1)/(3x-5)$ and not $(x+2)/(3x-5)$.
Let $$X=\frac{x+1}{3x-5},Y=\frac{2y}{3x-5}$$ so that $$x=\frac{5X+1}{3X-1},y=\frac{4Y}{3X-1}$$ and we are given $x^2+y^2=1$ so that $$(5X+1)^2+16Y^2=(3X-1)^2$$ ie $$X^2+Y^2+X=0$$ ie $$\left(X+\frac{1}{2}\right)^2+Y^2=\frac{1}{4}$$ which is a circle with center $(-1/2,0)$ and radius $1/2$.
In general it is better to avoid writing $z=x+iy$ and instead learn a bit of geometry on the complex plane directly. Here is a fact related to circles. If $a, b$ are distinct complex numbers then fixing the argument of $(z-a) /(z-b) $ gives us a circle (a point on a circle subtends the same angle from two given points on the circle). What is not so obvious is that fixing the modulus of $(z-a) /(z-b) $ also gives us a circle (or in special case of modulus $1$ a line the perpendicular bisector of line segment joining $a, b$). This is what is needed here as shown in one of answers.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the smallest maximum of $2|\sin x+a|+|\cos 2x+\sin x+b|$
Let $f(x)=2|\sin x+a|+|\cos 2x+\sin x+b|$ where $a,b \in \mathbb{R}$.
Denote the maximum of $f(x)$ as $M(a,b)$. Find the minimum value of
$M(a,b)$
Let $\sin x=t$. Then $f(x)=2|t+a|+|1-2t^2+t+b|=:g(t)$ where $t\in [-1,1]$. But how to go on with this?
| Okay, this is not exactly elegant, but here goes for a mechanical approach.
For any fixed values of $a$ and $b$, $\max_t g(t)$ is the max of:
*
*$g(1) = 2|a+1| + |b|$
*$g(-1)=2|a-1| + |b-2|$
*$\max_t \left\{2(t+a) + (-2t^2+t+1+b)\right\} = \frac{17}{8}+2a+b$
*$\max_t \left\{-2(t+a) + (-2t^2+t+1+b)\right\} = \frac{9}{8}-2a+b$
Can you see why this is true? It requires a few considerations. Looking at the possible values for the absolute values, we can reduce this to:
$$
\max_t g(t) = \max \begin{cases}
2+2a-b \\ 4-2a-b \\ \frac{17}{8}+2a+b \\ \frac{9}{8}-2a+b
\end{cases}
$$
From this we can work out that for a given value of $a$, the max is minimized with:
$$
\operatorname*{argmin}_b\max_tg(t) = \begin{cases}
\frac{23}{16} & a\le-\frac14 \\
\frac{15}{16}-2a & -\frac14\le a\le \frac12 \\
-\frac{1}{16} & \frac12 \le a
\end{cases}
$$
Plugging in these values of $b$, we get:
$$
\min_b \max_t g(t) = \begin{cases}
\frac{41}{16}-2a & a\le-\frac14 \\
\frac{49}{16} & -\frac14\le a\le \frac12 \\
\frac{33}{16} + 2a & \frac12 \le a
\end{cases}
$$
And from this we have:
$$
\min_{a,b}\max_t g(t) = \frac{49}{16},
$$
which is attained with $-\frac14\le a\le \frac12$, $b=\frac{15}{16}-2a$ at both $t=-1$ and $t=\frac34$. It is also attained at $(a, b, t)=(-\frac14,\frac{23}{16},-\frac14)$ and at $(a, b, t)=(\frac12,-\frac{1}{16},1)$.
I am sure there are better ways to do this. But at least this doesn't require any good ideas.
| {
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Bound of Chebyshev polynomials I'm learning about Chebyshev polynomial (First kind) $T_n$ and I'm confusing with this problem:
" Prove that with all $n>5$ we have this inequality:
$\vert(\sqrt 3)^{n}\cdot T_n(\dfrac{1}{\sqrt 3})\vert >1$ "
| The combinatorial facts needed are that if $n=2k$, ${n \choose 2}=2$ iff $n=2$.
This means that $\Re (1+i\sqrt 2)^{2N}=-1$ iff $N=2$ (all other terms in the expansion are divisible by $4$)
For $n=4m+1$, if $\nu_2(n-1)=k \ge 2$ ($\nu_2$ being the $2$ valuation so highest power of $2$ dividing the number) then $\nu_2{n-1 \choose 2^q}=k-q, q \le k$ so $2^{2^{q-1}}{n-1 \choose 2^q}$ divides by $2^{k+1}$ for $q \ge 3$ and then $2^r{n-1 \choose 2r}$ divides by $2^{k+1}$ for any $r \ge 3$ since${n-1 \choose 2^q}$ are local minima as is easily seen by writing the combinations out.
This means that if $n=4m+1$ the only way $\Re (1+i\sqrt 2)^{4N+1}=1$ iff $N=1,5$ (as all terms except $2{4N+1 \choose 2}, 4{4N+1\choose 4}$ divide by a higher power of $2$ hence to get $1$ either $4N+1=1$ or $2{4N+1 \choose 2}=4{4N+1\choose 4}$ so $4N+1=5$
Let $a_n=T_n(\dfrac{1}{\sqrt 3})$. The recurrence for $T_n(x)=\cos n\theta, x=\cos \theta, -1 \le x \le 1$ is $T_{n+1}(x)-2xT_n(x)+T_{n-1}(x)=0, T_0(x)=1, T_1(x)=x$ so,
$a_{n+1}-\frac{2}{\sqrt 3}a_n+a_{n-1}=0, a_0=1, a_1=\frac{1}{\sqrt 3}$
Solving the recurrence we get $a_n=\frac{1}{2}(c^n+\bar c^n), c=\frac{\sqrt 3 + i\sqrt 6}{3}$, so
$|\vert(\sqrt 3)^{n}\cdot T_n(\dfrac{1}{\sqrt 3})\vert =|\frac{1}{2}(b^n+\bar b^n)|=|\Re b^n|, b=1+i \sqrt 2$.
Now clearly $|\Re b^n|$ is an odd integer so we need to show that it is not $\pm 1$ when $n >5$ (note that it is $1$ for $n=5$)
If $\Re b^n=-1$ then $b^{2n}+2b^n+3^n=0$ so $x^2-2x+3/x^{2n}+2x^n+3^n$ and with $x=-1$ we get $n$ even hence the first paragraph shows $n=2$
If $\Re b^n=1$ then $b^{2n}-2b^n+3^n=0$ so $x^2-2x+3/x^{2n}-2x^n+3^n$ and with $x=-1$ we get $n$ odd hence the first paragraph shows $n=1,5$ (as ${n \choose 2}$ must be even since the next term in the expansion of $\Re (1+i\sqrt 2)^{n}$ is divisible by $4$, $n=4N+1$)
Done!
Notes: I would love an easier solution than the above - finding a solution is equivalent (by taking norms) to showing that the Diophantine equation $3^n=2q^2+1$ has only $(1,1),(2,2), (5,11)$ as roots in non zero positive integers, but I do not see a direct approach to that. Also as $T_n$ has $n$ roots at the appropriate rational multiples of $\pi$, the problem essentially requires showing that $\cos^{-1} \frac{1}{\sqrt 3}$ stays away far enough from those numbers, so the problem is definitely non-trivial.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $x^2+y^2 <1$ and $4y>x^2$, then $2x^2 + (2y-1)^2 <1$ I'm trying to prove the following two inequalities:
*
*If $x^2+y^2 >1$ and $4y>x^2$, then $2x^2 + (2y-1)^2 > 1$ .
*If $x^2+y^2 <1$ and $4y>x^2$, then $2x^2 + (2y-1)^2 < 1$ .
Both $x$ and $y$ are real numbers. The first inequality can be shown as follows: $$2x^2+(2y-1)^2>2(1-y^2)+(2y-1)^2=2(y-1)^2+1 \geq 1$$ (The condition $4y > x^2$ is superfluous)
For the second inequality, I tried: $2x^2 + (2y-1)^2 =2x^2 + (4y^2 - 4y + 1) <x^2 + 4y^2 + 1 $ (using $-4y < -x^2$). However, this seems not so useful.
It can be proved by plotting graph of ellipses $2x^2 + (2y-1)^2 = r$ for parameter $r$, and calculate the values of $r$ for which graphs are touched.
I seek an another proof of the second inequality. In particular, 'algebraic'(anyway) methods are welcomed.
| The second statement is actually false. It can be seen from the graph – it is false in the region that is inside the circle, above the parabola and outside the ellipse. Concretely, $(x,y)=(0.8,0.4)$ violates the second statement.
| {
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Show that ${2n \choose n} 2^{-2n} = (-1)^n {-\frac12 \choose n}$ Prove that
$${2n \choose n} 2^{-2n} = (-1)^n {-\frac12 \choose n},$$
$$\frac{1}n {2n -2 \choose n-1} 2^{-2n +1} = (-1)^{n-1} {\frac12 \choose n}.$$
The second part can be proved by replacing $n$ by $n-1$ in the first part. For the first part, I found that the right side is equal to ${n-1/2 \choose n}$, but when I expand the left side, I get something like
$$\frac{2n/4}{n}\frac{(2n-1)/4}{n-1}...\frac{(2n-n+1)/4}{1}$$
which does not look similar with ${n-1/2 \choose n}$.
I appreciate if you give some help.
| First, you have
$$\begin{equation}\begin{aligned}
n! & = 1(2)(3)\ldots (n-1)(n) \\
& = \left(\frac{2}{2}\right)\left(\frac{4}{2}\right)\left(\frac{6}{2}\right)\ldots\left(\frac{2n-2}{2}\right)\left(\frac{2n}{2}\right)
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Next, since $n - \frac{1}{2} = \frac{2n - 1}{2}$, note you have
$$\begin{equation}\begin{aligned}
{\frac{2n - 1}{2} \choose n} & = \left(\frac{1}{n!}\right)\left(\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)\left(\frac{5}{2}\right)\cdots\left(\frac{2n - 1}
{2}\right)\right) \\
& = \left(\frac{1}{(n!)^2}\right)\left(\frac{1}{2}\right)\left(\frac{2}{2}\right)\left(\frac{3}{2}\right)\left(\frac{4}{2}\right)\left(\frac{5}{2}\right)\cdots\left(\frac{2n - 1}
{2}\right)\left(\frac{2n}
{2}\right) \\
& = \left(\frac{1}{(n!)^2}\right)\frac{(2n)!}{2^{2n}} \\
& = 2^{-2n}\left(\frac{(2n)!}{(n!)^2}\right) \\
& = {2n \choose n} 2^{-2n}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
In the second line, I divided by the LHS of \eqref{eq1A} to get $\frac{1}{(n!)^2}$ and multiplied by the RHS of \eqref{eq1A} in the set of factors after that.
| {
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Prove that $1\cdot \frac{1}{2^2}\cdot ...\cdot \frac{1}{n^n}< \left(\frac{2}{n+1}\right)^\frac{n(n+1)}{2}$ Prove that $$1\cdot \frac{1}{2^2}\cdot ...\cdot \frac{1}{n^n}< \left(\frac{2}{n+1}\right)^\frac{n(n+1)}{2}$$ where $n$ is a positive integer.
My book suggests using AM-GM, but I couldn't do it. I just applied AM-GM to the numbers in the LHS, but it looks like I need some more upper bounds.
| Also induction works well here.
$$1\cdot \dfrac{1}{2^2}\cdots \dfrac{1}{n^n}\cdot\dfrac{1}{(n+1)^{n+1}} \le \left(\dfrac{2}{n+1}\right)^\frac{n(n+1)}{2}\cdot \dfrac{1}{(n+1)^{n+1}} \\= \Big[ \Big(\dfrac{2}{n+1}\Big)^{\frac{n}{2}}\cdot \dfrac{1}{n+1} \Big]^{n+1} \le(?)\ \left(\dfrac{2}{n+2}\right)^\frac{(n+1)(n+2)}{2}=\Big[ \Big(\dfrac{2}{n+2}\Big)^{\frac{n+2}{2}} \Big]^{n+1}.$$
To conclude you have to show that
$$\Big(\dfrac{2}{n+1}\Big)^{\frac{n}{2}} \cdot \dfrac{1}{n+1} \le \Big(\dfrac{2}{n+2}\Big)^{\frac{n+2}{2}}=\Big(\dfrac{2}{n+2}\Big)^{\frac{n}{2}}\cdot \dfrac{2}{n+2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3638388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Why is $2^{n}-1=\sum\limits_{k=0}^{n-1}2^k$? Recently I discovered that $2^{n}-1=\sum\limits_{k=0}^{n-1} 2^k$. As I don't have a math background, please tell me what this is called and a proof of why this is the case.
| As Hayden has pointed out, the correct expression is $2^n-1=\sum_{k=0}^{n-1} 2^k.$
So to see this, let’s first name the sum $S:=1+2^1+\dots+2^{n-2}+2^{n-1}.$
The ‘trick’ is to multiply the sum by $1$ in a clever way such that we can cancel some terms. Since $2=2-1$, we can say this:
$$
S=1+2^1+\dots+2^{n-2}+2^{n-1}\\
\implies\\
1\times S = \left(2-1\right)\times \left(1+2^1+\dots+2^{n-2}+2^{n-1}\right)\\
S=2\times\left(1+2^1+\dots+2^{n-2}+2^{n-1}\right)\\
-1\times\left(1+2^1+\dots+2^{n-2}+2^{n-1}\right)\\
= \left(2^1+2^2+\dots+2^{n-1}+2^{n}\right)\\
-\left(1+2^1+\dots+2^{n-2}+2^{n-1}\right)\\
= 2^n-1.
$$
Hope this helps!
Stay safe
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3640876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
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what is the value of $\binom{n}{1}+\binom{n}{4}+\binom{n}{7}+\binom{n}{10}+\binom{n}{13}+\dots$ what is the value of $$\binom{n}{1}+\binom{n}{4}+\binom{n}{7}+\binom{n}{10}+\binom{n}{13}+\dots$$ in the form of number,
cos, sin
attempts : I can calculate the value of $$\binom{n}{0}+\binom{n}{3}+\binom{n}{6}+\binom{n}{9}+\binom{n}{12}+\dots=\frac{1}{3}\left(2^n+2\cos \frac{n\pi}{3}\right)$$ by use primitive $3^\text{rd}$ root of the unity
but this problem i cant solve it.
| Binimial Series:
$$(1+x)^n=\sum_{k=0}^{n} {n \choose k} x^k~~~~(1)$$
$w^3=1, 1+w+w^2=0$, let $x=1$ in (1)
we get
$$2^n=\sum_{k=0}^n {n \choose k}~~~(2)$$
Let $x=w$ in (1) and miltiply it by $w^2$, to get
$$w^2(1+w)^n=(-1)^n w^{2n+2}=\sum_{k=1}^{n} w^{k+2} {n \choose k}~~~~~(3)$$
Let $x=w^2$ in (1) and multiply by $w$, to get
$$w(1+w^2)^n=(-1)^n w^{n+1}=\sum_{k=0}^{n} w^{2k+1} {n \choose k}~~~~(4)$$
Now add (2), (3), (4), to get
$$\sum_{k=0}^{n} [1+w^{k+1}+w^{2k+1}] {n \choose k}=2^n+(-1)^n[w^{2n+2}+w^{n+1}]$$
Whenever $k=3m+1$, $[1+w^{k+2}+w^{2k+1}]=[1+w^3+w^3]=3$, otherwise it vanishes as $[1+w+w^2]=0$ when $k\ne 2m+1$
So we get $$\sum_{m=0}^{n} {n\choose 3m+1}= \frac{1}{3}\left(2^n+(-1)^n[w^{2n+2}+w^{n+1}\right)=\frac{1}{3}(2^n+2\cos[(n-2)\pi/3])$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3644225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Calculating the volume of a restaurant take-away box that is circular on the bottom and square on the top Having a bit of a problem calculating the volume of a take-away box:
I originally wanted to use integration to measure it by rotating around the x-axiz, but realised that when folded the top becomes a square, and the whole thing becomes rather irregular. Since it differs in circumference I won't be able to measure it like I planned.
Is there any method or formula that can be used to measure a shape like this, or do I just have to approximate a cylinder and approximate a box and add those two together?
| A different approximation could be to take the cross-section at each height not as a linear interpolation between top and bottom surface, but as squares with rounded corners. This fits the photograph more closely, as the linear interpolation approach has a discontinuity in the radius of curvature at the bottom: The point that at the top is an edge of the square is modelled a crease along the entire side, while in the photograph, no such crease can be seen.
If we take the cross-section to have the above shape of a rounded square, we can calculate its area by using the formulas for circles and rectangles:
$$A = \pi \cdot r^2 + 2 \cdot a \cdot b - b^2$$
Using $b=a-2\cdot r$, we can eliminate that variable:
$$A = a^2 + (\pi -4) \cdot r^2$$
Now we assume that those variables vary linearly between top and bottom: $r$ goes from $r_0$ at the bottom to $0$ at the top, and $a$ goes from $2r_0$ at the bottom to $a_0$ at the top, giving us:
$$r(h) = \left(1-\frac{h}{h_0}\right)\cdot r_0$$
$$a(h) = \left(1-\frac{h}{h_0}\right)\cdot 2r_0 + \frac{h}{h_0} \cdot a_0$$
and thus:
$$A(h) = \left(1-\frac{h}{h_0}\right)^2 \cdot \pi \cdot r_0^2 + \left(1-\frac{h}{h_0}\right) \cdot \frac{h}{h_0} \cdot 4 \cdot r_0 \cdot a_0 + \left(\frac{h}{h_0}\right)^2 \cdot a_0^2$$
or, better ordered for integration:
$$A(h) = \pi \cdot r_0^2 + \frac{h}{h_0} \cdot \left( 4\cdot r_o \cdot a_0 - 2 \cdot \pi \cdot r_0^2 \right) + \left(\frac{h}{h_0}\right)^2 \cdot \left( a_0^2 - 4\cdot r_0 \cdot a_0 + \pi \cdot r_0^2\right) $$
Which gives us then:
$$ V = \int_0^{h_0} A(h) \mathrm{d}h= h_0 \cdot \pi \cdot r_0^2 + \frac{1}{2} \cdot \frac{h_0^2}{h_0} \cdot \left( 4\cdot r_o \cdot a_0 - 2 \cdot \pi \cdot r_0^2 \right) + \frac{1}{3}\cdot\frac{h_0^3}{h_0^2} \cdot \left( a_0^2 - 4\cdot r_0 \cdot a_0 + \pi \cdot r_0^2\right)$$
$$=h_0\cdot\left(\frac{1}{3}\pi r_0^2 + \frac{2}{3} r_0 a_0 + \frac{1}{3} a_0^2\right)$$
And here is a quick render of what this looks like in 3d:
This result can be obtained classically without any integrals as well, using only the formulas for the volume of conic solid and for cuboids. To do this, we split the solid into nine parts:
The central, yellow part is simply a square pyramid, and has a volume of $\frac{1}{3}\cdot h_0\cdot a_0^2$. The four magenta pieces are oblique cones with a quarter-circle as base. Again, using the formula for conic solids, their volume is each $\frac{1}{12}\cdot h_0\cdot\pi\cdot r_0^2$
And the four cyan pieces are irregularly shaped tetrahedrons, but we can determine their volume by adding a few pieces to see how they fit into a cuboid of the measurements $a_0\times r_0 \times h_0$:
In the more general case of $\frac{a_0}{2} \neq r_0$, there will be a fourth green piece needed (which would go in front and block our view of everything). However, the green pieces together form a prisma of height $a_0$ and a top surface area of $\frac{1}{2} \cdot r_0 \cdot h_0$, and the two blue-grey pieces are two pyramids with height $h_0$ and rectangular bases of size $\frac{1}{2} a_0 \times r_0$. Therefore, the volume of the tetrahedron is:
$$a_0 \cdot h_0 \cdot r_0 - a_0 \cdot \frac{1}{2} \cdot r_0 \cdot h_0 - 2 \cdot \frac{1}{3} \cdot h_0 \cdot \frac{1}{2} a_0 \cdot r_0 = \frac{1}{6} h_0 \cdot r_0 \cdot a_0$$
Putting it all together, we get the volume as:
$$V=\frac{1}{3}\cdot h_0\cdot a_0^2 + 4\cdot \frac{1}{12}\cdot h_0\cdot\pi\cdot r_0^2 + 4 \cdot \frac{1}{6} h_0 \cdot r_0 \cdot a_0$$
$$=h_0\cdot\left(\frac{1}{3}\pi r_0^2 + \frac{2}{3} r_0 a_0 + \frac{1}{3} a_0^2\right)$$
A more smooth model along the same lines would be to interpolate superellipses between the bottom and top, which similarly would give a creaseless change in the radius of curvature beneath the corners. However, superellipse areas have a formula involving the gamma function, and are therefore not easy to integrate again to get a volume.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3645139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "86",
"answer_count": 5,
"answer_id": 3
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Not understanding how answer to vector question is arrived at I'm going through Multivariable and Vector Calculus by Sarhan Musa and David Santos, and I'm stuck on exercise 1.1.9.
The question says:
Let A,B be two points on the plane. Construct two points I and J such that $$\overline{IA}=-3\overline{IB}$$ $$\overline{JA} = -\frac{1}{3}\overline{JB}$$ and then demonstrate that for any arbitrary point M on the plane $$\overline{MA} + 3\overline{MB} = 4\overline{MI}$$ and $$3\overline{MA} + \overline{MB} = 4\overline{MJ}$$
I've drawn the question out and I can see vaguely geometrically how that could be the case. But I don't see how the author arrives at the answer (from the back of the book):
$$\begin{align}\overline{MA} + 3\overline{MB} &= 3\overline{MI}+\overline{MI}+\overline{IA}+3\overline{IB} \tag{1}\\&= 4\overline{MI} + \overline{IA} + 3\overline{IB}\tag{2}\\&= 4\overline{MI}\tag{3} \end{align}$$ and $$\begin{align}3\overline{MA} + \overline{MB} &= 3\overline{MJ} + 3\overline{JA}+\overline{MJ}+\overline{JB} \tag{4}\\ \tag{5}&=4\overline{MJ} + 3\overline{JA} + \overline{JB} \\ \tag{6}&= 4\overline{MJ}\end{align}$$
Specifically, how does the author arrive at the right hand side of lines 1 and 4?
| For (1), we write:
\begin{align*}\overline{MA} + 3\overline{MB}
&= \left(\overline{MI}+\overline{IA} \right) + 3\left(\overline{MI}+\overline{IB}\right)\\
&= \overline{MI}+\overline{IA} + 3\overline{MI} + 3\overline{IB}\\
&= 3\overline{MI} + \overline{MI}+\overline{IA} + 3\overline{IB}
\end{align*}
The same computation with $J$ instead of $I$ yields (4)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3647403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Prove $\sum_{n\geq1}\frac{2^n (1-\cos\frac{x}{2^n})^2}{\sin\frac{x}{2^{n-1}}}=\tan\frac{x}{2}-\frac{x}{2}$ How to prove for $|x|<\pi$:
*
*$\sum_{n\geq1}\frac{2(1-\cos(\frac{x}{2^n}))}{\sin(\frac{x}{2^{n-1}})}=\tan(\frac{x}{2})$
*$\sum_{n\geq1}\frac{2^n (1-\cos(\frac{x}{2^n}))^2}{\sin(\frac{x}{2^{n-1}})}=\tan(\frac{x}{2})-\frac{x}{2}$
Any help will be appreciated.
| Note
$$ \begin{align}
(1-\cos\frac{x}{2^n})^2&= 1-2\cos\frac{x}{2^n} + \cos^2\frac{x}{2^n}\\
&=\sin^2\frac{x}{2^n}-2\cos\frac{x}{2^n}(1-\cos\frac{x}{2^n}) \\&=\sin^2\frac{x}{2^n}-4\cos\frac{x}{2^n}\sin^2\frac{x}{2^{n+1}}
\end{align}$$
Then
$$\frac{(1-\cos\frac{x}{2^n})^2}{\sin\frac{x}{2^{n-1}}}
= \frac{\sin^2\frac{x}{2^n} }{2\cos\frac{x}{2^{n}} \sin\frac{x}{2^{n}}} - \frac{4\cos\frac{x}{2^n}\sin^2\frac{x}{2^{n+1}}}{4\cos\frac{x}{2^{n}} \cos\frac{x}{2^{n+1}} \sin\frac{x}{2^{n+1}}}
=\frac12\tan \frac{x}{2^n}- \tan\frac{x}{2^{n+1}}
$$
Thus
\begin{align}
\sum_{n\geq1}\frac{2^n (1-\cos(\frac{x}{2^n}))^2}{\sin(\frac{x}{2^{n-1}})}
&= \sum_{n\geq1}\left( 2^{n-1} \tan \frac{x}{2^n}- 2^2\tan\frac{x}{2^{n+1}}\right) \\
&= \tan\frac{x}{2}-\frac{x}{2} \lim_{n\to \infty} \frac{\tan\frac{x}{2^{n+1}}}{\frac{x}{2^{n+1}}} = \tan\frac{x}2-\frac{x}{2}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3651555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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System of Equations: Distances from the Torricelli Point to the Vertices of a Triangle I am calculating the length between the Torricelli point and the triangle apex. With the law of cosines I get this system:
$$\left\{\begin{array}{cc}
x^2+y^2+xy&=a^2\,,\\x^2+z^2+xz&=b^2\,,\\z^2+y^2+zy&=c^2\,.\end{array}\right.$$
Please help me solve it.
| In order for me to remember which variable corresponds to which geometric object. I will relabel them as follows:
Let $A,B,C$ be the 3 vertices and $a,b,c$ be the 3 sides of a triangle in usual notation. Let $u, v, w$ be the distance
between the Torricelli point and $A,B,C$ respectively.
Let $\Delta$ be the area of the triangle. The problem at hand becomes:
Given $$\begin{align} \alpha &\stackrel{def}{=} a^2 = v^2 + vw + w^2\\
\beta &\stackrel{def}{=} b^2 = w^2 + wu + u^2\\ \gamma
&\stackrel{def}{=} c^2 = u^2 + uv + v^2 \end{align}$$
How to solve for $u,v,w$?
Let $\delta = uv + vw + wu$, it is not hard to verify
$$\begin{array}{rcrcrcrl} \alpha &+& \beta &+& \gamma &+& 3\delta &= 2(u+v+w)^2\\
-\alpha &+& \beta &+& \gamma &+& \delta &= 2u(u+v+w)\\
\alpha &-& \beta &+& \gamma &+& \delta &= 2v(u+v+w)\\
\alpha &+& \beta &-& \gamma &+& \delta &= 2w(u+v+w)
\end{array}$$
This implies
$$(u,v,w) = \left(\frac{P-2a^2}{Q}, \frac{P-2b^2}{Q}, \frac{P-2c^2}{Q}\right)$$
where $\;P = a^2 + b^2 + c^2 + \delta\;$ and $\;Q = \sqrt{2(a^2 + b^2 + c^2 + 3\delta)}\;$.
Everything comes down to the computation of $\delta$.
Notice the area of triangle can be expressed in terms of $\delta$,
$$\Delta = \frac12(uv + vw + wu)\sin\frac{2\pi}{3} = \frac{\sqrt{3}}{4}\delta$$
With help of Heron's formula, $\delta$ can be computed as follows:
$$\delta = \frac{4}{\sqrt{3}}\Delta =
\frac{1}{\sqrt{3}}\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3653184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Geometry circle problem and the curves $y= \pm \frac{1}{x}$
A circle touches the $y-$axis at the origin and the curves $y= \frac{1}{x}$, $y= -\frac{1}{x}$ when $x>0$. Determine the radius of the circle.
I managed to do the question, but it got pretty messy and terrible. By first denoting the center as $C(x_0, y_0)$ we can deduce that the radius is simply $x_0$. Then defining the points $A(a, \frac{1}{a})$ and $B(b, -\frac{1}{b})$ and drawing the tangents from the center to the points where the curves touch the circle I could algebraicly solve this and here it went messy. I had to use the derivative and the property that the product of two gradients equal $-1$ in order to first find point $a$ from what I could use to find $x_0$. Is there some clean way to do this or is it just one of those problems where it gets messy anyway?
| It is quite easy. By the symmetry of the problem, we know the circle's center must lie along the $x$-axis. So we only need one coordinate to describe it, which we will call $c$. Then the circle must have equation
$$(x - c)^2 + y^2 = r^2$$
Moreover, since the left hand extremity of the circle should touch the $y$-axis, then a line from the center to that touching point is a radius, so $r = c$. Thus
$$(x - c)^2 + y^2 = c^2$$
and we can expand out to get
$$\begin{align}
(x^2 - 2cx + c^2) + y^2 &= c^2\\
x^2 + y^2 - 2cx + c^2 &= c^2\\
x^2 + y^2 &= 2cx
\end{align}$$
So we now have a growable circle coming off the y-axis, that we can let the little thing grow up (by letting $c$ do so) just as much as need be so he juuust bumps/kisses the curve (❤︎#mehhr). Hence, we must find the limiting case where only a single point of the circle lies upon the curve. So take $y = \pm \frac{1}{x} = \pm x^{-1}$ in the above, to get
$$x^2 + x^{-2} = 2cx$$
Times $x^2$, you get
$$x^4 + 1 = 2cx^3$$
so we want to find which $c$ makes
$$x^4 - 2cx^3 + 1 = 0$$
have only a single solution (coordinate of point of intersection) in $x$. This is most easily done with calculus, though there are purely algebraic methods but a bit nontrivial since this is a quartic (degree 4) polynomial equation. Differentiating, we get that the minimum point is when $4x^3 - 6cx^2 = 0$ so $4x - 6c = 0$ and $x = \frac{6c}{4} = \frac{3c}{2}$. The height of the minimum is then
$$\begin{align}
\left(\frac{3c}{2}\right)^4 - 2c\left(\frac{3c}{2}\right)^3 + 1 &= \frac{81c^4}{16} - 2c\left(\frac{27c^3}{8}\right) + 1\\
&= \frac{81c^4}{16} - \frac{54c^4}{8} + 1\\
&= \frac{81c^4}{16} - \frac{108c^4}{16} + \frac{16}{16}\\
&= \frac{-27c^4 + 16}{16}\end{align}$$
So this height is 0 when $16 = 27c^4$ or $c = r = \sqrt[4]{\frac{16}{27}}$.
Thus the answer is the circle at $\left(\sqrt[4]{\frac{16}{27}}, 0\right)$ and with radius $\sqrt[4]{\frac{16}{27}}$. For scale, this center and radius are approximately 0.877 units, which looks to be about right from your graph given the center is just a bit to the left of 1 on the x-axis. This number seems to be what everybody else is getting, just expressed in different ways.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3653711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
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What is this period of this Markov chain? We have the following markov matrix.
\begin{bmatrix}0&1&0\\1/2&0&1/2\\0&1&0\end{bmatrix}
So we have states 1, 2, and 3. State 2 always return to 2 in 2 steps. States 1 and 3 can return to themselves in a multiple of 2 steps. So I am assuming that the period is 2? I know the strict definition of periodicity in markov chains, but I am trying to develop an intuition for the meaning.
| Indeed the period of this chain is $d=2$, and the classes are $\{1, 3\}$ and $\{2 \}$. Starting from state $2$ at time $n=0$, the chain will be in state $2$ at any even time $n=2p$ with probability 1; and at odd times, the chain will belong to the set $\{1, 3\}$ with probability 1.
Generally speaking, for an irreducible chain, a unique partition of the state space $S$ into $d$ sets $E_0, E_1, \dots, E_{d-1}$ can be found such that, for all $k \in \{0, 1, \dots, d-1 \}$ and for any $i \in E_k$,
$$\sum_{j \in E_{k+1}} p_{i,j} = 1,$$
where by convention $E_d = E_0$, and where $d$ is maximal. Therefore the chain moves from one class to the other at each transition, and this cyclically. In other words, the transition matrix $P$ can be written with blocks as follows:
$$
\begin{array}{c c} &
\begin{array}{c c c c c} \, E_0 \,\,\,\,& E_1 \,& E_2 & \dots & E_{d-1} \end{array}\\
\begin{array}{c} E_0\\ E_1 \\ E_2\\ \vdots\\ E_{d-1} \end{array} &
\begin{pmatrix}
0 & \mathbf{A}_0 & 0 & \cdots & 0 \\
0 & \ddots & \mathbf{A}_1 & 0 & 0\\
\vdots & \ddots & \ddots & \ddots &0\\
0 & & 0& 0& A_{d-2}\\
\mathbf{A}_{d-1}& 0 & \cdots & \cdots & 0
\end{pmatrix}
\end{array}
$$
Note that if the diagonal of the transition matrix is not zero, then the chain is acyclic, i.e. $d=1$ (The converse is not true). Moreover it can be checked that $P^d$ is block-diagonal.
In our case, the order of the states can be changed to $\{1, 3, 2\}$, so that the transition matrix $P$ has the form described above:
$$
\begin{array}{c c} &
\begin{array}{c c c} 1 & 3 & 2 \end{array}\\
\begin{array}{c} 1 \\ 3 \\ 2 \end{array} &
\begin{pmatrix}
0 & 0 & 1 \\
0 & 0 & 1 \\
\frac{1}{2} & \frac{1}{2} & 0
\end{pmatrix}
\end{array}
$$
with $E_0 = \{1, 3\}$, $E_1 = \{2\}$, $\mathbf{A_0} = \begin{pmatrix}1 \\ 1\end{pmatrix}$, $\mathbf{A_1} = \begin{pmatrix}\frac{1}{2} & \frac{1}{2}\end{pmatrix}$. Moreover, we can check that the transition matrix raised to the power $d=2$ is block-diagonal:
$$P^2 = \begin{pmatrix}
\frac{1}{2} & \frac{1}{2} & 0 \\
\frac{1}{2} & \frac{1}{2} & 0 \\
0 &0 &1
\end{pmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3653849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"If $a+a^3$ is irrational, then $a+a^2$ is also irrational". Given $a\in\mathbb{R}$, define $p=a+a^2$ and $q=a+a^3$. Show that the following statement is false:
"If $q$ is irrational, then $p$ is also irrational".
My approach: I just was lucky to find a counterexample setting $p=1$. Then, I found out that for both $a$ that satisfy the equation $q$ is irrational.
I wonder if someone here could come up with a more neat solution.
| Brainstorming. Given
$$ab=q$$
$$a+b=p+1$$
where $b=a^2+1$, we obtain $$a+\frac{q}{a}=p+1 \Rightarrow a^2-(p+1)a+q=0$$
with (recalling Vieta's)
$$a,b=\frac{p+1\pm\sqrt{(p+1)^2-4q}}{2}$$
One counter-example satisfying all of the above, is $a=\frac{\sqrt{2}-1}{2}$ where $a^2+1=\frac{7}{4}-\frac{\sqrt{2}}{2}$ and $p=\frac{1}{4}\in\mathbb{Q}$, whether $q=\frac{9\sqrt{2} - 11}{8}$ - irrational.
In fact, this reveals many more counter-examples like $a=\frac{k\sqrt{2}}{2}-\frac{1}{2} \Rightarrow a^2+1=\frac{2k^2+5}{4}-\frac{k\sqrt{2}}{2}$ and $p=\frac{2k^2-1}{4}\in\mathbb{Q}$, $q=\frac{(2k^3+7k)\sqrt{2} - 6k^2-5}{8}$ - irrational, for $k$ - integer.
Or even more for $a=\frac{k\sqrt{m}}{2}-\frac{1}{2} \Rightarrow a^2+1=\frac{mk^2+5}{4}-\frac{k\sqrt{m}}{2}$ and $p=\frac{mk^2-1}{4}\in\mathbb{Q}$ ... where $m$ - not a perfect square, covering the $p=1$ ($k=1$, $m=5$) case as well.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why $\sqrt{23-\sqrt{17}}-2\sqrt{7-\sqrt{17}}=\sqrt{71-17\sqrt{17}}$ is true? Easy to show this identity after squaring twice of the both sides.
But why it turned out true?
For example, if we want to prove that
$$\sqrt{23-3\sqrt{5}}-2\sqrt{3-\sqrt{5}}=\sqrt{3+\sqrt{5}},$$
we can do it without squaring:
$$\sqrt{23-3\sqrt{5}}-2\sqrt{3-\sqrt{5}}=\frac{1}{\sqrt{2}}\left(\sqrt{46-6\sqrt{5}}-2\sqrt{6-2\sqrt{5}}\right)=$$
$$=\frac{1}{\sqrt{2}}\left(\sqrt{(3\sqrt{5}-1)^2}-2\sqrt{(\sqrt5-1)^2}\right)=\frac{1}{\sqrt{2}}\left(3\sqrt{5}-1-2(\sqrt{5}-1)\right)=$$
$$=\frac{1}{\sqrt{2}}(\sqrt{5}+1)=\frac{1}{\sqrt{2}}\sqrt{6+2\sqrt{5}}=\sqrt{3+\sqrt{5}}.$$
But this way does not work for the starting identity.
How to prove the starting identity without squaring?
Thank you!
| Note that
$$\sqrt{23-\sqrt{17}}=\sqrt{7-\sqrt{17}}\sqrt{\frac{9+\sqrt{17}}2}=\frac{1+\sqrt{17}}{2}\sqrt{7-\sqrt{17}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3656053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Showing that $\sum_{m=1}^{n}{\sigma(m)}=\sum_{k=1}^{n}{k\cdot \left\lfloor \frac n k\right\rfloor}$ Let $n \in \mathbb{N}$.
$\sigma$ is an arithmetic function and $\sigma(n)$ is the sum of the (positive) divisors of $n$.
How to show that $\sum_{m=1}^{n}{\sigma(m)}=\sum_{k=1}^{n}{k\cdot \left\lfloor \frac n k\right\rfloor}$?
My idea was to show it by induction:
$n=1:$
$\sum_{m=1}^{1}{\sigma(m)}=\sigma(1)=\sum_{d\vert1}{d}=1=\sum_{k=1}^{1}{1\cdot \left\lfloor \frac 1 1\right\rfloor}$
So it holds for $n=1$.
$n \mapsto n+1:$
$\sum_{m=1}^{n+1}{\sigma(m)}$
$=\sigma(n+1)+\sum_{m=1}^{n}{\sigma(m)}$
$=\sigma(n+1)+\sum_{k=1}^{n}{k\cdot \left\lfloor \frac n k\right\rfloor}$
$=\sum_{d\vert n+1}{d}+\sum_{k=1}^{n}{k\cdot \left\lfloor \frac n k\right\rfloor}$
$=\sum_{k=1}^{n+1}{k\cdot \left\lfloor \frac n k\right\rfloor}$
So it holds for all $n \in \mathbb{N}$.
I'm not sure if this is completely correct. Especially in the last step I don't know how to argue that the equality holds. Is there something to improve?
| To prove the induction step, consider
$$\begin{equation}\begin{aligned}
& \sum_{k=1}^{n+1}k\cdot \left\lfloor \frac{n+1}{k}\right\rfloor - \sum_{k=1}^{n}k\cdot \left\lfloor \frac{n}{k}\right\rfloor \\
& = (n + 1)\cdot \left\lfloor \frac{n+1}{n+1}\right\rfloor + \sum_{k=1}^{n}k\cdot \left\lfloor \frac{n+1}{k}\right\rfloor - \sum_{k=1}^{n}k\cdot \left\lfloor \frac{n}{k}\right\rfloor \\
& = (n + 1) + \sum_{k=1}^{n}k\cdot \left(\left\lfloor \frac{n+1}{k}\right\rfloor - \left\lfloor \frac{n}{k}\right\rfloor\right)
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
The part in the summation brackets is
$$m = \left\lfloor \frac{n+1}{k}\right\rfloor - \left\lfloor \frac{n}{k}\right\rfloor \tag{2}\label{eq2A}$$
For each $k$, the division algorithm states there's a unique positive integer $j$ and non-negative integer $r$ such that
$$n = jk + r, \; 0 \le r \lt k \tag{3}\label{eq3A}$$
With $0 \le r \lt k - 1$, \eqref{eq2A} gives $m = j - j = 0$. Only when $r = k - 1$, so $n + 1 = (j + 1)k$, does \eqref{eq2A} give a non-zero value, i.e., $m = (j + 1) - j = 1$. This means the only non-zero terms in the summation in the last line of \eqref{eq1A} are of $k$ and occur when $k$ is a factor of $n + 1$, so the summation itself is of all positive factors of $n + 1$ less than or equal to $n$. Adding $n + 1$ then gives the sum of all positive factors of $n + 1$. Thus, replacing the last line in \eqref{eq1A} with this and using the induction hypothesis gives
$$\begin{equation}\begin{aligned}
& \sum_{k=1}^{n+1}k\cdot \left\lfloor \frac{n+1}{k}\right\rfloor - \sum_{k=1}^{n}k\cdot \left\lfloor \frac{n}{k}\right\rfloor = \sum_{d\,\mid\, n+1}{d} \\
& \sum_{k=1}^{n+1}k\cdot \left\lfloor \frac{n+1}{k}\right\rfloor - \sum_{m=1}^{n}\sigma(m) = \sigma(n + 1) \\
& \sum_{k=1}^{n+1}k\cdot \left\lfloor \frac{n+1}{k}\right\rfloor = \sum_{m=1}^{n + 1}\sigma(m)
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3656581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Sum of all integers from $1$ to $20$ except one is a multiple of $20$
Sam was adding the integers from $1$ to $20$. In his rush, he skipped one of the numbers and forgot to add it. His final sum was a multiple of $20$. What number did he forget to add?
My idea was to use Gauss's trick to find this relatively simply so I proceeded as follows.
We have $S=1+2+3+ \dots+ 18+19+20$. Using Gauss's trick we get $\frac{n(n+1)}{2} = \frac{20(21)}{2} = 210$. Since we want this to equal some multiple of $20$ we have that $210 = 20n$, but solving for $n$ results in $\frac{21}{2} = 10.5$.
The correct answer for this was $10$, but it seems that I'm missing something?
|
but it seems that I'm missing something?
Oh, I can't resist.
You aren't missing anything. That's the problem.
(....laughs in the corner to himself for hours....)
You were supposed to skip a number but you included them all.
The numbers add to $1+2+3 + 4+.... + 20 = 210$ and you did that correctly. But Sam did not do it correctly. Sam left out a number. So Sam did not get $210$.
So what did Sam get. If the number he skipped was $k$ then same got $1 + 2 + 3 + .... + (k-1) + (k+1) + .... +20 = 210 - k$.
So we have $210 - k = 20n$. Now $k$ can be as small as $1$ and is $210-k = 20n$ can be as big as $209$. And $k$ can be as big as $20$ so $210-k = 20n$ can be as small as $190$. So $190 \le 210-k = 20n \le 209$. Then only number in that range that is divisible by $20$ is ... $200$. So $20n = 210 -k = 200$. And that means $k$ is .... $10$.
And indeed, if $1 + 2 + 3 + 4 + ..... + 9 + 11 + 12 + 13 + .... + 20 = $
$(1 + 2+ .... + 9) + (11+12 + .... + 20) = $
$(\underbrace{1 + \underbrace {2 + .... +8}+9}) + (\underbrace{11 + \underbrace {12 + .... +19}+20})=$
$(4*10+5) + (5*31) =$
$45 + 155 = 200$.
Then $9+8+7+4 +..... + 1 + 20 + 12 + 13 + .... + 20 = N$ and
$10+10+10+ .... + 10 + 31 + 31+.... + 31 = N+N$
$90 + 310 = 2N$
$400 = 2N$
$N = 200$.
....
More elegant, if you know modular arithmetic is:
$20$ divides $(1+2+3+ ..... + 20) -k$ so
$(1+2+3 + ..... + 20) -k \equiv 0 \pmod {20}$
So $210 -k \equiv 0 \pmod {20}$
$k \equiv 210 \equiv 10 \pmod {20}$.
And as $1\le n \le 20; k \equiv 10 \pmod{20}$ then $k =10$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3657106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Metric on semi-Riemannian space Here we have some (blue) curves $\ln(x)\ln(y)=s$ for several values of $s$ and some (green) curves $\ln(1-x)\ln(y)=s$ for several values of $s.$ The (red) is the multiplication of the green and blue.
The blue and green curves are the invariant "hyperbola."
Let $(M_1,g_1)$ be the blue semi-Riemannian manifold and $(M_2,g_2)$ the green semi-Riemannian manifold with the following metrics:
$g_1=\frac{dxdy}{xy}$ and $g_2=\frac{dudv}{v-uv}.$
Upon taking the Cartesian product of blue curves and green curves, what metric can be put on this new space (of red curves)? Will it be some sort of combination of $g_1$ and $g_2$ or something completely different?
| For two semi-Riemannian manifolds $(M_{1}, g_{1})$ and $(M_{2}, g_{2})$, we have the product metric tensor $g_{} = g_{1} \bigoplus g_{2}$. Before doing that, however, we need to obtain $g_{1}$ and $g_{2}$ in terms of $x$, $y$, $dx$, and $dy$. We now have that
\begin{equation}
ds_{1}^{2} = \frac{dx dy}{xy} \implies xy = \frac{dx dy}{ds_{1}^{2}}
\end{equation}
\begin{equation}
ds_{2}^{2} = \frac{dx dy}{x - xy} \implies x - xy = \frac{dx dy}{ds_{2}^{2}}
\end{equation}
We can now add them and find a suitable metric
\begin{equation}
xy + x - xy = x = \frac{dx dy}{ds_{1}^{2}} + \frac{dx dy}{ds_{2}^{2}} = \frac{ds_{2}^{2} dx dy}{ds_{1}^{2}ds_{2}^{2}} + \frac{ds_{1}^{2} dx dy}{ds_{1}^{2}ds_{2}^{2}} = \frac{ds_{2}^{2} dxdy + ds_{2}^{1} dxdy}{ds_{1}^{2}ds_{2}^{2}}
\end{equation}
\begin{equation}
\implies ds_{1}^{2}ds_{2}^{2} = \frac{ds_{2}^{2} dxdy + ds_{2}^{1} dx dy}{x} = dxdy \frac{ds_{2}^{2} + ds_{2}^{1}}{x} = dxdy \left( \frac{dxdy}{x^{2}y} + \frac{dx dy}{x^{2} - x^{2}y}\right)
\end{equation}
by substituting in the original metrics $ds_{1}^{2}$ and $ds_{2}^{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3657866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding the pdf of $U=X+Y$ $(X,Y)$ has the following joint pdf:
$f_{X,Y}(x,y)=x+y$ if $0<x<1, 0<y<1$
If $U=X+Y$, find the marginal pdf of $U$.
I have tried to do it using transformation.
I have considered the transformation $(X,Y)\rightarrow (U,Y)$ where $U=X+Y$.
Clearly, $0<U<2$.
Now, $x=u-y$.
The jacobian is $J(\frac{x,y}{u,y})=\frac{\delta x}{\delta u}=1$, so $|J|=1$.
So, joint pdf of $(U,Y)$ is:
$f_{U,Y}(u,y)=f_{X,Y}(u-y,y)|J|$
But $-1<u-y<2$, whereas $0<x<1$, so I don't think I can write $f_{X,Y}(u-y,y)=u-y+y=u$. I am getting stuck here.
Please anyone help me solve it. Thanks in advance.
| $x, y\in [0,1]$ so $u=x+y\in [0,2]$.
\begin{eqnarray*}
F_U(u)
&=&
P(X+Y \leq u)\\
&=&
\int_{x=0}^{x=1}\int_{y=0}^{y=1}(x+y){\bf 1}_{\{x+y \leq u\}}(x,y)dydx\\
&=&
\int_{x=0}^{x=1\wedge u}\int_{y=0}^{y=1\wedge (u-x)}(x+y)dydx\\
\end{eqnarray*}
When $u\in [0,1]$ the integral becomes
\begin{eqnarray*}
F_U(u)
&=&
P(X+Y \leq u)\\
&=&
\int_{x=0}^{x=u}\int_{y=0}^{y=u-x}(x+y)dydx\\
&=&
\int_{x=0}^{x=u}\Big(xy+\frac{1}{2}y^2\Big)\Big|_{y=0}^{y=u-x}dx\\
&=&
\int_{x=0}^{x=u}\Big(x(u-x)+\frac{1}{2}(u-x)^2\Big)dx\\
&=&
\int_{x=0}^{x=u}\Big( xu-x^2+\frac{1}{2}u^2-ux+\frac{1}{2}x^2\Big)dx\\
&=&
\frac{1}{2}\int_{x=0}^{x=u}\Big(u^2-x^2\Big)dx\\
&=&
\frac{1}{2}\Big(xu^2-\frac{1}{3}x^3\Big)\Big|_{x=0}^{x=u}\\
&=&
\frac{1}{2}\Big(u^3-\frac{1}{3}u^3\Big)\\
&=&
\frac{1}{3}u^3\\
\end{eqnarray*}
Thus, for $u\in [0,1]$ $f_U(u)=F'_U(u)=u^2$.
When $u\in [1,2]$ the integral becomes
\begin{eqnarray*}
F_U(u)
&=&
P(X+Y \leq u)\\
&=&
\int_{x=0}^{x=1}\int_{y=0}^{y=1\wedge (u-x)}(x+y)dydx\\
&=&
\int_{x=0}^{x=u-1}\int_{y=0}^{y=1}(x+y)dydx
+\int_{x=u-1}^{x=1}\int_{y=0}^{y=u-x}(x+y)dydx\\
&=&
\int_{x=0}^{x=u-1}\Big(xy+\frac{1}{2}y^2\Big)\Big|_{y=0}^{y=1}dx
+\int_{x=u-1}^{x=1}\Big(xy+\frac{1}{2}y^2\Big)\Big|_{y=0}^{y=u-x}dx\\
&=&
\int_{x=0}^{x=u-1}\Big(x+\frac{1}{2}\Big)dx
+\int_{x=u-1}^{x=1}\Big(x(u-x)+\frac{1}{2}(u-x)^2\Big)dx\\
&=&
\Big(\frac{1}{2}x^2+\frac{1}{2}x\Big)\Big|_{x=0}^{x=u-1}
+\int_{x=u-1}^{x=1}\Big(xu-x^2+\frac{1}{2}u^2-xu+\frac{1}{2}x^2\Big)dx\\
&=&
\Big(\frac{1}{2}(u-1)^2+\frac{1}{2}(u-1)\Big)
+\frac{1}{2}\int_{x=u-1}^{x=1}\Big(u^2-x^2\Big)dx\\
&=&
\frac{1}{2}u^2-u+\frac{1}{2}u
+\frac{1}{2}\Big(xu^2-\frac{1}{3}x^3\Big)\Big|\int_{x=u-1}^{x=1}\\
&=&
\frac{1}{2}u^2-\frac{1}{2}u
+\frac{1}{2}\Big(u^2-\frac{1}{3} - (u-1)u^2+\frac{1}{3}(u-1)^3\Big)\\
&=&
u^2-\frac{1}{3}-\frac{1}{3}u^3
\end{eqnarray*}
Thus, for $u\in [1,2]$ $f_U(u)=F'_U(u)=2u-u^2$.
Hence, overall, the density is $f_U(u) = u^2{\bf 1}_{\{0\leq u\leq 1\}}+(2u-u^2){\bf 1}_{\{1< u\leq 2\}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3658971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving the system with $x^2 -xy -1=0$ and $2xy-4y^2 +3=0$ I want to solve the following system of quadratic equations:
$$\begin{align}
x^2 -xy -1 &=0 \\ 2xy-4y^2 +3 &=0
\end{align}$$
How can I do it?
My attempt:
$$xy = 1 -x^2$$
$$2x^2=4y^2-1$$
$$x = \sqrt{2y^2 - 0.5}$$
After plugging $x$, I get an equation that I can't solve.
| $$xy=x^2-1$$
$$2(x^2-1)=4y^2-3$$
$$2x^2=4y^2-1$$
$$4y^2=2x^2+1\tag{3}$$
From the first equation,
$$x^2-1=xy$$
Let's square both sides and use $(3)$,
$$(x^2-1)^2=x^2y^2=x^2\left(\frac12 x^2 + \frac14\right)$$
$$x^4-2x^2+1=\frac12 x^4 + \frac14x^2$$
$$\frac12 x^4-\frac94x^2+1=0$$
$$2x^4-9x^2+4=0$$
$$(2x^2-1)(x^2-4)=0$$
I will leave the rest to you as an exercise.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3660205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
On integer solutions to $x\sqrt{y}+y\sqrt{x} = a, x+y = b $ A question in quora
asked to
find real solution(s) to
$x\sqrt{y}+y\sqrt{x} = 6,
x+y = 5.
$
I showed that the solution
with $x \le y$ is
$x = 1, y = 4
$.
This naturally
brings up the question
for
which positive integers $a$ and $b$
do the equations
$x\sqrt{y}+y\sqrt{x} = a,
x+y = b
$
have integer $x$ and $y$
as solutions.
It is clear that
for any integers
$1 \le p \le q$,
$a = pq(p+q), b = p^2+q^2$
has the solution
(with $x \le y$) of
$x = p^2, y = q^2$.
The original question has
$p=1, q=2$.
My question is:
are there any other
integral $a$ and $b$
for which the
equations have
integral solutions?
Note:
If we just
try to solve for $x$,
this happens
(with the help of Wolfy):
$\begin{array}\\
y
&= b-x\\
a
&=x\sqrt{y}+y\sqrt{x}\\
&=x\sqrt{b-x}+(b-x)\sqrt{x}\\
a-x\sqrt{b-x}
&=(b-x)\sqrt{x}\\
a^2-2ax\sqrt{b-x}+x^2(b-x)
&=x(b-x)^2\\
&=x(b^2-2bx+x^2)\\
2ax\sqrt{b-x}
&=x^3-2bx^2+b^2x
-a^2-(bx^2-x^3)\\
&=2x^3-3bx^2+b^2x-a^2\\
4a^2x^2(b-x)
&=(2x^3-3bx^2+b^2x-a^2)^2\\
0
&=a^4 - 2 a^2 b^2 x + 2 a^2 b x^2 + b^4 x^2 - 6 b^3 x^3 + 13 b^2 x^4 - 12 b x^5 + 4 x^6\\
\text{with real roots}\\
x
&= \dfrac12 \left(b \pm \sqrt{b^2 - 4 \left(-\dfrac{r}{12} + \dfrac{(24 a^2 b - b^4)}{12 r} + \dfrac{b^2}{12}\right)}\right)\\
\text{where}\\
r
&=\left(-216 a^4 + 36 a^2 b^3 + 24 \sqrt{3} \sqrt{27 a^8 - a^6 b^3} - b^6\right)^{1/3}\\
\end{array}
$
I don't know
how much help
this is.
| Let $\sqrt x=s,\sqrt y=t$
$$s^2+t^2=b$$
$$st(s+t)=a$$
$$b=(s+t)^2-2st=(a/st)^2-2st$$
$$2(st)^3+b(st)^2-a^2=0$$ which is a cubic equation in $st$
Once $st$ is known, use $s+t=\dfrac a{st}$
Now we form a quadratic equation whose roots are $s,t$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3664241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\frac{1}{1 - \sqrt{ab}} + \frac{1}{1 - \sqrt{bc}} + \frac{1}{1 - \sqrt{ca}} \leq \frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c}$ Given that $0 < a , b , c < 1$.
Prove that $\frac{1}{1 - \sqrt{ab}} + \frac{1}{1 - \sqrt{bc}} + \frac{1}{1 - \sqrt{ca}} \leq \frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c}$.
I tried using modified C.S. and brute-force. But , it demands a lot of calculation. So , I want some better solution than this. Thank you.
| We have $\sqrt{ab}\leq \frac{a+b}{2}$ so $1-\sqrt{ab}\geq 1-\frac{a+b}{2}>0$, which implies
$$\dfrac{1}{1-\sqrt{ab}}\leq \dfrac{2}{2-a-b}$$
And using the simple inequality $\frac{4}{x+y}\leq \frac{1}{x}+\frac{1}{y}$ for all $x,y>0$ you get
$$\dfrac{1}{1-\sqrt{ab}}\leq \dfrac{2}{2-a-b}\leq \dfrac{1}{2(1-a)}+\dfrac{1}{2(1-b)}$$
Summing up gives the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3664739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
If $A=\begin {bmatrix} a & b \\ 0 & 1 \end{bmatrix}$, find $A^n$ for positive integer $n$
If $A=\begin {bmatrix} a & b \\ 0 & 1 \end{bmatrix}$, find $A^n$ for positive integer $n$.
Now my method of solving these were to take $n=1,2,3..$, find the respective value and analyse the options to find the right one. It works well, but I never quite figured out how to do this properly, in case options aren’t given. Thanks!
| Another way if you're interesting.
$A=B+C$ where $B=\begin {bmatrix} a & 0 \\ 0 & 1 \end{bmatrix}$ and $C=\begin {bmatrix} 0 & b \\ 0 & 0 \end{bmatrix}$
We have $B^n=\begin {bmatrix} a^n & 0 \\ 0 & 1 \end{bmatrix}$, $C^2=0$ and $BC=aCB=\begin {bmatrix} 0 & ab \\ 0 & 0 \end{bmatrix}$
So $A^n=(B+C)^n=B^n+\displaystyle\sum_{i=0}^{n-1}B^i C B^{n-1-i}$
We have $B^i C B^{n-1-i}=aB^{i-1}C B^{n-i}=\cdots=a^iCB^{n-1}=a^i \begin {bmatrix} 0 & b \\ 0 & 0 \end{bmatrix} $ so
$$A^n=B^n+\displaystyle\sum_{i=0}^{n-1}a^i C B^{n-1}=\begin {bmatrix} a^n & 0 \\ 0 & 1 \end{bmatrix}+\displaystyle\sum_{i=0}^{n-1}a^i \begin {bmatrix} 0 & b \\ 0 & 0 \end{bmatrix}=\begin {bmatrix} a^n & b\displaystyle\sum_{i=0}^{n-1}a^i \\ 0 & 1 \end{bmatrix}$$
If $a=1$ then $A^n=\begin {bmatrix} 1 & nb \\ 0 & 1 \end{bmatrix}$
If $a\neq 1$ then $A^n=\begin {bmatrix} a^n & b\frac{a^n-1}{a-1} \\ 0 & 1 \end{bmatrix}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3666817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
If $\alpha, \beta$ are the roots of the equation $x^2-(p+1)x+1=0.$ show that $\alpha^n + \beta^n$ is not divisible by $p$ $(p \ge3)$ Let $p \ge 3$ be an integer and $\alpha, \beta$ are the roots of the equation $x^2-(p+1)x+1=0.$ Using mathematical induction show that $\alpha^n + \beta^n$
(i) is an integer
(ii) is not divisible by $p$
First part is fairly easy, as we can write $(\alpha^n+\beta^n)(\alpha+\beta)= \alpha^{n+1}+\beta^{n+1}+\alpha^{n-1}+\beta^{n-1}$ and then we can show that $P(n+1)\in\Bbb Z$.
For the second part, we can show patterns of remainder in $P(1)$, $P(2),\dots$ but I was looking for general proof for the same.
| Hint. From Vieta's
$$\alpha+\beta =p+1$$
$$\alpha\cdot\beta =1$$
or
$$\alpha+\frac{1}{\alpha}=p+1 \equiv \color{red}{1} \pmod{p}$$
$$\alpha^2+\frac{1}{\alpha^2}=(p+1)^2-2 \equiv \color{red}{-1} \pmod{p}$$
and by induction
$$\alpha^{n+1}+\frac{1}{\alpha^{n+1}}=
\left(\alpha^n+\frac{1}{\alpha^n}\right)\cdot\left(\alpha+\frac{1}{\alpha}\right)-\left(\alpha^{n-1}+\frac{1}{\alpha^{n-1}}\right) \tag{1}$$
Using $(1)$
$$\alpha^{3}+\frac{1}{\alpha^{3}}=
\left(\alpha^2+\frac{1}{\alpha^2}\right)\cdot\left(\alpha+\frac{1}{\alpha}\right)-\left(\alpha+\frac{1}{\alpha}\right) \equiv \color{red}{-2} \pmod{p}$$
$$\alpha^{4}+\frac{1}{\alpha^{4}}=
\left(\alpha^3+\frac{1}{\alpha^3}\right)\cdot\left(\alpha+\frac{1}{\alpha}\right)-\left(\alpha^{2}+\frac{1}{\alpha^{2}}\right) \equiv -1 \pmod{p}$$
$$\alpha^{5}+\frac{1}{\alpha^{5}}=
\left(\alpha^4+\frac{1}{\alpha^4}\right)\cdot\left(\alpha+\frac{1}{\alpha}\right)-\left(\alpha^{3}+\frac{1}{\alpha^{3}}\right) \equiv 1 \pmod{p}$$
$$\alpha^{6}+\frac{1}{\alpha^{6}}=
\left(\alpha^5+\frac{1}{\alpha^5}\right)\cdot\left(\alpha+\frac{1}{\alpha}\right)-\left(\alpha^{4}+\frac{1}{\alpha^{4}}\right) \equiv 2 \pmod{p}$$
$$\alpha^{7}+\frac{1}{\alpha^{7}}=
\left(\alpha^6+\frac{1}{\alpha^6}\right)\cdot\left(\alpha+\frac{1}{\alpha}\right)-\left(\alpha^{5}+\frac{1}{\alpha^{5}}\right) \equiv \color{red}{1} \pmod{p}$$
$$\alpha^{8}+\frac{1}{\alpha^{8}}=
\left(\alpha^7+\frac{1}{\alpha^7}\right)\cdot\left(\alpha+\frac{1}{\alpha}\right)-\left(\alpha^{6}+\frac{1}{\alpha^{6}}\right) \equiv \color{red}{-1} \pmod{p}$$
$$\alpha^{9}+\frac{1}{\alpha^{9}}=
\left(\alpha^8+\frac{1}{\alpha^8}\right)\cdot\left(\alpha+\frac{1}{\alpha}\right)-\left(\alpha^{7}+\frac{1}{\alpha^{7}}\right) \equiv \color{red}{-2} \pmod{p}$$
The reminder "period" starts revealing, given that $R(\alpha, n) = \alpha^{n}+\frac{1}{\alpha^{n}} \pmod {p}$
$$R(\alpha, n)=R(\alpha, n-1)-R(\alpha, n-2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3668453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding area between two curves using double integral Find the area between two curves:
*
*$x \ge 0$
*$(x^2+y^2)^2=x^2-y^2$
using double integrals.
| Here's the area:
Solve $(x^2 + y^2)^2 = x^2 − y^2$ for $y(x)$ and get four solutions, restrict to positive $x$ to find:
$$y(x) = \pm \sqrt{2} \sqrt{-1 - 2 x^2 + \sqrt{1 + 8 x^2}}$$
so:
$$\int\limits_{x=0}^1 2 \sqrt{2} \sqrt{-1 - 2 x^2 + \sqrt{1 + 8 x^2}}\ dx = \frac{1}{2}$$
(I have no idea why there might be a need for a double integral.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3673453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b} +\frac{81abc}{4(a+b+c)^2} \geqq \frac{7}{4} (a+b+c)$ For $a,b,c>0$. Prove that$:$
$$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b} +\frac{81abc}{4(a+b+c)^2} \geqq \frac{7}{4} (a+b+c)$$
My proof:
We have$:$ $$\text{LHS}-\text{RHS} =\frac{g(a,b,c)}{4abc(a+b+c)^2} \geqq 0$$
Where
$g(a,b,c) =\frac{1}{16} \left( a+b \right) ^{2} \left( 2\,a+2\,b-c \right) ^{2} \left(
a+b-2\,c \right) ^{2}$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\frac{1}{64} \left( a-b \right) ^{2} \cdot \Big[ \left( 2\,c-a-b \right) ^{3} \left(
119\,a+119\,b+30\,c \right)$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\left( a+b-2\,c \right) ^{2} \left( 343\,{a}^{2}+346\,ab+343\,{b}^{2}
\right) $
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+24\, \left( 2\,c-a-b \right) \left( a+b \right) \left( 16\,{a}^{2}+a
b+16\,{b}^{2} \right) $
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+36\, \left( 4\,{a}^{2}-5\,ab+4\,{b}^{2} \right) \left( a+b \right) ^{
2} \Big] \geqq 0$
which is clearly true for $c=\max\{a,b,c\}$
I wish to see another proof without $uvw$! Thanks for a real lot!
You can see also here.
| I found a proof by Titu's Lemma$:$
Let $$\text{LHS} -\text{RHS} \equiv \frac{g(a,b,c)}{4abc(a+b+c)^2}$$
But we have
\begin{align*} g(a,\,b,\,c) &=\frac{1}{2} \sum\limits_{cyc} c^2(a+b-5c)^2 (a-b)^2-9(a-b)^2(b-c)^2(c-a)^2 \\&\geq \frac{1}{2(a^2+b^2+c^2)} \Big[\sum\limits_{cyc} c^2(a+b-5c)(a-b)\Big]^2-9(a-b)^2(b-c)^2(c-a)^2\\&={\frac { \left( a-b \right) ^{2} \left( b-c \right) ^{2} \left( c-a
\right) ^{2} \left( 7\,{a}^{2}+50\,ab+50\,ac+7\,{b}^{2}+50\,bc+7\,{c}
^{2} \right) }{2\,(a^2+b^2+c^2)}} \geq 0\end{align*}
So we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3678594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $p, q, r$ and $s$ are in continued proportion. Then find the minimum value of $\frac{p-s}{q-r}.$ My approach: We have, $\frac pq=\frac qr=\frac rs.$ Which implies $q^2=pr$ and $r^2=qs.$ On dividing we get:
$\frac{q^2}{r^2}=\frac{pr}{qs}$
$\implies \frac{q^3}{r^3}=\frac{p}{s}$
$\implies \frac{q^3-r^3}{r^3}=\frac{p-s}{s}$
\begin{align}
\implies\frac{p-s}{q-r}&=\frac{s(q^2+qr+r^2)}{r^3}\\\\
&=\frac{s(q^2+qr+r^2)}{qs\cdot r},\text{ as }r^2=qs\\\\
&=\frac{q^2+qr+r^2}{qr}\quad ... (1)
\end{align}
Now $(q-r) ^2\ge 0,$ true for all real $q$ and $r$. This gives
$q^2+r^2\ge 2qr\implies q^2+qr+r^2\ge 3qr.$
If $qr>0, $ then from $(1),$
$\min\left(\frac{p-s}{q-r}\right)=3.$
But I'm confused how to approach the same if $qr<0.$ Please suggest... (No additional informations were given in the question and I don't know the answer. I found this question on Quora, no one has answered it.)
| Let
$$\frac pq = \frac qr = \frac rs := a$$
We can express ratios in terms of $a$:
$$\frac pq = a \qquad \qquad \frac rq = \frac 1a \qquad \qquad \frac sq = \frac sr \frac rq = \frac{1}{a^2}$$
Hence, the expression becomes
$$\frac{p-s}{q-r} = \frac{\frac pq - \frac sq}{1 - \frac rq} = \frac{a - \frac{1}{a^2}}{1-\frac 1a} = \frac{a^3-1}{a^2-a} = \frac{a^2+a+1}{a}$$
assuming that $q \neq 0$, $a \neq 0$, and $a-1\neq 0$. Differentiating,
$$\frac{d}{da}\frac{a^2+a+1}{a}=0 \iff 1-\frac{1}{a^2} =0 \iff a = \pm 1$$
We compute
$$\frac{a^2+a+1}{a} \bigg|_{a=1} = 3 \qquad \qquad \frac{a^2+a+1}{a} \bigg|_{a=-1} = -1$$
Assuming that the ratio must be positive the minimum value is indeed $3$.
If the ratio is allowed to be negative, then clearly
$$\lim_{a\rightarrow -\infty} \frac{a^2+a+1}{a} = -\infty$$
so there is no minimum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3683756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $a+b+c=0$ find $\frac{a^2+b^2+c^2}{b^2-ca}$ I tried finding the value of $a^2+b^2+c^2$ and it is $-2(ab+bc+ca)$ but that doesn't have any common factor with $b^2-ca$ so that didn't help.
I squared on both sides of $a+b=-c$ to get $a^2+b^2-c^2+2ab=0$. But I didn't proceed any further.
| It's simple algebra to verify$$\frac{a^2+b^2+(a+b)^2}{b^2+a(a+b)}=2$$provided the denominator is nonzero, but it's $(a+b/2)^2+3b^2/4$, so that only requires $a,\,b$ to not both be $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3686798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Sum of two cubes equal to prime square If $a,b\in \mathbb{N}$ find all primes $p$ such that $a^3+b^3=p^2$
My approach-
$a^3+b^3=(a+b)(a^2-ab+b^2)=p^2$ suppose $a+b=x$ and $a^2-ab+b^2=y$ then there are two cases- $(x,y)=(p^2,1),(p,p)$
Now I am struggling for case 02 where $(x,y)=(p,p)$
| I think there's a big mistake in the two cases conclusion.
We have $(x,y) \in \{(p^2,1)(p,p)(1,p^2)\}$
So let's divide them, (noting that $a,b \ne 0$)
Case 1:
$$a+b=a^2-ab+b^2=p$$
$$(a+b)^2-3ab=p \implies p^2-3ab=p \iff 3ab=p(p-1)$$
$$p=3 \implies ab=2 \implies (a,b) \in \{(1,2),(2,1)\}$$
$$p \ne3, 3| \ p(p-1) \implies p \equiv 1 \pmod 3 \implies p=3k+1 \implies ab=kp \tag{for some $k \in \mathbb{N^*}$}$$
$$a+b=p \iff a=p-b \implies b(p-b)=kp \iff b^2=p(b-k)$$
$$p\ |\ b^2 \implies p \ | \ b \implies b=mp \implies b^2=m^2p^2 \implies m^2p=mp-k $$
for some positive integer $m$, now this rewrites to
$$mp(1-m)=k$$
which is of course impossible for positive $k$.
Case 2: $a+b=1$, $$a^2-ab+b^2=p^2 \iff (a+b)^2-3ab=p^2 \implies 1-3ab=p^2$$
which is, also, impossible
Case 3: $a+b=p^2$
$$a^2-ab+b^2=1 \iff (a+b)^2-3ab=1 \implies 3ab=p^4-1$$
$$a+b=p^2 \implies a=p^2-b \implies 3b(p^2-b)=p^4-1$$
$$\iff 3b^2-3bp^2+p^4-1=0$$
$$\implies b=\frac{1}{6} \left(3p^2 \pm \sqrt{9p^4-12(p^4-1)}\right)$$
and since the discriminant is less than zero, it's also impossible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3688628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Given that $n^4-4n^3+14n^2-20n+10$ is a perfect square, find all integers n that satisfy the condition So, I tried solving that by
$$n^4-4n^3+14n^2-20n+10=x^2\\10=x^2-a^2, a^2=n^4-4n^3+14n^2-20n+10\\10=(x+a)(x-a)$$ but I couldn't find any integers when I solved it
| Let f(n) = n^4 - 4*n^3 + 14*n^2 - 20*n + 10
As Daniel Fischer mentioned you can apply the substitution $m = n-1$ to get f(m+1) = m^4 + 8*m^2 + 1
complete the square, we want to solve $$(m^2 + 4)^2 - 15 = a^2.$$ subtract one from both sides an use the difference of squares factorization $$m^2 (m^2 + 8) = (a - 1)(a + 1)$$
lets put $b = a-1$ and $h = n^2$, we can work on this factored form of the problem: $$h (h + 8) = b (b+2)$$
Clearly there are 4 solutions attained by making both sides zero $(h,b) = (0,0), (0,-2), (-8,0), (-8,-2)$.
Let's restrict to positive integer solutions from now on since $h$ is a square it must be positive. Since $8 > 2$ A solution must have $h < b$. Put $L(h) = h (h + 8)$ and $R(b) = b (b+2)$.
*
*$L(h) < B(h)$
*$L(h) > B(h+6)$
therefore any solution will have $b = h+1$ or $h+2$ or $h+3$ or $h+4$ or $h+5$. With this in mind let's substitute in $b = h+k$:
So we only need to find small $k$ integer solutions of the following linear equation now
$$0 = (-2k + 6)h + (-k^2 - 2k + 1)$$
checking each case we quickly find there are no solutions at all.
In conclusion the only solutions are the four listed from making both sides zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3689925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Angle of Inclination if Range is at Maximum If the horizontal range of a projectile launched, without air resistance is given by: $$R = \frac{v_0\cos\theta}{g}\left(v_0\sin\theta +\sqrt{v_{0}^{2}\sin^{2}\theta + 2S_0g}\right)$$ where $S_0$ is the initial height of the projectile, $v_0$ is the initial velocity, and $g$ is the acceleration due to gravity. Consider the range as a function of $\theta$. Show that the projectile launched is a maximum for the angle of inclination $$\theta = \cos^{-1}\sqrt{\frac{2S_0g + v_{0}^{2}}{2S_0g + 2v_{0}^{2}}}.$$
This problem is from the book "Advanced Engineering Mathematics" (6th ed.) by Dennis Zill. After taking the derivative of the Range with respect to $\theta$, and equating it to zero, I'm stuck with $$\sqrt{\sin^2\theta + \frac{2S_0g}{v_{0}^2}}\cos2\theta + \cos2\theta\sin\theta -\frac{2S_0g\sin\theta}{v_{0}^{2}} = 0.$$
| I tried to change my approach in finding the derivative of the Range. Since the Range is a maximum, it implies that $\frac{\partial R}{\partial \theta} = 0$.
\begin{eqnarray*}
\ln{R} &=& \ln{v_0} - \ln{g} + \ln{\cos\theta} + \ln{\left(v_0\sin\theta + \sqrt{v_{0}^{2}\sin^2\theta + 2S_0g}\right)}\\
\frac{\partial R}{\partial \theta} &=& \frac{\partial}{\partial \theta}\left[\ln{v_0} - \ln{g} + \ln{\cos\theta} + \ln{\left(v_0\sin\theta + \sqrt{v_{0}^{2}\sin^2\theta + 2S_0g}\right)}\right]\\
\frac{R'}{R}&=& \frac{-\sin\theta}{\cos\theta} + \frac{v_0\cos\theta + \frac{v_{0}^{2}\sin\theta\cos\theta}{\sqrt{v_{0}^{2}\sin^2\theta + 2S_0g}}}{v_0\sin\theta + \sqrt{v_{0}^{2}\sin^2\theta + 2S_0g}}\\
\end{eqnarray*}
Since Range is a maximum, therefore $R' = 0$.
\begin{eqnarray*}
0 &=& -\tan\theta + \frac{\cos\theta + \frac{\sin\theta\cos\theta}{\sqrt{\sin^2\theta + \frac{2S_0g}{v_{0}^{2}}}}}{\sin\theta + \sqrt{\sin^2\theta + \frac{2S_0g}{v_{0}^{2}}}}\\
\sin\theta\left(\sin\theta + \sqrt{\sin^2\theta + \frac{2S_0g}{v_{0}^{2}}}\right)&=& \cos^2\theta \left(1 + \frac{\sin\theta}{\sqrt{\sin^2\theta + \frac{2S_0g}{v_{0}^{2}}}}\right)\\
\sin\theta\sqrt{\sin^2\theta + \frac{2S_0g}{v_{0}^{2}}} &=& \cos^2\theta
\end{eqnarray*}
Squaring both sides will yield,
\begin{eqnarray*}
\cos^4\theta &=& \sin^4\theta + \frac{2S_0g}{v_{0}^{2}}\sin^2\theta\\
\cos^4 - (1 - \cos^2\theta)^2 &=& \frac{2S_0g}{v_{0}^{2}} - \frac{2S_0g}{v_{0}^{2}}\cos^2\theta \\
2\cos^2\theta + \frac{2S_0g}{v_{0}^{2}}\cos^2\theta &=& \frac{2S_0g}{v_{0}^{2}} + 1 \\
\cos^2\theta &=& \frac{2S_0g + v_{0}^{2}}{2S_0g + 2v_{0}^{2}}
\end{eqnarray*}
Taking the root,
\begin{eqnarray*}
\cos\theta &=& \sqrt{\frac{2S_0g + v_{0}^{2}}{2S_0g + 2v_{0}^{2}}}\\
\therefore \theta &=& \cos^{-1} \sqrt{\frac{2S_0g + v_{0}^{2}}{2S_0g + 2v_{0}^{2}}}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3690648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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