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Compute average distance between numbers $0$ to $10$ I'd like to calculate the average difference between two numbers, each between $0$ and $10$. I calculated this for integers and came up with an average distance of $4$.
My method: there are $10$ ways to obtain a difference of $1$, $9$ ways to obtain a difference of $2$, $3$ of $8$, $4$ of $7$, $5$ of $6$, $6$ of $5$, etc. I took the weighted average of all the possibilities and I ended up with $\frac{220}{55} = 4$.
But I actually have float values between $0$ and $10$. If I did the integer average distance correctly, is the average distance between $0$ and $10$ continuous still $4$?
|
Every pair of numbers between $0$ and $10$ can be associated with a unique point $(x,y)$ in the square $[0,10] \times [0, 10]$. For such a point, the function $f(x,y)=|x-y|$ measures the difference between the two numbers. To find the average value of that function, one computes
$$ \frac{1}{100}\int_0^{10} \int_0^{10} |x-y| dx dy $$
The integral is the continuous analog of "summing" all of the integer values; the $100$ measures the area of the square region, which is the continuous analog of the "number" of integer values.
|
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|
Prove that $(2+ \sqrt5)^{\frac13} + (2- \sqrt5)^{\frac13}$ is an integer When checked in calculator it is 1. But how to prove it?
Also it is not a normal addition like $x+ \frac1x$ which needs direct rationalization. So I just need something to proceed.
|
Not a full solution but if you use the equality: $(x+y)^3 = x^3 + y^3 + 3x^2y + 3y^2x$, you can reach the equation $z^3 + z = 4$ where $z$ is the value of the above expression. This leads to $z = 1$ as a solution. I cubed your formula and simplified. I do not, however, know how to solve $z^3 + z = 4$. Cubing both sides is safe only if $z$ is positive.
|
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|
Interpreting Division as subtraction Basically, Division is the inverse of multiplication, see this instance:
5 × 4,
This basically means, adding five four times,
5, 5, 5, 5 added together is 20;
If division is the inverse of multiplication and multiplication comes from repeated addition, how can we interpret division as repeated subtraction?
Consider:
20/4;
How can we interpret this as the statements in bold above?
That's the level 1 of confusion, the next is, while taking the word PEPPER:
Total arrangement is given by the formula:
6!/(3!2!)
How can we interpret the result as above, i.e. with the division interpreted as a series of repeated subtraction?
EDIT: And how do we relate this with the fact that division is breaking things up equally with respect to the denominator? And in this case:
Why isn't 6!-(3!2!)used?
|
If you want to interpret "$20$ divided by $4$" in terms of repeated subtraction, what you would do is repeatedly subtract $4$ from $20$ until you get $0$:
$$16, 12, 8, 4, 0$$
The number of times you had to subtract $4$ is exactly the quotient $\frac{20}{4}$. This same principle carries over to the quotient $\frac{6!}{3!2!}$: you repeatedly subtract $3!2!$ from $6!$ until you get zero, and the number of times you subtracted it is the quotient.
The same principle sort of works when $a, b$ are positive integers, but $\frac{a}{b}$ is not an integer. Let's say you want to interpret the division of $27$ by $4$ as repeated subtraction. What you would do is repeatedly subtract $4$ from $27$, until you eventually get a number between $0$ and $3$. The number of times $q$ that you subtracted $4$ to get to that point, and the number $r$ between $0$ and $3$ that you get, satisfy $q = 6$ and $r = 3$. Then
$$q + \frac{r}{4} = 6 + \frac{3}{4}$$
is equal to the quotient $\frac{27}{4}$.
|
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Screen reader help for "the region {(x, y)|x + y ≤ 1}" (Series of screen reader related questions on basic symbology)
"the probability that x, y are in A where A is the region {(x, y)|x + y ≤ 1}."
How should I write this out the bolded section to that a screen reader will read it correctly?
|
$A$ is
the region in the plane consisting of the points whose $x$- and $y$-coordinates satisfy the inequality $x+y\le 1$.
|
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How to find the convergence of this sequence? $x_n=1+\frac{1}{x_{n-1}}$ The sequence ($x_n$) is defined by $x_0=1$ and $x_n=1+\frac{1}{x_{n-1}}$ for $n\in ℕ^*$, then how could I find the limit of it? By trying some, I've found that the answer is $\frac{1+\sqrt5}{2}$, but I'm looking for a "legit" way to find it.
|
In the limit, you have the following continued fraction:
$$x = \lim\limits_{n \to \infty} x_n = 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \ddots}}}}.$$
If you're familiar with the golden ratio you'll recognise this is $\varphi$ immediately, but if not, notice that
$$x = 1+\frac{1}{x},$$
which can be rearranged as
$$x^2 -x - 1 = 0.$$
This has roots $$
\frac{1+\sqrt{5}}{2} \;\;\text{ and }\;\; \frac{1-\sqrt{5}}{2}
$$
But since the fraction is clearly positive, $x$ must be the first one.
Note: The limit of your sequence is independent of the value of $x_0$!
|
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Prove $\cos{\frac{\pi}{n}}\notin\mathbb Q$ for any postive integer $n\ge 4$ Let $n\ge 4$ postive integer,show that
$$\cos{\dfrac{\pi}{n}}\notin\mathbb Q$$
Now I have solve for a case:
Assmue that
$$\cos{\dfrac{\pi}{n}}=\dfrac{q}{p},(p,q)=1,p,q\in N^{+}$$ use Chebyshev polynomials?
$$T_{n}(\cos{x})=\cos{(nx)}$$
so we have
$$T_{n}\left(\dfrac{q}{p}\right)=-1$$
then we have
$$2^{n-1}\left(\dfrac{q}{p}\right)^n+a_{n-1}\left(\dfrac{q}{p}\right)^{n-1}+\cdots+a_{0}+1=0$$
so we have
$$q|\le a_{0}+1$$
since
$|a_{0}|=0,1,-1$,
(1):$a_{0}\neq -1$,so we have
$$\dfrac{q}{p}\le\dfrac{2}{3}$$,but
$$\cos{\dfrac{\pi}{n}}\ge\cos{\dfrac{\pi}{4}}=\dfrac{\sqrt{2}}{2}>\dfrac{2}{3}$$
contradiction。
(2):But for $a_{0}=-1$, I can't prove it
|
An elementary proof. From deMoivre's Theorem, if $k\in \Bbb N$ then $\cos kx=\sum_{(0\le 2j\le k)}\binom {k}{2j}(-1)^j(\cos x)^{k-2j}(1-\cos^2 x)^j.$ So if $\cos (\pi/n)\in \Bbb Q$ and $m|n$ then $\cos (\pi/m)\in \Bbb Q.$ So it suffices to show $\cos (\pi/m)\not \in \Bbb Q$ when (i)$\,m=4,$ or (ii) $\,m=6,$ or (iii) $\,m=9,$ or (iv) $\,m \ge 5$ is prime.
For $m=9,$ let $y=2\cos (\pi/9).$ Then $4(y/2)^3-3(y/2)=\cos (\pi/3)=1/2,$ so $y^3-3y-1=0.$ If $y=A/B$ with $A,B\in \Bbb N$ and $\gcd (A,B)=1$ then $A^3-3AB^2-B^3=0 ,$ so we have $$1\equiv 1+A^3-3AB^2-B^3\equiv 1+A+AB+B\equiv (1+A)(1+B) \pmod 2$$ implying $1+A$ and $1+B$ are odd, implying $A$ and $B$ are even, contrary to $\gcd(A,B)=1.$
For prime $m\ge 5,$ if $\cos \pi/m=A/B$ with $A,B\in \Bbb N$ and $\gcd(A,B)=1$ then by deMoivre's Theorem we have $$(\bullet)\quad -B^m=\sum_{j=0}^{(m-1)/2}\binom {m}{2j}A^{m-2j}(B^2-A^2)^j(-1)^j.$$ The RHS of $(\bullet)$ is divisible by $A$ so $A|(-B^m).$ But $\gcd(A,B)=1$ so we must have $A=1.$
$\quad$ Now $B^m\equiv B \mod m,$ as $m$ is prime. And if $m$ is an odd prime and $0< 2j<m$ then $m$ divides $\binom {m}{2j}.$ So from $(\bullet)$ we have $$-B\equiv -B^m\equiv A^m\equiv 1 \pmod m.$$ So $B\ge m-1.$
But now $1/2=\cos (\pi/3)<\cos (\pi/m)=A/B=1/B\le 1/(m-1)\le 1/4,$ a paradox.
|
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According to this $0!$ Should be not defined, but why it is not? According to my knowledge factorial means that the number is multiplied by the numbers preceding it up to $1$ . That's it .
$5!=5×4×3×2×1=120$
$4!=4×3×2×1=24$
$3!=3×2×1=6$
$n!=n×(n-1)×\cdots \times 2×1$
And similarly
$0!=0×(-1)×(-2)\times\cdots\times1$
But this is not possible as we have crossed 1 before and another 1 is not possible . So it should be ND.
Please tell me where I am wrong in this?
|
My favorite is the following. We define $n!$ to be the number of ways $n$ distinct objects can be ordered. Now, how many ways can you order $0$ objects? Well, you can either say that (1)"you can't", (2) "there are zero ways", or (3) "there is one way to organize zero objects". If you choose (1) you're just being defiant. Moreover, you would like to avoid zeroes as much as possible, since factorials do indeed appear in denominators so you try to find a way around (2). So we go with definition (3), because we choose to come to the agreement that there is one way to order zero objects.
|
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Interesting log series Calculate the sum of series:
$$\sum_{k=1}^{\infty} k\left(\log\left(\frac{k+1}{\sqrt{(k+1)^2+1}}\right)-\log\left(\frac{k}{\sqrt{k^2+1}}\right)\right)$$
|
We write
$$ k\left(\log\left(\frac{k+1}{\sqrt{(k+1)^2+1}}\right)-\log\left(\frac{k}{\sqrt{k^2+1}}\right)\right) = (k+1) \log\left(\frac{k+1}{\sqrt{(k+1)^2+1}}\right)-k \log\left(\frac{k}{\sqrt{k^2+1}}\right) +\log\left(\frac{k+1}{\sqrt{(k+1)^2+1}}\right).$$
The sum of the first two terms is telescoping and we obtain that your sum $S$ is given by
$$S= \overbrace{\lim_{k\to\infty} k \log\left(\frac{k}{\sqrt{k^2+1}}\right)}^{=0} - \overbrace{\lim_{k\to 1} k \log\left(\frac{k}{\sqrt{k^2+1}}\right)}^{=-\frac12 \ln2} - \overbrace{\sum_{k=1}^\infty \log\left(\frac{k+1}{\sqrt{(k+1)^2+1}}\right)}^{=S'}.$$
Now for $S'$, we use the rules of the logarithm, to obtain
$$S' = \lim_{n\to\infty} \sum_{k=1}^n \left(\log(k+1) -\frac12 \log\left((k+1)^2+1\right) \right).$$
The latter sum is evaluated as $S'=\frac12 \log\left(\frac{2 \pi}{\sinh{\pi}}\right), see below.$
Thus, we obtain the final result
$$S= \frac{1}{2} \log\left(\frac{\sinh \pi}{\pi}\right).$$
Edit:
we are interested in the value of $$
S' = \lim_{n\to\infty} \sum_{k=1}^n \left(\ln(k+1) -\frac12 \log\left((k+1)^2+1\right) \right) =
\lim_{n\to\infty} \left(\log \Gamma(n+2) - \frac12\sum_{k=1}^n\left[ \log\left(k+1+i\right) + \log\left(k+1-i\right) \right] \right)
= \lim_{n\to\infty} \left( \log\Gamma(n+2) - \frac12 \log\left[\Gamma(2+i +n)/\Gamma(2+i) \right]- \frac12 \log\left[\Gamma(2-i +n)/\Gamma(2-i) \right] \right) = \frac12 \left(\lim_{n\to\infty}
\log\left( \frac{\Gamma^2(n+2)}{\Gamma(2+i+n) \Gamma(2-i +n)} \right) + \log[\Gamma(2+i)\Gamma(2-i) ]\right). $$
Now the ratio of the $\Gamma$ function in the first term approaches 1 and thus this term vanishes. Whereas in the second term, we use Euler's reflection formula to obtain the result quoted above.
|
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Asymptotes and focus of a conic? This is the conic $$x^2+6xy+y^2+2x+y+\frac{1}{2}=0$$
the matrices associated with the conic are:
$$
A'=\left(\begin{array}{cccc}
\frac{1}{2} & 1 & \frac{1}{2} \\
1 & 1 & 3 \\
\frac{1}{2} & 3 & 1
\end{array}\right),
$$
$$
A=\left(\begin{array}{cccc}
1 & 3 \\
3 & 1
\end{array}\right),
$$
His characteristic polynomial is: $p_A(\lambda) = \lambda^2-2\lambda-8$
A has eigenvalues discordant $(\lambda = 4, \lambda = -2)$, so it's an Hyperbole.
Then i found that the center of the conic is: $(-\frac{1}{16}, -\frac{5}{16})$
Then with the eigenvalues i found the two lines passing through the center:
$$4x-4y-1=0$$
$$8x+8y+3=0$$
Now i want to find the focus and the asymptotes but i have no idea how to do it.There is a way to find These two things through the data I have now? or do i need the canonical form of the conical? Thanks
|
These two examples are from a book that does not assume linear algebra... Both good and bad, as they are giving concrete methods, which are a bit much to memorize. Worth doing both ways, really, get them to agree; this way, then linear algebra and coordinate changes (translation to center followed by rotation). They do asymptotes and foci below
|
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|
General solution of linear equation = particular + general homogeneous solution
Let $x_p$ be the particular solution of $Ax = b$ and $x_h$ be the solution to
the homogeneous system $Ax = O$. All the solutions of $Ax = b$ are of the form $x_p + x_h$
Proof:
Let $x$ be the solution of $Ax = b$, then $A(x − x_p) = Ax − Ax_p = b − b = 0 \to x_h = x − x_p \to x = x_p + x_h$
We need to show all the solutions are of this format $x_p + x_h$.
Let $x'$ be a solution of $Ax = 0$, then
$A(x + x') = Ax + Ax' = Ax + 0 = b + 0 = b$.
Hence $x + x'$ is a solution of $Ax = b$.
I understand the technical details of this proof, but I am not sure about the intent of the arguments.
I think the first part is saying if $x $ is the solution of $Ax = b$, then $x = x_p + x_h.$
Second part is saying that $x_p + x_h$ is a solution to every system $Ax= b.$
Does that make sense?
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Yes, presented more clearly
$$\overbrace{{\rm if}\ \ ax_p = b}^{{\rm particular\ solution\ }{\large x_p}}\!\! {\rm then}\ \ ax=b \!\iff\! a(x\!-\!x_p) = 0\!\iff\! \ \underbrace{x-x_p = x_h\ \ {\rm and} \overbrace{ a\, x_h = 0}^{\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{\rm homogeneous\ solution}\ \large x_h}}_{\large \bbox[5px,border:1px solid #c00]{\begin{align}x\ &=\ \text{general solution }\\ &=\ \rm particular + homogeneous\end{align}}}\qquad$$
Remark $\ $ More generally this hold for any operator $A$ that is $\,\rm\color{#c00}{L}inear,\,$ i.e.
$\quad $ if $\,\ \color{#0a0}{Ax_p = b}\,\ $ then $\,\ Ax=b \!\iff\! A(x\!-\!x_p) = 0\!\iff\! \ x= x_p + x_h\,$ and $\, Ax_h = 0$
because $\ \ \ \smash[t]{ A(x-x_p)\overset{\rm\color{#c00}{\large L}} = Ax - \color{#0a0}{Ax_p} = Ax\! -\!\color{#0a0}b} $
therefore $\ A(x-x_p)\, =\, 0 \ \ \iff\ \ Ax = b$
Thus this relationship between general, particular and homogeneous solutions also holds for linear differential and difference equations (recurrences), linear Diophantine equations (Bezout scalings), and many other common linear equations. In linear algbra such solutions spaces of inhomogenous equations are abstracted in the study of affine spaces.
|
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|
finding the center of a circle (elementary geometry) I am considering the problem of finding the center of a given circle C of radius r. I currently know 3 elementary ways (ruler+compass construction) of doing this :
- choose 2 points on the circle, draw the bisector, and draw another bisector involving a third point on the circle
- choose 2 points on the circle, draw the bisector, get a diameter then find the middle point
- choose 2 points on the circle, draw a chord, then a perpendicular line and get a diameter (Thales theorem).
I was looking for other elementary method (not involving trigonometry/analytic geometry). Then I heard something quite confusing for me : with two points on the circle C, call them A and B and "close enough", draw the circle C1 of center A and radius AB, and the circle C2 of center B and radius AB, they met at D('inner' wrt the circle C) and E('outer' wrt the circle C). Call F the other intersection of the circles C and C1, then draw (FD) which intersects the circle C at G; it seems that the circle of center G and radius r contains the three points: B, D and the center of the circle C.
I have no idea if this is possible to prove this property using only elementary things. Any hint please? Comments are welcome (and please excuse my not-so-good english :/ )
|
Based on the described construction and using the OP's notations, we have the following figure:
It is enough to demonstrate that $DG=r$, the radius of the circle.
By the construction, the triangle $ABD$ is equilateral and then $AB=BD=AF$. $BDG$ is an isosceles triangle because the triangles $ABG$ and $ADG$ are congruent. $\angle AGD = \angle BGA$ and $AB=AD=AF$.
Now, connecting $A$ and $F$ with the center $O$ we get the green triangle $AFO$. To show that $DG=FO=r$, it is enough to prove that the triangles $BDG$ and $AFO$ are congruent.
We have already seen that the bases of these triangles equal. What remains is to show that $\angle AOF = \angle BGD$.
This is easy. Just look at the figure and you will see that the $\angle BDG =2\times\angle AGD = \angle AOF$. (See the Wiki article called Inscribed angle.)
Finally, this is how to use the construction at stake to find the center of the circle:
*
*Construct $E$, $F$, and $D$ and then $G$ as described.
*The straight $ED$ and the circle centered at $G$ through $D$ will meet in $O$.
Note
This construction is not that bad compared to the most frequently used one in the case of which you construct two bisecting perpendiculars. In the case of this present construction it is enough to draw the line through $F$ and $D$ which already exist after constructing the first bisecting perpendicular.
|
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|
Summation symbol with just one letter below it? How to verbally state.
I'm used to seeing summation capital sigmas with a "start" at bottom and a "stop" at top and would say the sum from i to 100 or whatever. But in this case, there is just a y at bottom.
How would I write this out so a screen reader would be able to read it aloud and get the correct idea across to the listener?
|
It depends on context. Sometimes it means "sum over all possible values of $y$" (where which values of $y$ are allowed should be clear from the context), but sometimes it means an indefinite sum or "anti-difference", analogous to an indefinite integral or antiderivative:
$$ \sum_y f(x,y) = g(x,y)\ \text{where}\ g(x,y+1) - g(x,y) = f(x,y) $$
|
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Simple inequality ( or false) Given $a$, $b$ and $c > 0$ such that $abc=1$, prove that
$$\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b} \leq 8abc$$
I don't know if its true or not but i need help in both cases.
Thanks
|
If $a$ and $b$ are "small" and very very close together (but $a$ slightly larger), then $c$ is "large" and $c/(a-b)$ can be made arbitrarily large.
$a/(b-c)$ is negative but a it can be a "small" negative (as $c$ is "large" compared to $a$ and $b$. $b/(c-a)$ is positive (but small). So that sum can be arbitrarily large.
Example:
Fix $M$ to be a "large" number.
Let $a = 1/M$ and $b = 1/2M$ are two small numbers close together so.....
Then $c = 1/ab = 2M^2$
Then $\frac a{b-c} = \frac 1{M(\frac 1{2M} - 2M^2)}=\frac 1{\frac 12 - 2M^3} \approx 0$
And $\frac b{c-a} = \frac{1}{2M(2M^2 - \frac 1M}=\frac{1}{4M^3- 1} \approx 0$
And $\frac c{a-b} = \frac{2M^2}{\frac {1}{2M}} =4M^3$
Then $\frac a{b-c} + \frac b{c-a} + \frac c{a-b} \approx 4M^3 > 8 = 8abc$.
$M$ doesn't even have to be that large. Say $M=2$ so $a= 1/2$ and $b=1/4$ and $c = 8$.
Then $\frac a{b-c} = -\frac 1{2(7\frac 34)}= -\frac {2}{31}$
$\frac b{c-a} = \frac 1{4(7 \frac 12)} = \frac 1{30}$
$\frac c{a-b} = \frac 8{1/4} = 32$ and
$-\frac {2}{31}+ \frac 1{30} + 32 = 32 - \frac {2*30 - 31}{31*30}=32 - \frac{29}{930} = 31 \frac{901}{930} > 8*\frac 12*\frac 14*8 = 8 $
|
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|
Solving $\lim\limits_{x \to \infty} \frac{\sin(x)}{x-\pi}$ using L'Hôpital's rule
I know how to solve this using the squeeze theorem, but I am supposed to solve only using L'Hôpital's rule
$$\lim\limits_{x \to \infty} \frac{\sin(x)}{x-\pi}$$
I tried:
$$\lim\limits_{x \to \infty} \frac{\sin(x)}{x-\pi} = \lim\limits_{x \to \infty} \frac{d/dx[\sin(x)]}{d/dx[x-\pi]} = \lim\limits_{x \to \infty} \frac{\cos(x)}{1}$$
From here I am stuck because the rule no longer applies and using $\infty$ for $x$ doesn't not help to simplify.
Logically the limit is $0$ because $\sin(x)$ can only be $-1$ to $1$, but this is using squeeze theorem.
Is still there any way to solve this without using the squeeze theorem?
|
I bet that your problem is, in fact, to compute
$$\lim _{x \to \color{red} \pi} \frac {\sin x} {x - \pi} ,$$
so either you have mistyped $\infty$ instead of $\pi$, or there is a typo in the text where you took this from.
In this case, L'Hospital's theorem could be used, but it's not necessary, because
$$\lim _{x \to \pi} \frac {\sin x} {x - \pi} = \lim _{x \to \pi} \frac {\sin x - 0} {x - \pi} = \lim _{x \to \pi} \frac {\sin x - \sin \pi} {x - \pi} = (\sin ') (\pi) = \cos \pi = -1 .$$
|
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|
Does an equilibrium solution have to be a constant? I'm having difficulty understanding equilibrium solutions. Consider trying to find the equilibrium solution(s) of the following ODE:
\begin{equation}
y'=x^2-y^4
\end{equation}
Firstly, I set $y'=0$ and solved for y:
\begin{align}
0&=x^2-y^4 \\
y^4&=x^2 \\
y&=\pm\sqrt{x}
\end{align}
Now my lecturer/workbook specify that there is only an equilibrium solution when $y=c$, for some constant $c$. This would mean that solutions for $y$ that are functions of another variable are not equilibrium solutions. What then is the meaning of this solution? Just that it isn't one?
|
What you found out is not a solution. Just try to substitute in the equation.
By definition, an equilibrium point is a constant solution of the equation. In this case, substituting $y=c$ leads to $x^2=c^2$ and so there are no solutions (note that by definition a solution is a $C^1$ function defined on some open interval, not on a point).
|
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|
Trigonometry and Quadratics If $\tan A$ and $\tan B$ are the roots of $x^2+px+q=0$, then prove that
$$\sin^2(A+B)+p \sin(A+B) \cos(A+B) + q \cos^2(A+B) = q$$
My Attempt:
Using the sum and product formulae we have,
$q=\tan A\tan B, $ $-p=\tan A+\tan B$
And,
$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B} \Rightarrow \tan (A+B)=\frac{-p}{1-q}$
|
You can proceed easily after finding $\tan (A+B)$.
So we have $$p =\tan (A+B)(q-1) $$ So substituting into the LHS we get, $$\sin^2 (A+B)+ (q-1)\tan (A+B)\sin (A+B)\cos (A+B) +q \cos^2 (A+B) = \sin ^2 (A+B) +(q-1)\sin ^2(A+B) +q\cos ^2 (A+B) =q\sin ^2 (A+B) +q\cos ^2 (A+B) =q $$ Hope it helps.
|
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|
What's wrong with this sequential reasoning for a formula for a_n? The following sequence is given. Write a formula for $a_n$.
$\displaystyle\{1,2,4,7,11,16\}$
Note the arithmetic formula.
$\displaystyle a_n=a_1+\left(n-1\right)d$
Define $d$ as $d_n$, where $d_n$ is the difference between term $n$ and its succeeding term, $n+1$.
$\displaystyle a_n=a_1+\left(n-1\right)d_n$
$\displaystyle d_n=d_1+\left(n-1\right)g$
Where $g$ is the common difference for $d_n$.
$\displaystyle a_n=a_1+\left(n-1\right)\left(d_1+\left(n-1\right)g\right)$
$a_n\in\{1,2,4,7,11,16\}$
$d_n\in\{1,2,3,4,5\}$
$g=1$
$\displaystyle a_n=a_1+\left(n-1\right)\left(d_1+\left(n-1\right)\right)$
$\displaystyle a_n=1+\left(n-1\right)\left(1+\left(n-1\right)\right)$
$\displaystyle a_n=n^2-n+1$
But this doesn't yield the correct sequence.
|
Following your notations :-
$d_n$ is the difference between the $(n+1)th$ and $nth$ term $\implies d_n=a_{n+1}-a_n$.
Also, $d_n=d_1+(n-1)g$ and $d_1=1$ and $g=1\implies d_n=n$.
Thus $a_{n+1}-a_n=n\implies a_{n+1}=a_n+n, \ n\geq 1$.
The mistake you were making is to assume that $a_n=a_1+(n-1)d$, which is true with a fixed common difference.
|
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|
Why does $\log x$ go to $-\infty$ so slowly? While checking convergence of integrals, I always have to remind myself that $\log x$ goes to infinity very slowly. This is almost like and axiom to me. I don't intuitively get why it goes to infinity so slowly. On a graph, it sure doesn't look that way. Any thoughts?
More precisely:
$$\forall\delta>0: \lim_0x^\delta\log x=0$$
|
Because $\log(x^{\delta})=\delta\log(x)$, which is essentially $\log(x)$, and
$$\log(\log(x))\leftrightarrow \log(x)\leftrightarrow x\leftrightarrow e^x\leftrightarrow e^{e^x}$$
truly represent different growth behaviors.
|
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|
Doing definite integration $\int_0^{\pi/4}\frac{x\,dx}{\cos x(\cos x + \sin x)}$ We have to solve the following integration
$$
\int_0^{\pi/4}\frac{x\,dx}{\cos x(\cos x + \sin x)}
$$
I divided both in Nr and Dr by $\cos^2 x$.
But after that I stuck.
|
\begin{align}
\int_0^{\pi/4}\frac{x\,dx}{\cos x(\cos x + \sin x)} &= \int_0^{\pi/4}\frac{x\,dx}{\cos^2(x)+\cos(x)\sin(x)}\\ &=\int_0^{\pi/4}\frac{2x\,dx}{\cos(2x)+\sin(2x)+1} \\&=\frac{1}{2}\int_0^{\pi/2}\frac{t\,dt}{\cos(t)+\sin(t)+1}
\end{align}
By $t \to 1-t$
$$I =\frac{1}{2}\int_0^{\pi/2}\frac{(\pi/2-t)\,dt}{\cos(t)+\sin(t)+1} $$
Separate the integrals
$$I =\frac{1}{2}\int_0^{\pi/2}\frac{(\pi/2)\,dt}{\cos(t)+\sin(t)+1} -I$$
$$I = \frac{\pi}{8}\int_0^{\pi/2}\frac{\,dt}{\cos(t)+\sin(t)+1} = \frac{\pi}{8}\log(2)$$
The last integral could be solved using Weierstrass substitution.
|
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|
Quadratic equation system $A^2 + B^2 = 5$ and $AB = 2$ Given a system of equations
$A^2 + B^2 = 5$
$AB = 2$
what is the correct way to solve it?
I see immediately that the answers are
*
*$A=1, B=2$
*$A=2, B=1$
*$A=-1, B=-2$
*$A=-2, B=-1$
but I don't understand the correct way of getting there. I have tried to isolate one of the variables and put the resulting expression into one of the equations, but this didn't get me anywhere. What is the correct way of solving this problem?
|
My favorite way:
Multiply the first equation by $A^2$ ($\ne0$):
$$A^4+A^2B^2=5A^2.$$
As $A^2B^2=4$, you get a biquadratic equation:
$$A^4-5A^2+4=0.^*$$
Then
$$A^2=\frac{5\pm3}{2}=1\text{ and }4,$$ giving the four solutions
$$A=-2,-1,1,2$$
and corresponding $B=2/A$:
$$B=-1,-2,2,1.$$
$^*$Following @Hurkyl's argument, you can stop there, because a biquadratic equation cannot have more than four roots, and by $AB=2$, there is a single $B$ for a given $A$.
|
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|
Why does taking the derivative with respect to time of an integral (with respect to tau) result in a function of time?
I am having trouble understanding why line 3 to line 4 works, specifically when the derivative of u with respect to time is taken.
u is an integral of a function of tau with respect to tau, but taking the derivative with respect to time results in f(t).
How is this possible?
|
For any continuous function $f$, we have that
$$\frac{d}{dx} \left(\int_a^x f(y)dy \right) = f$$
this is called the Fundamental theorem of calculus.
In your case, you have $t$ in the upper bound of the function. for every $t$, $\int_0^t f(\tau)d\tau$ is some number, so you have a function of $t$ (not of $\tau$.
|
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|
Is the following proof of convergence valid? A problem from Stephen Abbott's Understanding Analysis:
Exercise 2.5.4.: Assume $(a_n)$ is a bounded sequence with the property that
every convergent subsequence of $(a_n)$ converges to the same limit $a \in \mathbb{R}$. Show that $(a_n)$ must converge to $a$.
Is the following direct proof valid?
Since all subsequences converge, it means that for any particular $\varepsilon>0$ there exists some number $J\in\mathbb{N}$ in every subsequence, such that, given $j\geq J$ and $n_j\geq n_J$, $\left|a_{n_{j}}-a\right|<\varepsilon$. Let $N$ be equal to the greatest element from the set of specific $n_J$'s from all subsequences and $n\geq N$. Thus, for any $\varepsilon>0$, $\left|a_n-a\right|<\varepsilon$. $\square$
|
Your proof has two problems: first, the problem does not state that every subsequence converges, but rather that if a given subsequence has a limit then this limit must be $a$.
Second, since there are infinitely many subsequences of $\{a_n\}$, there are infinitely many $n_J$, and so the $N$ in your proof is not necessarily defined.
Instead of your approach, I suggest trying to prove that $\limsup_{n\to\infty}a_n=a=\liminf_{n\to\infty}a_n$, using the fact that there are subsequences of $\{a_n\}$ converging to the limsup and to the liminf.
|
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|
A trick to simplify $z\ln z+\overline{z}\ln\overline{z}$?
Q: Given complex conjugates $z = a+bi$ and $\overline z = a-bi$, what would be substitution needed to find $R$,
$$R= z\ln z+\overline{z}\ln\overline{z}\tag1 $$
such that $R$ is an expression without imaginary numbers?
This question arose as a special case of this post. Given $x^3-x-1=0$, let $x_1\approx1.3247$ be its real root (the plastic constant) and $x_2,x_3$ its complex roots. Then
$$\begin{aligned}\sum\limits_{n=0}^\infty\frac{n!\,(2n)!}{(3n+2)!}
&= x_1\ln(1+x_1)+x_2\ln(1+x_2)+x_3\ln(1+x_3)\\
&= \frac32\,x_1\ln(1+x_1)-\frac{1}{2}\sqrt{\frac{3-x_1}{x_1}}\arccos\Big(\frac{x_1-6}{6x_1+2}\Big)\\[2mm]
&= 0.5179778\dots\end{aligned}$$
The first is by yours truly, while the second is a simplified version of a result by Reshetnikov. How do we show the equality of the first two lines? In general, how do we get rid of the imaginary unit in $(1)$, and can we always express it as the sum of $\ln$ and $\arccos$ of real numbers?
|
With a notation that should be obvious,
$$R=(x+iy)(\log r+i\theta)+(x-iy)(\log r-i\theta)=2(x\log r-y\theta)=2\left(x\log\sqrt{x^2+y^2}-y\,\arccos\frac xr\right).$$
|
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|
Decrease rate of the shadow
If a man of height $6ft$ moves with $5ft/sec$ velocity towards a lamp hanging at $15ft$ height, then at what rate his shadow will decrease?
My Work: Let the initial distance between the man and the lamp post be $x$ and the length of shadow be $y$.
Then, $\frac{x}{9} = \frac{y}{6} \implies y = \frac{2x}{3}$
Now how do I handle the $5ft/sec$ velocity and the decrease rate of the shadow?
I think the rest part is so easy. But I am missing somewhere.
Note: This is a problem from BdPhO 2016 but completely math related. So I posted it here.
|
Anatoly has given a simple solution,
but in case you are expected to use calculus, here's another simple solution.
Starting from $y = \dfrac23 x,\;\; \dfrac{dy}{dt} = \dfrac23 \dfrac{dx}{dt} = \dfrac 23\times -5 = -\dfrac{10}{3}$ ft/s,
the negative sign showing decreasing length of shadow.
|
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|
Linear transformation and matrix representation in $\mathbb R^{2\times 2}$ After spending some time on the next problem, I still don't get it. It consists of a,b and c and I think I solved a (though I might have done something wrong so any corrections are appreciated)
The real trouble starts at b, I really don't know how to solve it
c I don't really know yet, since I failed at b
Here's the question:
Let $M$ and $N$ be two $2\times 2$ matrices. In this problem we will deal with the matrix equation $MX+XM=N$
(a) Let $L:\mathbb R^{2\times 2} \to \mathbb R^{2\times 2} $ given by $L(X)=MX+XM$
Show L is a linear transformation.
My attempt:
$L(\alpha X)=M(\alpha X)+(\alpha X)M=\alpha (MX)+\alpha (XM)=\alpha (MX+XM)=\alpha L(X)$
$L(X+Y)=M(X+Y)+(X+Y)M=MX+MY+XM+YM=(MX+XM)+(MY+YM)=L(X)+L(Y)$
So L is a linear transformation.
(b) Take $M=\begin{bmatrix}a & b \\c & d \end{bmatrix}$
Write down the matrix representation of L using the following basis for $\mathbb R^{2\times 2}$
$\begin{bmatrix}1 & 0 \\0 & 0 \end{bmatrix}$ $\begin{bmatrix}0 & 1 \\0 & 0 \end{bmatrix}$ $\begin{bmatrix}0 & 0 \\1 & 0 \end{bmatrix}$ $\begin{bmatrix}0 & 0 \\0 & 1 \end{bmatrix}$
My attempt:
I think I have to split it in $L(e_1),L(e_2),L(e_3)$ and $L(e_4)$
(c) Find all values of $a,b,c,d$ such that the matrix equation ($MX+XM=N$) has a unique solution for every $N\in \mathbb R^{2\times 2}$
Thanks in advance :)
|
When $M = \pmatrix{a&b\\c&d}$, we can find $L$ by multiplying out $MX$ and $XM$ and adding the results. Thus
$$
MX + XM = \pmatrix{a&b\\c&d}\pmatrix{x_{11}&x_{12}\\x_{21}&x_{22}} +
\pmatrix{x_{11}&x_{12}\\x_{21}&x_{22}}\pmatrix{a&b\\c&d} \\
LX= \pmatrix{ax_{11}+bx_{21} & ax_{12}+bx_{22}\\cx_{11}+dx_{21} & cx_{12}+dx_{22}}
+\pmatrix{ax_{11}+cx_{12} & bx_{11}+dx_{12}\\ax_{21}+cx_{22} & bx_{21}+dx_{22}}\\
LX = \pmatrix{2ax_{11}+bx_{21}+cx_{12} & ax_{12}+bx_{22}+bx_{11}+dx_{12}\\
cx_{11}+dx_{21} + ax_{21}+cx_{22}& cx_{12}+bx_{21}+2dx_{22}}
$$
From this, we can read off where the terms involving each $x_{ij}$ appear:
$$
L\pmatrix{1&0\\0&0} = \pmatrix{2a&b\\c&0}\\
L\pmatrix{0&1\\0&0} = \pmatrix{c&a+d\\0&c}\\
L\pmatrix{0&0\\1&0} = \pmatrix{b&0\\a+d&b}\\
L\pmatrix{0&0\\0&1} = \pmatrix{0&b\\c&2d}
$$
That solves part (b).
For part (c) you can read off the equations in $x_{ij}$ that would need to be solved to solve $MX + XM = N$. The coefficients, ordering the $x_{ij}$ as $\{x_{11}, x_{12}, x_{21}, x_{22}\}$ form the matrix
$$
C =\pmatrix{2a&c&b&0\\ b&a+d& 0& b\\ c&0&a+d&c\\ 0&c&b&2d}
$$
The equations will have a unique solution for any $N$ provided $\det C \neq 0$.
Thus the condition on $a,b,c,d\,$ is
$$
\det C = 4(a+d)^2(ad-bc) \neq 0 \\
(ad-bc) \neq 0 \mbox{ and } a+d \neq 0
$$
Which tells you that the equation has a unique solution for all $N$ unless
$\det M = 0$ or $M$ is traceless.
|
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|
Finding the numbers $n$ such that $2n+5$ is composite. Let $n$ be a positive integer greater than zero. I write
$$a_n =
\begin{cases}
1 , &\text{ if } n=0 \\
1 , &\text{ if } n=1 \\
n(n-1), & \text{ if $2n-1$ is prime} \\
3-n, & \text{ otherwise}
\end{cases}$$
The sequence goes like this $$1,1,2,6,12,-2,30,42,-5,72,90,-8,132,-10,-11,\ldots$$ I would like to prove the following two claims.
claim 1 : If $a_n>0$ and
${a_n \above 1.5 pt 3} \notin \mathbb{Q}$ then $\sqrt{4a_n+1}$ is prime.
The table below illustrates what I am seeing:
\begin{array}{| l | l | l | l }
\hline
n & a_n & {a_n \above 1.5 pt 3} & \sqrt{4a_n+1}\\ \hline
0 & 1 & .333333.. & 2.2360679.. \\
1 & 1 & .333333.. & 3 \\
2 & 2 & .666666.. & 3 \\
3 & 6 & 2 & 5 \\
4 & 12 & 4 & 7 \\
6 & 30 & 10 & 11 \\
7 & 42 & 14 & 13 \\
9 & 72 & 24 & 17 & \\
10 & 90 & 30 & 19 \\
12 & 132 & 44 & 23 \\
15 & 210 & 70 & 29 \\
16 & 240 & 80 & 31 \\
19 & 342 & 114 & 37 \\
21 & 420 & 140 & 41 \\
22 & 462 & 154 & 43 \\
\hline
\end{array}
claim 2: If $a_n<0$ then $2a_n+5$ is composite
|
Ignore $n = 0, 1$ since they're kind of irrelevant. Then $a_n \leq 0$ for all $n$ unless $2n - 1$ is prime, by the definition. In that case $a_n = n(n - 1)$. So,
$$4a_n + 1 = 4(n^2 - n) + 1 = (2n-1)^2$$
So, then $\sqrt{4a_n + 1} = 2n - 1$, which is prime.
|
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|
Analytical solution to linear coupled ODEs? I have this set of equations and I am trying to find $X(t)$ and $Y(t)$ analytically with initial values known such as $X(0)=X_0$ and $Y(0)=Y_0$. How should I approach to solve it?
$dX/dt=a \times X(t) + b \times Y(t)$
$dY/dt=c \times Y(t) + d \times X(t)$
|
Write your system of equations in matrix form:
$$\begin{pmatrix} \frac{dX}{dt} \\ \frac{dY}{dt} \end{pmatrix} =\begin{pmatrix} a & b \\ d & c \end{pmatrix} \begin{pmatrix} X \\ Y \end{pmatrix}$$
You can find the two eigenvalues $\lambda_1$ and $\lambda_2$ by letting $\det(A-\lambda I)=0$, and then evaluate corresponding eigenvectors $\vec{v_1}$ and $\vec{v_2}$ of the matrix $A=\begin{pmatrix} a & b \\ d & c \end{pmatrix}$ to give you the general solution to $X(t)$ and $Y(t)$.
Depending on your values of your eigenvalues $\lambda_1$ and $\lambda_2$ (which ultimately depend on the values of your constants $a,b,c,d$), there will be different general solutions.
There are 3 different solution cases:
Case 1: Real eigenvalues: $\lambda \in \mathbb{R}$
The solution will be of the form:
$$\begin{pmatrix} X(t) \\ Y(t) \end{pmatrix} =k_1 \vec{v_1} e^{\lambda_1 t}+k_2 \vec{v_2} e^{\lambda_2 t}$$
Case 2: Complex eigenvalues.
We denote a complex number by $\lambda=a+bi$.
Since $\lambda_{1,2}$ will both be complex, using $e^{i \theta}=\cos{\theta}+i\sin{\theta}$, we obtain the general solution:
$$\begin{pmatrix} X(t) \\ Y(t) \end{pmatrix}=k_1 \vec{v_1} e^{at} \cos{(bt)} + k_2 \vec{v_2} e^{at} \sin{(bt)}$$
Case 3: Real, repeated roots (i.e. $\lambda_1=\lambda_2$)
$$\begin{pmatrix} X(t) \\ Y(t) \end{pmatrix}=k_1 \vec{v_1} e^{\lambda t}+k_2 \vec{v_2} \cdot t e^{\lambda t}$$
Your initial values can be used to evaluate the constants $k_1$ and $k_2$.
Here is an Youtube video showing an example solve for Case 1.
|
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|
Linear Algebra: coordinates and basis I'm asked to show that if $f,g \in \mathrm{Sig}_3$ (sig stands for signal) have the coordinates $x,y \in \mathbb{R^4}$ with respect to the basis $(v_0, v_1, v_2, v_3)$, then
$$\left<f,g\right>=\pi(x_1y_1+x_2y_2+x_3y_3+x_4y_4)$$
where
$$v_0=\frac{1}{\sqrt2},
v_1=\cos x,
v_2=\cos(2x),
v_3=\cos(3x)$$
and
$$C^0([0;2\pi])$$
Any idea how to show this? I'm completely stuck. All of the examples online use vectors with numbers, but I'm just given functions.
|
The question is a bit hard to understand because it's written in a weird order. Usually the question goes (environment) (specifics) (statement to be proved). If I understand your problem well, I think it would be better written this way:
Let $f$ and $g$ be elements of $\mathrm{Sig}_3$, which has inner product
$$
\left<f,g\right> = \int_0^{2\pi} f(x)g(x)\,dx
$$
[I don't actually know what $\mathrm{Sig}_3$ is, even with the information that Sig stands for signal. But I'm trying to work from the context.] Let
\begin{align*}
v_0 &= \frac{1}{\sqrt{2}} \\
v_1 &= \cos(x) \\
v_2 &= \cos(2x) \\
v_3 &= \cos(3x)
\end{align*}
Suppose that with respect to the basis $(v_0, v_1, v_2, v_3)$, $f$ and $g$ have coordinates $x$ and $y$ in $\mathbb{R}^4$. Show that
$$
\left<f,g\right> = \pi(x_1 y_1 + x_2 y_2 + x_3 y_3 + x_4 y_4)
$$
Please correct me if that's not what your problem is. But notice how it starts with the environment of the vector space $\mathrm{Sig}_3$, along the definition of the inner product. Then comes the specific information: for these two elements, we have these two coordinate vectors. Finally, the statement we are to prove.
So I think you will be home free if you understand the meaning of this sentence:
Suppose that with respect to the basis $(v_0, v_1, v_2, v_3)$, $f$ and $g$ have coordinates $x$ and $y$ in $\mathbb{R}^4$.
What does this tell you about the formulas for $f(x)$ and $g(x)$? Once you know this, you should be able to compute $\left<f,g\right>$ directly.
|
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Proving $a\cos x+b\sin x=\sqrt{a^2+b^2}\cos(x-\alpha)$
Show that $a\cos x+b\sin x=\sqrt{a^2+b^2}\cos(x-\alpha)$ and find the correct phase angle $\alpha$.
This is my proof.
Let $x$ and $\alpha$ be the the angles in a right triangle with sides $a$, $b$ and $c$, as shown in the figure. Then, $c=\sqrt{a^2+b^2}$. The left-hand side is $a\cos x+b\sin x=\frac{ab}{c}+\frac{ab}{c}=2\frac{ab}{c}$. The right-hand side is $\sqrt{a^2+b^2}\cos(x-\alpha)=c\left(\cos{x}\cos{\alpha}+\sin{x}\sin{\alpha}\right)=c\left(\frac{ab}{c^2}+\frac{ab}{c^2}\right)=2\frac{ab}{c}$.
Is my proof valid? Is there a more general way to prove it?
For the second part of the question, I think it should be $\alpha=\arccos\frac{a}{\sqrt{a^2+b^2}}=\arcsin\frac{b}{\sqrt{a^2+b^2}}$. Is this correct?
|
We can prove this by using properties of complex numbers:
$$ \begin{align}a\cos(x) + b\sin(x) &= a \cos(x) + b \cos\left(x-\frac{\pi}{2}\right)\\
&=a \operatorname{Re}\left\{e^{ix}\right\} + b \operatorname{Re}\left\{e^{i(x-\pi/2)}\right\}\\
&= \operatorname{Re}\left\{ae^{ix} + be^{i(x-\pi/2)}\right\}\\
&= \operatorname{Re}\left\{e^{ix}(a-bi) \right\} \\
&= \operatorname{Re}\left\{e^{ix}\sqrt{a^2+b^2}e^{i\cdot\operatorname{atan2}(-b,a)} \right\}\\
&= \operatorname{Re}\left\{\sqrt{a^2+b^2}e^{i(x+\operatorname{atan2}(-b,a))}\right\}\\
&= \sqrt{a^2+b^2} \cos(x+\operatorname{atan2}(-b,a))
\end{align}
$$
where $\operatorname{atan2}(y_0,x_0)$ is the angle of the complex number $x_0+iy_0$.
This formula works for all real numbers $a$ and $b$ except when $a=b=0$.
Generalization
Following the same procedure, we can find a general formula for the sum of two sinusoids with same frequency but (possibly) different phases:
$$ \begin{align}a\cos(x+\alpha) + b \cos(x+\beta) &= \operatorname{Re}\left\{ae^{i(x+\alpha)} + be^{i(x+\beta)}\right\}\\
&= \operatorname{Re}\left\{e^{ix}\left(ae^{i\alpha}+be^{i\beta}\right) \right\}\\
&= \operatorname{Re}\left\{e^{ix}ce^{i\gamma} \right\}\\
&= \operatorname{Re}\left\{ce^{i(x+\gamma)} \right\}\\
&= c \cos(x+\gamma)
\end{align}
$$
Where $ce^{i\gamma} = ae^{i\alpha}+be^{i\beta}$.
We have
$$\begin{align}ce^{i\gamma}&=a (\cos(\alpha)+i\sin(\alpha)) +b (\cos(\beta)+i\sin(\beta))\\
&= (a\cos(\alpha)+b\cos(\beta)) + (a\sin(\alpha)+b\sin(\beta))i.\\
\end{align}
$$
So
$$c=\sqrt{(a\cos(\alpha)+b\cos(\beta))^2+(a\sin(\alpha)+b\sin(\beta))^2}$$
and
$$\gamma=\operatorname{atan2}\big(a\sin(\alpha)+b\sin(\beta),a\cos(\alpha)+b\cos(\beta)\big).$$
Sidenotes:
*
*This result is equivalent to
phasor addition.
*We can get similar formula involving $\sin$ by taking the imaginary
part.
|
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|
Transformation between random variables If given realization of X and Y, which are i.i.d. Gaussian random variables.
We need to generate random variables C and D which are jointly Gaussian with zero mean and covariance matrix like
$
\begin{bmatrix}
2 & -1 \\
-1 & 2
\end{bmatrix}$
using X and Y. How to provide a transformation that can be used to obtain C and D from X and Y?
|
from the covariance matrix $\sigma^2_C=2, \sigma^2_D=2, \sigma_{CD}=\sigma_{DC}=-1$.
Hence, $\rho_{CD}=\frac{-1}{2}$. The desired transform is
$$\begin{align}
C&=\sigma_CX\\
D&=\sigma_D(\rho_{CD}X+\sqrt{1-\rho^2_{CD}}Y)
\end{align}$$
It is assumed that $X$ and $Y$ are zero-mean. Otherwise, the same transformation should be used with $\tilde{X}=X-\mu_X$ and $\tilde{Y}=Y-\mu_Y$.
|
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|
Find coordinates of point $C$ lying on both the planes such that the perimeter of $\triangle ABC$ is minimum $A(3\mid −1\mid −1)\quad$ is a point in the plane $x+y+z=1.$
$B(3\mid 1\mid 0)\quad$ is a point in the plane $2x−y−z=5.$
Find coördinates of point C lying on both the planes such that the perimeter of △ABC is minimum.
My Attempt: Had the question been of $2D$-geometry. I would have easily calculated using the concept of reflection.
Here we have to rotate one of the planes say $x=y+z=1$ so that it lies on the other plane. Then we need to find cooresponding co-ordinates of the given point $A(3,-1,-1)$ say $A'$on the rotated plane(which is the plane $2x-y-z=5)$.
Now ,we need to find equation of line joining $A'$ and $B$ and solve it with equation of line of intersection of the two given planes(which happens to be $\frac{x-2}{0}=\frac{y+1}{1}=\frac{z}{-1}$).
So please tell me how to find $A'$.
Minkowski In equality does give the answer but I am more interested in the Reflection principle.
How is Minkowski inequality derived.
|
The third point is
$C(2\mid t\mid -1-t).$
Using the distance formula, calculate $\overline{AC}\text{ then }\overline{BC}.$
Their sum is a function in t. $\sqrt{2+2t+2t^2}+\sqrt{3+2t^2}=y$
Calculate $\frac{dy}{dt}=\frac{1+2t}{\sqrt{2+2t+2t^2}}+\frac{2t}{\sqrt{3+2t^2}}=0$
$t=-\frac{2-\sqrt2}2$
$C\left(2\mid-\frac{2-\sqrt2}2\mid-\frac{\sqrt2}2\right)$
|
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|
Intuition behind $\zeta(-1)$ = $\frac{-1}{12}$ When I first watched numberphile's 1+2+3+... = $\frac{-1}{12}$ I thought the sum actually equalled $\frac{-1}{12}$ without really understanding it.
Recently I read some wolframalpha pages and watched some videos and now I understand (I think), that $\frac{-1}{12}$ is just an associative value to the sum of all natural numbers when you analytically continue the riemann-zeta function. 3Blue1Brown's video really helped. What I don't really understand is why it gives the value $\frac{-1}{12}$ specifically. The value $\frac{-1}{12}$ seems arbitrary to me and I don't see any connection to the sum of all natural numbers. Is there any intuition behind why you get $\frac{-1}{12}$ when analytically continue the zeta function at $\zeta(-1)$?
EDIT(just to make my question a little clearer):
I'll use an example here. Suppose you somehow didn't know about radians and never associated trig functions like sine to $\pi$ but you knew about maclaurin expansion. By plugging in x=$\pi$ to the series expansion of sine, you would get sine($\pi$) = 0. You might have understood the process in which you get the value 0, the maclaurin expansion, but you wouldn't really know the intuition behind this connection between $\pi$ and trig functions, namely the unit circle, which is essential in almost every branch of number theory.
Back to this question, I understand the analytic continuation of the zeta function and its continued form for $s < 0$ $$\zeta(s)=2^s\pi^{s-1}\sin\frac{\pi s}2\Gamma(1-s)\zeta(1-s)$$ and how when you plug in s = -1, things simplify down to $\frac{-1}{12}$ but I don't see any connection between the fraction and the infinite sum. I'm sure there is a beautiful connection between them, like the one between trig functions and $\pi$, but couldn't find any useful resources on the internet. Hope this clarified things.
|
The values of $\zeta$ for negative integers can be directly calculated from the Bernoulli numbers, from:
$$\zeta(-n)=(-1)^n\frac {B_{n+1}}{n+1}$$
and $B_2=\dfrac 16$.
|
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|
Game Theory- altruistic preferences How would I answer this question?
John is an altruistic person who cares about his own income as well as the income of his best
friend Ryan. The value he attaches to each unit of his own income is the same as the value of
3 units of Ryan’s income. For example, he is indifferent between a situation where his income
is 1 and Ryan’s is 0 and one where his income is 0 and Ryan’s is 3. How would John rank
the following outcomes (2, 10), (3, 10), (5, 7) and (5, 20), where the first component is John’s
income and the second is Ryan’s? Provide a utility function consistent with these preferences.
Thanks in advance.
|
Your question steps out of game theory at the moment you say "John is an altruistic person"
However, I think you can make your way back to game theory by rewriting an "altruistic income". Take the usual income, and then recalculate the one of John by applying the rule of your choice (in that case, 3 times original income plus other person's income). Then you have a game in the classical sense (with selfish players) that takes into account John's altruism.
|
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|
Can we find a subfield $F$ of $\mathbb C$, such that ${ F}^{*p}={ F}^{*}$ for any prime $p\neq 5$ and ${F}^{*5}\neq F^{*}$? We know for field $\mathbb R$, ${\mathbb R}^{*p}={\mathbb R}^{*}$, for any prime $p\neq 2$ and ${\mathbb R}^{*2}\neq {\mathbb R}^{*}$. (${\mathbb R}^{*}$ is the multiplicative group of $\mathbb R$.) Can we find a subfield $F$ of $\mathbb C$ such that ${ F}^{*p}={ F}^{*}$ for any prime $p\neq 5$ and ${F}^{*5}\neq F^{*}$?
|
Here's a possible construction.
If $F$ is a countable subfield of $\mathbb{C}$, we can form the field $F'$ by splitting all polynomials of the form $X^p-a$ for $p\neq 5$ a prime and $a\in F$. There are countably many such polynomials, so order them in some way and split them successively.
Now, iterate this procedure. Starting with $F_0=\mathbb{Q}$, define $F_{n+1}=F_n'$ and $F_\infty = \cup F_n$. Then $F_\infty$ is a subfield of $\mathbb{C}$, and any of its elements contains all of its $p$th roots for $p\neq 5$ a prime. (This is because any such element belongs to some $F_n$, so all its roots are already in $F_{n+1}$.)
I claim that no fifth root of $2$ is in $F_\infty$. (Thanks to Yuval Dor for this argument). In fact, let's even extend $F_\infty$ a little bit: in the first step we constructed $F_1$ by adding, among other things, all roots of unity except those of order 5. Let's throw in the roots of $X^5-1$ as well, so now $F_1$ actually contains all roots of unity of prime order.
No fifth root of $2$ can belong in this extended $F_1$. If any root did, then all of them would, since $F_1$ contains the 5th roots of unity. But the splitting field of $X^5-2$ is a non-abelian extension of $\mathbb{Q}$, so it cannot be contained in a cyclotomic field.
Finally, by Kummer theory, splitting polynomials of the form $X^p-a$ over a field that has the necessary roots of unity can only result in a trivial extension of an extension of order $p$. So the order $5$ can never show up, meaning $\sqrt[5]{2}$ can't appear.
|
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|
How to find the $f(3)=y$ with given points? If an quadratic equation $f(x)$ has max value $4$ when $x = 0$ and $f(-2)=0$, what is the value of $f(3) =$ ?
|
making the ansatz $$f(x)=ax^2+c$$ we get easily $$c=4$$ from $$f(-2)=0$$ we get $$a=-1$$ and our searched function is $$f(x)=-x^2+4$$
|
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|
Prove by tableau method - Axioms of equality Example:
Let $T^*$ be a theory with axioms of equality. Prove by tableau method that
(a) $T^* \vDash x=y \rightarrow y=x \qquad $ (symmetry of =)
(b) $T^* \vDash (x=y \, \land \,y=z) \rightarrow x=z \qquad$ (transitivity of =)
Hint: To show (a) apply the axiom of equality (iii) for $x_1=x,\; x_2=x,\; y_1=y,\; y_2=x$, to show (b) apply (iii) for $x_1=x,\; x_2=y,\; y_1=z$
(a) Should the root of the tableau looks like this?
Then I use substitution and I get $F(y_1=x_1) -T(y_1=x_1)-\circ$
Then it is proved right?
|
You have to use both equality axioms :
$\forall x \ (x=x)$ --- reflexivity
$\forall x \ \forall y \ (x=y) \to (\phi \to \phi')$ --- where $\phi'$ is obtained from $\phi$ by replacing $x$ in zero or more places by $y$ : replacement axiom.
Specifically, the instance of the replacement axiom needed is :
$\forall x \ \forall y \ ((x=y) \to [ (x=x) \to (y=x)])$.
The tableau must start with the axioms and the negation of the conclusion :
$\text F \ [ \ \forall x \ \forall y \ ((x=y) \to (y=x)) ]$.
Thus, you have first to "unpack" the leading quantifiers, correctly using $y_1$ and $x_1$ new, to get:
$\text F \ ((x_1=y_1) \to (y_1=x_1))$
and then :
$\text T \ (x_1=y_1)$
$\text F \ (y_1=x_1)$.
But this does not close.
1) $\text T \ \forall x \ (x=x)$ --- axiom
2) $\text T \ \forall x \ \forall y \ ((x=y) \to [ (x=x) \to (y=x)])$ --- axiom
3) $\text F \ \forall x \ \forall y \ ((x=y) \to (y=x))$ - negation of conclusion
4) $\text T \ (x_1=y_1)$ --- from 3)
5) $\text F \ (y_1=x_1)$ --- from 3)
6) $\text T \ (x_1=x_1)$ --- from 1)
7) $\text T \ ((x_1=y_1) \to [ (x_1=x_1) \to (y_1=x_1)])$ --- from 2).
Left branch:
8) $\text F \ (x_1=y_1)$ --- closes with 4).
Right branch :
9) $\text T \ ((x_1=x_1) \to (y_1=x_1))$.
Left branch :
10) $\text F \ (x_1=x_1)$ --- closes with 6).
Right branch :
11) $\text T \ (y_1=x_1)$ --- closes with 5).
|
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Find all positive integer solutions to the equation $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$ Find all positive integers to the equation $\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{c^2}$
Multiply both sides with $(abc)^2$ to get $(bc)^2 + (ac)^2 = (ab)^2$.
I then tried some pythagorean triples and nothing worked so I assumed that there arent any solutions. Thank you for your responses.
|
Rewrite the equation as
$$
\frac{c^2}{a^2}+\frac{c^2}{b^2}=1
$$
and set $c/a=x/z$, $c/b=y/z$, where $x$ and $y$ are coprime.
Then $(x,y,z)$ is a primitive Pythagorean triple, so, without loss of generality,
$$
x=u^2-v^2,\quad y=2uv,\quad z=u^2+v^2
$$
where $\gcd(u,v)=1$, one among $u$ and $v$ being even.
Thus
$$
\frac{c}{a}=\frac{u^2-v^2}{u^2+v^2},\quad
\frac{c}{b}=\frac{2uv}{u^2+v^2}
$$
Therefore, for some positive integers $h$ and $k$,
$$
c=h(u^2-v^2)=2kuv,\quad
a=h(u^2+v^2),\quad
b=k(u^2+v^2)
$$
Note however that $u^2-v^2$ and $2uv$ are coprime, so $h=2muv$ and $k=m(u^2-v^2)$, so
$$
a=2muv(u^2+v^2),\quad
b=m(u^4-v^4),\quad
c=2muv(u^2-v^2)
$$
It's easy to see that any choice of $u$ and $v$ as before and any $m$ produces a solution of the original equation.
|
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Analyzing a function with cube roots I am analyzing this function:
\begin{align}
f(x) = \sqrt[3]{ x } - \sqrt[3]{x+1}
\end{align}
I proved it has a horizontal asymptote at y=0. However, on the graph, it has a minimum for x=-0.5 but I can't find that. First derivative has no zeroes, so the function has no stacionary points. Second derivative also has no zeroes. How do I find that minimum?
|
You must've taken the first derivative wrong:
$$f'(x)=\frac13(x^{-2/3}-(x+1)^{-2/3})$$
Setting $f'(x)=0$,
$$0=x^{-2/3}-(x+1)^{-2/3}$$
$$x^{-2/3}=(x+1)^{-2/3}$$
$$\pm x=x+1$$
$$x=-1/2$$
|
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Direct proof of the level four eta identity Of Somos' over 6200 eta identities, the one of smallest level (maybe the simplest?) is the eta identity of level $4$ $$\eta(z)^{16} \eta(4z)^8 + 16 \eta(4z)^{16} \eta(z)^8 = \eta(2z)^{24}.$$
As with all identities among modular forms, we can prove it by checking the first few coefficients. I wonder if it is possible to give a "real" proof of this straight from the product $$\eta(z) = q^{1/24} \prod_{n=1}^{\infty} (1 - q^n)$$ in the style of this answer. The higher exponents in this problem seem to make it harder to use Jacobi's triple product.
|
I am surprised at the idea that this needs "real" proof. The identity is the very first one in my list and the very first formulation of it is the "aequatio identica satis abstrusa" straight from Jacobi's Fundamenta Nova Theoriae Functionum Ellipticarum, page 90 equation (14) published in 1829. There is a reference to Bruce Berndt's editing of Ramanujan's Notebooks, Part III, page 40, Entry 25 (vii) which is exactly the same that is used in the parametrization of the Arithmetico-Geometric Mean by squares of the theta null functions. That is, $\;\theta_3^4(q) = \theta_4^4(q) + \theta_2^4(q)\;$ is the $a^2=b^2+c^2$ of the AGM and this goes back to Gauss even before Jacobi.
|
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|
Normalizer of upper triangular group in ${\rm GL}(n,F)$ The following question has already appeared on mathstack:
If $B$ is the subgroup of ${\rm GL}(n,F)$ consisting of upper triangular matrices then normalizer of $B$ in ${\rm GL}(n,F)$ is $B$ itself.
I know a proof of this using Bruhat decomposition of ${\rm GL}(n,F)$.
Question: Can we prove above theorem without using Bruhat decomposition?
Why came to this question: Consider the general linear Lie algebra $L=\mathfrak{gl}(n,F)$; in it, let $T=\mathfrak{t}(n,F)$ be the upper triangular sub-algebra. Then normalizer of $T$ in $L$ is $T$ itslef, and this can be proved just by considering a very simple decomposition of ${\mathfrak gl}(n,F)$: write any element as sum of upper triangular matrix and lower triangular matrix whose diagonal is $0$.
But then for problem above, is it necessary to use Bruhat decomposition?
|
$\DeclareMathOperator{\GL}{GL}$$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$$\newcommand{\Set}[1]{\left\{ #1 \right\}}$Let $e_0, e_1, \dots, e_{n-1}$ be a basis with respect to which $B$ is upper-triangular, and write
$$
V_i = \Span{ e_j : j \ge i}.
$$
Allow me to use row vectors, so that the group $G = \GL(n, F)$ acts on the right.
Then
$$
B = \Set{b \in G : V_i b \subseteq V_i \text{ for each $i$}}.
$$
Moreover,
$V_{n-i}$ is the unique subspace $W$ of dimension $i$ such that $W B \subseteq W$.
This is proved by induction on $i$. For $i = 1$ we have that $V_{n-1}$ is the unique common eigenspace for the elements of $B$ (just consider the element of $B$ which is a single Jordan block of size $n \times n$ and eigenvalue $1$, say), then pass to $V / V_{n-1}$ and use induction.
Let $g \in N_{G}(B)$. Then for each $b \in B$ we have $g b g^{-1} \in B$, that is for all $i$
$$
V_{i} g b g^{-1} \subseteq V_i
$$
or
$$
(V_{i} g) b \subseteq V_i g.
$$
It follows from the above that $V_{i} g = V_{i}$, so that $g \in B$.
|
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|
How to integrate $\tan(x) \tan(2x) \tan(3x)$? We have to integrate the following
$$\int \tan x \tan 2x \tan 3x \,\mathrm dx$$
In this I tried as
First split it into sine and cosine terms
Then used $\sin 2x =2\sin x \cos x$
But after that got stuck
|
Notice
$$\tan3x=\frac{\tan2x+\tan x}{1-\tan2x\tan x}\Rightarrow \tan3x\tan2x\tan x=\tan3x-\tan2x-\tan x$$
then
$$\int \tan3x\tan2x\tan x \mathrm{d}x = \int \tan3x - \tan2x - \tan x \ \mathrm{d}x$$
the answer will follow.
|
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Followup question on first-order formulas in first-order languages and open formulas. This is a followup question to my question here. I will reproduce the contents of my original question as follows.
For any first-order formula $X$ in the first-order language $\langle 0, S, \le\rangle$ (possibly with free variables) does there necessarily exist another open formula $Y$ such that the equivalence $X \equiv Y$ is true on the set of all nonnegative integer numbers?
Noah Schweber gives the following answer here, which I will reproduce as follows.
Yes, this is true. The structure $(\mathbb{N}; 0, S, \le)$ admits quantifier elimination, and the proof of this is via induction on the complexity of $X$. For an example of such a proof, see this, which goes through the proof of quantifier elimination for a version of Presburger arithmetic.
My new question is, does a similar statement hold for the language $\langle 0, \le\rangle$?
|
No, it does not. The set $\{1\}$ is definable in that structure - it is the set of $x$ such that $$x\not=0\wedge \forall y(y\le x\implies [y=0 \vee y=x]).$$ However, it's not hard to show that it is not definable by a quantifier-free formula (there aren't very many sets which are definable by a quantifier-free formula in this structure; see if you can find them all, and prove that your list is exhaustive!). Roughly speaking, the reason is that this language does not have enough terms.
Incidentally, note that the successor function is definable in this language ($S(x)=y$ iff $x\le y$ and $\neg y\le x$ and $\forall z[x\le z\wedge z\le y\implies x=z\vee z=y]$); combining this with the observation from the previous question that $(\mathbb{N}; 0, \le, S)$ does have quantifier elimination, we get two interesting facts:
*
*Definable expansions of theories without quantifier elimination might "gain" quantifier elimination. (By contrast, it's not hard to show that once you have quantifier elimination, you don't lose it by passing to a definable expansion.)
*Since the successor function is definable by a formula with one quantifier, every definable relation on $(\mathbb{N}; 0, \le)$ is definable by a Boolean combination of existential formulas; so, while we don't have full quantifier elimination, the quantifier hierarchy does collapse - just not right away.
|
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|
How to calculate $\lim\limits_{n \to \infty } \left(\frac{n+1}{n-1}\right)^{3n^{2}+1}$? I have $\lim_{n \to \infty } \biggr(\dfrac{n+1}{n-1}\biggr)^{3n^{2}+1}$
and I use this way: $\lim_{n \to \infty } \frac{n+1}{n-1}=|\frac{\infty}{\infty}|=\lim_{n \to \infty } \frac{1+\frac{1}{n}}{1-\frac{1}{n}}=1$ and $\lim_{n \to \infty } 3n^{2}+1=\infty$
Then $\lim_{n \to \infty } \frac{n+1}{n-1}^{3n^{2}+1}=|1^{\infty}|$
Continue using the formula $\lim_{n \to \infty } (1+\frac{1}{n})^{n}=e$, I decide to limit and get an answer: $\infty$, and it is right.
But for some reason I can imagine $\lim_{n \to \infty } \frac{n+1}{n-1}^{3n^{2}+1}$ how $\lim_{n \to \infty } \frac{n+1}{n-1}=|\frac{\infty}{\infty}|=\lim_{n \to \infty } \frac{1+\frac{1}{n}}{1-\frac{1}{n}}=1$ and $\lim_{n \to \infty } 3n^{2}+1=\infty$ ?
I can't find such a property of limits.
|
To apply your formula, notice that
$$\left(\frac{n+1}{n-1}\right)^{3n^2+1}=\left(1+\frac2{n-1}\right)^{3n^2+1}$$
Let $u=\frac{n-1}2$ or $n=2u+1$,
$$=\left(1+\frac1u\right)^{12u^2+12u+4}=\left(\left(1+\frac1u\right)^u\right)^{12(u+1)}\left(1+\frac1u\right)^4$$
$$\to e^{12(u+1)}\to e^\infty=\infty$$
Thus, the limit is infinite.
|
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|
Maximum value of $|x-y|$ Given $x,y\in\mathbb R$ such that
$$5x^2+5y^2-6xy\ =\ 8$$
find the maximum value of $|x-y|$.
My attempt
$5x^2 - 6yx + (5y^2-8)\ =\ 0$
$x\ =\ \dfrac{6y\pm\sqrt{(6y)^2-4(5)(5y^2-8)}}{10} = \dfrac{6y\pm\sqrt{160-64y^2}}{10} = \dfrac{3y\pm2\sqrt{10-4y^2}}{10}$
$5y^2 - 6xy + (5x^2-8)\ =\ 0$
$y\ =\ \dfrac{3x\pm2\sqrt{10-4x^2}}{10}$
$|x-y|\ =\ \dfrac{\left|3(y-x)\pm2\sqrt{10-4y^2}\mp2\sqrt{10-4x^2}\right|}{10}$
What do I do from here?
|
You tried to solve for both $x$ in terms of $y$ and $y$ in terms of $x$.
Try expressing everything as either a function of $x$ or of $y$.
|
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|
Calculate the center point between three pointsTh I have three positions, with an $X$ and $Y$ value and a value that indicates the distance to of each of the individual points (not in the same measurement). So to clarify my question:
I have three $XY$-positions and three values indicating the distance to the point (not in the same measurement). The lower the value is, the closer it is to the point.
For example:
Point $1(x:5,y:8,$strength$:43)$
Point $2(x:22,y:23,$strength$:21)$
Point $3(x:26,y:52,$strength$:132)$
I don't know if the values are actually usable/make sense as I made them up just now...
The main goal is to calculate a point in the "center" of the three points, though it can be outside
Picture to explain it further
|
Weighted ( by reciprocal strength $s$) average for each vertex
$$ \bar{ x} = \dfrac {x_1 s_1 + x_2 s_2 +x_3 s_3}{s_1+s_2+s_3} $$
$$ \bar{ y} = \dfrac {y_1 s_1 + y_2 s_2 +y_3 s_3}{s_1+s_2+s_3} .$$
where $$ s_i= \dfrac{1}{Strength_i} $$
|
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|
Prove $\int_{B(x,r)}|\nabla u|^2\leq \frac{C}{r^2}\int_{B(x,2r)}|u|^2$ if $-\Delta u(x)+f(x)u(x)=0.$ Let $\Omega $ a smooth domain of $\mathbb R^d$ ($d\geq 2$), $f\in \mathcal C(\overline{\Omega })$. Let $u\in \mathcal C^2(\overline{\Omega })$ solution of $$-\Delta u(x)+f(x)u(x)=0\ \ in\ \ \Omega .$$
Assume that $f(x)\geq 0$ for $x\in \Omega $. Prove that $$\int_{B(x,r)}|\nabla u|^2\leq \frac{C}{r^2}\int_{B(x,2r)}|u|^2,$$
for all $x\in \Omega $ and $r>0$, $B(x,2r)\subset \subset \Omega $ for somme $C\geq 0$ independant of $u,f,x$ and $r$.
My attempts
Using divergence theorem and that $\Delta u=fu$ in $\Omega $, I have that $$\int_{B(x,r)}|\nabla u|^2=\int_{B(x,r)}\text{div}(u\nabla u)-\int_{B(x,r)}u\Delta u=\int_{\partial B(x,r)}u\nabla u\cdot \nu-\int_{B(x,r)}fu^2.$$
But I can't do better. Any help would be welcome.
|
Suppose that $\varphi : \bar{B}(x,2r) \to [0,\infty)$ is Lipschitz and $\varphi =0$ on $\partial B(x,2r)$. Note that since $\varphi$ is Lipschitz, it is differentiable almost everywhere by Rademacher's theorem. Then
$$
0 \le \int_{B(x,2r)} \varphi^2 f u^2 = \int_{B(x,2r)} \varphi^2 u \Delta u = \int_{B(x,2r)} - \nabla (\varphi^2 u) \cdot \nabla u
$$
since $\varphi$ vanishes on $\partial B(x,2r)$. Since
$$
\nabla(\varphi^2 u) = 2 \varphi \nabla \varphi u + \varphi^2 \nabla u
$$
we find, upon plugging in above, that
$$
0 \le -\int_{B(x,2r)}2 \varphi u \nabla \varphi \cdot \nabla u + \varphi^2 |\nabla u|^2,
$$
and so
$$
\int_{B(x,2r)} \varphi^2 |\nabla u|^2 \le \int_{B(x,2r)} -2 \varphi u \nabla \varphi \cdot \nabla u \le 2 \left(\int_{B(x,2r)} \varphi^2 |\nabla u|^2\right)^{1/2} \left(\int_{B(x,2r)} u^2 |\nabla \varphi|^2 \right)^{1/2}.
$$
From this we then see that
$$
\int_{B(x,2r)} \varphi^2 |\nabla u|^2 \le 4 \int_{B(x,2r)} u^2 |\nabla \varphi|^2.
$$
With the last inequality in hand we can prove the result. Set
$$
\varphi(y) =
\begin{cases}
1 & \text{if } |x-y| \le r \\
2 - |x-y|/r & \text{if } r < |x-y| \le 2r.
\end{cases}
$$
It's easy to see that $\varphi \ge 0$, $\varphi$ vanishes on the boundary, and $\varphi$ is Lipschitz. Also
$$
|\nabla \varphi(y)| =
\begin{cases}
0 & \text{if } |x-y| \le r \\
1/r & \text{if } r < |x-y| \le 2r.
\end{cases}
$$
Plugging this in above then gives the inequality
$$
\int_{B(x,r)} |\nabla u|^2 \le \int_{B(x,2r)} \varphi^2 |\nabla u|^2 \le 4 \int_{B(x,2r)} u^2 |\nabla \varphi|^2 = \frac{4}{r^2} \int_{B(x,2r) \backslash B(x,r)} u^2.
$$
This is actually stronger than the desired inequality.
|
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|
Solving the integral of a function with modulus I would like to solve the following:
$$\int e^{-|x|}dx$$
The integral $\int e^{-x}dx = -e^{-x} +C$ is trivial, but the one with the modulus on $x$ seems to have a catch, as typing it into the integral calculator re-writes the integral and gives:
$$\frac{x}{|x|}\int e^{-|x|}\frac{|x|}{x}dx =-\frac{xe^{-|x|}}{|x|}$$
Why was the integral re-written like that? What would a general solution to future problems like this be?
|
This integral $$\int e^{-|x|}dx$$ is an even integral, such that we could define it as the amalgam of two integrals, each defined in a specific domain.
$$I_{1}=\int e^{-x}dx, x\geq 0 $$
$$I_{2}=\int e^{x}dx, x\leq 0 $$
As the question states, both these integrals are trivial.
$$I_{1}=- e^{-x} , x\geq 0 $$
$$I_{2}=e^{x}, x\leq 0 $$
We would like to write these two solutions as a single solution to the original integral. For this purpose, we need to appeal to some notation.
To begin with, we remember that, since we are on the Euclidean metric in one dimension, the magnitude the variable $x$ and the variable itself always have the same weight, such that
$$|x|= x , x\geq 0$$
$$|x|= -x , x\leq 0$$
In fact, we mae use of this very fact in order to establish that the integral was even.
But this property of the Euclidean Norm on the 1-D space means that
$$\frac{x}{|x|}= 1 , x> 0$$
$$\frac{x}{|x|}= 1 , x< 0$$
We then also note that
$$\lim_{x\rightarrow 0}\frac{x}{|x|}= 1$$
We can now return to the two integrals $I_{1}$ and $I_{2}$ and notice that they are both of the form
$$I = C e^{-|x|} $$
where $C = -1 $ when $x \geq 0$ and $ C = 1$ when $ x < 0 $. Thankfully, we already have a devise which has this property, so we employ is and claim that
$$I = -\frac{x}{|x|} e^{-|x|} +c_{1} $$
where we have also employed the arbitrary constant $c_{1}$ to highlight the fact that our integral is indefinite.
|
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|
Using the pigeon principle I was given the next question:
I need to prove that any subset with $25$ elements of the set $\{1,2,....150\}$, contain least $4$ elements, $x$, $y$, $z$, $t$ such that $x+y=z+t$.
Now my idea is to use the pigeon principle and set the pigeons to be the number of the possible pairings and the 'cages' to be the number of the possible sums of two elements from the set $\{1,2,....,150\}$
But I feel as if I'm missing something out and I'm not so sure about my answer.
I'd like to hear your thoughts on this, thank you very much!
|
Your idea is fine.
We have ${25\choose2}=300$ pigeons and as the sums can only range from $1+2=3$ to $149+150=299$, there are only $297$ cages. Hence there are two different pigeons $\{x,y\}$ and $\{z,t\}$ in the same cage, i.e., $x+y=z+t$. Note that the pairs cannot overlap as for example $y=z$ would imply $x=t$ contradicting the distinctiveness of the pigeons.
|
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|
Heine-Borel theorem. I'm interested in one question about Heine-Borel theorem.
We know that if S is bounded and closed then it's compact.
Standard proof using fact that some segment is compact. And because of S is bounded it's have left and right bound of segment. But my question is my S should be closed? We could take open , bounded S and some segment which cover out set. Why it's so necessary ?
|
Let $S=(0,1)$ and consider the cover $\{(1/n,1) : n \ge 2\}$. The union of any finite subset of the cover is of the form $(1/n,1)$ for some $n$, so no finite subset can cover $S$.
|
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|
Find largest dividing integer within range Is there a neat way to find the largest integer that divides another integer fully, within a range. As an example, I would like to find the largest integer smaller than 1131 that divides 3500 completely.
So far I have just tried by breaking up 3500 into its prime components and guessing, coming to 875, but is there a more structured way?
EDIT: I guess the problem is somewhat equivalent to get all dividing integers, and then just pick the largest within my range?
|
You could look for the smallest integer greater than $\dfrac{3500}{1131}\approx 3.09$ that divides $3500$ exactly
This is obviously $4$ (there is no smaller integer greater then $3.09$ and $4$ does divide $3500$ as it divides $100$) so the answer to your original question is $\dfrac{3500}{4}=875$
|
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|
Hyperbolic substitution for $\int\frac{dx}{x\sqrt{1-x^2}}$
Substitute $x=\sinh\theta$, $\cosh\theta$ or $\tanh\theta$. After integration change back to $x$.$$\int\frac{dx}{x\sqrt{1-x^2}}$$
Substituting $x=\tanh\theta$, we have
$$\begin{align}
\int\frac{dx}{x\sqrt{1-x^2}}&=\int\frac{\text{sech}^2\,\theta\,d\theta}{\tanh\theta\sqrt{1-\tanh^2\theta}}\\
&=\int\text{csch }\theta\,d\theta\\
&=-\ln|\text{csch }\theta+\text{coth }\theta|+C
\end{align}$$
Here comes the problem. We can substitute back $\text{coth }\theta=\frac1x$ but what do we have for $\text{csch }\theta$? We have $\text{csch}^2\,\theta=\coth^2\theta-1$, but neither $\sqrt{\coth^2\theta-1}$ nor $-\sqrt{\coth^2\theta-1}$ seems to be a one-off solution for $\text{csch }\theta$. How should we proceed?
|
For positive $x$, another substitution that does not involve integrating $\csc{x}$ or $\operatorname{csch}{x}$ is a hyperbolic secant substitution (similar to a secant/cosecant substitution for $x^2-1$), where $x=\operatorname{sech}{\theta}$, $dx=-\operatorname{sech}{\theta}\tanh{\theta}\,d\theta$, and $$\sqrt{1-x^2}=\sqrt{1-\operatorname{sech}^2{\theta}}=\sqrt{\tanh^2{\theta}}=\tanh{\theta}$$ since $\theta\geq0$ and $\tanh{x}$ is positive for that interval. $$\int{\frac{dx}{x\sqrt{1-x^2}}}=-\int{\frac{\operatorname{sech}{\theta}\tanh{\theta}}{\operatorname{sech}{\theta}\tanh{\theta}}d\theta}=-\int{d\theta}=-\theta+C$$ Substituting $x$ back gives $$-\operatorname{arsech}{x}+C\qquad x>0$$
For negative $x$, the substitution $x=-\operatorname{sech}{\theta}$ and $dx=\operatorname{sech}{\theta}\tanh{\theta}\,d\theta$ give $$-\operatorname{arsech}{\left(-x\right)}+C\qquad x<0$$ Combining the two branches gives $$-\operatorname{arsech}{\left|x\right|}+C$$
|
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|
how to convert /understand /reason the summation conversion? How to interpret / reason / understand the summation conversion for the below equation ?
$$\sum_{m = 1}^\infty 2^{-2m} = \frac{1}{4}\sum_{m = 0}^\infty 4^{-m}$$
|
There are two steps here. First, pulling down the $-2$ in the exponent:
$$\sum_{m=1}^\infty 2^{-2m} = \sum_{m=1}^\infty (\frac{1}{4})^m$$
Next, shifting the lower index to zero (by replacing $m$ everywhere with $m+1$) and pulling out a factor of $\frac{1}{4}$:
$$\sum_{m=1}^\infty (\frac{1}{4})^m = \sum_{m=0}^\infty (\frac{1}{4})^{m+1}
=\frac{1}{4}\sum_{m=0}^\infty (\frac{1}{4})^m=\frac{1}{4}\sum_{m=0}^\infty 4^{-m}$$
|
{
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|
initial conditions, pde Together with a differential equation and initial conditions arise the notion of a well-posed problem. My question is, how does one choose the type of initial conditions to be implemented in the problem or is it due to the nature of the differential equation? Any reference to this answer will be welcomed.
|
To be specific, let us consider the differential equation $x''(t)+x(t)=0$ for the SHM; for which in order to start oscillations you need to release the mass bob (tied to a thin thread with other end clamped) from extreme left i.e. $x(0)=-a$. Moreover, assuming that no external force is provided at the start i.e. $x'(0)=0$.
Edit: In case of PDE, considering the well known example of wave-equation is useful.
$$\frac{1}{c^2}\frac{\delta^2y}{\delta t^2}=\frac{\delta^2y}{\delta x^2}$$
where $c$ is a constant, $y(x,t)$ is the displacement along $Y-$ axis at position $x$ on $X-$ axis and time $t$. The initial conditions are as follows:
$1$. No displacement at the ends of string (of length l) at any time $t$ i.e. $y(0,t)=y(l,t)=0$.
$2$. No initial velocity at any point $x$ in the beginning $t=0$ i.e. $\frac{\delta y}{\delta t}(x,0)=0$
$3$. To initiate the vibrations in the string, suppose the string is stretched in the form of curve $y=f(x)$ i.e. $y(x,0)=f(x)$.
|
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|
If $f(x+1)+f(x-1)= \sqrt{2}\cdot f(x)$, find the period of $f(x)$. If $f(x+1)+f(x-1)= \sqrt{2}\cdot f(x)$, then the period of $f(x)$ is?
I tried replacing $x$ with $x-1$ and stuff, but didn't lead me to anything...
|
Note that $\sqrt{2} = \frac{2}{\sqrt{2}}$. Hence,
$$f(x + 1) + f(x - 1) = \frac{1}{\sqrt{2}}f(x) + \frac{1}{\sqrt{2}}f(x)$$
$$\begin{align}f(x + 1) &= \frac{1}{\sqrt{2}}f(x) + \frac{1}{\sqrt{2}}\left(f(x) - \sqrt{2}f(x - 1)\right)\\
&=\frac{1}{\sqrt{2}}f(x) + \frac{1}{\sqrt{2}}\left(-f(x - 2)\right)\\
&= \frac{1}{\sqrt{2}}\left(f(x) - f(x - 2)\right)\end{align}$$
$$\sqrt{2}f(x + 1) = f(x) - f(x - 2)$$
$$\sqrt{2}f(x) = f(x - 1) - f(x - 3)$$
Hence,
$$f(x - 1) - f(x - 3) = f(x + 1) + f(x - 1)$$
$$f(x + 1) = -f(x - 3)$$
$$f(x) = -f(x - 4)$$
Therefore,
$$f(x) = -f(x - 4) = -(-f(x - 8)) = f(x - 8)$$
thus the period is $8$.
|
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|
Prove that triangle is isosceles, triangle that is inscribed in a circle. $AD$ is a median to side $BC$ in triangle $ABC$ that is inscribed in a circle that his center is in $(5,6)$.
Given: $ D (9,2)$ and the centroid of the triangle is $(6,5$).
There is not so much information given about the circle, thus I can't draw very good the circle (for example I don't know points in the circle, a point of the triangle or radius of the circle). There is my problem. I feel I have few given information and don't know how to start.
Can someone help me how to prove that ABC is isosceles?
|
Sketch of the proof
Let $G$ be the centroid and $O$ the circumcentre. We know that $A$, $G$ and $D$ are collinear since they are in the median to the side $BC$.
We know that the line through $O$ and $D$ is the perpendicular bisector of the side $BC$.
Since $D$ $G$ and $O$ are collinear (the three points are in the line $y = -x + 11$), then the median to the side $BC$ and the perpendicular bisector of the side $BC$ must collide.
That means that $\angle{ADB} = \angle{ADC} = \pi/2$. One can deduce that the triangles $ADB$ and $ADC$ are congruent and so $\overline{AB} = \overline{AC}$. This completes the proof.
|
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|
Using big Oh notation. How to show that $O(x^{1/2} e^{ (log (\frac{x}{c}))^{1/2}})=O(x^{1/2+\epsilon})$ for any $\epsilon>0$?
|
Let $f(x)=e^{(\log(x/c))^{1/2}}=e^{\sqrt{\log(x/c)}}$
Then we have $f(x)^{\sqrt{\log(x/c)}}=e^{(\log(x/c))}=x/c$
So we can see that $f(x)^{\sqrt{\log(x/c)}}=x/c$ which means $f(x)=(x/c)^{\frac{1}{\sqrt{\log(x/c)}}}<x^{\epsilon}$ since $\frac{1}{\sqrt{\log(x/c)}}\rightarrow 0$.
Combining all together yields $x^{1/2}e^{(\log(x/c))^{1/2}}<x^{1/2+\epsilon}$
which gives immediately what yu want.
|
{
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|
If $n={P_1}^{r_1}.{P_2}^{r_2}.\dots.{P_k}^{r_k}$ then $f(n)={r_1}^{P_1}.{r_2}^{P_2}.\dots.{r_k}^{P_k}$then which one is true? If $n={P_1}^{r_1}.{P_2}^{r_2}. \dots .{P_k}^{r_k}$ then $f(n)={r_1}^{P_1}.{r_2}^{P_2}.\dots.{r_k}^{P_k}$then which one is true?
1.The function is one-to-one.
2.The function is surjective.
3.If there are $a,b$ that $f(a)=m$,$f(b)=n$ then there is a $c$ that $f(c)=mn$.
4.$f(m).f(n) \ge f(mn)$
5.$f(m).f(n) \le f(mn)$
My attempt:The first one is wrong because both $5^2*11^2$ and $3^2*13^2$ give $2^{16}$.The second one is also wrong because we cannot make primes.The last one is also wrong because consider $m=5^3*3^2$ and $n=5^2*2^3$ which will give $f(5^3*3^2).f(5^2*2^3) \le f(5^5*3^2*2^3)$ which gives $3^5*2^3*2^5*3^2\le 2^3*5^5*3^2$ which gives $6^5 \le 5^5$ which is wrong.But I am stuck in choosing between $3$ and $4$.
|
(4) is wrong. Note that primes $p$ are mapped to $1$, now let $p$ a prime and $m = n = p$. Then
$$ f(m)f(n) = f(p)f(p) = 1 \not\ge f(p^2) = 2^p $$
|
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Prove that the sequence $x_{n+1}=\frac{1}{2}\left(x_n+\frac{1}{x_n}\right)$ is not increasing Let $x_1\ge1$ and $x_{n+1}=\frac{1}{2}\left(x_n+\frac{1}{x_n}\right)$. Prove that the sequence $\{x_n\}$ is not increasing, that is, $x_{n+2}\le x_{n+1}$.
My attempt was as follows: we have to prove that $1\le \frac{x_{n+1}}{x_{n+2}}$. Now:
$$\frac{x_{n+1}}{x_{n+2}}=\frac{x_n+\frac{1}{x_n}}{x_{n+1}+\frac{1}{x_{n+1}}}=\frac{x_n+\frac{1}{x_n}}{\frac{1}{2}\left(x_n+\frac{1}{x_n}\right)+\frac{1}{\frac{1}{2}\left(x_n+\frac{1}{x_n}\right)}}$$
let $a=x_n+\frac{1}{x_n}$ for the sake of convenience, then:
$$ \frac{x_n+\frac{1}{x_n}}{\frac{1}{2}\left(x_n+\frac{1}{x_n}\right)+\frac{1}{\frac{1}{2}\left(x_n+\frac{1}{x_n}\right)}}=\frac{a}{\frac{1}{2}a+\frac{2}{a}}$$
so for the claim that $1\le \frac{x_{n+1}}{x_{n+2}}$to hold, we must have:
$$ 1\le\frac{a}{\frac{1}{2}a+\frac{2}{a}}\rightarrow \frac{1}{2}a+\frac{2}{a}\le a\rightarrow \frac{1}{2}+\frac{2}{a^2}\le 1\rightarrow a^2\ge 4\rightarrow a\ge 2\rightarrow x_n+\frac{1}{x_n}\ge 2$$
which follows by AM-GM.
My concern is, is this reasoning correct? I skipped some cases, assuming that we are dealing with positive numbers only - is my reasong rigorous? Or is there a serious flaw?
Either way, maybe there is a cleaner and perhaps more clever way to prove the claim?
|
The reasoning seems correct until the end when you write
$$x_n+\frac{1}{x_n}\geq 2$$ Now here you should prove that $$x_n+\frac{1}{x_n}\geq 2$$
Which is trivial by AM-GM.
Here's a similar approach note that by AM-GM
$$x_{n+1}=\frac{1}{2}(x_n+\frac{1}{x_n})\geq 1$$
hence we have that $x_n\geq 1$.
Now $$\frac{x_n}{x_{n+1}}=\frac{2x_n}{x_n+\frac{1}{x_n}}$$
But we have that $x_n\geq\frac{1}{x_n}$ because $x_n\geq 1$ that means $$\frac{x_n}{x_{n+1}}=\frac{2x_n}{x_n+\frac{1}{x_n}}\leq 1$$
|
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|
Compactness of the open and closed unit intervals In the article by Tao it's explained that the compactness can be formulated in the most general way as:
(All open covers have finite subcovers) If $`V_\alpha:\alpha\in\mathcal{a}`$ is any collection of open sets which covers $X$, then there must exist a finite sub-collection $V_{\alpha_1},V_{\alpha_2}...V_{\alpha_k}$ of these sets which still cover $X$.
Question: How is it possible to show (both intuitively and rigorously) that such coverage by a finite collection of open sets is possible for $X=[0,1]$ and not possible for $X=(0,1)$?
Edit: a related question and another one on open covers.
|
showing it is not possible for $(0,1)$ is easy, just take the open cover $(0,1-\frac{1}{n})$.
Showing it is possible for $[0,1]$ is not so easy, you may want to look here for proofs that do not go through the whole Heine-Borel theorem.
|
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When is a spherical polygon convex suppose we have a polygon on the sphere $\mathbb{S}^2$, that is, a domain with piecewise geodesic boundary. Is it true that this domain is convex if and only if all angles of the polygon are in $(0,\pi)$?
A subset $U\subset\mathbb{S}^2$ is called convex if for any pair of points $p,q\in U$ there exist a geodesic from p to q which is contained in $U$. (For example a hemisphere is convex.)
For polygons in the euclidean plane this is known to be true. But I do not know the answer for the sphere.
Best wishes
|
Is it true that this domain is convex if and only if all angles of the polygon are in $(0,π)$?
No, according to your definitions, this is not true. Take three quarter slices of a sphere. That gives you an inner angle of $\frac34\cdot2\pi>\pi$, but for every pair of points the plane spanned by these two points and the center of the sphere will intersect the sphere in a circle with some continuous portion missing. There is still a continouous portion which will contain a connecting geodesic, though.
|
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Is there anything better than an exact equivalence? It's a basic principle of category theory that any "good" concept should be preserved by equivalence of categories. However, since Abelian categories are by definition $\mathbf{Ab}$-enriched, we require a bit more structure for our functors – we almost always like them to be additive, and it should preserve at least some form of exactness.
But are there any concepts in Abelian categories that are not preserved by exact equivalences? Or is there some way of proving that any good concept will be preserved by exact equivalences?
On another hand, is it enough to require the functor to be simply an equivalence, in order to guarantee that it is additive and exact?
|
To see that any equivalence of abelian categories is additive and exact, use the following observations:
*
*If $F\colon \mathcal{A} \to \mathcal{B}$ and $G\colon \mathcal{B}\to \mathcal{A}$ give an equivalence, then $F$ is both left- and right-adjoint to $G$.
*Any left adjoint functor is right exact (as it preserves all small colimits), and any right adjoint functor is left exact (as it preserves all small limits).
*In particular, any left/right adjoint functor preserves binary coproducts/products (which coincide in an additive category), hence it is additive.
|
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|
Why can't we separate fraction powers in euler's formula? I am sure there is something wrong i am doing but i just want someone to point it to me.
Why can't we say that $e^{\frac {\pi i}{3}} = \left(e^{\pi i}\right)^{\frac {1}{3}} = (-1)^{\frac {1}{3}} = -1$
Why do we have to calculate it by the formula as it is (which produces different value)
Another question is why can't we say that $ (-1)^{\frac {1}{2}} = \left(\left (-1\right)^{2}\right)^{\frac {1}{4}} = 1^{\frac {1}{4}} = 1$
Again i have to say that i know this is all wrong i just want to know why it is wrong
|
It is simply the case that $(a^b)^c$ is not always equal to $a^{bc}$ if $a$ is not positive or if $b$ and/or $c$ are complex.
The problems here are that
$$x=(-1)^{1/2}\implies x^2=-1\implies x^4=1$$
However, when you say that $1^{1/4}=1$, you cause a misconception. Usually, this is perfectly fine, but when you think about it, why not $1^{1/4}=-1$? Indeed, both values are solutions to the equation $x^4=1$, but neither are equal to $(-1)^{1/2}$. However, upon factoring, you could see that
$$x^4=1\implies x^4-1=0\implies(x+1)(x-1)(x+i)(x-i)=0$$
One of these is the correct solution in our context, though you happen to choose the wrong one.
Similarly, you state that $(-1)^{1/3}=-1$, but
$$x^3=-1\implies x^3+1=0\implies(x+1)(x^2-x+1)=0$$
And $x=e^{\pi i/3}$ is one such possible solution. If you want more information, please see the link above.
|
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Finding the coefficient of $x^m$ in $(x+x^2+x^3+.....+x^i)^n$ EDIT:
My doubt may be silly but if the expansion of $(1-x)^{-n}$ is infinite, how come wolfram displayed 11 terms?
Wolfram result
Can anybody help to solve the whole equation?
I'm trying to solve this question.
I have an equation $(x+x^2+x^3+x^4+x^5+x^6)^2$ [ Taken 6-sided die x 2].
I want to find the coefficient of $x^5$.
I have reduced the equation to $x^2(1-x^6)^2(1-x)^{-2}$.
For $(1-x)^{-2}$,
$$(1-x)^{-2}=\sum_{k\ge0}(-1)^k\binom{-2}{k}x^k=\sum_{k\ge0}\binom{2+k-1}{k}x^k $$
I got -> $1+2x+3x^2$
For $(1-x^6)^2$
$$(1-x^6)^2=\sum_{k\ge0}(-1)^k\binom{2}{k}x^{6k}$$
I got -> $1-2x^6+x^{12}$
So my equation will be$$x^2(1+2x+3x^2)(1-2x^6+x^{12})$$
Am I correct till here? Because when I multiply all the terms I'm getting
$$x^2+2x^3+3x^4-2x^8-4x^9-6x^{10}+x^{14}+2x^{15}+3x^{16}$$
Where am I going wrong?
|
This GitHub repository solved my problem.
|
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Evaluate $\int_0^\infty \frac{\ln\left ( 2x+1 \right )}{x\left ( x+1 \right )}\mathrm dx$ How to evaluate
$$\int_{0}^{\infty }\frac{\ln(2x+1)}{x(x+1)}\,\mathrm dx?$$
I tried
$$\frac{\ln(2x+1)}{x(x+1)}=\ln(2x+1)\left (\frac{1}{x}-\frac{1}{1+x} \right)$$
but I don't know how to go on.
|
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\int_{0}^{\infty}{\ln\pars{2x + 1} \over x\pars{x + 1}}\,\dd x &
\,\,\,\stackrel{2x + 1\ \mapsto\ x}{=}\,\,\,
\int_{1}^{\infty}{\ln\pars{x} \over
\bracks{\pars{x - 1}/2}\bracks{\pars{x - 1}/2 + 1}}\,{\dd x \over 2} =
2\int_{1}^{\infty}{\ln\pars{x} \over x^{2} - 1}\,\dd x
\\[5mm] & \stackrel{x\ \mapsto\ 1/x}{=}\,\,\,
-2\int_{0}^{1}{\ln\pars{x} \over 1 - x^{2}}\,\dd x
\,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\,
-\,{1 \over 2}\int_{0}^{1}{x^{-1/2}\,\ln\pars{x} \over 1 - x}\,\dd x
\\[5mm] & =
\left.{1 \over 2}\,\totald{}{\mu}\int_{0}^{1}{1 - x^{\mu} \over 1 - x}\,\dd x\,
\right\vert_{\ \mu\ =\ -1/2} =
\left.{1 \over 2}\,\totald{\bracks{\Psi\pars{\mu + 1} + \gamma}}{\mu}
\right\vert_{\ \mu\ =\ -1/2}\label{1}\tag{1}
\end{align}
where $\ds{\Psi}$ is the $Digamma\ Function$ and $\gamma$ is the
$Euler\mbox{-}Mascheroni\ Constant$: See $\mathbf{6.3.22}$ in A&S Table.
From expression \eqref{1}:
$$
\int_{0}^{\infty}{\ln\pars{2x + 1} \over x\pars{x + 1}}\,\dd x =
{1 \over 2}\,\Psi\, '\pars{1 \over 2} = \bbx{\ds{\pi^{2} \over 4}}
$$
See $\mathbf{6.4.4}$ in
A&S Table.
|
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Basic proof that $GL(2, \mathbb{Z})$ is not nilpotent I need to show that $GL(2,\mathbb{Z})=\left\{ \begin{pmatrix} a&b \\ c&d \end{pmatrix} \mid a,b,c,d \in \mathbb{Z}, ad-bc = \pm 1\right\}$ is not nilpotent. I have seen this question but the answer given there is too advanced for where I am currently in my studies.
In order to show something is not nilpotent, all I have at my disposal is to show that the upper central series does not terminate or that it is not solvable (using a derived series that does not terminate or a subnormal series that either does not terminate or whose factors are not all abelian.)
This seems a little messy, and I've never used a direct approach to show that something was not nilpotent.
What is the best approach to this proof without appealing to $p$-groups, Sylow Groups, Galois Theory, or free abelian groups?
Thank you.
|
One possibility is to show that there is a proper, self-normalizing subgroup,
$$
\left\{ \begin{pmatrix} a&b \\ 0&d \end{pmatrix} : a, d = \pm 1, b \in \mathbb{Z}\right\}
$$
being perhaps the easiest candidate.
|
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Equation of a plane $Ax+By+Cz=D$ I have question abous parametrization and plane equation.How can I write the equation of a plane in the form $Ax+By+Cz=D$ if it is given as
$$x=2s+3t$$
$$y=s-2t$$
$$z=3s+t$$.
|
hint: $x - 2y = 7t, 3y - z = -7t \implies (x-2y) + (3y-z) = 0$
|
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Find the matrix $(X^{-1} A X)^n$ Given that $$A=\begin{bmatrix}1&1\\ 0&1\end{bmatrix}$$ $$X=\begin{bmatrix}2&1\\ 3&2\end{bmatrix}$$
I computed that $$X^{-1} A X = \begin{bmatrix}7&4\\ \:\:\:\:-9&-5\end{bmatrix}=Y$$
Now the usual way would be to find the eigenvalues and corresponding eigenvectors and then calculate the nth degree for this matrice, however it turns out that this matrix has only one eigenvalues and one eigenvector so I can't use that method.
Instead I thought of using the Cayley-Hammilton theorem and after finding the characteristic polynomial we have:
$Y^2=2Y-E$ which ultimately yields $Y^n = nY-(n-1)E$ is my approach correct?
|
Hint:
first prove, by induction, that
$$
A^n=\begin{bmatrix}
1&n\\0&1
\end{bmatrix}
$$
Then use the fact (easy to prove):
$$
(X^{-1}AX)^n=X^{-1}A^nX
$$
|
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Number of ordered triples with given condition We are given that there are $6250k$ ordered triples $(m,n,p)$ where $m,n,p$ are integers such that $$1≤m≤100,\;1≤n≤50,\;1≤p≤25\quad \& \quad 3\,|\,2^m+2^n+2^p$$ Find $k$.
With the given range of values of $m, n,p$, I guess the possible numbers of ordered triples without any conditions would be $25\times (50-1)\times (100-2)$, i.e., $25\times 49\times 98$. But after this I am unable to detect the relationship between m, n and p for the given divisibility condition.
Please help.
|
Given that $1≤m≤100,\;1≤n≤50,\;1≤p≤25$ and $ 3\,|\,2^m+2^n+2^p$ where $m,n,p$ are integers.
Now if $2^m+2^n+2^p$ is divisible by $3$, then each of $2^m$, $2^n$, $2^p$ is divided by $3$.
Remainder of ${2^m}$ when it is divisible by $3$ is $(-1)^m$.
So remainder of ${2^m+2^n+2^p}$ when it is divisible by $3$ is ${(-1)^m+(-1)^n+(-1)^p}$. Now if ${(-1)^m+(-1)^n+(-1)^p}$ is divisible by $3$, the we are done.
Case I: If $m$ is odd and $n$ and $p$ are even, then remainder of ${2^m+2^n+2^p}$ when it is divisible by $3$ is ${(-1)^m+(-1)^n+(-1)^p}={-1+1+1}={1}$ which is not divisible by $3$.
Case II: If $m$ is even and $n$ and $p$ are odd, then remainder of ${2^m+2^n+2^p}$ when it is divisible by $3$ is ${(-1)^m+(-1)^n+(-1)^p}={1-1-1}={-1}$ which is not divisible by $3$.
Case III: If $m$, $n$ and $p$ are even, then remainder of ${2^m+2^n+2^p}$ when it is divisible by $3$ is ${(-1)^m+(-1)^n+(-1)^p}={1+1+1}={3}$ which is divisible by $3$.
Case IV: If $m$, $n$ and $p$ are odd, then remainder of ${2^m+2^n+2^p}$ when it is divisible by $3$ is ${(-1)^m+(-1)^n+(-1)^p}={-1-1-1}={-3}$ which is divisible by $3$.
So we are interested in the last two cases.
Now $1≤m≤100,\;1≤n≤50,\;1≤p≤25$
In the given range, the number of possible choice that all $m, n, p$ are even is $50, 25, 12$ respectively and the number of possible choice that all $m, n, p$ are odd is $50, 25, 13$ respectively.
So the total number of ordered triples $(m,n,p)$ where $m,n,p$ are integers that $ 3\,|\,2^m+2^n+2^p$ is $(50 \times 25 \times 12)+(50 \times 25 \times 13)=50 \times 25 \times 25= 31250 = 5\times 6250$.
So the possible value of $k=5$ .
|
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Quaternion Algebras and isomorphisms I was reading a book on quaternion algebras and found the following exercise and was unable to do it and it has been really frustrating me,
Let $(a,b)_F$ be a quaternion algebra with $i^2=a$ , $j^2=b$ and $ij=-ji$
Prove that $(a,b)_F$ , $(b,a)_F$ and $(ac^2,b)_F$ are isomorphic and prove that $(a^2,b)_F$ is not a division algebra.
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Let's show that $(a,b)_F$ and $(b,a)_F$ are isomorphic. Let's say $(a,b)_F$ is generated by $i,j$ with
$$ i^2=a, \quad j^2=b, \quad ij=-ji $$
and $(b,a)_F$ is generated by $k$ and $\ell$ satisfying
$$ k^2=b, \quad \ell^2=a, \quad k\ell=-\ell k. $$
To define an isomorphism $\phi:(b,a)_F\to(a,b)_F$ it suffices to specify where $\phi$ sends the two generators $k$ and $\ell$. We must send $k$ and $\ell$ to two elements of $(a,b)_F$ where $\phi(k)$ squares to $b$, $\phi(\ell)$ squares to $a$, and $\phi(k)$ and $\phi(\ell)$ anticommute. Can you see two such elements of $(a,b)_F$ to send $k$ and $\ell$ to with $\phi$?
For showing $(a^2,b)_F$ is not a division algebra, try and find a nonzero element that is not invertible; one way this may occur is with zero divisors. Look at the new relation $i^2=a^2$. What can you do to this equation to exhibit zero divisors?
|
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A polynomial question. I came across this question in my weekly test paper and didn't know how to tackle it. The question goes as follows:
$a_0,a_1,a_2....a_{34}$ are the coefficients of $x^0,x^1,x^2......x^{34}$ of the polynomial obtained on opening the parenthesis of $(1+x+x^2)^{17}$, then which is true?
$1)~a_1+a_2+.......+ a_{34}$ is even;
$2)~a_1=17$;
$3)~a_{33}=17$;
$4)~a_2=153$.
(One or more is correct)
|
Hint: if you plug $x=1$ in, you will get the sum of the coefficients.
The coefficients may be represented using Viete's formulas.
|
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|
Find the sum of first $n$ terms of the series: $\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\cdots +\frac{1}{n\times(n+1)}$
I have the series $$\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\cdots +\frac{1}{n\times(n+1)}$$
I know the following formulas: $$1+2+3+\cdots +n=\frac {n (n+1)}{2}\tag1$$ $$1^2+2^2+3^2+\cdots +n^2=\frac{n(n+1)(2n+1)}{6}\tag2$$ $$1^3+2^3+3^3+\cdots +n^3=\left(\frac{n(n+1)}{2}\right)^2\tag3$$ But none of $(1)(2)$ and $(3)$ worked.
Please help___.
|
Try to observe that $$\frac{1}{n\times (n+1)}=\frac{1}{n}-\frac{1}{n+1}$$ $\therefore$ The given series can be written as $$1-\frac12+\frac12-\frac13+\frac13+\cdots -\frac{1}{n}+\frac1n-\frac{1}{n+1}$$ Each term will cancel out other term except $1$ and $\frac{1}{n+1}$ .
$\therefore$ $$=1-\frac1{n+1}$$ $$=\frac{n}{n+1}$$ Hope it helps!!!
|
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|
Difficulty solving problem regarding 2 parallel vectors I have been asked to solve the following question, and I am not sure where to begin with it :
Given two vectors $a$ and $b$ , where the magnitude of $b$ is double the
magnitude of $a$ and the angle between the two vectors is $60°$, we may
define two further vectors: $c = a − 2b$ and $d =14a − 2b$ .
Show that the vectors c and d are perpendicular to each other.
I am aware that $$a\cdot b = \vert a\vert\cdot \vert b\vert\cdot cos $$
However im not sure how to apply it in this question.
Any help in solving this question and the methodology to solve it would be greatly appreciated
|
$$c\cdot d=(a-2b)\cdot (14a-2b)=14|a|^2-30a\cdot b+4|b|^2$$
Use that
$$|b|=2|a| \quad (1)$$
and then
$$a\cdot b=|a||b|\cos 60º \rightarrow a\cdot b=2|a|^2\cdot1/2=|a|^2 \quad (2)$$
Remember that $c \perp d \Leftrightarrow c \cdot d=0$.
Using $(1)$ and $(2)$ we get:
$$c\cdot d=14|a|^2-30a\cdot b+4|b|^2=14|a|^2-30|a|^2+4\cdot 4|a|^2=0$$
|
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Find the numbers in A.P. those sum is $24$ and product is $440$
If the sum of three numbers in A.P. is $24$ and their product is $440$, find the numbers.
My Approach: Let the numbers be $a,a+d,a+2d$
So, according to question $$3a+3d=24$$ $$a+d=8$$ and $$a(a+d)(a+2d)=440$$ $$8a+ad=55$$ I can’t proceed from here.
Please help.
|
Let the numbers be : $a-d, a, a+d$
Then According To Question. $$a-d+a+a+d=24$$ $$a=8$$ and $$(a-d)(a)(a+d)=440$$ $$(8-d)(8)(8+d)=440$$ $$(8-d)(8+d)=55$$ $$64-d^2=55$$ $$-d^2=-9$$ $$d=\pm3$$ $\therefore$ the A.P. is $$5,8,11$$ $$**OR**$$ $$11,8,5$$
|
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How to factorize this $\sqrt{8 - 2\sqrt{7}}$? When I was at high school, our teacher showed us a technique to simplify square roots like this $\sqrt{8 - 2\sqrt{7}}$ that I forgot.
It was something like 8 = 7+1; 7 = 7*1; and using them we could represent $\sqrt{8 - 2\sqrt{7}}$ in simpler form. I would be happy if you can show how it works, and how this technique is called.
|
You could approach this by setting $\sqrt{8 - 2\sqrt{7}} = \sqrt a \pm \sqrt b$ for some $a,b$. Then, squaring both sides, we have:
$$8 - 2\sqrt{7} = a \pm 2\sqrt{ab} + b,$$
so that $8=a+b$ and $-2\sqrt7 = \pm 2\sqrt{ab}$.
In other words, the $\pm$ sign must be $-$, and we now have a system of two equations $\{8=a+b, 7=ab\}$ which is readily solved. Either $a=7$ and $b=1$ (which works), or $a=1$ and $b=7$ (which we reject because $1-\sqrt 7 < 0$).
|
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Covering compact subsets of plane with certain rectangles Suppose $\{F_n\}$ is sequence of compact subsets of the half-plane $\mathbb H=(0,+\infty)\times(-\infty,+\infty)$ with positive area such that $F_n\subset F_{n+1}$ and $\cup_n F_n=\mathbb H$. Do there exist a sequence of positive numbers $\{a_n\}$ and positive constants $\delta,\delta',\delta''$ such that
$$[\frac{1}{a_n},a_n]\times[0,a_n]\subset F_n\subset[\frac{\delta}{a_n},\delta'a_n]\times[-\delta'' a_n,\delta''a_n]$$
for $n>n_0$?
|
No. Let $c_n$ be a monotone increasing sequence. For convenience let $d_n = \frac{c_n}{c_n-1}$. Then set
$$F_n = \bigg( [c_n^{-1},c_n]\times[-c_n, c_n] - [d_n^{-1},d_n]\times[-d_n, d_n] \bigg) \cup \{1\}\times [-c_n, c_n] $$
Each $F_n$ is closed, bounded, so compact. It looks like a large rectangle with a smaller rectangle punched out and a line strung across the removed smaller rectangle. (The line there means that the $F_n$ indeed converge to $\Bbb{H}$.) As $n$ increases, the large rectangle lets larger and the removed rectangle gets smaller.
Because the inner rectangle you propose has $(1,\varepsilon)$ in its interior for small $\varepsilon$, whereas $(1,\varepsilon)$ is on the boundary of $F_n$ for each $n$, containment can never hold.
|
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Proving $ \sum_{n=1}^{\infty} nz^{n} = \frac{z}{(1-z)^2}$ for $z \in (-1, 1)$ I do not know where to start, any hints are welcome.
|
Without differentiation
$$\sum_{n=1}^\infty nz^{n-1} = 1+2z+3z^2+\cdots \tag 1$$
$$\sum_{n=1}^\infty nz^{n} = z+2z^2+3z^3+\cdots \tag 2$$
Subtract (2) from (1)
$$\sum_{n=1}^\infty nz^{n-1}(1-z) = 1+z+z^2+\cdots$$
$$\sum_{n=1}^\infty nz^{n}= \frac{z}{(1-z)^2}$$
|
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Limit I don't know how to start solving. This is the limit:
$$\lim_{n\to\infty}\frac{2^{n-1}-4^n}{3^n\sin n + 4^{n+2}}$$
I have a solution and the steps and I still haven't understood how it's done, here's the proposed solution:
$$\lim_{n\to\infty}\frac{2^{n-1}-4^n}{3^n\sin n + 4^{n+2}}=\lim_{n\to\infty}\frac{\frac12(\frac24)^n-1}{(\frac34)^n\sin n+16}=-\frac1{16}$$
WolframAlpha says this is correct but I haven't understood from where did all the fractions come from...
|
Some intuition for how to come up with this solution:
As $n$ gets very large, $4^n$ is much much bigger than $2^n$. A common trick to "prove" this, or tease this fact out, is to divide the numerator and denominator by the highest order term. This is legal since it amounts to multiplying by 1, and $4^n$ is never 0 for any $n$.
Doing this makes the result easier to see, as in your example. Once you divide by $4^n$ you end up with
$$
\frac{(\text{a bunch of stuff that goes to 0})-1}{(\text{a bunch of stuff that goes to 0})+4^{2}}\rightarrow-1/16
$$
as $n\rightarrow \infty$.
|
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Wolfram Alpha gives different answers for doing the same thing to both sides of this equation. I have the equation $10^x=2i$ (I'm doing this because I wanted to take the log of $2i$) I was trying to solve it by hand first, but I was running into errors, so I wanted to make sure when I square both sides of the equation, I wasn't doing anything wrong.
First I entered $10^x=2i$, I got
$$x = \frac {4 i π n + i π + 2 \log2}{2 \log2 + \log5},\ \ \ n\in\mathbb Z $$
(log is the natural logarithm in the solution above)
When I did $(10^x)^2=(2i)^2$, it said that is was equivalent to $10^{2x} = -4$ which was to be expected. However, it gave me the following output:
$$x = \frac {2i π n + i π + 2 \log2}{2 \log2 + \log5},\ \ \ n\in\mathbb Z $$
Is this a glitch in wolfram alpha? If so, why does this happen? It should give equivalent answers if I did the same thing to both sides of the equations.
If this isn't a glitch, please explain why it isn't a glitch, as I don't understand why this wouldn't be one.
|
It is not WolframAlpha's fault. To analyze the problem, let us look at a far simpler problem:
$$e^x=-1$$
We all know the solution to this should be given by $x=(2n+1)\pi i$ for $n\in\mathbb Z$. But upon squaring both sides,
$$e^{2x}=(-1)^2=1$$
The solution to $e^{2x}=1$ is different, as it includes, say, $x=0$, which the original did not.
This is why you want to be careful when manipulating complex numbers :-)
|
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$4x^2-5xy+4y^2=19$ with $Z=x^2+y^2$, find $\dfrac1{Z_{\max}}+\dfrac1{Z_{\min}}$ I have no idea how to approach this question at all. I've tried to find the maximum and minimum of the quadratic but i am too confused on what to do afterwards.
|
Without Lagrange multipliers (but with trigonometry). Use polar coordinates:
$x=\sqrt Z\cos\theta$
$y=\sqrt Z\sin\theta$
$xy=Z\sin\theta\cos\theta=\frac{Z}{2}\sin(2\theta)$
From your equation you get $\frac{1}{z}=\frac{4}{19}-\frac{5\sin(2\theta)}{38}$
As $\sin$ takes values between $-1$ and $1$:
So $\frac{1}{z_{min}}+\frac{1}{z_{max}}=2\cdot\frac{4}{19}=\frac{8}{19}$
|
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Maximum of $f(x) =(1+\cos x)\cdot \sin(\frac{x}{2})$ on $x \in (0, \pi)$ I've attempted to solve this question by using $$f(x) = 2\sin\frac{x}{2}\cos^2\frac{x}{2} \leq \frac{(2\sin\frac{x}{2}+\cos^2\frac{x}{2})^2}{2}$$ but it results in the wrong answer every time. Is there another way to solve this question and is there anything I can do to change my method to make it work.
|
By AM-GM $$\left(1+\cos{x}\right)\sin\frac{x}{2}=2\cos^2\frac{x}{2}\sin\frac{x}{2}=2\left(\sin\frac{x}{2}-\sin^3\frac{x}{2}\right)=$$
$$=-2\left(-\sin\frac{x}{2}+\sin^3\frac{x}{2}+\frac{1}{3\sqrt3}+\frac{1}{3\sqrt3}\right)+\frac{4}{3\sqrt3}\leq$$
$$\leq-2\left(-\sin\frac{x}{2}+3\sqrt[3]{\sin^3\frac{x}{2}\cdot\frac{1}{3\sqrt3}\cdot\frac{1}{3\sqrt3}}\right)+\frac{4}{3\sqrt3}=\frac{4}{3\sqrt3}.$$
THe equality occurs for $\sin\frac{x}{2}=\frac{1}{\sqrt3}$, which says that the answer is $\frac{4}{3\sqrt3}$.
|
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Is the Euclidean valuation on a Euclidean domain unique? I have just started studying Euclidean domain which is defined as follows :
An integral domain $R$ is called a Euclidean domain if there exists a function $\delta : R \setminus {\{0\}} \longrightarrow \mathbb {N_{0}}$ which satisfies the following properties.
(1) $\delta (a) \leq \delta (ab)$, for all $a,b \in R \setminus {\{0\}}$.
(2) For any $a,b \in R$ with $b \neq 0$ there exist $q,r \in R$ such that $a = bq + r$, where either $r = 0$ or $\delta (r) < \delta (b)$.
Here $q$ is called quotient and $r$ is called remainder.
The function $\delta$ is called a Euclidean norm function (or Euclidean valuation) on $R$.
Now my question is :
Is such Euclidean valuation on a Euclidean domain unique?
I am in a fix.Please tell me.
Thank you in advance.
|
Another example (to show non-uniqueness): On the ring of integers, define $\delta (x) = x^2$.
|
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How to show that this algorithm needs $\Omega(n \log n)$ comparisons How can I show, that every algorithm based on pairwise comparisons for sorting a total ordered set consisting of $n$ elements requires in the $\Omega(n \log n)$ comparisons in the average case.
The hint I was already given is to show that for all $x \in (0,1)$ the inequality $$x \log_2(x) + (1-2) \log_2(1-x) \geq -1 $$ holds, but I have no clue how to begin.
|
Suppose that we have $n$ elements that need to be sorted and we can only do pairwise comparisons. There are $n!$ ways to arrange these elements, and if we assume the elements are unique, then there is only one way to sort them.
Let's think of each pairwise comparison as a yes/no question. Each pairwise comparison rules out some set of the $n!$ permutations that cannot be the correctly sorted order. Before answering yes or no, there is a set of possible permutations assuming the answer is yes and there is a set of possible permutations assuming the answer is no. These sets are disjoint because the elements being compared are unique, and have only one correct ordering.
If the "yes" set is bigger than the "no" set or vice versa, then we can define our ordering so that it answers the pairwise comparison that gives us the bigger set. Therefore, our algorithm can at best choose a comparison that makes the "yes" set and "no" sets equal. Upon the answer, our choice of permutations is cut in half.
This implies that the best comparison based algorithm runs in $\Omega(\log_2(n!))$ time, because there are $n!$ permutations and each comparison eliminates at best half of the choices in worst-case time. Notice that
$$ \log_2(n!) = \sum_{i=1}^n \log_2(i) \le \sum_{i = \lfloor n/2 \rfloor}^n \log_2(i) \le \lceil n/2 \rceil \log_2(\lfloor n/2 \rfloor) \le \frac{n \log(n-1)}{2} - 1 \in \Omega(n\log(n))$$
|
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Find the value of $ 99^{50}-\binom{99}{1}(98)^{50}+\binom{99}{2}(97)^{50}-\cdots \cdots +99$ Find the value of $\displaystyle 99^{50}-\binom{99}{1}(98)^{50}+\binom{99}{2}(97)^{50}-\cdots \cdots +99$
Binomial identity:
$\displaystyle (1-x)^{99} = \binom{99}{0}-\binom{99}{1}x+\binom{99}{2}x^2-\cdots \cdots -\binom{99}{99}x^{99}$
I want be able to go further, could some help me with this, thanks
|
It is simply zero. Let $\delta$ be the (difference) operator mapping $p(x)$ into $p(x)-p(x-1)$: if $p(x)$ is a polynomial with degree $d\geq 1$, then $(\delta p)(x)$ is a polinomial with degree $d-1$. Since
$$(\delta^n p)(x) = \sum_{k=0}^{n}(-1)^k \binom{n}{k} p(x-k) \tag{1}$$
if $p(x)$ is a polynomial with degree $<n$, the RHS of $(1)$ is constantly zero.
The given sum is the RHS of $(1)$ in the case $p(x)=x^{50}$, $n=99=x$.
|
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Domino game maximum achievable points Given is a domino parts set with typical 28 parts from [ | ] to [6|6].
[ | ] [ |1] [ |2] [ |3] [ |4] [ |5] [ |6]
[1|1] [1|2] [1|3] [1|4] [1|5] [1|6] [2|2]
[2|3] [2|4] [2|5] [2|6] [3|3] [3|4] [3|5]
[3|6] [4|4] [4|5] [4|6] [5|5] [5|6] [6|6]
As every domino player knows, one can only place a part after one of the 2 ends of the dominoes chain; and the end placed part free half must match the half of the part to be placed that will be snapped to it!
The rules for scoring are: every time a part is placed, if the sum of both ends of the dominoes chain is a multiple of 5, you add it to you current score.
example: [4|2] [2|1] makes 5 points
A double (every part where the two halfes are equal) is always placed rotated by 90º; so for accounting purposes, unlike other pieces, both sides ot the part are taken into account.
Example:
___ ___
|6| |4|
——— [6|4] ———
|6| |4|
——— ———
makes for 20 points.
One single part placed accounts as the two tips.
Example [2|3] accounts as 5 points
And
___
|5|
———
|5|
———
accounts as 10 points.
So what is the chain that achieves the maximal number of points and what is that score?
|
I made a program to simulate the game and got the following result:
The middle row shows the placement of the dominoes, the bottom row shows the order of placement of each domino and the upper row shows the total score as each domino is placed. The total score after the placement of the last domino can be seen to be $175$.
The result comes with several caveats. A full blown simulation with all $28$ dominoes is simply too large to be finished in reasonable time (except perhaps on a super-computer). So I made some shortcuts:
*
*To reduce the number of computations needed, I decided to only start a chain with one of the five dominoes which, when placed first, gives points, i.e.: $[0,5], [1,4], [2,3], [4,6]$ or $[5,5]$
*To further reduce computations (and this is the big one) I decided to do the simulation incrementally in the following way:
a. Run the simulation with all $28$ dominoes, but cap the number of links (dominoes) in the chain at $N \lt 28$. Keep increasing $N$ until it really hurts (effectively a full day of computation, which was at $N=11$).
b. Check if domino placements for chains of length $n<N$ seem to stabilize in the chain length interval between $n$ and $N$. I.e. check if the same $n$ dominoes are placed in the same order repeatedly, for chains of length $n$ to $N$. If so, assume these $n$ dominoes will always be placed in this order, no matter how large the chains get.
c. Fixate the first $n$ dominoes to be these dominoes in that order. I.e. assume that no matter how long the chain gets, the first $n$ domino placements will be these $n$ dominoes. Continue the search for the best chain given that the first $n$ dominoes must be the ones found above and thus excluding these dominoes from participating in the further search.
d. Go to b.
With these caveats, I submit that the answer is $175$.
|
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Function linear in one variable and quadratic concave on another, is concave in jointly? I have a function $f(x,y,z)$. $f$ is convex with respect to $x$, and linear with respect to $y$. Also, it is quadratic concave with respect to $z$. I want to solve the following optimization problem:
\begin{align}
\min_{x\in C} \max_{y\in D_y} \max_z f(x,y,z)
\end{align}
My question is can I assume that this problem is concave with respect to $(y,z)$?
Let $g(y,z) = \min_{x\in C} f(x,y,z)$. Function $g(u)$, where $u=(y,z)$ is concave?
What if $f(x,y,z) = f_1(x,y) + f_2(x,z)$?
|
No, you may not. Suppose $f(x,y,z) = -yz^2$, with $D_y = \mathbb{R_+}$, then the Hessian is:
$$\begin{pmatrix}0 & -2z \\ -2z & -2y\end{pmatrix}$$
which is indefinite for $z \neq 0$, since the eigenvalues are $-y \pm \sqrt{y² + 4z²}$
|
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Approximating a function in $H^1\cap L^\infty$ by $H^1_0\cap L^\infty$ functions $\Omega$ is a bounded smooth domain. Take $u \in H^1(\Omega) \cap L^\infty(\Omega)$. Is it possible to find a sequence $u_n$ such that $u_n \in H^1_0(\Omega) \cap L^\infty(\Omega)$ and $u_n \to u$ in $L^\infty(\Omega)$?
It seems so, since we can take a smoothed version of $u\chi_{\Omega_n}$, where $\Omega_n$ is a subset of $\Omega$ that increases to $\Omega$ as $n \to \infty$.
Is it possible to get other norm convergence too? I think not $H^1$ norm.
|
*
*One cannot approximate $u \in H^1(\Omega) \cap L^\infty(\Omega)$ in $H^1$ norm by $H^1_0(\Omega)$ functions unless $u$ is already in $H^1_0(\Omega)$. This is because $H^1_0(\Omega)$ is a closed subspace of $H^1(\Omega)$.
*The function $u\equiv 1$ is in $ H^1(\Omega) \cap L^\infty(\Omega)$ but cannot be approximated by $H^1_0(\Omega)$ functions in $L^\infty$ norm. Indeed, if $u_n\to u$ in $L^\infty$, then for sufficiently large $n$ we have $u_n\ge 1/2$ a.e. in $\Omega$, and such a function is not in $H^1_0(\Omega)$.
*Approximation is possible with respect to $L^p$ norm for $1\le p<\infty$. Indeed, let $K$ be a compact subset of $\Omega$ such that the measure of $\Omega\setminus K$ is small (it can be made arbitrarily small). Then $u\chi_K$ is close to $u$ in $L^p$ norm. Furthermore, mollification of $u\chi_K$ produces a sequence of smooth functions with compact support in $\Omega$ which converge to $u \chi_K$ in $L^p$.
|
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Prove by contradiction that any prime number is either in the from of $ab+1$ or $ab+5$. It says:
Prove by contradiction that any prime number is either of the form of $ab+1$ or $ab+5$.
And this was all. But it seems both belong to $\mathbb{N} \cup \{0\}$. Because otherwise it wouldn't hold for $2$ which is prime.
What the problem says is to say:
$$(p\neq ab+1)\wedge (p\neq ab+5)\implies p\text{ is not prime}$$
I tried to show it would then be written as the product of two numbers and thus not prime. But up to what number should I do that?
Stuck.
|
HINT: Consider modulus 6. A number $p $ taken modulus 6 can only be $0, 1, 2, 3, 4$ or $5$. What happens if $p $ is either $0, 2$ or $4$ mod 6? What if it is $3$ mod 6?
|
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Is recursion a type of differential equations Unfortunately, this question has a very bland answer; however, it occurred to me that recursive functions, like ones to program and calculate factorials for instance, are very similar to differential equations whose results depend on initial conditions. Is recursion essentially just discrete DEs, versus the typical continuous DEs learned at school?
|
Discrete Differential equations are called "Difference Equations" and are mostly expressed by $$k_{n+1} - k_{n} = f(n, k) \quad n\in \mathbb{N}$$
These give the difference between two consecutive values.
Recursive Expressions ware mostly stated by $$k_{n+1} = f(n, k)$$ It is always possible to make this into a difference equation by just subtracting $k_n$ on both sides though.
|
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How to tackle this question related to gcd. Suppose $a,b$ be two positive integers such that $a = {p_{1}}^{r_{1}}{p_{2}}^{r_{2}}...{p_{n}}^{r_{n}}$ and $b = {p_{1}}^{s_{1}}{p_{2}}^{s_{2}}...{p_{n}}^{s_{n}}$, where $p_{i}$ are distinct primes and $r_{i} , s_{i} \in \mathbb N_{0}$ for $i = 1,2,...,n$ where $\mathbb N_{0} = \mathbb N \cup \{0\}$. Then show that $d = {p_{1}}^{t_{1}}{p_{2}}^{t_{2}}...{p_{n}}^{t_{n}}$ is the $gcd (a,b)$ where $t_{i} = \min \{r_{i} , s_{i}\}$ for $i = 1,2,...,n$.
It is easy to see that $d|a$ and $d|b$.But I find difficulty to show the second part i.e. if $x$ be any common divisor of $a$ and $b$.Then $x|d$.How can I show this?Please help me.
Thank you in advance.
|
Let $x$ be a common divisor.
Let $q$ be prime and $q|x$ then $q|a$ and $q|b$. The only prime factors of $a$ and $b$ are $p_i$. So $q = p_i$ for some $i$.
Let $k$ be the highest power of $q=p_i$ that divides $x$. That is $q^k|x$ but $q^{k+1} \not \mid x$. Then $q^k|a$ and $q^k|b$. If $k > r_i$ then $p_i^k\not \mid a$. If $k > s_i$ then $p_i^k \not \mid b$. So $k \le t_i = \min(r_i,s_i)$.
Let $x = \prod q_l^{k_l}$ be the prime factorization. We have shown that each $q_l = $ some $p_i$ and that each $k_l \le t_i$. So we know $\prod q_l^{k_l}|\prod p_i^{t_i}$. So $x|d$.
So $d$ is gcd.
|
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Proving that a series diverges Let $(a_n)$ positive sequence where $$\dfrac{n-1}{n} \leq \dfrac{a_{n+1}}{a_n} \leq \dfrac {n}{n+1}$$
Show that $\sum_{n=1}^{\infty} a_n$ diverges.
I already found out that
$\sum_{n=1}^{\infty} (-1)^na_n$ is convergent and that $\lim a_n =0$.
I'm not sure how to approach this one.
|
For all $n \ge 1$,
$$a_{n+1} \ge \frac{n-1}{n}a_n \ge \frac{(n-1)(n-2)}{n(n-1)}a_{n-1}\ge \cdots \ge \frac{(n-1)(n-2)\cdots 2\cdot 1}{n(n-1)\cdots 3\cdot 2}a_2 = \frac{1}{n}a_2$$
Since $\sum \frac{1}{n}a_2$ diverges, by comparison the series $\sum a_n$ diverges.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
"answer_count": 1,
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|
Is finding the minmax easier than finding the Nash Equilibrium? In one of the asnwers in game theory, I came across the answer that finding a Nash Equilibrium in a bimatrix game has a complexity level of PPAD-complete.
However I couldn't find whether finding the minimax(or maximin for that matter) for a game is a lower level of complexity. Correct me if im wrong, but if this were true, it would make life easier to find the Nash Equilibria for zero sum games computationally 'easier' (since then maximin=minimax=NE in zero sum games)?
The problem is, during my google searches, I found many algorithms that help solve m x n matrix games, however I have not found any algorithms that aim to find the minimax. Does anyone know of these as well?
|
The problem of finding maxmin or minmax strategies is (presumably) easier than PPAD. They can be found, and thus zero-sum games can be solved, by Linear Programming, which can be done (provably) in polynomial time, by the Ellipsoid method, or more practically by Interior Point methods (or non-provably but practically efficiently by the Simplex method).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2092459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Compute ${11 \choose 1} + {11 \choose 3} + ... + {11 \choose 11}$? I did this with brute force, and got 1024.
What is the faster method of solving this?
|
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{k = 0}^{5}{11 \choose 2k + 1} & =
\sum_{k = 0}^{10}{11 \choose k + 1}{1 + \pars{-1}^{k} \over 2} =
\sum_{k = 1}^{11}{11 \choose k}{1 - \pars{-1}^{k} \over 2}
\\[5mm] & =
{1 \over 2}\underbrace{\sum_{k = 0}^{11}{11 \choose k}}_{\ds{2^{11} = 2048}}\ -\ {1 \over 2}\underbrace{\sum_{k = 0}^{11}{11 \choose k}\pars{-1}^{k}}_{\ds{=\ 0}}\ = \bbx{\ds{1024}}
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2092649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 5
}
|
Calculate $\int_{0}^{\infty }e^{-x^{2}-x^{-2}}dx$ How to calculate
$$\int_{0}^{\infty }e^{-x^{2}-x^{-2}}dx$$
I have no idea where to start.Is it connect with the euler-poisson integral?
|
In fact,let see the more general form
$$I=\int_{0}^{\infty }e^{-\alpha ^{2}\left ( x^{2}+x^{-2} \right )}\,\mathrm{d}x$$
let $x\rightarrow x^{-1}$, we have
$$I=\int_{0}^{\infty }e^{-\alpha ^{2}\left ( x^{2}+x^{-2} \right )}\,\mathrm{d}x=\int_{0}^{\infty }x^{-2}e^{-\alpha ^{2}\left ( x^{2}+x^{-2} \right )}\,\mathrm{d}x$$
hence
\begin{align*}
I=\frac{1}{2}\int_{0}^{\infty }\left ( 1+x^{-2} \right )e^{-\alpha ^{2}\left ( x^{2}+x^{-2} \right )}\,\mathrm{d}x&=\frac{1}{2}\int_{-\infty }^{\infty }e^{-\alpha ^{2}\left ( x^{2}+x^{-2} \right )}\,\mathrm{d}\left ( x-x^{-1} \right ) \\
&=\frac{1}{2}\int_{-\infty }^{\infty }e^{-\alpha ^{2}\left [ \left ( x-x^{-1} \right )^{2}-2 \right ]}\,\mathrm{d}\left ( x-x^{-1} \right ) \\
&=\frac{1}{2}e^{-2\alpha ^{2}}\int_{-\infty }^{\infty }e^{-\alpha ^{2}t^{2}}\,\mathrm{d}t \\
&=e^{-2\alpha ^{2}}\cdot \frac{\sqrt{\pi }}{2\alpha }
\end{align*}
now let $\alpha =1$ and the answer will follow.
|
{
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"url": "https://math.stackexchange.com/questions/2092754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
}
|
Find probability that the sum of two dice is not 6 and not 5? When rolling two fair dice, what is P(sum of two dice is not 6 and not 5)?
Calculation:
First, I found the probability of two numbers that would roll a sum of 6:
(1,5) (2,4) (3,3) (4,2) (5,1) = 5/36 (each probability is 1/36)
Then, I found the probability of two numbers that would roll a sum of 5:
(1,4) (2,3) (3,2) (4,1) = 4/36
My first method was to add the probabilities and then subtract from one to give the probability of NOT rolling a sum of 6 or 5:
(5/36) + (4/36) = 9/36
1 - 9/36 = 27/36 = 0.75
This turned out to be the accepted answer of my online homework. But then, I realized that the original question asked for P(not sum of 6 AND not sum of 5).
So, I recalculated:
1 - 5/36 = 31/36
1 - 4/36 = 32/36
(31/36)(32/36) = 992/1296 ~ 0.7654 (using multiplication rule)
When I entered this fraction in, it was incorrect. But since the questions asks for "AND" and not "or", wouldn't the second probability be the actual correct answer?
|
The probability that the sum of two dice is 6 and 5 is 9/36.
So the probability that the sum of two dice is not 6 and 5 is 1- 9/36=27/36.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2092879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
}
|
Which number is bigger than the others? Which number is bigger?
$2^{431},3^{421},4^{321},21^{43},31^{42}$
My attempt:
$4^{321}=2^{642}>2^{431},4^{321}=2^{642}>2^{640}=32^{128}>31^{42}$
$3^{421}>3^{420}=27^{140}>21^43$
But I don't know how to compare $4^{321}$ and $3^{421}$ Any hints?
|
$4^{321}=2^{642}=2^{11*58+4}=16*2048^{58}<27*2187^{58}=3^3*3^{7*58}=3^{409}<3^{421}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2092994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Solving irrational inequalities: $\sqrt{x^2-7x+10}+9\log_4{x/8}\geq 2x+\sqrt{14x-20-2x^2}-13$
Solving the following inequalities: $$\sqrt{x^2-7x+10}+9\log_4{x/8}\geq 2x+\sqrt{14x-20-2x^2}-13$$
I have been solving some questions on inequalities lately and I have come across this question, but I can't figure out a way to solve this...it contains both log and square root. Here's how I approached the question:
$\sqrt{x^2-7x+10}+9\log_4{\frac{x}{8}}\geq 2x+\sqrt{14x-20-2x^2}-13$
$\sqrt{x^2-7x+10}+9\log_4{x}-3/2\geq 2x+\sqrt{(-2)(x^2-7x+10)}-13$
$(1-2i)\sqrt{x^2-7x+10}+\geq 2x-9\log_4{x}+3/2-13$
$(1-2i)\sqrt{(x-5)(x-2)}+\geq 2x-9\log_4{x}-23/2$
I don't know how to proceed further. It would be great if I could get a hint.
|
If $x \in \mathbb R $, then we have that the expressions $\sqrt {x^2-7x-10} \text { and } \sqrt {20+14x-2x^2}$ should be positive, that is, $\geq 0$. But one expression is $\sqrt {2}i $ times the other implying that both expressions are zero, that is, $x=2$ and $x=5$ can satisfy this inequality.
But however $x=5$ does not satisfy the inequality giving us $\boxed {x=2} $ is the only solution. Hope it helps.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2093159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Polynomial factorization into irreducibles over $\mathbb{Q}[x]$ I need to find irreducible factors of $f(x)=x^4+3x^3+2x^2+1$ in $\mathbb{Q}[x]$ and explicitely prove that these factors are indeed irreducible.
I believe we can't reduce $f(x)$ any further but I have to prove that this is the case.
I have already shown that there can't be any linear factors if there was such a factorization using the rational root theorem. But another possibilty is a factorization into two polynomials of degree $2$. So something like: $f(x)=(ax^2+bx+c)(mx^2+nx+p)$. I tried writing it out but since we're working with rationals I find it difficult to find or rule out such polynomials.
What do I have to do in this case? And more general: is there a way to find such second degree factors more easily in $\mathbb{Q}$?
|
$f$ has degree $4$ and assumes prime values for these $9 > 2 \cdot 4$ points and so must be irreducible:
$$
\begin{array}{rl}
n & f(n) \\
-8 & 2689 \\
-7 & 1471 \\
-4 & 97 \\
-3 & 19 \\
1 & 7 \\
3 & 181 \\
5 & 1051 \\
6 & 2017 \\
7 & 3529 \\
\end{array}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2093272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
}
|
When is $\int_{0}^1|x^{a-b-1}\cos(x^{-b})|\ dx<\infty$?
Let $a,b>0$. Consider the function
$$
f(x)=x^{a-b-1}\cos(x^{-b}),\quad x\in(0,1].
$$
When is $f$ absolutely integrable on $[0,1]$?
If $a>b>0$, it is not hard to show that $f$ is absolutely integrable on $[0,1]$. For the case $0<a\leq b$, is it true that
$$
\int_0^1|x^{a-b-1}\cos(x^{-b})|\ dx=\infty?
$$
By change of variables, one has
$$
\int_1^\infty y^{-a/b}|\cos y|\ dy\tag{1}
$$
One can show that $\int_1^\infty y^{-a/b}\ dy=\infty$ since $-a/b\geq -1$. How would you estimate (1)?
|
$\left|\cos y\right|$ is a continuous, bounded, $\pi$-periodic, non-negative function with mean value $\frac{2}{\pi}$.
It follows that
$$ \int_{1}^{M} y^{-a/b}\left|\cos y\right|\,dy \approx \frac{2}{\pi}\int_{1}^{M} y^{-a/b}\,dy \tag{1} $$
and the RHS of $(1)$ converges as $M\to +\infty$ iff $\frac{a}{b}>1$, i.e. $\color{red}{a>b}$.
In such a case, we may write
$$\int_{1}^{+\infty}y^{-a/b}\left|\cos(y)\right|\,dy = \frac{2b}{\pi(a-b)}-\frac{4}{\pi}\sum_{k\geq 1}\frac{(-1)^k}{4k^2-1}\int_{1}^{+\infty} y^{-a/b}\cos(2ky)\,dy\tag{2} $$
by exploiting the Fourier cosine series of $\left|\cos y\right|$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2093423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Where is the hypothesis that $X$ and $Y$ are path-connected being used?
The above is from Hatcher's book.
I am curious exactly where is the hypothesis that $X$ and $Y$ are path-connected being used in the proof? I don't see it being invoked explicitly, nor can I see where it is implicitly used.
Also, a second question would be are there any counter-examples where the proposition does not hold if $X$ or $Y$ are not path-connected?
Thanks!
|
There is confusion in the literature between a space and a space with base point. Grothendieck wrote to me in 1983:"... the choice of a base point, and the $0$-connectedness assumption, however innocuous they
may seem at first sight, seem to me of a very essential nature. To make an analogy, it would
be just impossible to work at ease with algebraic varieties, say, if sticking from the outset (as
had been customary for a long time) to varieties which are supposed to be connected. Fixing
one point, in this respect (which wouldn’t have occurred in the context of algebraic geometry)
looks still worse, as far as limiting elbow-freedom goes!...." He was also keen on the use of the fundamental groupoid $\pi(X,C)$, or $\pi_1(X,C)$, on a set $C$ of base points, chosen according to the geometry of a situation.
The particular issue of products is dealt with in Topology and Groupoids by showing that the natural morphism $\pi(X \times Y ) \to \pi(X) \times \pi(Y)$ is an isomorphism, where $\pi(X)$ denotes the fundamental groupoid. What do you expect for $\pi(X \times Y, C\times D)$?
The general issue of using groupoid (and higher groupoid) structures in algebraic topology is discussed as part of this preprint Modelling and Computing Homotopy Types:I. In particular, $\pi_1(X,C)$ is useful for modelling and computing homotopy $1$-types.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2093508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
How can I solve this? $$ \int_{-\infty}^{\infty} \frac{x^2e^{-\alpha x^2}}{x^2+b^2} dx $$
|
A 'sort-of' elementary way to proceed [without complex analysis] is as follows. Write
$$
\frac{1}{x^2+b^2}=\int_0^\infty ds\ e^{-s(x^2+b^2)}
$$
and exchange the order of integrals to get
$$
I= \int_0^\infty ds\ e^{-s b^2}\int_{-\infty}^\infty dx\ x^2 e^{-(\alpha+s)x^2}
$$
$$
=\partial_\alpha\int_0^\infty ds\ e^{-s b^2}\int_{-\infty}^\infty dx\ e^{-(\alpha+s)x^2}
$$
$$
=\partial_\alpha\int_0^\infty ds\ e^{-s b^2}\sqrt{\frac{\pi}{\alpha+s}}
$$
and changing variables $\alpha+s=u$
$$
=\sqrt{\pi}\partial_\alpha \left[e^{\alpha b^2}\int_\alpha^\infty du\frac{e^{-b^2 u}}{\sqrt{u}}\right]\ .
$$
The integrand admits a quite simple antiderivative
$$
\int du \frac{e^{-b^2 u}}{\sqrt{u}}=\frac{\sqrt{\pi }\ \text{erf}\left(b \sqrt{u}\right)}{b}+C\ ,
$$
in terms of the error function. The finishing touch should be easy.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2093579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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|
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