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Factor $x^5-5x^3+4x$ I am trying to factor$x^5-5x^3+4x$ so that I can find the roots. I know from the answers section that the roots are where $x = 0, 1, -1, 2$ and $-2$.
I'm stuck, here's as far as I got:
$$
x^5-5x^3+4x =
x(x^4-5x^2+4)
$$
Let $u = x^2$ and just focus on the term on the right (drop the first $x$ for now):
$$x^4-5x^2+4 = u^2-5u+4x.$$
Master term is $a \times c = 1 \times 4 = 4$.
Seeking a pair of numbers that sum to the middle term $-5$ and whose product is $4$:
\begin{align}
1 \times -4 &= -4, &\text{sum} &= -3 \\
4 \times -1 &= -4, &\text{sum} &= 3 \\
2 \times -2 &= -4, &\text{sum} &= 0
\end{align}
???
I'm not sure how to proceed since I cannot find a pair of numbers that satisfy the condition.
Have I gone wrong somewhere? How can I factor $u^2-5u+4x$?
| You have to factor $u^2 - 5u+4$, not $u^2 - 5u +4x$. Since $1\cdot 4 = 4$ and $1+ 4 = 5$, $u^2-5u+4=(u-4)(u-1)=(x^2-4)(x^2-1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3757792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Let $p$ be a prime number for which also $p^2+2$ is a prime. Show that then $p^3+2$ is also a prime.
Let $p$ be a prime number for which also $p^2+2$ is a prime. Show that then $p^3+2$ is also a prime.
Computing few first primes I got:
$p=2$: $2^2+2=6$ (not satisfying the condition)
$p=3$: $3^2+2=11$ (satisfying the condition)
$p=5$: $5^2+2=27$ (not satisfying the condition)
$p=7$: $7^2+2=51$ (not satisfying the condition)
so I would have a reason to believe that only the case $p=3$ will satisfy this condition, but how would I go about showing this rigorously?
| Except $3$ all the primes are of the form $3k+1$ or $3k+2$.
If $p$ is of the form $3k+1$ then, $p^2+2$ is divisible by $3$. If $p$ is of the form $3k+2$, then $p^2+2$ is again divisible by $3$. Hence there are no primes satisfying both conditions, except $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3758777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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sums and differences of perfect powers We have
$1=3^2-2^3$
$2=3^3-5^2$
$3=2^7-5^3$
$4=5^3-11^2$
$5=3^2-2^2$
and it is unknown if $6$ is representable as a difference of two perfect powers. Next such undecided example is $14$. More: http://oeis.org/A074981
However, I found that
$6=64-49-9=2^6-7^2-3^2$
and
$6=27+4-25=3^3+2^2-5^2$
Similarly
$14=27-9-4=3^3-3^2-2^2$
and
$14=9+9-4=3^2+3^2-2^2$
My question:
Is every positive integer representable in the form: $a_1^{n_1}+a_2^{n_2}-a_3^{n_3}$ or/and in the form $a_1^{n_1}-a_2^{n_2}-a_3^{n_3}$ where $a_1,a_2,a_3,n_1,n_2,n_3$ are natural numbers greater than $1$ with $a_2=0$ also acceptable ?
Are these things known?
The question is based on my own investigation.
| There are trivial solutions with squares:
$$n=a^k+b^2-c^2\iff(c+b)(c-b)=a^k-n$$
Choose $a$ to have opposite parity from $n$, so that $a^k-n\pm1$ is even, and then let $c+b=a^k-n$ and $c-b=1$ and solve for $c$ and $b$, i.e.,
$$n=a^k+\left(a^k-n+1\over2\right)^2-\left(a^k-n-1\over2 \right)^2$$
Remark: You could eliminate these trivial solutions by asking that the three powers all be different. Adding such a requirement, though, suggests doing the same for the original problem, and seeing which numbers can be written as a difference of two perfect powers of different degree. So $5=3^2-2^2$ is no longer allowed, but $5=2^5-3^3$ is.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3759403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Cubes as the sum of odd integers It is well known that
$1^3=1$
$2^3=3+5$
$3^3=7+9+11$
$4^3=13+15+17+19$
$5^3=21+23+25+27+29$
and so on. This is typically proven using induction. I have come up with a proof and I'm wondering what you guys think or if you have seen this solution before :)
We will consider the array
\begin{align*}
\begin{matrix}
1\\
3 & 5\\
7 & 9 & 11\\
13& 15 & 17 & 19\\
&&&&\ddots
\end{matrix}
\end{align*}
and in the fashion of matrices, we let $A_{ij}$ denote the entry in row $i$ and column $j$. To be clear, $A_{11}=1, A_{21}=3, A_{22}=5$, etc. Then it suffices to show that $\sum_{j=1}^i A_{ij}=i^3$. Let us consider our array up to row $i$.
\begin{align*}
\begin{matrix}
1\\
3 & 5\\
7 & 9 & 11\\
13& 15 & 17 & 19\\
\vdots \\
A_{(i-1)1}&...&A_{(i-1)(i-1)}\\
A_{i1}&...&A_{ij} &...&A_{ii}
\end{matrix}
\end{align*}
It is clear to see that for $i \geq 2$ we have $A_{ii}=A_{(i-1)(i-1)}+2i$ as row $i$ consists of the $i$ odds following $A_{(i-1)(i-1)}$. We can solve for $A_{(i-1)(i-1)}$ by iteration.
\begin{align*}
A_{(i-1)(i-1)}&=A_{(i-2)(i-2)}+2(i-1)\\
&=A_{(i-3)(i-3)}+2(i-1)+2(i-3)\\
&=A_{(i-4)(i-4)}+2(i-1)+2(i-3)+2(i-4)\\
&...\\
&=1+2(i-1)+2(i-3)+2(i-4)+...+2(3)+2(2)\\
&=(i-1)i-1.
\end{align*}
Remarking that $A_{ij}=A_{(i-1)(i-1)}+2j$, we conclude that $A_{ij}=(i-1)i-1+2j$. Making use of this formula, it follows that $\sum_{j=1}^i A_{ij}=i^3$ as desired.
Let me know if there is any clarification necessary!
| There are identitity's given below:
For even cube:
$p^3=(p^2-p+1)+\cdots+(p^2-5)+(p^2-3)+(p^2-1)+(p^2+1)+(p^2+3)+(p^2+5)+\cdots+(p^2+p-1)$
For odd cube:
$q^3=(q^2-q+1)+\cdots+(q^2-6)+(q^2-4)+(q^2-2)+(q)^2+(q^2+2)+(q^2+4)+(q^2+6)+\cdots+(q^2+p-1)$
For, $p=8$ we get:
$8^3=(57+59+61+63+65+67+69+71)$
For, $q=9$ we get:
$9^3=(73+75+77+79+81+83+85+87+89)$
| {
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"url": "https://math.stackexchange.com/questions/3760340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Simplify $\tan^{-1} ( \frac{x-\sqrt{1-x^2}}{x+\sqrt{1-x^2}} )$ with trigonometric substitution I will explain my approach, help me with the last step please!
$$ \tan^{-1} {\left(\frac {x - \sqrt {1-x^2}}{x + \sqrt {1-x^2}}\right)}$$
substituting x = $\sin \theta$ (as learnt from book) and solving 1-$\sin^2 \theta$ = $\cos^2 \theta$
$$ \tan^{-1} {\left(\frac {\sin \theta - |\cos \theta|}{\sin \theta + |\cos \theta| }\right)}$$
For solving modulus, it was important to determine range of $\theta$ , therefore I defined it (as it is my variable,i can define it my way) for [-$\pi$/2 , $\pi$/2] so that sine covers all values from $-1$ to $1$ (as , $ -1 \le x \le 1 \,$ , from domain ) and $\cos \theta$ is positive , and hence $|\cos \theta| = \cos \theta$.
$$ \tan^{-1} {\left(\frac {\sin \theta - \cos \theta}{\sin \theta + \cos \theta }\right)}$$
= dividing by $\cos \theta$ $$ \tan^{-1} {\left(\frac {\tan \theta - 1}{\tan\theta + 1 }\right)}$$
= by formula of $\tan (\theta - \pi/4)$ $$ \tan^{-1}( \tan{\left(\theta - \pi/4\right)})$$
That's where I am stuck ,as according to the identity,$\quad$ $tan^{-1} ( \tan \alpha) = \alpha$ $\quad$ only when $\, -\pi/2 <\alpha < \pi/2$ . But here $$ -3\pi/4 \le \,(\theta-\pi/4) \, \le \pi/4 $$
Therefore, I am not going to get ($ \,\theta - \pi/4 $) out of the expression. What i get will be based on that graph of $\bf {\tan^{-1} (\tan x)}$ .
$$ (\theta - \pi/4) +\pi \,$$ for $\,-3\pi/4 \le \, (\theta -\pi/4) \, < -\pi/2 \,\,$ and
$$\theta -\pi/4$$
for $\,-\pi/2 < \, (\theta -\pi/4) \, \le \pi/4 \,\,$
My teacher just cancelled arctan and tan and wrote $\theta - \pi/4$ and he didn't even include that modulus function over $\cos \theta$.
So what will be the exact answer because if everyone decide $\theta$ as per they like then there will not be a finite answer. Everyone will have their own answers and in each answer they have multiple cases as I just discussed above.
So please help me, very hopefully I signed up in stackexchange!
Found Solution :-
I was confused because I was thinking that there can be many solutions differing person to person, but even if you choose any value of $\theta$ , you are going to get two solutions which are in the asked question above. The problem resolves when we write $\theta$ in terms of $sin^{-1} x$ as then we would not simply write like $$ \theta = \sin^{-1} x $$
we would write an equation,$$ \sin^{-1} x = \sin^{-1} (\sin \theta)$$,
now if $\theta$ is not in range of $-\pi/2$ and $\,\pi/2$ , then there would be some constant in $\pi$ (like , $\pi/4 , 2\pi$ etc. we would have to add or subtract according to the graph of 'sin inverse sin' and when we would put that value of $\theta$ , we would end with the solutions as answered by people.
(I write the answer in this edit to help anyone who will reach here after searching web , thanks to everyone for answers)
| I have another answer :
let $x \in (-1,1)\setminus\left\{-\frac{1}{\sqrt{2}}\right\}$. Then the derivative of the formula with respect to $x$ is :
\begin{align}
\frac{d}{dx}\left(\arctan\left(\dfrac{x-\sqrt{1-x^2}}{x+\sqrt{1-x^2}}\right)\right) &= \frac{d}{dx}\left(\dfrac{x-\sqrt{1-x^2}}{x+\sqrt{1-x^2}}\right)\times \dfrac{1}{1+\left(\dfrac{x-\sqrt{1-x^2}}{x+\sqrt{1-x^2}}\right)^2}
\end{align}
We compute the first term :
\begin{align}
\frac{d}{dx}\left(\dfrac{x-\sqrt{1-x^2}}{x+\sqrt{1-x^2}}\right)&= \dfrac{\left(1 +\dfrac{x}{\sqrt{1-x^2}}\right)\left(x + \sqrt{1-x^2}\right) - \left(x - \sqrt{1-x^2}\right)\left(1-\dfrac{x}{\sqrt{1-x^2}}\right) }{\left(x+\sqrt{1-x^2}\right)^2} \\
&= \dfrac{\dfrac{\sqrt{1-x^2}+x}{\sqrt{1-x^2}}\left(x+\sqrt{1-x^2}\right)+\left(x-\sqrt{1-x^2}\right)\dfrac{x-\sqrt{1-x^2}}{\sqrt{1-x^2}}}{\left(x+\sqrt{1-x^2}\right)^2} \\
&= \dfrac{\left(x+\sqrt{1-x^2}\right)^2 + \left(x-\sqrt{1-x^2}\right)^2}{\sqrt{1-x^2}\left(x+\sqrt{1-x^2}\right)^2} \\
&= \dfrac{x^2 + 2x\sqrt{1-x^2} + 1-x^2 + x^2 -2x\sqrt{1-x^2} + 1-x^2}{\sqrt{1-x^2}\left(x+\sqrt{1-x^2}\right)^2} \\
&= \dfrac{2}{\sqrt{1-x^2}\left(x+\sqrt{1-x^2}\right)^2}
\end{align}
From there, we have :
\begin{align}
\frac{d}{dx}\left(\arctan\left(\dfrac{x-\sqrt{1-x^2}}{x+\sqrt{1-x^2}}\right)\right) &= \dfrac{2}{\sqrt{1-x^2}\left(x+\sqrt{1-x^2}\right)^2} \times \dfrac{1}{1+\left(\dfrac{x-\sqrt{1-x^2}}{x+\sqrt{1-x^2}}\right)^2} \\
&= \dfrac{2}{\sqrt{1-x^2}}\times \dfrac{1}{\left(x+\sqrt{1-x^2}\right)^2 + \left(x-\sqrt{1-x^2}\right)^2} \\
&= \dfrac{1}{\sqrt{1-x^2}}
\end{align}
Therefore, the solution is an antiderivative of $x\in (-1,1)\setminus\left\{-\frac{1}{\sqrt{2}}\right\}\mapsto \frac{1}{\sqrt{1-x^2}}$. It follows that there exists two constant $C$ and $C'$ such that:
\begin{align}
\forall x \in \left(-1,-\dfrac{1}{\sqrt{2}}\right) ,~ \arctan\left(\dfrac{x-\sqrt{1-x^2}}{x+\sqrt{1-x^2}}\right) &= \arcsin x + C \\
\forall x \in \left(-\dfrac{1}{\sqrt{2}},1\right) ,~ \arctan\left(\dfrac{x-\sqrt{1-x^2}}{x+\sqrt{1-x^2}}\right) &= \arcsin x + C'
\end{align}
To get the value of $C$ and $C'$, evaluate the limits at $-1$ and $1$ :
\begin{align}
C &=\arctan 1- \arcsin( -1)\\
&= \frac{\pi}{4}+\frac{\pi}{2} \\
&= \frac{3\pi}{4} \\
C' &=\arctan 1- \arcsin 1\\
&= \frac{\pi}{4}-\frac{\pi}{2} \\
&= -\frac{\pi}{4}
\end{align}
It seems like your teacher forgot to check what happens around $-\frac{1}{\sqrt2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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Two players playing a coin game - probability of winning Alex and Beth take turns flipping a pair of coins. The first person to flip a pair of heads wins the game. Alex flips first. Beth eventually wins. What is the probability she flipped a pair of heads on her second turn?
Can someone verify if this reasoning is correct or not:
Question can be converted to Bayes rule:
P(someone wins given that they don't flip first) = 3/7 (solved the equation p=(.75)(.25) + (.75)(.75)(p))
Desired probability: P(Beth wins on 2nd turn | she wins and Alex flips first) = (.75)(.75)(.75)(.25)/(3/7)
This equals approximately 1/4.
| my approach
$$
\begin{cases}
A & ... & A & B \\
!2H & ... & !2H & 2H \\
{3\over 4} & ... & {3\over 4} & {1\over 4}
\end{cases}
$$
$$
X: \text{B flipped a pair of heads on her second turn} \\
Y: \text{Alex flips first. Beth eventually wins.} \\
\Pr[X|Y] = {\Pr[X\cap Y] \over \Pr[Y] } = { ({3\over 4})^3 {1\over 4} \over \Pr[Y] } \\
\Pr[Y] = \sum_{n=1}^{\infty} ({3\over 4})^{2n-1} {1\over 4} = {1\over 3} \sum_{n=1}^{\infty} ({9\over 16})^n = {1\over 3} {9\over 7} = {3\over 7}\\
\Pr[X|Y] = { ({3\over 4})^3 {1\over 4} \over {3\over 7} } = {63\over 4^4}
$$
your approach
$$
... !2H, !2H \begin{cases}
(A) & (B) \\
!2H & !2H & {3\over 4} {3\over 4} \Pr[Y] \\
!2H & 2H & {3\over 4} {1\over 4} \cdot 1 \\
2H & & {1\over 4} \cdot 0 \\
\end{cases} \\
\Pr[Y] = {3\over 4} {3\over 4} \Pr[Y] + {3\over 4} {1\over 4} \cdot 1 + {1\over 4} \cdot 0 \\
\Pr[Y] = {3\over 7}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3761034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Distance between set and point, confused of partial derivatives. Let $H = \{(x,y,z)\ \in \mathbb{R}^{3}: x^2+y^2 - z^2 + 4 = 0$
Compute the shortest distance between H and point $p=(2,4,0)$.
I am a bit confused because I tried a direct approach.
$$ x^2+y^2 + 4 = z^2$$
Let $D(H,p) = \sqrt{(2-x)^{2}+(4-y)^{2} + x^{2} + y^{2}+4}$
So I tried compute $$\frac{\partial D}{\partial x} = (\sqrt{2} (-1 + x))/\sqrt{12 - 2 x + x^2 - 4 y + y^2}$$
$$\frac{\partial D}{\partial y} = (\sqrt{2} (-2 + y))/\sqrt{12 - 2 x + x^2 - 4 y + y^2}$$
It seems not nice to compare with zero.
Do you have another idea?
| You have found
$$\frac{\partial D}{\partial x} = (\sqrt{2} (-1 + x))/\sqrt{12 - 2 x + x^2 - 4 y + y^2}$$
$$\frac{\partial D}{\partial y} = (\sqrt{2} (-2 + y))/\sqrt{12 - 2 x + x^2 - 4 y + y^2}$$
The first of these is zero only when $-1+x=0$, i.e. $x=1$.
The second is zero only when $-2+y=0$, i.e. $y=2$.
Since $x^2+y^2 - z^2 + 4 = 0$, these two being true simultaneously imply $z=\pm3$.
The distance from $(2,4,0)$ to $(1,2,3)$ and to $(1,2,-3)$ is $\sqrt{14}$ and so this is the shortest distance (if you wish, you can use second derivatives to show it is a minimum).
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that for every integer $x$, if $x$ is odd then there exists an integer $y$ such that $x^2=4y+1$.
Prove that for every integer $x$, if $x$ is odd then there exists an integer $y$ such that $x^2=4y+1$.
Let $x$ be an odd integer. Then, there exists an integer m such that $x=2m+1$. But $x^2=4m^2+4m+1=4(m^2+m)+1$.
Case 1: $m$ is odd. Then there exists a $k$ such that $m=2k+1$. Since $m^2=4k^2+4k+1$, $m^2$ is odd. Hence, $m^2+m$ is even and $m^2+m=2y$ for some $y$. Thus, $4(m^2+m)+1=4(y)+1$ as required.
| As stated in the comment, you can take $y=m^2+m$ and you are done.
Alternatively, we just have to illustrate that $x^2-1$ is a multiple of $4$. Notice that $$x^2-1=(x-1)(x+1)$$
If $x$ is odd, both $x-1$ and $x+1$ are even, hence the product is a multiple of $4$.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\le 1$, prove that $(1+a^2)(1+b^2)(1+c^2)\ge 125$.
QUESTION: Let $a,b,c$ be positive real numbers such that
$$\cfrac{1}{1+a}+\cfrac{1}{1+b}+\cfrac{1}{1+c}\le 1$$
Prove that $$(1+a^2)(1+b^2)(1+c^2)\ge 125$$ When does equality hold?
MY APPROACH: Firstly, let's try to squeeze out all the information we can from what is given. $$\frac{(1+b)(1+c)+(1+a)(1+c)+(1+a)(1+b)}{(1+a)(1+b)(1+c)}≤1$$ Multiplying this out, we get $$3+2(a+b+c)+(ab+bc+ca)≤1+(a+b+c)+(ab+bc+ca)+abc$$ $$\implies 2+(a+b+c)≤abc$$ Also, since $$1≥\sum_{cyc}\frac{1}{1+a}$$ Therefore by AM-GM,
$$1≥\sum_{cyc}\frac{3}{\sqrt[3]{(1+a)(1+b)(1+c)}}$$
$$\implies (1+a)(1+b)(1+c)≥27$$
That's all I ended up in.. At first, I thought Hölder's inequality could be employed, but that too requires the sum of the powers to be $=1$.. and that is not going to be useful in $(1+a^2)(1+b^2)(1+c^2)$ , since here the sum of powers add up to $3$..
I don't know what to do next.. Any help will be much appreciated..
| We need to prove that:
$$\sum_{cyc}(\ln(a^2+1)-\ln5)\geq0,$$ for which it's enough to prove that
$$\sum_{cyc}\left(\ln(a^2+1)-\ln5-\frac{36}{5}\left(\frac{1}{3}-\frac{1}{a+1}\right)\right)\geq0,$$ which is true because for any $a>0$ easy to show that:
$$\ln(a^2+1)-\ln5-\frac{36}{5}\left(\frac{1}{3}-\frac{1}{a+1}\right)\geq0.$$
Indeed, $$\left(\ln(a^2+1)-\ln5-\frac{36}{5}\left(\frac{1}{3}-\frac{1}{a+1}\right)\right)'=\frac{2(a-2)(5a^2+2a+9)}{5(a^2+1)(a+1)^2},$$ which gives $a_{min}=2$ and we are done!
I got the coefficient $-\frac{36}{5}$ by the following way.
Let $f(x)=\ln(a^2+1)-\ln5+\lambda\left(\frac{1}{3}-\frac{1}{a+1}\right).$
We know that $f(2)=0$ and we'll choose a value of $\lambda$ such that $2$ would be a double point,
for which we need $f'(2)=0,$ which gives $\lambda=-\frac{36}{5}.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Initial value problem $xyz_x + x^2z_y = x^2+y^2-yz$ with initial curve Problem
Solve the initial value problem in the domain $x<0$:
$$
xyz_x + x^2z_y = x^2+y^2-yz
$$
where $z=x$ on the line $y=1$.
Attempt
Edit: As JJacquelin pointed out in the answers, my first integral $u_2$ is incorrect which would explain the problem. It should be $x(y-z)$.
I solved for two functionally independent first integrals $u_1(x,y):=x^2-y^2+1$ and $u_2(x,y,z):=\dfrac{y-z}{x}$ of the vector field $V(x,y,z)=(xy,x^2,x^2+y^2-yz)$. We can parameterize the initial curve as follows: $x=t, y=1, z=t$ where $t<0$.
Evaluating $u_1$ and $u_2$ on the initial curve, we get
$$
\begin{align}
&U_1 = t^2\\
&U_2 = \dfrac{1-t}{t}
\end{align}
$$
Here is where I have some confusion. I know we need to eliminate $t$, but there seems to be a subtlety about how to go about it. I am tempted to write $t=\pm\sqrt{U_1}$ so that $U_2=\dfrac{1\mp\sqrt{U_1}}{\pm\sqrt{U_1}}$. Then, we can write
$$
\dfrac{y-z}{x}=\dfrac{1\mp\sqrt{x^2-y^2+1}}{\pm\sqrt{x^2-y^2+1}}
$$
When $y=1$, we must have $x=z$, so this determines the choice of signs above as:
$$
\dfrac{y-z}{x}=\dfrac{1-\sqrt{x^2-y^2+1}}{\sqrt{x^2-y^2+1}}
$$
Solving for $z$, we then have
$$
z=x+y-\dfrac{x}{\sqrt{x^2-y^2+1}}.
$$
But when I plug this back into the PDE, it fails to satisfy the equation. My suspicion for what went wrong is how I eliminated $t$. Certainly, I can also get rid of $t$ by writing $U_1U_2^2=(1\pm\sqrt{U_1})^2$. If we follow the same procedure, we get a more complicated formula for $z$ with more terms. What is the correct way to proceed, and where did I go wrong in my attempt?
| This is not a direct answer to your question because I am not quite certain of your first integrals $u_1$ and $u_2$. You should edit the detailed calculus so that one could check it. I think that might be $u_2=x(y-z)$ instead of $\frac{y-z}{x}$ .
Nevertheless I hope that the comparison with the characteristic equations below would help to clarify.
$$xyz_x+x^2z_y=x^2+y^2-yz$$
The Charpit-Lagrange system of characteristic ODEs is :
$$\frac{dx}{xy}=\frac{dy}{x^2}=\frac{dz}{x^2+y^2-yz}$$
A first characteristic equation comes from $\frac{dx}{xy}=\frac{dy}{x^2}$ which solving leads to :
$$x^2-y^2=c_1$$
A second characteristic equation comes from $\frac{dy}{x^2}=\frac{dz}{x^2+y^2-yz}$ which solving leads to :
$\frac{dy}{c_1+y^2}=\frac{dz}{(c_1+y^2)+y^2-yz}\quad\implies\quad\frac{dz}{dy} =\frac{c_1+2y^2-yz}{c_1+y^2}$ which is a linear first order ODE easy to solve.
$$(z-y)\sqrt{c_1+y^2}=c_2$$
$$(z-y)x=c_2$$
The general solution of the PDE expessed on the form of implicite equation $c_2=F(c_1)$ is :
$$(z-y)x=F(x^2-y^2)$$
with arbitrary function $F$, to be determined later according to the specified condition.
$$\boxed{z(x,y)=y+\frac{1}{x}F(x^2-y^2)}$$
Condition :
$$z(x,1)=x=1+\frac{1}{x}F(x^2-1)\quad\implies\quad F(x^2-1)=x^2-x$$
Let $X=x^2-1\quad;\quad x=\pm\sqrt{X+1}$
$$F(X)=X+1\pm\sqrt{X+1}$$
The function $F$ is determined. We put it into the above general solution where $X=x^2-y^2$.
$$z(x,y)=y+\frac{1}{x}\left(x^2-y^2+1\pm\sqrt{x^2-y^2+1} \right) $$
The sign is determined so that $z(x,1)=x$
$$\boxed{z(x,y)=\begin{cases}
x+y+\frac{1}{x}\left(1-y^2-\sqrt{x^2-y^2+1} \right) \quad\text{if }x>0 \\
x+y+\frac{1}{x}\left(1-y^2+\sqrt{x^2-y^2+1} \right) \quad\text{if }x<0
\end{cases}}$$
| {
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What is the probability for $x$ to be positive only? If $x,y,z\in \mathbb R$ and $x+y+z=5,\, xy+yz+zx=3$, what is the probability that $x>0$ ? $$(a)\quad\frac3{16}\qquad (b)\quad\frac5{16}\qquad (c)\quad\frac{13}{16}\qquad (d)\quad \frac{15}{16}$$
I've tried forming a cubic equation and then trying to analyze its roots.
How to approach these type of questions?
| The solutions of your two equations form a circle in $xyz$ space, the intersection of the sphere $x^2 + y^2 + z^2 = 19$ with the plane $x+y+z=5$, which can be parametrized as
$$ \eqalign{x &= \frac{5}{3} - \frac{8}{3} \cos(t)\cr
y &= \frac{5}{3} + \frac{4}{\sqrt{3}} \sin(t) + \frac{4}{3} \cos(t)\cr
z &= \frac{5}{3} - \frac{4}{\sqrt{3}} \sin(t) + \frac{4}{3} \cos(t)\cr} $$
It's not at all clear how to interpret "probability" in this context, but there is no $t$ for which $x > 0$ while $y \le 0$ and $z \le 0$. So if $x > 0$ only means $x > 0$ while $y \le 0$ and $z \le 0$, the answer must be $0$. On the other hand, if you asked for the probability that $x > 0$, one possible interpretation would be the probability that $x>0$ if you took a random $t$ uniformly from $[-\pi,\pi]$.
Then you'd get $x > 0$ iff $\cos(t) < 5/8$, i.e. with probability $1 - \frac{\arccos(5/8)}{\pi} \approx 0.7149010415$. This is an irrational number. So again "none of the above".
| {
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Help in finding $\int \frac{x+x\sin x+e^x \cos x}{e^x+x\cos x-e^{x} \sin x} dx$ I want to find
$$\int \frac{x+x\sin x+e^x \cos x}{e^x+x\cos x-e^{x} \sin x} dx.$$
But since algebraic, exponential and trigonometric functions are involved I am not able to solve it. Please help in finding it by hand.
| Note the law of equal peopoetions, we have: $$\frac{1+\sin x}{\cos x}=\frac{\cos x}{1-\sin x} \implies \frac{x(1+\sin x)}{x \cos x}=\frac{e^x \cos x}{e^x-e^x \sin x}=\frac{x+x\sin x+e^x \cos x}{x\cos x+ e^{x}-e^{x} \sin x}$$
So $$\int \frac{x+x\sin x+e^x \cos x}{e^x+x\cos x-e^{x} \sin x} dx= \int \frac{\cos x}{1-\sin x}dx=-\ln (1-\sin x)+C$$
| {
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One number is removed from the set of integers from $1$ to $n.$ The average of the remaining numbers is $163/4.$ Which integer was removed?
One number is removed from the set of integers from $1$ to $n.$ The average of the remaining numbers is $\dfrac{163}4$. Which integer was removed?
Source. British Mathematical Olympiad 2010/11, Round 1, Problem 1
I was hoping if someone could spot the flaw in my working for this question.
Attempt.
I began by letting the integer that was removed be $x$.
Then: $$\frac{1 + 2 + \cdots + (x-1) + (x+1) +\cdots + n} {n-1} = \frac{163}{4}$$
There is two arithmetic sums in the denominator, the first from 1 to $x$ and the second from $x+1$ to $n$.
These are equal to $\frac{x(x-1)}{2}$ and $\frac{(n-x)(n+x+1)}{2}$, and subbing in to first equation this gives:
$$\frac{x(x-1) + (n-x)(n+x+1)}{2(n-1)} = \frac{163}{4}$$
which reduces to:
$$\frac{n^2 + n - 2x}{2(n-1)} = \frac {163}{4}$$
And then:
$$2(n^2 + n -2x) = 163(n-1)$$
At first I thought you could consider factors, as 163 was prime then:
$n-1 = 2$ giving $n = 3$ and $n^2 + n - 2x = 163$, which using $n=3$ gives $x= -75.5$ which isn't our positive integer.
I then tried considering a quadratic in $n$ and using the discriminant but again that just looked to give a negative value of $x.$ I would be grateful for any help
| Say you're removing $x$ from the set $1, 2, \ldots, n$. The average of the resulting numbers will be at least $n/2$ (if you remove $n$) and at most $(n+2)/2$ (if you remove $1$.) So we have
$$ n/2 \le 163/4 \le (n+2)/2 $$
or, multiplying through by 4,
$$ 2n \le 163 \le 2n+2. $$.
So $n = 81$. You don't have to explicitly work out the number being removed, but it's $(1 + 2 \ldots + 81) - (163/4) \times 80$ = $(81 \times 82)/2 - (163/4) \times 80 = 61$.
| {
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Finding largest delta value given epsilon for delta-epsilon limit The formal definition of the limit is that $\lim_{x\to c} f(x)=L$ if and only if for any $\varepsilon >0$ there exists a $\delta >0$ such that $|x-c| <\delta \rightarrow |f(x)-L| < \varepsilon$. Use this definition for $\lim_{x \to 5} (x^2 -15x+50)$ to find the largest value for delta that satisfies epsilon equal to one.
Source: FAMAT State Convention 2019, Limits&Derivatives #29
Attempted solution: I set $|x^2-15x+50 - 0| < 1$ and solved for $x$, obtaining the following inequalities: $\frac{15-\sqrt{29}}{2} < x < \frac{15+\sqrt{29}}{2}$ and $x<\frac{15-\sqrt{21}}{2} $ or $ x > \frac{15+\sqrt{21}}{2}$
Next, I tried getting this in the form $|x-c| <\delta$ by changing each inequality as follows:
$\frac{5-\sqrt{29}}{2} < x-5 < \frac{5+\sqrt{29}}{2}$ and $x-5<\frac{5-\sqrt{21}}{2} $ or $ x-5 > \frac{5+\sqrt{21}}{2}$
I don't know how to translate this result to $\delta$ because it does not follow a neat $-\delta<x-c<\delta \rightarrow |x-c| <\delta$
Answer provided by the competition:
$\frac{5-\sqrt{21}}{2}$
Edit: See comment to where I'm currently at
| Wait! You know that when x= 5, $x^2- 15x+ 50= 0$? And when you solved for x you got $\frac{15\pm\sqrt{21}}{2}$? NO! Since $x^2- 15x+ 50= 0$ when x= 5 you know that x- 5 is a factor! And it is easy to see that $x- 10$ is the other.
$x^2- 15x+ 50= (x- 5)(x- 10)$ so $|x^2- 15x+ 50|< \epsilon$ gives |(x- 5)(x- 10)|= |x- 5||x- 10|< \epsilon. As long as x is not 10 we can write $|x- 5|< \epsilon/|x-10|$.
| {
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Prove that $q(x,y)=Ax^2+2Bxy+Cy^2$ is positive definite I want to prove the following:
$q(x,y)=Ax^2+2Bxy+Cy^2$
is positive definite if and only if $A>0$ and $b^2-ac<0$.
I have seen a similar exercise but I want to know why $b^2-ac<0$ must be negative, thanks!
| Consider the following simplification:
$$q(x,y) = Ax^2+2Bxy+Cy^2 = A(x^2+\frac{2B}{A}xy)+Cy^2$$
Then, completing the square, we have:
$$x^2+\frac{2B}{A}xy = (x+\frac{B}{A}y)^2-\frac{B^2}{A^2}y^2$$
So, we can rewrite $q(x,y)$ as follows:
$$q(x,y) = A(x+\frac{B}{A}y)^2+Cy^2-\frac{B^2}{A}y^2$$
$$q(x,y) = A(x+\frac{B}{A}y)^2+\frac{CA-B^2}{A}y^2$$
If $A > 0$ and $B^2-AC < 0$, then $A > 0$ and $AC-B^2 > 0$. So, the second term of the expression above is always positive when $y \neq 0$.
In particular, even if the first term may go to $0$, the second term is never $0$ unless $y = 0$. Obviously, this implies that $q(x,y) = 0 \iff x = y = 0$. That proves that the given quadratic form is positive definite under those conditions.
| {
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Identify the function represented by $\displaystyle \sum_{k=2}^\infty \frac{x^k}{k(k-1)}$ So first I wrote it out in the terms, and I got
$\displaystyle \sum_{k=2}^\infty \frac{x^k}{k(k-1)} = \frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{12}+\frac{x^5}{20}+...$
I know the power series for $\displaystyle ln(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}$ which is similar to the derivative of the power series from above, as
$\displaystyle \frac{d}{dx}(\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{12}+\frac{x^5}{20}+...)=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+...$
My question is, where do I go from here? How would I make it so the series is similar ln(1+x)? Or am I even going down the right route when it comes to solving this problem? Any help would be appreciated.
| What you could have done is to differentiate twice to get
$$\left( \sum_{k=2}^\infty \frac{x^k}{k(k-1)}\right)''=\sum_{k=2}^\infty x^{k-2}=\frac 1 {1-x}$$making
$$\left( \sum_{k=2}^\infty \frac{x^k}{k(k-1)}\right)'=-\log(1-x)$$
$$ \sum_{k=2}^\infty \frac{x^k}{k(k-1)}=x+(1-x) \log (1-x)$$
| {
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Given that $x^2 + y^2 = 2x - 2y + 2$ , find the maximum value of $x^2 + y^2 + \sqrt{32}$ .
Given that $x^2 + y^2 = 2x - 2y + 2$ , find the maximum value of $x^2 + y^2 + \sqrt{32}$ .
What I Tried :- Since $x^2 + y^2 = 2x - 2y + 2$ , we have $2x - 2y + 2 + \sqrt{32}$
=> $2(x - y + 1 + 2√2)$ . From this step I am not sure how to move forward .
Also I tried to express $x^2 + y^2 + \sqrt{32} \leq S$ , so that in that way we can say that $x^2 + y^2 + \sqrt{32}$ is maximum at $S$ , but I couldn't do it .
Can anyone help me ? Some hints or suggestions to this problem will be appreciated !!
| Hint: $x^2+y^2=2x-2y+2$ is equivalent to $(x-1)^2+(y+1)^2=4$, which is the equation of a circle with the centre in $C=(1,-1)$ and radius $2$. To maximise $x^2+y^2+\sqrt{32}$ it is enough to first maximise $\sqrt{x^2+y^2}$, i.e. the distance from the origin $O=(0,0)$.
Now, which point in that circle is the farthest from the origin? It will be the intersection of half-line $[OC)$ with the circle, i.e. the point $(x,y)=(1+\sqrt{2},-1-\sqrt{2})$, giving you the maximum value $x^2+y^2+\sqrt{32}=6+8\sqrt{2}$.
| {
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Prove $\sum_{n=0}^{\infty} \frac{\Gamma(n+(1/2))}{4^n(2n+1)\Gamma(n+1)}=\frac{\pi^{3/2}}{3}$ Prove $$\sum_{n=0}^{\infty} \frac{\Gamma\left(n+\frac{1}{2}\right)}{4^n\left(2n+1\right)\Gamma\left(n+1\right)}=\frac{\pi^{\frac{3}{2}}}{3}$$
The original sum is multiplied by $\frac{\sqrt{\pi}}{2}$ and so it equals $\frac{\pi^2}{6}$ but I pulled the constant out because the actual series troubles me. I dont know how to evaluate this. I think maybe the Gammas and $4^n$ simplify and leave some constant divide by $2n+1$ which is the familiar arctan series. Wolfram can't help simplify it, just compute it. Any help please?
| Note that $$\Gamma\left(n+\frac{1}{2}\right)=\frac{(2n)!}{4^nn!}\sqrt{\pi}, \, \, \Gamma(n+1)=n!$$ and our sum get simplified to $$\sqrt{\pi}\sum_{n=0}^{\infty}\frac{(2n)!}{16^n(2n+1) (n!)^2}=\sum_{n=0}^{\infty}\frac{\sqrt{\pi}}{16^n(2n+1)}{2n\choose n} $$ Now recall the ordinary generating function of central binomial coefficients for $|x|<\frac{1}{4}$ , that is $$\sum_{n=0}^{\infty}{2n\choose n} x^n=\frac{1}{\sqrt{1-4x}}\cdots(1)$$ now replacing $x$by $\frac{x^2}{16}$ in $(1)$ we get $$ \sum_{n=0}^{\infty}\frac{1}{16^n}{2n\choose n}x^{2n} =\frac{2}{\sqrt{4-x^2}}\cdots(2)$$ Now we integrating $(2)$ from $0$ to $1$ gives $$\sum_{n=0}^{\infty}\frac{1}{16^n(2n+1)}{2n\choose n} =\int_0^1\frac{2}{\sqrt{4-x^2}}=2\int_0^1\frac{d}{dx}\sin^{-1}\left(\frac{x}{2}\right)dx=2\sin^{-1}\left(\frac{x}{2}\right)\bigg|_0^1=\frac{\pi}{3}\cdots(3)$$ now we multiply by the factor $\sqrt{\pi}$ in $(3)$ giving us the desired closed form
$$\sum_{n=0}^{\infty}\frac{\sqrt{\pi}}{16^n(2n+1)}{2n\choose n} =\frac{\pi^{\frac{3}{2}}}{3}$$
| {
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Let $\frac{1}{2}<\cos2A<1$ and $6\tan A-6\tan^3A=\tan^4A+2\tan^2A+1$, find $\tan 2A$ Let $\dfrac{1}{2}<\cos2A<1$ and $6\tan A-6\tan^3A=\tan^4A+2\tan^2A+1$, find $\tan 2A$
My attempt:
\begin{align*}
6\tan A(1-\tan^2A)&=\tan^4A+2\tan^2A+1\\
12\tan^2A&=\tan2A\tan^4A+2\tan2A\tan^2A+\tan2A\\
0&=\tan2A(\tan^4A)+(2\tan2A-12)\tan^2A+\tan2A\\
\because\tan2A&\in\mathbb{R}\\
\therefore \tan2A&\leqslant3
\end{align*}
From $\dfrac{1}{2}<\cos2A<1$ gives $0\leqslant\tan2A<\sqrt{3}$
Alfter using 2 inequality, I still can't find the exact value of $\tan2A$
| We obtain:
$$6\tan{A}(1-\tan^2A)=(1-\tan^2A)^2+4\tan^2A$$ or
$$\frac{6\tan{A}}{1-\tan^2A}=1+\frac{4\tan^2A}{(1-\tan^2A)^2}$$ or
$$\tan^22A-3\tan2A+1=0,$$ which gives
$$\tan2A=\frac{3+\sqrt5}{2}$$ or $$\tan2A=\frac{3-\sqrt5}{3}.$$
Also, $$\tan^22A=\frac{1}{\cos^22A}-1<4-1=3,$$ which fives $$\tan2A=\frac{3-\sqrt5}{2}.$$
| {
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How to prove distance from foci on an ellipse is equal to twice the semi-major axis (for specific ellipse) Prove that for any point (x,y) on the conic, the sum of the distances to the two foci is always twice the semi-major axis.
I know that this can be proven in general for all ellipses but the practice question specifically asks for this to be proven for $\frac{x^2}{9}$ + $\frac{y^2}{4}$ = 1. I feel like I'm really close but I've managed to math myself into a corner somehow.
Let the foci ($\sqrt{5}$, 0) and (-$\sqrt{5}$, 0) be denoted as F and F'. Let the point on the conic be denoted P(x,y). We are required to show PF + PF' = 2a. In this case, since a = 3, 2a = 6.
PF = $\sqrt{(x-\sqrt{5})^2 + y^2}$ and PF' = $\sqrt{(x+\sqrt{5})^2 + y^2}$
By rearranging the equation for the ellipse, we get y$^2$ = 4 - $\frac{4}{9}$x$^2$.
Substitute this into PF and PF' to get:
PF = $\sqrt{(x-\sqrt{5})^2 + 4 - \frac{4}{9}x^2}$ = $\sqrt{\frac{5}{9}x^2 - 2\sqrt{5}x + 9}$ = $\sqrt{(x - \frac{9\sqrt{5}}{5})^2}$ = x - $\frac{9\sqrt{5}}{5}$
PF' = $\sqrt{(x+\sqrt{5})^2 + 4 - \frac{4}{9}x^2}$ = $\sqrt{\frac{5}{9}x^2 + 2\sqrt{5}x + 9}$ = $\sqrt{(x + \frac{9\sqrt{5}}{5})^2}$ = x + $\frac{9\sqrt{5}}{5}$
Therefore PF + PF' = 2x
And then I got stuck
| An ellipse is by definition the locus of the points such that the sum of their distances to the foci is constant.
From this property, the result is immediate:
$$PQ=FQ+QG.$$
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Calculate:$\int_{0}^{\infty}\frac{\ln x}{(x+1)^{3}}\mathrm{d}x$ with contour integration Calculate: $$\int_{0}^{\infty}\frac{\ln x}{(x+1)^{3}}\mathrm{d}x$$
My try:
Keyhole integration:
$\displaystyle \frac{\pi i\ln R\cdot e}{(Re^{\theta i}+1)^{3}}\rightarrow 0$ (we take $r$ as large as we want)
and here is the confusion :
around the circle the residue is $0$:
as $\displaystyle \frac{x\ln x}{(1+x)^{3}}\rightarrow0$ when we approach to $0$. Therefore, the residue is $0$, and the whole integration of the keyhole is $0$. which leads that the result is $0$.
but if we take the pole in the keyhole, $x=-1$ this is $3$rd order pole, and its residue is $\displaystyle \left. -\frac{1}{x^{2}}\right|_{x=-1}=-1$ meaning that the whole integral is $-2\pi i$ which means that that the result should be $-\pi i$.
Both of the results are incorrect. Can you spot my mistakes?
| The slow & steady way (without contour integration) is integration by parts, then partial fractions, to find an antiderivative; then, taking limits to evaluate the improper integrals.
Integration by parts:
\begin{align*}
\text{Let } u = \ln(x), &\text{ and } dv = (x+1)^{-3}; \\
\text{Then } du = x^{-1}, &\text{ and } v = -(x+1)^{-2}/2. \
\end{align*}
We get $$\int \frac{\ln(x)}{(x+1)^3} dx = \frac{-\ln(x)}{2(x+1)^2} - \int \frac{-dx}{2x(x+1)^2} = \frac{-\ln(x)}{2(x+1)^2} + \int \frac{dx}{2x(x+1)^2}.$$
Partial fractions:
$$\int \frac{dx}{2x(x+1)^2} = \int \left[ \frac{A}{x} + \frac{B}{x+1} + \frac{C}{(x+1)^2} \right]dx,$$
and clearing denominators, we find $$1 = 2A(x+1)^2 + 2Bx(x+1) + 2Cx.$$ Plugging in $x = 0$ gives us $A = 1/2$, plugging in $x = -1$ gives us $C = -1/2$, and plugging in the numerical values of $A, C,$ we find $$1 = (x+1)^2 + 2Bx(x+1) - x = x^2 + x + 1 + 2Bx^2 + 2Bx,$$ which implies $B = -1/2$ as well. So $$\int \frac{dx}{2x(x+1)^2} = \int \left[ \frac{1}{2x} - \frac{1}{2(x+1)} - \frac{1}{2(x+1)^2} \right]dx = \frac{1}{2} \left[\ln(x) - \ln(x+1) + \frac{1}{x+1} \right],$$ therefore
\begin{align*}
\int \frac{\ln(x)}{(x+1)^3} dx &= \frac{1}{2} \left [\frac{-\ln(x)}{(x+1)^2} + \ln(x) - \ln(x+1) + \frac{1}{x+1} \right] \\
&= \frac{1}{2} \left [\frac{-\ln(x)}{(x+1)^2} + \ln \left(\frac{x}{x+1}\right) + \frac{1}{x+1} \right] + C. \
\end{align*}
Improper integrals:
We evaluate
\begin{align*}
\int_0^\infty \frac{\ln(x)}{(x+1)^3} dx &= \lim_{a \rightarrow 0^+} \left( \lim_{N \rightarrow \infty} \int_a^N \frac{\ln(x)}{(x+1)^3} dx \right) \\
&= \frac{1}{2} \lim_{a \rightarrow 0^+} \left( \lim_{N \rightarrow \infty} \left [\frac{-\ln(x)}{(x+1)^2} + \ln \left(\frac{x}{x+1}\right) + \frac{1}{x+1} \right]_a^N \right) \\
&= \frac{1}{2} \lim_{a \rightarrow 0^+} \left [\frac{\ln(a)}{(a+1)^2} - \ln \left(\frac{a}{a+1}\right) - \frac{1}{a+1} \right] \\
&= \frac{-1}{2} + \frac{1}{2} \lim_{a \rightarrow 0^+} \left [\frac{\ln(a)}{(a+1)^2} - \ln \left(\frac{a}{a+1}\right) \right] \\
&= \frac{-1}{2} + \frac{1}{2} \lim_{a \rightarrow 0^+} \left [\frac{\ln(a)}{(a+1)^2} - \ln(a) + \ln(a+1) \right] \\
&= \frac{-1}{2} + \frac{1}{2} \lim_{a \rightarrow 0^+} \left [\frac{\ln(a)}{(a+1)^2} - \ln(a) \right], \\
\end{align*}
and by L'Hopital's Rule,
\begin{align*}
\lim_{a \rightarrow 0^+} \ln(a) \left[ \frac{1}{(a+1)^2} - 1 \right] &= \left( \lim_{a \rightarrow 0^+} \frac{1}{(a+1)^2} \right) \lim_{a \rightarrow 0^+} \left[ \frac{\ln(a)}{1/(-2a - a^2)} \right] \\
&= 1 * \lim_{a \rightarrow 0^+} \frac{1/a * (-2a - a^2)^2}{-(-2 - 2a)} \\
&= 0, \
\end{align*}
and we finally get $$\int_0^\infty \frac{\ln(x)}{(x+1)^3} dx = \frac{-1}{2} + \frac{1}{2} (0) = \frac{-1}{2}.$$
| {
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"url": "https://math.stackexchange.com/questions/3782895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $\sum_{cyc}\sqrt{a^4+a^2b^2+b^4}\geq \sum_{cyc} a\sqrt{2a^2+bc}$ for non-negative $a$, $b$, $c$ I was trying this question with factorization and other similar methods,
Let $a, b, c \geq 0$. Prove that
$$\begin{array}{c}
\sqrt{a^4+a^2b^2+b^4}+\sqrt{b^4+b^2c^2+c^4}+\sqrt{c^4+c^2a^2+a^4} \\[4pt]
\geq a\sqrt{2a^2+bc}+b\sqrt{2b^2+ca}+c\sqrt{2c^2+ab}
\end{array}$$
This is one of Hoojoo-Lee's Inequality. This seems very intuitive at first as if we square each term, $$ 2\sum_{cyc}{a^4}+\sum_{cyc}{a^2b^2} \geq 2\sum_{cyc}{a^4}+\sum_{cyc}{a^2bc} \Rightarrow \sum_{cyc}{a^2b^2} \geq \sum_{cyc}{a^2bc} $$ which is quite clear. I noticed it but can not exploit it. May be taking the square on each side could help? But I couldn't find a solution.
Please help!
| By C-S $$\sum_{cyc}\sqrt{a^4+a^2b^2+b^4}=$$
$$=\sqrt{\sum_{cyc}\left(a^4+a^2b^2+b^4+2\sqrt{(a^4+a^2b^2+b^4)(a^4+a^2c^2+c^4)}\right)}\geq$$
$$\geq\sqrt{\sum_{cyc}(a^4+a^2b^2+b^4+2(a^4+a^2bc+b^2c^2))}=\sqrt{\sum_{cyc}(4a^4+3a^2b^2+2a^2bc)}.$$
In another hand, by AM-GM
$$\sum_{cyc}a\sqrt{2a^2+bc}=\sqrt{\sum_{cyc}(2a^4+a^2bc+2ab\sqrt{(2a^2+bc)(2b^2+ac)})}\leq$$
$$\leq\sqrt{\sum_{cyc}(2a^4+a^2bc+ab(2a^2+bc+2b^2+ac))}=\sqrt{\sum_{cyc}(2a^4+2a^3b+2a^3c+3a^2bc)}.$$
Id est, it's enough to prove that:
$$\sum_{cyc}(2a^4-2a^3b-2a^3c+3a^2b^2-a^2bc)\geq0$$ or
$$2\sum_{cyc}(a^4-a^3b-a^3c+a^2bc)+3\sum_{cyc}(a^2b^2-a^2bc)\geq0,$$ which is true by Schur and Muirhead.
| {
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"timestamp": "2023-03-29T00:00:00",
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We have $a,b,c$ and $d$ are real numbers such that $\frac{b + c + d}{a} = \frac{a + c + d}{b} = \frac{a + b + d}{c} = \frac{a + b + c}{d} = r$.
If $a,b,c$ and $d$ are real numbers such that $\frac{b + c + d}{a} = \frac{a + c + d}{b} = \frac{a + b + d}{c} = \frac{a + b + c}{d} = r$ , find the sum of all the possible values of $r$ .
What I Tried :- First of all, when $a = b = c = d$ , the $4$ equations hold and we get $r = 3$ as $1$ solution .
For other solutions I simplified to get $6$ expressions as :-
$b^2 + bc + bd = a^2 + ac + ad$
$bc + c^2 + cd = a^2 + ab + ad$
$bd + cd + d^2 = a^2 + ab + ac$
$ac + c^2 + cd = ab + b^2 + bd$
$ad + cd + d^2 = ab + b^2 + bc$
$ad + bd + d^2 = ac + bc + c^2$
Now I have no idea how to start finding solutions for $r$ from here . Can anyone help?
| We obtain: $$\frac{b+c+d}{a}+1=r+1$$ or
$$\frac{a+b+c+d}{a}=r+1.$$
Now, $a+b+c+d=0$ gives $r=-1$, but for $a+b+c+d\neq0$ we obtain:
$$\frac{a}{a+b+c+d}=\frac{1}{r+1}$$ and
$$\frac{4}{r+1}=\sum_{cyc}\frac{a}{a+b+c+d}=1,$$ which gives $$r=3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3784199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find minimum value of $f(b)$ where $f(b)$ denotes the maximum value of $g(x)$ where $g(x)=\left|\sin x+\frac{2}{3+\sin x}+b\right|$.
For any real number $b$ let $f(b)$ denote the maximum value of $g(x)$ where $$g(x)=\left|\sin x+\frac{2}{3+\sin x}+b\right|$$
Find the minimum value of $f(b)$
$f(b)$ appears to be lying between $b$ and $b+\frac{3}{2}$. So what can the minimum value of $f(b)$
| Let $t=3+\sin x$, then $2\leq t\leq 4$ so $$3\leq t+{2\over t} \leq {9\over 2}$$ and $$g(x)= \underbrace{|t-3+{2\over t}+b|}_{=:h(t)}$$
If $b\geq 0$ then $$h(t) = t+{2\over t}+b-3 \leq {3\over 2}+b = f(b) \implies \min f(b) = {3\over 2}$$
If $b\leq -{3\over 2}$ then $$h(t) = -t-{2\over t}-b+3 \leq -b = f(b) \implies \min f(b) = {3\over 2}$$
If $-{3\over 2}<b<0$ then $$f(b)= \max \{h(2),h(4)\} = \max \{-b,{3\over 2}+b \}$$
$$={{3\over 4} +|b+{3\over 4}| } \implies \boxed{\min f(b) = {3\over 4}}$$
and eqaulity is achieved at $b=-{3\over 4}$
Notice: I use this formula for max at the end
$$ \max\{a,b\} = {a+b+|a-b|\over 2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplifying $\frac{b^5-a^5-5b^3a^2+5b^2a^3}{b^4+a^4-2a^2b^2}$ into $\frac{b^3-a^3-2a^2b+2ab^2}{a^2+b^2+2ab}$ I got two algebraic expression, the first should be simplified, with some manipulation I think, to be the second. It's similar to binomial expansion, even though coefficients are a mess.
$$\frac{b^5-a^5-5b^3a^2+5b^2a^3}{b^4+a^4-2a^2b^2} \tag 1$$
$$\frac{b^3-a^3-2a^2b+2ab^2}{a^2+b^2+2ab} \tag 2$$
It's not exactly a problem related to my studies, but something I'm interested in. I remember something similar, when we add/subtract some terms till we got full expansion or so, but I couldn't do anything. Actually, I'm not sure if this problem is correct, so if there's a proof for this problem as invalid one, it's welcome.
| Hint: Notice that substituting $a=b$ gives us numerator as zero for both the expressions. Hence, $(a-b)$ is one of the factors.
$$1.\ -\frac{(a-b)^3(a^2+3ab+b^2)}{(a^2-b^2)^2}\\ 2.\ -\frac{(a-b)(a^2+3ab+b^2)}{(a+b)^2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I study the convergence of the improper integral $\int_{2}^{\infty} \frac{\arctan(x+1)+x}{2^x+3^x}\, \mathrm dx$? I need to study the convergence of the following improper integral:
$$\int_{2}^{\infty} \dfrac{\arctan(x+1)+x}{2^x+3^x}\, \mathrm dx$$
I did the following:
$$ -\dfrac{\pi}{2} < \arctan(x+1) < \dfrac{\pi}{2} \\
\implies -\dfrac{\pi}{2} + x < \arctan(x+1) +x < \dfrac{\pi}{2} +x \\
\implies \dfrac{-\dfrac{\pi}{2} + x}{2^x+3^x} < \dfrac{\arctan(x+1) +x}{2^x+3^x} < \dfrac{\dfrac{\pi}{2} + x}{2^x+3^x} \\
$$
I planned to integrate the inequality and then using the comparison criterion to proof its convergence. However, the idea did not work for me.
| The function $x\mapsto \lvert\arctan(x+1)\rvert$ is globally bounded by $\frac\pi2$ so your integral is bounded in absolute value by $$\int_0^\infty \frac{\frac\pi2+x}{2^x+3^x}\,\mathrm dx<\infty.$$
Note that the last integral is finite because $$\int_0^\infty \frac{P(x)}{b^x}\,\mathrm dx$$ is finite for all polynomials $P$ and numbers $b>1$. This can be proven for example by noting that $$b^x\geq 1+C x^{\mathrm{deg}(P)},$$ where $C=\frac{\ln(b)^{n+2}}{(n+2)!}$ and $\mathrm {deg}(P)$ is the degree of the polynomial $P$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Two methods are giving two different answers to the this differential equation : $\frac{dy}{dx}=\frac{1}{2} \frac{d(\sin ^{-1}(f(x))}{dx}$
$f(x)=\left(\sin \left(\tan ^{-1} x\right)+\sin \left(\cot ^{-1} x\right)\right)^{2}-1,\ |x|>1$
If $\displaystyle\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} x}\left(\sin ^{-1}(f(x))\right)$ and
$y(\sqrt{3})=\frac{\pi}{6}$, then $y(-\sqrt{3})$ is equal to :
Options:
$1. \quad-\frac{\pi}{6}\\
2. \qquad \frac{2 \pi}{3}\\
3. \qquad \frac{5 \pi}{6}\\
4. \qquad\frac{\pi}{3}$
Now I am getting two Answers in two methods. Can anyone tell me which method is wrong and why?
Method - 1 $f(x) = [\sin(\tan^{-1}x) + \sin(\cot^{-1}x) ]^2 -1 $.
Let $\tan^{-1}x = \theta $. So $f(x) = [\sin(\theta) + \sin(\frac{\pi}{2} - \theta) ]^2 -1 = \sin 2\theta$.
So $f(x) = \sin 2\theta = \frac{2\tan \theta}{1+ \tan^2 \theta} = \frac{2x}{1+x^2}\tag 1$.
Now $\frac{d}{dx} \sin^{-1} \frac{2x}{1+x^2} = \frac{2(1-x^2)}{\sqrt{(1-x^2)^2}(1+x^2)} = \frac{2(1-x^2)}{(1+x^2)(x^2 -1)}$ [Since $|x| > 1$].
Now $\frac{dy}{dx} = \frac{1}{2}\frac{d(\sin^{-1}f(x)}{dx}$ . So $\frac{dy}{dx} = \frac{-1}{1+x^2}$. So $y= - \tan^{-1} x + C$. Now as $y(\sqrt 3) = \frac{\pi}{6}$ , $C = \frac{\pi }{2}$.
So $y = -\tan^{-1} x + \frac{\pi}{2}$.
So $ \displaystyle y(-\sqrt 3) = \frac{\pi}{3} + \frac{\pi}{2} = \frac{5\pi}{6}$
Method -2 - $\displaystyle\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} x}\left(\sin ^{-1}(f(x))\right)$. So $y = \frac{\sin^{-1}f(x)}{2 } + C$.
Now $f(x) = \sin (2\tan^{-1} x)$.
SO $\displaystyle y = \frac{\sin^{-1}(\sin (2\tan^{-1} x))}{2 } + C$ . Now as $y(\sqrt 3) = \frac{\pi}{6}$ , $C = 0$.
So $\displaystyle y = \frac{\sin^{-1}(\sin (2\tan^{-1} x))}{2 } $ .
So $ \displaystyle y(-\sqrt 3) = \frac{-\pi}{6}$
I am really confused. Why I am getting two answers? Can anyone please help me out?
| So we can write $f(x) = \sin (2 \tan^{-1} x)$ where $|x | > 1$.
$\frac{dy}{dx} = \frac{1}{2}\frac{d(\sin^{-1}f(x)}{dx}$ . So $2y = \sin^{-1} f(x) +C$ which is nothing but $2y = \sin^{-1}(\sin (2 \tan^{-1} x)) + C$.
Now $y(\sqrt 3) = \frac{\pi}{6}$ . So $C = 0$ if $x> 1$.
$\therefore 2y = \sin^{-1}(\sin (2 \tan^{-1} x)) $ if $x> 1$
So the solution to the differential equation is $$
y(x)=\begin{cases}
\frac{1}{2}\sin^{-1}(\sin (2 \tan^{-1} x)) & x>1 \\[6px]
\frac{1}{2} \sin^{-1}(\sin (2 \tan^{-1} x))+ d & x<-1
\end{cases}
$$
Here $d \in \mathbb R$.
So $y(-\sqrt 3)$ will be any element of the set $\{\frac{1}{2} \sin^{-1}(\sin (2 \tan^{-1} (-\sqrt 3)))+ d : d \in \mathbb R\}$ which is nothing but $\mathbb R$.
So all answers are correct.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Hard Geometry Question Problem:
In triangle $ABC$, $D$ and $E$ are two points on side $BC$ such that $BD = CE$ and $\angle BAD = \angle CAE$. Prove that triangle $ABC$ is isosceles.
I am having a lot of trouble with this question; what I have done so far is figure out that I need to use a translation that takes $B$ to $E$ and $D$ to $C$. Solutions would be greatly appreciated.
|
Let $b:=\overrightarrow{AB}$, $c:=\overrightarrow{AC}$ and $A$ be the origin. Let $\frac{BD}{BC}=\frac{CE}{BC}=x$. Then
$$\overrightarrow{AD}=xc+(1-x)b,\quad
\overrightarrow{AE}=xb+(1-x)c,\\
\cos\angle BAD=\frac{
(\overrightarrow{AB}\cdot \overrightarrow{AD})}{
|\overrightarrow{AB}|\cdot |\overrightarrow{AD}|},\quad
\cos\angle CAE=\frac{
(\overrightarrow{AC}\cdot \overrightarrow{AE})}{
|\overrightarrow{AC}|\cdot |\overrightarrow{AE}|},
$$
thus (equating the cosine squares)
$$\frac{(b(xc+(1-x)b))^2}{b^2(xc+(1-x)b)^2}=
\frac{(c(xb+(1-x)c))^2}{c^2(xb+(1-x)c)^2}$$
$$(b(xc+(1-x)b))^2c^2(xb+(1-x)c)^2=(c(xb+(1-x)c))^2b^2(xc+(1-x)b)^2$$
$$c^2(xbc+(1-x)b^2)^2(x^2b^2+2x(1-x)bc+(1-x)^2c^2)=\\
b^2(xbc+(1-x)c^2)^2(x^2c^2+2x(1-x)bc+(1-x)^2b^2)$$
$$c^2(x^2(bc)^2+2x(1-x)(bc)b^2+(1-x)^2b^4)(x^2b^2+2x(1-x)bc+(1-x)^2c^2)=\\
b^2(x^2(bc)^2+2x(1-x)(bc)c^2+(1-x)^2c^4)(x^2c^2+2x(1-x)bc+(1-x)^2b^2)$$
However, I'm more convinient with feeding this thing to WolframAlpha at this point (scalars $A:=bc,\,B:=b^2,\,C:=c^2$), see here:
$$(x - 1) x^2 (-(b^2 - c^2)) (c^2 b^2 - (bc)^2) (-2 (bc) x + c^2 x - c^2 + b^2 x - b^2)=0$$
So it's either
*
*$x=1$ but then $D=C,\,E=B$ or
*$x=0$ but then $D=B,\,E=C$ or
*$AB^2=AC^2$ or
*$\cos^2\angle CAB=
\frac{
((\overrightarrow{AB}\cdot \overrightarrow{AC}))^2}{
|\overrightarrow{AB}|^2\cdot |\overrightarrow{AC}|^2}=
\frac{(bc)^2}{b^2c^2}=1$ so $\cos\angle CAB=\pm 1$ and the triangle $\triangle ABC$ is degenerate or
*$x(b-c)^2=b^2+c^2$ but then the cosines ($\cos\angle CAE$, $\cos\angle DAB$) signs are different so the angles can't be equal. This thing did come up because we equated the squared cosines, not the cosines themselves.
So out of all the possibilities it's left only $AB^2=AC^2$ i.e. $|AB|=|AC|$, QED.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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How to integrate this "$\int_0^\infty \frac{\frac{5}{4}x^3-\frac{93}{32}x^2+\frac{95}{24}x}{1+e^{\beta (x-x_0)}}$" I do some project about harmonic oscillator potential in QM and need to find this term, but I don't know how to integrate this one. When, $\beta >0$
| This is not an easy problem which requires special function since, for $\beta >0$ and $n \geq0$,
$$\int_0^\infty \frac{x^n}{1+e^{\beta (x-a )}}\,dx=- \frac{\Gamma (n+1) }{ \beta ^{n+1}}\,\,\text{Li}_{n+1}\left(-e^{ \beta a}\right)$$
If $a=0$ this would give for the integral
$$\frac{190 \pi ^2 \beta ^2-2511 \beta \zeta (3)+42 \pi ^4}{576 \beta ^4}$$
If the integral was for $a$ to $\infty$, letting $x=y+a$ would lead to integrals
$$\int_0^\infty \frac{y^n}{1+e^{\beta y}}\,dy=\frac{ \left(2^n-1\right) \zeta (n+1) \Gamma (n+1)}{2^n\,\beta^{n+1} }$$ and your numerator would be
$$\frac{1}{96} \left(120 a^3-279 a^2+380 a\right)+\frac{1}{48} \left(180 a^2-279
a+190\right) y+\left(\frac{15 a}{4}-\frac{93}{32}\right) y^2+\frac{5
}{4}y^3$$ and the result of the integral would be
$$\frac{a \left(120 a^2-279 a+380\right) \log (2)}{96 \beta }+\frac{\pi ^2 \left(180
a^2-279 a+190\right)}{576 \beta ^2}+\frac{9 (40 a-31) \zeta (3)}{64 \beta
^3}+\frac{7 \pi ^4}{96 \beta ^4}$$
Edit
Concerning the term
$$\text{Li}_{n+1}\left(-e^{ \beta a}\right)$$ if $a$ is small, we could expand it as
$$-\left(1-2^{-n}\right) \zeta (n+1)-a \beta 2^{-n} \left(2^n-2\right) \zeta (n)-a^2
\beta ^2 2^{-n-1} \left(2^n-4\right) \zeta (n-1)-\frac{1}{3} a^3 \left(\beta ^3
2^{-n-1} \left(2^n-8\right) \zeta (n-2)\right)+O\left(a^4\right)$$
| {
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How does $x^2-8x+17=0$ have nonreal solutions?
The solutions of $x^2-8x+17=0$ are $4 + i$ and $4 - i$.
Well, I calculated and the results are different.
$$\begin{align}
x^2-8x+17 &= 0 \\
x^2-8x &=17 \\
x(x-8) &= 17
\end{align}$$
So the roots are $x=17$ or $x=17+8=25$.
Why $i$ comes from the problem? Could you please explain about it?
| If you take $x = 17$ then $x^{2} - 8x + 17 = 289 - 56 + 17 = 362 \not=0$
Neither hold if you take $x = 25$
You can use Bhaskara's formula. $x = \frac{8+\sqrt{64 - 68}}{2} = 4 + i$ or $x = \frac{8-\sqrt{64 - 68}}{2} = 4 - i$
| {
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Is it possible that $2^{2A}+2^{2B}$ is a square number?
Let A and B be two positive integers greater than $0$. Is it possible that $2^{2A}+2^{2B}$ is a square number?
I am having trouble with this exercise because I get the feeling the answer is no, but I cannot elaborate on the proof. So far what I thought was to assume that there is some integer $C>0$ such that $2^{2A}+2^{2B}=C^2$. Then
$$(2^A+2^B)^2=C^2+2^{A+B+1}$$
I was trying to see if the previous expression could hold a contradiction but I got stuck. All I could find is that $C$ needs to be an even number but that doesn't seem to get me anywhere. I'd appreciate any help.
Thanks in advance!
| Just do it.
Assume without loss of generality that $A \le B$ so
$2^{2A} + 2^{2B}=$
$2^{2A} (1 + 2^{2B-2A})=$
$(2^A)^2 [1 + 2^{2B-2A}]=$
$(2^A)^2 [(2^{B-A})^2 + 1]$.
So if that is a perfect square then we must have $(2^{B-A})^2 + 1$ being a perfect square.
But $(2^{B-A})^2$ is a perfect square so we have two consecutive perfect squares. It should be easy to convince yourself that the only time that ever occurs is $0^2$ and $1^2$. (Proof as addendum).
So the only way this can happen is if $(2^{B-A})^2 = 0$ and $(2^{B-A})^2 + 1=1$.
But $2^{B-A} = 0$ is not possible.
====
Addendume: Then only two consecutive squares are $0$ and $1$.
Proof: Suppose $m^2 = n^2 + 1$. where $m,n$ are non-negative integers. $n^2 < m^2 = n^2 + 1 \le n^2 + 2n + 1= (n+1)^2$ so $n < m \le m+1$. But the only integers between $n$ (exclusive) and $n+1$ (inclusive) is $n+1$ so $m = n+1$. And so $n^2 + 1 = m^2 = (n+1) = n^2 + 2n + 1$ so $2n = 0$ and $n = 0$ and $m =1$.
| {
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Eliminating $m$ from the system $x=\frac{\sqrt{a^2-b^2}+\sqrt{a^2+b^2m^2}}{\sqrt{1+m^2}}$, $y=\frac{m\sqrt{a^2-b^2}+\sqrt{a^2m^2+b^2}}{\sqrt{1+m^2}}$ I have 2 equations and need to eliminate the variable $m$ from them, however the equations quickly get drastically complex:
$$x = \frac{\sqrt{a^2-b^2}+\sqrt{a^2+b^2m^2}}{\sqrt{1+m^2}}$$
$$y = \frac{m\sqrt{a^2-b^2}+\sqrt{a^2m^2+b^2}}{\sqrt{1+m^2}}$$
I have wasted hours on my written approach and also tried many online equations solving websites, but they are no good to me.
In case when $a=2$ and $b=1$ the following relation should be obtained:
$$x^2 + y^2 + \frac{1}{x^2} + \frac{1}{y^2} = 16$$
Just to make sure I calculated the above quantity using desmos and changing the variable $m$, keeping $a$ and $b$ as $2$ and $1$ respectively does indeed had no effect on it.
Please help me out or just let me know if finding a general solution is really possible here, any kind of help will be greatly appreciated.
Here are some extra details, where this question actually comes from:
I was trying to find the locus of focus of an ellipse which rotates in the first quadrant such that it always touches the coordinate axes.
To start with, I wrote an equation of ellipse having it's major and minor axes as $mx-y+c=0$ and $x+my+d=0$ respectively, applying the given condition gives
$$c = \frac{-m\sqrt{a^2+b^2m^2} + \sqrt{b^2+a^2m^2}}{\sqrt{1+m^2}}$$
$$d = -\frac{m\sqrt{b^2+a^2m^2} + \sqrt{a^2+b^2m^2}}{\sqrt{1+m^2}}$$
Now the above given value of $x,y$ is actually the coordinates of one of it's foci, which we can find by shifting the coordinate axes to given major and minor axes, from which eliminating $m$ should give the desired locus.
| $$x^2+y^2+\frac{b^4}{x^2}+\frac{b^4}{y^2}=4a^2$$
Let $A=a^2m^2+b^2$, $B=a^2+b^2m^2$, $M=m^2+1$ and $D=a^2-b^2$. Then
$$x=\sqrt{\frac BM}+\sqrt{\frac DM}\\
\frac{b^2}x=\sqrt{\frac BM}-\sqrt{\frac DM} \\
x-\frac{b^2}x=2\sqrt{\frac DM}\\
y-\frac{b^2}y=2m\sqrt{\frac DM}\\
\left(x-\frac{b^2}x\right)^2+\left(y-\frac{b^2}y\right)^2=4D$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3792188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
let $\mathbf a$ and $\mathbf b$ be 3D vectors. Find a $3\times3$ matrix $\mathbf R$ such that $\mathbf {Ra} = \mathbf a_{\bot \mathbf b}$. Hi as the title says im trying to find this.
let $\mathbf a$ and $\mathbf b$ be 3D vectors. Find a $3\times3$ matrix $\mathbf R$ such that $\mathbf {Ra} = \mathbf a_{\bot \mathbf b}$.
according to my exercises the answer is
$$ R = \frac{1}{b^2}
\begin{bmatrix}
b^2_y+b^2_z & -b_xb_y & -b_xb_z \\
-b_xb_y & b^2_x+b^2_z & -b_yb_z \\
-b_xb_z & -b_yb_z & b^2_x+b^2_y \\
\end{bmatrix}
$$
I haven't been able to get to this solution and i've managed to get as far as
$$
a_{\bot b} = a - a_{||b} = a - \frac{a\cdot b}{b^2}b
$$
and i can substitute $ a_{||b} $ for its expression as a matrix product $$
a_{||b} = \frac{1}{b^2}bb^{\mathrm T}a $$
and this is an outer product so it becomes
$$a_{\bot b} = \frac{1}{b^2}\begin{bmatrix}
b^2_x & b_xb_y & b_xb_z \\
b_xb_y & b^2_y & b_yb_z \\
b_xb_z & b_yb_z & b^2_z+b^2_y \\
\end{bmatrix}\begin{bmatrix}
a_x \\
a_y \\
a_z \\
\end{bmatrix}$$
from this I can get
$$
Ra = a - \frac{1}{b^2}\begin{bmatrix}
b^2_x & b_xb_y & b_xb_z \\
b_xb_y & b^2_y & b_yb_z \\
b_xb_z & b_yb_z & b^2_z+b^2_y \\
\end{bmatrix}\begin{bmatrix}
a_x \\
a_y \\
a_z \\
\end{bmatrix}
$$
This is as far as i was able to get and im not sure the steps needed to get the last equation to the first.
Thanks for any insight anyone can provide.
| The last few steps will be
$$ \begin{align*} Ra &= a - \frac{1}{b^2}\begin{bmatrix}b_x^2 & b_xb_y & b_xb_z\\ b_xb_y & b_y^2 & b_yb_z\\ b_xb_z & b_yb_z & b_z^2\end{bmatrix}a\\ &= \frac{1}{b^2}\Bigg(b^2I - \begin{bmatrix}b_x^2 & b_xb_y & b_xb_z\\ b_xb_y & b_y^2 & b_yb_z\\ b_xb_z & b_yb_z & b_z^2\end{bmatrix}\Bigg)a \\ \end{align*}$$
Notice that $b^2 = b_x^2 + b_y^2 + b_z^2$. So
$$\begin{align*} Ra &= \frac{1}{b^2}\Bigg(\begin{bmatrix}b_x^2 + b_y^2 + b_z^2& 0 & 0\\ 0 & b_x^2 + b_y^2 + b_z^2 & 0\\ 0 & 0 & b_x^2 + b_y^2 + b_z^2\end{bmatrix}-\begin{bmatrix}b_x^2 & b_xb_y & b_xb_z\\ b_xb_y & b_y^2 & b_yb_z\\ b_xb_z & b_yb_z & b_z^2\end{bmatrix}\Bigg)a \\ &= \frac{1}{b^2}\begin{bmatrix}b_y^2 + b_z^2 & -b_xb_y & -b_xb_z\\ -b_xb_y & b_x^2 + b_z^2 & -b_yb_z\\ -b_xb_z & -b_yb_z & b_x^2 + b_y^2\end{bmatrix}a\end{align*} $$
Hence
$$ R = \frac{1}{b^2}\begin{bmatrix}b_y^2 + b_z^2 & -b_xb_y & -b_xb_z\\ -b_xb_y & b_x^2 + b_z^2 & -b_yb_z\\ -b_xb_z & -b_yb_z & b_x^2 + b_y^2\end{bmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3792305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Where is the error in this "proof" that 3=0? I saw this video (link at bottom), with a supposed "proof" that $3=0$. It goes as follows:
Let $x$ be a solution of $$x^2+x+1=0 \tag1$$
Since $x\neq0$, we can divide both sides by $x$:
$$\frac{x^2+x+1}{x}=\frac0x\implies x+1+\frac1x=0 \tag2$$
From $(1)$,
$$x^2+x+1=0\implies x+1=-x^2$$
Substitute $x+1=-x^2$ into $(2)$
$$\begin{align*}
-x^2+\frac1x&=0 \tag3\\
\frac1x&=x^2\\
1&=x^3\implies x=1 \tag4
\end{align*}$$
Substitute $x=1$ into $(1)$
$$\begin{align*}
1^2+1+1&=0\\
3&=0
\end{align*}$$
The explanation given in the video is
Substituting $x+1=-x^2$ into $(2)$ creates the extraneous solution $x=1$ which is not a solution to the original equation $(1)$, $x^2+x+1=0$.
Equations $(1)$ and $(2)$ have solutions $\frac{-1\pm i\sqrt3}{2}$, but after the substitution, equation $(3)$ has these two solutions and $1$.
Basically, it is saying that the issue is substituting $x+1=-x^2$, but I'm not sure if this is actually the problem. How can a substitution cause an issue if everything before the substitution is correct?
After reading the comments, I realised many of them say that the real issue is $(4)$, because $1=x^3$ could also mean that $x=\frac{-1\pm i\sqrt3}{2}$. Not considering these solutions is the issue with the "proof". One also needs to check these solutions before making conclusions, and "pick" whichever one is correct.
So, my question is, what is the issue with the above "proof" that $3=0$?
Video: "Prove" 3 = 0. Can You Spot The Mistake? https://www.youtube.com/watch?v=SGUZ-8u1OxM.
| Substitution is able to cause an extraneous root because it's an irreversible step. That is, it is clear that if $x^2 + x + 1 = 0$, then we have $x + 1 + 1/x = 0$, $x+1 = -x^2$, and by substitution,
$$
-x^2 + 1/x = 0.
$$
However, the reverse is not true: if $-x^2 + 1/x = 0$, then it does not necessarily hold that $-x^2 = x+1$, from which it would follow that $x^2 + x + 1 = 0$.
Indeed, we see that this is how the solution $x = 1$ fits in: it satisfies $-x^2 + 1/x = 0$, but not $-x^2 = x+1$.
Another perspective: the substitution can be summarized with the following multiplication:
$$
x^2 + x + 1 = 0 \implies\\
(-1 + 1/x)(x^2 + x + 1) = 0 \implies\\
-(x^2 + x + 1) + \frac 1x(x^2 + x + 1) = 0 \implies\\
-x^2 + 1/x = 0.
$$
Multiplying $x^2 + x + 1$ by another factor has given the polynomial another root.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3793678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 1
} |
Evaluating $\left| \frac{\tan40^\circ + \tan100^\circ + \tan160^\circ}{\tan20^\circ\tan40^\circ\tan80^\circ} \right| $
How do I find the value of the following expression?
$$
\left| \frac{\tan40^\circ + \tan100^\circ + \tan160^\circ}{\tan20^\circ\tan40^\circ\tan80^\circ} \right|
$$
I tried writing the numerator as $\tan 40^\circ - \tan80^\circ -\tan20^\circ,$ but then the expression was getting complicated.
| Notice that each $\newcommand{\degree}{{\lower{.5pt}\Large\circ}}x\in\left\{20^\degree,-40^\degree,80^\degree\right\}$ satisfies
$$
\begin{align}
\sqrt3
&=\tan(3x)\\
&=\frac{3\tan(x)-\tan^3(x)}{1-3\tan^2(x)}\tag1
\end{align}
$$
Thus,
$$
\tan^3(x)-3\sqrt3\tan^2(x)-3\tan(x)+\sqrt3=0\tag2
$$
Vieta says that the sum of the roots is the negative of the coefficient of $\tan^2(x)$. That is,
$$
\tan\left(20^\degree\right)-\tan\left(40^\degree\right)+\tan\left(80^\degree\right)=3\sqrt3\tag3
$$
and the product of the roots is the negative of the constant term. That is,
$$
-\tan\left(20^\degree\right)\tan\left(40^\degree\right)\tan\left(80^\degree\right)=-\sqrt3\tag4
$$
Therefore,
$$
\begin{align}
\frac{\tan\left(40^\degree\right)+\tan\left(100^\degree\right)+\tan\left(160^\degree\right)}{\tan\left(20^\degree\right)\tan\left(40^\degree\right)\tan\left(80^\degree\right)}
&=\frac{\tan\left(40^\degree\right)-\tan\left(80^\degree\right)-\tan\left(20^\degree\right)}{\tan\left(20^\degree\right)\tan\left(40^\degree\right)\tan\left(80^\degree\right)}\\[6pt]
&=-3\tag5
\end{align}
$$
Just take the absolute value of $(5)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3796320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How would you calculate a derivative of $ f(x)= \frac{\sqrt{x+1}}{2-x}$ by the limit definition? I have a function defined as follows:
$$
f(x)= \frac{\sqrt{x+1}}{2-x}
$$
I tried to calculate the derivative using the limit definition using four methods, but I was unsuccessful in any. Could someone help me calculate it and explain the method?
$$
1) \lim_{h\to 0} =\frac{\frac{\sqrt{(x+h)+1}}{2-(x+h)}-\frac{\sqrt{x+1}}{2-x}}h
$$
$$
2)\lim_{z\to x} =\frac{\frac{\sqrt{z+1}}{2-z}-\frac{\sqrt{x+1}}{2-x}}{z-x}
$$
$$
3)\;f(x)= \frac{\sqrt{x+1}}{2-x}; u=\sqrt{x+1}
$$
$$
\lim_{h\to 0} =\frac{\frac{u+h}{3-(u+h)^2}-\frac{u}{3-u^2}}h
$$
$$
4)\;f(x)= \frac{\sqrt{x+1}}{2-x}; u={x+1};
$$
$$
\lim_{h\to 0} =\frac{\frac{\sqrt{u+h}}{3-(u+h)}-\frac {\sqrt{u}}{3-u}}h
$$
| Note that we have
$$\begin{align}
\frac{\sqrt{x+1+h}}{2-x-h}-\frac{\sqrt{x+1}}{2-x}&=\frac{\sqrt{x+1+h}-\sqrt{x+1}}{2-x-h}+\frac{h\sqrt{x+1}}{(2-x-h)(2-x)}\\\\
&=\frac{h}{(2-x-h)(\sqrt{x+1+h}+\sqrt{x+1})}+\frac{h\sqrt{x+1}}{(2-x-h)(2-x)}\\\\
\end{align}$$
Now divide by $h$ and let $h\to 0$ to find
$$\frac{d}{dx}\left(\frac{\sqrt{x+1}}{2-x}\right)=\frac{1}{2(2-x)\sqrt{x+1}}+\frac{\sqrt{x+1}}{(2-x)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3798018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Find all complex solutions (real and non real) of $2x^3-3x^2+32x+17$ I am to find the solutions of $2x^3-3x^2+32x+17$.
My textbook says the solutions are $\frac{-1}{2}$, $1\pm4i$
I got $\frac{-1}{2}$ and $1\pm i \sqrt{17}$
First I used the fundamental theorem of algebra to find candidate zeros and verified using synthetic division that $\frac{-1}{2}$ is a zero.
I then had:
$(x+\frac{1}{2})(2x^2-4x+34)$
Then, using the quadratic formula with $(2x^2-4x+34)$ to find the zeros:
$a=2$
$b=-4$
$c=34$
$$\frac{4\pm\sqrt{4^2-4(2)(34)}}{4}$$
$$\frac{4\pm\sqrt{-272}}{4}$$
$$\frac{4\pm4i\sqrt{17}}{4}$$
$$1\pm\sqrt{17}i$$
Where did I go wrong and how can I arrive at $1\pm4i$?
| Your problem is expression under root is $16-8*34 = -256 \ne -272$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3798800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
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Given four real numbers $a,b,c,d$ so that $1\leq a\leq b\leq c\leq d\leq 3$. Prove that $a^2+b^2+c^2+d^2\leq ab+ac+ad+bc+bd+cd.$
Given four real numbers $a, b, c, d$ so that $1\leq a\leq b\leq c\leq d\leq 3$. Prove that
$$a^{2}+ b^{2}+ c^{2}+ d^{2}\leq ab+ ac+ ad+ bc+ bd+ cd$$
My solution
$$3a- d\geq 0$$
$$\begin{align}\Rightarrow d\left ( a+ b+ c \right )- d^{2}= d\left ( a+ b+ c- d \right ) & = d\left ( 3a- d \right )+ d\left ( \left ( b- a \right )+ \left ( c- a \right ) \right )\\
& \geq b\left ( b- a \right )+ c\left ( c- a \right ) \\
& \geq \left ( b- a \right )^{2}+ \left ( c- a \right )^{2} \\
& \geq \frac{1}{2}\left ( \left ( b- a \right )^{2}+ \left ( c- a \right )^{2}+ \left ( c- a \right )^{2} \right )\\
& \geq \frac{1}{2}\left ( \left ( b- a \right )^{2}+ \left ( c- b \right )^{2}+ \left ( c- a \right )^{2} \right )\\
& = a^{2}+ b^{2}+ c^{2}- ab- bc- ca
\end{align}$$
How about you ?
| It's wrong.
Try $$(a,b,c,d)=(1,1,1,4).$$
For these values we need to prove that $$19\leq15,$$ which is not so true.
The following inequality is true already.
let $\{a,b,c,d\}\subset[1,3].$ Prove that:
$$a^2+b^2+c^2+d^2\leq ab+ac+bc+ad+bd+cd.$$
We can prove this inequality by the Convexity.
Indeed, let $f(a)=ab+ac+bc+ad+bd+cd-a^2-b^2-c^2-d^2$.
Thus, $f$ is a concave function, which says that $f$ gets a minimal value for an extreme value of $a$,
id est, for $a\in\{1,3\}$.
Similarly, for $b$, $c$ and $d$.
Thus, it's enough to check our inequality for $\{a,b,c,d\}\subset\{1,3\}$, which gives that our inequality is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3801666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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} |
Given two real numbers $x,y$ so that $x^{2}+y^{2}+xy+4=4y+3x$. Prove that $3\left(x^{3}-y^{3}\right)+20x^{2}+2xy+5y^{2}+39x\leq 100$.
Given two real numbers $x, y$ so that $x^{2}+ y^{2}+ xy+ 4= 4y+ 3x$. Prove that
$$3\left ( x^{3}- y^{3} \right )+ 20x^{2}+ 2xy+ 5y^{2}+ 39x\leq 100$$
I used derivative and Wolfram|Alpha but only the minimum value found
$$\min\{3\left ( x^{3}- y^{3} \right )+ 20x^{2}+ 2xy+ 5y^{2}+ 39x\}\Leftrightarrow \left\{\begin{matrix} x\cong 0.0320241\\ y\cong 2.16078\\ \left ( z\cong 2.19235 \right ) \end{matrix}\right.$$
where
*
*$z$ is a root of $472z^{3}- 449z^{2}- 689z- 1305= 0$,
*$y$ is a root of $27468y- 11800z^{2}+ 17833z- 41733= 0$,
*$x$ is a root of $27468x+ 3304z^{2}- 11167z+ 7722= 0$.
Why is unsuccessful ? Here are two examples of my claim.
| Alternative solution
$x^2 + y^2 + xy + 4 = 4y + 3x$ is written as
$$(y + \tfrac{x}{2} - 2)^2 = \tfrac{1}{4}x(4-3x).\tag{1}$$
From $\frac{1}{4}x(4-3x) \ge 0$, we have $0 \le x \le \frac{4}{3}$.
If $x = 0$, then $y = 2$ and $3(x^3 - y^3) + 20x^2 + 2xy + 5y^2 + 39x = -4 < 100$.
If $0 < x \le \frac{4}{3}$, let $x = \frac{4}{t^2 + 3}$. Then from (1), we have
$$y = \frac{2t^2 - 2t + 4}{t^2 + 3}, \quad \mathrm{or}, \quad y = \frac{2t^2 + 2t + 4}{t^2 + 3}.$$
Denote $F = 3(x^3 - y^3) + 20x^2 + 2xy + 5y^2 + 39x$. There are two possible cases:
*
*If $y = \frac{2t^2 - 2t + 4}{t^2 + 3}$, we have
$$F = \frac{-4t^4+32t^3+128t^2+900}{(t^2+3)^2}$$
and
$$100 - F = \frac{8t^2(13t^2-4t+59)}{(t^2+3)^2} \ge 0.$$
*If $y = \frac{2t^2 + 2t + 4}{t^2 + 3}$, we have
$$F = \frac{-4t^4-32t^3+128t^2+900}{(t^2+3)^2}$$
and
$$100 - F = \frac{8t^2(13t^2 + 4t + 59)}{(t^2+3)^2} \ge 0.$$
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3803949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to evaluate $\lim _{x\to 2^+}\left(\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}\right)$ (without L'Hopital)? I am trying to evaluate the following limit:
$$ \lim _{x\to 2^+}\left(\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}\right)$$
Approach #1
$ \frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}} = \\ \sqrt{x+2} + \frac{\sqrt{x}-\sqrt{2}}{\sqrt{x-2}} \cdot \frac{\sqrt{x}+\sqrt{2}}{\sqrt{x}+\sqrt{2}} =\\ \sqrt{x+2} + \frac{x-2}{\sqrt{x^2-2x}+\sqrt{2x-4}} =\\ \sqrt{x+2} + \frac{x-2}{\sqrt{x^2-2x}+\sqrt{2x-4}}\cdot\frac{\sqrt{x^2-2x}-\sqrt{2x-4}}{\sqrt{x^2-2x}-\sqrt{2x-4}} = \\ \sqrt{x+2} + \frac{\sqrt{x^2-2x}-\sqrt{2x-4}}{x-2} $
But I still end up at the indefinite form $\frac12 + \frac{0}{\infty}$
My Approach #2
$\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}} = \frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}} \cdot\frac{\sqrt{x^2-4}-(\sqrt{x}-\sqrt{2})}{\sqrt{x^2-4}-(\sqrt{x}-\sqrt{2})}= \frac1{\sqrt{x-2}}\frac{x^2-x+2\sqrt{2x}-2}{\sqrt{x^2-4}-(\sqrt{x}-\sqrt{2})}$
Which also seems to be a dead end.
Any ideas on how to evaluate this?
| As regards your first approach, from the first line, as $x\to 2^+$,
$$\frac{\sqrt{x^2-4}+\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}=\underbrace{\color{blue}{\sqrt{x+2}}}_{\to 2}+\underbrace{\frac{\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}}_{0/0}.$$
Now note that the $0/0$ indeterminate term is
$$\frac{\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}=\frac{\sqrt{x}-\sqrt{2}}{\sqrt{x-2}}\cdot \frac{\sqrt{x}+\sqrt{2}}{\sqrt{x}+\sqrt{2}}=
\frac{x-2}{\color{blue}{\sqrt{x-2}(\sqrt{x}+\sqrt{2})}}=\frac{\sqrt{x-2}}{\sqrt{x}+\sqrt{2}}\to 0.$$
Therefore the required limit is $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3806082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Prove that $\sum_{cyc} \sqrt{\frac{a}{b+c}+\frac{b}{c+a}}\ge 2+\sqrt{\frac{a^2+b^2+c^2}{ab+bc+ca}}$ For $a,b,c\geq 0$, no two of which are $0$, prove that:
$$\sqrt{\dfrac{a}{b+c}+\dfrac{b}{c+a}}+\sqrt{\dfrac{b}{c+a}+\dfrac{c}{a+b}}+\sqrt{\dfrac{c}{a+b}+\dfrac{a}{b+c}}\geq 2+\sqrt{\dfrac{a^2+b^2+c^2}{ab+bc+ca}}$$
This inequality actually came up as an accident when I tried to combine 2 known results, and after many testings on computer it still remains true, but there's still no original proof yet. Hope everyone enjoy and have some good ideas for it.
Here's that 2 known results:
$$\dfrac{a^2+b^2+c^2}{ab+bc+ca}\geq \prod \left(\dfrac{a}{b+c}+\dfrac{b}{c+a}\right)$$
$$\sqrt{\dfrac{a}{b+c}+\dfrac{b}{c+a}}+\sqrt{\dfrac{b}{c+a}+\dfrac{c}{a+b}}+\sqrt{\dfrac{c}{a+b}+\dfrac{a}{b+c}}\geq 2+\sqrt{\prod \left(\dfrac{a}{b+c}+\dfrac{b}{c+a}\right)}$$
The second one can be proved by direct Karamata's inequality, but it may also inspire some ideas for the original one too.
See the following links:
https://artofproblemsolving.com/community/u410204h2218857p16854913
https://artofproblemsolving.com/community/c6h487722p5781880
https://artofproblemsolving.com/community/u414514h2240506p17302184
| Here is a solution with some Matlab help for the analysis, but with a clear manual proof path.
Due to homogeneity, we can demand $a^2+ b^2 + c^2 = 1$. Define $m$ to be the mean of $a,b,c$, i.e. $a + b+c = 3m$. Then note that
$$
9 m^2 = (a+b+c)^2 = a^2+ b^2 + c^2 + 2 (ab + bc + ca) = 1 + 2(ab + bc + ca)
$$
Hence the claim can be written
$$
\sum_{cyc} \sqrt{\frac{a}{b+c}+\frac{b}{c+a}}\ge 2+\sqrt{\frac{2}{9m^2-1}}
$$
Now turn to the LHS. Write $a = m +x$, $b = m + y$, $c = m+z$ with $x+y+z=0$ and $1 = a^2 + b^2 + c^2 = 3 m^2 + x^2 + y^2 + z^2$ which gives two conditions for $(x,y,z)$. W.l.o.g. $(x,y,z)$ can then be expressed as
$$
x = \sqrt\frac23 \sqrt{1 - 3m^2}\cos(\phi-2\pi/3)\\
y = \sqrt\frac23 \sqrt{1 - 3m^2}\cos(\phi-4\pi/3)\\
z = \sqrt\frac23 \sqrt{1 - 3m^2}\cos(\phi)
$$
Hence the claim can be written, with these $(x,y,z)$, as
$$
\sum_{cyc} \sqrt{\frac{m+x}{2m-x}+\frac{m+y}{2m-y}}\ge 2+\sqrt{\frac{2}{9m^2-1}}
$$
The LHS is now a function of $\phi$ whereas the RHS is not. For any $m$, a free (unbounded) minimum w.r.t. $\phi$ of the LHS occurs at $\phi = \pi$ which can be shown by varying $\phi$ about $\pi$. [For bounded minima see below.] So we have to inspect the LHS at that minimum and show that
$$
\lim_{(\phi = \pi)} \sum_{cyc} \sqrt{\frac{m+x}{2m-x}+\frac{m+y}{2m-y}}- 2-\sqrt{\frac{2}{9m^2-1}} \ge 0
$$
Since $(a,b,c)$ should be nonnegative, this requires that $c = m + z = m - \sqrt\frac23 \sqrt{1 - 3m^2} > 0$ or $m > \sqrt2 / 3$, this bound corresponds to $(a,b,c) = (\frac{1}{\sqrt2},\frac{1}{\sqrt2},0)$. On the other hand, the maximum possible $m$ occurs when $a = b = c = m$ or, since $a^2+b^2 + c^2 = 1$, at $m = 1/\sqrt3$.
Let's look at the two extreme values for $m$. Indeed we have (using Matlab) that
$$
\lim_{(m = \sqrt2 / 3)} \lim_{(\phi = \pi)} \sum_{cyc} \sqrt{\frac{m+x}{2m-x}+\frac{m+y}{2m-y}}- 2-\sqrt{\frac{2}{9m^2-1}} = 0 \\
\lim_{(m = 1 / \sqrt3)} \lim_{(\phi = \pi)} \sum_{cyc} \sqrt{\frac{m+x}{2m-x}+\frac{m+y}{2m-y}}- 2-\sqrt{\frac{2}{9m^2-1}} = 0
$$
and for all values of $m$ in between the $> 0 $ holds. Below is a plot which illustrates this.
The minimum of the LHS may as well be bounded by the fact that $(a,b,c)$ should be nonnegative. In that case, the bound arises when the smallest variable, say $c$, is zero, and it must be inspected, while keeping $c=0$, until another variable becomes zero. So that bound is given by $0 = c = m + \sqrt\frac23 \sqrt{1 - 3m^2}\cos(\phi)$ or $m = \sqrt{\frac{\frac23 \cos^2(\phi) }{1 + 2 \cos^2(\phi) }}$ and must be inspected for $\frac43 \pi > \phi > \frac23 \pi$ since at $\frac23 \pi$ (or $\frac43 \pi$ ) we have that also $b =0$ or $a =0$ (then the terms diverge, and this case was excluded by the OP). That means we have to look at (with the $(x,y,z)$ as above)
$$
\lim_{m = \sqrt{\frac{\frac23 \cos^2(\phi) }{1 + 2 \cos^2(\phi) }}} \sum_{cyc} \sqrt{\frac{m+x}{2m-x}+\frac{m+y}{2m-y}}- 2-\sqrt{\frac{2}{9m^2-1}}
$$
which is a function of $\phi$. Variation of $\phi$ about $\pi$ already shows local positivity. Here is a plot (where $\phi$ was denoted $x$) which illustrates the overall behavior:
This proves the claim. $\qquad \Box$
| {
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"question_score": "7",
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Limit of $(x+1+\sqrt{(x+1)^2 +1})$ as $x\to-\infty$ This seemed easy at first, but then I confronted my result with the one I got from WolframAlpha and they are different:
$$
\lim_{x\to-\infty}(x+1+\sqrt{(x+1)^2 +1})
$$
WolframAlpha says the limit is equal to 0, though for me it seems to be negative infinity. I tried some simple algebraic manipulation, namely multiplying each factor by 1:
$$
\frac{x\sqrt{(x+1)^2+1}}{\sqrt{(x+1)^2+1}}+\frac{\sqrt{(x+1)^2+1}}{\sqrt{(x+1)^2+1}}+[(x+1)^2+1]*\frac{1}{\sqrt{(x+1)^2+1}}
$$
The last part is 0, because the fraction goes to 0:
$$
\lim_{x\to-\infty}([(x+1)^2+1]*\frac{1}{\sqrt{(x+1)^2+1}})=0
$$
The middle part is 1 and the first part should be negative infinity, as the only factor left standing is x and it goes to negative infinity. What am I missing?
| Most of the time when you want to get rid of a root expression you use the 3rd binomial formula. So you would expand into a fraction with $(x+1-\sqrt{(x+1)^2+1})$
So we get:
$\frac{(x+1)^2-(x+1)^2-1}{x+1-\sqrt{(x+1)^2+1}}=\frac{-1}{x+1-\sqrt{(x+1)^2+1}}=\frac{1}{\sqrt{(x+1)^2+1}-(x+1)}$
We have $\sqrt{(x+1)^2+1}-(x+1)\to\infty$ as $x\to -\infty$. So the fraction goes to $0$.
| {
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Does a value for $\sqrt{x+\sqrt{x+\sqrt{x+...}}}$ actually exist? I have seen questions of this type being solved as follows :
$\sqrt{x+\sqrt{x+\sqrt{x+...}}}$'s value does not change if we add an $x$ to the expression and square root it. Let the value of this expression be $y$. So
$$\sqrt{x+y} = y \implies x+y = y^2 \implies y^2-y-x=0$$
Using the quadratic formula, we obtain the value of $y$ as :
$$\dfrac{-(-1)\pm\sqrt{(-1)^2-4(1)(-x)}}{2(1)} = \dfrac{1\pm\sqrt{1+4x}}{2} = \sqrt{x+\sqrt{x+\sqrt{x+...}}}$$
This has been used to solve $\sqrt{6+\sqrt{6+\sqrt{6+...}}}$ in my Mathematics textbook.
Now, this method would work perfectly, assuming that a value for $\sqrt{x+\sqrt{x+\sqrt{x+...}}}$ exists. If a value for this expression does not exist, this would be similar to Numberphile's popular $\displaystyle\sum_{n=1}^\infty n = \dfrac{-1}{12}$ which is doubtlessly wrong.
So, does a value for $\sqrt{x+\sqrt{x+\sqrt{x+...}}}$ exist? Why/why not?
Thanks!
| Define the sequence
$$
u_0=0\tag1
$$
$$
u_{n+1}=\sqrt{x+u_n}\tag2
$$
For $x\gt0$, $u_1\gt u_0$, and then by induction and $(2)$, we have
$$
u_{n+1}\gt u_n\tag3
$$
Suppose that $u_n\le\frac{1+\sqrt{1+4x}}2$, then
$$
\begin{align}
u_{n+1}
&\le\sqrt{x+\frac{1+\sqrt{1+4x}}2}\\
&=\frac{1+\sqrt{1+4x}}2\tag4
\end{align}
$$
Since $u_0\le\frac{1+\sqrt{1+4x}}2$, we must have
$$
u_n\le\frac{1+\sqrt{1+4x}}2\tag5
$$
for all $n\ge0$.
Thus, $u_n$ is an increasing sequence that is bounded above. Therefore, the limit exists.
| {
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Find the limit $\lim_{x\to0}\left(\frac{1}{\arcsin x}-\frac{1}{\sin x}\right)$
Find the limit $$\lim_{x\to0}\left(\frac{1}{\arcsin x}-\frac{1}{\sin
x}\right)$$
\begin{align}
\lim_{x\to 0}\left(\frac{1}{\arcsin x}-\frac{1}{\sin x}\right)&=\lim_{x\to 0}\left(\frac{\sin x-\arcsin x}{\sin x\times \arcsin x}\right)\\
&=\lim_{x\to 0}\frac{x}{\arcsin x}
\cdot\frac{x}{\sin x}\cdot \frac{\sin x-\arcsin x}{x^2}
\end{align}
Here's where I'm stuck. I know from L'Hopital's Rule that $\frac{x}{\sin x}\to 1$, but what about $\frac{x}{\arcsin x}$? Or is there any other way than using L'Hopital to solve this question?
| $$L=\lim_{x\to 0} \left( \frac{1}{\sin^{-1} x}-\frac{1}{\sin x}\right)$$
Note that if $|x|$ is very small then $\sin^{-1} x=x+x^3/6+..., \sin x=x-x^3/6+... (1+x)^{k}=1+kx+...$
Then $$L=\lim_{x \to 0} \left( \frac{1}{x+x^3/6}-\frac{1}{x-x^3/6} \right)$$
$$\implies L=\lim_{x \to 0}\frac{1}{x}[(1+x^2/6)^{-1}-(1-x^2/6)^{-1}]=\lim_{x \to 0}\frac{1}{x}[(1-x^2/6)-(1+x^2/6)]=0$$
| {
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Sum of squares of consecutive triangular numbers Find a general formula for:
$t^2_1 + t^2_1 ... t^2_n$.
I tried evaluating:
$1/4[1(2)^2+2^2(3)^2 ... (n)(n+1)]$ then I used sum of cubes to evaluate this.
$(1^3) +(1^3 +2^3) ... (1^3 + 2^3 ... + n^3)$
$n(1^3) + (n-1)(2^3) ... + n^3(1)$
I also tried grouping terms like:
$n(1)(1^2+n^2) +(n-1)(2)((n-1)^2+2^2)) ...$
but I didn't see anything else
| Hint:
Using Proving $1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$ using induction
$$\sum_{r=1}^n(1^3+2^3+\cdots+r^3)$$
$$=\dfrac14\cdot\left(\sum_{r=1}^nr^4+2\sum_{r=1}^nr^3+\sum_{r=1}^nr^2\right)$$
Use
How to get to the formula for the sum of squares of first n numbers?
and
Sum of fourth powers in terms of sum of squares
| {
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Solving a system of two equations in polar coordinates Below is a problem I did. I believe the answer is right. Is it? However, I am not sure my reasoning is correct. I am also interested in comments about my style.
Problem:
Find the points of intersection of the follow two pairs of curves.
\begin{align*}
r &= a(1 + \cos \theta) \\
r &= a( 1 - \sin \theta )
\end{align*}
Answer:
\begin{align*}
a \left(1 + \cos \theta \right) &= a \left( 1 - \sin \theta \right) \\
1 + \cos \theta &= 1 - \sin \theta \\
\cos^2 \theta &= \sin^2 \theta = 1 - \cos^2 \theta \\
2 \cos^2 \theta &= 1 \\
\cos \theta &= \pm \frac{1}{ \sqrt{2}}
\end{align*}
Consider $\theta = \frac{\pi}{4}$ as a solution. This corresponds to an $r$ value
of $a\left( 1 + \frac{ \sqrt{2}}{2} \right)$. However, this value does not satisfy the second
equation so it must discarded. Now, we consider $\theta = \frac{3\pi}{4}$. For the first equation, I get:
$$ r = a \left( 1 - \frac{1}{ \sqrt{2}} \right) $$.
For the second equation, I get:
$$ r = a \left( 1 - \frac{1}{ \sqrt{2}} \right) $$
Hence, one of the points of intersection is:
$\left( a\left( 1 + \frac{ \sqrt{2}}{2} \right), \frac{ 3 \pi }{4} \right)$
Now consider $\theta = \frac{5\pi}{4}$ as a solution. In this case, I have:
\begin{align*}
a(1 + \cos\left( \frac{5\pi}{4} \right) ) &= a( 1 - \frac{ \sqrt{2}}{2} ) \\
a(1 - \sin\left( \frac{5\pi}{4} \right) ) &= a( 1 - \frac{ \sqrt{2}}{2} )
\end{align*}
Hence, one of the points of intersection is:
$$ \left( a( 1 - \frac{ \sqrt{2}}{2} ) , \frac{5\pi}{4} \right) $$
Now, we need to consider $\theta = - \frac{5\pi}{4} $
In this case, I have:
\begin{align*}
a(1 + \cos\left( - \frac{5\pi}{4} \right) ) &= a( 1 - \frac{ \sqrt{2}}{2} ) \\
a(1 - \sin\left( - \frac{5\pi}{4} \right) ) &= a( 1 + \frac{ \sqrt{2}}{2} )
\end{align*}
Hence $\theta = - \frac{5\pi}{4}$ is not a solution of the original equation. We picked this solution when we squared both sides.
Now I claim that $(0,a)$ and $(0,b)$ are the same point for all real numbers $a$ and $b$. Hence,
we have to consider the fact that both equations go through the orgin. Therefore, I claim that
$(0,0)$ is the third point of intersection even though $(0,0)$ is not on either curve.
| In a Cartesian coordinate system two points $(x_1,y_1)$ and $(x_2, y_2)$ are the same only if $x_1=x_2$ and $y_1=y_2$. In the polar coordinate system, two points are identical if $r_1=r_2$ and either $\theta_1=\theta_2$ or $r_1=r_2=0$. The origin is special. Since both curves contain the origin, it is not necessary to have $\theta_1=\theta_2$ at that point. It is only for $r\ne0$ you solve the way you did.
You could have simplified a little bit your calculations. You know $$\cos\theta=-\sin\theta$$
You squared this, and you got two answers for $\cos\theta$, then you checked all four quadrants where $|\cos\theta|=|\sin\theta|=\frac 1{\sqrt 2}$. You could have immediately skipped first and third quadrant, where both sine and cosine have the same sign.
| {
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Proving $\frac{{a}^3+{b}^3+ac(a-c)+bc(b-c)-5abc}{{(a+b+c)}^3} \geqslant -\frac{1}{5}$ For $a,b,c\geqslant 0.$ Prove$:$
$$\dfrac{{a}^3+{b}^3+ac(a-c)+bc(b-c)-5abc}{{(a+b+c)}^3} \geqslant -\dfrac{1}{5}$$
I found an AM-GM proof.
Since $$P+\frac{1}{5}\geqslant 0\Leftrightarrow 6\,{a}^{3}+6\,{b}^{3}+8\,{a}^{2}c-2\,a{c}^{2}+8\,{b}^{2}c-2\,b{c}^{2}-19\,abc+3\,{a}^{2}b+3\,a{b}^{2}+{c}^{3} \geqslant 0$$
And by AM-GM$:$
$$2\,a{c}^{2}\leqslant 6{a}^{3}+\frac49{c}^{3},$$
$$2\,b{c}^{2}\leqslant 6{b }^{3}+\frac49{c}^{3},$$
$$19\,abc\leqslant \frac19{c}^{3}+3a{b}^{2}+3{a}^{2}b+8 \,{a}^{2}c+8\,{b}^{2}c.$$
So we are done!
Is there another nice proof$?$ Thanks for a real lot!
| The LHS is homogenous of degree $0$. Hence we can replace $a=\frac{cx}{3}$ and $b=\frac{cy}{3}$ with $x,y\geq 0$ and obtain the equivalent objective $$2x^3 + x^2y + xy^2 + 2y^3 + 8x^2 - 19xy + 8y^2 - 6x - 6y + 9 \geq 0 \\
\text{or} \\ (x-y)^2 \left(2x+2y+11\right) + 3(x-1)(y-1)\left(x+y+3\right) \geq 0 \, .$$
The last equation is symmetric with respect to $x$ and $y$ and can only be negative if the second term is negative, that is when WLOG $x > 1$ and $y < 1$. In this case the LHS is larger than $$\left(x-1\right) \left[ (x-y)(2x+2y+11)+3(y-1)(x+y+3) \right] \\
= (x-1)\left[2x^2+8x+3xy + y^2 - 5y - 9\right] > (x-1) (y-1)^2 > 0 \, .$$
| {
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Why isn't $x^2+x+1$ a factor of $x^{12}+x^6+1$? When we solve the equation
$$x^{12}+x^6+1=0$$ we obtain $2$ solutions that also satisfy $$x^2+x+1=0$$
namely $-\frac{1}{2}\pm i\frac{\sqrt3}{2}$.
Shouldn't this imply that $x^2+x+1$ is a factor of $x^{12}+x^6+1$? However, the fully factorsied form of
$x^{12}+x^6+1$ is $$(x^6-x^3+1)(x^6+x^3+1)$$
The reason I think that it should be a factor is that when we have, for a function $f(x)$, (here a polynomial,) $f(a)=0$ then we know that $(x-a)$ is a factor of $f(x)$.
If we multiply $(x-(-\frac{1}{2}+ i\frac{\sqrt3}{2}))$ by $(x-(-\frac{1}{2}- i\frac{\sqrt3}{2}))$ which are both factors of $x^{12}+x^6+1$ we get $x^2+x+1$. It seems to me that this should apply to the polynomials above. I have a feeling that my logic is specious, but I'm not entirely certain why; I think the answer to my problem may be that $x^2+x+1$ is factor of $x^{12}+x^6+1$, but only if the other factor has complex coefficients, but I'm not sure. Thank you for your help.
EDIT
Oh wow, I'm so sorry for making such a stupid mistake everyone, thanks for correcting me (see numerous comments and answers). I was looking at a few polynomials simultaneously and got mixed up. Thanks for your help.
| You are correct that if all of the roots of a polynomial $f(x)$ are roots of a polynomial $g(x)$, then $f(x)$ is a factor of $g(x)$.
However $x^2 + x + 1$ is not a factor of $x^{12} + x^6 + 1$. The reason is because $- \frac{1}{2} \pm i \frac{\sqrt{3}}{2}$ are not roots of that polynomial. This is because
$$ \left( - \frac{1}{2} \pm i \frac{\sqrt{3}}{2} \right)^3 = 1,$$
so plugging it into $x^{12} + x^6 + 1$ gives $(x^3)^4 + (x^3)^2 + 1 = 1 + 1 + 1 = 3 \neq 0$.
| {
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Inequality with a, b, c about finding minimal and maximal value Find the minimal and maximal value (if they exist) of ${\sqrt{\frac{a(b+c)}{b^2+c^2}}} +{\sqrt{\frac{b(a+c)}{a^2+c^2}}} +{\sqrt{\frac{c(b+a)}{b^2+a^2}}}$ if are non-negative real numbers, such that at least two of them are positive.
My attempts for the case where all variables are positive:
I tried applying AM-GM on ${\sqrt{a(b+c)}} $ and etc. and then applying $b^2+c^2=>(b+c)^2/2$,but the inequality I received was false. I also tried rewriting $a(b+c)/(b^2+c^2)=(b/a+c/a)/((b/a)^2+(c/a)^2)$ and etc. and then letting $a/b=x$, $b/c=y$ and $c/a=z$ and etc but I was stuck from there. I also tried applying Cauchy Schwarz by squaring the whole expression.
| For $c=0$ and $a=b=1$ we obtain a value $2$.
We'll prove that it's a minimal value.
Indeed, by Holder:
$$\left(\sum_{cyc}\sqrt{\frac{a(b+c)}{b^2+c^2}}\right)^2\sum_{cyc}\frac{a^2(b^2+c^2)}{b+c}\geq(a+b+c)^3.$$
Thus, it's enough to prove that:
$$(a+b+c)^3\geq4\sum_{cyc}\frac{a^2(b^2+c^2)}{b+c},$$ which is obvious after full expanding:
$$\sum_{sym}(a^5b-a^3b^3+5a^4bc+11a^3b^2c+5a^2b^2c^2)\geq0.$$
The maximal value does not exist
because for $b=c=1$ and $a\rightarrow+\infty$ our expression close to $+\infty$.
| {
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prove $F(p)$ is non-negative for all $p$ Let $p>0$. Define, $$ \displaystyle F(p) = 1- \,\dfrac{\dfrac{p^N}{N!}\,(N+1)\,\left(\sum_{n=0}^N\, \dfrac{p^n}{n!}\right)- \,\dfrac{p^{N+1}}{N!}\,\sum_{n=0}^{N-1}\,\dfrac{p^n}{n!}}{\left(\sum_{n=0}^N\, \dfrac{p^n}{n!}\right)^2}\,\,\,-(*)$$
I need to prove that $F(p)>0$ for all $p>0$. The result holds when $p<1$ but I'm having difficulty proving it for $p\geq 1$.
My work (Assume $N\geq 3$ since the result holds for $N=1,2$):
Taking the common denominator and further simplifying the numerator of $F(p)$,
$$\sum_{n=0}^{N-1}\,\dfrac{p^n}{n!}\left(\sum_{n=0}^{N}\,\dfrac{p^n}{n!}-\dfrac{p^N\,(N+1)}{N!}\,+\dfrac{p^{N+1}}{N!}\right)\,+ \dfrac{p^N}{N!} \, \sum_{n=0}^{N}\,\dfrac{p^n}{n!}\,-\left(\dfrac{p^N}{N!}\right)^2\,(N+1) $$
$$= \sum_{n=0}^{N-1}\,\dfrac{p^n}{n!}\left(\sum_{n=0}^{N-1}\,\dfrac{p^n}{n!} -\dfrac{p^N}{(N-1)!}\,+\dfrac{p^{N+1}}{N!}\right)\,+ \dfrac{p^N}{N!} \, \sum_{n=0}^{N-1}\,\dfrac{p^n}{n!}\,-\dfrac{p\,^{2N}}{N!(N-1)!}$$
$$= \sum_{n=0}^{N-1}\,\dfrac{p^n}{n!}\left(\sum_{n=0}^{N-1}\,\dfrac{p^n}{n!}-\dfrac{p^N}{(N-1)!}\right)\,+ \dfrac{p^{N+1}}{N!}\,\sum_{n=0}^{N-2}\, \dfrac{p^n}{n!}+ \dfrac{p^N}{N!} \, \sum_{n=0}^{N-1}\,\dfrac{p^n}{n!}\,\, (**)$$
I'm not sure how to proceed any further. Any help would be appreciated.
| Let $x = \sum_{n=0}^N \frac{p^n}{n!}$. It suffices to prove that
$$x^2 - \frac{p^N}{N!}(N+1)x + \frac{p^{N+1}}{N!}\left(x - \frac{p^N}{N!}\right) > 0$$
which is quadratic in $x$.
It suffices to prove that
$$x > \frac{p^N}{2\cdot N!} (N + 1 - p + \sqrt{N^2 - 2Np + p^2 + 2N + 2p + 1})$$
or
$$\sum_{n=0}^N \frac{p^n}{n!}
> \frac{p^N}{2\cdot N!} (N + 1 - p + \sqrt{N^2 - 2Np + p^2 + 2N + 2p + 1})$$
or
$$\mathrm{e}^{-p}\sum_{n=0}^N \frac{p^n}{n!}
> \frac{\mathrm{e}^{-p} p^N}{2\cdot N!} (N + 1 - p + \sqrt{N^2 - 2Np + p^2 + 2N + 2p + 1}).$$
Let $f(p) = \mathrm{LHS} - \mathrm{RHS}$, for fixed $N$.
By using $(\mathrm{e}^{-p}\sum_{n=0}^N \frac{p^n}{n!})' = - \frac{1}{N!}p^N\mathrm{e}^{-p}$, we have
\begin{align}
f'(p) &= - \frac{1}{N!}p^N\mathrm{e}^{-p}\\
&\quad - \frac{\mathrm{e}^{-p} p^N(N-p)}{2p\cdot N!}(N + 1 - p + \sqrt{N^2 - 2Np + p^2 + 2N + 2p + 1})\\
&\quad - \frac{\mathrm{e}^{-p} p^N}{2\cdot N!}\left(-1 + \frac{p-N + 1}{\sqrt{N^2 - 2Np + p^2 + 2N + 2p + 1}}\right)\\
&= - \frac{\mathrm{e}^{-p} p^N}{2pQ\cdot N!}\Big((N^2-2Np+p^2+N)Q + B \Big)
\end{align}
where $Q = \sqrt{N^2 - 2Np + p^2 + 2N + 2p + 1}$
and $B = N^3-3N^2p+3Np^2-p^3+2N^2-Np-p^2+N$.
Since $(N^2-2Np+p^2+N)^2Q^2 - B^2 = 4Np^2 > 0$,
we have $(N^2-2Np+p^2+N)Q + B > 0$ and thus $f'(p) < 0$.
Also, clearly, $\lim_{p\to \infty} f(p) = 0$. Thus, $f(p) > 0$. We are done.
| {
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Is this correct even if I get two different answers for slope-intercept form? Writing an Equation for a Linear Function Given Two Points
If $f$ is a linear function, with $f(3)=−2$, and $f(8)=1$, find an equation for the function in slope-intercept form.
We can write the given points using coordinates.
\begin{align*}
f(3) & = −2 \to (3,−2)\\
f(8) & = 1 \to (8,1)
\end{align*}
We can then use the points to calculate the slope.
\begin{align*}
m & = \frac{y_2 - y_1}{x_2 - x_1}\\
& = \frac{1 - (-2)}{8 - 3}\\
& = \frac{3}{5}
\end{align*}
Substitute the slope and the coordinates of ONE OF THE POINTS into the point-slope form.
*The book decided to use $(3, -2)$ whereas I decided to use $(8, 1)$.
\begin{align*}
y - y_1 & = m(x - x_1)\\
y - (-2) & = \frac{3}{5}(x - 3)
\end{align*}
The book goes further in their example.
We can use algebra to rewrite the equation in the slope-intercept form.
\begin{align*}
y + 2 & = \frac{3}{5}(x - 3)\\
y + 2 & = \frac{3}{5}x - \frac{9}{5}\\
y & = \frac{3}{5}x - \frac{19}{5}
\end{align*}
The points I chose to use $(8, 1)$, which then gave me the answer
\begin{align*}
y - 1 & = \frac{3}{5}(x - 8)\\
y & = \frac{3}{5}x - 5
\end{align*}
Because of choosing different points I get what seems a different slope-intercept even when it states "Substitute the slope and the coordinates of ONE OF THE POINTS into the point-slope form.". Is this ok?
| It doesn't matter which point one picks, you will get the same solution.
$$y-1=\frac35 \left( x-8\right)$$
$$y=\frac35x - \frac{24}5+1=\frac35x-\frac{24-5}{5}=\frac35x-\frac{19}{5}$$
which is consistent with the other solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the value of the sum $600\sum_{a = 1}^\infty \sum_{b = 1}^\infty \sum_{c = 1}^\infty \frac{ab(3a + c)}{4^{a + b + c} (a + b)(b + c)(c + a)}$
The value of the sum
$$600\sum_{a = 1}^\infty \sum_{b = 1}^\infty \sum_{c = 1}^\infty \frac{ab(3a + c)}{4^{a + b + c} (a + b)(b + c)(c + a)}$$
is ______
My attempt:
This looks very symmetric, except for the numerator. We can resolve it into two sums:
$$\begin{align}
S_1(a,b,c) &= 600\sum_{a = 1}^\infty \sum_{b = 1}^\infty \sum_{c = 1}^\infty\frac{abc}{4^{a+b+c}(a+b)(b+c)(c+a)} \\
S_2(a,b,c) &= 1800\sum_{a = 1}^\infty \sum_{b = 1}^\infty \sum_{c = 1}^\infty\frac{a^2b}{4^{a+b+c}(a+b)(b+c)(c+a)}
\end{align}$$
I don't know how to evaluate these two sums. The function does not 'separate' cleanly into different summations. I also read the Symmetric sums article on AoPS but coudn't get how to apply that concept here. Any hints/solutions are appreciated.
| We should consider the sum
$$S(n)=\sum_{a = 1}^n \sum_{b = 1}^n \sum_{c = 1}^n \frac{ab(3a + c)}{4^{a + b + c} (a + b)(b + c)(c + a)}$$
Exchange $a, b, c$ and we get:
$$S(n)=\sum_{a = 1}^n \sum_{b = 1}^n \sum_{c = 1}^n \frac{ac(3a + b)}{4^{a + b + c} (a + b)(b + c)(c + a)}$$
and
$$S(n)=\sum_{a = 1}^n \sum_{b = 1}^n \sum_{c = 1}^n \frac{ab(3b + c)}{4^{a + b + c} (a + b)(b + c)(c + a)}$$
etc.
We eventually get:
$6S(n)\\=\sum_{a = 1}^n \sum_{b = 1}^n \sum_{c = 1}^n \frac{ab(3a+c) + ac(3a+b)+ab(3b+c)+bc(3b+a)+ac(3c+b)+bc(3c+a)}{4^{a+b+c} (a+b)(b+c)(c+a)}\\=\sum_{a = 1}^n \sum_{b = 1}^n \sum_{c = 1}^n \frac{3(a+b)(b+c)(c+a)}{4^{a+b+c} (a+b)(b+c)(c+a)}\\=\sum_{a = 1}^n \sum_{b = 1}^n \sum_{c = 1}^n \frac{3}{4^{a+b+c}}\\=3 \sum_{a = 1}^{n} \frac{1}{4^a} \sum_{b = 1}^{n} \frac{1}{4^b} \sum_{c = 1}^{n} \frac{1}{4^c}$
When $n \longrightarrow \infty$,
$$600\lim_{n \to \infty}S(n)=600\cdot\frac{1}{2} \left(\frac{\frac{1}{4}}{1-\frac{1}{4}}\right)^3=\frac{100}{9}$$
Edit :
A similar question was found here, in which we have to evaluate $$\sum_{a=1}^6\sum_{b=1}^6\sum_{c=1}^6\frac{ab(3a+c)}{2^a2^b2^c(a+b)(b+c)(c+a)}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Trying to solve: $y'' + 2y' = \frac{(y')^2}{y + 1} + \frac{y'}{x} \ln(\frac{y+1}{y'})$, $y(1) = 1$, $y'(1) = 2/e$. The problem is the following: Solve
$$y'' + 2y' = \frac{(y')^2}{(y + 1)} + \frac{y'}{x} \ln\left(\frac{y+1}{y'}\right),$$
given that $y(1) = 1$, $y'(1) = \frac{2}{e}$.
My solution:
After noting that
$$\left(\ln\left(\frac{y+1}{y'}\right)\right)' = \frac{y'}{y + 1} - \frac{y''}{y'}$$
we divide the original equation by $y'$, rearrange and get
$$\frac{y'}{y + 1} - \frac{y''}{y'} + \frac{1}{x}\ln\left(\frac{y + 1}{y'}\right) = 2$$
Then we substitute $t(x) = \ln\left(\frac{y + 1}{y'}\right)$. Now we have
$$t' + \frac{t}{x} = 2$$
Particular solution to this equation is $t = x$ and for $t' + \frac{t}{x} = 0$ we have $t = \frac{C}{x}$. So finally we have $t = x + \frac{C}{x}$.
Now we note that $t(1) = \ln\left(\frac{e(1 + 1)}{2}\right) = 1$ and conclude that $t = x$ and then solve for $y$.
For $y$ we have
$$\frac{y + 1}{y'} = e^x$$
Solution for this equation is $\ln|y + 1| = -e^{-x} + C$. After some manipulations we get
$$y = \frac{C}{e^{e^{-x}}} - 1$$
and following the initial condition $y(1) = 1$ we get $C = 2e^{e^{-1}}$. So the answer will be
$$y = 2e^{e^{-1} - e^{-x}}$$
I have 2 questions:
$\qquad 1)$ Is my solution correct? The answer looks terrible, so maybe I had an error somewhere.
$\qquad 2)$ Is there any easier solution? I thought about finding $y''(1)$, multiplying by $x/y'$, taking derivative to remove $\ln$-function and solve the resulting equation, but it seemed to be much harder than what I did in the solution above.
| $$y'' + 2y' = \frac{(y')^2}{(y + 1)} + \frac{y'}{x} \ln\left(\frac{y+1}{y'}\right),$$
$$(\ln y')'+2=(\ln (y+1))'+\dfrac 1x (\ln( y+1))-\dfrac 1x \ln y'$$
$$x(\ln y')'+2x=x(\ln (y+1))'+ (\ln (y+1))- \ln y'$$
$$x(\ln y')'+\ln y'+2x=x(\ln (y+1))'+ \ln (y+1)$$
The integration is obvious.
$$x\ln y'+x^2=C+x \ln (y+1)$$
$$\ln (\dfrac {y'}{y+1})=\dfrac Cx-x$$
$$\ln ((\ln (y+1))')=\dfrac Cx-x$$
Your solution looks good. I see no mistakes.
| {
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In a triangle ABC, $\begin {vmatrix} a&b&c \\b&c&a \\c&a&b \end {vmatrix}=0$, then find $\sin A\sin B+\sin B\sin C+\sin C\sin A$ From the given determinant,
$$a^3+b^3+c^3-3abc=0$$
Which implies $a+b+c=0$ or $a^2+b^2+c^2 -ab-bc-ac=0$
Since the former isn’t possible,
$$a^2+b^2+c^2=ab+bc+ac$$
$$\sin A\sin B+\sin B\sin C +\sin C\sin A=\sin^2A +\sin^2B+\sin ^2C$$
The answer for this is either $\cos^2A+\cos^2 B+\cos^2C$ or one of numbers $0, 1, \frac 94$
How do I proceed from here?
| The cubic need not be factorised, as side lengths are positive. By AM-GM the triangle is equilateral, leading to @AnindyaPrithvi's calculation of the desired trigonometry.
| {
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Let $z=x+iy$, where $x,y\in \mathbb{Z}$. Find the area of the octagon whose vertices are roots of $(z\bar z)|z^2-\bar z^2|=1200$ $$|z^2||z-\bar z||z+\bar z|=1200$$
$$(x^2+y^2)(|2x|)(|2y|)=1200$$
$$(x^2+y^2)(|xy|)=300$$
I don’t think there is any realistic way to obtain $x$ and $y$ other than hit and trial. I am asking this question to know if there is one. Thanks!
If not possible, then any integers which satisfy this can be provided as an answer.
Answer is 62
Please note: $x,y$ are integers.
| We have $$|z^2-\bar z^2| = |x^2 + 2ixy - y^2 -(x^2 - 2ixy - y^2)| = |4ixy| = 4xy$$ And $$z \bar z = (x+iy)(x-iy) = x^2+y^2$$So $$(x^2+y^2)|4xy| = 1200 \implies (x^2+y^2)|xy| = 300$$ Note that if $(x,y)$ is solution then $(x , -y) , (-x ,y) , (-x , -y), (y , x) , (-y , x) , (y ,-x)$ and $(-y , -x)$ are also solution as well. Because $x , y \in \mathbb{Z}$ we have $x^2 + y^2 \in \mathbb{N}$ and $|xy| \in \mathbb{N}$. The positive divisors of $300$ is as follows: $$1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 25, 30, 50, 60, 75, 100, 150, 300$$
We see that $12 \times 25 = 300$ and also $x =3 , y = 4$ works, so we have all $8$ points because the other points are $(4,3) , (-3 , -4) , (-4 , -3) , (-3 , 4) , (3 , -4) , (4,-3) , (-4 , 3)$.
You can compute the area by adding the areas of rectangle and two trapezoids $6\times 8 + 2\times(\frac{(6 + 8)1}{2}) = 48 + 14 = 62$.
| {
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Cesàro sum of $1+ 0 - 1 + 1 + 0 - 1 + \dots$ I am trying to compute the Cesàro sum of $1+ 0 - 1 + 1 + 0 - 1 + \dots$. When I compute the Cesàro means, I get the following sequence
$$\left(1, 1, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{4}{6}, \frac{5}{7}, \frac{6}{8}, \frac{6}{9}, \frac{7}{10}, \frac{8}{11}, \frac{8}{12}, \cdots\right)$$
Where does this sequence converge to? Is it $\frac{2}{3}$? I fail to see the pattern of this sequence. If I can just rewrite the sequence into a more general form, then I might be able to compute the limit.
| Lets say the first term is the $0$th term. If $s_n,n=0,1,2,3,\dots$ are the partial sums,
$$ s_{3n} = 1,\\s_{3n+1}=1,\\s_{3n+2}=0.$$
Consequently
$$ \sum_{n=0}^{3k} s_n = 2k+1, \\\sum_{n=0}^{3k+1} s_n= 2k+2, \\\sum_{n=0}^{3k+2}s_n = 2k+2. $$
The Cesàro sums are the average of the partial sums $c_k = \frac1{k+1}\sum_{n=0}^k s_n$, and its easy to check via squeeze theorem that $c_{3k}\to 2/3$, $c_{3k+1}\to 2/3$, $c_{3k+2}\to 2/3$. Hence $c_{k}\to 2/3$.
| {
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Proving a polynomial is positive How do I prove that the polynomial
$$ f(x) = x^5-x^3/2-x+4/5 $$
is positive for all real numbers $x > 0$? One obvious way is to show that
$$ f'(x) = 5x^4-3x^2/2-1 $$
has a unique positive root $x_0$ for which $f(x_0) > 0$ and $f''(x_0) > 0$, so the minimum of $f$ must be $f(x_0) > 0$, but the computation of the root and then inserting it into $f, f''$ seems so inelegant to me. I tried completing the square but to no avail. Any elegant ideas?
| For $x\ge1$ the result is trivial since
$$f'(x) = 5x^4-\frac 32x^2-1 \ge 5x^2-\frac 32x^2-1=\frac 7 2x^2-1>0$$
then for $0<x=\frac1y<1$ with $y>1$ we have
$$x^5-x^3/2-x+4/5>0 \iff 8y^5-10y^4-5y^2+10>0$$
and by subsuquent elementary completion of the squares
$$8y^5-10y^4-5y^2+10=(8y^5-16y^4+8y^3)+6y^4-8y^3-5y^2+10=$$
$$=8y^3(y-1)^2+(6y^4-12y^3+6y^2)+4y^3-11y^2+10=$$
$$=8y^3(y-1)^2+6y^2(y-1)^2+(4y^3-8y^2+4y)-3y^2-4y+10=$$
$$=8y^3(y-1)^2+6y^2(y-1)^2+4y(y-1)^2-3(y-1)^2-10(y-1)+3>$$
$$>18y(y-1)^2-3(y-1)^2-10(y-1)+3=(18y-3)(y-1)^2-10(y-1)+3>$$
$$>15(y-1)^2-10(y-1)+3>0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Can we find $ \lim_{n \to \infty } n\left ( \frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \frac{1}{n+3} + ... \right ) $? I have got one method,
If we consider $ a_{n} = \int_{0}^{1} \frac{nx^{n-1}}{1+x} \ dx $
Then, $ \lim_{n \to \infty } n\left ( \frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \frac{1}{n+3} + ... \right ) = \lim_{n \to \infty }a_{n} = \frac{1}{2} $
But can anyone attack this problem in a different & more standard way?
| Like @sansae
$$L= \lim_{n \to \infty } n\left ( \frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \frac{1}{n+3} + ... \right )$$
$$\implies L=\lim_{n \to \infty}n\left(\frac{1}{n(n+1)}+\frac{1}{(n+2)(n+3)}+\frac{1}{(n+4)(n+5)}+...+\frac{1}{(n+k)(n+k+1)}+...+\right)$$
But conver the limit to integral as
$$\implies L= \lim_{n \to \infty}\frac{1}{n} \sum_{k=0}^{n} \frac{1}{(1+k/n)(1+(k+1)/n)}=
\int_{0}^{1} (1+x)^{-2} dx=\frac{1}{2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Alternative approach for proving that for any $x\in\mathbb R^+$, $x^2+3x+\frac{1}{x} \ge \frac{15}{4}$. Let $x \in \mathbb{R^+}$. Prove that:
$$x^2+3x+\frac{1}{x} \ge \frac{15}{4}.$$
While this is indeed easily proven using derivatives, where the minimum is obtained when $x=\frac{1}{2}$, is it possible to prove it through other means, say by AM-GM? Any hints would be much appreciated.
| A bit late but I thought worth mentioning it:
Consider the equivalent inequality
$$4x^2+12x+ \frac 4x = (2x)^2 + 6(2x)+\frac 8{2x}\geq 15$$
Setting $y=2x$, you get
\begin{eqnarray*} y^2+6y+\frac 8y
& = & y^2 - 2y +8\left(y+\frac 1y\right)\\
& = & (y-1)^2-1+8\left(y+\frac 1y\right) \\
& \geq & 0-1+8\cdot 2 = 15
\end{eqnarray*}
with equality if and only if $y=1 \Leftrightarrow x=\frac 12$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Two ways to find the number of arrangements of MISSISSIPPI such that $S$'s are not placed next to each other? In how many arrangements of MISSISSIPPI the $S$'s are not placed next to each other?
Assume we only have the word MIIIPPI,there are $8$ places for which a S can be placed in,on the other hand there are permutations of the first word,follows the number of such arrangements is :$$\binom{8}{4}\frac{7!}{2!\cdot4!}$$
I tried to use another method,the number of such arrangements is the total arrangements minus the number of the arrangements for which such $S$ are all consecutive,so :
$$\frac{11!}{4!4!2!}-8\frac{7!}{4!2!}$$
But the answers are not the same,so where was I wrong?
| The first part is fine. We have
\begin{align*}
\binom{8}{4}\frac{7!}{4!2!1!}=\color{blue}{7\,350}\tag{1}
\end{align*}
The number of all different words which can be built from the letters in $MISSISSIPPI$ is
\begin{align*}
\frac{11!}{4!4!2!1!}=34\,650\tag{2}
\end{align*}
From these words we have to subtract the words which contain consecutive $S$. We consider the $8$ positions where we can place the $4$ S.
\begin{align*}
\,_{1}\,M\,_{2}\,I\,_{3}\,I\,_{4}\,I\,_{5}\,P\,_{6}\,P\,_{7}\,I\,_{8}:
\end{align*}
We have the following ways to group $4$ S with the following number of placements per group
\begin{align*}
&(1,1,1,1)&&(S,S,S,S)\\
&(1,1,2)&&(S,S,SS)\ \ \to3\\
&(2,2)&&(SS,SS)\ \ \ \ \to1\\
&(1,3)&&(S,SSS)\ \ \ \ \to2\\
&(1)&&(SSSS)\ \ \ \ \ \ \to1
\end{align*}
Since only the first group $(1,1,1,1)$ representing $4$ S-runs of length $1$ is admissible, we have to subtract from (2) the number of words built from the other groups.
*
*The size of the group: $|(S,S,SS)|=3$ for instance. So, we have $\binom{8}{3}$ ways to place the members of the group and $3$ different ways to arrange the three members within a placement.
We obtain
\begin{align*}
&\frac{11!}{4!4!2!1!}-\frac{7!}{4!2!1!}\left(3\binom{8}{3}
+\binom{8}{2}+2\binom{8}{2}+\binom{8}{1}\right)\\
&\qquad=34\,650-105\left(56\cdot3+28\cdot1+28\cdot2+8\cdot 1\right)\\
&\qquad=34\,650 -26\,670\\
&\qquad\,\,\color{blue}{=7\,350}
\end{align*}
in accordance with (1).
| {
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How to think of factorising $x^7+x^2+1$ to $(x^2+x+1)(x(x-1)(x^3+1)+1)$ (Thales 2016) I was just doing the following question:
Find a prime number which divides the number $A=14^7+14^2+1$.
I solved it by finding the result which is $A=105413504+196+1=105413701$ and then trying out all prime numbers till I found that 211 divides it. However, obviously this is extremely tedious. I hence looked at the solution which says that $x^7+x^2+1=(x^2+x+1)(x(x-1)(x^3+1)+1)$ and from here by saying that $x=14$ we get the solution. However I can't seem to think of how to intuitively turn $x^7+x^2+1$ into $(x^2+x+1)(x(x-1)(x^3+1)+1)$. I realize that from the question it is obvious to go looking for factors of A and hence trying to factorize $14^7+14^2+1$, but I can't work out how to go about factorizing it, what are the steps which you need to take in order to factorize a given polynomial. Could you please explain to me how to go about factorizing such an expression and how to intuitively think of each step?
| Hint: $x^2+x+1$ always divides $x^a+x^b+x^c$ if $\{a, b, c) \equiv \{0, 1, 2\} \mod 3$. Hence it is easy to note $14^7+14^2+1 \equiv 0 \pmod {14^2+14+1}$, and all that remains is checking primality of $211$.
In case you wish to find the other factor(s), in this case, it may be simpler to consider dividing by $x^3-1$, like so:
$$x^7+x^2+1 = x^4(x^3-1) + x(x^3-1)+(x^2+x+1)$$
$$\implies \frac{x^7+x^2+1}{x^2+x+1} = x^4(x-1)+x(x-1)+1=(x-1)(x^4+x)+1$$
| {
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What is complex number z if $z^8+16z^4+256=0$? So far, I have set y to equal to $z^4$ and used the quadratic equation to solve $y = -8+8\sqrt{3}i$ or $-8-8\sqrt{3}i$. How do I determine the 8 different values of $z$?
| $$z^8+16z^4+256=0$$
$$(\frac{z}{2})^8+(\frac{z}{2})^4+1=0$$
let $x=\frac{z}{2}$
$$x^8+x^4+1=0$$
$$(x^4+1)^2-x^4=0$$
$$(x^4-x^2+1)(x^4+x^2+1)=0$$
$$((x^2+1)^2-3x^2)((x^2+1)^2-x^2)=0$$
so
$$(x^2-\sqrt{3}x+1)(x^2+\sqrt{3}x+1)(x^2-x+1)(x^2+x+1)=0$$
then you can use the quadratic formula to find all the complex roots
| {
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Show that $\binom{n}{3} = \binom{2}{2} + \binom{3}{2} + \binom{4}{2}+...+\binom{n - 1}{2}$ I cam across a question in The Book of Proof that states:
Show that $$\binom{n}{3} = \binom{2}{2} + \binom{3}{2} + \binom{4}{2}+...+\binom{n - 1}{2}$$
Which I found the answer to be:
Assume n ≥ 3. Then:
$$\binom{n}{3} = \binom{n - 1}{3} + \binom{n - 1}{2} = \binom{n - 2}{3} + \binom{n-2}{2}+ \binom{n - 1}{2} = \binom{2}{2} + \binom{3}{2}+...+\binom{n-1}{2}$$
I have just learned about the binomial theorem, and I am not sure how we got to that answer.
I understand that $\binom{n}{3} = \binom{n - 1}{3} + \binom{n - 1}{2}$ because that is the sum of the two previous rows in the Pascal triangle. But I can't find a simple explanation for the steps that follow or for why we assumed $n \geq 3$
| $3$ is arbitrary here. You can apply the same idea of this proof to obtain that
$$\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-2}{k-1} + \dots + \binom{k-1}{k-1} $$ with an anologous proof.
Also it's helpful to see the identity $$\binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1} $$ in a more combinatorial sense.
The number of ways of choosing $k$ objects out of $n$ is equal to the number of ways of choosing $k$ objects out of the first $n-1$ or choosing $k-1$ out of the first $n-1$ (therefore forcing your k^th choice to be the n^th object).
| {
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Is the sequence always divergent Consider the sequence
$x_n = [(\frac{2}{n}+1)^r+(\frac{2}{n}-1)^r]n^{r-1}$, $n=1,2,3,..., r=1,2,3...$
Is it always true that $x_n \rightarrow\infty$ as $n\rightarrow\infty$? Or is it true only when $r$ is even number and does it converge to $0$ when $r$ is odd?
If instead the sequence was
$y_n= [1+(-1)^r]n^{r-1}$, $n=1,2,3,..., r=1,2,3,...$ then what would be the case? Would $y_n$ be divergent regardless of whether $r$ is even or odd?
| The first sequence $\{x_n\}$ tends to $0$ for $r=1$, and diverges to $\infty$ for all other values of $r$. The second one $\{y_n\}$ is identically $0$ for odd $r$, and diverges to $\infty$ for even $r$. I've left the solution for $\{x_n\}$ below.
If $r=1$, then $x_n=\left[\left(\frac{2}{n}+1\right)^1+\left(\frac{2}{n}-1\right)^1\right]n^{1-1}=\frac{2}{n}+1+\frac{2}{n}-1=\frac{4}{n}$, which approaches $0$. For all other values of $r$, we'll need to give separate arguments for even and odd $r$. We'll make use of the following facts:
$$\lim\limits_{n\to\infty}\left[\left(\frac{2}{n}+1\right)^r+\left(\frac{2}{n}-1\right)^r\right]=1+(-1)^r$$
$$\lim\limits_{n\to\infty}n^{r-1}=\infty$$
Case 1: $r$ is even
Here, $1+(-1)^r=1+1=2$. Since the product of a convergent sequence with nonzero limit and one that tends to $\infty$ also tends to $\infty$, we have that
$$\lim\limits_{n\to\infty}x_n=\lim\limits_{n\to\infty}\left(\left[\left(\frac{2}{n}+1\right)^r+\left(\frac{2}{n}-1\right)^r\right]n^{r-1}\right)=\infty$$
Case 2: $r$ is odd
Note that we have already argued the case $r=1$, so $r$ is at least 3 (this will be important later).
Here, $1+(-1)^{r}=1-1=0$, so we can't just multiply $\left(\frac{2}{n}+1\right)^r+\left(\frac{2}{n}-1\right)^r$ with $n^{r-1}$ and take the limit. Instead, we'll consider the function $f(x)=\left[\left(\frac{2}{x}+1\right)^r+\left(\frac{2}{x}-1\right)^r\right]x^{r-1}$ so we can apply L'Hôpital's. This is equivalent to
$$f(x)=\frac{\left(\frac{2}{x}+1\right)^r+\left(\frac{2}{x}-1\right)^r}{x^{1-r}}$$
Both the numerator and denominator approach $0$ as $x\to\infty$, so we differentiate.
\begin{align*}
\frac{\frac{d}{dx}\left[\left(\frac{2}{x}+1\right)^r+\left(\frac{2}{x}-1\right)^r\right]}{\frac{d}{dx}x^{1-r}} &= \frac{r\left(\frac{2}{x}+1\right)^{r-1}\cdot\frac{-2}{x^2}+r\left(\frac{2}{x}-1\right)^{r-1}\cdot\frac{-2}{x^2}}{(1-r)x^{-r}}\\
&= -\frac{2rx^{-2}}{1-r}\cdot x^{r}\cdot \left[\left(\frac{2}{x}+1\right)^{r-1}+\left(\frac{2}{x}-1\right)^{r-1}\right]\\
&= -\frac{2rx^{r-2}}{1-r}\left[\left(\frac{2}{x}+1\right)^{r-1}+\left(\frac{2}{x}-1\right)^{r-1}\right]
\end{align*}
Notice that $\left(\frac{2}{x}+1\right)^{r-1}+\left(\frac{2}{x}-1\right)^{r-1}\to 1+(-1)^{r-1}$ as $x\to\infty$. Because $r$ is odd, $r-1$ is necessarily even, so $1+(-1)^{r-1}=2$. It's clear that $x^{r-2}\to\infty$ because $r$ is at least $3$. Noting that $-\frac{2r}{1-r}>0$ because $r\geq 3$, we get
$$\lim\limits_{x\to\infty}-\frac{2rx^{r-2}}{1-r}\left[\left(\frac{2}{x}+1\right)^{r-1}+\left(\frac{2}{x}-1\right)^{r-1}\right]=\infty$$
so
$$\lim\limits_{x\to\infty}f(x)=\lim\limits_{x\to\infty}\left(\left[\left(\frac{2}{x}+1\right)^r+\left(\frac{2}{x}-1\right)^r\right]x^{r-1}\right)=\infty$$
Thus,
$$\lim\limits_{n\to\infty}x_n=\lim\limits_{n\to\infty}\left(\left[\left(\frac{2}{n}+1\right)^r+\left(\frac{2}{n}-1\right)^r\right]n^{r-1}\right)=\infty$$
This completes the solution.
| {
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"url": "https://math.stackexchange.com/questions/3845249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\frac{(a-b)(c-d)}{(b-c)(d-a)} = \frac{2016}{2017}$ , find $\frac{(a-c)(b-d)}{(a-b)(c-d)}$ .
If $\frac{(a-b)(c-d)}{(b-c)(d-a)} = \frac{2016}{2017}$ , find $\frac{(a-c)(b-d)}{(a-b)(c-d)}$ .
What I Tried :- First I thought for a moment and found out that I can write this :
$$\frac{(a-c)(b-d)}{(b-c)(d-a)} = \frac{(a-c)(b-d)}{(a-b)(c-d)} * \frac{2016}{2017}$$
But how is it going to help?
Then I thought maybe cross-multiply everything and try to factor it? I get :-
$$2017(a-b)(c-d) = 2016(b-c)(d-a)$$
$$\rightarrow 2017(a-b)(c-d) - 2016(b-c)(d-a) = 0$$
$$\rightarrow 2017(ac - ad - bc + bd) - 2016(bd - ab - cd - ac) = 0$$
$$\rightarrow 2017ac + bd + 2016ab + 2016cd + 2016ac - 2017ad - 2017bc - 2016bd = 0$$
I am hopeless at this point, how do you even factorise this?
Can anyone help?
| Quite similar approach:
$$\frac{2016}{2017}=1-\frac{1}{2017}$$
$$\frac{(a-b)(c-d)}{(b-c)(d-a)}=1-\frac{1}{(b-c)(d-a)}$$
After reducing we get:
$$(a-c)(b-d)=-1=-\frac{(a-b)(c-d)}{(a-b)(c-d)}$$
Or:
$$\frac{(a-c)(b-d)}{(a-b)(c-d)}=\frac{-1}{(a-b)(c-d)}=-\frac{1}{2016}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3846338",
"timestamp": "2023-03-29T00:00:00",
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Prove that $\left(1-\frac{1}{2}\right)\left(1+\frac{1}{4}\right)\left(1-\frac{1}{8}\right)\dots$ is convergent Prove that $\left(1-\frac{1}{2}\right)\left(1+\frac{1}{4}\right)\left(1-\frac{1}{8}\right)\dots$ is convergent:
Let us rewrite this into
$$\prod_{n=1}^{\infty}\left(1+\frac{(-1)^n}{2^n}\right)$$
This product converges if
$$\sum_{n=1}^{\infty}\ln \left(1+\frac{(-1)^n}{2^n}\right)$$
does. I write now
$$\sum_{n=1}^{\infty}\ln \left(1+\frac{(-1)^n}{2^n}\right)<\sum_{n=1}^{\infty}\ln \frac{(-1)^n}{2^n}$$
I'm not sure about this step. Can someone help me with this please?
| From $$\sum_{n=1}^{\infty}\ln \left(1+\frac{(-1)^n}{2^n}\right)$$
you can use the alternating series test. Specifically, the sum is equal to $$\sum_{n=1}^{\infty}\left(-1\right)^{n}\ln\left(\left(1+\frac{\left(-1\right)^{n}}{2^{n}}\right)^{\left(-1\right)^{n}}\right)$$
Then it is enough to show that $$\ln\left(\left(1+\frac{\left(-1\right)^{n}}{2^{n}}\right)^{\left(-1\right)^{n}}\right) > \ln\left(\left(1+\frac{\left(-1\right)^{n+1}}{2^{n+1}}\right)^{\left(-1\right)^{n+1}}\right)$$
Exponentiating both sides yields $$\left(1+\frac{\left(-1\right)^{n}}{2^{n}}\right)^{\left(-1\right)^{n}} > \left(1+\frac{\left(-1\right)^{n+1}}{2^{n+1}}\right)^{\left(-1\right)^{n+1}}$$
Split this into two cases: $n$ odd and $n$ even. For $n$ odd, the inequality would simplify to $$\left(1-\frac{1}{2^{n}}\right)^{-1} > \left(1+\frac{1}{2^{n+1}}\right)^{1}$$
This is easy to show since $\left(1-\frac{1}{2^{n}}\right)\left(1+\frac{1}{2^{n+1}}\right) < 1-\frac{1}{2^{n+1}} < 1$. Similarly for $n$ even, the inequality would be $$\left(1+\frac{1}{2^{n}}\right)^{1} > \left(1-\frac{1}{2^{n+1}}\right)^{-1}$$
This is also easy to show since $\left(1+\frac{1}{2^{n}}\right)\left(1-\frac{1}{2^{n+1}}\right) = 1+\frac{1}{2^{x}}\left(\frac{1}{2}-\frac{1}{2^{\left(n+1\right)}}\right) > 1$. Therefore by the alternating series test, the sum would converge.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
} |
Boolean algebra transformation I read online that
$${\displaystyle (A\cdot {\overline {B}})+({\overline {A}}\cdot B)\equiv (A+B)\cdot ({\overline {A}}+{\overline {B}})}$$
and I can verify this, but I'm not quite sure how to basically take the left-hand side and transform it into the right-hand side. Any help would be appreciated, thank you.
| *
*As right hand can be transformed in way:
$$(A+B)\cdot ({\overline {A}}+{\overline {B}}) \equiv \\
A \cdot \overline {A}+A \cdot\overline {B} + B \cdot \overline {A} + B \cdot \overline {B } \equiv \\
A\cdot {\overline {B}}+{\overline {A}}\cdot B$$
and knowing, that equivalence works in both direction, you can now write this equivalences in opposite direction to reach desired.
*For second way let's use $\overline{A\cdot B}=\overline{A} + \overline {B}$ and $\overline{A+ B}=\overline{A} \cdot \overline {B}$:
$$A\cdot {\overline {B}}+{\overline {A}}\cdot B \equiv \overline { \overline{A}+B}+\overline {A+\overline {B}} \equiv \\
\overline {(\overline{A}+B) \cdot (A+\overline {B})} \equiv \overline {A\cdot B + \overline{A} \cdot \overline {B}} \equiv \overline{A\cdot B} \cdot \overline{\overline{A} \cdot \overline {B}} \equiv \\
(A+B)\cdot ({\overline {A}}+{\overline {B}})
$$
| {
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Solve $\frac{x^3-4x^2-4x+16}{\sqrt{x^2-5x+4}}=0$ Solve $$\dfrac{x^3-4x^2-4x+16}{\sqrt{x^2-5x+4}}=0.$$
We have $D_x:\begin{cases}x^2-5x+4\ge0\\x^2-5x+4\ne0\end{cases}\iff x^2-5x+4>0\iff x\in(-\infty;1)\cup(4;+\infty).$ Now I am trying to solve the equation $x^3-4x^2-4x+16=0.$ I have not studied how to solve cubic equations. Thank you in advance!
| $$\frac{x^3-4x^2-4x+16}{\sqrt{x^2-5x+4}}=\frac{(x-4)(x-2)(x+2)}{\sqrt{(x-4)(x-1)}}==\frac{\sqrt{x-4}(x-2)(x+2)}{\sqrt{x-1}}$$
The fraction will be $=0$ if the numerator is $0$ and the denominator is not $0$. That is true for $x=2,-2,4$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proof $f(x)>x$ for $x>0$ I'm a little stuck at proving that for these function:
$$\ f(x) = \frac{x}{\sqrt{1+x^2}} + \sqrt{1+x^2}\cdot\ln(1+x^2) $$
$f(x)>x$ for every $x>0$. Another question: is there any $a$ that $f(x)>x^2$ for every $x>a$?
My steps
I have computed the derivative:
$$\ f'(x) = \frac{1}{\sqrt{1+x^2}} - x^2 + \frac{\sqrt{1+x^2}}{1+x^2} + x\ln(1+x^2) $$
then compared to zero:
$$\ f'(x) = \frac{1}{\sqrt{1+x^2}} - x^2 + \frac{\sqrt{1+x^2}}{1+x^2} + x\ln(1+x^2) > 0 $$
and calculated that:
$$ 4>x^2(1+x^2)(x-\ln(1+x^2))^2 $$
Each of the components is $>0$ I can see from the primary function that the root is 0, but I cannot write it in more precise way than the equation above. I am thinking of Mean Value Theroem, but I cannot see, how to apply it in this case. I was considering something like:
$$ f'(c) = \frac{f(x) - f(0)}{x} = \frac{f(x)}{x} > 0 $$
Since the function gives positive values for $x>0$, but it has leaded me to nowhere. I would really appreciate your help.
EDIT
As Andrei suggested, I have corrected the derivative:
$$ \frac{x\ln(x^2+1)}{\sqrt{x^2+1}} + \frac{2x}{\sqrt{x^2+1}} + \frac{1}{(x^2+1)^{3/2}} $$
Then calculated:
$$ 1>(x^2+1)(\sqrt{x^2+1}-(2x+x\ln(x^2+1)) $$
Since
$$ \sqrt{x^2+1}<(2x+x\ln(x^2+1)/^2 \\
x^2+1<4x^2+4x^2\ln(x^2+1)+x^2\ln(x^2+1)^2 $$
The positive and negative multiplication is smaller than 1, thus the inequality is true.
However, what about $f(x)>x^2$? Do I have to calculate the second derivative?
I have continued Andrei method and for $f'(x)-2x>0$ obtained:
$$ 1>(x^2+1)(2x\sqrt{x^2+1}-(2x+x\ln(x^2+1)) $$
$$ 2x\sqrt{x^2+1}<(2x+x\ln(x^2+1)/^2 $$
And reduced to:
$$ x^4<\frac{3}{2}\ln(x^2+1) /^e(\cdot) \\
e^{x^4} < (x^2+1)\sqrt{x^2+1} $$
The order of growth indicates that this is false (only even power of x) and there is no such a that fulfil these requirements.
Is it correct?
| We need to prove that: $$\sqrt{\frac{x}{x+1}}+\sqrt{1+x}\ln(1+x)>\sqrt{x},$$ where $x>0$, or $g(x)>0,$ where
$$g(x)=\ln(1+x)-\frac{\sqrt{x}(\sqrt{x+1}-1)}{1+x}$$ and easy to show that $g'(x)>0,$ which gives $g(x)>g(0)=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3859536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate the sum of series of arctan and find limit I try to find the value of following limit:
$$
\lim _{n \rightarrow \infty} 2^{-n/2} \sum_{j=2^{n}+1}^{2^{n+1}} \tan ^{-1}\left(\frac{1}{\sqrt{j-1}}\right)
$$
But I cannot evaluate the summation. Can anybody help?
| METHODOLOGY $1$: Simple Bounds and Using the Squeeze Theorem
Since the arctangent is monotonically increasing, $\arctan(1/\sqrt{x-1})$ is monotonically decreasing. Hence we can assert that
$$\int_{N}^{M+1}\arctan\left(\frac1{\sqrt{x-1}}\right)\,dx\le \sum_{j=N}^M\arctan\left(\frac1{\sqrt{j-1}}\right)\le \int_{N-1}^{M}\arctan\left(\frac1{\sqrt{x-1}}\right)\,dx\tag1$$
The antiderivative of $\arctan\left(\frac1{\sqrt{x-1}}\right)$ can be written as
$$\int \arctan\left(\frac1{\sqrt{x-1}}\right)\,dx=\sqrt{x-1}+x\arctan\left(\frac1{\sqrt{x-1}}\right)+C\tag2$$
For $N=2^n+1$ and $M=2^{n+1}$ in $(2)$, we find that
$$\begin{align}
\int_{2^{n}+1}^{2^{n+1}+1} \arctan\left(\frac1{\sqrt{x-1}}\right)\,dx&=2^{n/2}(\sqrt 2-1)\\\\
&+(2^{n+1}+1)\arctan\left(\frac1{\sqrt{2}2^{n/2}}\right)\\\\
&-(2^n+1)\arctan\left(\frac1{2^{n/2}}\right)\tag3
\end{align}$$
Dividing $(3)$ by $2^{n/2}$ and letting $n\to\infty$ reveals that
$$\liminf_{n\to\infty}2^{-n/2}\sum_{j=2^n+1}^{2^{n+1}}\arctan\left(\frac1{\sqrt{j-1}}\right)\ge 2(\sqrt 2-1)\tag4$$
Similarly, evaluation of the right-hand side of $(1)$ reveals
$$\limsup_{n\to\infty}2^{-n/2}\sum_{j=2^n+1}^{2^{n+1}}\arctan\left(\frac1{\sqrt{j-1}}\right)\le 2(\sqrt 2-1)\tag5$$
Finally, putting $(4)$ and $(5)$ together yields the coveted limit
$$\lim_{n\to\infty} 2^{-n/2}\sum_{j=2^n+1}^{2^{n+1}}\arctan\left(\frac1{\sqrt{j-1}}\right)=2(\sqrt 2-1)$$
METHODOLOGY $2$: Use on the Euler-Maclaurin Summation Formula
From the Euler-Maclaurin Summation formula we have
$$\begin{align}
2^{-n/2}\sum_{j=2^n+1}^{2^{n+1}}\arctan\left(\frac1{\sqrt{j-1}}\right)&=2^{-n/2}\int_{2^n+1}^{2^{n+1}}\arctan\left(\frac1{\sqrt{x-1}}\right)\,dx+o(1)\\\\
&=2^{-n/2}\left(\sqrt{2^{n+1}-1}-\sqrt{2^n} \right)\\\\
&+2^{n/2+1}\arctan\left(\frac1{\sqrt{2^{n+1}-1}}\right)\\\\
&-(2^{n/2}+1)\arctan\left(\frac1{\sqrt{2^{n}}}\right)+o(1)\tag6
\end{align}$$
Letting $n\to \infty$ in $(6)$, we find that
$$\lim_{n\to\infty} 2^{-n/2}\sum_{j=2^n+1}^{2^{n+1}}\arctan\left(\frac1{\sqrt{j-1}}\right)=2(\sqrt 2-1)$$
| {
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"source": "stackexchange",
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The given matrix has three linearly independent eigenvectors, then $x+y=0$. The question asked is ::
If the matrix
$$
A=\left(\begin{array}{lll}
0 & 0 & 1 \\
x & 1 & y \\
1 & 0 & 0
\end{array}\right)
$$
has three linearly independent eigenvectors, then show that $x+y=0$.
solving for eigenvalues from the characteristic polynomial:
$$\left|\begin{matrix}
0-\lambda & 0 & 1 \\
x & 1-\lambda & y \\
1 & 0 & 0-\lambda
\end{matrix}\right| =-λ^3+λ^2+λ-1$$
$$=-(λ-1)*(λ^2-1)=-(λ-1)*(λ-1)=-(λ-1)^2*(λ+1)$$
So eigenvalues are $λ_1=1$ and $λ_2=-1$, Independent of the values of $x$ and $y$.
Now solving for eigenvectors I got
$\left(\begin{matrix}
0 \\
1 \\
0
\end{matrix}\right)$ and $\left(\begin{matrix}
-1 \\
\frac{x-y}{2} \\
1
\end{matrix}\right)$
From here how to show that if there are three linearly independent eigenvectors, then show that $x+y=0$.
| The condition is equivalent to the condition that the eigenspace $E_1$ has dimension $2$, which in turn is equivalent to
$\:\dim(\ker(A-I))=2$. Now
$$A-I=\begin{pmatrix}-1&0&1\\x&0&y\\1&0&-1\end{pmatrix},$$
and the kernel has dimension $2$ if and only if this matrix has rank $1$, which means columns 2 and 3 are collinear. This yields $x=-y$, or equivalently,$x+y=0$.
| {
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"url": "https://math.stackexchange.com/questions/3861757",
"timestamp": "2023-03-29T00:00:00",
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Prove that $\sum\limits_{j=0}^k\,j\,\binom{n}{j}\,\binom{n-j}{2k-2j}\,2^{2k-2j}=n\binom{2n-2}{2k-2}$ I encountered this in my homework. I derived two ways to solve the problem and the answer which I have tested using programming, seem to be the same, but I am not sure how to prove this equation.
Let $n$ and $k$ be nonnegative integers with $k\leq n$. Prove that $$\sum\limits_{j=0}^k\,j\,\binom{n}{j}\,\binom{n-j}{2k-2j}\,2^{2k-2j}=n\binom{2n-2}{2k-2}\,.$$
The original problem is the following:
A shoe rack has n pairs of shoes. Of those, 2k individual shoes are chosen at random, k ≤ n. Calculate the expected number of matching shoes among 2k chosen shoes.
The left hand side is from directly calculating expectation, while the right hand side is using sum of indicator variables of each pair being chosen. The expectation is just the equation divided by $\binom{2n}{2k}$.
| Let's take the RHS term, multiply it by $x^{2k}$, and sum over $k$
$$
\eqalign{
& F_R (x^{\,2} ,n) = \sum\limits_{0\, \le \,k}
{n\left( \matrix{ 2n - 2 \cr 2k - 2 \cr} \right)x^{\;2k} } = \cr
& = n\,x^{\,2} \,\sum\limits_{0\, \le \,\left( {1\, \le } \right)\,k}
{\left( \matrix{ 2\left( {n - 1} \right) \cr 2\left( {k - 1} \right) \cr} \right)
\left( {x^{\,2} } \right)^{\;k - 1} } = \cr
& = n\,x^{\,2} \,\sum\limits_{0\, \le \,\left( {1\, \le } \right)\,k}
{\left( \matrix{ 2\left( {n - 1} \right) \cr 2k \cr} \right)\left( {x^{\,2} } \right)^{\;k} } = \cr
& = {{n\,x^{\,2} } \over 2}\left( {\left( {1 + x} \right)^{2n - 2} + \left( {1 - x} \right)^{2n - 2} } \right) \cr}
$$
Then we do the same on the LHS
$$
\eqalign{
& F_L (x^{\,2} ,n) = \sum\limits_{0\, \le \,k} {\sum\limits_{0\, \le \,j}
{j\left( \matrix{ n \cr j \cr} \right)
\left( \matrix{ n - j \cr 2k - 2j \cr} \right)2^{\,2k - 2j} x^{\;2k} } } = \cr
& = n\sum\limits_{0\, \le \,k} {\sum\limits_{0\, \le \,j}
{\left( \matrix{ n - 1 \cr j - 1 \cr} \right)\left( \matrix{ n - j \cr 2\left( {k - j} \right) \cr} \right)
\left( {2^{\,2} x^{\,2} } \right)^{\,\left( {k - j} \right)} x^{\;2j} } } = \cr
& = n\sum\limits_{0\, \le \,j}
{\left( \matrix{ n - 1 \cr j - 1 \cr} \right)x^{\;2j} \sum\limits_{0\, \le \,\left( {k - j} \right)}
{\left( \matrix{ n - j \cr 2\left( {k - j} \right) \cr} \right)
\left( {2^{\,2} x^{\,2} } \right)^{\,\left( {k - j} \right)} } } = \cr
& = {n \over 2}\sum\limits_{0\, \le \,j}
{\left( \matrix{ n - 1 \cr j - 1 \cr} \right)x^{\;2j}
\left( {\left( {1 + 2x} \right)^{n - j} + \left( {1 - 2x} \right)^{n - j} } \right)} = \cr
& = {{nx^{\;2n} } \over 2}\sum\limits_{0\, \le \,j}
{\left( \matrix{ n - 1 \cr n - j \cr} \right)x^{\;2j - 2n}
\left( {\left( {1 + 2x} \right)^{n - j} + \left( {1 - 2x} \right)^{n - j} } \right)} = \cr
& = {{nx^{\;2n} } \over 2}\sum\limits_{0\, \le \,n - j}
{\left( \matrix{ n - 1 \cr n - j \cr} \right)
\left( {\left( {{{1 + 2x} \over {x^{\;2} }}} \right)^{n - j}
+ \left( {{{1 - 2x} \over {x^{\;2} }}} \right)^{n - j} } \right)} = \cr
& = {{nx^{\;2n} } \over 2}\left( {\left( {1 + {{1 + 2x} \over {x^{\;2} }}} \right)^{n - 1}
+ \left( {1 + {{1 - 2x} \over {x^{\;2} }}} \right)^{n - 1} } \right) = \cr
& = {{nx^{\;2} } \over 2}\left( {\left( {1 + x} \right)^{\,2\left( {n - 1} \right)}
+ \left( {1 - x} \right)^{\,2\left( {n - 1} \right)} } \right) \cr}
$$
The two polynomials are equal, so must be their coefficients.
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 2
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How to compute the limit as $x\to 3$ of a $\textit{complicated}$ product and quotient of trigonometric functions $$\lim_{x\rightarrow 3}\frac{ \tan\frac{x-3}{x+3}\sin(9\sin(x-3)) }{ \sin(x-3)\sin(x^3-27))}$$
I substituted $x-3$ for $u$ and got as far as
$$\frac{1}{6} \lim_ {u\to 0} \frac{\sin(9 \sin u)}{\sin((u+3)^(3) -27)}.$$
This is where I get stuck. Should I try a different approach?
| We multiply top and bottom by $9$ to get a $\frac{\sin x}x$ part:
$$\lim_{x\to3}\frac{9\tan\frac{x-3}{x+3}\sin(9\sin(x-3)) }{ 9\sin(x-3)\sin(x^3-27)}=\lim_{x\to3}\frac{9\tan\frac{x-3}{x+3}}{\sin(x^3-27)}$$
Now use small-angle approximations:
$$=\lim_{x\to3}\frac{9\frac{x-3}{x+3}}{x^3-27}=\lim_{x\to3}\frac{9/(x+3)}{x^2+3x+9}=\frac9{6(9+9+9)}=\frac1{18}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3865740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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} |
Proving $\lim_{x \to 1}\frac{x+1}{x-2} + x = -1$ using definition I got a question regarding my answer of proving limit using epsilon-delta, here's the question
Prove $\lim_{x \to 1}\frac{x+1}{x-2} + x = -1$
Here's the answer I've come up so far
let $f(x) = \frac{x+1}{x-2} + x$
by algebra manipulation, we get
$|f(x) - (-1)| = |\frac{x+1}{x-2} + x +1| $ $=|\frac{x^2 - 1}{x-2}|$ $=|\frac{(x-1)(x+1)}{x-2}|$
let $|x-1| < 1$, by triangle inequality we get $|x| < 2$, then
$|x + 1| < 3$ and $|x - 2| < 1$
now, using the definiton of limit,
for every $\epsilon > 0$, there exist $\delta = min\{1, \frac{\epsilon}{3}\}$ such that
if $0 < |x - 1| < \delta$ then,
$|f(x) - (-1)| = |\frac{x+1}{x-2} + x +1|$ $=|\frac{(x-1)(x+1)}{x-2}|$ $=|\frac{1 \cdot 3}{1}|$ $< \epsilon$
Is this correct? honestly I'm not sure on getting the upper bound of $|x-2|$, so I used the assumption of $|x-1| < 1$
Any tips would help, thanks beforehand.
| Note that by definition, let $f(x):=\frac{x+1}{x-2}+x$ so, we need to prove that
\begin{eqnarray}
\boxed{\lim_{x\to 1} f(x)=-1 \iff \forall \epsilon>0, \exists \delta>0; \forall x: 0<|x-1|<\delta \implies |f(x)-(-1)|<\epsilon}
\end{eqnarray}
Now, we need to analyse (this part is not the formal prove) the $(*)$ here:
\begin{eqnarray}
\lim_{x\to 1} f(x)=-1 \iff \forall \epsilon>0, \exists \delta>0; \forall x: 0<|x-1|<\delta \implies \underbrace{|f(x)-(-1)|<\epsilon}_{*}
\end{eqnarray}
Note that
\begin{eqnarray}
|f(x)-(-1)|&=&\left|\frac{x+1}{x-2}+x-(-1) \right|\\
&=&\left|\frac{x+1}{x-2}+x+1 \right|\\
&=& \left|\frac{x+1}{x-2}+(x+1)\left(\frac{x-2}{x-2}\right) \right|\\
&=& \left|\frac{x+1+(x+1)(x-2)}{x-2} \right|\\
&=&\left|\frac{x+1+x^{2}-2x+x-2}{x-2} \right|\\
&=&\left|\frac{x^{2}-1}{x-2} \right|\\
&=&\left|\frac{(x-1)(x+1)}{x-2} \right|\\
&=&\left|x-1\right|\cdot\left|\frac{x+1}{x-2}\right|
\end{eqnarray}
Now, we want that:
\begin{eqnarray}
\text{si} \quad 0<|x-1|<\delta \implies \left|x-1\right|\cdot \left| \frac{x+1}{x-2}\right|<\epsilon
\end{eqnarray}
Now, note that if we can find a constant $C$ such that $\left|\frac{x+1}{x-2}\right|<C$, so we have $$\left|x-1\right|\cdot \left|\frac{x+1}{x-2}\right|<C|x-1|$$
and we can do $C|x-1|<\epsilon$ taking $|x-1|<\epsilon/C=\delta$
We can find a number $C$ such that we restrict $x$ to some interval centered on $1$.
In fact, we are interested only in values close to $1$, so it's reasonable to assume that $x$ is within a distance $1$ of $1$, that is $|x-1|<1$. So, we can see that $$|x+1|=|x+1-1+1|\leq |x-1|+|2|=1+2=3 \implies |x+1|\leq 3$$
and using reverse triangle inequality and since $|x-1|>0$, we have $$|x-2|=|(x-1)-1|\geq ||x-1|-|1||>|0-1|=|-1|=1 \implies |x-2|>1$$
Note that $$|f(x)-(-1)|=\left|x-1\right|\left| \frac{x+1}{x-2}\right|<\delta\cdot \frac{3}{1}<\epsilon$$
To ensure that all inequalities are satisfied, we take $ \delta $ as the smallest of the numbers $1$ and $\epsilon/3$. The notation for this is $\delta = \min \{1, \epsilon/ 3\}$
Can you continue from here?
You need to write the formal prove.
| {
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"url": "https://math.stackexchange.com/questions/3867689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Limit of the sequence using intermediate value theorem $\lim_{x\to\infty}[\frac{1}{\sin(\frac{1}{x+\psi(x)})}-\frac{1}{\sin(\frac{1}{x+\phi(x)})}]$ Find the limit of:
$$\lim_{x\to\infty}[\frac{1}{\sin(\frac{1}{x+\psi(x)})}-\frac{1}{\sin(\frac{1}{x+\phi(x)})}]$$
where:
$$\psi(x)=(1+\frac{1}{x})^x, ~~~~~~~\phi(x)=\sqrt[x]{x}$$
I used Lagrange theorem for the intermediate value for $f(x)=\frac{1}{\sin(x)}$, which is a recommended way of solving the problem, but I am stuck now and I would ask you for some help.
$$\frac{f(b)-f(a)}{b-a}=f'(c),~~~~c\in(a,b)\\ f(b)-f(a)=f'(c)\cdot(b-a) \\ \lim_{x\to\infty}\frac{\frac{1}{\sin(\frac{1}{x+\psi(x)})}-\frac{1}{\sin(\frac{1}{x+\phi(x)})}}{x+\psi(x)-x-\phi(x)}=\lim_{x\to\infty}\frac{\cos(\frac{1}{c})}{\sin^2(\frac{1}{c})\cdot x^2}\cdot (\psi(x)-\phi(x))$$
That's the end of my story. I have also tried $\frac{1-\cos(2\alpha)}{2}=\sin^2(\alpha)$, but it leaded me to nowhere. I would really appreciate your hints.
| Assuming that $c\approx x$, I have used Taylor expansion for the:
$$ \frac{\cos(\frac{1}{c})}{\sin^2(\frac{1}{c}) \cdot x^2} =\\$$
$$\frac{1-\frac{1}{2x^2}+\frac{1}{4! x^4} + o(\frac{1}{x^4})}{(\frac{1}{x^2}-\frac{1}{3 x^4} + \frac{1}{36 x^6}+o(\frac{1}{x^6}))\cdot x^2} =\\$$
$$\frac{1-\frac{1}{2x^2}+\frac{1}{4! x^4} +o(\frac{1}{x^4})}{1-\frac{1}{3 x^2} + \frac{1}{36 x^4}+o(\frac{1}{x^4})}$$
After division by $\frac{1}{x^4}$ and taking out $x^4$, the limit is:
$$ \frac{x^4}{x^4}\cdot\frac{(1-\frac{1}{2x^2}+\frac{1}{4!x^4})}{(1-\frac{1}{3x^2}+\frac{1}{36x^4})} = \frac{1-\frac{1}{2x^2}+\frac{1}{4!x^4}}{1-\frac{1}{3x^2}+\frac{1}{36x^4}}$$
for $x\to\infty$ it tends to $1$, thus:
$$\lim_{x\to\infty}\frac{\cos(\frac{1}{c})}{\sin^2(\frac{1}{c}) \cdot x^2}(\psi(x)-\phi(x))=e-1$$
since $(1+\frac{1}{x})^x\to e$ and $\sqrt[x]{x}\to1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3871223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Minimum value of magnitude of complex expression
If $|2z_1 + \bar z_2| = 2\sqrt2$ and $|1 + 2z_1z_2 | = 3 $ then minimum value of $(|z_1|^2 + 4|z_2|^2)$, is:-
Now
$|2z_1 + \bar z_2|^2 = 8$
$(2 z_1 + \bar z_2)(2\bar z_1 + z_2) = 8$
$4z_1\bar z_1 + 2 z_1z_2 + 2\bar z_1\bar z_2 + z_2\bar z_2 = 8$
$4|z_1|^2 + |z_2|^2 + 2 z_1z_2 + 2\bar z_1\bar z_2 = 8.......(1)$
Now
$|1 + 2z_1z_2 |^2 = 9 $
$(1 + 2z_1z_2)(1 + 2\bar z_1\bar z_2) = 9 $
$1 + 4|z_1|^2|z_2|^2 + 2z_1z_2 + 2\bar z_1\bar z_2= 9........(2)$
$(1) - (2)$
$4|z_1|^2 + |z_2|^2 = 4|z_1|^2|z_2|^2$
I am not able to make any progress after this step.
| $$4|z_1|^2 + |z_2|^2 = 4|z_1|^2|z_2|^2$$
$$\dfrac{1}{|z_1|^2} + \dfrac{4}{|z_2|^2} = 4$$
By Cauchy-Schwarz,
$$\Bigl(\dfrac{1}{|z_1|^2} + \dfrac{4}{|z_2|^2} \Bigr)\Bigl(|z_1|^2 + 4|z_2|^2\Bigr) \ge (1 + 2\cdot2)^2 = 25 $$
$$|z_1|^2 + 4|z_2|^2 \ge \dfrac{25}{4} $$
with equality for
$$\dfrac{1}{|z_1|}\cdot2|z_2| - 2\dfrac{1}{|z_2|}\cdot|z_1|=0$$
$$\Rightarrow |z_1|=|z_2|=\dfrac{\sqrt5}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3872067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solving a differential equation using an integrating factor
I'm trying the solve the following equation: $ \left\{\begin{matrix}
(x^B+y^B)(xdy-ydx)=(1+x)x^9dx \\ y(-1)=A \end{matrix}\right. $
for $A=1$ and $A=0$.
$B\in2\mathbb{N}_0+1$
My solution is the following, but I got stuck:
$(x^B+y^B)(xdy-ydx)=(1+x)*x^9dx$
$x^B*xdy-y*x^Bdx+y^B*xdy-y*y^Bdx-(1+x)*x^9dx=0$
$x^B*xdy+y^B*xdy-y*x^Bdx-y*y^Bdx-(1+x)*x^9dx=0$
$xdy(x^B+y^B)+[-y(x^B+y^B)-(1+x)*x^9]dx=0$
where:
$Q(x,y)=x(x^B+y^B)dy$
$P(x,y)=[-y(x^B+y^B)-(1+x)*x^9]$
Then I made partial derivation:
$\frac{∂Q}{∂x}:(x^B+y^B)x(Bx^{B-1})$
$\frac{∂P}{∂y}: -(x^B+y^B)-y(By^{B-1})$
And then subtract:
$\frac{∂P}{∂y}-\frac{∂Q}{∂x}=-(x^B+y^B)-y(By^{B-1})-(x^B+y^B)-x(Bx^{B-1})$
But probably I did something wrong and
I'm stuck and not sure where made I mistake.. Can you help me please?
| $(x^B+y^B)(xdy-ydx)=(1+x)x^9dx$
$xdy-ydx = \frac{(1+x)x^9}{(x^B+y^B)}dx$
$x \frac{dy}{dx} - y = \frac{(1+x)x^9}{(x^B+y^B)}$
$\displaystyle \frac{dy}{dx} - \frac{y}{x} = \frac{(1+x)x^8}{(x^B+y^B)}$
Based on the form of LHS, take integrating factor of $ \displaystyle \frac{1}{x}$.
$\displaystyle \frac {1}{x} \frac{dy}{dx}- \frac{y}{x^2} = \frac{(1+x)x^7}{(x^B+y^B)}$
$\displaystyle \frac{d}{dx} (\frac{y}{x}) = \frac{(1+x)x^{(7-B)}}{1+ (\frac{y}{x})^B}$
Substitute $u = \frac{y}{x}$
$\displaystyle ({1+ u^B}) du = ((1+x)x^{(7-B)}) dx$
Can you please take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3879739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluate the limit $\lim_{x\rightarrow \infty}\sqrt[]{n^3}(\sqrt{n+1}+\sqrt{n+2}-\sqrt{n}-\sqrt{n+3})$ Evaluate the limit:
$$\lim_{n\rightarrow \infty}\sqrt[]{n^3}(\sqrt{n+1}+\sqrt{n+2}-\sqrt{n}-\sqrt{n+3})$$
Using the fact that ${(1 + x)^{1/2} \approx 1 + x/2}$ for "small" x,
I have that $\sqrt{n+1}\approx\sqrt{n}(\frac{1}{2n}+1)$ then $n\rightarrow \infty$. However, following this procedure I end up with the following limit: $\lim_{n\rightarrow \infty}2n^2=\infty$, but the answer is $\frac{1}{2}$. I would be thankful for any help.
| Another way to multiply by conjugates:
$$\begin{align}
\sqrt{n+1}+\sqrt{n+2}-\sqrt n-\sqrt{n+3}&={(n+1+2\sqrt{(n+1)(n+2)}+n+2)-(n+2\sqrt{n(n+3)}+n+3)\over\sqrt{n+1}+\sqrt{n+2}+\sqrt n+\sqrt{n+3}}\\
&={2(\sqrt{n^2+3n+2}-\sqrt{n^2+3n})\over\sqrt{n+1}+\sqrt{n+2}+\sqrt n+\sqrt{n+3}}\\
&={4\over(\sqrt{n+1}+\sqrt{n+2}+\sqrt n+\sqrt{n+3})(\sqrt{n^2+3n+2}+\sqrt{n^2+3n})}\end{align}$$
so
$$\begin{align}\sqrt{n^3}(\sqrt{n+1}+\sqrt{n+2}-\sqrt n-\sqrt{n+3})
&={4\over\displaystyle\left({\sqrt{n+1}+\sqrt{n+2}+\sqrt n+\sqrt{n+3}\over\sqrt n}\right)\left({\sqrt{n^2+3n+2}+\sqrt{n^2+3n})\over n}\right)}\\
&\to{4\over(1+1+1+1)(1+1)}={1\over2}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3881097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Application of Triple Integrals Any easy way to solve this integration?
$\int_{0}^{3} \int_{0}^{\frac{6-2x}{3}} \int_{0}^{\frac{6-3y-2x}{2}} xyz \ dzdydx$
My work
$\int \int \int_D xyz \ dV = \dfrac{1}{2}\int_{0}^{3} \int_{0}^{\frac{6-2x}{3}} xy(z^2 )\biggr|_{0}^{\frac{6-3y-2x}{2}} \ dy \ dx$
$\int \int \int_D xyz \ dV = \dfrac{1}{8}\int_{0}^{3} \int_{0}^{\dfrac{6-2x}{3}} xy(6-3y-2x)^2 \ dy \ dx$
$\int \int \int_D xyz \ dV = \dfrac{1}{8}\int_{0}^{3} x \left [y\dfrac{(6-3y-2x)^3}{3(-3)} -\dfrac{(6-3y-2x)^4}{4(-3)3(-3)} \right ]\biggr|_{0}^{\dfrac{6-2x}{3}} \ dx$
$\int \int \int_D xyz \ dV = \dfrac{1}{8}\int_{0}^{3} x \left [y\dfrac{(6-3y-2x)^3}{3(-3)} -\dfrac{(6-3y-2x)^4}{4(-3)3(-3)} \right ]\biggr|_{0}^{\dfrac{6-2x}{3}} \ dx$
The terms are expanding, any way to find a solution in easy case. please help
| The domain of integration is $D$, the set of $(x,y,z)$ with $x,y,z\ge 0$ and $2x+3y+2z\le 6$. Well, to have a simpler, symmetric formula, we can change variables. For instance $X=2x/6$, $Y=3y/6$, $Z=2z/6$. Then the domain $D$ is transformed into the simplex (cone) $E$ of all $(X,Y,Z)$ with $X,Y,Z\ge 0$ and $X+Y+Z\le 1$. Formally, $x=3X$, $y=2Y$, $z=3Z$, so $dx\; dy\; dz=3\cdot 2\cdot 3\; dX\; dY\; dZ$. We can compute in a few easy lines, explicitly, the integral $J$ (notation) from the OP:
$$
\begin{aligned}
J &=\iiint_D xyz\; dx\; dy\; dz\\
&=\iiint_E (3X)(2Y)(3Z)\cdot (3\cdot 2\cdot 3)\; dX\; dY\; dZ\\
&=324\iiint_{\substack{0\le X,Y,Z\\X+Y+Z\le 1}}XYZ\; dX\; dY\; dZ\\
&=324\int_0^1X\;dX\int_0^{1-X}Y\; dY\int_0^{1-X-Y}Z\; dZ\\
&=324\int_0^1X\;dX\int_0^{1-X}Y\; dY\cdot\left[\frac 12Z^2\right]_0^{1-X-Y}\\
&=324\cdot \frac 12\int_0^1X\;dX\int_0^{1-X}Y(1-X-Y)^2\; dY\\
&=324\cdot \frac 12\int_0^1X\;dX\int_0^{1-X}\Big(\ (1-X)(1-X-Y)^2-(1-X-Y)^3\ \Big)\; dY\\
&=324\cdot \frac 12\int_0^1X\;dX\left[
-\frac 13(1-X)(1-X-Y)^3+\frac 14(1-X-Y)^4\right]_0^{1-X}
\\
&=324\cdot \frac 12\int_0^1X\;dX\left(
\frac 13(1-X)^4-\frac 14(1-X)^4\right)
\\
&=324\cdot \frac 12\int_0^1X\cdot\frac 13\cdot \frac 14(1-X)^4\;dX
\\
&=324\cdot \frac 12\cdot\frac 13\cdot \frac 14\int_0^1(1-(1-X))\cdot(1-X)^4\;dX
\\
&=324\cdot \frac 12\cdot\frac 13\cdot \frac 14\int_0^1\Big((1-X)^4-(1-X)^5\Big)\;dX
\\
&=324\cdot \frac 12\cdot\frac 13\cdot \frac 14\left(\frac 15-\frac 16\right)\;dX
\\
&=324\cdot \frac 12\cdot\frac 13\cdot \frac 14\cdot \frac 15\cdot \frac 16\\
&=\frac{18^2}{6!}=\frac 9{20}\ .
\end{aligned}
$$
$\square$
Note: I saw no quick way to proceed using Stokes...
Homework: Try to find a generalization. What is for instance the integral of $XYZW$ w.r.t. volume $dX\; dY\; dZ\;dW$ on the simplex $0\le X,Y,Z,W\le X+Y+Z+W\le 1$?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding all integer solutions of $\frac{1}{x} + \frac{1}{y} - \frac{1}{z} = \frac12$ If we have integers $x,y,z$ such that $x,y,z \ge 3,$ find all solutions to $$\frac{1}{x} + \frac{1}{y} - \frac{1}{z} = \frac12.$$
I was thinking of first expanding out this, and then simplifying from there, but the equation got very messy. Is there a better method?
| WLOG let $x\leq y$. Then, we have that $x = 3$ because we are given that $x\geq 3$, and if $x\geq 4$, $\frac{1}{x}+\frac{1}{y}\leq \frac{1}{2}$.
Then, we must solve $\frac{1}{y}-\frac{1}{z} = \frac{1}{6}$. Note that $y = 3,4,5$ because they are the only values such that $y\geq 3$ and $\frac{1}{y}>\frac{1}{6}$. Then, we can go into casework:
For $y = 3$, we get $\frac{1}{z} = \frac{1}{6}$ and thus $z = 6$.
For $y = 4$, we get $\frac{1}{z} = \frac{1}{12}$ and thus $z = 12$.
Finally, for $y = 5$ we get $\frac{1}{z} =\frac{1}{30}$ and thus $z = 30$.
We must permute $x$ and $y$ when they are different to get all of the solutions because of the earlier WLOG.
Thus, the only solutions are $\boxed{(3,3,6),\ (3,4,12),\ (3,5,30),\ (4,3,12),\text{ and }(5,3,30).}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3885152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $\frac{1-\cos2x+\sin2x}{1+\cos2x+\sin2x} = \tan x$ Show that $$\frac{1-\cos2x+\sin2x}{1+\cos2x+\sin2x} = \tan x$$
I have substituted the expansions for $\cos2x$ and $\sin2x$ and gotten, after simplification:
$$\frac{1-\sin x\cos x + 2\sin^2x}{1+\sin x\cos x-2\sin^2x}$$
I'm not sure how to carry on. I factored out the $\sin x$, but ended up with
$$\frac{1+\sin x}{1-\sin x}$$
I haven't been taught that as equal to $tan x$.
| $$\frac{1-\cos2x+\sin2x}{1+\cos2x+\sin2x} = \frac{1- (2\cos^2x-1)+2\sin x .\cos x}{1+ 2\cos^2x-1+2\sin x .\cos x}$$
$$ = \frac{2- 2\cos^2 x +2\sin x .\cos x}{ 2\cos^2 x +2\sin x .\cos x} = \frac{1- \cos^2 x +\sin x .\cos x}{ \cos^2 x + \sin x .\cos x}$$
$$ = \frac{\sin^2 x +\sin x .\cos x}{ \cos^2 x + \sin x .\cos x} = \frac{\sin x (\sin x + \cos x )}{ \cos x (\sin x + \cos x )} = \frac{\sin x }{ \cos x } = \tan x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3885812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Find z s.t. $z+\frac{1}{z}$ is real. Is my solution good? I must find all z s.t. $z+\frac{1}{z}$ is real. I know that $z = a + bi$ is real when the Imaginary part is 0. So, there we go:
$$z+\frac{1}{z}= \frac{z^2+1}{z}=\frac{(a+bi)^2+1}{a+bi}=\frac{[(a+bi)^2+1](a-bi)}{(a+bi)(a-bi)}$$
$$\frac{[(a+bi)^2+1](a-bi)}{(a+bi)(a-bi)} = \frac{(a^2 +2abi-b^2+1)(a-bi)}{(a^2+b^2)}$$
$$\frac{(a^2 +2abi-b^2+1)(a-bi)}{(a^2+b^2)}=\frac{a^3 +2a^2bi-b^2a+a-ba^2i+2ab^2+b^3i-bi}{(a^2+b^2)}$$
$$2a^2b-ba^2+b^3-b=0$$
$$a^2b+b^3-b=0$$
$$b_1=0$$
$$a^2+b^2-1=0$$
$$b^2=1-a^2$$
$$b_2=\sqrt{1-a^2}$$
$$b_3=-\sqrt{1-a^2}$$
so, the solutions are:
$$z_1=a+0i$$
$$z_1=a+\sqrt{1-a^2}i$$
$$z_1=a-\sqrt{1-a^2}i$$
I am not sure if that should be done like I did it.
| Your solution is correct if you restrict $a \ne 0$ in the case
$$
z = a + 0i
$$
and $|a| \le 1$ in the case
$$
z = a \pm i \sqrt{1-a^2} \, .
$$
So for nonzero $z$, $z + 1/z$ is real if $z$ is real or $|z| = 1$.
As alternative solution you can use that a complex number is real if and only if it equal to its complex conjugate:
$$
z + \frac 1 z \in \Bbb R \\
\iff z + \frac 1 z = \overline z + \frac{1}{\overline z} \\
\iff z^2 \overline z + \overline z = z \overline z^2 + z \\
\iff (z - \overline z)(|z|^2 - 1) = 0 \, .
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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"answer_id": 1
} |
Find all integer polynomials $f(p)$ that divide $2^p–2$ for every prime $p>2$ Find all polynomials $f(p)$ with integer coefficients that divide $2^p–2$ for every prime $p>2$
From Fermat's little theorem we can find $2p, p, -2p, -p$ (and per comments, some constant polynomials). But are there others, or can we prove higher degrees won't work?
| $(p-1)$ is even, let $p-1=2k$, we can write:
$2^p-2=2(2^{p-1}-1)=2(2^k-1)(2^k+1)=2(2^{\frac{p-1}{2}}-1)(2^{\frac{p-1}{2}}+1)$
If $k=\frac{p-1}2=2m+1$ is odd we have:
$m=\frac{p-3}2$
Then:
$2^k-1=(2-1)(2^{2m}+2^{2m-1}+2^{2m-2}+ \cdot\cdot\cdot)=(2^{p-3}+2^{p-4}+2^{p-5}\cdot \cdot \cdot)$
Also:
$2^k+1=(2+1)(2^{2m}-2^{2m-1}+2^{2m-2}-2^{2m-3} \cdot\cdot\cdot)=3(2^{p-3}-2^{p-4}+2^{p-5}-2^{p-7}\cdot \cdot \cdot)$
Hence the factors of $2^p-2$ can at most be ±2, ±p,3,$A=(2^{p-3}+2^{p-4}+2^{p-5}\cdot \cdot \cdot)$ and $B=(2^{p-3}-2^{p-4}+2^{p-5}-2^{p-7}\cdot \cdot \cdot)$.
Now we construct an equation it's roots are these factors:
$[x-(2^n-2)](x±2)(x±p)(x-3)(x-A)(x-B)=0$
We can rewrite this equation as:
$f(x,p)=(±2)(3)(±p)(2^n-2)\cdot A\cdot B$
RHS of this equation divides $2^p-2$, therefore polynomial $f(x, p)$, on LHS divides it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3891942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Expected Value Problem with a 10 Sided Die I have a $10$ sided die numbered $1-10$. I keep rolling it until it lands on a prime number. Let $X$ be the number of times I roll the die.
What is the expected value $E(X)$?
What I have so far is the probability of rolling a prime is $P(\text{prime}) = \frac{4}{10}$.
How do I calculate the expected value? I tried the formula: $$\sum_{s\in S} P(s) \cdot X(s) = P(1)X(1) + P(2)X(2) +P(3)X(3)+P(4)X(4)$$
Where $P(s)$ is the probability of rolling a prime and $X(s)$ is the number of rolls. I plug in $\frac{4}{10}$ for each $P$ and $X$ is incremented from $1$. I am getting $4$ and the answer is $\frac{10}4$. Can someone guide me to what is going wrong here?
| Hint
$$E(X) = \frac{4}{10} + 2 \cdot \frac{6}{10} \cdot \frac{4}{10} + 3 \cdot \left(\frac{6}{10}\right)^2 \cdot \frac{4}{10} +.... + (n+1) \cdot \left(\frac{6}{10}\right)^n \cdot \frac{4}{10} + ... \infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3893540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
prove that $5<\sqrt{5}+\sqrt[3]{5}+\sqrt[4]{5}$
prove that $$5<\sqrt{5}+\sqrt[3]{5}+\sqrt[4]{5}$$
.A little use of calculator shows that $\sqrt{5}+\sqrt[3]{5}+\sqrt[4]{5}=5.44$.Thus the inequality is indeed true.
Generalising this result with $$f(x)=x-\sqrt{x}+\sqrt[3]{x}+\sqrt[4]{x}<0$$ does not help as we see that $$8>\sqrt{8}+\sqrt[3]{8}+\sqrt[4]{8} \tag !$$
Repeated efforts of Bernoullis inequality have failed
| By hand, you can compute :
$$2^2=4$$
$$1.7^3 =4.913$$
$$1.3^4=2.8561$$
So $$\sqrt{5} + \sqrt[3]{5} + \sqrt[4]{5} > \sqrt{4} + \sqrt[3]{4.913} + \sqrt[4]{2.8561}= 2 + 1.7 + 1.3 = 5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3894934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Prove that $x ^ {1 - x} (1 - x) ^ {x} \le \frac {1}{2}$ How do I prove $x ^ {1 - x}(1 - x) ^ {x} \le \frac{1}{2}$, for every $x \in (0, 1)$.
For $x = \frac {1}{2}$ the LHS is equal to one half. I tried studying what happens when $x \lt \frac {1}{2}$ and its correspondent, but to no result.
| let $a=x,b=1-x$,
$a+b=1$,
By AM-GM$$\frac{1}{2}=\frac{{(a+b)}^2}{2}\ge 2ab=\frac{ab+ba}{a+b}\ge \sqrt[a+b]{a^b b^a}=a^bb^a=x^{1-x}{(1-x)}^{x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3895314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\int_{0}^{\pi/2} \int_{0}^{\pi/2} \frac{\theta\cot\theta-\varphi\cot\varphi}{\cos\theta-\cos\varphi} \text{d}\varphi\text{d}\theta = \pi\ln2$ Month ago I encounter a nice result numerically checked by Mathematica
$$
\int_{0}^{\pi/2} \int_{0}^{\pi/2} \frac{\theta\cot\theta-\varphi\cot\varphi}{\cos\theta-\cos\varphi} \mathrm{d}\varphi\mathrm{d}\theta = \pi\ln2
$$
where the integrated function is actually well-defined even around its singularity $\theta=\varphi=0$.
At my first sight, I thought it might be a trivial conclusion derived from a kind of typical integral like
$$
\int_{0}^{\pi} \frac{\cos n\theta}{\cos\theta-\cos\varphi} \mathrm{d}\theta = \pi\frac{\sin n\varphi}{\sin\varphi}
$$
just using a proper series expansion. However, when I review it in detail, the result over $(0,\pi/2)$ will be awkwardly complicated. I realize this double integrals may not be done directly, or I may lack some essential insight to solve it.
So I question it here for some further discussion, and thanks in advance for any suggestion.
| The following illustrates an approach that looks promising, but
that I could not yet complete: wish you and the community here might help
to finalize it.
Starting from the integrand
$$
f(\theta ,\varphi )
= {{\theta \cot \theta - \varphi \cot \varphi } \over {\cos \theta - \cos \varphi }}
= {{\theta {{\cos \theta } \over {\sin \theta }} - \varphi {{\cos \varphi } \over {\sin \varphi }}}
\over {\cos \theta - \cos \varphi }}
= f(\varphi ,\theta )
$$
it's structure suggests the idea that Green's Theorem might be applied.
Formulated as above, it is difficult to find the needed potential function.
But applying the transformation
$$
\left\{ \matrix{
\theta = x + y \hfill \cr
\varphi = y - x \hfill \cr} \right.\quad \Leftrightarrow \quad \left\{ \matrix{
x = {{\theta - \varphi } \over 2} \hfill \cr
y = {{\theta + \varphi } \over 2} \hfill \cr} \right.
$$
we get
$$
\eqalign{
& g(x,y) = f(x + y,\;y - x) = \cr
& = {{\left( {y - x} \right){{\cos \left( {y - x} \right)} \over {\sin \left( {y - x} \right)}}
- \left( {y + x} \right){{\cos \left( {y + x} \right)}
\over {\sin \left( {y + x} \right)}}}
\over {\cos \left( {y - x} \right) - \cos \left( {y + x} \right)}} = \cr
& = {{y\left( {{{\cos \left( {y - x} \right)} \over {\sin \left( {y - x} \right)}}
- {{\cos \left( {y + x} \right)} \over {\sin \left( {y + x} \right)}}} \right)
- x\left( {{{\cos \left( {y - x} \right)} \over {\sin \left( {y - x} \right)}}
+ {{\cos \left( {y + x} \right)} \over {\sin \left( {y + x} \right)}}} \right)}
\over {2\sin x\sin y}} = \cr
& = {{y\sin \left( {2x} \right) - x\sin \left( {2y} \right)}
\over {2\sin x\sin y\sin \left( {y - x} \right)\sin \left( {y + x} \right)}} = \cr
& = y{{\sin \left( {2x} \right)} \over {2\sin x\sin y\sin \left( {y - x} \right)\sin \left( {y + x} \right)}}
- x{{\sin \left( {2y} \right)} \over {2\sin x\sin y\sin \left( {y - x} \right)\sin \left( {y + x} \right)}} = \cr
& = y{{\cos x} \over {\sin y\left( {\sin ^2 y\cos ^2 x - \cos ^2 y\sin ^2 x} \right)}} - x{{\cos y}
\over {\sin x\left( {\sin ^2 y\cos ^2 x - \cos ^2 y\sin ^2 x} \right)}} = \cr
& = y{{\cos x} \over {\sin y\left( {\cos ^2 x - \cos ^2 y} \right)}}
- x{{\cos y} \over {\sin x\left( {\cos ^2 x - \cos ^2 y} \right)}} = \cr
& = g( - x,y) \cr}
$$
Putting:
$$
\eqalign{
& M(x,y) = - {y \over {2\sin ^2 y}}\ln \left( {{{ - \sin x + \sin y}
\over {\sin x + \sin y}}} \right) \cr
& L(x,y) = {x \over {2\sin ^2 x}}\ln \left( {{{ - \sin x + \sin y}
\over {\sin x + \sin y}}} \right) \cr}
$$
then we have the expression required for applying the Green's theorem
$$
g(x,y) = {\partial \over {\partial x}}M(x,y) - {\partial \over {\partial y}}L(x,y)
$$
With the help of this sketch
we have
$$
\eqalign{
& I = \int_0^{\pi /2} {\int_0^{\pi /2} {f(\theta ,\varphi )\,d\varphi \,d\theta } } = \cr
& = \int\!\!\!\int\limits_{square} {f(\varphi ,\theta )\,d\varphi \,d\theta }
= 2\int\!\!\!\int\limits_{triang} {f(\varphi ,\theta )\,d\varphi \,d\theta } = \cr
& = 4\int\!\!\!\int\limits_{triang} {g(x,y)\,dx\,dy}
= 4\int\!\!\!\int\limits_{triang} {\left( {{\partial \over {\partial x}}M
- {\partial \over {\partial y}}L} \right)\,dx\,dy} = \cr
& = 4\oint\limits_{triang} {Ldx + Mdy} \cr}
$$
that is
$$
\eqalign{
& {I \over 4} = \oint\limits_{triang} {Ldx + Mdy} = \cr
& = \left( \matrix{
\int_{y = 0}^{\pi /4} {\left( {L\left( {y,y} \right) + M(y,y)} \right)dy} + \hfill \cr
+ \int_{y = \pi /4}^{\pi /2} {\left( { - L\left( {\pi /2 - y,y} \right) + M(\pi /2 - y,y)} \right)dy}
+ \hfill \cr
- \int_{y = 0}^{\pi /2} {M(0,y)dy} \hfill \cr} \right) = \cr
& = I_{\,1} + I_2 + I_3 \cr
}
$$
Now
$$
I_3 = - \int_{y = 0}^{\pi /2} {M(0,y)dy}
= \int_{y = 0}^{\pi /2} {{y \over {2\sin ^2 y}}\ln \left( {{{\sin y} \over {\sin y}}} \right)dy}
= 0
$$
$I_{\,1}$ instead, is not null as it might appear.
If the path is taken to approach $y=x$ while maintaining positive the difference $y-x$ then a numerical
computation gives a value of $\approx -1.233688 \ldots$, which together with the computation of $I_{\, 2}$ returns the
expected value for $I$.
I am blocked at this point: analytically establish and justify the path for $I_{\, 1}$ and find a close expression for it and for $I_{\, 2}$.
| {
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"url": "https://math.stackexchange.com/questions/3896469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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Is this a valid linear transformation? (1) Which of the following is a linear transformation?
$ \quad $ (a) $ T ( \begin{pmatrix} x \\ y \end{pmatrix} )= \begin{pmatrix} 0 \\ x \\y \end{pmatrix}$, $ \quad $ (b) $ T ( \begin{pmatrix} x \\ y \\ z \end{pmatrix} ) = \begin{pmatrix} y \\ x \end{pmatrix}$ $ \quad $ (c) $ T (\begin{pmatrix} x \\ y \end{pmatrix} )= \begin{pmatrix} x^2 \\ y^2 \\ 0 \end{pmatrix}$
$ \quad $ (d) $ T (\begin{pmatrix} x \\ y \end{pmatrix}) = \begin{pmatrix} 1 \\ x \\y \end{pmatrix}$ $ \quad $ (e) $ T ( \begin{pmatrix}\ x \\ y \\ z \end{pmatrix} ) = x \begin{pmatrix} 1 \\ 1 \\ z \end{pmatrix}$
I am new to this topic, am I correct in saying that only (a) is a linear transformation?
(2) $ T (\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}) = \begin{pmatrix} 2 \\ 2 \\2 \end{pmatrix}$ , $ \quad $ $ T (\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}) = \begin{pmatrix} 1 \\ -1 \\ -1 \end{pmatrix}$, $ \quad $ $ T (\begin{pmatrix} 0 \\ 1 \\ -2 \end{pmatrix}) = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$ $ \quad $
$ \quad $$ M = \begin{pmatrix}1 & 1 & 0 \\ 2 & 1 & 1 \\ 3 & 2 & -2 \end{pmatrix}. $
$ \quad $ Then, $ T (\begin{pmatrix} x \\ y \\ z \end{pmatrix}) = \begin{pmatrix} 2 & 1 & 0 \\ 2 & -1 & 0 \\ 2 & -1 & 0 \end{pmatrix} M^{-1} \begin{pmatrix} x \\ y \\ z \end{pmatrix} $ ?
| $(a)$ and $(b)$ are. The others do not possess the properties that $T(cv)=cT(v)$ and $T(v+w)=T(v)+T(w)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3898504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can anyone explain why expressions of the form $\sqrt[3]{x-\sqrt{y}}+\sqrt[3]{x+\sqrt{y}}$ can be rational? $$\sqrt[3]{2-\sqrt{5}}+\sqrt[3]{2+\sqrt{5}}=1$$
Can anyone explain to me how this works? I don't understand why two irrational numbers cube rooted and added together return the number $1$.
I'm trying to find all the positive integer values for $x$ and $y$ in this formula
$$\sqrt[3]{x-\sqrt{y}}+\sqrt[3]{x+\sqrt{y}}=1$$
Sorry if this is a stupid question.
| Note that
$$\sqrt[3]{2-\sqrt{5}}+\sqrt[3]{2+\sqrt{5}}=1$$
is a special case $(m=1)$ of the identity
$$s=\sqrt[3]{\left(3m-1\right) -\sqrt{ (8m-3)m^2}}
+ \sqrt[3]{\left(3m-1\right) +\sqrt{ (8m-3)m^2}} =1
$$
It can be verified by taking cubic power of both sides, i.e.
$$s^3-3(1-2m)s +2-6m=(s-1)(s^2+s+6m-2)=0$$
which leads to $s=1$. Then, write the identity in the form
$$\sqrt[3]{x-\sqrt{y}}+\sqrt[3]{x+\sqrt{y}}=1$$
$$x= 3m-1,\>\>\>\>\>y=(8m-3)m^2$$
$x$ and $y$ are both positive integer for $m=1,2,3...$. and there are infinite possibilities. The first few examples are listed below.
\begin{align}
\sqrt[3]{2-\sqrt{5}} + \sqrt[3]{2+\sqrt{5} } & =1 ,\>\>\>m=1\\
\sqrt[3]{5-\sqrt{52}} + \sqrt[3]{5+\sqrt{52} } & =1 ,\>\>\> m=2\\
\sqrt[3]{8-\sqrt{189}} + \sqrt[3]{8+\sqrt{189} } & = 1 ,\>\>\>m=3\\
\sqrt[3]{11-\sqrt{464}} + \sqrt[3]{11+\sqrt{464} } & = 1 ,\>\>\>m=4\\
\sqrt[3]{14-\sqrt{925}} + \sqrt[3]{14+\sqrt{925} } & = 1 ,\>\>\>m=5\\
...&= 1,\>\>\>...
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Example of bijective function from $(-1,1)$ to $\Bbb R$ I'm looking for a bijective function $f : (-1,1) \to \mathbb{R}$. I'm having trouble finding an example.
| For a different example, $$f(x) = \frac{1}{x - 1} + \frac{1}{x + 1}$$ has this property:
*
*It is injective: if $x,y\in(-1,1)$ and $f(x) = f(y)$, then we have $$\frac{1}{x-1}+\frac{1}{x+1} = \frac{1}{y-1} + \frac{1}{y+1}.$$ Multiplying by $(x-1)(x+1)(y-1)(y+1)$ gives
$$(x+1)(y-1)(y+1) + (x-1)(y-1)(y+1) = (x-1)(x+1)(y+1) + (x-1)(x+1)(y-1).$$ Now, we can simplify that down to $$x(y - 1)(y + 1) = y(x - 1)(x + 1).$$
At this point, we look to be stuck with expanding the brackets, so we'll do that, obtaining
$$xy^2 - x = yx^2 - y,$$
or
$$xy(y - x) = x - y.$$
If $x = y$, we're done, so suppose not (so that we can divide by $x - y$). Then we have $xy = -1,$ so $|xy| = 1$. But we know that $|x| < 1$ and $|y| < 1$, so $|xy| = |x||y| < 1$, a contradiction. Thus, $f$ is injective.
*It is surjective: for this, we shall simply show that it is unbounded above and below, and appeal to a version of the intermediate value theorem. First, note that, for $x = 1 - \varepsilon$ and $\varepsilon < 1$, we have $$f(x) = \frac{1}{-\varepsilon} + \frac{1}{2 + \varepsilon} = \frac{2}{-\varepsilon(2 + \varepsilon)}\leq \frac{-2}{3\varepsilon} \to -\infty$$ as $\varepsilon \to 0$ from above. Similarly, for $x = -1 + \varepsilon$, we have $f(x) \geq \frac{2}{3\varepsilon} \to +\infty$ as $\varepsilon \to 0$ from above.
Thus, for any $y \in \mathbb{R}$, there are some $x_0, x_1$ such that $f(x_0) < y < f(x_1)$ (and, indeed, we see that $x_0 > x_1$). Applying the intermediate value theorem (after noting that $f$ is continuous on $(-1, 1)$ as a sum of rational functions whose denominators have no zeroes in that interval) to $[x_1, x_0]$ then gives us some $x_2 \in [x_1, x_0] \subset (-1, 1)$ such that $f(x_2) = y$, and so $f$ is surjective.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\sqrt[3]{ab}+\sqrt[3]{ac}+\sqrt[3]{bc} <0$, where $a,b,c$ are roots of $x^3 +3x^2-24x +1 = 0$ I am working on this problem:
$a, b$ and $c$ are the roots of $x^3 +3x^2-24x +1 = 0$. Prove that
$$S = \sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0$$
After I used Vietta's Formula,expanded $S^3$ and used many algebraic tricks I have
$S[S^2 -3(a'b' +a'c +b'c')] = 0$, wich leads to $S = 0 $ or $ S^2 -3(a'b' +a'c +b'c') = 0$, where $a' =\sqrt[3]{a}$, $b' =\sqrt[3]{b}$ and $c' =\sqrt[3]{c}$
So the final step is to prove the impossibility of the second case by proving that
$$a'b' +a'c +b'c' < 0$$
(which I verified numerically). It's here that I have trouble.
I also noticed that I didn't use the fact that $ab+ac+bc = -24$ (given by Vietta's Formula), so I guess it should be used here.
| Rewrite the equation $x^3 +3x^2-24x +1 = 0$ as
$$(x+1)^3-27x=0$$
Use $a^3-b^3=(a-b)(a^2+ab+b^2)$ and $ a^2+ab+b^2>0$ to obtain
$$x-3\sqrt[3]x+1=0
$$
which is a cubic equation in $\sqrt[3]x$. Thus, given its roots $\sqrt[3]{a}, \sqrt[3]{b} $ and $\sqrt[3]{c} $, the Vieta’s formulas produces
$$\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0$$
and
$$\sqrt[3]{ab}+\sqrt[3]{ac}+\sqrt[3]{bc} =-3<0$$
Note that the latter is actually not needed to prove the former.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding a rational number in between two rationals Can anyone give me some hints or show me how to solve this problem?
Problem Find a rational number in between $\frac{9}{10}$ and $\frac{10}{11}$ which may be written in the form $\frac{m}{2^n}$ where $m$ is an integer and $n$ is a non negative integer.
I just do not know where to even start or what theorems I need to use to approach this. I would appreciate hints or guidance to solving these types of problems.
| I'll show you a method that works for any base. I'll do the question but with base $5$ instead of $2$, and leave base $2$ as an exercise.
$\frac{10}{11} - \frac{9}{10} = \frac{100}{110} - \frac{99}{110} = \frac{1}{110}$, i.e. the distance between $\frac{10}{11}$ and $\frac{9}{10}$ is $\frac{1}{110}$.
$\frac{1}{5^2} > \frac{1}{110}\ $, but $\frac{1}{5^3} < \frac{1}{110}$.
So $ \exists k \in \mathbb{Z}$ such that $\frac{k}{5^3} \in (\frac{10}{11}, \frac{9}{10})$, otherwise $\exists j \in \mathbb{Z}\ $ such that both $\frac{j}{5^3} < \frac{99}{110}$ and $\frac{j+1}{5^3} > \frac{100}{110}$, contradicting the fact that $\frac{1}{5^3} < \frac{1}{110}$.
$\frac{9}{10} = \frac{112.5}{125}\ \text{and} \ \frac{10}{11} = \frac{113.6...}{125} ,\ \therefore \frac{113}{125} \ $ does the trick.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof by $\epsilon - \delta$ definition that $\lim_{x \to 1} \frac{x^2-1}{x^3-1} = \frac{2}{3}$ $$\lim_{x \to 1} \frac{x^2-1}{x^3-1} = \frac{2}{3}$$
I want to know if my proof is correct. If not, please show me how to fix it, if possible.
Let $\epsilon > 0$.
$\left| \dfrac{x^2-1}{x^3-1}-\dfrac23 \right|=\left| \dfrac{3x^2-3-2x^3+2}{3x^3-3} \right|=\left| \dfrac{-2x^3+3x^2-1}{3x^3-3}\right|=\left|\dfrac{(x-1)^2(2x+1)}{3(x-1)(x^2+x+1)} \right|=\dfrac{|x-1||2x+1|}{3|x^2+x+1|}$
Assume $\delta < 1$. Then $|x-1|<1$ implies:
(i) $-2 < 0 < x < 2 \implies |x| < 2$
(ii) $-5<-3<2x+1<5 \implies |2x+1|<5$
(iii) $-3<-1<x+1<3 \implies |x+1|<3$
(iv) $|x^2+x+1| = |x(x+1)+1| \le |x||x+1|+1 < 2.3+1 = 7 $
So, make $\delta = \min \{1, \dfrac{21\epsilon}{5} \}$. Then $|x-1|<\delta \implies \dfrac{|x-1||2x+1|}{3|x^2+x+1|}<\epsilon$.
| Looks good to me.
I would say, though, that if you were writing this on a homework, the first part of this answer which contains the majority of the work is simply the scratch work, and you should try to flesh out the proof on the bottom a little bit more, in the standard way for these types of proofs.
| {
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Converting $\sin(A)+\cos(B)$ to a product? Similar to,
$$ \sin(u) + \sin(v) = 2 \sin \left(\frac{u+v}{2}\right) \cos \left(\frac{u-v}{2}\right)$$
$$ \cos(u) + \cos(v) = 2 \cos \left(\frac{u+v}{2}\right) \cos \left(\frac{v-u}{2}\right)$$
Are there combination laws for expressions such as:
$$ \sin(A) + \cos(B)$$
where $A,B$ are some real angles?
| From the comment by @Andrei
$$ \sin(A) + \cos(B) = \sin(A) + \sin( \frac{\pi}{2} -B)= 2 \sin \left( \frac{A-B}{2} + \frac{\pi}{4} \right) \cos \left( \frac{A+B}{2} - \frac{\pi}{4} \right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3905455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding $\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x$ I need to compute a limit:
$$\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x$$
I tried to apply the L'Hôpital rule, but the emerging terms become too complicated and doesn't seem to simplify.
$$
\lim_{x \to 0+}(2\sin \sqrt x + \sqrt x \sin \frac{1}{x})^x \\
= \exp (\lim_{x \to 0+} x \ln (2\sin \sqrt x + \sqrt x \sin \frac{1}{x})) \\
= \exp (\lim_{x \to 0+} \frac
{\ln (2\sin \sqrt x + \sqrt x \sin \frac{1}{x})}
{\frac 1 x}) \\
= \exp \lim_{x \to 0+} \dfrac
{\dfrac {\cos \sqrt x} {x} + \dfrac {\sin \dfrac 1 x} {2 \sqrt x}
- \dfrac {\cos \dfrac 1 x} {x^{3/2}}}
{- \dfrac {1} {x^2} \left(2\sin \sqrt x + \sqrt x \sin \frac{1}{x} \right)}
$$
I've calculated several values of this function, and it seems to have a limit of $1$.
| Hint.
For $x>0$ small we have
$$
\left(2\sin(\sqrt{x})-\sqrt{x}\right)^x\le \sigma(x)\le \left(2\sin(\sqrt{x})+\sqrt{x}\right)^x
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3905629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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proof with induction $2^n + 4 > n^2 + 2n$ prove w/ full induction : $2^n + 4 > n^2 + 2n$.
Let's skip the other steps for a moment, I'm just getting stuck with the proof
the claim is: $2^{n+1} + 4 > (n+1)^2 + 2(n+1)$
and now the proof $$2^{n+1} + 4 > 2 * 2^n + 4$$ $$> 2 * (n^2 + 2n)$$ $$> 2n^2 + 4n$$
i am stuck here
| Please note this is not true for low values of $n \in \mathbb{Z+}$ up to $4$.
Just your induction step -
If $2^{n} + 4 > n^2 + 2n$ is true
we need to prove $\, 2^{n+1} + 4 > (n + 1)^2 + 2(n+1)$
$2^{n+1} + 4 = 2 \times 2^{n} + 8 - 4 = 2 (2^{n} + 4) - 4$
$ \gt 2n^2 + 4n - 4 = n^2 + (n+1)^2 + 2(n+1) - 7 = (n+1)^2 + 2(n+1) + (n^2 - 7)$
As $(n^2 - 7) \gt 0$ for $n \geq 3$, this is true.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is this $\epsilon$-$\delta$ limit correct? I have to show that $\lim_{x\to2}x^2+x+1=7$ with the $\epsilon$-$\delta$ definition of limit. Here is how I have done it: $\left\lvert x^2+x-6\right\rvert=\lvert x−2\rvert\lvert x+3\rvert<\epsilon$ and since we are close to $2$, we can assume that the $\delta$-neighborhood of $c=2$ must be have a radius of max $\delta=1$ which implies that : $\lvert x-2\rvert < \frac{\epsilon}{6}$.
We now choose $\delta = \min\{3,\frac{\epsilon}{6}\}$ and we can conclude that if $\lvert x−2\rvert<\delta$, it follows that $\left\lvert x^2+x+1-7\right\rvert < 6\frac{\epsilon}{6}=\epsilon$. Is this calculation correct? Do I miss something? Or some details?
| We put $f(x) = x^2 +x+1$
We can prove that :
$|f(x) - l|<\delta $ $ \Rightarrow $ $ |x-a|<\alpha $
$\alpha , \delta > 0$
$|f(x) - l|= |x^2 +x+1-7|=|x^2 +x-6|=|x-2||x+3|$
$|f(x) - l|<\delta$
$\Rightarrow $ $|x-2||x+3|<\delta$
Suppose $x\in [\frac{3}{2}, \frac{5}{2}] $
$\Rightarrow $$\frac{9}{2}\leq x+3\leq\frac{11}{2}$
$\Rightarrow $$|x+3|\leq \frac {11}{2} $
$\Rightarrow $$|x-2||x+3|\leq \frac{11}{2} |x-2|$
we know that :
$|x-2||x+3|<\delta$
So:
$\frac{11}{2} |x-2|<\delta$
$\Rightarrow $ $|x-2|<\frac{2} {11}\delta$
We put $\alpha=\frac{2\delta} {11}$
Finally :
After the definition of limite we proved $\lim_{x\to 2} f(x) =7$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $\arctan(\operatorname{arctanh}(x))\geq \operatorname{arctanh}(\arctan(x))$ for $0\leq x <1$ It's inspired by a general result proposed in a book (Dictionary of inequalities second edition by Peter Bullen p.101) related to composition of function :
In fact we have $x\in[0,1)$:
$$\arctan(\operatorname{arctanh}(x))\geq \operatorname{arctanh}(\arctan(x))\quad (1)$$
To prove it I have introduced a function :
$$f(x)=\arctan(\operatorname{arctanh}(x))-\operatorname{arctanh}(\arctan(x))$$
We differentiate to get:
$$f'(x)= -\frac{1}{((x^2-1)(\operatorname{arctanh}^2(x)+1))}+\frac{1}{((x^2+1)(\arctan^2(x)-1))}$$
And then I use :
Let $0\leq x <1$ then we have :
$$\operatorname{arctanh}(x)\geq x \geq \arctan(x)$$
Then I cannot conclude directly .
Question :
How to prove $(1)$ ?
Thanks in advance !
| Fact 1: $\frac{-3x^3+12x}{-x^3-x^2+12x+12}\sqrt{\frac{1+x}{1-x}} \ge \operatorname{arctanh}(x)$ for all $x \in [0, 1)$.
(Hint: Use $u^2 = \frac{1+x}{1-x}$ and then take derivative.)
Fact 2: $\frac{64x^5+735x^3+945x}{225x^4+1050x^2+945} \ge \arctan x$ for all $x \in [0, 1)$.
(Hint: Take derivative.)
Let $f(x) = \arctan(\operatorname{arctanh}(x)) - \operatorname{arctanh}(\arctan(x))$.
By Facts 1-2, we have, for all $x$ in $[0, 1)$,
\begin{align}
f'(x) &= \frac{1}{(1-x^2)(\operatorname{arctanh}^2(x)+1)} - \frac{1}{(x^2+1)(1 - \arctan^2(x))}\\
&\ge \frac{1}{(1-x^2)(\frac{(-3x^3+12x)^2}{(-x^3-x^2+12x+12)^2}\frac{1+x}{1-x}+1)}
- \frac{1}{(x^2+1)(1 - (\frac{64x^5+735x^3+945x}{225x^4+1050x^2+945})^2)}\\
&\ge 0.
\end{align}
Also, $f(0) = 0$. Thus, $f(x) \ge 0$ for all $x$ in $[0, 1)$.
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3911318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
} |
Infinitely many positive integers $n$ such that $n$ divides $2^{2^{n}+1}+1$ and $n$ doesn’t divide $2^{n}+1$ Show that there are infinitely many positive integers $n$ such that $n$ divides $2^{2^{n}+1}+1$ and $n$ does not divide $2^{n}+1$
I think we should increase $n$ by induction:
$x_{k+1}$=$2^{x_{k}}+1$, then we have $x_{k}$ doesn’t divide $x_{k+1}$ and divide $x_{k+2}$.
Next we should show that if it allows for $k=1,2,3,..m$, then allows for $k=m+1$.
I am not getting if this approach is right.
Thanks in advance.
| I don't see any way to solve the problem using induction. Instead, with the Lifting-the-exponent lemma, using $x = 2$, $y = 1$ and $p = 3$, then since $3 \mid 2 + 1$, for any positive integer $j$ and odd integer $k$ where $3 \not\mid k$,
$$\nu_3(2^{k(3^j)} + 1) = \nu_3(2 + 1) + \nu_3(k(3^j)) = 1 + j \tag{1}\label{eq1A}$$
which means $3^j \mid 2^{k(3^j)} + 1$. Next, for any integer $j \ge 2$ and odd integer $k$,
$$2^9 + 1 = 513 = 19(3^3) \implies 19 \mid 2^{k(3^j)} + 1 \tag{2}\label{eq2A}$$
We also have
$$2^{19} + 1 = 524289 = 3(174763) \tag{3}\label{eq3A}$$
Using the multiplicative order, let
$$\operatorname{ord}_{174763}(2) = m \tag{4}\label{eq4A}$$
Since $2^{2(19)} \equiv 1 \pmod{174763}$, then $m \mid 2(19)$. As $m \neq 1, 2, 19$, then $m = 2(19)$. Thus, for a positive integer $q$ and some non-negative integer $r$,
$$2^q \equiv -1 \pmod{174763} \iff q = 19 + rm \tag{5}\label{eq5A}$$
i.e., $q$ is an odd integral multiple of $19$.
For any integer $s \ge 2$, with $n = 174763(3^s)$, then since $19 \not\mid n$ (actually, $174763$ is a prime number), \eqref{eq5A} indicates $174763 \not\mid 2^n + 1$, so $n \not\mid 2^n + 1$. However, \eqref{eq2A} shows $19 \mid 2^n + 1$, so \eqref{eq5A} gives that $174763 \mid 2^{2^{n} + 1} + 1$. Also, as shown in \eqref{eq1A}, $3^s \mid 2^{n} + 1$ so $3^s \mid 2^{2^{n} + 1} + 1$. Thus, $n \mid 2^{2^{n} + 1} + 1$. This shows there's an infinite number of $n$ which meet the question conditions.
| {
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"url": "https://math.stackexchange.com/questions/3912378",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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How can I find all $x$ such that $1+(x+1)2^x$ is a perfect square? How can I find all $x$ such that $1+(x+1)2^x$ is a perfect square?
I only find that when $x = 4$, the value is $81$ is the square of $9$, but I cannot get a general solution for this. Could anyone help me with that?
| Write $(x+1)2^x=n^2-1$ for integers $n,x$.
For $x \leq -1$, notice that $2^{-x}(n^2-1)=x+1\leq 0$, which implies $n^2-1 \leq 0$ (since $2^{-x}>0$) and so $n=0,1$, from which only $n=1,x=-1$ works in the original problem.
For $0 \leq x \leq 4$, we can see that only $x=4$, $n=9$ works (simply inspect the five cases by hand).
For $x \geq 5$, notice that $4 \mid (n-1)(n+1)$, and also $\gcd(n-1,n+1) \mid 2$, so exactly one of terms $n-1$ or $n+1$ is divisible by $2^{x-1}$ (and possibly higher power, but that is not important here) and the other term is divisible by $2$ (and no by higher power of $2$). So either $n-1=d2^{x-1}$ or $n+1=d2^{x-1}$ where $d \geq 1$ is an integer. Either way, $$(x+1)2^x = n^2-1 \geq (d2^{x-1}-2)(d2^{x-1}),$$
and after division by $2^x$ and using $d \geq 1$ we get
$$x+1 \geq (d2^{x-2}-1)d \geq 2^{x-2}-1,$$
impossible for $x \geq 5$ (you can prove the opposite inequality by induction).
| {
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"timestamp": "2023-03-29T00:00:00",
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