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Are axioms assumed to be true in a formal system? In a logical system, there is assignment of truth values to the sentences in the language, and axioms are assigned the true value.
A logical system is a formal system.
In a formal system, there is no truth value assignment, but there are still axioms. Does that imply that axioms in a formal system are not assumed to be true?
THanks.
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For a formal system to be of much interest, it needs to be consistent -- it needs to have at least one model. In that model, the axioms of the system will be true. It doesn't make much sense to talk about the truth of axioms apart from models.
The formal systems of the most interest -- such as Peano arithmetic and Zermelo-Fraenkel set theory - come with particular "intended" models in which the axioms are true. Indeed, a common way of coming up with axioms is to begin with a model and then try to isolate some of its properties as axioms.
|
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Separation of variables: when to have exponential solution and when sinusoidal? In separation of variables, one can assume a solution of V(x,y) = X(x)Y(y) and after plugging this into Laplace's equation which is: ${{\partial^2 V} \over {\partial x^2}}$ + ${{\partial^2 V} \over {\partial y^2}}$ = 0 we can get:
${d^2X \over dx^2}$ = ${k^2X}$ which gives a solution ${X(x) = Ae^{kx} - Be^{-kx}}$?
and
${d^2Y \over dy^2}$ = ${-k^2Y}$ which gives solution ${Y(y) = C\sin(ky) - D\cos(ky)}$
(where k is some constant) However, I can't understand, why does positive ${k^2}$ give a solution with exponents and ${-k^2}$ has sinusoidal solution? Is it always so?
The book that I am referencing this from is Griffith 3rd edition of "Intro to Electrodynamics". He does mention this there: "If X were sinusoidal, we could never arrange for it to go to zero at infinity, and if Y were exponential we could not make it vanish at both zero and a."
But I don't quite see this? Also, the image used is:
Ps: I'm sorry if this has been asked before on this site. If it has, could someone please direct me to the answer, since I could not find it?
|
If you solve the differential equation $\frac{d^2 X}{dx^2} - k^2 X = 0$, then let $X=e^{mx}$ so that you get the auxiliary equation $m^2-k^2=0$. That auxiliary equation has roots $m_1=k,m_2=-k$, which are real. If you plug the roots to the formula
$$y=A e^{m_1 x}+ B e^{m_2 x}$$ you will get a solution with exponents.
If you solve the differential equation $\frac{d^2 X}{dy^2} + k^2 X = 0$, then let $X=e^{my}$ so that you get the auxiliary equation $m^2+k^2=0$. That auxiliary equation has roots $m_3=ki,m_4=-ki$, which are imaginary. If you plug the roots to the formula
$$y=C e^{m_3 x}+ D e^{m_4 x}$$ you will get a sinusoidal solution.
|
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invertibility of self adjoint operators prove that if $T$ is a self adjoint operator and $a^2$ is less than $4b$ Then $T^2$+$aT$+$bI$ is invertible.
Where $a$ and $b$ are scalars and $I$ is the identity operator
Not: please dont use determinants because the book i am using didn't define determinants yet
|
This question originally posited the condition $a < 4b$ sufficient for $T^2 + aT + bI$ to be invertible, provided $T$ is self-adjoint; however, I belive that Byron Schmuland was correct to edit this condition to $a^2 < 4b$, and I hope to make my reasons for this clear in what follows, both by proving the corrected proposition and by exhibiting counterexamples to the original one.
Here I shall assume that $T$ is of finite size $n$; that is, that $T$ operates on some vector space $V$ of dimension $n$. These assumptions are consistent with the approach taken in the second edition of Axler's text, which is the one I have; in a comment, ferrr mentions Axler as his current source.
These things being said, we observe that the eigenvalues of $T$ are all real, as is well-known: indeed, if $\lambda$ is an eigenvalue of $T$, then there is a nonzero $v \in V$ with
$Tv = \lambda v ; \tag{1}$
for such $v$ we have
$\lambda \langle v, v \rangle = \langle v, \lambda v \rangle = \langle v, Tv \rangle$ $=\overline{\langle Tv, v \rangle} = \overline{\langle v, T^\dagger v \rangle}$
$=\overline{\langle v, Tv \rangle} = \overline{\langle v, \lambda v \rangle}$
$= \overline {\lambda \langle v, v \rangle} = \bar \lambda \langle v, v \rangle, \tag{2}$
which, since $\langle v, v \rangle \ne 0$, implies that
$\lambda = \bar \lambda; \tag{3}$
note we used $T = T^\dagger$ in establishing (2), (3). We next invoke the also well-known fact that self-adjoint operators are diagonalizable, or, to put it in more immediate terms, they are possessed of a complete basis of eigenvectors of $V$; that is, there exists a set of $n$ linearly independent vectors $v_i \in V$ such that each satisfies
$Tv_i = \lambda_i v_i \tag{4}$
for some real $\lambda_i$, $1 \le i \le n$; note no $v_i = 0$: we have $V = \text{span}\{v_1, v_2, \ldots, v_n \}$.
From (4) we see that
$T^2 v_i = T(Tv_i) = T(\lambda_i v_i) = \lambda_i T(v_i) = \lambda_i^2 v_i, \tag{5}$
and likewise
$(aT + bI)v_i = (a\lambda_i v_i + b) v_i, \tag{6}$
and hence
$(T^2 + aT + bI) v_i = (\lambda_i^2 + a\lambda_i + b) v_i \tag{7}$
holds for all $v_i$. We have thus shown that the eigenvalues of $T^2 + aT + bI$ consist of the real numbers $\lambda_i^2 + a\lambda_i + b$, and that the $v_i$ form a complete eigenbasis for the operator $T^2 + aT + bI$.
To proceed further, we examine the function $y: \Bbb R \to \Bbb R$, $y = x^2 + ax + b$. This is a quadratic polynomial whose first two derivatives are given by
$y'(x) = 2x + a, \tag{8}$
$y''(x) = 2; \tag{9}$
by (8) and (9), $y(x)$ has a global minimum at
$x = -\dfrac{a}{2}, \tag{10}$
and the value of $y$, $y_{\text{min}}$ at this $x$ is
$y_{\text{min}} = (-\dfrac{a}{2})^2 - a(\dfrac{a}{2}) + b = b - \dfrac{a^2}{4}. \tag{11}$
We see that $y_{\text{min}} > 0$ precisely when $a^2 < 4b$ or $(a^2/4) < b$, and since $y(x) \ge y_{\text{min}}$ for all $x \in \Bbb R$, we have shown that every eigenvalue of $T^2 + aT + bI$ is positive. This in turn implies that $T^2 + aT + bI$ is invertible, since $\ker (T^2 + aT + bI) = \{0\}$ (recall that $\ker (T^2 + aT + bI) = \{0\}$ is spanned by all eigevectors with zero eigenvalue; but are there are none to do the spanning in this case!). Alternatively, we may explicitly construct $S = (T^2 + aT + bI)^{-1}$ as follows: since the $v_i$ are linearly independent, we may define $S$ on $\{v_1, v_2, \ldots, v_n\}$ and extend it to $V$ by linearity; we set
$S(v_i) = (\lambda_i^2 + a \lambda_i + b)^{-1}v_i \tag{12}$
for $1 \le i \le n$. We have
$S(T^2 + aT + bI)(v_i) = (T^2 + aT + bI)S(v_i) = v_i \tag{13}$
for all $v_i$, or
$(T^2 + aT + bI)S = S(T^2 + aT + bI) = I; \tag{14}$
$T^2 + aT + bI$ is thus invertible with inverse $S$. QED.
Note: What happens when we hypothesize that $a < 4b$? Well, if take $1 < a < 4b \le a^2$ these conditions are met, but now $b \le a^2/4$ so $y_{\text{min}} \le 0$, and the parabola $y(x)$ opens upwards; it has real zeroes. If we define a self-adjoint operator $T_1$ to have eigenvalues at the zeroes of $y(x)$, then $\ker(T_1^2 + aT_1 + bI) \ne 0$, hence there will no inverse for this operator. So we do need $a^2 < 4b$ to make her fly, Wilbur, for any self-adjoint $T$. End of Note
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!
|
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tank problem Differential equation A tank is partially filled with 100 gallons of coffee in which 10 lbs of sugar is dissolved. Coffee containing 1/3 lb of sugar per gallon is pumped into the tank at rate 3 gal/min. The yummy well-mixed solution is then pumped out at a slower rate of 1 gal/min.
A- What is the rate at which the tank is increasing before the tank is full?
B- Set up a differential equation for finding the number of pounds A(t) of sugar in the tank at anytime?
for part A i'm not sure but for part B i think that it should be something like that
dA(t)/t= (1/3)(3)-(A(t)/100)(3) but I'm not sure how to finish this
|
Let $A(t)$ be the number of pounds of coffee in the tank at time $t$. We find an expression for $A'(t)$.
There is a standard pattern for setting up the appropriate differential equatiom. We look separately at the rate sugar is (i) entering the tank and (ii) leaving the tank.
Entering: Liquid is entering at $3$ gallons per minute, and each gallon has $\frac{1}{3}$ pound of sugar. Thus sugar is entering at the rate $\frac{1}{3}\cdot 3$, that is, $1$.
Leaving: If we set $t=0$ at the beginning, then the amount of liquid in the tank at time $t\ge 0$ is $100+2t$. The concentration of sugar at time $t$ is $\frac{A(t)}{100+2t}$. Since liquid is leaving at rate $1$, sugar is leaving at rate $\frac{A(t)}{100+2t}$.
A suitable differential equation for $A(t)$ is therefore
$$A'(t)=1-\frac{A(t)}{100+2t}.$$
One needs to write down an appropriate initial condition.
The differential equation now can be solved in any of the usual ways, for example by first considering the related homogeneous equation.
|
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maximum area of a rectangle inscribed in a semi - circle with radius r.
A rectangle is inscribed in a semi circle with radius $r$ with one of its sides at the diameter of the semi circle. Find the dimensions of the rectangle so that its area is a maximum.
My Try:
Let length of the side be $x$,
Then the length of the other side is $2\sqrt{r^2 -x^2}$, as shown in the image.
Then the area function is
$$A(x) = 2x\sqrt{r^2-x^2}$$
$$\begin{align}A'(x) &= 2\sqrt{r^2-x^2}-\frac{4x}{\sqrt{r^2-x^2}}\\
&=\frac{2}{\sqrt{r^2-x^2}} (r^2 - 2x -x^2)\end{align}$$
setting $A'(x) = 0$,
$$\implies x^2 +2x -r^2 = 0$$
Solving, I obtained:
$$x = -1 \pm \sqrt{1+r^2}$$
That however is not the correct answer, I cannot see where I've gone wrong? Can someone point out any errors and guide me the correct direction. I have a feeling that I have erred in the differentiation.
Also how do I show that area obtained is a maximum, because the double derivative test here is long and tedious.
Thanks!
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hint :$x\sqrt{r^2-x^2}=\sqrt{x^2(r^2-x^2)}\le \dfrac{x^2+(r^2-x^2)}{2}=\dfrac{r^2}{2}$
|
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How to show that $(a+b)^p\le 2^p (a^p+b^p)$ If I may ask, how can we derive that $$(a+b)^p\le 2^p (a^p+b^p)$$ where $a,b,p\ge 0$ is an integer?
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Using Jensen's Inequality, we get for $p\ge1$ or $p\le0$,
$$
\left(\frac{A+B}2\right)^p\le\frac{A^p+B^p}2
$$
Which, upon multiplication by $2^p$, yields
$$
(A+B)^p\le2^{p-1}(A^p+B^p)
$$
|
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|
Examples of interesting integrable functions with at least 2 fixed points and an explicit inverse What are some interesting functions I can use to demonstrate this integration trick:
$$\int_a^b [f(x)+f^{-1}(x)]=b^2-a^2$$
I would like to know of some interesting functions where this trick is not obvious. EDIT: The functions I am receiving are the obvious ones like $f(x)=x$, which I don't want. :)
|
I'll throw one in for kicks. This is the first one that came to mind that isn't directly given in the form you stated. It's nothing mind blowing but here we go.
$\int 2xdx = x^2$, obviously. Now let $f(x) = x$. Then we get
\begin{align*}\int_a^b 2xdx & = \int_a^b 2f(x)dx \\
& = \int_a^b (f(x) + f(x))dx \\
& = \int_a^b (f(x) + f^{-1}(x))dx \\
& = b^2 - a^2 \text{,}
\end{align*}
and of course this simply coincides with exactly what we'd expect from $\int_a^b 2xdx$.
|
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Proof of an $\iff$ statement on binary trees
Let $x$ and $y$ be two nodes of a binary tree $B$.
Prove that $x$ is an ancestor of $y$ $\iff$ $x$ stands before $y$ in the pre-order traversal of $B$ and $x$ stands after $y$ in the post-order traversal of $B$.
I can do the $\implies$ part without trouble, but I can't deal with the converse with enough rigour.
Thanks for any kind of rigorous proof.
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For the converse I would proceed by contrapositive with the following ideas.
(1) Assume $x$ and $y$ are vertices for which neither is an ancestor of the other.
(2) Argue that there is a lowest common ancestor (I am picturing the tree drawn with the root at the top and going downwards) of $x$ and $y$, say $a$.
(3) Argue that one of $x$ and $y$, say $x$ is a left descendant of $a$ and the other is a right descendant of $a$. (If both are on the same side of $a$, then there is a lower common ancestor.)
(4) Then in preorder traversal, these three vertices will be in order $a$...$x$...$y$, and in postorder traversal $x$...$y$...$a$.
|
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Open and closed equivalence relations I am looking for canonical examples of open and closed equivalence relations, especially ones that are generated by a continuous functions. Intuitively I think that an open /closed continuous function $f:X \rightarrow Y$ should also generate a closed/open equivalence relation by $x R y :\Leftrightarrow f(x)=f(y)$. Is this true? Or are there other ways to construct typical candidates of such relations by using continuous functions?
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Every $G_\delta$ equivalence relation $E$ on a standard Borel space $X$ is so called smooth, which means that there is a Borel function $f : X \rightarrow X$ such that $x \ E \ y$ if and only if $f(x) = f(y)$.
All closed and open sets are $G_\delta$ hence if you weaken continuous function to Borel functions, in some sense the method you described above is the only way to get such equivalence relations.
|
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Minimum Value of $x_1+x_2+x_3$ For an Acute Triangle $\Delta ABC$
$$\begin{align}x_n=2^{n-3}\left(\cos^nA+\cos^nB+\cos^nC\right)+\cos A\,\cos B\,\cos C\end{align}$$ Then find the least value of $$x_1+x_2+x_3$$
My Approach: I have found $x_1$, $x_2$ and $x_3$
$$\begin{align}x_1=\frac{1}{4}\left(\cos A+\cos B+\cos C\right)+\cos A\,\cos B\,\cos C\\
=\frac{1}{4}\left(1+4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\right)+\cos A\,\cos B\,\cos C \tag{1}\end{align}$$
$$\begin{align}x_2=\frac{1}{4}\left(3+\cos 2A+\cos 2B+\cos 2C\right)+\cos A\,\cos B\,\cos C=\frac{1}{2} \tag{2}\end{align}$$
$$x_3=\frac{1}{4}\left(3\cos A+3\cos B+3\cos C+\cos 3A+\cos 3B+\cos 3C\right)+\cos A\,\cos B\,\cos C$$ $$\implies x_3=\frac{1}{2}+x_1+\frac{1}{4}\sum \cos 3A+2\prod \sin\frac{A}{2}\\
$$
$$\implies x_3=\frac{1}{2}+x_1-\prod \sin\frac{3A}{2}+2\prod \sin\frac{A}{2}\\
\tag{3}$$
$$\text{So}\;\;\;\;\;\;\;\begin{align}x_1+x_2+x_3=\frac{3}{2}+4\prod \sin\frac{A}{2}-\prod \sin\frac{3A}{2}+2\prod \cos A \end{align}$$
I cannot proceed any further.
|
Use AM-GM inequality,we have
$$\cos^3{x}+\dfrac{\cos{x}}{4}\ge 2\sqrt{\cos^3{x}\cdot\dfrac{\cos{x}}{4}}=\cos^2{x}$$
then we have
$$x_{1}+x_{3}\ge\cos^2{A}+\cos^2{B}+\cos^2{C}+2\cos{A}\cos{B}\cos{C}=2x_{2}$$
so
$$x_{1}+x_{2}+x_{3}\ge 3x_{2}=\dfrac{3}{2}$$
because we have use this follow well know
$$\cos^2{A}+\cos^2{B}+\cos^2{C}+2\cos{A}\cos{B}\cos{C}=1$$
|
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Borel measure supported on $\mathbb{Q}$ Let $\mu$ be a Borel measure supported on $\mathbb{Q} \subset \mathbb{R}$. Must $\mu$ be a sum of Dirac measures?
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Yes, because by definition of the support (depending on your definition), you have $\mu ( \Bbb{R} \setminus \Bbb{Q}) \leq \mu(\Bbb{R} \setminus {\rm supp}(mu)) = 0$ and thus for every measurable set $M$:
$$
\mu(M) = \mu(M \cap \Bbb{Q}) = \sum_{q \in \Bbb{Q} \cap M} \mu({q}) = (\sum_{q \in \Bbb{Q}} \mu({q}) \delta_q ) (M),
$$
so that $\mu$ is a (countable, nonnegative) linear combination of Dirac measures.
If you meant a finite sum, this is not correct, take e.g. $\sum_n 1/n^2 \delta_{1/n}$.
|
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Continuity of a function defined by an integral, when the variable is in the region of integration Hi everyone: Suppose that $f$ is locally integrable in $\mathbb{R}^{n}$, $(n\geq2)$; $B(a,r)$ is the ball of center $a$ and radius $r>0$ , and $\lambda$ is the $n$-dimensional Lebesgue measure. It seems clear that the function $$r\mapsto\int_{B(a,r)}f(t)d\lambda(t)$$ is continuous on $(0,+\infty)$. But how would you write a rigorous proof for it? Thanks for your reply.
|
Assume without loss of generality that $a=0$ and let $B(0,r)=B_r$. We have to prove the following: given $\epsilon>0$ there exists $\delta>0$ such that
$$
0<r<R<\delta\implies\Bigl|\int_{B_R\setminus B_r}f(t)\,d\lambda(t)\Bigr|\le\epsilon.
$$
Since $\lambda(B_R\setminus B_r)\to0$ as $r\to R$, the result is a consequence of the following general fact of Lebesgue measure (or more generally, of measure theory.)
Let $A\subset\mathbb{R}^n$ bemeasurable and $f\colon A\to\mathbb{R}$
integrable. Given $\epsilon>0$ there exists $\delta>0$ such that $$
B\subset A \text{ measurable
},\lambda(B)<\delta\implies\Bigl|\int_Bf(t)\,d\lambda(t)\Bigr|\le\epsilon.
$$
This is clear if $f$ is bounded. For the general case we may assume $f\ge0$. Since $f$ is integrable there is a simple function $s\colon A\to\mathbb{R}$ such that $0\le s(t)\le f(t)$ and$\int_A(f(t)-s(t))d\lambda(t)<\epsilon/2$. Since $s$ is simple it is bounded, say by $C>0$. Then for any $B\subset A$ measurable with $\lambda(B)\le\epsilon/(2\,C)$
$$
\int_Bf=\int_B(f-s)+\int_Bs\le\frac\epsilon2+C\,\lambda(B)\le\epsilon.
$$
|
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Prove that if the square of a number $m$ is a multiple of 3, then the number $m$ is also a multiple of 3. I'd like to prove that if ${m}^{2}$ is a multiple of $3$, then ${m}$ is also a multiple of $3$. Similarly, I'd like to disprove that if ${n}^{2}$ is a multiple of $4$, then ${n}$ is also a multiple of $4$.
Per the comment from @thisismuchhealthier, the context is that I'm studying the proof of the elementary theorem from analysis that there is no rational number whose square is $2$ and the related statements that $\sqrt{3}$ and $\sqrt{6}$ are both irrational, but there is a rational number whose square is $4$.
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Hint $\ 3\mid (m\!-\!1)m(m\!+\!1)=\color{#c00}{m^3\!-m},\ $ so $\ 3\mid\color{#0a0}{m^3}\,\Rightarrow\, 3\mid \color{#0a0}{m^3}\!-(\color{#c00}{m^3\!-m}) = m$
|
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Problem regarding Euler's Theorem: $a^{\phi(n)}\equiv 1 \bmod n$ Here's the problem:
If $n \geq 2$, and if $p$ is a prime number s.t $p|n$ but $p^2$ is not a factor of $n$, then $$p^{\phi(n)+1}\equiv p \mod n$$
So since we're dealing with Euler's Phi function, I figured that this was an application of Euler's Theorem (please correct me if I am wrong). An attempt (though miniscule):
Attempted Proof
We know that $p|n$, so $\exists q \in \mathbb{Z}$ s.t
$$n=pq.$$
However, since $p^2$ is not a factor of n, we know that these integers are relatively prime to eachother. Thus, $\exists r,t \in \mathbb{Z}$ s.t
$$1=p^2r+nt$$
Now if we consider the integers modulo n, we see that $p^2$ is invertible since it is relatively prime to $n$ and that $r$ happens to be our inverse. Thus, we can say that
$$1=p^2r + nt$$
$$\implies 1-p^2r=nt$$
$$\implies [1-p^2r]_n=[0]_n$$
$$\implies [p^2]_n[r]_n=[1]_n$$
So $r$ must be the inverse of $p^2$.
Okay so I'm sure I've gone off into a tangled direction. So my question is: how am I to finally get to the $\phi$ function? I mean, considering how Euler's Theorem goes, we know that we have
$$a^{\phi(n)} \equiv 1 \mod n$$
So wouldn't it follow naturally that
$$a^{\phi(n)+1}\equiv a \mod n$$
????
For our problem, we of course have $a=p$.
|
I think I've got it now:
We know that $p|n$ $\implies$ $\exists \gamma \in \mathbb{Z^{+}}$ s.t $n=p\gamma$. We are given the $p^2$ is not a factor of $n$ $\implies p^2 > n$ and so $p$,$\gamma$ share no common divisors. Thus, $\gcd(p,\gamma)=1$. Then by Euler's Theorem, we have
$$p^{\phi(\gamma)} \equiv 1 \mod p$$
Since $\phi(n)=\phi(p)\phi(\gamma)$, it follows that
$$p^{\phi(p) \cdot \phi(\gamma)} \equiv 1 \mod p \implies (p^{\phi(\gamma)})^{\cdot \phi(p)} \equiv 1 \mod \gamma$$
$$\implies p^{\phi(n)} \equiv 1 \mod \gamma$$
So it follows that $\gamma | (p^{\phi(n)} - 1)$ $\implies$ $\gamma \cdot p |(p^{\phi(n)+1}-p)$ $\implies$ $n|(p^{\phi(n)+1}-p)$. Hence
$$p^{\phi(n)+1} \equiv p \mod n$$
|
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"url": "https://math.stackexchange.com/questions/872997",
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|
On a basic tensor product question I am trying to show that
$$
\mathbb{Z} / (10) \otimes \mathbb{Z} / (12) \cong \mathbb{Z}/(2)
$$
by defining a map
$$
h([a]_{10} \otimes [b]_{12}) = [ab]_2
$$
and extend it linearly. I am having trouble trying to prove that this map is
well defined on $\mathbb{Z} / (10) \otimes \mathbb{Z} / (12)$. (I am using the
constructive definition of tensor products for exercise). I would greatly appreciate any help!
Thanks!
|
try to show in general that: let $A$ a commutative ring and $I$ , $J$ two ideals, then $A/I\otimes A/J\simeq A/(I+J)$. And remark that $12\mathbb{Z}+10\mathbb{Z}=2\mathbb{Z}$.
|
{
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|
Websites for math tests/quizzes Next semester I'm taking calculus at college and I was looking for websites that have quizzes/test for things like trigonometry, trig formulas, pre-calculus, calculus readiness, etc. so I can get ready this summer. I found the "MU Math Tests Homepage"
http://mathonline.missouri.edu/mucgi-bin/munew.cgi?variable=
The questions at MU aren't that challenging, but I'm looking for sites similar. Any ideas? (other than doing problems from library books).
I know this is isn't a conventional math stack question. I checked with the meta site to see if this was appropriate, so please don't downvote. Thank you:
Asking for websites
|
There are a couple of resources that I enjoy using. First, is Brilliant.org:
http://brilliant.org/
There is a subscription fee but it is worth it. It starts with simple questions on the topics you mentioned and progressively gets harder.
Another site that is good is this quiz page written by Terry Tao:
http://scherk.pbworks.com/w/page/14864181/FrontPage
I can recommend plenty of good books to get you started or advance your knowledge on these topics if you like?
|
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|
holomorphic function with nonvanishing derivative on unit disk $D$ Let $f$ be a holomorphic function on the unit disk $D$. Suppose for any $z\in D$, $f'(z)\neq 0$. Then does $f$ have to be a conformal map from $D$ to $f(D)$?
|
Yes: This is true for functions which are smooth enough, and in fact (if $f$ is smooth) we have
$$\int |\cos kx| f(x) dx \to A \int f(x) dx$$
where $A$ is the average of $|\cos x|$ over a single cycle.
Now given $f \in L^1$, choose a sequence of smooth enough $f_n$ converging to $f$ in the $L^1$ norm, and then note
$$\left|\int |\cos kx| f(x) dx - \int |\cos kx| f_n(x) dx\right| \le \int |f - f_n| \to 0$$
|
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|
What is the ratio of the side length of a regular hepatgon to the side length of the internal heptagon? Given a regular heptagon with side length 1, create a star heptagon by connecting every vertice.
Note that removing the "points" of the star yields a similar heptagon. I want to know the side length of this internal heptagon (blue sides) in relation to the side length of the original heptagon.
The blue lines: $$ \rho = 2cos(\pi/7)
$$
The green lines: $$ \sigma=4cos^2(\pi/7) - 1
$$
[Golden Fields: A Case for the Heptagon, Peter Steinbach, Albuquerque Technical-Vocational Institute, Mathematics Magazine, Vol. 70, No. 1, February 1997]
No idea where to start :(
|
Refer to the following diagrams:-
|
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|
Equality of sums with fractional parts of the form $\sum_{k=1}^{n}k\{\frac{mk}{n}\}$ I recently encountered the following equality ($\{\}$ denotes fractional part):
$$\sum_{k=1}^{65}k\left\{\frac{8k}{65}\right\}=\sum_{k=1}^{65}k\left\{\frac{18k}{65}\right\}$$
and found it very interesting as most of the individual summands on one side of the equation do not have a corresponding match on the other side. Investigating further, I found several other similar equalities:
$$\sum_{k=1}^{77}k\left\{\frac{9k}{77}\right\}=\sum_{k=1}^{77}k\left\{\frac{16k}{77}\right\}$$
$$\sum_{k=1}^{77}k\left\{\frac{17k}{77}\right\}=\sum_{k=1}^{77}k\left\{\frac{24k}{77}\right\}$$
$$\sum_{k=1}^{85}k\left\{\frac{7k}{85}\right\}=\sum_{k=1}^{85}k\left\{\frac{22k}{85}\right\}$$
Does anyone have any idea what general principle/pattern these arise from?
|
Given $n$, let $a$ and $b$ be integers coprime with $n$. Then:
$$\sum_{k=1}^n k \left\{ \frac{ak}{n}\right\} = \sum_{k=1}^n k \left\{ \frac{bk}{n}\right\}$$
As Michael Stocker commented:
$$f(a,n) = \sum_{k=1}^n k \left\{ \frac{ak}{n}\right\} = \sum_{k=1}^n k \frac{ak \mod n}{n} = \frac{1}{n}\sum_{k=1}^n k (ak \mod n)$$
$$f(b,n) = \sum_{k=1}^n k \left\{ \frac{bk}{n}\right\} = \sum_{k=1}^n k \frac{bk \mod n}{n} = \frac{1}{n}\sum_{k=1}^n k (bk \mod n)$$
I've been able to prove that, as Thomas Andrews said, if $ab \equiv 1 \pmod{n}$ then $f(a,n) = f(b,n)$:
Let $k' := ak \mod n$ then $k' (bk' \mod n) = k (ak \mod n)$. Here's the proof:
$k'[b(ak \mod n) \mod n] = k'(abk \mod n) = k'(k \mod n) = k'k = k(ak \mod n)$.
Note that $\lbrace (ak \mod n) \vert 1\leq k \leq n\rbrace = \lbrace 0, 1,2\cdots ,n-1\rbrace$.
So now $$f(a,n) = \sum_{k=1}^n k \left\{ \frac{ak}{n}\right\} = \sum_{k'=1}^n k' \left\{ \frac{bk'}{n}\right\} = f(b,n)$$
Now for the cases like $f(8,65) = f(18,65)$ where $ab \not\equiv 1$ I've found that $$k(ak \mod n) - k(bk\mod n) = \\(n-k)(b(n-k)\mod n) - (n-k)(b(n-k)\mod n)$$
But so far I've been unable to caracterize the pairs $(a,b)$ with that property.
|
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|
integrate $ \int \frac {x dx}{\sqrt {1+x^{10}} } $ This is a tough one. Thanks.
$$\int \frac {x dx}{\sqrt {1+x^{10}} } $$
This is not a homework problem. I spend 10 hours over the course of 3 days on this. I tried:
1) substituting u for x^5 to get a tangent-like quantity in the denominator
2)subsituting u for 1+x^10
3) substituting u for sqrt(1+x^10)
4) integration by parts. This leaves me with $$\int \frac { dx}{\sqrt {1+x^{10}} } $$ which I still find difficult.
5). Multiplying denominator and numerator by sqrt(1+x^10), and then integrating.
|
As mentioned in the comments, you can at least simplify this integral somewhat via the substitution $u=x^2$:
$$\int\frac{x\,\mathrm{d}x}{\sqrt{1+x^{10}}}=\frac12\int\frac{\mathrm{d}u}{\sqrt{1+u^5}}.$$
Next, substituting $t=-u^5$ puts the integral in a form recognizable as the definition of an incomplete beta function: for non-negative $t$,
$$\frac12\int\frac{\mathrm{d}u}{\sqrt{1+u^5}}=\frac{\sqrt[5]{-1}}{10}\int\frac{t^{-\frac45}}{\sqrt{1-t}}\mathrm{d}t=\frac{\sqrt[5]{-1}}{10}\operatorname{B}{\left(t;\frac15,\frac12\right)}+\text{const}.$$
|
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|
Finding Cauchy principal value for: $ \int_1^\infty \frac{ a x^2 + c }{x^4 - b x^2 - c} \mathrm{d}x $ I need to solve the integral
$ \displaystyle \mathcal{P} \int_1^\infty \frac{ a x^2 + c }{x^4 - b x^2 - c} \mathrm{d}x $,
where $\mathcal{P}$ is the Cauchy principal value, $ - 1 \leq c \leq 1$ and $a, b$ are both real, but can be arbitrarily large, positive or negative.
I'm not sure, whether this integral is solveable, but any hints or ideas are very welcome.
|
I actually ended up using a different solution, I found more direct and intuitively.
It is completely equivalent with @Yves method, but I just state it for completeness.
$$ \mathcal{P} \int_{1}^{\infty} \mathrm{d} x
\frac{a x^{2} + c }
{
x^{4} - b x^{2} - c
}
=
\mathcal{P} \int_{1}^{\infty} \mathrm{d} x
\frac{a x^{2} + c }{ (x^{2} - d)^{2} - g^{2}}
$$
$$
= \frac{1}{2g}
\Bigg\{
\mathcal{P} \int_{1}^{\infty}
a x^{2}
\left[
\frac{1}{ x^{2} - d - g}
-
\frac{1}{ x^{2} - d - g}
\right]
\mathrm{d} x
$$
$$
\qquad \quad +
\mathcal{P} \int_{1}^{\infty}
c
\left[
\frac{1}{ x^{2} - d - g}
-
\frac{1}{ x^{2} - d - g}
\right]
\mathrm{d} x
\Bigg\}
$$
with
$$ d = - \frac{b}{2}, \qquad g = \sqrt{c + d^2}. $$
Then dividing each fraction in two once again, we end with two integrals to solve
$$
\mathcal{P} \int_{1}^{\infty} \mathrm{d} x \frac{x}{x \pm \sqrt{d \pm g}}
=
\lim_{\epsilon \rightarrow \infty} \left[ x \mp \sqrt{r} \ln( x \pm \sqrt{r} ) \right]_{x = 1}^{x = \epsilon},
$$
$$
\mathcal{P} \int_{1}^{\infty} \mathrm{d} x \frac{1}{x \pm \sqrt{d \pm g}}
=
\lim_{\epsilon \rightarrow \infty} \left[ \sqrt{r} \ln( x \pm \sqrt{r} ) \right]_{x = 1}^{x = \epsilon}.
$$
Then, putting it all together, we end up with
$$
\frac{1}{8 g} \left[
\frac{2 a (d + g) + c}{\sqrt{d + g}} \ln \left( \frac{1 + \sqrt{d+g}}{1 - \sqrt{d+g}} \right)
-
\frac{2 a (d - g) + c}{\sqrt{d - g}} \ln \left( \frac{1 + \sqrt{d-g}}{1 - \sqrt{d-g}} \right)
\right],
$$
which can then be rewritten in terms of $\tan^{-1}$ or $\tanh^{-1}$ depending on the properties of the quantitiy $\sqrt{d \pm g}$.
|
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|
Reverse Fatou's lemma on probability space Let $(\Omega, \mathcal{F},\mathbb{P})$ be probability space and $E_{n \in \mathbb{N}}$ be $\mathcal{F}$-measurable sets.
Show example that reverse Fatou's Lemma, $\mathbb{P}(\limsup_n E_n) \geq \limsup_n \mathbb{P}(E_n)$, meets inequality strictly.
I understand this inequality of inf. However, I cannot solve this.
I want to remember the direction of inequality. However, I found that this question's answer on the Internet is only inf version.
My try
I understand $\mathbb{E}[\limsup_{n} E_n] \geq \limsup_n \mathbb{E}[E_n]$
and
$E_{2k-1}(\omega)=1$ if $\omega \in (0,1/2)$
$E_{2k-1}(\omega)=0$ if $\omega \in [1/2,1)$
$E_{2k}(\omega)=0$ if $\omega \in (0,1/2)$
$E_{2k}(\omega)=1$ if $\omega \in [1/2,1)$
so
$1=\mathbb{E}[\limsup_{n} E_n] > \limsup_n \mathbb{E}[E_n]=0$
However, I cannot show this question on probability space.
I think that I use the relationship, $\mathbb{P}(A)=\mathbb{E}[1_A]$ , right?
|
The correct example: $\Omega=[0,1]$, $\mathcal F=\mathcal B(\Omega)$, $P=\mathrm{Leb}$, $E_{2n}=[0,1/2]$ and $E_{2n+1}=(1/2,1]$ for every $n$, then $\limsup E_n=\Omega$ and $P(E_n)=1/2$ for every $n$ hence $$P(\limsup E_n)=1\gt1/2=\limsup P(E_n).$$
|
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|
How many of the 9000 four digit integers have four digits that are increasing? How to find the number of distinct four digit numbers that are increasing or decreasing?
The correct answer is $2{9 \choose 4} + {9 \choose 3} = 343$. How to get there?
|
The analysis that was used goes as follows:
Not using $0$: We choose $4$ non-zero digits. Once we have done that, we can arrange them in increasing order in $1$ way, and in decreasing order in $1$ way, for a total of $2\binom{9}{4}$.
Using $0$: They can only be decreasing. And we need to choose $3$ non-zero digits to go with the $0$. This can be done in $\binom{9}{3}$ ways.
I prefer Juanito's approach. Note that the sum is not $343$, so if the book got that, there is a computational error.
|
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|
Word problem regarding system of linear congruences... Full problem:
A hoard of gold pieces ‘comes into the possession of’ a band of $15$ pirates.
When they come to divide up the coins, they find that three are left over.
Their discussion of what to do with these extra coins becomes animated,
and by the time some semblance of order returns there remain only $7$
pirates capable of making an effective claim on the hoard. When, however,
the hoard is divided between these seven it is found that two pieces are left
over. There ensues an unfortunate repetition of the earlier disagreement,
but this does at least have the consequence that the four pirates who remain
are able to divide up the hoard evenly between them. What is the minimum
number of gold pieces which could have been in the hoard?
So the information that I picked out from this was, using equivalence classes
$$[x]_{15}=[3]_{15}$$
$$[x]_7=[2]_7$$
$$[x]_4=[0]_4$$
I solved the system and obtained $[-390]_{420}=[30]_{420}$. The answer to our problem is supposedly $408$, but from the point that I've gotten to, I am not quite sure how to get this number. Did I overlook something big? Or am I on the right track?
EDIT:
I re-calculated and obtained $[12]_{420}$. Or $x \equiv 12 \mod 420$. If only it were $-12$...
EDIT 2:
Nevermind. I've got the answer. I had a sign error and indeed my answer is $[-12]_{420}$.
|
Have
$$x \equiv 3 \mod 15$$
$$x \equiv 2 \mod 7$$
$$x \equiv 0 \mod 4$$
Using Chinese Remainder Theorem, we first solve
$$x \equiv 3 \mod 15$$
$$x \equiv 2 \mod 7$$
Since $15,7$ are relatively prime, we have
$$15(1)+7(-2)=1$$
$$\implies 15(1)(2)+7(-2)(3)=30-42=-12$$
$$\implies x \equiv 93 \mod 105$$
Now we need to solve the system
$$x \equiv 93 \mod 105$$
$$x \equiv 0 \mod 4$$
Since $105,4$ are relatively prime, we have
$$105(1)+4(-26)=1$$
$$\implies 105(1)(0)+4(-26)(93)=(-104)(93)=-9672$$
$$\implies x \equiv -9672 \mod 420$$
Note that $420\cdot 23=9660$, so $-9672+9660=-12$. Further, observe that $420-12=408$. Thus,
$$x \equiv 408 \mod 420$$
So the minimum number of gold pieces is 408.
|
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|
How to find derivative of an integral of this type $$f(x) = \int _x^{e^x}\:\left(\sin t^2\right)\,dt$$
How to find the derivative $f'(x)$
Attempt: $\sin (e^{x^2}) e^x$
|
Use Fundamental theorem of calculus, let $F$ be antiderivative of $\sin t^2$, then you have:
$$f(x)=F(e^x)-F(x)$$
So:
$$f'(x)=e^xF'(e^x)-F'(x)=e^x \cdot \sin ((e^{x})^2)-\sin (x^2)$$
|
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|
A transcendental number from the diophantine equation $x+2y+3z=n$ Let $\displaystyle n=1,2,3,\cdots.$ We denote by $D_n$ the number of non-negative integer solutions of the diophantine equation
$$x+2y+3z=n$$
Prove that
$$
\sum_{n=0}^{\infty} \frac{1}{D_{2n+1}}
$$
is a transcendental number.
|
If I have not mistaken something,
$$ [x^{2n+1}]\frac{1}{(1-x)(1-x^2)(1-x^3)} = \frac{(n+1)(n+3)}{3}, $$
when $n\equiv 0,2\pmod{3}$, and
$$ [x^{2n+1}]\frac{1}{(1-x)(1-x^2)(1-x^3)} = \frac{(n+1)(n+3)+1}{3} $$
when $n\equiv 1\pmod{3}$, hence:
$$\begin{eqnarray*}\sum_{n=0}^{+\infty}\frac{1}{D_{2n+1}}&=&\sum_{n=0}^{+\infty}\frac{3}{(n+1)(n+3)}-\!\!\!\!\sum_{n\equiv 1\!\!\pmod{\!\!3}}\frac{3}{(n+1)(n+3)(n+2)^2}\\&=&\frac{9}{4}-\frac{1}{3}\left(\frac{9}{2}-\frac{\sqrt{3}\,\pi}{2}-\frac{\pi^2}{6}\right)\\&=&\frac{1}{36}\left(27+2\pi\left(\pi+3\sqrt{3}\right)\right),\end{eqnarray*}$$
and we only need to show that $u=\pi(\pi+3\sqrt{3})$ is a trascendental number.
But if $u$ were algebraic, then
$$\pi = \frac{-3\sqrt{3}+\sqrt{27+4u}}{2}$$
would be algebraic too, contradiction.
|
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Quotient Gaussian Integers Following Quotient ring of Gaussian integers, their extended conclusion is $\mathbb{Z}[i]/(a-ib) \cong \mathbb{Z}/(a^{2}+b^{2})\mathbb{Z}$. However it does not convince me, at least, one example below:
Let $a=2,b=0$, I cannot find explicit isomorphism between $\mathbb{Z}[i]/2\mathbb{Z}[i]$ and $\mathbb{Z}/4\mathbb{Z}$.
The coset leaders of $\mathbb{Z}[i]/2\mathbb{Z}[i]=\{0,1,i,1+i\}$, and the coset leaders of $\mathbb{Z}/4\mathbb{Z}=\{0,1,2,3\}$.
I appreciate if anyone could give a bijection mapping between the two quotient rings.
I am not sure I am right or not but will be happy to discuss with anyone who is interested in.
From my point of view, $\mathbb{Z}/4\mathbb{Z}=\{0,1,2,3\}$ (I know it is not good to give this expression, just for convinence) has 2 units $1,3$ where the sum is $0$, but $\mathbb{Z}[i]/2\mathbb{Z}[i]$ has two units, say, $i,1$ and their sum is not zero. The structure of the two quotient rings are different, and hence the conclusion $\mathbb{Z}[i]/(a-ib) \cong \mathbb{Z}/(a^{2}+b^{2})\mathbb{Z}$ is not correct.
Anyone here could make things clearer. Thanks a lot.
|
You'll never find such an isomorphism, because $x+x=0$ for all $x\in\mathbb Z[i]/2\mathbb Z[i]$, but not for all $x\in\mathbb Z/4\mathbb Z$.
The theorem you state is only true if $\gcd(a,b)=1$.
See this answer from the same question.
|
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|
Differentiability at a point Let $f:\mathbb{R}^{2}\mapsto\mathbb{R}\mathbb{}^{2}$ be given
by
$$f(x,y) = \left(\begin{array}{c}
x^{2}y+2y-x\\
3xy+4y
\end{array}\right)$$
Find a open set containing (0,0) where f has a differentiable inverse?.
I know the inverse function theorum guarentees there exists a neigbourhood
(open ball around)
$(0,0)$ on which an inverse exists (since f has continuous partial derivatives) but in the sample solution for the example
it has an additional point $\nabla f_{1}$ points in a direction in
the second quandrant and $\nabla f_{2}$ points in a direction in
the first quadrant,
and so the level curves of $f_{1}$ and $f_{2}$ cannot cross twice
on this ball.
Hence $f$ has an inverse on $B((0,0), 1/2)$- Why is is important that
the level curves do not cross (is it a theorum?) and how are they
checking they $(f_1,f_2)$ point in different directions? .
|
Let $U$ be an open neighborhood of $(0,0)\in{\mathbb R}^2$ for which $d{\bf f}(x,y)$ is regular at all points $(x,y)\in U$. When ${\bf f}$ is injective on $U$ then ${\bf f}$ maps $U$ bijectively onto an open neighborhood $V$ of $(0,0)$, and the inverse map ${\bf f}^{-1}:\ V\to U$ is $C^1$, by the inverse function theorem.
The restriction ${\bf f}\restriction U$ is not injective iff there are two different points $(x_1,y_1)$, $(x_2,y_2)\in U$, satisfying
$$f_1(x_1,y_1)=f_1(x_2,y_2)=:u_0,\quad f_2(x_1,y_1)=f_2(x_2,y_2)=:v_0\ .$$
But this means that the two level lines
$$f_1(x,y)=u_0,\qquad f_2(x,y)=v_0$$
intersect in two different points, both lying in $U$.
For a quantitative study of the ${\bf f}$ at hand we compose it with the linear map
$$\bigl(d{\bf f}(0,0)\bigr)^{-1}=\left[\matrix{-1&{\textstyle{1\over2}}\cr 0&{\textstyle{1\over4}}\cr}\right]$$
and obtain the new map
$${\bf g}:\quad (x,y)\mapsto\left\{\eqalign{u&=x+{\textstyle{3\over2}}xy-x^2y\cr v&=y+{\textstyle{3\over4}}xy\cr}\right.$$
with $$d{\bf g}(0,0)=\left[\matrix{1&0\cr 0&1\cr}\right]\ .$$
Compute
$$\nabla g_1(x,y)=\bigl(1+{\textstyle{3\over2}}y-2xy,{\textstyle{3\over2}}x-x^2\bigr),\qquad \nabla g_2(x,y)=\bigl({\textstyle{3\over4}}y, 1+{\textstyle{3\over4}}x\bigr)\ .$$
When $|x|, \>|y|<{1\over20}$ (this bound should do the job), then the vector $\nabla g_1$ includes an angle $<45^\circ$ with the $x$-axis. This implies that the level lines of $g_1$ (resp., their tangents) include an angle $<45^\circ$ with the $y$-axis. In a similar way the level lines of $g_2$ include an angle $<45^\circ$ with the $x$-axis. It should then be clear that (a) the two gradients are always linearly independent, and that (b) a $g_1$- and a $g_2$-level-line can intersect in at most one point. It follows that ${\bf f}$ is injective in the open square $\>\bigl]-{1\over20},{1\over20}\bigr[\times\bigl]-{1\over20},{1\over20}\bigr[\>$.
By the way: The equations $f_1(x,y)=u$, $\>f_2(x,y)=v$ can be solved explictly for $(x,y)$ in the neighborhood of $(x,y)=(u,v)=(0,0)$ in terms of square roots.
|
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Solving a master equation with linear coefficients I have the following PDE:
$$
\partial_t P(x,y,t)=x\partial_xP(x,y,t)+(y-1)\partial_yP(x,y,t)+2P(x,y,t).
$$
Mathematica suggests that the solution is
$$
\dfrac{f((y-1)/x,t+\log x)}{x^2},
$$
where $f$ is an arbitrary function.
My question is: How to solve this equation? Or put it in other words, how can I arrive at the solution given by Mathematica?
|
$\partial_tP(x,y,t)=x\partial_xP(x,y,t)+(y-1)\partial_yP(x,y,t)+2P(x,y,t)$
$\partial_tP(x,y,t)-x\partial_xP(x,y,t)+(1-y)\partial_yP(x,y,t)=2P(x,y,t)$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$
$\dfrac{dx}{ds}=-x$ , letting $x(0)=x_0$ , we have $x=x_0e^{-s}=x_0e^{-t}$
$\dfrac{dy}{ds}=1-y$ , letting $y(0)=y_0$ , we have $y=(y_0-1)e^{-s}+1=(y_0-1)e^{-t}+1$
$\dfrac{dP}{ds}=2P$ , letting $P(0)=f(x_0,y_0)$ , we have $P(x,y,t)=f(x_0,y_0)e^{2s}=f(xe^t,(y-1)e^t+1)e^{2t}$
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|
What's wrong with my aproach to solving this equation with multiple logarithms? A question I was faced with asked "For which $x$ is $\log_{10}(x)^{\log_{10}(\log_{10}(x))}= 10,000$?"
My instincts tell me I can say $$\log_{10}(x)=10$$ and $$\log_{10}(\log_{10}(x))=4$$
However, this leads to an incorrect answer. Instead, the solution posits I let $a=\log_{10}(x)$, and take the $\log$ base ten of both sides. Why is my answer wrong?
|
Taking $\log_{10}$ of both sides and using the rule $\log_{10}(a^b)=b\log_{10}a$ gives
$$(\log_{10}(\log_{10}(x))(\log_{10}(\log_{10}(x))=\log_{10}10000=4\ ,$$
that is,
$$(\log_{10}(\log_{10}(x)))^2=4\ .$$
Hence
$$\log_{10}(\log_{10}(x))=2\quad\Rightarrow\quad
\log_{10}(x)=100\quad\Rightarrow\quad x=10^{100}$$
or
$$\log_{10}(\log_{10}(x))=-2\quad\Rightarrow\quad
\log_{10}(x)=\frac{1}{100}\quad\Rightarrow\quad x=10^{1/100}\ .$$
|
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|
Find all integer solutions of $1+x+x^2+x^3=y^2$ I need some help on solving this problem:
Find all integer solutions for this following equation:
$1+x+x^2+x^3=y^2$
My attempt:
Clearly $y^2 = (1+x)(1+x^2)$, assuming the GCD[$(1+x), (1+x^2)] = d$, then if $d>1$, $d$ has to be power of 2. This implies that I can assume: $1+x=2^s*a^2, 1+x^2=2^t*b^2$. If $t=0$ then it is easy to finish. Considering $t>0$, we can get $t=1$ (simple steps only), so I come up with a "Pell-related" equation .. Then I get sticking there. It has a solution $x=7$, so I guess it's not easy to find the rest.
Please help.
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The solution is given in Ribenboim's book on Catalan's conjecture, where all Diophantine equations $$y^2=1+x+x^2+\cdots +x^k$$ are studied.
For $k=3$, only $x=1$ and $x=7$ are possible.
|
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|
The probability of getting a certain image by random pixelation Well, seeing that I'm terribly bad at math I don't know how to solve this, I'll try to explain, excuse me if I sound dumb.
Just suppose that I've got a photo/image with 320x240 resolution and 24 bit color depth (16,777,216 colors) and suppose that I made a computer program that sets pixels on 320x240px screen with random color values of truecolor (24 bits), so what is the probability for this program to draw my photo, or let's say any image I have with this particular resolution?
I know it's almost impossible but I want to know the exact proportion. The result will also give the answer to the question how many pictures we would have if we wanted to have all the pictures that can ever exist (on certain values of course), in which you can pick up even the craziest ones beyond imagination.
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The chance of 1 pixel would be 1 in 16777216.
The chance of 2 pixels would be (1/16777216) * (1/16777216).
So the chance of all of them would be (1/16777216) ^ (320*240).
Not very likely at all :) I'd stick with the lottery...
|
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|
Trace of power of stochastic matrix I would like to know if this statement is true.
Having a stochastic matrix (rows sum up to 1), with a positive (non-negative) diagonal, then it holds that
$$\text{trace}({W^2})\leq \text{trace}({W}),$$
(or more generally, if $p\geq q$, then $\text{trace}({W^p})\leq \text{trace}({W}^q)$.)
In other words, does it hold that $\sum_{i=1}^N\lambda_i^2\leq \sum_{i=1}^N\lambda_i$ ? Note that $\lambda_i$ may be also negative (it is not a positive-definite, or symmetric matrix)!
My intuition says it should hold, since with the powers of a stochastic matrix ${W}$, the spectrum gets smaller and smaller, eventually leading to $\lambda_1=1$ and $\lambda_i=0$, for $i\ne 1$. But is also the SUM of eigenvalues getting smaller with growing power? Thanks.
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This is (still) trivially false if you just take a small epsilon
$p_{12}=p_{21}=1-\epsilon$ and $p_{11}=p_{22}=\epsilon$
The trace of $P^2$ is $2((1-\epsilon)^2+\epsilon^2)>2\epsilon$, if take $\epsilon$ to be, say, smaller than 0.2
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Why are duals in a rigid/autonomous category unique up to unique isomorphism? I'm having trouble understanding the following statement:
"In a rigid category, duals are unique up to unique isomorphism."
It seems to me that this isomorphism is not unique.
Let me try to give a counterexample: Let $(X,Y,\epsilon: X \otimes Y \to I,\eta: I: Y \otimes X)$ be a dual pair (satisfying the snake identities).
Now, let $f:Y \to Y$ be any isomorphism other than the identity. We define
$$h := \left(f^{-1} \otimes 1_X\right) \circ \eta$$
$$e := \epsilon \circ (1_X \otimes f)$$
Claim: $(X, Y, e, h)$ is a dual pair. Let's prove the snake identities (in a strict monoidal category):
$$(e \otimes 1_X)\circ(1_X \otimes h) = \left((\epsilon \circ \left(1_X \otimes f\right)) \otimes 1_X) \circ \left(1_X \otimes \left(\left(f^{-1} \otimes 1_X\right) \circ \eta\right)\right)\right)\\
=(\epsilon \otimes 1_X) \circ (1_X \otimes f \otimes 1_X) \circ \left(1_X \otimes f^{-1} \otimes 1_X\right) \circ (1_X \otimes \eta)\\
=(\epsilon \otimes 1_X) \circ (1_X \otimes \eta) = 1_X$$
Similarly:
$$(1_Y \otimes e) \circ (h\otimes 1_Y) = (1_Y \otimes (\epsilon \circ (1_X \otimes f))) \circ \left(\left(\left(f^{-1} \otimes 1_X\right) \circ \eta\right)\otimes 1_Y\right)\\
= (1_Y \otimes (\epsilon \circ (1_X \otimes f))) \circ \left(\left(\left(f^{-1} \otimes 1_X\right) \circ \eta\right)\otimes 1_Y\right)\\
= (1_Y \otimes \epsilon) \circ (1_Y \otimes 1_X \otimes f) \circ \left(f^{-1} \otimes 1_X \otimes 1_Y\right) \circ (\eta \otimes 1_Y)\\
= f^{-1} \circ (1_Y \otimes \epsilon) \circ (1_Y \otimes 1_X \otimes f) \circ (\eta \otimes 1_Y)\\
= f^{-1} \circ (1_Y \otimes \epsilon) \circ (\eta \otimes 1_Y) \circ f\\
= f^{-1} \circ f = 1_Y
$$
I'm sorry for the lengthy formulae, it is an easy exercise when done in graphical calculus.
So if $Y$ has more than one automorphism (which will frequently be the case), we can define any number of different dualities. It seems that duals may be unique up to isomorphism, but this isomorphism itself is not unique. Am I wrong?
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Thanks to Zhen Lin to bring me on the right track.
The misunderstanding is the following:
The dual object ($Y$ in my example) is unique up to isomorphism, but this isomorphism is not unique. However, a dual is more than merely the object, it's the triple $(Y,\epsilon,\eta)$. This one is indeed unique up to unique isomorphism, and the calculation in the question shows exactly this.
So the same dual object $Y$ can be part of different duals.
|
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What is the remainder when the number below is divided by $100$? What is the remainder when the below number is divided by $100$?
$$
1^{1} + 111^{111}+11111^{11111}+1111111^{1111111}+111111111^{111111111}\\+5^{1}+555^{111}+55555^{11111}+5555555^{1111111}+55555555^{111111111}
$$
How to approach this type of question? I tried to brute force using Python, but it took very long time.
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HINT:
$$(1+10n)^{1+10n}=1+\binom{1+10n}1(10n)\pmod{100}\equiv1+10n$$
and $$(5+50n)^{1+10n}=5^{1+10n}+\binom{1+10n}1(50n)5^{10n}\pmod{100}$$
Now, $$5^{m+2}-5^2=5^2(5^m-1)\equiv0\pmod{100}\implies5^{m+2}\equiv25\pmod{100}$$ for integer $m\ge0$
$$\implies5^{1+10n}+\binom{1+10n}1(50n)5^{10n}\equiv25+(1+10n)(50n)25\pmod{100}$$
$$\equiv25+1250n$$ for $n\ge1$
For odd $n,$ $$(5+50n)^{1+10n}\equiv25+50\pmod{100}$$
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Question on Green's Theorem Consider the vector field $\textbf{f}(x,y)=(ye^{xy}+y^2\sqrt{x})\textbf{i}+(xe^{xy}+\frac{4}{3}yx^{\frac{3}{2}})\textbf{j}$.
Use Green's Theorem to evaluate $\int_C\textbf{f} \dot d\textbf{r}$, where $C$ is the ellipse given by $(x-1)^2+\frac{y^2}{9}=1$, oriented counterclockwise.
$\int_C\textbf{f} \dot d\textbf{r} = \int \int_R (\frac{\partial Q}{\partial x} -\frac{\partial P}{\partial y}) \ dx \ dy$
$=\int \int_R (xe^{xy}+ye^{xy}+2yx^\frac{1}{2})-(ye^{xy}+xe^{xy}+2\sqrt{x}y) \ dx \ dy$
$=\int \int_R 2y\sqrt{x}-2\sqrt{x}y \ dx \ dy$
What would the bounds be on the integral? I think I should parametrize to polar coordinates, but I'm not sure how to do that when the ellipse's center isn't the origin. Or would I just use $0\leq x \leq 2$ and $-\sqrt{9(1-(x-1)^2)} \leq y \leq +\sqrt{9(1-(x-1)^2)}$ ?
Thanks.
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After fixing the partial derivative typo, we have
$$\int \int_R 2y\sqrt{x}-2\sqrt{x}y \ dx \ dy$$
$$\int \int_R 0 \ dx \ dy$$
It does not matter what the bounds are, the answer is zero.
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Distinguishability problem / How many ways are there to put 6 balls in 3 boxes if the balls are distinguishable but the boxes are not?
I'm not quite sure how to approach it, $\frac{3^6}{3!}$ is not an integer. Thanks.
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I'm not the best at combinatorics but here's a go. I think I remember problems like this in statistical mechanics. $B$ for box.
$B\mid\quad B\mid\quad B\mid \qquad ways$
$6\mid\quad 0\mid\quad 0\mid \qquad 1 \qquad$ all in one box.
$5\mid\quad 1\mid\quad 0\mid \qquad6\qquad$ one of the six on its own
$4\mid\quad 2\mid\quad 0\mid \qquad {6\choose2}\qquad$ choose two of the six for one box
$4\mid\quad 1\mid\quad 1\mid \qquad \frac{6*5}{2}\qquad$ pick one of the six, then one of the 5 and divide by 2 ways of doing this.
$3\mid\quad 3\mid\quad 0\mid \qquad \frac{1}{2}{6\choose3}\qquad$ six choose 3 but divide by two because the boxes are indistinguishable.
$3\mid\quad 2\mid\quad 1\mid \qquad 6*{5\choose2}\qquad$ pick one of the six then two of the five remaining
$2\mid\quad 2\mid\quad 2\mid \qquad \frac{1}{3!}{6\choose2}\cdot{4\choose2}\qquad$ two of the six, then two of the remaining four, and there are 6 ways these can be ordered so divide out by this.
Thanks to Ned and JMac31 for the help
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Number of ways distribute 12 identical action figures to 5 children Need a little help with this problem.
Use generating functions to determine the number of different ways 12 identical action figures can be given to five children so that each child receives at most three action figures.
So far I have that we are looking for the coefficient of $x^{12}$ and the generating function is $G(x) = (1 + x + x^2 + x^3)^{5}$ so this is equal to the form $G(x) = 1/(1-x)^n$ which is equal to $(1-x)^{-n}$ and then I'm trying to use the formula $C(n+k-1,k)x^k$ and using $k=12$ and $n=5$ to come out with $C(16,12)$ but I'm not sure if that is correct.
I don't know if I'm doing this correctly or messing up a step along the way. Any help greatly appreciated.
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You have $\displaystyle G(x)=(1+x+x^2+x^3)^5=\big(\frac{1-x^4}{1-x}\big)^5=(1-x^4)^5(1-x)^{-5}$
$\displaystyle=\big(1-5x^4+\binom{5}{2}x^8-\binom{5}{3}x^{12}+\cdots\big)\big(\sum_{k=0}^{\infty}\binom{k+4}{4}x^4\big)$,
so now you just have to find the coefficient of $x^{12}$ in this expression.
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Existence of a function with certain integral properties Does there exist a non-negative Borel-measurable function $g:\mathbb [1,\infty)\to[0,\infty)$ such that
\begin{align*}
\int_1^{\infty}g(y)^2\,\mathrm dy<&\,\infty,\\
\int_1^{\infty}\frac{g(y)}{\sqrt{y}}\,\mathrm dy=&\,\infty?
\end{align*}
$g(y)=1/\sqrt{y}$ “almost” works, but yet it doesn't. In fact, no function of the form $g(y)=1/y^{m}$ with $m>0$ works, because the first condition would imply that $2m>1$ and the second would require that $m+1/2\leq1$, and these two inequalities are incompatible.
Intuitively, $g$ must decline “much faster” than $1/\sqrt{y}$ so as to make the integral of $g(y)^2$ convergent, yet not so fast as to make the integral of $g(y)/\sqrt{y}$ convergent.
Does anyone have any ideas?
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Got it. $$g(y)=\frac{\chi_{[2,\infty)}(y)}{\sqrt{y}\log y}.$$
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Linear Maps from a finite space to an infinite space Suppose V is finite dimensional with dim V > 0. Prove that if W is infinite dimensional then $L(V, W)$ is infinite dimensional.
Help? I really have no idea how to go about this one?
I'm assuming I need to use the fact that if a space is infinite dimensional then there is no list that spans it.
Maybe something like if dim V = n, then there are infinitely many maps $Tv_j = w_j$ that map to different linearly independent lists of length n in W, assuming that $v_1,...v_n$ is a basis for V.
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Maybe it's easier to prove the contrapositive: if $L(V,W)$ is finite-dimensional, then $W$ is finite-dimensional.
Let $f_1,...,f_n : V \rightarrow W$ be linear functions that span $L(V,W).$ Fix any nonzero $v \in V$. Then $f_1(v),...,f_n(v)$ span $W$, so $W$ is finite-dimensional.
The reason for this is that, for any $w \in W,$ you can find a linear function $f : V \rightarrow W$ such that $f(v) = w,$ and you can write $f$ as a combination of the $f_1,...,f_n.$
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Is it possible to create a bigger square using distinct smaller ones? Another user just inquired about possible solutions to the famous $70$x$70$ square puzzle. When I encountered that many years ago and the first idea that came to my mind as to why I wouldn't think it was possible to solve had to do with the $1$x$1$ square. Once you place this square, it appears that it creates a problem and you seem to end up building around that piece endlessly (results don't really change even if you hold off on placing the $1$x$1$).
This got me thinking so I started drawing a few pictures. I couldn't come up with a way to use smaller distinct squares (can't use the same square twice) to create a bigger one. I tried working with a few Pythagorean Triples as they share a similar idea of taking smaller 'squares' and putting them together to make bigger ones, but that didn't offer me anything.
Does anyone know of a example? Or, if it is impossible, a proof to support why not?
I apologize if this is obvious/trivial. Also, I didn't have a good idea how to tag this if someone could be so kind as to correct any misgivings.
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I commented on the previous post as well!
There is a good reason you couldn't find an example by hand: the smallest example of what's called a perfect squared square is a $112\times 112$ (link).
There is much more research here.
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Joint density distribution and Variance I was wondering if there is a way to calculate the joint distribution of two fully correlated variables, both with known distributions, expected value and variance, without knowing the conditional distribution?
If this is not possible, is there a way of finding Var$(X,Y)$ = E$[(XY)^2]$ - E$[XY]^2$ when knowing that Cor$(X,Y) = 1$? I can't seem to find an expression for E$[(XY)^2]$...
Thanks!
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With slightly more general details than in the
original version of André Nicolas's answer, it must be that $Y = aX+b$ where
$$a = \sqrt{\frac{\operatorname{var}(Y)}{\operatorname{var}(X)}}
\quad \text{and}\quad b = E[Y] - aE[X].\tag{1}$$
There is no joint density of $X$ and $Y$ in the sense that $X$ and $Y$ are not jointly
continuous random variables. Thus, to find $E[g(X,Y)]$, simply substitute $aX+b$ for $Y$ and find the expectation of this function of $X$ alone via the law of the unconscious statistician. In other words, $$E[g(X,Y) = E[g(X,aX+b)] = E[h(X)]$$
where $h(x) = g(x,ax+b)$ with $a$ and $b$ are as given in $(1)$.
Note: you say that you know the marginal distributions of $X$ and $Y$ (and
their means and variances). Be aware that the assumption of perfect correlation
with correlation coefficient $+1$ means that it must be that
$$F_Y(z) = F_X\left(\frac{z-b}{a}\right), \quad f_Y(z) = \frac{1}{a}f_X\left(\frac{z-b}{a}\right).\tag{2}$$
If the distributions that are known to you (or given to you) do not satisfy $(2)$, then the problem you are trying to solve has contradictory assumptions, and has no
meaningful answer.
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if $\frac{1}{(1-x^4)(1-x^3)(1-x^2)}=\sum_{n=0}^{\infty}a_{n}x^n$,find $a_{n}$ Let
$$\dfrac{1}{(1-x^4)(1-x^3)(1-x^2)}=\sum_{n=0}^{\infty}a_{n}x^n$$
Find the closed form $$a_{n}$$
since
$$(1-x^4)(1-x^3)(1-x^2)=(1-x)^3(1+x+x^2+x^3)(1+x+x^2)(1+x)$$
so
$$\dfrac{1}{(1-x^4)(1-x^3)(1-x^2)}=\dfrac{1}{(1-x)^3(1+x+x^2+x^3)(1+x+x^2)(1+x)}$$
then I feel very ugly,can you someone have good partial fractions methods by hand?
because I take an hour to solve this problem.
ago I have solve $$x+2y+3z=n$$ the number of the positive integer solution $a_{n}$ I found
$$a_{n}=\left[\dfrac{(n+3)^2}{12}\right]$$
Thank you
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Hints :
*
*First, prove that
$$\frac{1}{(1-x^2)(1-x^3)(1-x^4)} = \frac{7}{32(x+1)}-\frac{59}{288(x-1)}+\frac{1}{8(x-1)^2}+\frac{1}{16(x+1)^2}-\frac{1}{24(x-1)^3}+\frac{x+2}{9(x^2+x+1)}+\frac{1-x}{8(x^2+1)}.$$
*Then, use that
$$ \frac{1-x}{x^2+1} = -\frac{1+i}{2(x-i)}+\frac{-1+i}{2(x+i)}.$$
and
$$ \frac{x+2}{x^2+x+1} = \frac{j}{x+j}+\frac{\overline{j}}{x+\overline{j}}$$ where $j = \frac{1+i\sqrt{3}}{2}$.
*Finally, use the classical series
$$ \sum_{n=0}^{+\infty} x^k = \frac{1}{1-x},$$
$$ \sum_{n=0}^{+\infty} (k+1)x^k = \frac{1}{(1-x)^2},$$
and
$$ \sum_{n=0}^{+\infty} \frac{1}{2}(k+1)(k+2)x^k = \frac{1}{(1-x)^3}.$$
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Inequality involving Jensen (Rudin's exercise)
Exercise (Rudin, R&CA, no. 3.25). Suppose $\mu$ is a positive measure on the space $X$ and let $f \colon X \to (0,+\infty)$ be such that $\int_X f \, d\mu=1$.
Then for every $E \subset X$ with $0<\mu(E)<\infty$ we have
$$
\int_E \log f \, d\mu \le \mu(E) \log \frac{1}{\mu(E)}.
$$
This is not homework, it is self-studying. I think I should use Jensen's inequality, but I cannot get it.
I thought considering the positive probability measure $\nu$ given by $\nu:=f\mu$ but I do not see which are the right functions to play Jensen's inequality with. Since $\log$ is concave, I would have
$$
\int_E f \log f \, d\mu \le \log \int_E f^2 \, d\mu
$$
but this is not helpful. Than I can try $\log \frac{1}{x}$ which is convex and I would have
$$
\int_E f \log\left( \frac{1}{f}\right) \, d\mu \ge \log \int_E 1\, d\mu = \log \mu(E)
$$
which looks nicer since it gives
$$
\int_E f \log f \, d\mu \le \log\left(\frac{1}{\mu(E)}\right)
$$
but now I do not know how to handle the LHS.
Any hint, please? Thanks in advance.
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Write the inequality as
$$\frac{1}{\mu(E)} \int_E \log f\,d\mu \leqslant \log \frac{1}{\mu(E)}.$$
Jensen's inequality gives you
$$\exp \left(\frac{1}{\mu(E)}\int_E \log f\,d\mu\right) \leqslant \frac{1}{\mu(E)}\int_E f\,d\mu.$$
The remaining part should be clear.
|
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What are some conceptualizations that work in mathematics but are not strictly true? I'm having an argument with someone who thinks it's never justified to teach something that's not strictly correct. I disagree: often, the pedagogically most efficient way to make progress is to iteratively learn and unlearn along the way.
I'm looking for examples in mathematics (and possibly physics) where students are commonly taught something that's not strictly true, but works, at least in some restricted manner, and is a good way to understand a concept until one gets to a more advanced stage.
|
What about all the basic rules of weight & motion--aren't they just simplifications of terribly complex rules that generally work as long as you don't deal with anything too small or going too fast?
It seems that EVERY problem in early physics/calculus is simplified to eliminate most of the variables because the problem would become impossibly complex if you added them? For instance, falling object calculations don't generally take into consideration wind resistance, and if they do they don't take wind and varying pressure into account. A ball rolling down a ramp always considers only perfect surfaces. A draining tub doesn't take into account the speed difference of the funnel created by the flow or how long it will take to form?
Most things are simplified so we can fit them into our heads in one way or another. I've always had lots of luck imagining electricity as water--I know it's inaccurate for many reasons, but it works extremely well even for things like induction (inertia), even though it's obviously "Wrong".
I also believe every description of what is going on in quantum mechanics is a best guess at this point, doesn't mean it isn't helpful though.
|
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|
Need help with this Geometric sequence problem First, sorry if Im not using the right syntax, im translating the problem and im not sure if im supposed to say "Sequence" or "series", and also thanks to who ever tries to help.
The sum of a geometric sequence is 20, and the sum of its squared terms is 205.
find how many terms are in the sequence if the first term is $\frac{1}{2}$.
$a$1 + $a$2 + $a$3 +...+$a$$n$ = 20
$a$1 2 + $a$2 2 + $a$3 2 +...+$a$$n$ 2 = 205
$a$1 = $\frac{1}{2}$
find $n$.
$20=\frac{a\left(q^n-1\right)}{q-1}$ ==> $205=\frac{10.25a\left(q^n-1\right)}{q-1}$
$205=\frac{a^2\left(q^{2n}-1\right)}{q^2-1}$ = $\frac{a^2\left(q^n-1\right)\left(q^n+1\right)}{\left(q-1\right)\left(q+1\right)}$
$\frac{10.25a\left(q^n-1\right)}{q-1}$ =$\frac{a^2\left(q^n-1\right)\left(q^n+1\right)}{\left(q-1\right)\left(q+1\right)}$
$10.25a\left(q^n-1\right)\left(q+1\right)$ = $a^2\left(q^n-1\right)\left(q^n+1\right)$
$20.5\left(q+1\right)\space =\space \left(q^n+1\right)$
This is basically where I tell myself i did something wrong.
|
You have found that $q^n+1=20.5(q+1)$.
But since the sum of the terms is $20$, we have $q^n-1=40(q-1)$. (In essence you had written down this equation also.)
Subtract, and find $q$, and then $n$. The numbers are disappointingly small, $3$ and $4$ respectively.
|
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How to evaluate $\lim_{x \to \infty}\left(1 + \frac{2}{x}\right)^{3x}$ using L'Hôpital's rule? I'm stuck on how to evaluate the following using L'Hôpital's rule:
$$\lim_{x \to \infty}\left(1 + \frac{2}{x}\right)^{3x}$$
This is a problem that I encountered on Khan Academy and I attempted to understand it using the resources there. Here are the tips given for the problem; the portion that I'm having trouble understanding is highlighted:
I also attempted to use this video (screenshot following) to help; I understand the concepts in the video but it seems like there are some missing steps in the tips above.
I also attempted to use WolframAlpha's step-by-step solution but it was indecipherable to me.
Any help is greatly appreciated.
|
$(1+{2 \over n})^{3n} = ((1+{2 \over n})^{n})^3$.
We have $\lim_{n \to \infty} (1+ {\alpha \over n})^n = e^\alpha$.
To see the latter using l'Hôpital, let $a_n = (1+ {\alpha \over n})^n$.
Then $\log a_n = n \log(1+ {\alpha \over n})= { \log(1+ {\alpha \over n}) \over {1 \over n}}$.
Note that $\lim_{n \to \infty} { \log(1+ {\alpha \over n}) \over {1 \over n}} = \lim_{x \to 0} { \log(1+ \alpha x) \over x}$. Using l'Hôpital, we see that the limit is $\alpha$, so we have $\lim_{n \to \infty} \log a_n = \alpha$,from which we get $\lim_{n \to \infty} a_n = e^\alpha$.
|
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|
Why is a densely defined symmetric operator $T$ extended by its adjoint $T^*$? This is a result I've seen stated a few times, but I can't seem to come up with a proof!
Suppose $T$ is a densely defined linear operator with domain $D(T)\subset H$, where $H$ is a Hilbert space then if $T$ is symmetric i.e. $\langle Tx,y\rangle = \langle x,Ty\rangle$ for all $x,y \in D(T)$, it follows that the adjoint $T^*$ extends $T$, i.e. $T \subset T^*$.
What I'm having trouble showing is why $D(T)\subset D(T^*)$. Where $$D(T^*)=\{y\in H:x\mapsto \langle Tx,y\rangle \text{ is a bounded linear functional on } D(T)\}.$$
|
For $y \in D(T)$, we have, due to the symmetry of $T$,
$$x \mapsto \langle Tx,y\rangle = x \mapsto \langle x, Ty\rangle,$$
and the latter is easily seen to be continuous, hence $D(T) \subset D(T^\ast)$.
|
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Frobenius method, why is it an issue when the roots of the indicial equation differ by an integer When solving second-order differential equations by the Frobenius method at a regular singular point, you are supposed to use the two roots of the indicial equation to give you two independent solutions.
If there is only one root, it makes sense that you would need another method to get the second independent solution. However, many texts say that you also need to do this when the roots differ by an integer.
Why?
Is it that when the roots differ by an integer, the two matching solutions are not independent? If so, why must they be independent?
Is it that sometimes they will be independent and sometimes they won't? If so, when will they be independent and when won't they?
Is it that there is something that prevents calculating one of the solutions? If so, why?
|
If you look at http://math.creighton.edu/nielsen/DE_Fall_2010/Series%20Solutions/Series_Solutions_Beamer.pdf, they write the resulting recurrence for one of the solutions as $a_n F(n+r) = E$, where $F(r)$ is the indicial equation, and I'm writing $E$ to abbreviate a complicated expression which depends on a variety of things, including $n$. If $F(n+r)=0$ for some $n$ and $E \neq 0$, then this isn't solvable.
|
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How can I calculate the total number of possible anagrams for a set of letters? How can I calculate the total number of possible anagrams for a set of letters?
For example:
"Math" : 24 possible combinations.
math
maht
mtah
mtha
mhat
mhta
amth
amht
atmh
athm
ahmt
ahtm
tmah
tmha
tamh
tahm
thma
tham
hmat
hmta
hamt
hatm
htma
htam
Total: 24
I generated this by actually generating each combination, one by one, but I want to know the formula for calculating the total possible number of combinations of any group of letters.
|
I worked out a formula for calculating the number of anagrams for an a-letter word where b letters occur c times, d letters occur e times, f letters occur g times, etc.
a!/((c!^b) * (e!^d) * (g!^f)...)
|
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|
A question on the proof of 14 distinct sets can be formed by complementation and closure In Munkres, problem 20 of Section 2-6, it says that 14 distinct sets can be formed by complementation and closure. I see only five so far.
Let f be the function of closure mapping and g be the function of complementation mapping.
It is clear, f,g, fg,gf, and gfg are the 5 of 15 distinct sets. What are the rests? Was there any topological argument associated with it? How can I understand this intuitively and pictorially?
|
The following page lists the rest, and since it lets you experiment in real time, may also accelerate your intuitive and pictorial understanding:
http://www.maa.org/sites/default/files/images/upload_library/60/bowron/k14.html
Though the 14-set theorem is more algebraic than topological, constructing a Kuratowski 14-set in the reals will enrich your understanding of topology.
|
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|
Prove $\,17\mid 2a+3b \,\Rightarrow\, 17\mid 9a + 5b$ So, according to the book, for all $a, b, c$ that are elements of integers, it holds that $a|b$ implies $a|bx$ for all $x$ that is an element of integers. In other words it works for all ARBITRARY $x$ in the universe $Z$.
However, please consider this question:
When $2a + 3b$ is a multiple of $17$, prove that $17$ divides $9a + 5b$.
Proof(textbook):
We observe that $17|(2a + 3b) \implies 17|(-4)(2a + 3b)$ by the theorem where $a|b$ implies $a|bx$ for all $x$ that is an element of integers. Also since $17|17$ it follows that $17|[(17a + 17b) + (-4)(2a + 3b)]$ and consequently this simplifies to $17|(9a + 5b)$.
My problem with proof:
The book chooses the $x = -4$ for the arbitrary $x$ that is part of the universe $Z$, but I find that this isn't arbitrary at all because I'm pretty sure that if I used any other number for $x$ in the universe of $Z$ it wouldn't work with the proof. It would then seem that the book specifically chose it as $-4$ because the part where they include $17|17$ seems to be specific towards $-4$ being $x$ as well. Am I right in this? How would I solve questions like these?
|
HINT:
Eliminate one unknown
$$9(2a+3b)-2(9a+5b)=?$$
|
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|
Trigonometric functions of the acute angle Find the other five trigonometric functions of the acute angle A, given that:
\begin{align}
&\text{a)}\ \ \sec A = 2 \\[15pt]
&\text{b)}\ \ \cos A = \frac{m^2 - n^2}{m^2 + n^2}
\end{align}
Help me. I don't know how to solve this one. Thanks.
|
It is useful to know at least some basic trigonometric identities. See List of trigonometric identities at Wikipedia for a very complete list.
a) Since $\cos A=\frac1{\sec A}$, you get $\cos A=\frac12$. Can you get possible values of $A$ from there?
b) If $$\cos A=\frac{m^2-n^2}{m^2+n^2}=\left(\frac m{\sqrt{m^2+n^2}}\right)^2-\left(\frac n{\sqrt{m^2+n^2}}\right)^2,$$ then for an angle $B$ such that $\cos B=\frac m{\sqrt{m^2+n^2}}$ and $\sin B=\frac m{\sqrt{m^2+n^2}}$ you get $$\cos A=\cos^2B-\sin^2B=\cos2B.$$ Maybe this could help to express $A$ using $B$. You should be able to find $B$ using some inverse trigonometric functions. Double angle formulae should help you to get the values of $\sin A$, $\tan A$, etc for $A=2B$. (You should also think about the possibility $A=-2B$. Do you know why?)
|
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If $a_{n+1}=\cos(a_n)$ for $n\ge0$ and $a_0 \in [-\pi/2,\pi/2]$, find $\lim_{n \to \infty}a_n$ if it exists Let $a_{n+1}=\cos(a_n)$ for $n\ge0$ and $a_0 \in [0,\pi/2]$
Find $\lim_{n \to \infty}a_n$ if it exists.
I drew some sketches and it does seem like the limit exists, it's probably $x$ such that $\cos(x)=x$
I have no idea how to go about solving this, hints would really be appreciated.
Thank you for your time!
|
The "Fixed Point" method in numerical analysis says that if a function is continuously differentiable in a neighbourhood of a fixed point $x_0$ and
$$
|f'(x)| \lt 1
$$
it will converge to a fixed point
source: Fixed point
|
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|
Intuitive ways to get formula of cubic sum Is there an intuitive way to get cubic sum? From this post: combination of quadratic and cubic series and Wikipedia: Faulhaber formula, I get $$1^3 + 2^3 + \dots + n^3 = \frac{n^2(n+1)^2}{4}$$
I think the cubic sum is squaring the arithmetic sum $$1^3 + 2^3 + \dots + n^3 = (1 + 2 + \dots + n)^2$$
But how to prove it? Please help me. Grazie!
|
Here are some 'Proof without Words' for this identity;
As Hakim showed (but this one might be slightly clearer);
Here's another very clean illustration by Anders Kaseorg;
Here, the total top volume is undoubtedly $1^2+2^2+3^3+\cdots +n^2,$ while on the volume of the bottom part is;
$$1\left(1+2+\cdots+n\right)+2\left(1+2+\cdots+n\right)+\cdots +n\left(1+2+\cdots+n\right)=\left(1+2+\cdots+n\right)^2.$$
|
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Why is $\frac{d}{d \mu} \big|_{\mu=0} \, F^*_\mu F^*_\lambda t = F^*_\lambda \frac{d}{d \mu} \big|_{\mu=0} \, F^*_\mu t $? In the book "Manifolds, Tensor Analysis, and Applications" by Marsden, Ratiu, Abraham the following relation (see the proof of 6.4.1, third edition) is used:
$$\frac{d}{d \mu} \bigg|_{\mu=0} \, F^*_\mu F^*_\lambda t = F^*_\lambda \frac{d}{d \mu} \bigg|_{\mu=0} \, F^*_\mu t $$
Here $t$ is an arbitrary tensor field $t \in \mathcal T^r_s$, $F$ being a flow, $F^*$ its pullback. However I don't see why this is true. Particularly I'm interested in the simplest case of t being just a smooth function ($t \in \mathcal T^0_0$).
|
Result : If $\phi$ and $\psi$ are $1$-parameter groups s.t. $$ \psi'= X$$
then $$\phi^\ast (L_X\alpha) = \phi^\ast \lim_t \frac{\psi(t)^\ast
\alpha - \alpha }{t} = \lim \frac{(\phi^{-1} \circ \psi \circ \phi
)^\ast
(\phi^\ast \alpha )-\phi^\ast \alpha }{t} $$
Here recall $$ \frac{d}{ds}\bigg|_{s=0}
\phi^{-1}\circ \psi (s,\phi (z)) =d\phi^{-1} X_{\phi(z)} = \phi^\ast X $$ so that $$
\phi^\ast (L_X\alpha) = L_{\phi^\ast X} \phi^\ast\alpha $$
Returning to OP : Let $X:= \frac{\partial }{\partial \mu}\bigg|_{\mu =0
}F_\mu$. Then $$ \frac{d}{d\mu}\bigg|_{\mu=0} F_\mu^\ast
F^\ast_\lambda t=\frac{d}{d\mu}\bigg|_{\mu=0} F_\lambda^\ast
F^\ast_\mu t $$
and $$ F_\lambda^\ast
\frac{d}{d\mu}\bigg|_{\mu=0} F_\mu^\ast
t= F^\ast_\lambda L_X t=L_X F^\ast_\lambda t=\frac{d}{d\mu}\bigg|_{\mu=0} F_\mu^\ast
F^\ast_\lambda t $$
|
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Derivative of a matrix function with respect to a matrix I am trying to differentiate the following function, with respect to a matrix $X$:
$$
\operatorname{tr}(AX(X^TX)^{-1}B)
$$
where tr corresponds to the trace. Is there an easy way to see what the derivative will be? I've come across rules for $\operatorname{tr}(AXB)$ but not the form above with inverses etc .
$A$ and $B$ are known (constant) matrices.
|
Define $W \equiv (X'X)^{-1}$, then $f = BAX : W$ and its derivative is
$$
\frac {\partial f} {\partial X} = A'B'W - XW(BAX+X'A'B')W
$$
The algebra to arive at this result is tedious but straight-forward. The only tricky part is knowing that
$$ \eqalign {
dW &= - W\,\,dY\,\,W \cr
} $$
where $Y \!\equiv W^{-1}\!= X'X$
It's worth noting that both $Y$ and $W$ are symmetric.
Then you just expand the differential of $f$
$$ \eqalign{
df &= BA\,dX : W + BAX : dW \cr
&= dX : (BA)'W - BAX : W\big[dX'X + X'dX\big]W \cr
&= dX : A'B'W - WBAXW : \big[dX'X + X'dX\big] \cr
&= A'B'W : dX - WBAXW : \big[dX'X + X'dX\big] \cr
&= A'B'W : dX - WBAXW : dX'X - WBAXW : X'dX \cr
&= A'B'W : dX - WBAXWX' : dX' - XWBAXW : dX \cr
&= A'B'W : dX - XWX'A'B'W : dX - XWBAXW : dX \cr
&=[A'B'W - XWX'A'B'W - XWBAXW]: dX \cr
} $$
So the expression in brackets must be the derivative.
If you dislike the Frobenius product, you can carry out the above steps using the trace
$$
\text{tr}(A'B) = A:B
$$
|
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How to show if A is denumerable and $x\in A$ then $A-\{x\}$ is denumerable My thoughts:
If $A$ is denumerable then it has a bijection with $\mathbb{N}$
So therefore $A\rightarrow \mathbb{N}$. Then x is a single object in A and A is infinite.
So if a single object is removed from then $A$ is still infinite.
|
Your reasoning is informally correct. If you want to give a careful answer to this question you should demonstrate a bijection from $A \setminus \{x\}$ to $\mathbb{N}$.
You have a given bijection between $A$ and $\mathbb{N}$. Can you use this as a stepping stone to get from $A \setminus \{x\}$ to $\mathbb{N}$?
|
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Proof $e^n*n!$ is an asymptote of $(n+1)^n$ I would like to prove $\lim_{n\to \infty}e^nn!-(n+1)^n=0$.
All I have really done is show $(n+1)^n=\sum_{i=0}^n\frac{n!}{(n+1)^i(i!)(n-i)!}$
|
(This is too long for a comment, but neither is it yet a full answer.)
Consider the Taylor series expansion of $e^x$.
\begin{align*}
e^x &= \sum_{k = 0}^\infty \frac{x^k}{k!}\\
&= \lim_{n \rightarrow \infty} \sum_{k = 0}^n \frac{x^k}{k!}
\end{align*}
It follows
\begin{align*}
\lim_{n \rightarrow \infty} n! e^n &= \lim_{n \rightarrow \infty} n! \sum_{k = 0}^n \frac{n^k}{k!}\\
&= \lim_{n \rightarrow \infty} \sum_{k = 0}^n (n-k)!\binom{n}{k} n^k.
\end{align*}
On the other hand, the Binomial Theorem gives
\begin{align*}
\lim_{n \rightarrow \infty} (1+n)^n &= \lim_{n \rightarrow \infty} \sum_{k=0}^n \binom{n}{k}n^k.
\end{align*}
(At this point, I'm not quite sure what to say about that factor of $(n-k)!$ except that it is least significant for the largest values of $k$.)
|
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Polygons with equal area and perimeter but different number of sides? Let's say we have two polygons with different numbers of sides. They can be any sort of shape, but they have to have the same area, and perimeter.
There could be such possibilities, but can someone show me with pictures? I just need visualize it.
Sometimes in life you just have to know it, and sometimes we need a picture shown in our faces :).
|
┌─┐ ┌┐
│ │ │└─┐
│ └─┐ │ └┐
└───┘ └───┘
Edit: I like the above figures because they're easy to generalize to many sides. But if it's unclear that they have the same area, here's another pair: the L and T tetrominoes.
You can imagine sliding the square on the right side up and down relative to the 1x3 bar on the left side; this operation preserves both area and perimeter. Explicitly, both tetrominoes have area 4 and perimeter 10. The L has 6 sides, and the T has 8 sides.
|
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Null-homotopic covering space map I'm stuck with the following question, which looks quite innocent.
I'd like to show that if a covering space map $f:\tilde{X}\to X$ between cell complexes is null-homotopic, then the covering space $\tilde{X}$ must be contractible.
Since $f$ is null-homotopic there exists a homotopy $H_t:\tilde{X}\to X$ from $H_0=x_0$ to $H_1=f$ and I would like to use it to construct another homotopy $G:\tilde{X}\to \tilde{X}$ from $G_0=\tilde{x}_0$ to $G_1=Id_{\tilde{X}}$.
By the homotopy lifting property, $H_t$ lifts to a homotopy $\tilde{H}_t:\tilde{X}\rightarrow \tilde{X}$ such that $H_t(x)=f(\tilde{H}_t(x))$ and $\tilde{H}_0(x)=\tilde{x}_0$
So we have a homotopy $\tilde{H}_t:\tilde{X}\rightarrow \tilde{X}$ from $\tilde{H}_0(x)= \tilde{x}_0$ to $\tilde{H}_1(x)$ and besides $f(x)=H_1(x)=f(\tilde{H}_1(x))$.
If $f$ was injective we would be done, but in principle $\tilde{H}_1(x)$ could be any point in $f^{-1}(x_0)$ right?
|
Since $f$ is nullhomotopic, $f_*:\pi_n(\tilde X)\to \pi_n(X)$ are trivial for all $n$. Consequently $\pi_n(\tilde X)$ are all trivial. Whitehead theorem implies $\tilde X$ is contractible.
|
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|
In group theory, is it true that $f(X \vee Y) = f(X) \vee f(Y)$? ($\vee$ denotes join).
Let $G$ and $H$ denote Abelian groups, $X$ and $Y$ denote subalgebras of $G$, and let $f : G \rightarrow H$ denote a homomorphism. Then:
$$f(X \vee Y) = f(X+Y) = f(X)+f(Y) = f(X) \vee f(Y)$$
So $f(X \vee Y) = f(X) \vee f(Y)$.
Now suppose we relax our conditions, so that $G$ and $H$ are no longer assumed Abelian.
Question. Is it still provable that $f(X \vee Y) = f(X) \vee f(Y)$?
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That we have $f(X\vee Y)\subseteq f(X)\vee f(Y)$ should be clear (the image of a product of elements in $X$ or $Y$ is a product of elements in $f(X)$ or $f(Y)$).
For the other direction, let $g\in f(X)\vee f(Y)$. Then $g=g_1\cdot\cdots\cdot g_n$ where $g_i\in f(X)$ or $g_i\in f(Y)$. Thus each $g_i$ is of the form $f(x_i)$ where $x_i\in X$ or $x_i\in Y$. Thus we can rewrite $g=f(x_1\cdot\cdots\cdot x_n)$. Thus $g\in f(X\vee Y)$.
This approach can be generalized for arbitrary collections of subgroups since it didn't matter that $\{X,Y\}$ was a finite collection. The only thing that was important was that elements were a finite product of other elements. Thus for any collection of subgroups $\cal S$ of $G$, we have that $f(\bigvee_{S\in\cal S} S)=\bigvee_{S\in\cal S}f(S)$.
The dual question is however false. In general, we do not have that $f(X\wedge Y)=f(X)\wedge f(Y)$ for arbitrary subgroups $X$ and $Y$, specifically the inclusion $f(X)\wedge f(Y)\leq f(X\wedge Y)$ need not hold (the other inclusion always holds). I was having trouble finding a counterexample, so I asked here where Marcin Łoś gave a counterexample in the comments. Thus in general, a group homomorphism does not induce a lattice homomorphism between the subgroup lattices.
However, if $f$ is a quotient map (or surjective by the 1st isomorphism theorem), the 4th isomorphism theorem does give us some relation between subgroup intersections.
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|
Evaluating $\int_{-\infty}^\infty \frac{\sin x}{x-i} dx$ I would like to evaluate the integral
$$\int_{-\infty}^\infty \frac{\sin x}{x-i} dx,$$
which I believe should be equal to $\frac{\pi}{e}$. However, I cannot reproduce this result by hand. My work is as follows: first, we evaluate the indefinite integral.
\begin{align*}
\int \frac{\sin x}{x-i} dx &= \int \frac{\sin(u+i)}{u} du \text{ where }u=x-i \\
&= \int \frac{\sin u \cos i + \cos u \sin i}{u} du \\
&= \mathrm{Si}(u) \cos i + \mathrm{Ci}(u) \sin i \\
&= \mathrm{Si}(x-i) \cosh 1 + i\mathrm{Ci}(x-i) \sinh 1 \\
\end{align*}
Then, we insert the bounds.
\begin{align*}
\int_{-\infty}^\infty \frac{\sin x}{x-i} dx &= \mathrm{Si}(\infty-i) \cosh 1 + i\mathrm{Ci}(\infty-i) \sinh 1 \\
&\phantom{=}-\mathrm{Si}(-\infty-i) \cosh 1 - i\mathrm{Ci}(-\infty-i) \sinh 1 \\
&= \frac{\pi}{2}\cosh 1 + 0 +\frac{\pi}{2} \cosh 1 + \pi \sinh 1 \\
&= \pi (\cosh 1 + \sinh 1) \\
&= \pi e
\end{align*}
I assume I have made some mistake in manipulating the complex sine and cosine integrals, since the result would be correct if $\mathrm{Ci}(-\infty-i)$ were evaluated to $-i \pi$. However, I cannot pinpoint the error.
|
The mistake is simple--$\mathrm{Ci}$ has a branch cut across the negative real axis, so $\mathrm{Ci}(-\infty - i)$ should indeed evaluate to $-i \pi$ rather than $i \pi$.
|
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|
Evaluate $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-\frac{1}{2}(x^2-xy+y^2)}dx\, dy$ I need to evaluate the following integral:
$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-\frac{1}{2}(x^2-xy+y^2)}dx\, dy$$
I thought of evaluating the iterated integral $\displaystyle\int_{-\infty}^{\infty}dx\int_{-\infty}^{\infty}e^{-\frac{1}{2}(x^2-xy+y^2)}dy$, but because of the presence of $x^2$ and $y^2$ terms, I am not being able to do that. I tried substituting $x=r\cos \theta$ and $y=r\sin \theta$ but in that case I have some confusion regarding the limits of $r$ and $\theta$. Can I get some help?
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Recall:
$$x^2-xy+y^2=(x-\frac{1}{2}y)^2+\frac{3}{4}y^2$$
With that you get:
$$\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-\frac{1}{2}(x^2-xy+y^2)}dxdy=\\\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-\frac{1}{2}\left((x-\frac{1}{2}y)^2+\frac{3}{4}y^2\right)}dxdy=\\
\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-\frac{1}{2}\left(x'^2+\frac{3}{4}y^2\right)}dx'dy=\\
\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-\frac{1}{2}x'^2}dx'e^{-\frac{3}{8}y^2}dy=\\
\int_{-\infty}^\infty\sqrt{2\pi}e^{-\frac{3}{8}y^2}dy=\\
\sqrt{2\pi}\sqrt{\frac{8}{3}\pi}=\frac{4}{\sqrt{3}}\pi
$$
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|
$\int_\mathbb{R} \bigg( \frac{1}{h} \int_x^{x+h} |f(t)| dt\bigg) dx= ||f||_{L^1}$? $$\int_\mathbb{R} \bigg( \frac{1}{h} \int_x^{x+h} |f(t)| dt\bigg) dx= ||f||_{L^1} \;\;?$$
I worked out that the equality holds for each $\chi_{[a,b]}$, therefore it holds for each piecewise constant function. By a density argument, it must hold for all functions in $L^1(\mathbb{R})$.
*
*Is this correct?
*If it were true, I feel like there must be an easier argument for this inequality; I just can't quite see it. Maybe using Hardy-Littlewood maximal inequality or Markov's inequality, etc.
Thank you very much for the help!
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Your argument is correct, if you note that the left hand side of the equality is a continuous function of $f\in L^1$. Then the denseness perpetuates the equality from the simple functions to all of $L^1$.
But there is an easier argument, change the order of integration:
$$\begin{align}
\int_\mathbb{R} \left(\int_x^{x+h} \lvert f(t)\rvert\,dt\right)\,dx
&= \iint\limits_{x \leqslant t \leqslant x+h} \lvert f(t)\rvert\,dt\,dx\\
&= \iint_{t-h \leqslant x \leqslant t} \lvert f(t)\rvert\,dx\,dt\\
&= \int_\mathbb{R} h\lvert f(t)\rvert\,dt\\
&= h\lVert f\rVert_{L^1}.
\end{align}$$
Since the integrand is non-negative and measurable, the change of order of integration is unproblematic.
|
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|
Exercise 1.13 of chapter 1 of Revuz and Yor's This is the exercise 1.13 of chapter 1 of Revuz and Yor's.
*
*Let $B$ be the standard linear BM. Prove that $\varlimsup_{t\to\infty}(B_t/\sqrt{t})$ is a.s. $>0$ (it is in fact equal to $+\infty$ as will be seen in Chap. 2)
*Prove that $B$ is recurrent, namely: for any real $x$, the set $\{t: B_t=x\}$ is unbounded.
*Prove that the Brownian paths are a.s. nowhere locally Holder continuous of order $\alpha$ if $\alpha>\frac{1}{2}$.
[Hint: Use the invariance properties of Proposition (1.10).]
What I tried:
For 1. $P(\varlimsup_{t\to\infty}B_t/\sqrt{t}>0)=1-P(\varlimsup_{t\to\infty}B_t/\sqrt{t}\leq 0)$ and fix $\varepsilon>0$ there exist large enough $t_0$ s.t. $P(\varlimsup_{t\to\infty}B_t/\sqrt{t}\leq 0)\leq P(B_{t_0}/\sqrt{t_0}<\varepsilon)$<1/2, then $P(\varlimsup_{t\to\infty}B_t/\sqrt{t}>0)>1/2$. By Hewitt-Savage zero-one law, $P(\varlimsup_{t\to\infty}B_t/\sqrt{t}>0)=1$. But I am not very familier with HS 0-1 law, is there some other method to solve this problem? (also not using law of iterated logarithm)
For 2. By 1, $P(\varliminf_{t\to\infty}B_t/\sqrt{t}<0)=P(\varlimsup_{t\to\infty}(-B_t)/\sqrt{t}>0)=P(\varlimsup_{t\to\infty}B_t/\sqrt{t}>0)=1$, so $x=0$ is recurrent. But how to prove it when $x\neq0$?
For 3. It is sufficient to prove $P\left(\sup_{0\leq s,t\leq 1}\frac{|B_t-B_s|}{\sqrt{|t-s|}}=+\infty\right)=1$. It can be proved by the independent increment property and scaling property of Brownian Motion.
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Re 1., note that the random variable $X=\limsup\limits_{t\to\infty}B_t/\sqrt{t}$ is asymptotic hence, by Kolmogorov zero-one law, if $P(X\gt0)\lt1$, then $P(X\gt0)=0$, that is, $X\leqslant0$ almost surely. If this holds, then, by symmetry, $\liminf\limits_{t\to\infty}B_t/\sqrt{t}\geqslant0$ almost surely, thus, $B_t/\sqrt{t}\to0$ almost surely.
In particular, this would imply that $B_t/\sqrt{t}\to0$ in distribution, which is absurd since every $B_t/\sqrt{t}$ is standard normal. Hence $X\gt0$ almost surely.
|
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|
Hide my invoice number I'm not a mathematician, so please forgive any ignorance.
I have a small business - I'm generating invoices incrementally. I'm currently on about invoice number 4000.
I guess I don't want my customers knowing how much business I'm doing (i.e. if they get an invoice for 4500, they know I've been doing a lot of business lately. However, if they get an invoice for 4010, they know things are slow).
So, my question is: how can I map say, 4500, to a guaranteed-to-be unique, human-readable/human-rememberable sequence?
e.g.
4500 -> a48b82w
4501 -> b802aq2
4502 -> qi289a1
etc.
Is there a quick mathematical function that can do this? I have no idea...
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Why make such a simple problem so complex?
Most e-commerce and invoice software supports starting numbers and increments, so just pick a large number and increment by another odd number. So something like:
1034578 = first invoice number
32876 = increment amount
So your invoices would follow 1034578 + 32876X like so:
*
*1067454
*1100330
*1133206
*etc.
For any person to know your invoice total they would need to determine the initial invoice number (improbable), and determine your increment (fairly easy).
|
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|
What's the name of this algebraic property? [Complementary Subgroup Test] I'm looking for a name of a property of which I have a few examples:
$(1) \quad\color{green}{\text{even number}}+\color{red}{\text{odd number}}=\color{red}{\text{odd number}}$
$(2) \quad \color{green}{\text{rational number}}+\color{red}{\text{irrational number}}=\color{red}{\text{irrational number}}$
$(3) \quad\color{green}{\text{algebraic number}}+\color{red}{\text{transcendental number}}=\color{red}{\text{transcendental number}}$
$(4) \quad\color{green}{\text{real number}}+\color{red}{\text{non-real number}}=\color{red}{\text{non-real number}}$
If I were to generalise, this, I'd say that if we partition a set $X$ into two subsets $S$ and $S^c=X\setminus S$, then the sum of a member of $S$ and a member of $S^c$ is always in either $S^c$ or $S$.
My question is:
"Is there a name for this property (in these four cases) and is this property true in general?"
Also, does anyone have any more examples of this property?
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Don't prime numbers offer a counter-example to the general truth of this property?
Prime $+$ not-prime $=$ not-prime $==> 17 + 4 = 21$
Prime $+$ not-prime $=$ prime $==> 7+4=11$
|
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|
Polynomial $f(x)$ degree problem. Suppose the polynomial $f(x)$ is of degree $3$ and satisfies $f(3)=2$, $f(4)=4$, $f(5)=-3$, and $f(6)=8$. Determine the value of $f(0)$.
How would I solve this problem? It seems quite complicated...
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If you just want to use subtraction, you could take first differences
2 4 -3 8
2 -7 11
-9 18
27
then expand this to the left
-328 -137 -36 2 4 -3 8
191 101 38 2 -7 11
-90 -63 -36 -9 18
27 27 27 27
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|
Generalising integration by parts for the product of more than two functions Just as the product rule can be generalised to the product of more than two functions, i.e. $$\frac{d}{dx} \left [ \prod_{i=1}^k f_i(x) \right ]
= \sum_{i=1}^k \left(\frac{d}{dx} f_i(x) \prod_{j\ne i} f_j(x) \right)
= \left( \prod_{i=1}^k f_i(x) \right) \left( \sum_{i=1}^k \frac{f'_i(x)}{f_i(x)} \right),$$ is there a way to generalise integration by parts to evaluate $$\int f_1(x) f_2(x) \cdots f_n(x) dx \qquad ?$$
For contextual purposes only, I'm trying to evaluate by hand
$$\int x \cosh(x+1)e^x\sin(x) dx \quad
.
$$
I know that I could let $u=x \cosh(x+1)$ and $v \ '=e^x\sin(x)$ but that would require integration by parts to be performed at least three times.
Is there a more-efficient way to evaluate this integral using the proposed 'generalisation'?
Thanks!
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The Wikipedia article on integration by parts gives you the generalization you're looking for:
$$\Bigl[ \prod_{i=1}^n u_i(x) \Bigr]_a^b = \sum_{j=1}^n \int_a^b \prod_{i\neq j}^n u_i(x) \, du_j(x),$$
where $u_i(x)$ are your $n$ functions of $x$ that are terms of the product that comprise your integrand.
|
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|
How do we define the mirror image of a knot in general 3-manifolds How do we define the mirror image of a knot in general oriented 3-manifolds ? For instance for a knot in an irreducible integer homology sphere.
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One method would be to express the manifold as an open book decomposition (see http://en.wikipedia.org/wiki/Open_book_decomposition). Then project the knot onto one of the pages (fibers) of the decomposition recording over an under crossings. Switch over crossings to undercrossings and vice verse. You might want to see my paper Reidemeister's Theorem in Three Manifolds http://sci-prew.inf.ua/v110/2/S0305004100070353.pdf.
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Applications of Algebra in Physics Often I have heard about the link between Algebra (in particular Representations of Groups and Algebras) and some "indefinite" field of Physics.
I have a good preparation in Algebra and Representation Theory (in particular about Representations of Lie Algebras), and I'm fascinated with Physics. My idea is try to understand this link and eventually study it with more depth.
Hence I'm looking for an introductory book that emphasizes the applications of Algebra in Physics from a comprehensible and mathematical point of view.
Does anyone have an idea for a book with these requisites?
Thank you!
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Peter Woit, the author of the book "Not Even Wrong" and a blog by the same name, has been working on a book on quantum mechanics as described by representation theory. The latest draft may be found at the following link:
Quantum Theory, Groups and Representations:
An Introduction.
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|
Is there a way to calculate the area of this intersection of four disks without using an integral? Is there anyway to calculate this area without using integral ?
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Let $R$ be its radius and $D$ its diameter: $R = 5$, $D = 10$.
$$\begin{align}
\text{Area of big square} &= D^2 = 100 \\
\text{Area of circle} &= \frac{\pi D^2}{4} \approx 78.54 \\
\text{Area outside circle} &= 100 - 78.54 = 21.46 \\
\text{Area of 4 petals} &= 78.54 - 21.46 = 57.08 \\
\text{Area of single petal} &= \frac{57.08}{4} = 14.27 \\
\text{Area of small square} &= R^2 = 25
\end{align}$$
Let $x$ denote the area of the portion selected.
$$\begin{align}
\text{Area of 2 petals} &= 2 \cdot 14.27 = 28.54 \\
0 &= 25 - 28.54 + x \\
\text{Area of 8 petals} &= 2 \cdot 57.08 = 114.16 \\
\text{OR}\\
100 - 114.16 + 4 x &= 0
\end{align}$$
$$ x = 3.54 $$
Here we have a small area which needs to be added,
that I found out by modeling to be $4.34$
$$\begin{align}
\text{This gives us the desired area of}\\
\text{4.34+3.54} &\approx 7.88\\
\text{& a percentage of} &\approx 4 \times 7.88\\
\text{that is} &\approx 31.52\%
\end{align}$$
|
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|
Solutions to functional equation $f(f(x))=x$ Is there any more solutions to this functional equation $f(f(x))=x$?
I have found: $f(x)=C-x$ and $f(x)=\frac{C}{x}$.
|
If you don't make any niceness assumptions about $f$, there are lots. Partition $\Bbb{R}$ (or whatever you want $f$'s domain to be) into $1$- and $2$-element subsets, in any way you like. Then define $f(x)=y$, if $\{x, y\}$ is in your partition, or $f(x)=x$, if $\{x\}$ is in your partition.
Moreover, any such $f$ yields such a partition, into the sets $\{x, f(x)\}$. So this is a complete description of all solutions.
Of course, $f$ will be wildly discontinuous for most choices of partition.
|
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|
Difference isometrically isomorphism and homeomophism. What is the difference between isometrically isomorphism and homeomorphism?is an isometric mapping is continuous?
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An isometry is a map $f:X\to Y$ between metric spaces that preserves distances: $d_Y(fx, fy) = d_X(x, y)$. Such maps are automatically continuous (just use the $\delta$-$\epsilon$ definition of continuity) and injective, but they may not be surjective; an isometric isomorphism is one that's both a bijection and an isometry. The inverse $f^{-1}$ is then also an isometry and thus continuous, so $f$ is also a homeomorphism. Homeomorphisms aren't necessarily isometries; take $x\to x^3$ on $\mathbb{R}$ with the usual metric.
|
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Non integer derivative of $1/p(x)$ I need to find the $k$'th derivative of $1/p(x)$, where $p(x)$ is a polynomial and $k\in\mathbb{R}$
It dosen't have to be an explicit formula, an algorithm which finds a formula for some $k$ is fine.
|
Why don't you just use the formula?
For generic $f$ and for 0<$\alpha$<1 you have:
$$D^\alpha(f(x))=\frac{1}{\Gamma(1-k)}\frac{\partial}{\partial x}\int_0^x\frac{f(t)}{(x-t)^\alpha}\,\mathrm{d}t$$
So after deriving [k] (the integer part) times you have the remainder $\alpha\in[0,1]$:
$$D^\alpha(D^{[k]}\left(\frac{1}{p(x)}\right))=\frac{1}{\Gamma(1-k)}\frac{\partial}{\partial x}\int_0^x\frac{1}{(x-t)^k}D^{[k]}\left(\frac{1}{p(x)}\right)\,\mathrm{d}t$$
|
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How to find an angle in range(0, 360) between 2 vectors? I know that the common approach in order to find an angle is to calculate the dot product between 2 vectors and then calculate arcus cos of it. But in this solution I can get an angle only in the range(0, 180) degrees. What would be the proper way to get an angle in range of (0, 360)?
|
Before reading this answer - Imagine your angle in a 3D space - you can look at it from the "front" and from the "back" (front and back are defined by you). The angle from the front will be the opposite of the angle that you see from the back. So there is no real sense in a value in a range larger than $[0,180]$.
If you still want to read more, enjoy
In the 3D case, your two vectors would be on some plane (the plane that you can get its normal from the cross-product of the two vectors). Getting a $[0,180]$ degrees angle is of course possible by computing $arccos(\frac{\vec{a}*\vec{b}}{|\vec{a}| *|\vec{b}|})$.
I think that what you can do, is to fix the Z axis, such that the two vectors will only differ in X and Y. Then you solve a 2D geometry problem. You should be able to fix the Z axis by dividing the two vectors by their Z component. Since the Z component is just a scalar, the vector directions will remain the same (the trend might change though, if this scalar is negative). You should remember whether these two scalars were both positive, both negative or one positive and one negative. In the last case, you're final results will be flipped!
If the Z component of the two vectors is $0$, then this step can (and actually must) be skipped. However, if only one of them has $0$ in its Z component, then it is probably impossible to compute the very precise angle, but you can still compute an approximation, by dividing the other vector by a very large number.
Having that, you can reduce vector $\vec{a}$ from both vectors $\vec{a}$ and $\vec{b}$ and then add the vector $(1, 0, 0)$ to both. Thus, vector $\vec{a}$ will be $(1, 0, 0)$, while vector $\vec{b}$ will be $(b_x, b_y, 0)$.
Now, if $b_y$ is positive, than your angle is $arccos(\frac{\vec{a}*\vec{b}}{|\vec{a}| *|\vec{b}|})$. Else, it is $360 - arccos(\frac{\vec{a}*\vec{b}}{|\vec{a}| *|\vec{b}|})$. Remember to use the opposite result if only one of the original Z values was negative, as described above!
|
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Problem involving decomposition of measures Let $\mu$ be a signed measure. We wish to prove that $$\left| \int{f} \> d\mu \right| \leq \int{|f|} \> d|\mu|.$$
(We are given the following defintion: $\int{f} \> d\mu = \int{f} \> d\mu^{+} - \int{f} \> d\mu^{-}$.)
For the proof, I want to just write $$\left| \int{f} \> d\mu \right| = \left| \int{f} \> d\mu^{+} - \int{f} \> d\mu^{-} \right| \leq \left| \int{f} \> d\mu^{+} \right| + \left| \int{f} \> d\mu^{-} \right| \leq \int{|f|} \> d\mu^{+} + \int{|f|} \> d\mu^{-} = \int{|f|} \> d|\mu|.$$
But, I feel as though I am missing justification for the last (and, obviously, most important) equality. That is, I'm sure if that follows from the definition provided to me. Any help would be great!
|
The last equality rests on the fact that $$|\mu|=\mu^++\mu^-,\tag{1} $$ where $\mu= \mu^+-\mu^-$ (Jordan-Hahn decomposition), where $\mu^+$ and $\mu^-$ are positive and muutually singular.
Equality (1) is either the definition of $|\mu|$, or it canbe deduced from the definition of $|\mu|$ with partitions using the supports of $\mu^+$ and $\mu^-$.
|
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|
Finite additivity in outer measure Let $\{E_i\}_{i=1}^n$ be finitely many disjoint sets of real numbers (not necessarily Lebesgue measurable) and $E$ be the union of all these sets. Is it always true that
$$
m^\star (E)=\sum_{i=1}^N m^\star(E_n)
$$
where $m^\star$ denotes the Lebesgue outer measure? If not, please give a counterexample. The Vitali set is a counterexample in the countable case, but I am not sure whether it is false in finite case.
|
The sentence "$m^{*}$ is not finitely additive" is independent from the theory $ZF+DC$.
Fact 1. (ZF)+(AC). $m^{*}$ is not finitely additive.
Proof
Let $X$ be a subset of the real line $R$. We say that $X$ is a Bernstein set in $R$ if
for every non–empty perfect set $P \subseteq R$ both sets
$P \cap X$, $P \cap (R \setminus X)$
are non–empty.
It is well known that there exists a Bernstein subset $X$ of $R$.
Let consider two sets $[0,1]\cap X$ and $[0,1]\setminus X$. Then
(i) $m_{*}([0,1]\cap X)=m_{*}([0,1]\setminus X)=0$;
(ii) $m^{*}([0,1]\cap X)=m^{*}([0,1]\setminus X)=1$.
Indeed, since both $[0,1]\cap X$ and $[0,1]\setminus X$ (like $X$ and $R \setminus X$) no contain any non–empty perfect subset
we claim that that (i) holds true.
If we assume that $m^{*}([0,1]\cap X)<1$ then there will be a non–empty perfect subset $Y \subset [0,1]$ such that $Y \cap ([0,1]\cap X)=\emptyset$. The later relation implies that $R \setminus X$ like $[0,1]\setminus X$ contains a non–empty perfect set and we get a contradiction that $X$ is a Bernstein subset of $R$.
The proof of the equality $m^{*}([0,1]\setminus X)=1$ is similar.
Hence we have
$$m^{*}(([0,1]\cap X))\cup ([0,1]\setminus X))=m^{*}([0,1])=1 <1+1=m^{*}([0,1]\cap X)+m^{*}([0,1]\setminus X).
$$
Fact 2. (Solovay model= (ZF)+(DC)+(every subset of $R$ is Lebesgue measurable)). $m^{*}$ is finitely additive.
Proof. In Solovay model $m^{*}=m$. Hence $m^{*}$ is finitely additive since $m$ is finitely additive.
|
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|
Solving this Exponential Equation This is the equation that I need help with. The fact that there is that extra $1$ is throwing me off. If you move the $4^x$ term over and take the $\log$ of both sides, then you have a $\log$ with a polynomial inside.
$$5^x - 4^x = 1$$
|
The function $y(x)=5^x-4^x$ is negative if $x<0$ . So $x<0$ is excluded. For $x>0$ the function is strictly increasing from $0$ to $\infty$. So, there is only one value of $x$ so that $y=1$.
Obviously this value is $x=1$ because $y(1)=5^1-4^1=1$
|
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|
Mathematician who talked about the probability of a "good" graph? In my undergraduate years, one of my professors always talked about this one mathematician who was talking about "good" graphs and wondered about the existence of such a graph. Apparently this mathematician could not find such a graph, and then proceeded to show that the probability that such a graph existed was 1 by using a probability measure.
Does anyone know to whom my professor was referring? [I apologize if such a question is inappropriate for this site.]
|
Not sure without more context, but my best guess is the mathematician was Paul Erdős and technique you are talking about is the Probabilistic Method.
|
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|
$F/K$ algebraic and every nonconstant polynomial in $K[X]$ has a root in $F$ implies $F$ is algebraically closed. Let $F/K$ be an algebraic extension of fields in characteristic zero. If $F/K$ is normal, and every nonconstant polynomial $f \in K[X]$ has a root in $F$, then $F$ is algebraically closed. This is obvious, because if $u$ is algebraic over $F$ (let us agree to fix some algebraic closure $\overline{K}$ of $K$ containing $F$ with $u \in \overline{K}$), then it is algebraic over $K$ with minimal polynomial $f \in K[X]$, which by hypothesis must have one and therefore all of its roots in $F$. In particular $u \in F$.
I have heard that the above assertion remains true if we drop the assumption that $F/K$ is normal. But it seems to be a nontrivial result. Can anyone get me started on how to prove this?
Thoughts so far: It's enough to show that every algebraic extension of $K$ is $K$-isomorphic to a subfield of $F$. Certainly this is true for every finite extension $E$ of $K$; we're in characteristic zero, so $E = K(v)$ for some $v \in E$. If $f \in K[X]$ is the minimal polynomial of $v$ over $K$, then $f$ has some root $u \in F$, whence $E$ is $K$-isomorphic to the subfield $K(u)$ of $F$.
So, I'm thinking next use some kind of Zorn's Lemma argument?
|
Let $E/F$ be a finite extension (which is separable as the characteristic is zero), we will prove that $E=F$, hence $F$ is algebraically closed:
Let $a$ be a primitive element of $E/F$. Then $a$ is algebraic over $K$, so there exists a finite Galois extension $N/K$ containing $a$, in particular
$$
E\subseteq NF.
$$
(One can take $N$ to be the Galois closure of $K(a)$, for example.)
Let $f\in K[x]$ be the minimal polynomial of a primitive element of $N$ over $K$. Since $N/K$ is Galois, any of the roots of $f$ generates $N$ over $K$.
Now, by assumption $f$ has a root in $F$, so $N\subseteq F$, hence $E\subseteq NF=F$. This proves that $F$ has no non-trivial extensions, hence $F$ is algebraically closed.
|
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|
Proving that $f(x)=2^x$ is $O(x^2)$ Can someone help me with this problem? I don't really know what to do if the x is in exponential form.
|
To see that this is way false, we just observe:
$$2^x\le Mx^2\iff x\log 2\le \log M+2\log x$$
$$\iff x\le \log_2(M)+2\log_2(x)$$
But it is clear from L'Hôpital's rule (among other things) that this is false, since
$$\lim_{x\to\infty} {M\over x\log 2}+{2\log x\over x\log 2}=\lim_{x\to\infty} {2/x\over \log 2}=0$$
(in fact this would imply)
$$x=O(\log x)$$
|
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|
Smooth surfaces that isn't the zero-set of $f(x,y,z)$ The zero-set of any smooth function $f(x,y,z)$ with a non-vanishing gradient is a smooth surface. I was wondering if the reverse is true: is every smooth surface in $E^3$ the zero-set of some smooth function? If not, what do the counterexamples look like?
I was thinking that a plane with a hole may qualify as a counterexample, but I have yet to prove it.
|
Georges Elencwajg's "easy proof" is providing a "global solution" to your problem. Locally one can argue as follows: A smooth surface $S\subset{\mathbb R}^3$ is produced by a $C^1$-map
$${\bf g}:\quad(u,v)\mapsto\left\{\eqalign{x&=g_1(u,v)\cr y&=g_2(u,v)\cr z&=g_3(u,v)\cr}\right.$$ with $d{\bf g}(u,v)$ having rank $2$ at all points $(u,v)$ in sufficiently small neighborhoods $U$ of ${\bf p}$. We may assume that
$$\det\left[\matrix{g_{1.u}&g_{1.v}\cr g_{2.u}& g_{2.v}\cr}\right]\ne0\qquad\bigl((u,v)\in U\bigr)\ .$$
By the inverse function theorem it follows that there is a neighborhood $V$ of ${\bf q}=\bigl(g_1({\bf p}), g_2({\bf p})\bigr)$ and a $C^1$-inverse ${\bf h}'=(h_1,h_2)$ of ${\bf g}'=(g_1,g_2)$ mapping $V$ onto $U$. This mapping ${\bf h}'$ computes $(u,v)\in U$ for given $(x,y)$ in $V$. From this we can conclude that $S$ can be as well be presented by the map
$${\bf f}:\quad(x,y)\mapsto\bigl(x,y,\>g_3\bigl(h_1(x,y),h_2(x,y)\bigr)\bigr)\ .$$
Letting
$$g_3\bigl(h_1(x,y),h_2(x,y)\bigr)=:f(x,y)$$
the surface $S$ appears as solution set of the equation
$$z-f(x,y)=0\ .$$
|
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|
Lexicographical rank of a string with duplicate characters Given a string, you can find the lexicographic rank of the string using this algorithm:
Let the given string be “STRING”. In the input string, ‘S’ is the first character. There are total 6 characters and 4 of them are smaller than ‘S’. So there can be 4 * 5! smaller strings where first character is smaller than ‘S’, like following
R X X X X X
I X X X X X
N X X X X X
G X X X X X
Now let us Fix S and find the smaller strings staring with ‘S’.
Repeat the same process for T, rank is 4*5! + 4*4! +…
Now fix T and repeat the same process for R, rank is 4*5! + 4*4! + 3*3! +…
Now fix R and repeat the same process for I, rank is 4*5! + 4*4! + 3*3! + 1*2! +…
Now fix I and repeat the same process for N, rank is 4*5! + 4*4! + 3*3! + 1*2! + 1*1! +…
Now fix N and repeat the same process for G, rank is 4*5! + 4*4 + 3*3! + 1*2! + 1*1! + 0*0!
Rank = 4*5! + 4*4! + 3*3! + 1*2! + 1*1! + 0*0! = 597
Since the value of rank starts from 1, the final rank = 1 + 597 = 598
I would like to know how to find the rank if the string contains duplicate characters. When to divide by factorial of repeated occurrences ?
|
There's a similar process, complicated by counting permutations of strings with duplicates. For example, the number of permutations of AAABB is $5!/3!2!$.
With that in mind, here's how we could find the rank of BCBAC. We count the smaller permutations $s$ by considering the first position where $s$ is smaller. For example, if it's position 1, $s$ looks like A followed by a permutation of the remaining letters {BBCC}, of which there are $4!/2!2!$.
1: A + {BBCC} $\to$ 4!/2!2!
2: BB + {ACC} $\to$ 3!/1!2!, BA + {BCC} $\to$ 3!/1!2!
3: BCA + {BC} $\to$ 2!/1!1!
4: BCB? (not possible) $\to$ 0
So the answer is 6 + (3 + 3) + 2 + 0 = 14.
|
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|
Help with simple rotation on an x,y plane I'm a programmer, with too little background in mathematics, and I am currently faced with the challenge of rotating an object on a 2 axis plane.
Something that is hopefully quite easy for you guys.
Anyway, to be concrete,
what I need is, say I have an object (image) on an x,y plane and I know its exact coordinates, the corners, height, width, everything on that plane.
Now its rotated by say X degrees. Is there a formula to get the new position of the four corners ? or one corner from which I will deduce all the other data I need ?
Thanks in advance to any good helping soul!
Ariel
|
Take one of the corners, say it has coordinates $(x,y)$. Let $\alpha$ be the angle that $(x,y)$ makes with the positive $x$-axis and let $r=\sqrt{x^2+y^2}$. Then $x=r\cos\alpha$ and $y=r\sin\alpha$. Let's say you are rotating by an angle of $\theta$. Call this linear transformation $T_\theta:\mathbb{R}^2\to\mathbb{R}^2$. Then $T_\theta(x,y)$ makes an angle of $\alpha+\theta$ with the positive $x$-axis. Thus, $$T_\theta(x,y)=(r\cos(\alpha+\theta),r\sin(\alpha+\theta))=(x\cos\theta-y\sin\theta,x\sin\theta+y\cos\theta)$$
That is, $$T_\theta=\left[\begin{array}{c c} \cos\theta&-\sin\theta\\\sin\theta &\cos\theta\end{array}\right]$$
|
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|
Where can I find the proof of this Ramanujan result? I'm searching for a proof of one impressive Ramanujan result. Not one in particular, the only request I have is to be really impressive.
For example
$$
\sqrt{\phi+2}-\phi=\frac{e^{-2\pi/5}}{1+\frac{e^{-2\pi}}{1+\frac{e^{-4\pi}}{1+\cdots}}}
$$
where $\phi=\frac{1+\sqrt5}{2}$.
Or maybe
$$
\frac1{\pi}=\frac{2\sqrt2}{9801}\sum_{n=0}^{+\infty}\frac{(4n)!(1103+26390n)}{(n!)^4396^{4n}}\;.
$$
Can someone suggest me a precise reference where to find such a proof?
Thank you all
|
The series for $1/\pi$ is proved in J. M. Borwein and P. B. Borwein, Pi and the AGM; A Study in Analytic Number Theory and Computational Complexity, Wiley, New York, 1987.
See also Motivation for Ramanujan's mysterious $\pi$ formula
|
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$f$ is twice differentiable, $f + 2 f^{'} + f^{''} \geq 0$ , prove the following Let $ f : [0,1] \rightarrow R$. $f$ is twice diff.
and $f(0) = f(1) = 0$
If
$f + 2 f^{'} + f^{''} \ge 0$ , prove that $f\le 0$ in the domain.
Please don’t give complete solution, only hints.
|
Hint Let $g(x)=f(x)e^x$. Then
$$g''=(f+2f'+f'')e^x \geq 0 \,.$$
That means that $g$ is....
How does this solve the problem?
|
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|
Evaluation of Sum of $ \sum_{n=1}^{\infty}\frac{\sin (n)}{n}$. If $\displaystyle S = \sum_{n=1}^{\infty}\frac{\sin (n)}{n}.$ Then value of $2S+1 = $
Using Fourier Series Transformation I am Getting $2S+1=\pi$
But I want to solve it Using Euler Method and Then Use Logarithmic Series.
$\bf{My\; Try::}$ Using $\displaystyle \sin (n) = \left(\frac{e^{in}-e^{-in}}{2i}\right)$. So $\displaystyle S = \sum_{n=1}^{n}\frac{\sin (n)}{n} = \frac{1}{2i}\sum_{n=1}^{\infty}\frac{e^{in}}{n}-\frac{1}{2i}\sum_{n=1}^{\infty}\frac{e^{-in}}{n}$
Now Using $\displaystyle \ln(1-x) = -x-\frac{x^2}{2}-\frac{x^3}{3}...............\infty$
So Let $\displaystyle S = -\frac{1}{2i}\ln(1-e^{i})+\frac{1}{2i}\ln(1-e^{-i})$
Now How can I solve after that
Help me
Thanks
|
my attempt :
$$\ \ S=\sum_{n=1}^{\infty } \frac{sin(n)}{n}=\sum_{n=1}^{\infty }
= \int_{0}^{\infty } e^{-nw}\sin(n)dw\\ \\ \\$$
$$\therefore S=Im\int_{0}^{\infty }\sum_{n=1}^{\infty
}e^{-(w-i)n}dw=Im\ \int_{0}^{\infty
}\frac{1}{e^{w-i}}dw=Im\int_{0}^{\infty
}\frac{dw}{cos(1)e^{w}-isin(1)e^{w}-1}\\ \\ \\$$
$$\therefore S=Im\int_{0}^{\infty
}\frac{e^{w}cos(1)-1+isin(1)e^{w}}{(cos(1)e^{w}-1)^{2}+(sin(1)e^{w})^{2}}dw=\int_{0}^{\infty
}\frac{sin(1)\ e^{-w}}{sin^{2}(1)+(cos(1)-e^{-w})^{2}}dw\\ \\ \\$$
$$\therefore S=\lim_{w\rightarrow \infty }\ tan^{-1}\left (
\frac{cos(1)-e^{-w}}{sin(1)} \right )-tan^{-1}\left (
\frac{cos(1)-1}{sin(1)} \right )\\ \\ \\$$
$$\therefore S=tan^{-1}\left ( cot(1) \right )-tan^{-1}\left ( cot(1)-csc(1) \right )=\frac{\pi }{2}-1+\frac{1}{2}=\frac{1}{2}\left ( \pi -1 \right )$$
|
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Fundamental Theorem of Calculus and limit I've been reading through a paper and my question has essentially came down to this:
Let
$f(\beta) \to M$ as $ \beta \to 0$ and $f(\alpha) \to 0$ as $\alpha \to \infty.$ Prove that $M=-\int_{0}^\infty f'(x)dx$. I wanted to check this before going any further. It is probably trivial but here is my attempt.
My attempt:
$$M=\lim_{\beta \to 0}f(\beta)$$
$$M=\lim_{\beta \to 0 , \alpha \to \infty}\left[f(\beta)- f(\alpha)\right]$$
$$M=-\lim_{\beta \to 0,\alpha \to \infty}\left[\int_\beta^\alpha f'(s)dx\right]$$
$$M=-\displaystyle \int_0^{\infty} f'(x)dx$$
as required.
|
Right , just the signs are here and there.Rest is fine
|
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If one number is thrice the other and their sum is $16$, find the numbers If one number is thrice the other and their sum is $16$, find the numbers.
I tried,
Let the first number be $x$ and the second number be $y$
Acc. to question
$$
\begin{align}
x&=3y &\iff x-3y=0 &&(1)\\
x&=16-3y&&&(2)
\end{align}
$$
|
Let the first number be $x$.
Let the second number be $y$.
According to question
$$ \tag{1}
x+y=16
$$
$$
\tag{2}
x=3y
$$
So, $x-3y=0 \tag{2}$
Multiply equation $(1)$ by $3$.
Solve both equations:
$$\tag{1} 3x+3y=48$$
$$\tag{2} x-3y=0$$
$$\tag{1) + (2}4x=48$$
$$\tag{3}x=12$$
Putting in equation $(1)$:
$$\tag{1} x+y=16$$
$$\tag{1),(3} 12+y=16$$
$$\tag{4}y=16-12$$
$$\tag{5}y=4$$
|
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|
A closed form for $\int_{0}^{\pi/2}\frac{\ln\cos x}{x}\mathrm{d}x$? The following integrals are classic, initiated by L. Euler.
\begin{align}
\displaystyle \int_{0}^{\pi/2} x^3 \ln\cos x\:\mathrm{d}x & = -\frac{\pi^4}{64} \ln 2-\frac{3\pi^2}{16} \zeta(3)+\frac{93}{128} \zeta(5),
\\ \int_{0}^{\pi/2} x^2 \ln\cos x\:\mathrm{d}x & = -\frac{\pi^3}{24} \ln 2-\frac{\pi}{4} \zeta(3),
\\ \int_{0}^{\pi/2} x^1 \ln\cos x\:\mathrm{d}x & = -\frac{\pi^2}{8} \ln 2-\frac{7}{16} \zeta(3),
\\ \int_{0}^{\pi/2} x^0 \ln\cos x\:\mathrm{d}x & = -\frac{\pi}{2}\ln 2.
\end{align}
We may logically consider the case when the first factor of the integrand is $\displaystyle x^{-1} = \frac 1x $ leading to the following non classic convergent integral.
$$ \int_{0}^{\pi/2} \frac{\ln\cos x}{x}\:\mathrm{d}x \qquad (*)$$
I do not have a closed form for this integral.
My question is does someone have some references/results about $(*)$?
|
I've established some related explicit formulae.
Theorem 1.
Let $n$ be any positive integer.
Set
$$ I_{2n}:=\int_{0}^{\pi/2}\! \! x^{2n} \ln \cos x \: \mathrm{d}x $$
Then
$$
I_{2n} = - \frac{\pi^{2n+1}\ln 2}{2^{2n+1}(2n+1)} - (-1)^{n}\frac{(2n)!}{2^{2n+1}}\sum_{p=1}^{n} \frac{(-1)^p}{(2p-1)!}\pi^{2p-1}\zeta(2n-2p+3) \tag1
$$
Set
$$ I_{2n+1}:=\int_{0}^{\pi/2}\! \! x^{2n+1} \ln \cos x \:\mathrm{d}x $$
Then
$$
\begin{align}
I_{2n+1}=- \frac{\pi^{2n+2}\ln 2}{2^{2n+2}(2n+2)} - (-1)^n\left(1-\frac{1}{2^{2n+2}}\right)\frac{(2n+1)!}{2^{2n+2}}\zeta(2n+3) \\\\ - (-1)^n\frac{(2n+1)!}{2^{2n+2}} \sum_{p=0}^{n} \frac{(-1)^p}{(2p)!}\pi^{2p}\zeta(2n-2p+3) \tag2
\end{align}
$$
[To be continued]
|
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Summation of Infinite Geometric Series Determine the sum of the following series:
$$\sum_{n=1}^{\infty } \frac{(-3)^{n-1}}{7^{n}} $$
My work:
$$\sum_{n=1}^{\infty } \frac{(-3)^{n-1}}{7^{n}} = \sum_{n=1}^{\infty } \frac{-1}{7} (\frac{3}{7})^{n-1}$$
$$\sum_{n=1}^{\infty } ar^{n-1} = \frac{a}{1-r} = \frac{\frac{-1}{7}}{1-\frac{3}{7}} = -\frac{1}{4}$$
Why does this not work?
Sorry for the incorrect initial post!!!
Edit: -3 changed to (-3)
|
$$\begin{align}
\sum_{n=1}^{\infty } \frac{-3^{n-1}}{7^{n}}
& = - \frac{1}{7} \sum_{n=1}^{\infty } (\frac{3}{7})^{n-1}
\\
& = - \frac{1}{7} \frac{1}{1-\frac 3 7}
\\
& = - \frac 1 4
\\[2ex]
\sum_{n=1}^{\infty } \frac{(-3)^{n-1}}{7^{n}}
& = \frac{1}{7} \sum_{n=1}^{\infty } (-\frac{3}{7})^{n-1}
\\
& = \frac{1}{7} \frac{1}{1+\frac 3 7}
\\
& = \frac 1 {10}
\end{align}$$
|
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Relation between an unsatisfiable set and a tautology In mathematical logic, satisfiability and validity are elementary concepts of semantics. A formula is satisfiable if it is possible to find an interpretation (model) that makes the formula true. A formula is valid if all interpretations make the formula true. The opposites of these concepts are unsatisfiability and invalidity, that is, a formula is unsatisfiable if none of the interpretations make the formula true, and invalid if some such interpretation makes the formula false.
I read some Midterm exam on logic for computation course. I read this fact on my tutorial note. But I think it's false.
if set of {$\varphi_1,\varphi_2,...,\varphi_m$} is not satisfiable, then formulate $\neg\varphi_1 \wedge \neg\varphi_2 \wedge ...\wedge \neg\varphi_m$ is a TAUTOLOGY.
I think it's False, but in my lecture note is written True. I think it's false. Any hint and tutorial would be highly appreciated.
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It is not true. What is true is that $\lnot \varphi_1 \lor \lnot \varphi_2
\lor \cdots \lor \lnot \varphi_m$ is a tautology.
To see that the sentence with the $\land$ is not necessarily a tautology, let $m=2$, let $\varphi_1=\varphi$ and $\varphi_2=\lnot\varphi$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Rewriting the matrix equation $AX = YB$ as $Y = CX$? Is it possible in general, if $A,B,C,X,Y$ are square and of the same dimensions? If so, does it generalize to non-square matrices (using a pseudoinverse)? I'm doing some curve fitting in which I have to estimate the two independent polarizations of a signal given the data from multiple detectors and the scalar response function for each polarization (LIGO data analysis). Being able to rewrite the equation above in general would make it possible to express fitted values as seen from multiple detectors (Y) as an explicit function of X, the data seen at each detector, and would enable a generalized cross-validation calculation to choose a regularization parameter.
Thanks.
|
In general it is not going to be possible.
Example: Let
$
A = Y =
\begin{pmatrix}
1 & 1 \\
0 & 0
\end{pmatrix}
$
and
$
X = B =
\begin{pmatrix}
1 & 0 \\
1 & 0
\end{pmatrix}.
$
Then $AX=YB$, but no matter what $C$ you choose, $CX$ has $0$'s in its second column and thus cannot equal $Y$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How find this sum $\sum\limits_{i=0}^{2n}\binom{2n}{2i}\binom{2i}{i}y^{2i}$
Find the sum close form
$$f(x)=\sum_{i=0}^{2n}\dfrac{\binom{2n}{2i}\binom{2i}{i}x^{2i}}{2^{2i}}$$
if we let
$$\dfrac{x}{2}=y$$
then
$$f(y)=\sum_{i=0}^{2n}\binom{2n}{2i}\binom{2i}{i}y^{2i}$$
this PDF have this page 5
$$\sum_{k=j}^{n}\binom{n}{k}\binom{k}{j}=2^{n-j}\binom{n}{j}$$
the solution can see
page 5
Maybe my problem can use this mathods?Thank you
|
Given any formal Laurent series $\;(???) = \sum \alpha_{k_1 k_2 \ldots k_n} t_1^{k_1} t_2^{k_2} \cdots t_n^{k_n}$, we will use the notation $[ t_1^{k_1} t_2^{k_2} \cdots t_n^{k_n} ](???)$ to denote the coefficient $\alpha_{k_1 k_2 \cdots k_n}$ in front of corresponding monomial.
Instead of $f(y)$, let us denote the polynomial we wish to find a closed form as $p_{2n}(y)$. We have
$$\begin{align}
p_{2n}(y)
&= \sum_{i=0}^{n}\binom{2n}{2i}\binom{2i}{i} y^{2i}
= \sum_{i=0}^{n} \binom{2n}{2i} y^{2i}\bigg( [t^0](t + t^{-1})^{2i}\bigg)\\
&= \sum_{i=0}^{2n} \binom{2n}{i} \bigg( [t^0](y(t + t^{-1}))^i\bigg)
= [\;t^0\;] \bigg( 1 + y(t+t^{-1})\bigg)^{2n}
\end{align}
$$
Substitute $t$ by $e^{i\theta}$ in above formal expression and notice for any $k \in \mathbb{Z}$, we have
$$\frac{1}{2\pi}\int_0^{2\pi} e^{ik\theta} d\theta = \begin{cases}1,&k = 0\\0,&\text{ otherwise }\end{cases}$$
We obtain an integral representation for $p_{2n}(y)$,
$$p_m(y) = \frac{1}{2\pi}\int_0^{2\pi} (1+2y\cos\theta)^{m} d\theta\quad\text{ for }\quad m = 2n$$
Treat this as a definition for $p_m(y)$ for general $m \in \mathbb{N}$ and consider following generating function:
$$p(y,\rho) = \sum_{m=0}^\infty p_m(y)\rho^m$$
It is easy to see
$$
p(y,\rho)
= \frac{1}{2\pi}\int_0^{2\pi}\frac{d\theta}{1-\rho(1+2y\cos\theta)}
= \frac{1}{4\pi y\rho}\int_0^{2\pi}\frac{d\theta}{\frac{1-\rho}{2y\rho}-\cos\theta}
$$
Using the identity
$$\frac{1}{2\pi}\int_0^{2\pi}\frac{d\theta}{a - \cos\theta} = \frac{1}{\sqrt{a^2-1}}\quad\text{ for } a > 1$$
We get
$$\begin{align}
p(y,\rho) &= \frac{1}{2y\rho}\frac{1}{\sqrt{\left(\frac{1-\rho}{2y\rho}\right)^2 - 1}}
= \frac{1}{\sqrt{1-2\rho + (1- 4y^2)\rho^2}}\\
&= \frac{1}{\sqrt{1-2\frac{1}{\sqrt{1-4y^2}}(\rho\sqrt{1-4y^2}) + (\rho\sqrt{1-4y^2})^2}}\
\end{align}
$$
Compare this with the generating function for Legendre polynomials,
$$\frac{1}{\sqrt{1-2zt+t^2}} = \sum_{k=0}^\infty P_k(z) t^k$$
We find
$$p(y,\rho) = \sum_{k=0}^\infty P_k\left(\frac{1}{\sqrt{1-4y^2}}\right) \left(\rho\sqrt{1-4y^2)}\right)^k$$
This leads to the expression we claimed in comment:
$$p_{2n}(y) = (1-4y^2)^n P_{2n}\left(\frac{1}{\sqrt{1-4y^2}}\right)$$
For example, when $y = \frac12$, this leads to an interesting identity:
$$\begin{align}
\sum_{i=0}^{n} \frac{\binom{2n}{2i}\binom{2i}{i}}{2^{2i}}
&= p_{2n}\left(\frac12\right)
= \lim_{y\to\frac12^{-}} (1-4y^2)^n P_{2n}\left(\frac{1}{\sqrt{1-4y^2}}\right)\\
&= [ t^{2n} ] P_{2n}(t) = \frac{(4n-1)!!}{(2n)!}
\end{align}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Evaluate $\int_0^\infty\frac{dl}{(r^2+l^2)^{\frac32}}$ How to evaluate the following integral
$$\int_0^\infty\frac{dl}{(r^2+l^2)^{\large\frac32}}$$
The solution is supposed to look like this, unfortunately I can't derive it.
$$
\left[\frac{l}{r^2\sqrt{r^2+l^2}}\right]_{l=0}^\infty
$$
|
Let $l=r\tan{u}$, then $dl=r\sec^2{u} \ du$. The integral becomes
$$\frac{1}{r^2}\int^{\pi/2}_0\frac{\sec^2{u}}{\sec^3{u}}du=\frac{1}{r^2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/880415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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|
When the numerator of a fraction is increased by $4$, the fraction increases by $2/3$... When the numerator of a fraction is increased by $4$, the fraction increases by $2/3$. What is the denominator of the fraction?
I tried,
Let the numerator of the fraction be $x$ and the denominator be $y$.
Accordingly, $$\frac{x+4}y=\frac xy+\frac 23$$
I am not able to find the second equation.
|
Given,
$$\frac{n+4}{d}=\frac{n}{d}+\frac{2}{3}$$
So,
$$\frac{n}{d}+\frac{4}{d}=\frac{n}{d}+\frac{2}{3}$$
Or,
$$\frac{4}{d}=\frac{2}{3}$$
That, gives us $d=6$
|
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 3
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|
How to prove that $ 1- \frac{x^2}{n} \leq (1+\frac{x}{n})^n\cdotp(1-\frac{x}{n})^n$ How would I prove this inequality (assuming its true, its from a textbook)
$$1 - \frac{x^2}{n} \leq (1+\frac{x}{n})^n\cdotp(1+\frac{-x}{n})^n$$
if $n > |x|$, $x\in R$ and $n\in N$
I first rewrote the inequality to
$$1 - \frac{x^2}{n} \leq (1-\frac{x^2}{n^2})^n$$I then tried to manipulate the inequalities by saying the right hand side was greater than a smaller expression however I was unable to prove the above. I also tried induction where the base case works however I was unable to show that a case being true implies the next also being true.
Any help would be appreciated
|
Deriving both sides on $x$, $$-\frac{2x}{n}\le-n\frac{2x}{n^2}(1-x^2)^{n-1},$$ or
$$-1\le-(1-\frac{x^2}{n^2})^{n-1}.$$
The latter relation is obviously true for $|x|<n$, so that the LHS of the initial relation decreases faster than the RHS, while they are equal for $x=0$.
(If you prefer, $l'(x)\le r'(x)\implies l'(x)-r'(x)\le0\implies l(x)-r(x)$ is decreasing $\implies l(x)-r(x)\le l(0)-r(0)=0\implies l(x)\le r(x)$.)
|
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"timestamp": "2023-03-29T00:00:00",
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|
Combo Identity: How to prove this using Induction
$$ \sum_{n = 0}^{\infty} \binom{n + k}{k}x^n = \dfrac{1}{(1 - x)^{k + 1}} $$
Could someone suggest how I should get started to prove this using induction?
|
HINT:
$$\frac1{(1-x)^{m+1}}=\frac{1-x}{(1-x)^{m+2}}=\frac1{(1-x)^{m+2}}-\frac x{(1-x)^{m+2}}$$
$$\implies(1-x)^{-(m+1)}=(1-x)^{-(m+2)}-x(1-x)^{-(m+2)}$$
Assume that the formula is true for $k=m+2$ and establish the same for $k=m+1$
Alternatively, use $$(1-x)(1-x)^{-(m+2)}=(1-x)^{-(m+1)}$$
Assume that the formula is true for $k=m+1$ and establish the same for $k=m+2$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/880652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
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