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Find $\lim\limits_{x\to\frac\pi2}(1-\sin x)\tan^2 x.$ I'm stuck finding following limit.
$$\lim_{x\to\frac\pi2}(1-\sin x)\tan^2 x.$$
Attempt: I tried to use L'Hopital rule but I cant find the solution
|
I usually advise to “go at $0$”: do the substitution $x=\pi/2-t$, so the limit becomes
$$
\lim_{t\to0}(1-\cos t)\cot^2t=
\lim_{t\to0}\frac{1-\cos t}{\sin^2t}\cos^2t=
\lim_{t\to0}\frac{1-\cos t}{t^2}\frac{t^2}{\sin^2t}\cos^2t
$$
If you want to use l'Hôpital, do
$$
\lim_{x\to\pi/2}\frac{1-\sin x}{\cot^2x}
\overset{\mathrm{(H)}}{=}
\lim_{x\to\pi/2}\frac{-\cos x}{-2\cot x/\sin^2x}
$$
Now simplify.
|
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|
Describe the ring $\mathbb{Z}[x]/(x^2)$.
Describe the ring $\mathbb{Z}[x]/(x^2)$.
Attempt: $\mathbb{Z}[x]/(x^2)$ can be thought as the linear polynomials with integer coefficients.
So $\mathbb{Z}[x]/(x^2) = \{a + bx + (x^2) : a,b \in \mathbb{Z}\}$.
What are other things I could describe?
Thank you!
|
This is actually a very interesting ring:
You can consider 'x' an "infinitesimal" in a sense.
Let $A = a + x a'$ and $B = a + x a'$ then we have
*
*$A + B = (a+b) + x(a'+b')$
*$AB = (ab) + x(a'b + ab')$
The first component is the normal operation, but the second component follows the same formulas as differentiation does!
This has applications in automatically differentiating functions in programming.
|
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Direct Product and Direct sum of a Category (Integer divide)
For the following category, give a definition of the direct product and direct sum; determine if they exist; if they do, give an explicit description, and give an explicit description of the direct product and the direct sum over an empty set:
The category $D \mathbb{Z}$. In this category, the objects are the integers, and there are no morphisms from $m$ to $n$ unless $m$ divides $n$, in which case there is only one morphism.
$$Ob(D \mathbb{Z}) = \mathbb{Z} $$
\begin{equation*}
\text{Mor}(m,n) = \begin{cases}
\{*_{m,n}\} & \text{if } \; m|n \\
\emptyset & \text{otherwise}. \\
\end{cases}
\end{equation*}
Where we use the notation $m\mid n$ means $m$ divides $n$ if there exists $k \in \mathbb{Z}$ s.t. $mk=n$.
I think the category $D \mathbb{Z}$ is the set of ideals created by $d \in D$
E.g. $D=3$ then $D \mathbb{Z} = \{..., -6,-3,0,3,6, ...\}$
The direct product is the usual multiplication $(\times)$
$$if \; m|n \Rightarrow m|\lambda \times n, (\forall \lambda) $$
And direct is sum the usual addition $(+)$
$$if \; m|a \; and \; m|b \Rightarrow m|a+b$$
There is no direct product nor direct sum when we do not have an $m|n$ as there is no morphism.
When we do have $m|n$ then the direct product and direct sum over an empty set are vacuously empty.
|
(Direct sum) In the category $D\Bbb Z$, for $a,b\in \Bbb Z$,
the direct sum is
$$
a \coprod b := \operatorname{lcm}(a,b),
$$
the least common multiple of $a,b$ (which we'll take to be $\ge 0$).
Clearly, $a\!\mid\!\operatorname{lcm}(a,b)$ and $b\!\mid\!\operatorname{lcm}(a,b)$.
That is, there are morphisms
$$
a\longrightarrow \operatorname{lcm}(a,b) \longleftarrow b.
$$
Suppose now that $x$ is such that
$$
a\longrightarrow x \longleftarrow b,
$$
which means $a\!\mid\! x$ and $b \!\mid\! x$. Then by definition of "$\operatorname{lcm}$", we have $\operatorname{lcm}(a,b)\!\mid\! x$, which is to say, there is a (necessarily unique) morphism
$$
\operatorname{lcm}(a,b)\longrightarrow x,
$$
such that
*
*the (unique) morphism $a\to x$ equals the composition of the morphisms $\operatorname{lcm}(a,b)\to a\to x$, and
*the (unique) morphism $b\to x$ equals the composition of the morphisms $\operatorname{lcm}(a,b)\to b\to x$.
These two bullet points are trivial: "divides" is transitive, and for any $y,z\in \Bbb Z$ there is at most one morphism $y\to z$.
(Direct sum of the empty set)
$$
\coprod \emptyset = 0.\quad\text{(Why?)}
$$
(Direct product, direct product of the empty set) Exercise.
|
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Nonhomgeneous Linear Differential Equation: Harmonic Oscillator
Consider frictionless harmonic oscillator (w/ m = 1) driven by an external force $f(t) = A\sin{\omega t} $, so that $$\frac{d^2 x}{dt^2} + \omega_0^2x = A\sin{\omega t}. $$ Show that the particular solution for $\omega \neq \omega_0 $ is $$x_p(t) = \frac{A}{\omega_0^2 - \omega^2} \sin{\omega t} $$
Method of undetermined coefficients looks like the best choice. I Acknowledged this is a second order linear nonhomegeneous equation.
|
Hint:
Assume the particular solution $$y_p=C_1\cos w t+C_2\sin wt$$
$$x_p'=-wC_1\sin w t+wC_2\cos w t$$
$$x_p''=-w^2C_1\cos w t-w^2C_2\sin w t$$
then substitute them to get
$$-w^2C_1\cos w t-w^2C_2\sin w t+w_0^2(C_1\cos w t+C_2\sin wt)=A\sin w t$$
$$(-w^2C_1+w_0^2C_1)\cos w t+(-w^2C_2+w_0^2C_2)\sin w t=0\cos w t+A\sin w t$$
$$(-w^2C_1+w_0^2C_1)=0$$
$$C_1=0$$
$$(-w^2C_2+w_0^2C_2)=A$$
$$C_2=\frac{A}{w_0^2-w^2}$$
|
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|
Rewrite exponential in form $a+bi$ My professor lately has become really bad at giving examples and any help with an example would be great.
if I have $e^{(2+3i)}$ how do I rewrite it as $a+bi$? is it just $2+3i$ in this case?
|
No, it is not simply $2+3i$. Use Euler's formula, i.e.
$$e^{ix} = \cos x + i \sin x$$
so that
$$e^{a+bi} = e^ae^{bi} = e^a(\cos b + i \sin b).$$
|
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Find an integer n such that $U(n)$ is isomorphic to $\Bbb Z_2⊕\Bbb Z_4⊕\Bbb Z_9$ Find an integer n such that $U(n)$ is isomorphic to $Z_2⊕Z_4⊕Z_9$
I have gotten this far:
I know $Z_2$ is isomorphic to $U(4)$ and $Z_4$ is isomorphic to $U(5)$. However, I'm having trouble figuring out what $Z_9$ is isomorphic to in regards to the U-group. I remember proving somewhere that for all integers $n\geq 3$, $|U(n)|$ is even. Since $|Z_9|=9$, which is odd, I can't see to find a $U$ group that $Z_9$ is isomorphic to.
Thanks for all the help!
|
$\mathbb{Z}_2\oplus \mathbb{Z}_4 \oplus \mathbb{Z}_9\cong\mathbb{Z}_2 \oplus \mathbb{Z}_{36}.$
$2+1$ and $36+1$ are distinct primes, so $\mathbb{Z}_3\oplus \mathbb{Z}_{37}\cong \mathbb{Z}_{3.37}$ would work.
Similarly $\mathbb{Z}_2\oplus \mathbb{Z}_4\oplus \mathbb{Z}_9\cong \mathbb{Z}_4\oplus \mathbb{Z}_{18}$.
Then $4+1$ and $18+1$ are also distinct primes, so $\mathbb{Z}_5\oplus \mathbb{Z}_{19}\cong \mathbb{Z}_{5.19}$ would also work.
|
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Proving df is the sum of partial derivatives Wikipedia states:
Since any vector $v$ is a linear combination $\sum v_je_j$ of its components, $df$ is uniquely determined by $df_p(e_j)$ for each $j$ and each $p \in U$, which are just the partial derivatives of $f$ on $U$. Thus $df$ provides a way of encoding the partial derivatives of $f$. It can be decoded by noticing that the coordinates $x_1, x_2, \ldots, x_n$ are themselves functions on $U$, and so define differential 1-forms $dx_1, dx_2, \ldots, dx_n$. Since $\frac{\partial x_i}{\partial{x_j}} = \delta_{ij}$, the Kronecker delta function, it follows that
$$df = \sum_{i=1}^n \frac{\partial f}{\partial x^i} \, dx^i$$
Can anyone elaborate on this? In particular, how do we know what the components of $df$ are? What's the relationship between $x = (x_1, x_2, \ldots, x_n)$ and $p$? Where does the Kronecker function come in?
|
Let $A$ be open in $\Bbb{R}^{n}$; let $f: A \to \Bbb{R}$ be of class $C^{1}$; for every $x \in A$ let $df^{x}: v \mapsto \nabla f(x)\cdot v$ on $\Bbb{R}^{n}$. Then $df$, the differential of $f$, is a $1$-form on $A$ (provided that a $1$-form is defined as an alternating $1$-tensor).
Each elementary 1-form $dx_{i}$, which is the differential of the $i$th projection map on $\Bbb{R}^{n}$, is thus the map that assigns to every $x \in A$ the map $dx_{i}^{x}: v \mapsto \nabla x_{i}(x)\cdot v = v_{i}$ on $\Bbb{R}^{n}.$
If $x \in A$ and if $v \in \Bbb{R}^{n}$, then
$$
df^{x}(v) = \nabla f(x)\cdot v = \sum_{i=1}^{n}D_{i}f(x)v_{i} = \sum_{i=1}^{n}D_{i}f(x)dx_{i}^{x}(v);
$$
this result can be written succinctly as
$$
df = \sum_{i=1}^{n}(D_{i}f) dx_{i}.
$$
Note that it is a (conventional) "abuse" of notation to write the $i$th projection map as $x_{i}$; if we denote it by $p_{i}$, say, then it is clear that
$$
\frac{\partial p_{i}}{\partial x_{j}} = \frac{\partial x_{i}}{\partial x_{j}} = \delta_{ij},
$$
which simply says that $D_{j}x_{i} = 1$ if $i=j$ and $=0$ if $i \neq j$.
|
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How many ways can you get a bag of $12$ donuts if a shop sells $30$ kinds of donuts? A donut shop sells $30$ kinds of donuts, and there are at least $12$ donuts of each kind. How many ways can you:
*
*Get a bag of $12$ donuts?
*Get a bag of $12$ donuts if you want at least $3$ glazed donuts and at least $4$ chocolate donuts?
*Get a bag of $12$ donuts if you want exactly $3$ glazed donuts and exactly $4$ chocolate donuts?
For the first one I am thinking it is $C(30+12-1,12)=C(41,12)=7898654920$. This seems like a big number, but I think it's correct. I am not sure how I'd go about solving the second and third, though.
EDIT:
For 2, perhaps I do the following? I can choose $3$ glazed and $4$ chocolate in $\frac{7!}{3!4!}=35$ ways and that leaves me to choose $5$ donuts from the remaining $30$ kinds, so $C(30+5-1,5)=278256$. Then multiplying together I would get $9738960$ ways. Is this correct?
|
Problem 1 is standard Stars and Bars. You solved it correctly. The number of ways is $\binom{30+12-1}{12}$, or equivalently $\binom{30+12-1}{30-1}$.
For Problem 2, I think we are to assume that glazed is one of the $30$ kinds, and that all glazed doughnuts (from that store) are identical. Ditto for chocolate.
So if we want at least $3$ glazed and at least $4$ chocolate, put these in the bag. We need to get a further $5$ from the $30$ kinds. Solve as in the first problem. The number of ways is $\binom{30+5-1}{5}$. Note that your answer contained this number. The $\binom{7}{3}$ that you multiplied by should not be there.
For Problem 3, things are pretty much the same, except we must choose $5$ doughnuts from the $28$ kinds remaining, since we don't want any more glazed or chocolate.
|
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Improper integrals - where am I going wrong? The question asks:
$\int_0^\pi \frac{sinx}{\sqrt{|cosx|}}$
My attempt:
$\lim\limits_{t \to \frac{\pi}{2}^-} \int_0^t \frac{sinx}{\sqrt{|cosx|}}$ + $\lim\limits_{t \to \frac{\pi}{2}^+} \int_t^\pi \frac{sinx}{\sqrt{|cosx|}}$
$\int \frac{sinx}{\sqrt{|cosx|}} = -2\sqrt{|cosx|}$
$\lim\limits_{t \to \frac{\pi}{2}^-} [-2\sqrt{|cosx|}]_0^t = 2$
$\lim\limits_{t \to \frac{\pi}{2}^+} [-2\sqrt{|cosx|}]_t^\pi = -2$
$\lim\limits_{t \to \frac{\pi}{2}^-} [-2\sqrt{|cosx|}]_0^t + \lim\limits_{t \to \frac{\pi}{2}^+} [-2\sqrt{|cosx|}]_t^\pi = 0$
My final answer is $0$, but the marking scheme shows the final answer as $4$.
I also notice the following step from the marking scheme:
$\int_0^\pi \frac{sinx}{\sqrt{|cosx|}}$ = $2\int_0^\frac{\pi}{2} \frac{sinx}{\sqrt{|cosx|}}$
I do not understand how the marking scheme arrived to this step.
I have been struggling with this question for too long now. Could someone please help me get to the correct answer? I will appreciate that.
Thanks
|
Note that after $\frac {\pi} 2$, cosine is negative, so the absolute value function in your second integral will have a -cos(x) in the bottom. That'll add an extra - to your antiderivative of the second part, which flips it from -2 to 2
|
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Does an inverse exist for a set of complex numbers with operation of multiplication? I will give some context of my question: I was trying to solve this problem
Let $S$ be a finite non-empty set of non-zero complex numbers which is closed under multiplication. Show that $S$ is a subset of the set $\{z \in \mathbb C: |z|=1 \}$. Show that $S$ is a group, and deduce that for some $n \in \mathbb N$, S is the set of n-th roots of unity.
When trying to prove that $S$ is a subset of $\{z \in \mathbb C: |z|=1 \}$, I tried to argue that since $S$ is closed, $\forall a \in S, \ \exists \ \text{a corresponding} \ n \in \mathbb N^+ $ such that $$a^n = a \tag{1}$$ and hence $|a|=1$.
Now I want to prove that the identity element is in $S$, can I argue from $(1)$ that $a^{n-1}=e=1$?
|
To prove this in the order suggested:
*
*Let $a \in S$. If $|a|>1$, then $1 < |a| < |a|^2 < \cdots $ and the powers of $a$ would be an infinite subset of $S$. The same would follow if $|a|<1$. Thus, $|a|=1$.
*Let $S$ have $n$ elements. Consider $x \mapsto ax$. This is injective and so is surjective. Therefore $a_1 \cdots a_n = (a a_1) \cdots (a a_n) = a^n a_1 \cdots a_n$ and so $a^n=1$, which proves in one stroke that $S$ is a group formed by $n$-roots of unity. Since $n$ has $n$ elements, $S$ is the set of all $n$-roots of unity.
(This last argument is the proof of the little Lagrange theorem for finite abelian groups.)
|
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Calculating of $\lim_{x\to 0^+} x^{(x^x-1)}$ When I want calculate $\lim_{x\to 0^+} x^{(x^x-1)}$ , I have to calculate $\lim_{x\to 0^+}(x^x-1)\ln x$ . I calculat $\lim_{x\to 0^+}x^x=\lim_{x\to 0^+}e^{x\ln x}=1$ so $\lim_{x\to 0^+}(x^x-1)\ln x$ is $0\times \infty$ .
How can I calculate $\lim_{x\to 0^+}(x^x-1)\ln x$?
|
$$\lim_{x\to 0^+} \exp((x^x-1)\ln x) =$$
$$\exp(\lim_{x\to 0^+} (x^x-1)\ln x) = \exp(\lim_{x\to 0^+} \frac{x^x-1}{\frac1{\ln x}}) = \exp(\lim_{x\to 0^+} \frac{\exp(x\ln x)-1}{\frac1{\ln x}})$$
Using Hopital
$$\exp(\lim_{x\to 0^+} \frac{\exp(x\ln x)-1}{\frac1{\ln x}})=\exp(\lim_{x\to 0^+} \frac{(\ln x+1)\exp(x\ln x)}{\frac{-1}{x \ln^2 x}})=$$
$$=\exp(-\lim_{x\to 0^+} x \ln^2 x(\ln x+1)\exp(x\ln x))$$
Here we have $x \ln^3x$ and $x$ overcomes over $\ln^3 x$ the result is $\exp(0)=1$
|
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Integration by parts of Wirtinger derivative As a physics student, I am currently trying to use Wirtinger derivatives in everyday work.
While the basics are clear and so far seem to make life considerably simpler, I am having trouble to see if integration by parts is possible as in the real case.
Transforming to real two-dimensional coordinates, I have come to the following conclusions:
$$
\int_D \frac{\partial f}{\partial z} \, g \, \mathrm{d}^2 z
= \int_{\partial D} f \, g \, \mathrm{d} \bar z
- \int_D f \, \frac{\partial g}{\partial z} \, \mathrm{d}^2 z
$$
$$
\int_D \frac{\partial f}{\partial \bar z} \, g \, \mathrm{d}^2 z
= \int_{\partial D} f \, g \, \mathrm{d} z
- \int_D f \, \frac{\partial g}{\partial \bar z} \, \mathrm{d}^2 z
$$
Are these correct? I would be glad for any help, and especially for references that cover this topic, as I could not find any.
|
First of all, there is no way that your suggested formulas can be true. Your integrals over $D$ contain $1$-forms, which is nonsense since you're integrating over the 2-dimensional $D$.
On the other hand, Stokes' theorem is still valid in the complex setting:
$$
\int_{\partial D} \omega = \int_{D} d\omega.
$$
Furthermore, $d = \partial + \bar\partial$ so if $\omega = f(z)g(z)\,dz$ you get
\begin{align*}
\int_{\partial D} f(z)g(z)\,dz &= \int_{D} d\big( f(z)g(z)\,dz \big) \\
&= \int_{D} \bar\partial\big( f(z)g(z)\,dz \big) \\
&= \int_D \Big( \frac{\partial f}{\partial \bar z}(z) g(z) + f(z)\frac{\partial g}{\partial \bar z}(z) \Big) d\bar z \wedge dz,
\end{align*}
where the second equality it true since $dz \wedge dz = 0$ (the $\partial$-term picks up a $dz$).
If you want derivatives with respect to $\bar z$ in the integral over $D$, apply Stokes' to $\int_{\partial D} f(z)g(z)\,d\bar z$ instead.
|
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Conceptually, why does a positive definite Hessian at a specific point able to tell you if that point is a maximum or minimum? This is not about calculating anything. But can anyone tell me why this is the case?
So, from wikipedia:
If the Hessian is positive definite at x, then f attains a local minimum at x. If the Hessian is negative definite at x, then f attains a local maximum at x. If the Hessian has both positive and negative eigenvalues then x is a saddle point for f. Otherwise the test is inconclusive. This implies that, at a local minimum (resp. a local maximum), the Hessian is positive-semi-definite (resp. negative semi-definite).
Can someone explain, intuitively, why this is the case?
|
This is because of Taylor's formula at order $2$:
\begin{align*}f(x+h,y+k)-f(x,y)&=hf'_x(x,y)+kf'_y(x,y)\begin{aligned}[t]&+\frac12\Bigl(h^2f''_{x^2}(x,y)+2hkf''_{xy}(x,y)\\&+k^2f''_{y^2}(x,y)\Bigr)+o\bigl(\bigl\lVert(h,k)\bigr\rVert^2 \bigr)\end {aligned}\\
&=\frac12\Bigl(h^2f''_{x^2}(x,y)+2hkf''_{xy}(x,y)+k^2f''_{y^2}(x,y)\Bigr)+o\bigl(\bigl\lVert(h,k)\bigr\rVert^2 \bigr)
\end{align*}
If the quadratic form $\;q(h,k)=\frac12\Bigl(h^2f''_{x^2}(x,y)+2hkf''_{xy}(x,y)+k^2f''_{y^2}(x,y)\Bigr)$ is positive definite, the sign of the left-hand side is positive for all $\lVert(h,k)\bigr\rVert^2$ small enough, hence $f(x+h,y+k)-f(x,y)>0$, so we have a local minimum. If it is definite negative, for the same reasons, we have a local maximum.
|
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Proof: 1007 can not be written as the sum of two primes. The claim is: 1007 can be written as the sum of two primes. We want to prove or disprove it.
Edit: My professor provided this definition in his previous assignment: An integer $n \geq 2$ is called prime if its only positive integer divisors are $1$ and $n$.
I want to disprove it.
Here is my proof outline:
Claim: 1007 can not be written as the sum of 2 primes.
Lemma: An odd integer can not be written as the sum of 2 even integers, or the sum of 2 odd integers. This means that an odd integer can only be written as the sum of an odd integer and an even integer.
Proof for lemma:
Let $a, b, c, d$ be integers.
$2a$ is even, $2b$ is even, $2c+1$ is odd, and $2d+1$ is odd.
$2a+2b=2(a+b)$ is even.
$(2c+1)+(2d+1)=2(c+d+1)$ is even.
$2a+(2c+1)= 2(a+c)+1$ is odd.
Thus, we have proved our lemma.
Since 1007 is odd, it can only be written as the sum of an odd integer and an even integer.
This means that if $x+y=1007$, for some integers $x,y$, then $x$ must be even and $y$ must be odd, without loss of generality.
We will show with cases that $x$ and $y$ can never both be prime.
2 is the only even prime number.
Case 1: $x=2$: $2+y=1007$, $y=1005$. Since 1005 is divisible by 5, it is not prime.
Case 2: $x$= any even integer $> 2$. According to our lemma, if $x$ is even, and $x+y=1007$, then $y$ must be odd. Every even integer greater than 2 is not prime, and so $x$ will always not be prime.
Thus, 1007 can not be written as the sum of two primes.
Thus we have disproved the original claim.
1) Is this proof complete?
2) Am I over complicating this?
3) Is there a more efficient way to prove this?
Any help would be appreciated.
|
$1007$ is an odd number so it cannot be the sum of two odd numbers and it cannot be the sum of two even numbers. Therefore, it can only be the sum of an even and an odd number. Since $2$ is the only even prime it would have to be $2+1005$ and $1005$ is not prime.
|
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Largest irreducible representation of a finite non-commutative group Let $G$ be a finite non-commutative group of order $k$. Is there any way to determine a number $m$ such that there will necessarily exist an irreducible representation of $G$ of dimension $d \geq m$? If not, what kind of additional information could give an answer? Thanks!
|
As the comments indicate, the example of the dihedral groups shows that you cannot expect irreps of dimension greater than $2$ without additional hypotheses, and that bounding the size of the abelianization isn't a particularly helpful hypothesis.
One thing to say is that if a group $G$ has $c(G)$ conjugacy classes and $a(G) = |G/[G, G]|$, then it must also have $c(G)$ irreps (I am working over $\mathbb{C}$ throughout), $a(G)$ of which are $1$-dimensional. Since the dimensions $d_i$ of these irreps must satisfy $\sum d_i^2 = |G|$, you can conclude by the pigeonhole principle that $G$ must have an irrep of dimension at least
$$\sqrt{ \frac{|G| - a(G)}{c(G) - a(G)} }.$$
As you can see, $c(G)$ has a much bigger influence than $a(G)$ on this bound. Conversely, if the maximum dimension of an irrep is $d$, then
$$c(G) - a(G) \ge \frac{|G| - a(G)}{d^2}.$$
For example, the symmetric group $S_n$ satisfies $c(S_n) = p(n)$ (the partition number of $n$) and $a(S_n) = 2$ ($n \ge 2$), so we conclude that $S_n$ must have an irrep of dimension at least
$$\sqrt{ \frac{n! - 2}{p(n) - 2} }.$$
As a subexample, when $n = 7$ this bound is
$$\sqrt{ \frac{7! - 2}{p(7) - 2} } = \sqrt{ \frac{5038}{13} } \approx 19.7$$
so we conclude that $S_7$ has an irrep of dimension at least $20$. In fact it has irreps of dimensions $20, 21, 35$. (Of course, for the symmetric groups we know the dimensions of all of their irreps via the hook length formula, so we can say much more here.)
Another thing to say is that the dimensions $d_i$ of the irreps must also divide the size $|G|$ of $G$. This is a strong restriction if $G$ is, for example, a $p$-group, since it implies that the dimensions must themselves be powers of $p$.
|
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$S(x)=\frac{x^4}{2\cdot 4}+\frac{x^6}{2\cdot 4\cdot 6}+\frac{x^8}{2\cdot4\cdot6\cdot8}+\cdots$ Find the sum of $$S(x)=\frac{x^4}{2\cdot 4}+\frac{x^6}{2\cdot 4\cdot 6}+\frac{x^8}{2\cdot4\cdot6\cdot8}+\cdots$$
What I did so far:
It's trivial that
$$S'(x)-xS(x)=\frac{x^3}{2}-\lim_{n\to\infty}\frac{x^{2n}}{n!2^{{n(n+1)}/2}}$$
and $$\lim_{n\to\infty}\frac{x^{2n}}{n!2^{{n(n+1)}/2}}=0$$
Solve this ODE I got
$$S(x)=\frac{x^4}{8}e^x+Ce^x$$
And $S(x)=0$ , So $C=0$.
Finally, I got $S(x)=\frac{x^4}{8}e^x$
The problem is when I expand the function I got into power series, it looks different from the original one. So I may make some mistakes but I can't tell. Please help.
|
@labbhattacharjee provided a very efficient approach to evaluating the series.
The approach in the OP did not provide a way forward since the solution to the ODE $S'(x)-xS(x)=\frac12x^3$, $S(0)=0$ is given by
$$\begin{align}
S(x)&=\int_0^x \frac12x'^3 e^{(x^2-x'^2)/2}\,dx' \tag 1\\\\
&=e^{x^2/2}-\frac12x^2-1
\end{align}$$
where we used the integrating factor $\mu(x)=e^{-x^2/2}$ to facilitate generating the solution to the ODE and integration by parts to evaluate the integral in $(1)$.
|
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Intermediate value theorem: showing there exists $c \in [0,2/3]$ such that $f(x+1/3) = f(x)$ Intermediate value theorem: show there exists $c \in [0,2/3]$ such that $f(c+\frac{1}{3}) = f(c)$
Let $f: [0,1] \to [0,1]$ be continuous and $f(0) = f(1)$
This is pretty straight forward and the other ones ive done it was easier to find why IVT applies, namely find why it changes signs.
Letting $g(x) = f(x + \frac{1}{3}) - f(x)$ then this is only defined for the interval $[0,\frac{2}{3}]$ b/c not sure what $g(1) = f(\frac{4}{3}
) - f(1)$ may be.
Now, $g(0) = f(\frac{1}{3}) - f(0)$
and $g(\frac{2}{3}) = f(1) - f(\frac{2}{3}) = f(0) - f(\frac{2}{3})$.
I can't (at least I don't see why I can) conclude this is equal to $-g(0)$ which has been the case in other exercises, which is what I used to apply the IVT and get the wanted result.
|
HINT:
You were on the right track by defining $g(x)\equiv f(x+1/3)-f(x)$ for $x\in [0,2/3]$.
Now, note that $$g(0)+g(1/3)+g(2/3)=f(1)-f(0)=0$$Then either $g(0)=g(1/3)=g(2/3)=0$ or at least one of the three is positive and one of the three is negative. Now, use the IVT.
|
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Image of normal subgroup under surjective homomorphism as kernel Let $f: G \to H$ be a group homomorphism. Then the preimage $f^{-1}(N)$ of a normal subgroup $N$ of $H$ is normal in $G$: this preimage is the kernel of the composition of $f$ with the canonical projection $H \to H/N$.
Now, if $K$ is a normal subgroup in $G$, and if additionally $f$ is surjective, then the image $f(K)$ is normal in $H$, which is easy to see 'elementary' (on the level of elements)... but I would like to recognize the image as normal by recognizing it as some kernel (i.e. along the lines of the above argument). How?
|
We want to push, by some homomorphism, $f(K)$ to identity (and no other elements to identity).
How can we do this: first pull-back $f(K)$ via (inverse of) $f$: sine $\ker(f)=N$, we have
$$f^{-1}(f(K))=KN \,\,\,\,(\mbox{which is normal in $G$})$$
Next, push this $KN$ (and only that) to identity via a homomorphism; the most natural choice is
$$\varphi\colon G\rightarrow G/KN.$$
Thus, we define $\eta\colon H\rightarrow G/KN$ by $\eta=\varphi\circ f^{-1}$.
(You may try to show: $\eta$ is well defined and also a homomorphism).
Claim: $f(K)$ is the kernel of this homomorphism $\eta$.
Apply $\eta$ on $f(K)$:
$$\eta(f(K))=\varphi\circ f^{-1}(f(K))=\varphi(KN)=\{1\}.$$
So $f(K)\subseteq \ker\eta$. Converlely,
$$\eta(x)=1\Longleftrightarrow \varphi(f^{-1}(x))=1\Longleftrightarrow f^{-1}(x)\in \ker\varphi=KN\Longrightarrow x\in f(KN)=f(K).$$
|
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How to prove this bound of $L^\infty$ norm. A differentiable function $ f:\mathbb R\to \mathbb R$ satisfies such conditions,
$ $\begin{cases} \lim_{x\to\infty} f(x)=\lim_{x\to-\infty} f(x)=0, &\\
\int_{-\infty}^{\infty}|{f(x)}|^{2}dx<\infty, \ \ \int_{-\infty}^{\infty}|{f'(x)}|^{2}dx<\infty. &
\end{cases}$$
Prove
$$\max_{x\in \mathbb R}|f(x)|\leq \left(\int_{-\infty}^{\infty}|{f(x)}|^{2}dx<\infty\right)^{1\over4}\left(\int_{-\infty}^{\infty}|{f'(x)}|^{2}dx<\infty\right)^{1\over4}.$$
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I'll assume $f'$ is continuous. Let $a\in \mathbb {R}.$ Then for $h>0,$ the FTC gives
$$f(a+h)^2/2-f(a)^2/2 = \int_a^{a+h} f\,f'$$
and
$$f(a-h)^2/2-f(a)^2/2 = -\int_{a-h}^{a} f\,f'. $$
Add these together, then take absolute values to get
$$|f(a+h)^2/2-f(a-h)^2/2 -f(a)^2| \le \int_{a-h}^{a+h} |f|\,|f'| \le \int_{\mathbb R} |f|\,|f'|$$
Cauchy-Schwartz then shows
$$|f(a+h)^2/2-f(a-h)^2/2 -f(a)^2| \le (\int_{\mathbb {R}} |f|^2)^{1/2}(\int_{\mathbb {R}} |f'|^2)^{1/2}.$$
Let $h\to \infty$ and use the assumption $f(x)\to 0$ as $|x| \to 0$ to see
$$|f(a)|^2 \le (\int_{\mathbb {R}} |f|^2)^{1/2}(\int_{\mathbb {R}} |f'|^2)^{1/2}.$$
Taking square roots gives
$$\tag 1 |f(a)| \le (\int_{\mathbb {R}} |f|^2)^{1/4}(\int_{\mathbb {R}} |f'|^2)^{1/4}.$$
Since $a$ was an arbitrary point, $\max_{\mathbb {R}}|f|$ is less than or equal to the right side of $(1)$ as desired. (The assumption $f'(x) \to 0$ was not used.)
|
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Gradient vector at some point Suppose you are given some function $f(x, y, z) = xy^3z^2$ and you are told to find its the gradient vector normal to the surface at some point (-1, -1, 2).
Is it right think that the gradient vector is normal to the surface as shown in the figure?
Also, if you substitute the point you get a gradient vector at that point say $P$ ?
|
Because you have a function $f(x,y,z)$ of three variables, I assume you're talking about the level surface at a certain point, where given a constant $k$ the function has the value
$$f(x,y,z)=k$$
Then in that case the gradient vector
$$\nabla f(x,y,z) =\frac{\partial f}{\partial x}\hat{\imath}+\frac{\partial f}{\partial y}\hat{\jmath}+\frac{\partial f}{\partial z}\hat{k}$$
is always normal to the level surface. Substituting any point $P$ will give you another vector normal to the level surface at that point $P$, but it may not be the same level surface as the original one.
|
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Counting the number of words made of $2n$ letters
Compute the number of words made of $2n$ letters taken from the alphabet $\{a_1, a_2,\ldots,a_n\}$ such that each letters occurs exactly twice and no two consecutive letters are equal.
I started first of all thinking about the possibilities when we are not restricted by the fact that two consecutive letters are equal. Because we need to choose $2n$, and letters occur exactly twice, so that we can permute each letter pair $2!$ times, we will have something like this: $$\frac{2n!}{(2!)^n} = \frac{2n!}{2^n}$$
Now, I have to somehow exclude the possibilities when two consecutive letters are equal. I was stuck here, then I came across the answer to the problem (which I assume is the same problem), from Putnam and Beyond, page 326 in the PDF, and 309 of the book. I see that he has used inclusion-exclusion principle, but I didn't quite get how the numbers were obtained. Any explanation of that answer, or another solution is highly welcomed.
|
The strategy taken is to find the total number ways to arrange the letters $-$ the total number of ways to arrange the letter such that at least one pair of letters are together.
The total number of ways to arrange the letter as you have found is: $$\frac{(2n)!}{2^n}$$
Let $A_i$ be the number of ways to arrange the $2n$ letters such that $i^{th}$ pair are together. Therefore, we need to find:
$$|\overline{A_1 \cup A_2 \cup \dots \cup A_n}|=\frac{(2n)!}{2^n}-|A_1 \cup A_2 \cup \dots \cup A_n|$$
$\sum_{i=1}^n|A_i|$ is the number of ways to arrange the letters with any $1$ pair together and is given by:
$$\binom{n}{1}\frac{(2n-1)!}{(2!)^{n-1}}$$
This is because we arrange the remaining $2n-2$ letters and the $1$ grouped pair. After which we divide by $(2!)^{n-1}$ as there are $n-1$ remaining pairs.
Next, we have to find $\sum_{1 \leq i,j \leq n} |A_i \cap A_j|$, the case when any two pairs are together. Using a similar argument, we have:
$$\binom{n}{2}\frac{(2n-2)!}{(2!)^{n-2}}$$
Extending this to the $k^{th}$ case, to find $\sum_{1 \leq i,j \dots,k \leq n} |A_i \cap A_j\dots \cap A_k|$ we can extend the formula to:
$$\binom{n}{k}\frac{(2n-k)!}{(2!)^{n-k}}$$
Using the inclusion and exclusion principle,
$$|\overline{A_1 \cup A_2 \cup \dots \cup A_n}| =\frac{(2n)!}{2^n}-\sum_{i=1}^n|A_i|+\sum_{1 \leq i,j \leq n} |A_i \cap A_j|-\sum_{1 \leq i,j \dots,k \leq n} |A_i \cap A_j\dots \cap A_k|$$
$$\quad \quad \quad \quad \quad =\frac{(2n)!}{2^n}-\binom{n}{1}\frac{(2n-1)!}{(2!)^{n-1}}+\binom{n}{2}\frac{(2n-2)!}{(2!)^{n-2}}+\dots+(-1)^k\binom{n}{k}\frac{(2n-k)!}{(2!)^{n-k}}$$
$$=\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{(2n-k)!}{(2!)^{n-k}}$$
|
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How are theses functions called and what are their properties? We define a system of two equations with two variables and six parameters:
$\begin{cases}
a_1x+b_1y=c_1\\
a_2x+b_2y=c_2
\end{cases}$
We obtain a function $f:\mathbb{R^3\times\mathbb{R}^2\rightarrow\mathbb{R}^2}$,
$$\pmatrix{a_1 & a_2 \\ b_1 & b_2 \\ c_1 & c_2}\mapsto \pmatrix{x \\ y} $$
If we fix the five parameters $a_1$, $a_2$, $b_1$, $b_2$, $c_1$ and we vary the value of $c_2$. We obtain a function $f_{c_2}: \mathbb{R}\rightarrow\mathbb{R}^2$,
$c_2\mapsto\left(x,y\right)$. In the same manner we obtain the five other functions $f_{a_1}$, $f_{a_2}$, $f_{b_1}$, $f_{b_2}$, $f_{c_1}$.
My question is how are theses functions called and what are their properties?
|
The functions in OP are defined by the system:
$$\begin{cases}
a_1x+b_1y=c_1\\
a_2x+b_2y=c_2
\end{cases}$$
so that:
$$
x=\dfrac {\Delta_x}{\Delta} \qquad y=\dfrac {\Delta_y}{\Delta}
$$
where:
$$
\Delta=a_1b_2-a_2b_1 \quad \Delta_x=b_1c_2-b_2c_1 \quad \Delta_y=a_1c_2-a_2c_1
$$
so that the functions $f_{c_1}:\mathbb{R}\to\mathbb{R}^2$ and $f_{c_2}:\mathbb{R}\to\mathbb{R}^2$ are:
$$
f_{c_1}(t)=
\begin{bmatrix}
\frac{-b_2}{\Delta}t+\frac{b_1c_2}{\Delta}\\
\frac{-a_2}{\Delta}t+\frac{a_1c_2}{\Delta}\\
\end{bmatrix}
$$
and
$$
f_{c_2}(t)=
\begin{bmatrix}
\frac{b_1}{\Delta}t-\frac{b_2c_1}{\Delta}\\
\frac{a_1}{\Delta}t-\frac{a_2c_1}{\Delta}\\
\end{bmatrix}
$$
so these functions represent straight lines on $\mathbb{R}^2$.
The functions $f_{a_i}$ and $f_{b_i}$ are different. As example the function $f_{a_1}:\mathbb{R} \to \mathbb{R}^2$ is:
$$
f_{a_1}(t)=
\begin{bmatrix}
\dfrac{b_1c_2-b_2c_1}{b_2t-a_2b_1}\\
\dfrac{c_2t-a_2c_1}{b_2t-a_2b_1}
\end{bmatrix}
$$
that represent an homographic function in $\mathbb{R}^2$ whose graph is an hyperbola.
I dont know if all these function have some collective name (I never see it).
1.
|
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Structure of dense subset of $\mathbb{R}$ Here user Mario Carneiro formulated statement which I find really interesting and useful and I proved it and I would like to know is my proof correct?
If sequence $\{a_n\}_{n=1}^{\infty}$ implies the following conditions:
1) $a_1<a_2<\dots<a_{n}<\dots$
2) $a_n\to \infty$ as $n\to \infty$
3) $a_{n+1}-a_{n}\to 0$ as $n\to \infty$
Then $\{a_n+m: (n,m)\in \mathbb{N}\times \mathbb{Z}\}$ is dense in $\mathbb{R}^1$.
Proof: Taking $x\in \mathbb{R}^1$ and let $\epsilon>0$ be given. Then $\exists k\in \mathbb{N}$ such that $a_k<x<a_{k+1}$ (assuming that $x>a_1$). Then $\exists N:=N_{\epsilon}$ such that $n\geqslant N$ implies $a_{n+1}-a_n<\epsilon$.
1) If $k\geqslant N$ then $|x-a_k|<\epsilon.$
2) If $k<N$ then $\exists M\in \mathbb{N}$ such that $x+M\in(a_l, a_{l+1})$ where $l\geqslant N$ then $|x+M-a_l|<\epsilon.$
If $x\leqslant a_1$ then $\exists K\in \mathbb{N}$ such that $x+K>a_1$ and we can apply previous reasoning for $x+K$.
Q.E.D.
Remark: If $A$ is dense subset of $\mathbb{R}^1$ then: 1) for any $r>0$ the set $rA$ also dense subset of $\mathbb{R}$. 2) for any $c\in \mathbb{R}$ the set $c+A$ also dense.
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There are just two very small corrections that need to be made. First, there is a $k\in\Bbb N$ such that $a_k\color{red}{\le}x<a_{k+1}$. And secondly, in $(2)$ there is an $M\in\Bbb N$ such that $x+M\in\color{red}[a_\ell,a_{\ell+1})$ for some $\ell\ge N$.
|
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Two non-differentiable functions whose product is differentiable. So I was wondering while studying analysis if there is any case where two functions aren't differential at $0$ (kind of like $1/x$) but is differentiable at 0 when combined (i.e. $fg$).
I mean this for functions that are defined on $\mathbb{R}$.
|
On $\mathbb R$ let $f$ be the indicator of $[0,\infty)$ and $g$ the indicator of $(-\infty,0)$. Their product is $0$, whereas each of the functions is not differentiable at $0$.
Furthermore, you dont need your functions to be defined on $\mathbb R$ to say something about differentiability in $0$, since differentiability is a local property.
|
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Linear Transformations, Linear Algebra Let T:P3→P3 be the linear transformation such that $T(−2x^2)= −2x^2 − 2x$, $T(0.5x + 2)= 3x^2 + 4x−2$, and $T(2x^2 − 1)= 2x + 1$.
Find $T(1), T(x), T(x^2)$, and $T(ax^2 + bx + c)$, where a, b, and c are arbitrary real numbers.
I understand how to find $T(x^2)$ where you just divide the given $T(-2x^2)$ by $-2$ to get $T(x^2) = x^2 + x$.
I'm not sure sure how to proceed in calculating the other transformation functions. Please list as many of the steps as possible in solving for the three other functions.
Thank you!
|
We can write $1=-(-2x^2)-(2x^2-1), x=2(0.5x+2)-4$, so
$$T(1)=T(-(-2x^2)-(2x^2-1))=-T(-2x^2)-T(2x^2-1)=2x^2-1$$
$$T(x)=T(2(0.5x+2)-4)=2T(0.5x+2)-4T(1)=-2x^2+8x$$
Then we get
$$T(ax^2+bx+c)=a(x^2+x)+b(-2x^2+8x)+c(2x^2-1)$$
$$=(a-2b+2c)x^2+(a+8b)x-c$$
|
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Remainder of the polynomial division, knowing other remainders Let $p(x)$ be a polynomial of 3rd degree.
We know that the division of $p(x)$ by $x-4$ gives us a remainder of 2 and divided by $x+2$ gives us the remainder of 1.
What's the remainder of $p(x)$ by $(x-4)(x+1)$?
I've used the remainder theorem but I don't seem to get anywhere...
|
We can write
$$p(x)=(x-4)(ax^2+bx+c)+2.$$
Since $p(-2)=1$, we have
$$1=(-6)(4a-2b+c)+2\quad\Rightarrow\quad c=-4a+2b+\frac 16.$$
So, we can write
$$p(x)=ax^3+(b-4a)x^2+\left(-4a-2b+\frac 16\right)x+16a-8b+\frac 43.$$
Thus, we have
$$p(x)=(x-4)(x+1)(ax+b-a)+\color{red}{\left(-3a+b+\frac 16\right)x+12a-4b+\frac 43}.$$
The answer is the red part. Here, we cannot eliminate $a,b$.
|
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Does the order of operations matter with just addition and subtraction? Had a debate on whether you could do addition/subtraction in any order you want. Specifically, for the following:
$9 - 4 + 3$
We both agree that the answer is 8.
I argue that, by giving addition a higher priority than subtraction (rather than the same priority and going left-to-right), you would end up with $9 - 4 + 3 = 9 - 7 = 2$, which is an incorrect answer, and therefore it matters that addition and subtraction have the same priority.
The other person argues that the order of operations doesn't matter and that it can be done in any order, as by giving addition a higher priority than subtraction, you would end up with $9 - 4 + 3 = 9 + (-4 + 3) = 9 + - 1 = 8$, which is the correct answer, and therefore it wouldn't matter if addition/subtraction had different priorities rather than the same.
I'm arguing that bringing the $- 4$ inside the bracket and then performing it before the $+ 3$ wouldn't be done if addition had a higher priority. The fairly long, debate can be seen here if you want to read it, so I don't paraphrase it all and bias my side too much.
My overall question is who is correct; does the order of operations matter with just addition and subtraction? I'm willing to accept answers for either side, so long as they give a reason.
|
I would say that the order (for adding/subtracting) doesn't matter AND you need not elevate 'priorities' so long as you realize that subtracting is just adding negative numbers...
So the 9 - 4 + 3 isn't 9 - (4 + 3) but rather is 9 + ( -4 + 3 ).
Putting the negative sign in front of the parenthesis implies that you want to make everything inside negative and that's not what we actually wanted.
So just make everything a signed number, and do all processes as an addition in whatever order is easiest for you.
|
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$G$ is a group, if ∀a,b∈G, $a^2b=ba^2$, is $G$ Abelian? $G$ is a group, if ∀a,b∈G, $a^2b=ba^2$, how to make a counterexample show that $G$ is NOT Abelian?
What's the counterexample when $a^nb=ba^n$?
|
To get a counterexample, you need an element which is not the square of another element. A good example would be the quaternion group, where $i^2=j^2=k^2=-1$, and $(-1)^2=1$. $-1$ commutes with all the elements in the quaternion group.
Read the Wikipedia article here.
Since every element in the quaternion group raised to the $4$th power is the identity, it follows that for all $n|4$, your group does not need to be Abelian.
For $n\equiv2\mod4$, you can also use the quaternion group, as every element raised to the $n$th power would be either $-1$ or $1$, which commutes.
(This still leaves out the case when $n$ is odd.)
|
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Find the minimal polynomial for the given matrix. The first part is trivial. How to go about the second part?
|
You can check that
$$A^2=\begin{pmatrix}
0 &c &ac\\
0&b&ab+c\\
1&a&a^2+b
\end{pmatrix}
$$
Suppose that the minimal polynomial is $m(x) = x^2+\alpha x+\beta$. Then $m(A) = 0$. Notice that
$$(m(A))_{11} = 0+0+\beta=0\Rightarrow \beta=0$$
$$(m(A))_{21} = 0+\alpha+0=0\Rightarrow \alpha=0$$
Thus, $A^2=0$ which is clearly false.
By a simpler argument you can conclude that $m(x)$ is not a linear polynomial either. Hence, $m(x)$ must be at least of degree 3. But the characteristic polynomial is of degree 3 and we are done.
|
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|
Calculating syzygies with Macaulay2 I'm trying to calculate the syzygies of a set of elements on the polinomial ring of 6 variables. But I'm trying to specify the number of generator in each degree the syzygies have. I know that Macaulay2 can give me the syzygies of the sistem very fast, but it is returning some columns with mixed degree, and because of this I don`t know how to determinate how many generator of each degree I have.
I thought in define the set of polynomials as a map, and find a resolution for the kernel of this map, but it seems that even with Macaulay2, if I go that way I will have aplenty of calculations to do... My question is if that is any other faster way to solve this.
Thanks in any advance.
|
I think the problem here is understanding Macaulay2's notation. Let's take an example:
i1 : R = QQ[x,y,z]
o1 = R
o1 : PolynomialRing
i7 : I = ideal(random(2,R), random(3,R), random(2,R));
o7 : Ideal of R
i8 : betti res I
0 1 2 3
o8 = total: 1 3 3 1
0: 1 . . .
1: . 2 . .
2: . 1 1 .
3: . . 2 .
4: . . . 1
Here the last table gives us all the numerical information about the resolution of some ideal in $\mathbb Q[x,y,z]$. The way to read this is as follows: In the first syzygy $1$ there's two generators of degree $2$ and one of degree $3=1+2$. In the second syzygy there's one generator of degree $4=2+2$ and two of degree $5=3+2$.
In general, the number of generators of degree $i$ in the $j$th syzygy is found by reading the $(i,j-i)$ position in the Betti table.
|
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Area between three curves involves tangent line Let $R$ be the region enclosed by the $x$-axis, the curve $y=x^{2}$, and the tangent to the curve at $x=a$, where $a>0$. If the area of $R$ is $\frac{2}{3}$, then the value of $a$ is?
Is there any clue for the tangent equation and how's the integral equation will be?
Thanks!
|
Hints :
tangent is passing through $(a,a^2)$ where its slope would be $\displaystyle{\frac{dy}{dx}|_{x=a}}$
Area of R would be the difference between the region of curve and the region of tangent line enclosed by the x-axis. (from $0$ to $a$)
|
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Covering a rectangle with circles On a rectangle table with area A, n unit-radius circles are placed and it is not possible to place any extra circles without overlapping with some of the existing ones or without placing circle's center outside the rectangle which results in falling off of the table.
Notice that this placement and the number n are not unique for a given rectangle, but n, the rectangle, and the placement are given to us. Show that with 4n of these circles you can cover the rectangle entirely (that is any arbitrary point is covered by at least one circle).
|
WOLOG, choose the coordinate system so that the table cover the rectangle
$$R = [0,w] \times [0,h]\quad\text{ with }A = wh.$$
Let $C = \{ \vec{c}_1, \vec{c}_2, \ldots, \vec{c}_n \}$ be the centers of the $n$ circles.
For any $\vec{x} \in R$, let
$$d_C(\vec{x}) = \min\{ d(\vec{x},\vec{c}) : \vec{c} \in C\}$$
be the minimal distance of $\vec{x}$ to the centers. It is clear if there is a $\vec{x} \in R$ with $d_X(\vec{x}) > 2$, then the unit circle centered at $\vec{x}$ will not intersect any of the $n$ circles. This violates the condition we cannot add one more circles to the $n$ circles without overlapping. As a result, we have
$$\sup\{ d_C(\vec{x}) : \vec{x} \in R \} \le 2$$
Equivalently, we can cover $R$ by $n$ circles centered at $\vec{c}_i \in C$ with radius $2$. Scale everything down linearly by a factor of $2$, we can cover
$$\frac12 R \stackrel{def}{=} \left\{ \frac12\vec{x} : \vec{x} \in R \right\}$$
by $n$ unit circles centered at $\frac12\vec{c}_i$ with $\vec{c}_i \in C$.
Tiling this configuration in both $x$ and $y$ directions once, we can cover $R$ by $4n$ unit circles centered at
$$\frac12 \vec{c}_i + \vec{d}_j
\quad\text{ where }\quad
\vec{c}_i \in C\quad\text{ and }
\vec{d}_j =
\begin{cases}
(0,0),& j = 0\\
\left(\frac{w}{2}, 0\right),& j = 1\\
\left(0, \frac{h}{2}\right),& j = 2\\
\left(\frac{w}{2}, \frac{h}{2}\right),& j = 3.
\end{cases}$$
|
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|
permutation/combination There are six cards with numbers written on them: $1, 2, 3, 4, 5, 6$ .
Two of them are drawn at at time and put together to form fractions, e.g. $\dfrac45$
How many proper fractions can be formed?
What is correct equation to solve this type of questions?
thanks
|
If you think of all the possible fractions, exactly half of them will be proper (since you're always drawing cards of two different values), so there are $6 \times 5 \times \frac12$ possibilities.
|
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|
Symmetric group isomorphic to semidirect product of Alternating group and Z/2Z
I'm having a hard time understanding why $A_n \rtimes \mathbb{Z}_2 \cong S_n$.
I understand that $A_n$ is normal in $S_n$. But that's about it. What would the $\alpha$: $\mathbb{Z}$$_2$$\longrightarrow$Aut($A_n$) be?
|
Pick an odd element of $ S_n $ of order two; for example, a transposition. Then conjugation by that element is an order two automorphism of $ A_n $.
|
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Is every manifold a metric space? I'm trying to learn some topology as a hobby, and my understanding is that all manifolds are examples of topological spaces. Similarly, all metric spaces are also examples of topological spaces. I want to explore the relationship between metric spaces and manifolds, could it be that all manifolds are examples of metric spaces?
|
Assuming the usual definition of a topological manifold as a locally Euclidean space which is both Hausdorff and second-countable, it turns out that every manifold $M$ is a metrizable space. That is, you can put a metric on $M$ which induces the topology of $M$. This follows from example from Urysohn's metrization theorem. The metric is highly non-unique and a manifold doesn't come with a preferred metric which turns it into a metric space.
|
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Finding the value of this limit $$\lim_{x\to 0}\frac{\sin x^4-x^4\cos x^4+x^{20}}{x^4(e^{2x^4}-1-2x^4)}$$
The answer is given to be $1$. Can someone give any hint/s related to this problem?
More importantly, why am I not able to get the answer by simply using L hopital's rule and by basic substitution of trigonometric and exponential limits?
Any kind of help would be appreciated. I gave the problem simply as an example to reflect the above confusion.
EDIT: Even though tag says so, you are still allowed to use l hospital's rule to help evaluate the limit.
The question is edited now. It is e^(2x^4) in denominator.
Again, thanks a lot for taking your time to reply to my doubt.
|
Divide numerator and denominator by $x^4$, giving you
$$\lim_{x\to 0}\frac{\frac{\sin x^4}{x^4}-\cos x^4+x^{16}}{e^{2x^4}-1-2x^4}$$
Now expand the sine, cosine, and exponential into power series (Maclaurin series) with as many terms as needed.
|
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Find the minimum of $(u-v)^2+ (\sqrt{2-u^2}-\frac{9}{v})^2$ for $00$.
Find the minimum of $\displaystyle (u-v)^2+ \left(\sqrt{2-u^2}-\frac{9}{v}\right)^2$ for
$0<u<\sqrt{2}$ and $v>0$.
I think I have to use the Arithmetic and Geometric Means Inequalities. Or $\displaystyle \frac{1}{2}(u-v)^2+ \left(\sqrt{2-u^2}-\frac{9}{v}\right)^2 \geq \frac{1}{4}\left(u-v+\sqrt{2-u^2}-\frac{9}{v}\right)^2$ by Chebychev inequality. Is anyone is able to give me a hint how to finish it?
I think I have to use the Arithmetic and Geometric Means Inequalities.
|
You can simply think $(u-v)^2+ (\sqrt{2-u^2}-\frac{9}{v})^2$ is the square of distance between two points $a$ and $b$ in the plane, where
$$a=(u,\sqrt{2-u^2}),b=(v,\frac{9}{v}).$$
Clearly, $a$ satisfies $x^2+y^2=2,(0<x<\sqrt{2})$ and $b$ satisfies $y=\frac{9}{x},(x>0)$. Then from the geometry graph, we easily know the minimum distance between the two curves is the distance between $(1,1)$ and $(3,3)$. So the minimum distance is $\sqrt{8}$, and hence the minimum solution is 8 at $u=1,v=3$.
|
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How to prove D70 = {1, 2, 5, 7, 10, 14, 35, 70} is a Boolean algebra Prove that the set $D_{70}$ = {1, 2, 5, 7, 10, 14, 35, 70} of positive factors is a Boolean algebra under the operation (+), (.), (') defined by
$$x + y = lcm(x, y)$$
$$x . y = gcd(x, y)$$
$$x' = \frac{70}{x}$$
Attempt:
To Prove $D_{70}$ is a boolean algebra we have to satisfy below 4 properties of boolean algebra
*
*+ & . are commutative.
*+ is distributive over . and vice versa.
*$\exists$ two distinct identity element 0, I for ($D_{70}$, +) and ($D_{70}$, .) respectively.
*$\forall a \in D_{70}$, a + a' = I & a + a' = 0
Property 1: Satisfied since lcm, gcd are commutative
Property 2: ?
Property 3: I = 70, 0 = 1
Property 4: Satisfied
I'm having trouble proving the Property 2 can anybody help.
|
Write out the operation tables for "+", ".", and " ' ".
Then check that for all x, y, z in D70 under those operations that the distributive laws hold.
You could write a computer program for the last part, use OTTER, or treat those tables as an 8-valued logical system and write out truth tables for the computation.
|
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Arithmetic sequence to geometric sequence.
The numbers $a_1, a_2, a_3, . . .$ form an arithmetic sequence with $a_1 \ne a_2$. The three
numbers $a_1, a_2, a_6$ form a geometric sequence in that order. Determine all possible
positive integers $k$ for which the three numbers $a_1, a_4, a_k$ also form a geometric
sequence in that order.
This is from Euclid 2015:Problem 7B
I got that:
$a_n = a_1 + (n-1)d = r^{n-1} a_1$ where $d$ is difference, and $r$ the geom. ratio.
But this doesnt give many leads, HINTS ONLY PLEASE.
|
HINT :
Use the fact that
$$\text{$a,b,c$ form a geometric sequence}\Rightarrow b^2=ac$$
Thus, you have
$$a_2^2=a_1a_6\quad\text{and}\quad a_4^2=a_1a_k$$
Express these by $a_1$ and $d$.
|
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How can I change the variable in this integral? I have the following equality
$\int_{0}^{\pi}cos(nt)cos(mt)dt=0$ (if $m\neq n$)
This is in the context of Chebyshev polynomials and the book states the following to deduce ortoghonality.
"the orthogonality property is drawn by using the substitution t = arcos(x) in the interval [-1,1] relative to the weight function $(1-x^2)^{-\frac{1}{2}}$, which is verified by the system of polynomials {Tn(x)}:"
$\int_{-1}^{1}Tn(x)Tm(x)dt=0$, (if m $\neq$ n)
Can anybody explain (or point me to an example) how to do the above variable change in the initial integral and where does the weight function come from?
|
$t = \arccos{x} \implies dt = -(1-x^2)^{-1/2} \, dx$.
Also note that
$$T_m(x) = \cos{(m \arccos{x})} $$
|
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About the rank of submodules over PID
If $M$ is a free $R$-module of finite rank $n$ and $R$ is a PID, do
proper submodules of $M$ have strictly less rank than $M$?
I know that in this case, every submodule of $M$ is free and has finite rank $m\leq n$, but can we guarantee that if the submodule is proper, its rank will be strictly less than the rank of $M$? In other words,
does the rank of a submodule $N$ of $M$ equals $n$ if and only if
$N=M$?
I think this is not necessarily true, but I can't find a counter-example. Some help would be very appreciated. Thanks!
|
The rank of a submodule $N$ of $M$ will be equal to the rank of $M$ if and only if the quotient module $M/N$ is a torsion module, by the fundamental theorem of finitely generated modules over P. I. D.s. Hence its rank will be strictly less than the rank of $M$ if and only if $M/N$ contains non-torsion elements.
|
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Product of numbers $\pm \sqrt{1} \pm \sqrt{2} \pm \dots \pm \sqrt{100}$ is a perfect square Let $A$ be the product of $2^{100}$ numbers of the form
$$\pm \sqrt{1} \pm \sqrt{2} \pm \dots \pm \sqrt{100}$$
Show that $A$ is an integer, and moreover, a perfect square.
I found a similar problem here, but the induction doesn't seem to show that $A$ is a perfect square. And I think we can generalize the following problem: if $n$ is a perfect square then $A$ is a perfect square, too.
|
The same argument can be used to show that the product of the $2^{n-1}$ numbers $\sqrt{1} \pm \sqrt{2} \pm \cdots \pm \sqrt{n}$ is an integer.
Then your number is exactly the square of this product.
|
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Obtain the expression $T=\frac x2+\sum_{n=2}^\infty (-1)^{n-1} \frac{1*3*\ldots*(2n-3)}{n!}*\frac{x^n}{2^n}$ Obtain the expression
$$T=\frac x2+\sum_{n=2}^\infty (-1)^{n-1} \frac{1*3*\ldots*(2n-3)}{n!}*\frac{x^n}{2^n}$$
for one root of the equation
$$T^2+2T-x=0,$$
and show it converges so long as $|x| \lt 1$.
I have no idea where to start with this, can I have a hint on how to obtain the expression? and how do I find the other root? Can I use the ratio test on the expression to show that it converges?
|
First notice that
\begin{align}
T(x) &= \frac{x}{2} - \frac{1}{2} \, \sum_{n=2}^{\infty} (-1)^{n-1} \frac{(-1/2)_{n} \, x^{n}}{n!} \\
&= \frac{x}{2} + \frac{1}{2} \, \left[ (1+x)^{-1/2} - x -2 \right] \\
&= -1 + \sqrt{1+x}
\end{align}
Now the quadratic equation $T^{2} + 2 \, T - x = 0$ leads to
$$T(x) = - 1 \pm \sqrt{1+x}$$
and is of the same form as the series for $T(x)$ given in the problem.
As to the convergence it is not hard to show by any number of standard tests and is left do to.
|
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Irreducible finite-state Markov Chain Could anyone help me with this?
Let $X_n$ be an irreducible Markov chain on the state space $S=\{1,\ldots,N\}$. Show that there exist $C<\infty$ and $\rho <1$ such that for any states $i,j$,
$ P (X_m \neq j, m = 0, \ldots, n | X_0 =i) \leq C \rho^n.$
Show that this implies that $E(T)<\infty$, where $T$ is the first time that the Markov chain reaches $j$. (Suggestion: There is $\delta >0$ such that for all $i$, the probability of reaching $j$ in some moment of the first $N$ steps, starting at $i$, is major that $\delta$. Why?)
I`ve already got the suggestion but i do not know how to apply it.
|
Use this fact that if process has not hitted $j$ in $m$ steps, it's not hitted $j$ in $k$ steps where $k$ is greatest multiple of $n$ smaller than $m$.
In fact, if there is a $\delta$ such that beginning from each $i$, you have reached $j$ in $n$ (may be smaller) steps with probablity greater than $\delta$, then take $\rho = (1-\delta)^ {1/n} $ and $C = 1/\rho$. Then, the assertion just says that after $kn+r$ where $(0≤r<n)$ steps, the probablity of not reaching $j$, starting from $i$ is less than $\rho ^ {(k-1)n+r}$. and its obvious. In fact more is true; such probablity is less than $\rho ^ {kn}$ because of strong Markov property.
|
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Why does count of Z -Transform of sequence change? I was looking at the reference [1] below and noted the author defined the Z-transform for [1, 2, 3] as $$[6, \frac{11}{4}, 2]$$
I worked it out as follows:
$$X[z]=\sum_{n=0}^2x[n]z^{-n}$$
$$=x[0]z^0+\frac{x[1]}{z}+\frac{x[2]}{z^2}$$
$$=1+\frac{1}{z}+\frac{2}{z^2}$$
so from there the following results:
$$X[1]=1+\frac{2}{1}+\frac{3}{1}=6$$
$$X[2]=1+\frac{2}{2}+\frac{3}{2^2}=\frac{11}{4}$$
$$X[3]=1+\frac{2}{3}+\frac{3}{3^2}=2$$
Hence
$$X[z]=[6, \frac{11}{4}, 2]$$
However, in the 2nd part when plugging in for z it implies X[z] counts from 1 to 3 i.e. X[0] does not exist only X[1], X[2], and X[3]. Before the transform, the count of x[n] was from 0 to 2 so x[0], x[1], and x[2] exist while x[3] does not. Can anyone explain why the count changes? I've tried using n=1 to 3 to carry out the transform but a different result comes out. I've also changed z to z=0 to 2 but doesn't work either.
References
[1] http://algorithmicassertions.com/quantum/2014/04/27/The-Not-Quantum-Laplace-Transform.html
|
When I wrote that post I used 0-based indexing for the input because that's just what programmers do by default, and I used 1-based indexing for the output because of the division-by-zero issue. I don't know if it's more common for the indices to start at 1 for the input.
Basically, don't worry too much about it. It's just a convention, and it's not hard to convert between the various possibilities.
|
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How to solve $\max|x_1|\Sigma_{i=1}^n|x_i|$, with constraint $\Sigma_{i=1}^n|x_i|^2=1,n>2$ Alternatively, that making the upper bound for $|x_1|\Sigma_{i=1}^n|x_i|$ as tight as possible is also welcome. For me, if $\Sigma_{i=1}^n|x_i|=\sqrt{n}$, then $|x_1|=\sqrt{1/n}$. If $|x_1|=1$,then $\Sigma_{i=1}^n|x_i|=1$. In a word, I guess the objective function may be bounded by a constant.
|
WLOG, $x_i\geq 0$ for every $i$. From $\sum_{i>1}x_i\leq \sqrt{(n-1)\sum_{i>1}x_i^2}=\sqrt{(n-1)\left(1-x_1^2\right)}$, we have $x_1\sum_{i=1}^nx_i\leq x_1^2+x_1\sqrt{(n-1)\left(1-x_1^2\right)}$. By AM-GM, $\sqrt{ax_1^2\cdot b\left(1-x_1^2\right)}\leq \frac{b}{2}-x_1^2$, where $0\leq a<b$ satisfy $ab=n-1$ and $b-a=2$. Note that $a=\sqrt{n}-1$ and $b=\sqrt{n}+1$. Hence, $x_1\sum_{i=1}^nx_i\leq \frac{b}{2}=\frac{\sqrt{n}+1}{2}$. This bound is sharp for all positive integers $n$, i.e., by taking $x_1=\sqrt{\frac{b}{a+b}}=\sqrt{\frac{\sqrt{n}+1}{2\sqrt{n}}}$ and $x_2=\ldots=x_n=\sqrt{\frac{1}{b(a+b)}}=\sqrt{\frac{1}{2n+2\sqrt{n}}}$.
Using a similar argument, one can also maximize $\left(\sum_{i=1}^k\,x_i\right)\left(\sum_{i=1}^n\,x_i\right)$, where $k\in\{1,2,\ldots,n\}$. It is easy to show that $\left(\sum_{i=1}^k\,x_i\right)\left(\sum_{i=1}^n\,x_i\right)\leq \frac{\sqrt{kn}+k}{2}$ and the equality is achieved when $x_1=\ldots=x_k=\sqrt{\frac{\sqrt{n}+\sqrt{k}}{2k\sqrt{n}}}$ and $x_{k+1}=\ldots=x_n=\sqrt{\frac{1}{2\sqrt{n}(\sqrt{n}+\sqrt{k})}}$.
|
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|
What are the residues of $\frac{z^2 e^z}{1+e^{2z}}$? I hope to know the singularity and pole of $\frac{z^2 e^z}{1+e^{2z}}$.
I try $\frac{z^2 e^z}{1+e^{2z}} = \frac{z^2}{e^{-z}+e^{z}}$ and observe that the denominator seems like cosine function. So I think the singularites are i(2n+1)$\pi$/2. But when I try to evaluate the poles, I fail and different values of $n$ has different residues.
Could the residues be imaginary?
|
Since $\frac{z^2 e^z}{1+e^{2z}} = \frac{z^2}{2\cosh z}$, let's start with
$$f(z)=\frac{z^2}{2\cos iz}$$
Now both $z^2$ and $\cos iz$ are entire. The function can be singular only where $z_{0_n} = iz = \frac{\pi +2\pi n}{2}$ and since the zeroes of the denominators are first order, we have
$$\begin{align*}
Res f(z_{0_n}) & =\frac{z^2}{(2\cos iz)'}\bigg|_{z_{0_n}} \\ \\
& =\frac{\left(\frac{\pi +2\pi n}{2}\right)^2}{-2i\sin \left(\frac{\pi +2\pi n}{2}\right)} \\ \\
& =\frac{i\left(\pi +2\pi n\right)^2}{8\sin \left(\frac{\pi}{2} +\pi n\right)} \\ \\
& =\frac{i\pi^2\left(1 +2 n\right)^2}{8\cos\left(\pi n\right)} \\ \\
\end{align*}
$$
$$\boxed{Res f(z_{0_n}) =(-1)^n\frac{i\pi}{8}\left(1 +2 n\right)^2}$$
Compare with WolframAlpha...
|
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|
What is the positive divisors of $n(n^2-1)(n^2+3)(n^2+5)$ I want to find the positive divisors of $n(n^2-1)(n^2+3)(n^2+5)$
from $n(n-1)(n+1)$ 2 and 3 should divide this expression for all positive n. how can I find the rest? which python says $(2, 3, 6, 7, 9, 42, 14, 18, 21, 63,126)$
|
If $$n\equiv 0 \mod 3,$$ then the factor $$n^2+3\equiv 0 \mod 3.$$ Otherwise
$$n^2\equiv1 \mod 3$$ and thus $$n^2+5\equiv 0\mod3.$$
Therefore $$(n^2+3)(n^2+5)\equiv0\mod3.$$
Let us now look modulo $7$ and compute the number for all elements of $\mathbb Z_7$ (which means to check if the number is a multiple of $7$ for $n=0,\dots 6$. You can easily check that it is true. Therefore this number is a multiple of $2\times3\times3\times7=126$ and all its divisors. There are no others since for $n=2$, $n(n^2-1)(n^2+3)(n^2+5)=126\times3$ but for $n=3$, $n(n^2-1)(n^2+3)(n^2+5)=126\times 32$.
|
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Prove or Disprove: if $\lim\limits_{n \to \infty} (a_{2n}-a_n)=0,$ then $\lim\limits_{n \to \infty} a_n=0.$ Prove or Disprove: if $\lim\limits_{n \to \infty} (a_{2n}-a_n)=0,$ then $\lim\limits_{n \to \infty} a_n=0.$
I don't think that this is true. and I'm trying to think about counterexample, but couldn't figure out a mathematical form of the sequence that I thought about, its a sequence where certain terms that are multipliers of $2n$ and $n$ become more close to each other as $n$ grows.
any hints ?
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Just let $a_n = 1$ for all $n$, for example.
|
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|
Is there another kind of two-dimensional geometry? I learned that in two-dimensional geometry, there are Euclidean geometry, hyperbolic geometry and spherical geometry. These geometries are homogeneous and isotropic. Is there another kind of two-dimensional geometry that is homogeneous and isotropic?
|
In addition to the nine following a specific set of rules,
There are also more exotic versions such as projective geometry (think twisting the plane into a Mobius strip) and finite geometries.
If you relax the isotropic requirement you open up more possibilites such as the Manhattan metric aka Taxicab Geometry.
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How to prove that $\frac1{n\cdot 2^n}\sum\limits_{k=0}^{n}k^m\binom{n}{k}\to\frac{1}{2^m}$ when $n\to\infty$ I have solve following sum
$$\sum_{k=0}^{n}k\binom{n}{k}=n2^{n-1}\Longrightarrow \dfrac{\displaystyle\sum_{k=0}^{n}k\binom{n}{k}}{n\cdot 2^n}\to\dfrac{1}{2},n\to\infty$$
$$\sum_{k=0}^{n}k^2\binom{n}{k}=n(n+1)2^{n-2}\Longrightarrow \dfrac{\displaystyle\sum_{k=0}^{n}k^2\binom{n}{k}}{n\cdot 2^n}\to\dfrac{1}{2^2},n\to\infty$$
$$\sum_{k=0}^{n}k^3\binom{n}{k}=2^{n-3}n^2(n+3)\Longrightarrow \dfrac{\displaystyle\sum_{k=0}^{n}k^3\binom{n}{k}}{n\cdot 2^n}\to\dfrac{1}{2^3},n\to\infty$$
so I conjecture the following:
$$\dfrac{\displaystyle\sum_{k=0}^{n}k^m\binom{n}{k}}{n\cdot 2^n}\to\dfrac{1}{2^m},n\to\infty$$
$$\cdots\cdots$$
so I conjecture for $m$ be positive real number also hold.
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Let $(Z_i)$ be i.i.d. Bernoulli(1/2) random variables, and set $\bar Z_n={1\over n}\sum_{i=1}^n Z_i$ be the sample average. Then $\bar Z_n\to 1/2$ almost surely by the law of large numbers.
On the other hand, $X=\sum_{i=1}^n Z_i$ has a Binomial$(n,1/2)$ distribution, so
that, $\mathbb{P}(X=k)={n\choose k}(1/2)^n$ for $0\leq k\leq n$.
Therefore, for every real number $m\geq 0$
$$ \mathbb{E}(\bar Z_n^m)={1\over n^m}\sum_{k=0}^n {n\choose k} k^m\left({1\over 2}\right)^n \to {1\over 2^m},$$
by the bounded convergence theorem.
|
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Find the tight bound for the recurrence relation $T(n) = T(\frac{n}{3}) + 6^n $ $$T(1) = 2$$
$$T(n) = T(\frac{n}{3}) + 6^n \text{ For n > 1}$$
I tried using the substitution method to find it's closed form but I even from there, I could not figure out how to find its bound. My working:
$k = 1$
$$T(n) = T(\frac{n}{3}) + 6^n $$
$k = 2$
$$T(n) = T(\frac{n}{9}) + (6^{\frac{n}{3}})+(6^n) $$
$k = 3$
$$T(n) = T(\frac{n}{27})+ (6^{\frac{n}{9}}) + (6^{\frac{n}{3}})+(6^n) $$
$k = 4$
$$T(n) = T(\frac{n}{81}) + (6^{\frac{n}{27}}) + (6^{\frac{n}{9}}) + (6^{\frac{n}{3}})+(6^n) $$
From the pattern, I can guess that it has the following recurrence
$$T(n) = T(\frac{n}{3^k}) + (k)(6^n) $$
Solving for $k$ we have
$1 = \frac{n}{3^k}$
$3^k = n$
$\log{_3}{n} = k$
Substituting all the k we have
$T(n) = 2 + 6^n\log{_3}{n}$
I'm stuck over here. Can master theorem apply for this case? Is there an upperbound for $k^n$ ?
|
Hint:
$$6^n > n^{\log_3 2}$$
Can you take another look at the Master theorem cases ?
|
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Common methods for proving: Existence proof For every real number x with $x\neq -1$ there exists a real number y such that $ \frac{y}{y+1}=x $.
$ \forall x\in \Bbb R\setminus (1) \ \ \exists y \in \Bbb R : \frac{y}{y+1}=x \\, x \neq -1 $
$ So \ \ if \ \ I \ \ choose \ \ e.g. \ x= \frac{1}{2} \ \ then \ \ y=1 \ \ such \ \ that \ \ \frac{y}{y+1}=x $
Is this a correct way of proving this?
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Use the transformation of $y\rightarrow z-1$ to change it to for all $x\neq 1$ there exists $z$ such that $\frac{z-1}{z}=1-\frac{1}{z}=x$, which is easily solvable for $z$ in terms of $x$.
|
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Why the ordering of the quantifiers matters here? $\newcommand{\fool}{\operatorname{fool}}$I want to transfer the statement
You can fool all the people some of the time
into predicate logic.
Now I am using fool(X,T) to mean that I fool person X at time T.
Now I know that the correct form is $\exists T\ \forall X\ \fool(X,T)$.
But I want to understand why is it not correct if we change the ordering of the quantifiers to $\forall X\ \exists T\ \fool(X,T)$.
The first one is saying there exists a time that I can fool all people right ?
Ans the second one is saying for all the people out there, There exists a time that I can fool you , right ?
What is the difference ??
What makes the first once correct but not the second.
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To expand on AJ stas's comment, the first formulation, $\exists T \forall X fool(X,T)$, first you assert the existence of (at least one) single Time, T. Then you say that for EVERY person X, $fool(X,T)$, i.e, every single person can be fooled at that exact time.
When you reverse the quantifiers, $\forall X \exists T fool(X,T)$, you are asserting that for every single person, there exists (at least one) time (that depends on that person!) for which that person can be fooled. So, I might be fooled today and you might be fooled tomorrow.
In the first example, there's a time, like 3:17PM, where every person in the world can be fooled at once.
I'm pretty sure that the second version more accurately states the colloquial meaning of the phrase.
|
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What is $\langle a \rangle + \langle b \rangle$ where $a$ and $b$ are natural numbers? Let $\langle a \rangle$ and $\langle b \rangle$ be ideals in $\mathbb{Z}$, where a and b are natural numbers.
Define
$$S = \langle a \rangle + \langle b \rangle = \{x + y \mid x \in \langle a \rangle \text{ and } y \in \langle b \rangle\}.$$
What is $\langle a \rangle + \langle b \rangle$ ?
I know $S \supseteq d\mathbb{Z}$ where $d = \gcd(a,b)$ since there exist integers $m$ and $n$ such that $d = am + bn$. Then any integer multiple of $d$ can be expressed as the sum of an integer multiple of $a$ and $b$.
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For the other direction, let $x+y\in S$ where $s\in \langle a\rangle, y\in \langle b \rangle$. Then there are $h,k\in\mathbb{Z}$ such that $x = ha$, $y = kb$.
Then with $d = \gcd(a,b)$, $d$ is a divisor of $a$ and $b$, so there are $n$ and $m$ such that $a = nd$ and $b = md$. So $x + y = ha + kb = hnd + kmd = (hn + km)d \in d\mathbb{Z}$.
|
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once, twice, thrice differentiable So I am going through abbotts 2nd edition of Understanding analysis and I would like to know if this is correct.
Exercise:
(a) Let $g : [0, a]→R$ be differentiable, $g(0) = 0$, and $|g'(x)| ≤M$ for all $x∈[0, a]$. Show $|g(x)| ≤Mx$ for all $x∈[0, a]$.
(b) Let $h : [0, a]→R$ be twice differentiable, $h'(0) = h(0) = 0$ and $|h''(x)| ≤ M$ for all $x∈[0, a]$. Show $|h(x)| ≤Mx^{2}/2$ for all $x∈[0, a]$.
(c) Conjecture and prove an analogous result for a function that is differentiable three times on $[0, a]$.
My solution:
(a) $g : [0, a]→R$ and $g(0) = 0$ and $|g'(x)| \leq M$ for all $x \in [0,a]$.
For any $x \in (0,a]$ we have by the mean value theorem there exists $c \in (0,x)$ such that $g(x) - g(0) = (x-0) g'(c) $[By MVT on $[0,a]$].
$|g(x)| = |x g'(c)| $ [as $g(0) = 0$]
$|g(x)|= |x| \cdot |g'(c)|$. As $|g'(c)| \leq M$ and $x \in (0,a]$. Therefore $|x|=x$. Hence $|g(x) | \leq Mx$ for all $x \in (0,a]$ when $x=0$ the inequality holds trivially. Thus $|g(x)| \leq Mx$ for all $x \in [0,a]$.
(b) $h: [0,a] \rightarrow \mathbb{R}$ is twice differentiable, $h'(0)=h(0)=0$. And $|h''(a)| \leq M \forall x \in [0,a]$. For any $x \in [0,a]$ we have
$h(x) = h(0) + xh'(0) + (x^{2}/2!) h''(c)$ where $c \in (0,x)$. As $h(0)=h'(0)$, so $h(x)= (x^2/2) h''(c)$
$|h(x)| = |(x^2/2) h''(c)| \leq Mx^2/2$ [as $|h''(c)| \leq M$]
(c) Let $ f:[0,a] \rightarrow \mathbb{R}$ is three times differentiable and $f''(0)=f'(0)= f(0) =c$. Also $|f'''(x)| \leq M$ for all $x \in [0,a]$. For any $x \in [0,a]$, we have by the taylor series expantion of the given function $f(x)$about the point '$0$' in the interval [0,x]
$f(x) = f(0) + x f'(0) + (x^2/2!)f''(0) + (x^3/3!) f'''(c)$ where $c \in (0,x)$
So $|f(x)|= (x^3/3!) |f'''(c)|$, as $x\geq 0$
$|f(x)| \leq Mx^3/(3!)$ as $|f'''(c)| \leq M$. So the given statement is true in this case.
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I only read your proof for (a), and it is slightly confusing. I think it might have the general idea right, but it is very unclear because of the poor presentation. You need to fix these issues:
*
*Currently, you say that by MVT on $[0,a]$, you know there exists some $c\in(0,x)$ such that $g(x)-g(0) = (x-0)\cdot g'(c)$, but this is simply untrue. MVT on $[0,a]$ tells you there is some $c$ in $(0, a)$ such that $g(a)-g(x) - (a-x)\cdot g'(c)$. You need to correct your justification here.
*You say $|g(x)|= |n||g'(c)|$, but you did not define $n$. This statement is currently meaninless.
*You say "Therefore $|x|=|x|$". This is a tautology that is completely irrelevant.
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|
finding density of $1/Z$ when $Z$ is a standard random normal variable If $Z$ is a standard normal r.v., we know that its density is
$$f_Z(z)=\frac{1}{\sqrt{2\pi}}e^{-z^2/2},$$
where $-\infty \leq z\leq \infty$. I want to find what $f_{1/Z}(z)$ is.
I let $Y=1/Z$, so $Z=1/Y$, and then I used the above formula to obtain
$$f_{1/Z}(z)=\frac{1}{\sqrt{2\pi}}e^{-1/2y^2},$$
but my solution is wrong. I don't know what else to do. Some help?
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By the properties of a continuous density,
$$
\mathbb P(0 < Y \le b) = \int_0^b f_Y(y) \, dy.
$$
Therefore, for $Y = 1/Z$,
$$
\mathbb P(0 < Y \le y) = \mathbb P(Z \ge 1/y) = \int_{1/y}^{\infty} f_Z(z) \, dz = 1/2 - \int_0^{1/y} f_Z(z) \, dz.
$$
Differentiating with respect to $y$ to recover $Y$'s density,
$$
f_Y(y) = \frac d{dy} \left( 1/2 - \int_0^{1/y} f_Z(z) \, dz \right) = -f_Y(1/y) \cdot \frac{-1}{y^2} = \frac{f_Y(1/y)}{y^2}.
$$
All that is left to do now is plugging in. Note that we have only taken care of the case where $z \ge 0$ ; I leave the case $y \le 0$, it can be done similarly (hint : consider $\mathbb P(1/Z \le y)$ ; note that I considered $\mathbb P(0 < 1/Z \le y)$ because the density of $Y$ satisfies $f_Y(y) = \frac{d}{dy} \mathbb P(Y \le y)$).
Hope that helps,
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How does this definition of curl make sense? In our maths textbook, curl is defined as $\text{curl} \textbf{F} = \nabla \times \textbf{F}$. If the $\nabla$ operator isn't a vector, but an operation on a vector, how can it be an operand of a cross product?
$$\textbf{F} \in \mathbb R^3 \\ (\nabla) \in \mathbb R^3 \rightarrow \mathbb R^3 \\ (\times) \in \mathbb R^3 \rightarrow \mathbb R^3 \rightarrow \mathbb R^3 \\ \nabla \times \textbf{F} \in \mathbb R^3$$
This seems as silly to me as:
$$2 \in \mathbb R \\ \text{sin} \in \mathbb R \rightarrow \mathbb R \\ (+) \in \mathbb R \rightarrow \mathbb R \rightarrow \mathbb R \\ \text{sin} + 2 \in \mathbb R$$
Obviously, you can't treat sine as a number and add it to 2. How come you can treat functions as numbers in the definition of curl? Finding this cross product makes less sense, since you have the determinant of a matrix containing 3 vectors, 3 functions, and 3 scalars.
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$
\mathrm{\nabla}
$
can be regard as a vector
$
\left({\frac{}{\mathrm{\partial}{x}}\mathrm{,}\frac{}{\mathrm{\partial}{y}}\mathrm{,}\frac{}{\mathrm{\partial}{z}}}\right)
$
to participate in calculation
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Is every Banach space densely embedded in a Hilbert space? Can every Banach space be densely embedded in a Hilbert space? This is clear if the Banach space is actually a Hilbert space, but much can you relax this?
If the embedding exists, is the target Hilbert space unique?
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I will show that the answer is yes, for any separable Banach space $B$. The non-separable case is covered by George Lowther above.
So let $B$ be a separable Banach space, and consider some countable dense subset $\{x_n\}_{n\in \Bbb N} \subset B$. For each $n$ we apply the Hahn Banach theorem to find some $f_n \in B^*$ such that $f_n(x_n)=\|x_n\|$ and $\|f_n\|=1$.
We then define an inner product $\langle \cdot, \cdot \rangle_2$ on $B$ by setting $$\langle x,y \rangle_2 = \sum_{n=1}^{\infty} 2^{-n} f_n(x)f_n(y)$$ and we let $\|\cdot\|_2$ be the asociated norm.
To check that this is actually an inner product, the only nontrivial property we need to check is nondegeneracy (which means that $\|x\|_2=0 \implies x=0$). Define $g:=\sup_n f_n$. Note that $|g(x)| \leq \|x\|$ since $\|f_n\|= 1$ for all $n$. Also notice that $g$ is continuous, because for any fixed $N$ we have that $f_N(x)-\sup_nf_n(y) \leq f_N(x)-f_N(y)$ so taking the sup over all $N$ on both sides gives that $g(x)-g(y) \leq g(x-y) \leq \|x-y\|$. Symmetrically we can switch $x$ and $y$ so this actually shows that $|g(x)-g(y)| \leq \|x-y\|$, thus proving continuity. Also $g(x_n) \geq f_n(x_n)=\|x_n\|$ and thus $g(x_n)=\|x_n\|$ for all $n$. By density of the $x_n$ and continuity of $g$ it follows that $g(x) = \|x\|$ for all $x$. Hence $\|x\|=\sup_n f_n(x)$, so that if $\|x\|_2=0$ then $f_n(x)=0$ for all $n$ and thus $x=0$.
We denote by $B'$ the completion of $B$ with respect to $\|\cdot\|_2$. Then the inclusion map $i:B \to B'$ is a continuous embedding, because $$\|x\|_2^2 = \sum_n 2^{-n}f_n(x)^2 \leq \sum_n 2^{-n} \|x\|^2=\|x\|^2$$
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Proving the determinant of a matrix is positive How can I show that $\det A \leq 0 $,
if $A = \begin{pmatrix}
a & b & c\\
b & d & e\\
c & e & f
\end{pmatrix}$
and $a>0, ad-b^2 = 0$
|
We have $$|A|=adf-b^2f-ae^2-dc^2+2bce=-(ae^2+dc^2-2bce)=-(ae^2+dc^2\pm 2\sqrt{ad}ce)=-(e\sqrt{a}\pm c\sqrt{d})^2\le 0$$
Note that we need $a>0$ for the step $\sqrt{ad}=\sqrt{a}\sqrt{d}$.
|
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|
Arranging 6 different people in all possible ways in a 3 floor house Say we have a house that fits 6 people. All the rooms on the left side of the house have big rooms, all the rooms on the right side have small rooms
---------
Floor 3| A | B |
---------
Floor 2| A | B |
---------
Floor 1| A | B |
---------
We have 6 people to throw in this house now.
Say Jeremiah gets $Floor\ 3\ A$ bedroom and Todd gets $Floor\ 3\ B$. If they were to switch rooms, this would be considered a different layout for them, solely because the rooms are not equal.
The floors are not equal either. If the exact arrangement above is duplicated on $Floor\ 2$ that is different than if they were in $A$ and $B$ respectively, on $Floor\ 3$.
So with 6 people in this house with possible orderings counting, I calculate there being $6!$ different ways for the house to be filled out.
Say we go to another house where the rooms are equal but the floors are not.
---------
Floor 3| = | = |
---------
Floor 2| = | = |
---------
Floor 1| = | = |
---------
Now we know if Jeremiah and Todd are on $Floor\ 3$, there can only exist one way they can be on that floor since the rooms are equal and interchangeable. So it seems that since every floor's permutation counts as only one ordering, the answer for this house arrangement is $\frac{6!}{2!}$
Now say we go to another house where the rooms and floors are equal. This means if Jeremiah and Todd are together, in what rooms on what floors is entirely irrelevant. The only information we need to know is who is together as a pair and that is all. For this house arrangement I see $\frac{6!}{2!\times3!}$ different possibilities.
I am wondering if the way I am looking at narrowing down these orderings is correct or if I am missing something in my division or thought process. Thanks!
|
So it seems that since every floor's permutation counts as only one ordering, the answer for this house arrangement is $\frac{6!}{2!}$
This part of the answer is wrong. Since there are 3 floors, on each of which the permutations don't matter, the denominator will change as:
$$\frac{6!}{2!^3}$$
Similarly:
For this house arrangement I see $\frac{6!}{2!\times3!}$ different possibilities.
This will become:
$$\frac{6!}{2!^3*3!}$$
|
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|
$\mathbf{E}[X|Z=z]=\mathbf{E}[Y|Z=z]$ implies $X=Y$ a.s.? Suppose that $X,Y,Z$ are real-valued random variables defined on a probability space $(\Omega, \mathscr{F}, P)$ such that for all reals $z$ such that $P(Z=z)>0$ it holds
$$
\mathbf{E}[X|Z=z]=\mathbf{E}[Y|Z=z].
$$
What can be said about $X$ and $Y$? In particular, is it true that $X=Y$ a.s. $P\upharpoonright \sigma(Z)$?
|
Just consider $X, Z$ and $Y, Z$, to be independent, $X$ and $Y$ can have any probability distribution as long as they have equal mean. Then your equality holds for all $z$, but $X$ and $Y$ are not a.s. equal.
|
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|
Determine which of the following relation is a function? Given two set $ A = \{0, 2, 4, 6\}$ and $B = \{1, 3, 5, 7\}$, determine which of the following relation is a function?
$(a) \{(6, 3), (2, 1), (0, 3), (4, 5)\}$,
$(b) \{(2, 3), (4, 7), (0, 1), (6, 5)\}$,
$(c) \{(2, 1), (4, 5), (6, 3)\}$,
$(d) \{(6, 1), (0, 3), (4, 1), (0, 7), (2, 5)\}$.
My attempt :
"So far I know that for an equation to be functional, any x value cannot be repeated. You cannot have two x's. Y can repeat however much it wants to but despite that not being able to determine which are functional.
|
A function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output.
So, according definition of function, given relations with domain $A = \{0, 2, 4, 6\}$,
$(a) \{(6, 3), (2, 1), (0, 3), (4, 5)\} = \{ (0, 3),(2, 1), (4, 5),(6, 3)\}$ is function , since it defines for every element of function $(i.e.\{0, 2, 4, 6\})$ and has exactly one output for each element.
$(b) \{(2, 3), (4, 7), (0, 1), (6, 5)\} = \{(0, 1),(2, 3), (4, 7), (6, 5)\}$ is function , since it defines for every element of function $(i.e.\{0, 2, 4, 6\})$ and has exactly one output for each element.
$(c) \{(2, 1), (4, 5), (6, 3)\}$ is not a function, since it defines on element $0$ of element of function $(i.e.\{0, 2, 4, 6\})$.
$(d) \{(6, 1), (0, 3), (4, 1), (0, 7), (2, 5)\} = \{(0, 3),(0, 7),(2, 5),(4, 1),(6, 1) \}$ is not a function, since it not has exactly one output of element of function $(i.e.\{0, 2, 4, 6\})$. Element $0$ have two output $3$ and $7$.
Reference@Wikipedia.
|
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|
Homework: Sum of the cubed roots of polynomial Given $7X^4-14X^3-7X+2 = f\in R[X]$, find the sum of the cubed roots.
Let $x_1, x_2, x_3, x_4\in R$ be the roots. Then the polynomial $X^4-2X^3-X+ 2/7$ would have the same roots. If we write the polynomial as $X^4 + a_1X^3 + a_2X^2 +a_3X + a_4$ then per Viete's theorem:
$a_k = (-1)^k\sigma _k(x_1,x_2,x_3,x_4), k\in \{1,2,3,4\}$, where $\sigma _k$ is the $k$-th elementary symmetrical polynomial. Therefore:
$x_1+x_2+x_3+x_4 = 2$
$x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4 = 0\ (*)$
$x_1x_2x_3 +x_1x_2x_4+x_1x_3x_4+x_2x_3x_4 = 1$
$x_1x_2x_3x_4 = 2/7$
Now how to determine the sum of the cubed roots?
$2^3 = 8= (x_1+x_2+x_3+x_4)(x_1+x_2+x_3+x_4)^2 = (x_1+x_2+x_3+x_4)(x_1^2+x_2^2+x_3^2+x_4^2 + 2(*))$
Here's where things go out of hand:
$(x_1+x_2+x_3+x_4)(x_1^2+x_2^2+x_3^2+x_4^2) = (x_1^3 + x_2^3 + x_3^3+x_4^3) + x_1^2(x_2+x_3+x_4)+x_2^2(x_1+x_3+x_4)+x_3^2(x_1+x_2+x_4)+x_4^2(x_1+x_2+x_3) = 8$
What should I do here?
|
Hint: as we can have only cubic terms in the symmetric polynomial sums, the only terms which can be used are of form $(\sum x)^3, \sum x \sum xy$ and $\sum xyz$. Then it is a matter of testing $3$ coefficients...
$$\sum_{cyc} x_1^3 = \left(\sum_{cyc} x_1 \right)^3-3\left(\sum_{cyc} x_1 \right)\left(\sum_{cyc} x_1 x_2 \right)+3\sum_{cyc} x_1x_2x_3$$
|
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|
Laplace transform to bio heat equation This is the bio heat equation and I have several questions when trying to work with it.
$$
\rho c \frac{\partial u(x,t)}{\partial t}
= \nabla[k \nabla u(x,t)]
+ \omega_b \rho_b c_b [u_a - u(x,t)]
+ Q_m + Q_r(x,t)
$$
where the first expression on the right hand side describes the conduction of heat induced by temperature gradient.
Then by letting $k$ to be a constant through out the process this equation has been written as
$$
\rho c \frac{\partial u(\mathbf{x},t)}{\partial t}
= k \nabla^2 u(\mathbf{x},t)
+\omega_b \rho_b c_b[ u_a - u(\mathbf{x},t) ]
+ Q_m + Q_r(\mathbf{x},t).
$$
*
*Here does $\nabla[k \nabla u(x,t)]$ become $k \nabla^2 u(\mathbf{x},t)$ due to this constant $k$?
*When taking the Lapalace transform what is the Laplace transform of the term $k \nabla^2 u(\mathbf{x},t)$?
*The bio heat equation at the top of the post is for a 2-D case. That is $\mathbf{x} = (x_1,x_2)$. But if this equation is written for a 1-D case at steady state temperature does the bio heat equation become
$$
pc \frac{\partial u(x)}{\partial t}
= \frac{\partial^2 u(x)}{\partial^2 x}
+ \omega_b p_b c_b[ u_a - u(x) ]
+Q(m) + Q(x)?
$$
|
*
*I think yes if $$\nabla[k \nabla u(\underline x,t)] = \nabla \cdot [k \nabla u(\underline x,t)]:$$
By one of the product rules (),
$$\nabla[k \nabla u(\underline x,t)]$$
$$= \nabla k \nabla u(\underline x,t) + k \nabla \nabla u(\underline x,t)]$$
$$= (0) \nabla u(\underline x,t) + k \nabla \nabla u(\underline x,t)]$$
$$= (0) \nabla u(\underline x,t) + k \nabla \nabla u(\underline x,t)]$$
$$= k \nabla \nabla u(\underline x,t)]$$
But what does $$\nabla \nabla u(\underline x,t)$$
even mean?
By definition, $\nabla^2 u(\underline x,t) = \nabla \cdot \nabla u(\underline x,t)$.
Also for 'well-behaved' u(\underline x,t), we have:
$$\nabla \times \nabla u(\underline x,t) = 0$$
*Try (10) and (11) here.
*Steady state temperature means $t = 0$?
If so, I think so except I think the last terms should be: $Q_m + Q_r(x)$ based on (1) and (4) here. In the link, it seems that $Q_r(x) = 0$
|
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|
How to show that $a_{1}=a_{3}=a_{5}=\cdots=a_{199}=0$ Let $$(1-x+x^2-x^3+\cdots-x^{99}+x^{100})(1+x+x^2+\cdots+x^{100})=a_{0}+a_{1}x+\cdots+a_{200}x^{200}$$
show that
$$a_{1}=a_{3}=a_{5}=\cdots=a_{199}=0$$
I have one methods to solve this problem:
Let$$g(x)=(1-x+x^2-x^3+\cdots-x^{99}+x^{100})(1+x+x^2+\cdots+x^{100})$$
Note $$g(x)=g(-x)$$
so $$a_{1}=a_{3}=\cdots=a_{199}=0$$
there exist other methods?
|
Let $A (x) = (1-x+x^2-x^3+\cdots-x^{99}+x^{100})$ and $B (x) = 1+x+x^2+\cdots+x^{100}$. Then we have $g(x) = A (x) B (x)$. We have $(x + 1) A (x) = x^{101} + 1$ and $(x - 1) B (x) = x ^ {101} - 1$. Then $$g (x) = \frac {x^ {202} - 1} {x^2 - 1} = x^{200} + x^{198} + \cdots + 1,$$ as desired.
|
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|
Complicated sum with binomial coefficients I know how to prove, that $\frac{1}{2^{n}}\cdot\sum\limits_{k=0}^nC_n^k \cdot
\sqrt{1+2^{2n}v^{2k}(1-v)^{2(n-k)}}$ tends to 2 if n tends to infinity for $v\in (0,\, 1),\ v\neq 1/2$. This can be proved with the use of Dynamical systems reasonings, which are non-trivial and complicated. Is there quite good combinatiorial proof of this fact? How would you solve this problem as it is? Here $C_n^k$ denotes $\frac{n!}{k!(n-k)!}$.
|
You can rewrite the expression as
$$
\sum_{k=0}^n \sqrt{P_{1/2,n}(k)^2 + P_{v,n}(k)^2},
$$
where $P_{a,n} = {n\choose k} a^k (1-a)^k = P\big(\mathrm{Binom}(n,a) = k\big)$ is the binomial distribution with parameters $n$ and $a\in(0,1)$.
Now the statement is pretty clear, since by the law of large numbers the distributions $P_{1/2,n}$ and $P_{v,n}$ "sit" on disjoint sets: $P_{1/2,n}$ concentrates around $n/2$, $P_{v,n}$, around $nv$. Therefore, $\sqrt{P_{1/2,n}(k)^2 + P_{v,n}(k)^2}\approx P_{1/2,n} $ for $k\approx n/2$, and $\sqrt{P_{1/2,n}(k)^2 + P_{v,n}(k)^2}\approx P_{v,n} $ for $k\approx nv$, so the whole sum is $\approx \sum_{k\approx n/2} P_{1/2,n} + \sum_{k\approx nv} P_{v,n} \approx 1+ 1 = 2$.
Formally, assuming wlog that $v<1/2$ and taking any $b\in (v,1/2)$,
$$
2\ge \sum_{k=0}^n \sqrt{P_{1/2,n}(k)^2 + P_{v,n}(k)^2} \ge \sum_{k\le nb} P_{v,n}(k) + \sum_{k > nb} P_{1/2,n}(k) \\
= 2 - \sum_{k> nb} P_{v,n}(k) - \sum_{k\le nb} P_{1/2,n}(k) = 2 - P\big(\mathrm{Binom}(n,v) \ge nb\big) - P\big(\mathrm{Binom}(n,1/2) < nb\big) \\
\ge 2 - P\big(|\mathrm{Binom}(n,v)-nv| \ge n(b-v)\big) - P\big(|\mathrm{Binom}(n,1/2)-n/2| \ge n(1/2-b)\big)
\\\ge 2 - \frac{v(1-v)}{n(b-v)^2} - \frac{1}{4n(1/2-b)^2},
$$
where the last is by Chebyshev's inequality. Hence the result follows.
Note that by using some stronger result (Chernoff, Hoeffding, Bernstein) it is easy to show that the convergence is exponentially fast.
|
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|
Solve $x-\lfloor x\rfloor= \frac{2}{\frac{1}{x} + \frac{1}{\lfloor x\rfloor}}$ Could anyone advise me how to solve the following problem:
Find all $x \in \mathbb{R}$ such that $x-\lfloor x\rfloor= \dfrac{2}{\dfrac{1}{x} + \dfrac{1}{\lfloor x\rfloor}},$ where $\lfloor *\rfloor$ denotes the greatest integer function.
Here is my attempt:
Clearly, $x \not \in \mathbb{Z}.$
$\dfrac{x}{\lfloor x \rfloor} - \dfrac{\lfloor x \rfloor}{x}= 2$
$ \implies x^2 -2x\lfloor x \rfloor - \lfloor x \rfloor ^2 = 0$
$\implies x = (1 \pm \sqrt2 )\lfloor x \rfloor $
$\implies \{x\} = \pm\sqrt2 \lfloor x \rfloor$
Thank you.
|
Completing the square gives
$$
x^2-2x\lfloor x\rfloor+\lfloor x\rfloor^2=2\lfloor x\rfloor^2
$$
so
$$
(x-\lfloor x\rfloor)^2=2\lfloor x\rfloor^2
$$
Since $0\le x-\lfloor x\rfloor<1$, we conclude $\lfloor x\rfloor=0$ that's disallowed by the starting equation.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Convergence of a series with binomial coefficient Let $S_n$ be a series with binomial coefficients as follows:
$$ S_n = \sum_{k=1}^n \begin{pmatrix} n \\ k \end{pmatrix}\frac 1k\left(-\frac{1}{1-a} \right)^k\left(\frac{a}{1-a} \right)^{n-k},$$
where $0 < a < 1$. My question is: when $n\to +\infty$, does the series $S_n$ converge? And If it converges, can we find the limit $ \underset{n\to +\infty}{lim} S_n$ ? Thank you.
|
We have:
$$\begin{eqnarray*} S_n = \frac{1}{(1-a)^n}\sum_{k=1}^{n}\binom{n}{k}\frac{(-1)^k}{k}a^{n-k}&=&\frac{1}{(1-a)^n}\sum_{k=1}^{n}\binom{n}{k}a^{n-k}\int_{0}^{-1}x^{k-1}\,dx\\&=&\frac{1}{(1-a)^n}\int_{0}^{-1}\frac{dx}{x}\sum_{k=1}^{n}\binom{n}{k}x^k a^{n-k}\,dx\\&=&\frac{1}{(1-a)^n}\int_{0}^{-1}\frac{(a+x)^n-a^n}{x}\,dx\\&=&\left(\frac{a}{1-a}\right)^n \int_{0}^{-1/a}\frac{(1+x)^n-1}{x}\,dx\end{eqnarray*} $$
and by approximating the last integrand function with $n\cdot \exp\left(\frac{n-1}{2}x\right)$ (Laplace's method) we have that the limit is zero if $a<\frac{1}{2}$, $-2$ if $a=\frac{1}{2}$ and $-\infty$ if $a>\frac{1}{2}$.
|
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|
Determining whether a transformation is linear I was hoping someone would be able to explain the following to me as I do not understand it:
f is the reflection in the line x = 1.
The reflection f maps the origin to point (2, 0) so it does not fix the origin. Therefore f is not a linear transformation.
Would someone be able to explain how f maps the origin to the point (2, 0) and why it is not linear.
Thanks so much.
|
Here is a visualization:
How does the reflection transformation map the origin?
$f(P) = P'$ such that the line segment $PP'$ is halved by the mirror line.
The shortest distance to the mirror line $x = 1$ is the same for point $P$ and mirrored point $P'$.
Why is this transformation not linear?
Linear transformations have the properties
$$
T(x + y) = T(x) + T(y) \quad (x, y \in V) \\
T(\alpha x) = \alpha T(x) \quad (x \in V, \alpha \in F)
$$
for some vector space $V$ and field $F$.
If we use the second with $x = 0 = (0,0) \in \mathbb{R}^2$ and $\alpha = 2 \in \mathbb{R}$ we get
$$
T(2 \cdot 0) = 2\cdot T(0) \iff \\
T(0) = 2 \cdot T(0) \iff \\
0 = T(0)
$$
So a linear map has to map the origin $0$ to itself.
|
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|
Probability of symmetric difference inequality I think I am missing something easy here, but my book notes that
$$P\left\{\bigcup_{j=1}^\infty A_j \mathbin\triangle \bigcup_{j=1}^\infty B_j\right\} \leq \sum_{j=1}^\infty P\{A_j \mathbin\triangle B_j\}$$
where $A\mathbin\triangle B$ is the symmetric difference of the two sets $A,B$. It seems some what intuitive as the sets may overlap and the right hand side will be greater in that case, no? I have tried to write the symmetric difference of the unions as disjoint subsets of the union of symmetric differences, and then use additivity and $P(A) \leq P(B)$ for $A\subset B$ (despite this fact being proven after the original note, not that I think it depends on this) but did not succeed.
Any suggestions or hints would be helpful, thank you.
|
Well, let's look at the base case.
$$\begin{align}(A_1\cup A_2)\triangle(B_1\cup B_2)
& = ((A_1\cup A_2)\cap B_1^c \cap B_2^c)\cup(A_1^c\cap A_2^c\cap(B_1\cup B_2))
\\[1ex] & = (A_1\cap B_1^c\cap B_2^c)\cup(A_1^c\cap A_2^c\cap B_2)\cup (A_2\cap B_1^c\cap B_2^c)\cup (A_1^c\cap A_2^c\cap B_1)
\\[1ex] & \subseteq (A_1\cap B_1^c)\cup(A_1^c\cap B_1)\cup(A_2\cap B_2^c)\cup(A_2^c\cap B_2)
\\[2ex]\therefore (A_1\cup A_2)\triangle(B_1\cup B_2) & \subseteq (A_1\triangle B_1)\cup(A_2\triangle B_2)
\end{align}$$
Now let look at the inductive step.
$$\begin{align} \left(\bigcup_{j=1}^n A_j\;\cup A_{n+1}\right)\triangle\left(\bigcup_{j=1}^n B_j\;\cup B_{n+1}\right) & \subseteq \left(\bigcup_{j=1}^n A_j \triangle \bigcup_{j=1}^n B_j\right)\cup(A_{k+1}\triangle B_{k+1})
\end{align}$$
And we can clearly see where that is going. Thus we argue that because it is true for two, and inductive step also holds, therefore it shall be true for more, ad infinitum.
Then we apply measure theory to this result and reach the requisite conclusion.
$\Box$
|
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|
Are there any limit questions which are easier to solve using methods other than l'Hopital's Rule? Are there any limit questions which are easier to solve using methods other than l'Hopital's Rule? It seems like for every limit that results in an indeterminate form, you may as well use lHopital's Rule rather than any of the methods that are taught prior to the rule, such as factoring, rationalizing, trig limits, etc.
EDIT: These are great answers, thanks!
|
If you try to use L'Hospital's rule to evaluate
$$ \lim_{x\to\infty} \frac{2x}{x+\sin x} $$
you end up with
$$ \lim_{x\to\infty} \frac{2}{1+\cos x} $$
which spectacularly fails to converge. But the original limit does exist (it is $2$).
|
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|
Coordinates of a Change of Basis Let $\{x_1, x_2, x_3\}$ and $\{y_1, y_2, y_3\}$ be basis of of $\mathbb{R}^3$ with a basis transformation:
$$
y_1 = 2x_1 - x_2 - x_3
$$
$$
y_2 = -x_2
$$
$$
y_3 = 2x_2 + x_3
$$
What are all the vectors that have the same coordinates with respect to the two bases?
|
All you really want to do it find the values of $x_1,x_2,x_3$ such that you get back $x_1,x_2,x_3$ when you change the basis. So $y_1=x_2$, $y_2=x_2$ and $y_3=x_3$ which you then put into your system of equations:
$$ x_1=2x_1-x_2-x_3\\ x_2=-x_2 \\ x_3=2x_2+x_3 $$
So from that we can see $x_2=0$ so then we are just solving:
$$ x_1=2x_1-x_3\\ x_3=x_3 $$
which simplifies to $x_1=x_3$ so we can represent the solution as any vector in the form $(t,0,t)$ for all $t\in \mathbb{R}$. Which you can quickly check by putting in your system of equations is invariant under the base change.
|
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|
How to find the preimage of a set I'm trying to prove continuity of two different functions $f\colon A\to B$. I know that to find continuity, the inverse of all open subsets in $B$ must be open in $A$. I also know that to find the inverse of an open subset you have to find the preimage. I just don't know how to find the preimage. How do you find the preimage?
For example, if the set is the discrete topology of $\{1,2,3\}$, how would you find the preimage of $\{1\}$, $\{2\}$, etc. The same for the indiscrete topology. How would you find the preimage of the null set and the whole set? Thank you!
|
Let $f:A\to B$ be a map and let $S$ be a subset of $B$, then the preimage of $S$ under $f$ is
$$f^{-1}(S)=\{a\,|\,f(a)\in S\}.$$
For example if $A=\{1,2,3\}$, $B=\{a,b,c\}$ and $f$ is map defined by $f(1)=f(2)=a$, $f(3)=b$, then
$$f^{-1}(\emptyset)=\emptyset,\\
f^{-1}(\{a\})=\{1,2\},\
f^{-1}(\{b\})=\{3\},\
f^{-1}(\{c\})=\emptyset,\\
f^{-1}(\{a,b\})=\{1,2,3\},\
f^{-1}(\{a,c\})=\{1,2\},\
f^{-1}(\{b,c\})=\{3\},\\
f^{-1}(\{a,b,c\})=\{1,2,3\}.$$
|
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|
Energy method for one dimensional wave equation with Robin boundary condition Show that the initial-boundary value problem
\begin{align}
& {{u}_{tt}}={{u}_{xx}}\text{ }(x,t)\in \left( 0,l \right)\times \left( 0,T \right),\text{ }T,l>0 \\
& u\left( x,0 \right)=0,\text{ }x\in \left[ 0,l \right] \\
& {{u}_{x}}\left( 0,t \right)-u\left( 0,t \right)=0,\text{ }{{u}_{x}}\left( l,t \right)+u\left( l,t \right)=0,\text{ }t\in \left[ 0,T \right]\\
\end{align}
has zero solution only.
My attempt 2:
Previously I tried separation by variables but got stuck at the end. Inspired by BCLC, I try energy method this time.
Set
$$E\left( t \right)=\frac{1}{2}\int_{0}^{L}{\left( u_{x}^{2}\left( x,t \right)+u_{t}^{2}\left( x,t \right) \right)dx}.$$
The equation ${{u}_{tt}}={{u}_{xx}}$ and the Robin b.c. gives
$$\begin{align}
& \frac{dE}{dt}=\int_{0}^{L}{\left( {{u}_{x}}{{u}_{xt}}+{{u}_{t}}{{u}_{tt}} \right)dx} \\
& \text{ }=\int_{0}^{L}{\left( -{{u}_{t}}{{u}_{xx}}+{{u}_{t}}{{u}_{tt}} \right)dx}+\left. {{u}_{t}}{{u}_{x}} \right|_{0}^{L} \\
& \text{ }={{u}_{t}}\left( l,t \right){{u}_{x}}\left( l,t \right)-{{u}_{t}}\left( 0,t \right){{u}_{x}}\left( 0,t \right) \\
& \text{ }=-{{u}_{t}}\left( l,t \right)u\left( l,t \right)-{{u}_{t}}\left( 0,t \right)u\left( 0,t \right)\le 0\text{ }\left( ? \right) \\
\end{align}$$
Therefore, $E\left( t \right)\le E\left( 0 \right)$ for all $t\ge 0$. Since $E\left( t \right)\ge 0$ and , we obtain $E\left( 0 \right)=0 (?)$ for all $t\ge 0$, thus $E\equiv 0$ and hence $u\equiv 0$ .
Is the proof correct?
|
$\newcommand{\e}{\epsilon}$The first $(?)$ is indeed questionable. Instead, note that you have shown that
$$E'(t) = -u_{t}(l, t)u(l, t) -u_{t}(0, t)u(0, t). \tag{$1$}$$
Defining $\e : [0, \infty) \to \Bbb R$ by $$\e(t) := \frac{1}{2}[u^{2}(l, t) + u^{2}(0, t)],$$
we see that $(1)$ tells us that $(E + \e)' \equiv 0$.
Moreover, $E(0) = \e(0) = 0$. Thus, $E + \e$ is identically equal to $0$. Since $\e$ and $E$ are both nonnegative functions, it follows that $E \equiv 0$. Now, you can conclude $u \equiv 0$ as before.
In the above, I'm assuming that we are also given the data $u_t(x, 0) = 0$ for $x \in [0, l]$. That is how I'm concluding $E(0) = 0$.
|
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|
solving the limit of $(e^{2x}+x)^{1/x}$ I tried to solve for the following limit:
$$\lim_{x\rightarrow \infty} (e^{2x}+x)^{1/x}$$
and I reached to the indeterminate form:
$${4e^{2x}}\over {4e^{2x}}$$
if I plug in, I will get another indeterminate form!
|
I have no idea how you got $\lim\limits_{x\to\infty}\frac{4e^{2x}}{4e^{2x}}$, but as has been pointed out, that limit is easily evaluated: the fraction is identically $1$, so the limit is also $1$.
Let $L=\lim\limits_{x\to\infty}\left(e^{2x}+x\right)^{1/x}$; then
$$\ln L=\ln\lim_{x\to\infty}\left(e^{2x}+x\right)^{1/x}=\lim_{x\to\infty}\ln\left(e^{2x}+x\right)^{1/x}=\lim_{x\to\infty}\frac{\ln\left(e^{2x}+x\right)}x\;.$$
Now apply l’Hospital’s rule.
|
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|
Is it correct to think of the Laplacian as the divergence of a gradient field? Factoring out the notation, I see that
$$\nabla^2(\phi) = \nabla \cdot \nabla(\phi) = \nabla \cdot (\nabla(\phi)) $$
which looks something like the divergence of the gradient of phi.
Is it actually true?
Thanks,
|
Yes, it is correct. Indeed, let $\phi \colon \mathbb{R}^N \to \mathbb{R}$ be of class $C^2$, then
\begin{align}
\operatorname{div}(\nabla\phi) := \sum_{i = 1}^N\frac{\partial}{\partial x_i}(\nabla\phi)_i = \sum_{i = 1}^N\frac{\partial^2 \phi}{\partial x_i^2} =: \Delta\phi.
\end{align}
|
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|
Who determines if a mathematical proof is valid? I'm studying mathematics and as you all know the most important things in mathematics are proofs.
My question is, who determines if a proof that someone invents in mathematics is valid? Is there some mathematics professors who check all people's proofs in the world?
If I invented a new proof, where do I send it to? Can anyone invent their own mathematical proofs?
|
That's a very interesting question. Part of the answer is already included in the question itself - for it is not evident at all that the question should not start with the word "what".
As was pointed out -- correctly I believe -- a determination of the validity of a mathematical proof is a social process. This may come somewhat as a surprise -- especially for high priests of mathematics who believe with all their heart that mathematics equals truth.
The mathematical community is first and foremost a community. It has its institutions, governments, ambassadors, pundits, enthusiasts, cults and rebellious underground movements. The current state of affairs was concisely summarized by Clement Guarin in the comments above. Observe how structured is the social process -- first the proof inventors have to believe in their proof. Then they must present it in front of other people. These other people can be members of their faculty, or readers on a site like this one. Then they ought to present it yet in front of still other people -- presumably holding more distinguished posts than the first ones -- these could be journal editors or otherwise distinguished persona in widely accepted forums, such as some mathematicians who post on the real-mathematics sites. The purpose of this all is to expose the proof to as many eyes as possible, in the hope that if there were some mistake somewhere, it surely would have been found. But this has not always been the case. See the story of the Busemann-Petty problem.
|
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|
Evaluate $\lim_\limits{x \to 0}\frac{x}{\sqrt[n]{1+ax} \cdot \sqrt[k]{1+bx} -1}$ For all $n,k \in N a,b > 0$
$$\lim_{x \to 0}\frac{x}{\sqrt[n]{1+ax} \cdot \sqrt[k]{1+bx} -1} = \lim_{x \to 0}\frac{x}{(1+ \frac{ax}{n})(1+ \frac{bx}{k})- 1}= \lim_{x \to 0}\frac{x}{x(\frac{a}{n} + \frac{b}{k}) + \frac{ab}{nk}x^2} = \lim_{x \to 0}\frac{1}{\frac{a}{n} + \frac{b}{k}} = \frac{nk}{ka+nb}$$
Am I right?
|
The limit is fine. Perhaps with a view to formalizing the procedure is appropriate to introduce an infinite series , and then Landau notation.
$$O(g(x)) = \left\{\begin{matrix} f(x) : \forall x\ge x_0 >0 , 0\le |f(x)|\le c|g(x)| \end{matrix}\right\}$$
We will also use:
$$f_1=O(g_1)\wedge f_2=O(g_2)\implies f_1f_2=O(g_1g_2)\,$$
Recalling, from MacLaurin series:
$${(1+ax)^{1/n}}={\sum_{i=0}^{\infty} \binom{1/n}{i} (ax)^i }={ (1+\dfrac{ax}{n}+O(x^2))}\\~\\
{(1+bx)^{1/k}}={( 1+\dfrac{bx}{k}+O(x^2))}
$$
Now, we have:
$${(1+ax)^{1/n}}{(1+bx)^{1/k}}={ (1+\dfrac{ax}{n}+O(x^2))}{( 1+\dfrac{bx}{k}+O(x^2))}={(1+\dfrac{ax}{n} + \dfrac{bx}{k}+O(x^2))}
$$
And finally
$$
\lim_{x \to 0}\frac{x}{\sqrt[n]{1+ax} \cdot \sqrt[k]{1+bx} -1} = \lim_{x \to 0}\frac{x}{{(1+\dfrac{ax}{n} + \dfrac{bx}{k}+O(x^2))}-1}\\~\\= \lim_{x \to 0}\frac{(x^{-1})x}{{(x^{-1})(\dfrac{ax}{n} + \dfrac{bx}{k}+O(x^2))}} = \lim_{x \to 0}\frac{1}{\frac{a}{n} + \frac{b}{k} +O(x)} = \frac{1}{\frac{a}{n} + \frac{b}{k}}= \frac{nk}{ak+nb}
$$
|
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|
Show that $\int_{0}^{+\infty} \frac{\sin(x)}{x(x^2+1)} dx = \frac{\pi}{2}\left(1-\frac{1}{e}\right) $ I'm trying to show that
$$\int_{0}^{+\infty} \frac{\sin(x)}{x(x^2+1)} dx = \frac{\pi}{2}\left(1-\frac{1}{e}\right) $$
using Jordan's lemma and contour integration.
MY ATTEMPT: The function in the integrand is even, so I have:
$$\int_{0}^{+\infty} \frac{\sin(x)}{x(x^2+1)} dx =\frac{1}{2}\int_{-\infty}^{+\infty} \frac{\sin(x)}{x(x^2+1)} dx$$
There is a simple pole at $z=0$ and poles at $z=+i, z=-i$.
A method in the chapter I am working on (Ablowitz & Fokas sections 4.2 & 4.3) usually considers the integral
$$\int_{-\infty}^{+\infty} \frac{e^{ix}}{x(x^2+1)} dx=2\pi iRes\left(\frac{e^{ix}}{x(x^2+1)},z=i,-i,0\right)$$ Which when I compute results in $\dfrac{-(-1+e)^2\pi}{2e}$, which is close but not quite the answer. (Notice that factored in another way the answer is also equal to $\dfrac{(-1+e)\pi}{2e}$.
But I am not sure if this will work, instead another example builds a contour $C_r+C_e+(-R,-e)+(R,e)$ which avoids the poles and thus integrating over that yields zero and helps me get my answer. Unfortunately, this attempt does not give me the right value either.
Do any of you integration whizzes out there have anything for me? Many thanks.
|
Worked this out, it's a combination of both methods I mention. You need to go around the contour $C_r+C_e+(-R,-e)+(R,e)$. By the only theorem on section 4.3 of Ablowitz & Fokas the integral around $C_e$ will go to $i\pi$. By the approach I mentioned in the first part of my attempt, the integral around $C_R$ will go to $2\pi i(-\frac{1}{2e})$ adding these two solutions up, and remembering to divide by $2$ (notice the integrand is even) yields the answer.
|
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|
Why does the principle value of $z^{\frac{1}{2}} := \sqrt{r}e^{i\frac{\theta}{2}}$ need to be restricted to one rotation in order to be continuous? I understand that the function $f(z) = z^{\frac{1}{2}}$ is multi-valued, since it has exactly two solutions for every $z$. However if I take the principle value square root function, expressed as
$$f(r,\theta) = \sqrt{r}e^{i\frac{\theta}{2}}$$
Then why does a branch cut need to be taken here? From what I've read, if I want this function to be continuous then I should restrict $\theta$. But as far as I understand, $e^z$ is single-valued for any complex number $z$.
|
$f(r,\theta)$ is continuous, but the map $z\mapsto(r,\theta)$ is multi-valued, due to $Arg(z)$ is multivalued. If you want $z\mapsto(r,\theta)$ to be continuous, you have to select a single valued branch and restrict the range of $\theta$.
|
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Hahn-Banach separation theorem with a countable subset of functionals For a separable Banach space $X$, the unit sphere of $X^*$ always contains a countable set $D$ such that
$$ \left\Vert x \right\Vert = \sup_{f \in D} \left\vert f(x) \right\vert \qquad \mbox{ for every }x\in X.$$
Assume that $A\subset X$ is nonempty, bounded, convex, and $\inf_{a\in A}\left\Vert a \right\Vert > 1$. Then, by Hahn-Banach separation theorem, we can find $f\in X^*$, $\left\Vert f \right\Vert = 1$, such that $f(a) > 1$ for all $a\in A$.
Is it possible to find also $f\in D$ such that $f(a)>1$ (or $f(a)>1/2$) for all $a\in A$?
|
Not in general. Here is a counterexample.
Let $X = C([-1,1])$. For $t \in [-1,1]$, let $\delta_t$ denote the point mass / evaluation functional $\delta_t(x) = x(t)$. Let $D = \{ \delta_q : q \in [-1,1] \cap \mathbb{Q}\}$. Then $D$ is countable and we have $\|x\| = \sup_{f \in D} |f(x)|$ for every $x \in X$.
Let $$y(t) = \begin{cases} 4t, & -1 \le t \le 0 \\ 0, & 0 \le t \le 1 \end{cases}$$
and
$$z(t) = -y(-t) = \begin{cases} 0, & -1 \le t \le 0 \\ 4t, & 0 \le t \le 1 \end{cases}$$
Let $A = \{s y + (1-s)z : 0 \le s \le 1\}$ be the line segment connecting $y$ and $z$. $A$ is nonempty, convex and bounded (even compact). And for any $a \in A$ we have either $a(-1) \le -2$ or $a(1) \ge 2$, so $\inf_{a \in A} \|a\| = 2 > 1$. But for any $f = \delta_q \in D$, if $-1 \le q \le 0$ we have $f(z)=0$ and if $0 \le q \le 1$ we have $f(y) = 0$. So in all cases there exists $a \in A$ with $f(a) = 0$.
|
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|
Prove $\lim_\limits{n \to \infty}-\cos(\pi\sqrt{4n^2+10})$ exists and is equal to $-1$ My idea is to take two subsequence converging to different limits. On this step I have problem. How can I set equation of subsequence when $\cos$ has certain value and $n \to \infty$?
$$
\lim_{n \to \infty}-\cos(\pi\sqrt{4n^2+10})
$$
If I say that $\frac{5}{4n^2} \to 0$ than $n \to \infty$, I have $\lim_{n \to \infty}-\cos(2\pi n) = -1$. Is this right?
|
HINT:
The cosine is continuous and $$\pi\sqrt{4n^2+10}= 2\pi n\left(1+\frac5{4n^2}+O\left(\frac{1}{n^4}\right)\right)$$
|
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|
Problem on entire function which is reduced to constant Suppose $f$ is an entire function satisfying any one of the following conditions for all $z\in \mathbb C$
(1) im$f(z)$ has no zeros
(2)$|f(z)|\geq 1$.
Then f is constant.
My thought: For(2) since $|f(z)|\geq 1$ ,then $f$ has no zero in $\mathbb C$.
Define $g=1/f$, then $g$ is bounded entire function implies $g$ is constant implies $f$ is constant, am I right ? I have no idea about (1), please give some hints. Thanks.
|
Hints: 1) Show that ${\rm Im}(f(z))>0$ for all $z$, or ${\rm Im}(f(z))<0$ for all $z$; (if ${\rm Im}(f(z_1))>0$ and ${\rm Im}(f(z_2))<0$, put $g(t)={\rm Im}(f((1-t)z_1+tz_2))$ for $t\in [0,1]$; $g$ is clearly continuous.)
2) If you are in the first case, and if $h(z)=f(z)+i$, what can you say of $|h(z)|$ ?
|
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|
The set of traces of orthogonal matrices is compact
Is the following set compact: $$M = \{ \operatorname{Tr}(A) : A \in M(n,\mathbb R) \text{ is orthogonal}\}$$ where $\operatorname{Tr}(A) $ denotes the trace of $A$?
In order to be compact $M$ has to be closed and bounded.
$\|A\|=\sqrt {\sum_{i,j} {a_{ij}}^2}=\sqrt n $ and hence bounded. So $\operatorname{Tr}(A)<\sqrt n $. Hence $M$ is bounded.
Now we have to prove that $M$ is closed. Let $\operatorname{Tr}(A_n)$ be a sequence of matrices converging to $\operatorname{Tr}(A)$ where $A_n$ is a sequence of orthogonal matrices. The only thing remaining to show is that $A$ is orthogonal.
|
Hint: To show that $A$ is orthogonal, consider $\lim_{n \to \infty} \left\| A_n^TA_n - I \right\|$, noting that the function
$$
f(X) = \left\| X^TX - I \right\|
$$
is continuous.
|
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|
What does it mean to represent elements of an ideal? Say I have the polynomial $x^9 + 1$
Then: $x^9 + 1 = (x+1)(x^2 + x + 1)(x^6 + x^3 + 1)$
is a complete factorization over $GF(2)$ of $x^9 + 1$
The dimension of each ideal is: length $n - deg(ideal)$
So for $n=9$, dimension of $(x+1)$ = $9-1=8$
of $(x^2 + x + 1) = 9 - 2 = 7$
of $(x^6 + x^3 + 1) = 9 - 6 = 3$
So let's use $(x^6 + x^3 + 1)$ as the example. The dimension is 3. So there should be $2^3 = 8$ elements. How do I find those elements?
|
Let’s call $f=x^6+x^3+1$. You want three linearly independent elements of the ideal $(f)$ of the ring $R=\Bbb F_2[x]/(x^9+1)$. Since $(f)$ is just the set of multiples of $f$, you certainly have $1\cdot f$, $xf$, and $x^2f$. Notice that $x^3f=x^9+x^6+x^3=1+x^6+x^3=1\cdot f$, already counted. I’ll leave it to you to show that those three polynomials are $\Bbb F_2$-linearly independent in $R$.
|
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|
Matrix exponential of non diagonalizable matrix? I am currently self-learning about matrix exponential and found that determining the matrix of a diagonalizable matrix is pretty straight forward :).
I do not, however, know how to find the exponential matrix of a non-diagonalizable matrix.
For example, consider the matrix
$$\begin{bmatrix}1 & 0 \\ 1 & 1\end{bmatrix}$$
Is there any process of finding the exponential matrix of a non-diagonalizable matrix? If so, can someone please show me an example of the process? :). I am not looking for an answer of the above mentioned matrix (since I just made it up), but rather I'm interested in the actual method of finding the matrix exponential to apply to other examples :)
|
Laplace Transforms approach is an useful for finding $e^{A t}$, whether $A$ is diagonalizable or non-diagonalizable. It is a very useful method in Electrical
Engineering.
Basically, we use the formula
$$
e^{A t} = \mathcal{L}^{-1}\left[ (s I - A)^{-1} \right]
$$
As an illustration, consider the defective matrix
$$
A = \left[ \begin{array}{cc}
1 & 0 \\
1 & 1 \\
\end{array} \right]
$$
We find
$$
s I - A = \left[ \begin{array}{cc}
s - 1 & 0 \\
-1 & s - 1 \\
\end{array} \right]
$$
where $s$ is a complex variable.
We find that
$$
(s I - A)^{-1} = \left[ \begin{array}{cc}
{1 \over s - 1} & 0 \\[2mm]
{1 \over (s - 1)^2} & {1 \over s - 1} \\[2mm]
\end{array} \right]
$$
The inverse Laplace transform of $(s I - A)^{-1}$ is the state transition matrix, which is also the matrix exponential, $e^{A t}$.
Hence, we get
$$
e^{A t} = \mathcal{L}^{-1}\left[ (s I - A)^{-1} \right]
= \left[ \begin{array}{cc}
\mathcal{L}^{-1}\left( {1 \over s - 1} \right) & \mathcal{L}^{-1}(0) \\[2mm]
\mathcal{L}^{-1}\left( {1 \over (s - 1)^2} \right) & \mathcal{L}^{-1}\left( {1 \over s - 1} \right) \\[2mm]
\end{array} \right] = \left[ \begin{array}{cc}
e^{t} & 0 \\[2mm]
t e^t & e^t \\[2mm]
\end{array} \right]
$$
|
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|
Evaluate $\lim_{x \to 0} \left(\frac{ \sin x }{x} \right)^{\frac{1}{x^2}}$ $$\lim_{x \to 0} \bigg(\frac{\sin x}{x} \bigg)^{\frac{1}{x^2}}$$
The task should be solved by using Maclaurin series so I did some kind of asymptotic simplification
$$\lim_{x \to 0} \bigg(\frac{\sin x}{x} \bigg)^{\frac{1}{x^2}} \approx \lim_{x \to 0} \bigg(\frac{x - \frac{x^3}{6}}{x} \bigg)^{\frac{1}{x^2}} \approx \lim_{x \to 0} \bigg(1 - \frac{x^2}{6} \bigg)^{\frac{1}{x^2}}$$ How can we say answer that is $e$ in the power of $-\frac{1}{6}$. I want some proving of that fact.
Thank you.
|
$$
\lim_{x \to 0} \left( 1 - \frac {x^2}6\right)^{\frac 1{x^2}} = \lim_{x \to 0} \left [ \left( 1 - \frac {x^2}6\right)^{\frac 6{x^2}} \right ]^{\frac 16} = \left [ \left (e^{-1} \right ) \right ]^{\frac 16} = e^{-\frac 16}
$$
Here, I used somewhat modified limit regarding Euler's number
$$
\lim_{t \to 0} \left( 1 - t\right )^{\frac 1t} = e^{-1}
$$
More info, and proofs can be found here.
|
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|
Can I think of the conformal mapping w = (z+1/z) as a linear fractional transformation? The mapping
$$w = z + \frac{1}{z}$$
looks linear in $z$. However, it would not be in the form
$$\frac{Az+B}{Cz+D}$$
since putting the two terms together gives
$$\frac{z^2+1}{z}$$
So my question is: is this mapping a linear fractional transformation?
I am hoping it is, so that I can determine its action on a half disk, using the nice and familiar symmetry properties of LFTs.
If it is not an LFT, are there symmetry properties of this mapping to notice?
One specific nice property of LFTs that I have in mind is that symmetric points w.r.t. to a circle, under an LFT, are again symmetric points w.r.t. the image of that circle.
Thanks,
|
No: linear fractional transformations are bijective, and this map isn't: consider $z=2$ and $z=1/2$.
You can take a look at the graph here: http://davidbau.com/conformal/#z%2B1%2Fz
There is some nice symmetry in the fact that $f(z)=f(1/z)$, so it's "symmetric about the unit circle" (up to flipping across the real axis), in some sense. There are probably other nice properties, but this is the first that comes to mind.
|
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Prove Piecewise Function Integrable $$
f(x) =
\begin{cases}
-2, & \text{if }x < 0 \\
1, & \text{if }x > 0\\
0, & \text{if }x = 0
\end{cases}
$$
Hey guys I need some help showing that this function is integrable on the closed interval $[-1,2]$.
So far my idea has been to show $$U(f,P)-L(f,P) < \epsilon$$ for some $\epsilon>0$.
The only problem is the point $(0,0)$ on the function.
I don't understand how to handle that.
Can I just say that $U(f,P)$ for some partition will equal to $3$ and then find a partition $P$ for which $$3-L(f,P)<\epsilon?$$
|
Notice that for any $x$ at which $f$ is discontinuous, $x$ can be contained in an interval as small as you please. Make sure to include this small interval in your partition.
|
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Reducing a 2nd order system of ODEs to a 1st order system I need to numerically solve the following system of ODEs:
$$x''(t)=- \frac{3x}{(x^2+y^2)^{3/2}}$$
$$y''(t)=- \frac{3y}{(x^2+y^2)^{3/2}}$$
I know that I must convert to a system of 1st order equations but I am having trouble doing so...Could someone provide me with a starting point? This problem is called the "one-body problem" apparently but I cannot find much about it as all the hits I'm getting involve the 2 body problem...Help please.
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All that needs to be done to convert to a system of first order ODEs is to play some games with your notation, in this case it's particularly simple because you don't have any first derivatives.
So let's say we choose to define
\begin{align}
x_1 &\equiv x, \\
x_2 &\equiv x_1', \\
y_1 &\equiv y, \\
y_2 &\equiv y_1'.
\end{align}
So now your system system will consist of 4, first order, coupled ODEs in your 4 variables. It will look like the following
\begin{align}
x_1' &= x_2, \\
x_2' &= \frac{-3x_1}{\left( x_1^2 + y_1^2 \right)^{3/2}}, \\
y_1' &= ..., \\
y_2' &= ... .
\end{align}
Are you now able to write down the final two equations in your system?
Now you can just numerically solve these 4 ODEs for each of your 4 variables; $x_1, x_2, y_1, y_2$. You can then relate these back to your original $x, y$ through their respective definitions.
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|
How can I derive monotonically increasing polynomial functions with steep curves from [0,1] to [0,1]? For computer graphics reasons, I need a Taylor polynomial function on the interval [0,1] like the classic smoothstep function 3x^2-2x^3 which outputs a monotonically increasing value on the interval [0, 1] but that has a much steeper curve. It should have only one inflection point (not counting the beginning or the ends.) The beginning should be convex and the end should be concave.
I would also like to be able to tweak the function to my needs and make it steeper or not so steep with some parameters if possible.
How can I derive steep polynomial functions from [0,1] to [0,1]?
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I want a polynomial, probably low-degree.
So let's try a + bx + cx^2 + dx^3 + e^4.
I want the point at zero to be zero so a must be zero.
I want the point at one to be one so c + d + e must be one.
I want the start to be convex and curve up so the first derivative at zero must be zero and hence b must be zero.
I want the end to be concave and curve down so the first derivative at one must be zero and hence 2c + 3d + 4e = 0.
Then d = 4 - 2 c and e = 1 - (c + d) and c is left free.
I am unsure of how to determine if c is two low or too high. I know that values lower than -1 don't work and values higher than around 7 don't work but I don't know why.
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|
Solve the above program Consider the problem of covering the triangle with vertices at the points
$(0, 0), (0, 1),$ and $(1, 0)$ with a ball of smallest radius.
$$\min r$$
$$s. t. \> x ^2 + y ^2 ≤ r$$
$$(x − 1)^ 2 + y ^2 ≤ r$$
$$x^ 2 + (y − 1)^ 2 ≤ r.$$
Solve the above program
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There’s no need for fancy techniques. Clearly the diameter of the disk must be at least the distance between the points $\langle 0,1\rangle$ and $\langle 1,0\rangle$, which is $\sqrt2$, so let’ see what happens if we try a disk of radius $\frac{\sqrt2}2$ centred midway between those points, at $\left\langle\frac12,\frac12\right\rangle$; does it cover $\langle 0,0\rangle$?
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|
how to solve this counting problem? Each user on a computer system has a password, which is six to eight characters long, where each character is an uppercase letter or a digit. Each password must contain at least two digits. How many possible passwords are there?
Is the answer: $(36^6 - 26^5) + (36^7 - 26^6) + (36^8 - 26^7)$?
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I would use
$$\left(36^6 - 26^6 - \binom{6}{1}\cdot 26^5\cdot 10 \right) + \left(36^7 - 26^7 - \binom{7}{1}\cdot 26^6\cdot 10 \right) + \left(36^8 - 26^8 - \binom{8}{1}\cdot 26^7\cdot 10 \right).$$
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"timestamp": "2023-03-29T00:00:00",
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|
$z_1,...z_n$ are the $n$ solutions of $z^n =a$ and $a$ is real number, show that $z_1+...+z_n$ is a real number I was actually trying this simple question, might be just me being really rusty not doing maths for a very long time. I tried adding all the $\theta$ up on all zs but it doesn't seem to work.
Anyway as the title states,
$z_1,...z_n$ are the $n$ solutions of $z^n =a$ and $a$ is real number, show that $z_1+...+z_n$ is a real number.
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Use thae fact that $$z^n -a =z^n -(z_1 +...+z_n ) z^{n-1} +...+ z_1 \cdot z_2 \cdot ...\cdot z_n $$
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|
Positive equilibria for a system of eqautions I have the following system of equations
\begin{align}
\frac{dx}{d \tau} &= x \left(1-x-\frac{y}{x+b} \right) \\
\frac{dy}{d \tau} &= cy \left(-1+a\frac{x}{x+b} \right)
\end{align}
I am asked to show that if $a<1$, the only nonnegative equilibria are $(0,0), (1,0)$.
So first it is obvious that in order to the equations become $0$ is $(x,y)=(0,0)$
Then I decided $y=0$ and $1-x-\displaystyle \frac{y}{x+b}=0 \Leftrightarrow x=1$ , hence $(x,y)=(1,0)$
In the same way I decided $x=0$ and $-1+a\displaystyle \frac{x}{x+b}=0$, but there is no $y$.
Finally I decided $-1+a\displaystyle \frac{x}{x+b}=0$ and $1-x-\displaystyle \frac{y}{x+b}=0$, and this is difficult.
I can't figure out what to do now. What about the fact that $a<1$?
Can anyone help?
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If $x \ne 0$, We have
$$ 1 - a\frac{x}{x+b} = 0 $$
$$ a\frac{x}{x+b} = 1 $$
$$ ax = x + b$$
$$ (a-1)x = b$$
$$ x = \frac{b}{a-1} $$
If $a < 1$ and $b > 0$, $x < 0$ and therefore is not a non-negative solution
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|
What is the value of $ x(\log x)$ when $x=0$ and $x\not \to 0$? I know that $ x(\log x)\to 0$ when $x \to 0$, but some people say that since $\log x$ is not defined when $x=0$, so the value of $ x(\log x)$ can also not be found when $x=0$.
But isn't it true that if you multiply anything with 0, the answer comes out to be 0? According to me the value of $x(\log x)$ at $x=0$ should be $0$.
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Since $\ln(x)$ is not defined, $x\ln(x)$ is not defined. Difficult to multiply things that do not exist.
However, since the existing definitions do not cover the case $\ln(0)$, you could arbitrarily choose a value for $\ln(0)$: $0$, $1$, $e$, $1664$ or your mother's birthday... There is no problem doing that, but there is no interest since the properties that works for $\ln(x)$, when $x$ is positive, would no longer be true if you include $0$ in the domain of $\ln$: the function will no longer be continuous on its whole domain and the theorem $\ln(xy)=\ln(x)+\ln(y)$ is not true if one of the numbers $x$ or $y$ is equal to $0$.
The same apply to $0\ln(0)$. In that case, many people (for example, people working with entropy) defines $x\ln(x)$ as $0$ when $x=0$. They find it convenient since it makes the functions $x\ln(x)$ continuous on $[0,\infty)$ and it is all that matters for them (I am maybe exagerating here).
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"timestamp": "2023-03-29T00:00:00",
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|
upper bound $\sum_{i=1}^n a_i (a_i-1)/2$ using a function of $\sum_{i=1}^n a_i$ Given $a_i \in \mathbb{N}\cup\{0\}$ and
define
$$
A(n) = \sum_{i=1}^n a_i (a_i-1)/2
$$
and
$$
B(n) = \sum_{i=1}^n a_i
$$
Any ideas how to upper bound $A(n)$ as a "function" of $B(n)$? (the tighter, the better; form of this "function" does not matter).
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Let $f(x)=\dfrac{x(x-1)}{2}$. We
note first that
$$\forall\,x,y\ge0,\quad f(x+y)-f(x)-f(y)=xy\ge0$$
Thus, by induction, for nonnegative numbers $a_1,\ldots,a_n$ we have
$$\sum_{i=1}^n f(a_i)\le f\left(\sum_{i=1}^na_i\right)$$
Thus,
$$A(n)\le \frac{B(n)(B(n)-1)}{2}$$
With equality if $a_1=\cdots=a_{n-1}=0$ for example. So, this is an optimal inequality.
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{
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|
N marbles puzzle: find the heaviest among them. Suppose there are $N$ marbles and a two-pan balance used to compare the weight of 2 things. All of the marbles weigh the same except for one, which is heavier than all of the others. How would you find the heaviest marble in minimum number of comparisons.This link explains for $N=12$ .I have tried solving for $N$ but haven't come up with any solution.How it be can generalised for $N$ marbles? Explain in detail.
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Each weighing can reduce the number of possibilities to $1/3$ of that from before, using the same argument as in the example you linked. If you have $3^n$ marbles, then you need at least $n$ weighings to find the heavier marble. If you have $3^n < N \leq 3^{n+1}$ marbles you should need $n+1$ weighings--since $N$ is not a power of three, some weighings are not as 'efficient' as possible. (i.e. don't eliminate exactly $2/3$ of possibilities since possibilities are no longer divisible by 3. In the worst case scenario, $\lceil N/3 \rceil$ possibilities remain after an 'inefficient' weighing.) Thus the number of weighings needed is
$$\lceil \log_3(N) \rceil $$
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{
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"url": "https://math.stackexchange.com/questions/1536963",
"timestamp": "2023-03-29T00:00:00",
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|
Prove Graph G is in Vizing Class 2 if $\alpha$'(G) < |e(G)|/$\Delta$(G) G is a simple graph with e edges, maximum vertex degree $\Delta$ and edge independence number $\alpha$' which satisfies $\alpha$' < e/$\Delta$.
What does this inequality mean? How is it helpful in proving that G is Vizing Class 2?
I'm struggling to understand what property of G this inequality implies. This inequality obviously works for graphs known to be in Vizing Class 2, like odd cycle graphs, but I'm not sure if I should be reasoning backwards like that.
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I discovered the answer was simpler than I thought.
$\alpha$' < e/$\Delta$ --> $\Delta$ < e/$\alpha$'
$\chi$' >= e/$\alpha$'
$\chi$' > $\Delta$
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"timestamp": "2023-03-29T00:00:00",
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|
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