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Finding the average value of a cubic function
Let $p(x)=ax^{3}+bx^{2}+cx+d$. There exists real numbers $r$ and $s$(independent of $a,b,c$ and $d$) $0<r<s<1$. For which the average value of $p(x)$ on the interval $[0,1]$ is equal to the average value of $p(r)$ and $p(s)$. Find the product of $rs$ expressed as a fraction.
What I know is $$f_{\text{avg}}=\frac{1}{b-a}\int\limits_a^bf(x)dx$$
$$f_{\text{avg}}=\int\limits_0^1f(x)dx$$
From that I am having trouble finding $r$ or $s$. I would think they would be the same but they clearly can't be. I know how to plug in the right values and take the integral but after that unsure. I'm having trouble understanding the question or what to try. Thanks for any help.
|
The question probably misses a condition $a,b,c,d\ne0$.
We have
\begin{align}
p_{\mathrm{avg}} &= \tfrac14 a+\tfrac13 b+\tfrac12 c+d
=p(r)+p(s)
=
\tfrac12 (r^3+s^3)a
+
\tfrac12 (r^2+s^2)b
+
\tfrac12(r+s)c
+d
\end{align}
From this system we have
\begin{align}
\tfrac12 r+\tfrac12s &=\tfrac12
\\
\tfrac12 s^2+\tfrac12 r^2&=\tfrac13
\end{align}
Which gives $rs=\tfrac16$.
Moreover, we can find that
$r=\tfrac12-\tfrac16\sqrt{3}$,
$s=\tfrac12+\tfrac16\sqrt{3}$.
|
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|
Problem concerning eigenspace and rank of some matrix Problem: Let $N,P \in \mathbb{R}^{n \times n}$ be matrices, and let $P \neq 0$. Suppose that $P = NP$ and that $P$ is diagonalizable. Prove then that $N$ has an eigenspace with dimension greater than or equal to the rank of $P$.
Attempt: Let $E_{\lambda_i}$ be an eigenspace of $N$ corresponding to the eigenvalue $\lambda_i$. Then we have to prove that $\text{rank}(P) \leq \dim(E_{\lambda_i})$. Since $P$ is diagonalizable, there exists an invertible matrix $B$ such that $B^{-1} P B = D$ is a diagonal matrix. We know that the equation $n = \text{rank}(P) + \text{nullity}(P)$ always holds. Now, I tried to relate the rank of $P$ to the rank of $NP$, but I don't know how.
|
Take the first $m = \text{rank}(P)$ eigenvalues in $D$ to be nonzero. Then
$$
P = NP \Rightarrow BD = NBD \Rightarrow 1 \cdot (b_jd_j) = N (b_jd_j), \; j=1,\ldots,m
$$
where $b_j$ is the column $j$ of $B$ and so are independent. This says $N$ has an eigenspace associated with $\lambda = 1$ with at least rank $m$. I'm not sure if you need to explicitly show that it can indeed be strictly greater than $m$. I need to think about that part.
|
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|
Are $R_1=\mathbb{F}_5[x]/(x^2+2)$ and $R_2=\mathbb{F}_5[y]/(y^2+y+1)$ isomorphic rings?
Are $R_1=\mathbb{F}_5[x]/(x^2+2)$ and $R_2=\mathbb{F}_5[y]/(y^2+y+1)$ isomorphic rings? If so, write down an explicit isomorphism. If not, prove they are not.
My Try:
Since $x^2+2$ is irreducible in $\mathbb{F}_5[x]$, and $y^2+y+1$ is irreducible in $\mathbb{F}_5[y]$, both $R_1$ and $R_2$ are fields. Moreover, $O(R_1)=25=O(R_2)$. Since for a given prime $p$ and integer $n$ there is a unique field with $p^n$ elements, $R_1$ and $R_2$ are isomorphic. But how can I write an explicit isomorphism? Can somebody please help me to find it?
|
Hint: $(y + 3)^2 + 2 = y^2 + y + 1$ as elements of $\mathbb F_5[y]$.
|
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|
What does this use of little-o notation mean? I am currently going through the proof of Prime Number Theorem, as given in Hardy and Wright, and in it they define the following constant:
$$\alpha = \limsup_\limits{x \to \infty} \left|V(x) \right|$$
It then states:
$$|V(x)| \leq \alpha + o(1)$$
The book explains that $f(x)=o(g(x))$ means that $f(x)/g(x) \to 0$ as $x \to \infty$, and likewise $f(x)=o(1)$ simply means that $f(x) \to 0$ as $x \to \infty$. It also states that the use of $o(g(x))$ in an equation simply means "some unspecified function f(x) such that f(x)=o(g(x))". This definition makes sense to me when the equation is an equality, however I don't know how to interpret this statement, which is an inequality.
Does it mean that there is a specific function $f(x)$ such that $f(x)=o(1)$ and $|V(x)| \leq \alpha + f(x)$?
Does it mean that for all functions $f(x)$ such that $f(x)=o(1)$, the inequality $|V(x)| \leq \alpha + f(x)$ remains true (at least for sufficiently large $x$)?
|
The whole point of the big-oh and little-oh notation is to provide a way to write down that a particular function is negligible/is of the same order as another simpler function in a rigorous way. It allows us to deal with possibly disgusting and unpredictable functions by comparing them to much simpler functions of the same order.
The point is that once we've written $f(x)=o(g(x))$, we no longer care what $f(x)$ actually is; just that it is in a sense "smaller than $g(x)$".
With this in mind, whilst the way to read it would probably be $$|V(x)|\le \alpha +\text{a function that is equal to }o(1),$$
the take home message is
$$|V(x)|\le \alpha +\text{something which tends to $0$ as $x$ tends to $\infty$,}$$
or perhaps in this context
$$|V(x)|\le \alpha +\text{an error term that is negligible for large $x$.}$$
EDIT as per the comments.
If any function $f(x)\to \alpha$ as $ x\to \infty$, then by definition, $f(x)-\alpha=o(1)$, so $$f(x)= \alpha +o(1).$$
In the case of $\limsup$, we have$$\alpha =\limsup_{x\to\infty}f(x) =\lim_{x\to\infty}\big(\sup_{y\ge x}f(y)\big),$$
So for any $x$ we have $$f(x)\le\sup_{y\ge x}f(y)=\alpha + o(1).$$
|
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|
What do we call a "function" which is not defined on part of its domain? Before the immediate responses come in, I realize that a properly defined function means that it is defined for every value in its domain.
My question is this: if $f:A\to B$ has the property $f(a)=b_1$ and $f(a)=b_2$, then it is often still called a function, but one which is "not well-defined".
If there is $b$ in $B$ such that there is no pre-image under $f$ then we say $f$ is "not surjective".
So what would we call a "function" which has the property that $f(a)$ is not defined for some $a$ in $A$? It seems like there should be a word for this, other than just saying $f$ is not a function.
Edit: I realize that a function which is not well defined is not actually a function. I'm talking about informal speak, for example in class how we say "let's check if this function is well defined" as though it were a function even if it weren't well defined. I'm wondering if there is an analogous phrase for maps which aren't defined on their whole domain. This is all informal, which is why I tagged it a soft question.
|
It is not possible that $f(a) = b_1$ and $f(a) = b_2$ unless $b_1=b_2$. However, we can consider functions that output sets, so $f(a) = \{b_1,b_2\}$ and $f(c) = \{b_3\}$. If we did that we could think of $f$ as taking multiple values on $a$ but only a single value on $b$. This would be called a multivalued function (https://en.wikipedia.org/wiki/Multivalued_function). E.g. inverses of trigonometric functions.
|
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|
Question on Lebesgue point integration
If $f(x)$ is finite at $x$ and $\lim\limits_{h\to 0}\frac{1}{h}\int_x^{x+h} |f(t)-f(x)|dt = 0$ then $x$ is called a Lebesgue point of function $f$.
a) If $f$ is continuous at $x$ then $x$ is a Lebesgue point.
b) Give an example Lebesgue point is not a continuous point.
c) If $f$ is Lebesgue integrable on [a, b] then a.e points in [a, b] are Lebesgue points.
My attempt: c) For each $r \in Q$, if $E_r = \{lim_{h\to 0}\frac{1}{h} \int_x^{x+h} |f(t)-r|dt = |f(x)-r|\}$ then $\lambda (E_r) = b-a$. Set $\cap E_r$ is the answer to the question.
I have no idea whether this is true or not. Can any one give me some hints for part a) and b) also? Thank you in advance!
|
For c), it appears that you have the right idea. Let $r \in \mathbb{Q}$ and $f \in L^1$. From Lebesgue's differentiation theorem applied to the function $|f(t) - r|$, we have that for ae $x \in [a,b]$ that
$\frac{1}{h} \int_x^{x+h} |f(t) - r| dt \to |f(x) - r|$ as $h \to 0$.
In particular, let $E_r$ be the set of exceptional points at which the above fails to hold for each $r$. So $|E_r| = 0$, by assumption. Set $E = \bigcup_r E_r$. Suppose that $x \notin E$. Since $f(x)$ is finite, choose $r_n \to f(x)$, with $r_n \in \mathbb{Q}$. Since $x \notin E$, $x \notin E_{r_n}$ for any choice of n. With $\epsilon > 0 $ fixed, choose $n$ sufficiently large that $|r_n - f(x)| < \epsilon$. Now choose $\delta_n > 0$ so that when $0 < h < \delta_n$,
$| \frac{1}{h} \int_x^{x+h} |f(t) - r_n| dt - |f(x) - r_n| | < \epsilon$.
So, $\frac{1}{h} \int_x^{x+h} |f(t) - f(x)| dt = \frac{1}{h} \int_x^{x+h} |f(t) - r_n + r_n - f(x)| dt $
$\le \frac{1}{h} \int_x^{x+h} |f(t) - r_n| dt + |r_n - f(x)| < 3\epsilon $.
So $x$ is a Lebesgue point of $f$.
Edit: For completeness, here's a solution to a)
let $f \in L^1$ be continuous at $x$. Let $\epsilon > 0$. Choose $\delta > 0$ so that $|f(x) - f(t)| < \epsilon$ whenever $|x - t| < \delta$. Take $0 < h < \delta$.
Then $\frac{1}{h} \int_x^{x+h} |f(t) - f(x)| dt \le \frac{1}{h} \int_x^{x+h} \epsilon dt$
where the last inequality follows since $t \in [x, x+h]$. This shows a)
|
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|
Cohomology ring of $\mathbb RP^n$ with integral coefficient. I know cup product structure on $H^*(\mathbb{R}P^n;\mathbb{Z}_2)= \mathbb{Z}_2[\alpha]/(\alpha^{n+1})$. How to get $H^*(\mathbb{R}P^n;\mathbb{Z})$ from this? I have two cochain complexes for two coefficient rings.
Now my question is what will be the induced map between these two cochain complexes and what will be $H^*(\mathbb{R}P^n;\mathbb{Z})$?
|
This is a duplicate of this question, the answer to which is given in a comment. See also this website, which was the first thing I found when I searched for "cohomology of projective space".
|
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Continuous surjective mapping from half-open interval to the reals / the Cantor set 1) Is there a continuous surjective mapping of $(0, 1]$ to $\mathbb{R}$ ?
To me it seems not, but haven't been able to prove why it is false though. I was thinking of finding compact sets contained in (0, 1], and having the right endpoint fixed, e.g $[\frac{1}{2} ,1]$ $[\frac{1}{3}, 1]$, $[\frac{1}{4}, 1]$ etc. But then not sure if this is the right approach or what to do after this.
2) Is there a continuous surjective mapping of $(0, 1]$ to the Cantor Set ?
Again to me it seems not true because Cantor Set is Hausdorff space (to be precise it is Stone space) and we know that the pre-image of continuous functions should conserve compactness if we are mapping to a Hausdorff space, and Cantor Set is compact, but is there any other way to prove this without talking about Hausdorff space or compactness of the Cantor Set?
|
For 1) consider $\frac{1}{x} \sin\frac{1}{x}$
2) is false since the image of a connected set under a continuous map is connected.
|
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Why are the Kuratowski closure axioms so interesting? I never really understood the relevance of the Kuratowski closure axioms. My problem is this:
A user2520938 closure operator is an assignment $cl:\mathcal{P}(X)\to\mathcal{P}(X)$ s.t.
*
*$cl(X)=X$ and $cl(\emptyset)=\emptyset$
*For any collection of sets $U_\alpha\in \mathcal{P}(X)$ s.t. $cl(U_\alpha)=U_\alpha$ for all $\alpha$ we have $cl(\bigcap U_\alpha)=\bigcap U_\alpha$
*For any finite collection of sets $U_i\in \mathcal{P}(X)$ s.t. $cl(U_i)=U_i$ for all $i$ we have $cl(\bigcup U_i)=\bigcup U_i$
Then this will clearly also induce a topology by saying $U$ is closed iff $cl(U)=U$.
Now of course it's a bit silly to specify the operator in this fashion. However, this does make me wonder what's so special about the formulation of the Kuratowski closure axioms that they deserve a name and fame?
I suspect maybe the reason is that the Kuratowski formulation of the axioms makes it easy to verify that a given operator is a closure operator, since every axiom involves only $1$ or $2$ sets, while 'my' formulation here requires a lot more work?
|
A Kuratowski closure operator $f:\mathcal P(X)\to\mathcal P(X)$ has the property that there is a (unique) topology $\tau$ on $X$ such that $f$ is the closure operator for that topology, i.e., for every subset $U$ of $X$, $f(U)$ is the $\tau$-closure of $U$.
A "user2520938 closure operator" $f:\mathcal P(X)\to\mathcal P(X)$ in general will not have that nice property. For example, if $(X,\tau)$ is a topological space, and if I define $f:\mathcal P(X)\to\mathcal P(X)$ by setting $f(U)=U$ if $U$ is $\tau$-closed and $f(U)=\emptyset$ otherwise, then $f$ is a "user2520938 closure operator".
In other words, your conditions, besides being more complicated, are insufficient to characterize topological closure operators.
|
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Change of variable leads to contradiction for an elementary integral. Integral $I_1(\alpha,\beta)=\int_0^\infty t^\alpha \exp(-i t^{\beta}) ~dt$ converges for $-1<\alpha<\beta-1$. By introducting $u = t^\beta$ the integral is reduced to $\frac{1}{\beta}\int_0^\infty u^{\frac{\alpha+1}{\beta}-1} \exp(-i u)~du$ and the condition for the convergence is easily obtained. However, another change of variable by $t=1/\tau$ yields $I_2(\alpha,\beta)=\int_0^\infty \tau^{-\alpha-2} \exp(-i \tau^{-\beta}) ~d\tau$. Thus we have $I_2(\alpha,\beta) = I_1(-\alpha-2,-\beta)$. This implies the condition for the convergence is $-1<-\alpha-2<-\beta-1$. By rearranging terms, we get $\beta-1<\alpha<-1$. This range is totally dijoint with the original condition $-1<\alpha<\beta-1$. This is obviously a contradiction. What did I do wrong?
|
You can't plug $-\beta$ into your formula in place of $\beta$, since your formula only works for $\beta\gt0$. Working with a negative $\beta$ inside the exponential gives different results.
If $\beta\gt0$, then substituting $t\mapsto t^{1/\beta}$ gives
$$
\begin{align}
\int_0^\infty t^\alpha\exp\!\left(it^\beta\right)\,\mathrm{d}t
&=\frac1\beta\int_0^\infty t^{\frac{\alpha+1}\beta-1}\exp(it)\,\mathrm{d}t\tag{1}\\
&=\frac1\beta\,\exp\!\left(i\frac\pi2\frac{\alpha+1}\beta\right)\Gamma\!\left(\frac{\alpha+1}\beta\right)\tag{2}
\end{align}
$$
which converges when $0\lt\frac{\alpha+1}\beta\lt1$ which equivalent to $0\lt\alpha+1\lt\beta$.
However, if $\beta\lt0$, then $\exp\,\left(it^\beta\right)$ behaves quite differently. It doesn't oscillate near $\infty$ and oscillates quite quickly near $0$.
$$
\begin{align}
\int_0^\infty t^\alpha\exp\!\left(it^\beta\right)\,\mathrm{d}t
&=\int_0^\infty t^{-\alpha-2}\exp\!\left(it^{-\beta}\right)\,\mathrm{d}t\tag{3}\\
&=-\frac1\beta\int_0^\infty t^{\frac{\alpha+1}\beta-1}\exp(it)\,\mathrm{d}t\tag{4}\\
&=-\frac1\beta\,\exp\!\left(i\frac\pi2\frac{\alpha+1}\beta\right)\Gamma\!\left(\frac{\alpha+1}\beta\right)\tag{5}
\end{align}
$$
which converges for $0\lt\frac{\alpha+1}\beta\lt1$ which is equivalent to $\beta\lt\alpha+1\lt0$.
|
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Metric Isometry is always smooth? Let $M$ be a smooth manifold. Let $d$ be any metric on $M$ which induces the topology on $M$. Let $f:(M,d) \rightarrow (M,d) $ be an isometry (in the sense of metric spaces). Is it true that $f$ must be smooth?
(if the metric $d$ is induced by some Riemannian metric $g$ then the answer is positive by the Myers–Steenrod theorem)
|
No. Let $(X,d')$ be a metric space and let $f \colon X \rightarrow X$ be a homeomorphism. You can define a new metric on $X$ (which you might call the pullback metric) by $d(x,y) := d'(f(x),f(y))$. The topology induced by $d$ is the same as the topology induced by $d'$ and $\varphi \colon X \rightarrow X$ is an isometry for $(X,d')$ if and only if $f^{-1} \circ \varphi \circ f$ is an isometry for $(X,d)$. This gives you a construction that modifies isometries by possibly nonsmooth maps.
To generate a counterexample, take $M = \mathbb{R}$, $d'(x,y) = |x - y|$, $f = x^3$ (a homeomorphism but not a diffeomorphism of $\mathbb{R}$ with the usual smooth structure) and $\varphi(x) = x + 1$. Then $d(x,y) = |x^3 - y^3|$ is a metric on $\mathbb{R}$ inducing the standard topology and $\tilde{\varphi}(x) = (f^{-1} \circ \varphi \circ f)(x) = \left(x^3 + 1\right)^\frac{1}{3}$ is a nonsmooth isometry of $(\mathbb{R}, d)$.
|
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Help with trig identities to solve an AIME geometry question Quadrilateral $ABCD$ has side lengths $AB = 20$, $BC = 15$, $CD = 7$, and $AD = 24$, with diagonal length $AC = 25$. If we write $\angle ACB = \alpha$ and $\angle ABD = \beta$, then $\tan (\alpha + \beta)$ can be expressed in the form $-m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.
What would be the general steps to solving this question?
|
A meta observation. The comments discussion under the question shows how to solve this by recognizing right triangles in the picture, so that tangents of the two angles are known (and from that, the tangent of the sum of those angles).
The meta point is that had the lengths been different, we would only know $\cos \alpha$ and $\cos \beta$ to be rational numbers, and $\tan (\alpha + \beta)$ can be computed from those cosines, but there is no reason for it to be rational in that case.
So the problem is really about the trick of noticing the special Pythagorean triangles.
|
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Evaluate the limit $\lim_{x \to 0} \left(\frac{1}{x^{2}}-\frac{1}{\tan^{2}x}\right)$
Evaluate the limit $$\lim_{x \to 0}\left( \frac{1}{x^{2}}-\frac{1}{\tan^{2}x}\right)$$
My attempt
So we have $$\frac{1}{x^{2}}-\frac{\cos^{2}x}{\sin^{2}x}$$
$$=\frac{\sin^2 x-x^2\cos^2 x}{x^2\sin^2 x}$$
$$=\frac{x^2}{\sin^2 x}\cdot\frac{\sin x+x\cos x}{x}\cdot\frac{\sin x-x\cos x}{x^3}$$
Then I have $3$ limits to evaluate
$$\lim_{x \to 0}\frac{x^2}{\sin^2 x}=\left(\lim_{x \to o}\frac{x}{\sin x}\right)^2=1^2=1$$
$$\lim_{x \to 0}\frac{\sin x+x\cos x}{x}=\lim_{x \to 0}\left(\frac{\sin x}{x} + \cos x\right)=1+1=2$$
Now I'm having trouble with the last one which is
$$\lim_{x \to 0}\frac{\sin x-x\cos x}{x^3}=?$$
Thanks for any help.
|
Using L'Hospital's rule (since direct evaluation gives $\bigl(\frac{0}{0}\bigr)$ ), we have the following:
$$\lim_{x \to 0} \frac{\cos x-\cos x +x\sin x}{3x^2}= \lim_{x \to 0} \frac{\sin x}{3x}.$$
We take the derivative of the numerator and denominator again:
$$\lim_{x \to 0} \frac{\cos x}{3} = \frac{1}{3}.$$
|
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Algebra and Analysis I'm a math major at university and my tutor told me that for most people it's best to focus on either algebra or analysis, however I have trouble understanding the difference between them.
What are the actual differences between algebra and analysis, and how do I tell which is for me?
|
At an elementary level, algebra has shorter proofs; and sometimes a formula says all, whereas in geometry or topology a picture can say a lot. At a higher level, these areas begin to merge. For example, algebraic number theory very often uses analytic tools (see the "analytic class number formula", to give an example), and this holds for several other areas of modern mathematics.
|
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Quadratic formula - check my simplificaiton I am trying to solve this equation using the quadratic formula:
$$x^2 + 4x -1 = 0$$
I start by substituting the values into the quadratic formula:
$$x = {-(4) \pm \sqrt {(4)^2 - 4(1)(-1)} \over 2}$$
which becomes
$$x = {-4 \pm \sqrt{20} \over 2}$$
This is the answer the textbook that I am using gives but I would have thought I could have simplified this further to:
$$x = {-4 \pm \sqrt {(5)(2)(2)} \over 2}$$
which becomes
$$x = {-4 \pm 2 \sqrt 5 \over 2}$$
which becomes
$$x = -2 \pm \sqrt 5$$
Am I right and if so, why would the textbook not have simplified it further?
|
Note that
$$\frac{B+C}{A}=\frac{B}{A}+\frac{C}{A}$$
$$x=\frac{-4\pm 2\sqrt{5}}{2}=\frac{-4}{2}\pm\frac{2\sqrt{5}}{2}=-2\pm\sqrt{5}.$$
|
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Prove a matrix is invertible The $2 \times 2$ matrix ${A}$ satisfies
$A^2 - 4 {A} - 7{I} = {0},$
where ${I}$ is the $2 \times 2$ identity matrix. Prove that ${A}$ is invertible.
What is the best way to do this?
|
$${1\over7}(A-4I)A={1\over 7}(A^2-4A)={1\over 7}{7I}=I\\
A{1\over7}(A-4I)={1\over 7}(A^2-4A)={1\over 7}{7I}=I$$
|
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|
How do people perform mental arithmetic for complicated expressions?
This is the famous picture "Mental Arithmetic. In the Public School of S. Rachinsky." by the Russian artist Nikolay Bogdanov-Belsky.
The problem on the blackboard is:
$$
\dfrac{10^{2} + 11^{2} + 12^{2} + 13^{2} + 14^{2}}{365}
$$
The answer is easy using paper and pencil: $2$.
However, as the name of the picture implies, the expression ought be simplified only mentally.
My questions:
*
*Are there general mental calculation techniques useful for performing basic arithmetic and exponents?
*Or is there some trick which works in this case?
*If so, what is the class of problems this trick can be applied to?
|
If you know your squares out to $14$ (which students used to memorize) and do some simple three-digit arithmetic in your head, you can see that
$$100+121+144=365$$
and
$$169+196=365$$
|
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|
Algebraic Aproach For this word problem How can the followin question be solved algebraically?
A certain dealership has a total of 100 vehicles consisting of cars and trucks. 1/2 of the cars are used and 1/3 of the trucks are used. If there are 42 used vehicles used altogether, how many trucks are there?
|
$
\newcommand{\xu}{x^{\text{used}}}
\newcommand{\yu}{y^{\text{used}}}
\newcommand{\xn}{x^{\text{new}}}
\newcommand{\yn}{y^{\text{new}}}
$
The key in converting text problems into algebraic expressions is to write an expression for every sentence or phrase which contains quantifiable information.
For example, consider your problem
A certain dealership has a total of $100$ vehicles consisting of cars and trucks.
$1/2$ of the cars are used and $1/3$ of the trucks are used.
If there are $42$ used vehicles used altogether, how many trucks are there?
Let us disassemble it into set of statements working with on at a time:
*
*
A certain dealership has a total of $100$ vehicles consisting of cars and trucks.
Let $x, y$ be the total number of cars and trucks respectively.
Then we write the first equation
$$
x + y = 100
$$
*
$1/2$ of the cars are used and $1/3$ of the trucks are used.
Note that in this sentence we have two statemnts, so let us deal with them separately.
Denote $\xn$ and $\yn$ the number of new cars and trucks, $\xu$, $\yu$ – number of used cars and trucks, then we can write the second and the third equations
$1/2$ of the cars are used $(\dots)$
$$
\xu = \frac{1}{2} x
$$
$(\dots)$ and $1/3$ of the trucks are used
$$
\yu = \frac{1}{3} y
$$
*
If there are $42$ used vehicles used altogether, how many trucks are there?
The last sentence contains one quantitate statement and states the question for problem
If there are $42$ used vehicles used altogether, $( \dots ) $
$$\xu + \yu = 42$$
and states the question for problem
how many trucks are there?
which can be written in our notation as "find $y$""
$$
y \quad - \quad ?
$$
Finally, combining items $1$ to $3$, we write the system of equations, which is the algebraic formulation of the original text problem:
$
\begin{aligned}
\text{Find } y \text{ given } \qquad \qquad \qquad \qquad
\begin{cases}
x+ y = 100, \\
\xu = \dfrac{1}{2} x, \\
\yu = \dfrac{1}{3} y, \\
\xu + \yu = 42,
\end{cases}
\end{aligned}
$
Furthermore, we can simplify the system and write
$$
\begin{cases}
x + y = 100, \\
\dfrac{1}{2} x + \dfrac{1}{3} y = 42,
\end{cases}
$$
which is easy to solve:
$$
\begin{cases}
x + y = 100, \\
\dfrac{1}{2} x + \dfrac{1}{3} y = 42,
\end{cases}
\implies
\begin{cases}
\dfrac{1}{2} x + \dfrac{1}{2}y = 50, \\
\dfrac{1}{2} x + \dfrac{1}{3} y = 42,
\end{cases}
\implies
\begin{cases}
\left(\dfrac{1}{2} x - \dfrac{1}{2} x \right) + \left(\dfrac{1}{2} y - \dfrac{1}{3} y \right) = 50 - 42, \\
\dfrac{1}{2} x + \dfrac{1}{3} y = 42,
\end{cases}
\implies
\dfrac{1}{6}y = 8 \implies y = 48
$$
Thus, the final answer is
There are $48$ trucks at the dealership.
|
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|
Perpendicular Bisector of Made from Two Points For a National Board Exam Review:
Find the equation of the perpendicular bisector of the line joining
(4,0) and (-6, -3)
Answer is 20x + 6y + 29 = 0
I dont know where I went wrong. This is supposed to be very easy:
Find slope between two points:
$${ m=\frac{y^2 - y^1}{ x^2 - x^1 } = \frac{-3-0}{-6-4} = \frac{3}{10}}$$
Obtain Negative Reciprocal:
$${ m'=\frac{-10}{3}}$$
Get Midpoint fox X
$${ \frac{-6-4}{2} = -5 }$$
Get Midpoint for Y
$${ \frac{-0--3}{2} = \frac{3}{2} }$$
Make Point Slope Form:
$${ y = m'x +b = \frac{-10}{3}x + b}$$
Plugin Midpoints in Point Slope Form
$${ \frac{3}{2} = \frac{-10}{3}(-5) + b}$$
Evaluate b
$${ b = \frac{109}{6}}$$
Get Equation and Simplify
$${ y = \frac{-10}{3}x + \frac{109}{6}}$$
$${ 6y + 20x - 109 = 0 }$$
Is the problem set wrong? What am I doing wrong?
|
Notice, the mid=point of the line joining $(4, 0)$ & $(-6, -3)$ is given as $$\left(\frac{4+(-6)}{2}, \frac{0+(-3)}{2}\right)\equiv \left(-1, -\frac{3}{2}\right)$$ The slope of the perpendicular bisector
$$=\frac{-1}{\text{slope of line joining}\ (4, 0)\ \text{&}\ (-6, -3)}$$
$$=\frac{-1}{\frac{-3-0}{-6-4}}=-\frac{10}{3}$$
Hence, the equation of the perpendicular bisector: $$y-\left(-\frac{3}{2}\right)=-\frac{10}{3}(x-(-1))$$ $$6y+9=-20x-20$$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{equation of the perpendicular bisector:}\ 20x+6y+29=0}}$$
|
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|
evaluate the integral Evaluate the integral from:
$$\int_0^{\infty} \frac{x \cdot \sin(2x)}{x^2+3}dx$$
The way I approach this problem is
$$\int_0^{\infty} \frac{x \cdot \sin(2x)}{x^2+3}dx = \frac{1}{2}\int_{-\infty}^{\infty} \frac{z \cdot e^{i2z}}{(z - i\sqrt{3})(i+i\sqrt{3})}dz$$
and
$$ \text{Res}_{i\sqrt3}(f(z)) = \frac{e ^{-2\sqrt3}}{2\sqrt3}$$
Then, the integral will be:
$$\frac{1}{2}\int_{-\infty}^{\infty} \frac{z \cdot e^{i2z}}{(z - i\sqrt{3})(i+i\sqrt{3})}dz = \frac{1}{2} \cdot 2\pi i \cdot \frac{e^{-2\sqrt3}}{2\sqrt3} = \frac{\pi i e^{-2\sqrt3}}{2\sqrt3}$$
Is my approach correct? if not, can someone show me? Sorry because I just learn about residue theorem and don't know if my work is correct or not.
|
Here is another approach:
$$f(a)=\int_0^{\infty} \frac{x \cdot \sin(ax)}{x^2+3}dx$$ take a Laplace transform with respect to $a$ to obtain
\begin{align}
\mathcal{L}(f(a))&=\int_0^{\infty} \frac{x^2}{(x^2+3)(x^2+s^2)}dx\\
&=\frac{\pi}{2\sqrt3+2s}
\end{align}
now take an inverse Laplace to obtain
$$f(a)=\frac{\pi}{2} e^{-\sqrt{3} a}$$
Therefore $f(2)=\frac{\pi}{2} e^{-2\sqrt{3}}$.
|
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|
A question from Titchmarsh's The Theory of the Riemann Zeta-Function. On pages 35-36 here, we have that the integral
$$\frac{1}{2i\sqrt{y}}\int_{1/2-i\infty}^{1/2+i\infty}\phi(s-1/2)\phi(1/2-s)(s-1)\Gamma(1+s/2)\pi^{-s/2}\zeta(s)y^sds$$
equals for $\phi(s)=1$ to:
$$\frac{1}{i\sqrt{y}}\sum_{n=1}^\infty \int_{2-i\infty}^{2+i\infty} (\Gamma(2+s/2)-\frac{3}{2}\Gamma(1+s/2))\left(\frac{y}{n\sqrt{\pi}}\right)^sds$$
I don't see how to derive this? I assume it has to do with the previous section, but I took a long pause from it before returning to this section, so if someone can explain this to me, that would be excellent.
|
We move the contour to the line $\operatorname{Re}(s)=2$ then a residue at $s=1$ appears, but this residue is zero. Replacing $\phi(s)=1$
$$
\begin{align*}
I&:=\frac{1}{2i\sqrt{y}}\int_{2-i\infty}^{2+i\infty}(s-1)\Gamma(1+s/2)\pi^{-s/2}\zeta(s)y^sds\\
&=\frac{1}{2i\sqrt{y}}\sum_{n\geq 1}\int_{2-i\infty}^{2+i\infty}(s-1)\Gamma(1+s/2)\pi^{-s/2}\frac{1}{n^s}y^sds\\
&=\frac{1}{2i\sqrt{y}}\sum_{n\geq 1}\int_{2-i\infty}^{2+i\infty}(s-1)\Gamma(1+s/2)\left(\frac{y}{n\sqrt{\pi}}\right)^s ds\\
&=\frac{1}{i\sqrt{y}}\sum_{n\geq 1}\int_{2-i\infty}^{2+i\infty}\left(\frac{s}{2}\Gamma(1+s/2)-\frac{1}{2}\Gamma(1+s/2)\right)\left(\frac{y}{n\sqrt{\pi}}\right)^s ds\\
\end{align*}
$$
Finally, see that
$$
\frac{s}{2}\Gamma(1+s/2)=\left(\frac{s}{2}+1-1\right)\Gamma(1+s/2)=(1+s/2)\Gamma(1+s/2)-\Gamma(1+s/2).
$$
Then $(s/2)\Gamma(1+s/2)=\Gamma(2+s/2)-\Gamma(1+s/2)$. Replacing this we are done.
|
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|
loop invariant fibonacci fibonacci(int n)
if n < 0
return 0;
else if n = 0
return 0;
else if n = 1
return 1;
else x = 1;
y = 0;
for i from 2 to n {
t = x;
x = x + y;
y = t;
}
}
return x;
I'm trying to find a loop invariant for the above algorithm, but am not sure where to start. I know that the invariant must hold true immediately before and after completion of the loop.
How would I go about doing this?
|
Let $f(n)$ be the $n^{th}$ term of the Fibonacci sequence.
A correct invariant would be, for example:
" At the end of an iteration , $y=f(i-1)$ and $x=f(i)$ "
It is true for the first iteration ($i=2, y=1, x=1$), and stays true during the loop due to the definition of $f$ ($f(n)=f(n-1)+f(n-2)$ for $n > 1$).
Hence, at the end of the loop, $x=f(i)=f(n)$, so the function returns the correct value.
|
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|
Calculating Rotation from centroid I have a polygon as such:
where the green polygon is the rotated polygon and the purple is the extent of the polygon.
Is there a way to calculate the angle of rotation of the green polygon from the extent?
I am using python to do this as well, so if you have it in python code, that would be helpful too, but not needed.
I created the following code:
def distance(ptA, ptB):
""" calculates theta """
diffX = math.pow((ptB[0] - ptA[0]), 2)
diffY = math.pow((ptB[1] - ptA[1]), 2)
return math.sqrt((diffX + diffY))
def theta(p1, p2, p3):
"""
"""
radians = math.atan((distance(ptA=p2, ptB=p3)/distance(ptA=p1, ptB=p3)))
degree = radians * 360 / math.pi
return degree
p1,p2, and p3 are [X,Y] pairs.
Now I am using the following point:
The result doesn't seem to be the correct rotation value. Any suggestions?
Thank you
|
Rotated from what? Are we to assume that the lower left edge of the rectangle was initially at or parallel to the lower edge of the "extent"? If so then:
Set up a coordinate system with $(0, 0)$ at the lower left corner of the "extent", the $x$-axis along the lower edge, and the y-axis along the left edge. Let the $x$-coordinate where the rectangle touches the lower edge be $x_0$ and the $y$-coordinate where the rectangle touches the left edge be $y_0$. Then angle is given by $\operatorname{arccot}\left(\frac{y_0}{x_0}\right)$.
|
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|
Which pair of ratios form a proportion? My daughter is stuck on this question and we're not sure what the answer is. I'm rusty with math skills, so I don't understand how to help her, or what approaches she's tried. Please help us!
10.5/12 and 2/5
13/7 and 7/13
9/30 and 1.5/5
7/5 and 10/8
|
Two numbers $a,b$ are proportional when there is an integer $A \neq 0$ and such that $a\cdot A = b$ or $ a = A \cdot b$.
You can check that the only pair of numbers that you have listed satysfying the above requirement are $\frac{9}{30}$ and $\frac{1.5}{5}$ because $\frac{9}{30} = 1 \cdot \frac{1.5}{5}$.
|
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|
Question related to elliptical angles Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 $ be an ellipse and $AB$ be a chord. Elliptical angle of A is $\alpha$ and elliptical angle of B is $\beta$. AB chord cuts the major axis at a point C. Distance of C from center of ellipse is $d$. Then the value of $\tan \frac{\alpha}{2}\tan \frac{\beta}{2}$ is
$(A)\frac{d-a}{d+a}\hspace{1cm}(B)\frac{d+a}{d-a}\hspace{1cm}(C)\frac{d-2a}{d+2a}\hspace{1cm}(D)\frac{d+2a}{d-2a}\hspace{1cm}$
I dont know much about eccentric angles, so could not attempt this question. My guess is $\alpha+\beta=\frac{\pi}{2}$. I dont know this is correct or not. Please help me in solving this question.
|
Notice, we have the coordinates of the points $A$ & $B$ as follows $$A\equiv(a\cos\alpha, b\sin \alpha)$$
$$B\equiv(a\cos\beta, b\sin \beta)$$ Hence, the equation of the chord AB
$$y-b\sin\beta=\frac{b\sin\alpha-b\sin \beta}{a\cos\alpha-a\cos \beta}(x-a\cos \beta)$$
$$y-b\sin\beta=\frac{b}{a}\frac{\sin\alpha-\sin \beta}{\cos\alpha-\cos \beta}(x-a\cos \beta)$$ Since, chord AB intersects the x-axis at the point $C(d, 0)$, where $x=d$ & $y=0$ hence setting these values in the equation of the chord AB, we get $$0-b\sin\beta=\frac{b}{a}\frac{\sin\alpha-\sin \beta}{\cos\alpha-\cos \beta}(d-a\cos \beta)$$
$$\frac{\sin\alpha-\sin \beta}{\cos\alpha-\cos \beta}=\frac{a\sin\beta}{a\cos \beta-d}$$
Now, set the values of $\sin \alpha=\frac{2\tan \frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}$, $\sin \beta=\frac{2\tan \frac{\beta}{2}}{1+\tan^2\frac{\beta}{2}}$, $\cos \alpha=\frac{1-\tan^2 \frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}$, $\cos \beta=\frac{1-\tan^2 \frac{\beta}{2}}{1+\tan^2\frac{\beta}{2}}$
I hope you can take it from here.
|
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|
maximum value of $a+b+c$ given $a^2+b^2+c^2=48$? How can i get maximum value of this $a+b+c$ given $a^2+b^2+c^2=48$ by not using AM,GM and lagrange multipliers .
|
The Cauchy-Schwarz-inequality yields
$$|a + b + c|^2 \le (a^2 + b^2 + c^2)\cdot 3 = 144$$
and therefore $a + b + c \le 12$. Plugging in $a = b = c = 4$ shows that this value is actually the maximum.
Alternatively, you could use the convexity of the function $x \mapsto \sqrt{x}$. By Jensen's inequality we have
$$a + b + c \le |a| + |b| + |c| = 3\left(\frac{1}{3}\sqrt{a^2} + \frac{1}{3}\sqrt{b^2} + \frac{1}{3}\sqrt{c^2}\right) \le 3 \sqrt{\frac{1}{3}a^2 + \frac{1}{3}b^2 + \frac{1}{3}c^2} = 12$$
Again, plugging in $a = b = c = 4$ yields the result.
|
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|
Solving Second Order PDE with Dirac Delta I want to find the functional form of the Green function G(x,t) for a parabolic differential equation:
$$ \frac{\partial{}G(x,t)}{\partial{}t}=a\frac{\partial{}^2G(x,t)}{\partial{}x^2}+\delta(t)\delta(x)$$
Then, I'd like to write the general solution of the heat equation:
$$\frac{\partial{}T(x,t)}{\partial{}t}=a\frac{\partial{}^2T(x,t)}{\partial{}x^2}+f(x,t) $$
where f is a known source function.
I think I need to use a transform like that of Laplace or Fourier. How can I handle the Dirac delta function in this equation?
|
You can separate the dimensions. You start by trying to find a function $G(x,t)$ such that
*
*$G(x,t)$ satisfies the homogeneous heat equation in all of space and time, except formally at $x=t=0$;
*At $t=0$, $G(x,t)$ is zero forall non-zero values of $x$;
*For any nonzero real $(a,b)$, $\int_{-|a|}^{|b} G(x,0) dx = 1$.
I'm not saying yet that this function will be the Green's function you are looking for, but say you find such a function.
An example function would be something like
$$
G(x,t) = \frac{1}{4\pi\alpha} \frac{1}{\sqrt{t}} e^{-\frac{x^2}{4\alpha t}}
$$
where the powers of 2 and $\pi$ are chosen carefully such that for all non-zero $t = t_0$,
$$\int_{-\infty}^\infty G(x,t_0) dx = 1 $$
thus ensuring (at least plausibly) that when $t = 0$ the integral remains $1$. And when $t=0$, since outside of $x=0$ the function is zero, yet it's integral is $1$, that means it is a $\delta$-function in one dimension.
his would work if all your heat sources were instantaneous at $t=0$. But notice that the heat equation is time invariant, in the sense that you can make a substitution of $t' = t+\tau$ and the form remains the same.
So in terms of that $G(x,t)$ your general solution will be:
$$
h(x,y) = \int_{\tau = -\infty }^\infty \int_{x'=-\infty}^\infty f(x,\tau) G(x-x',t-\tau) dx' \, d\tau $$
There will be issues of solution stability as the problem is ill-posed if you try to find initial conditions that lead to arbitrary future heat distributions, of course.
|
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|
$\lim _{ x\to 1 }{ \frac { \sqrt { x } +\sqrt [ 3 ]{ x } -2 }{ x-1 } } $ I have been trying to algebraically solve this limit problem without using L'Hospital's rule but was whatsoever unsuccessful:
$$\lim _{ x\to 1 }{ \frac { \sqrt { x } +\sqrt [ 3 ]{ x } -2 }{ x-1 } } $$
Thanks in advance!
|
Or without using L'Hospital, you can do this:
$$
\lim _{ x\rightarrow 1 }{ \frac { \sqrt { x } +\sqrt [ 3 ]{ x } -2 }{ x-1 } } \quad =\quad \lim _{ x\rightarrow 1 }{ \frac { \sqrt { x } -1+\sqrt [ 3 ]{ x } -1 }{ x-1 } } \\ \qquad \qquad \qquad \qquad \qquad \quad =\quad \quad \lim _{ x\rightarrow 1 }{ \frac { \sqrt { x } -1 }{ x-1 } } +\frac { \sqrt [ 3 ]{ x } -1 }{ x-1 } \\ \qquad \qquad \qquad \qquad \qquad \quad =\quad \lim _{ x\rightarrow 1 }{ \frac { \sqrt { x } -1 }{ \left( \sqrt { x } -1 \right) \left( \sqrt { x } +1 \right) } } +\frac { \sqrt [ 3 ]{ x } -1 }{ \left( \sqrt [ 3 ]{ x } -1 \right) \left( \sqrt [ 3 ]{ x } +{ \left( \sqrt [ 3 ]{ x } \right) }^{ 2 }+1 \right) } \\ \qquad \qquad \qquad \qquad \qquad \quad =\quad \quad \lim _{ x\rightarrow 1 }{ \frac { 1 }{ \sqrt { x } +1 } } +\frac { 1 }{ \sqrt [ 3 ]{ x } +{ \left( \sqrt [ 3 ]{ x } \right) }^{ 2 }+1 } \\ \qquad \qquad \qquad \qquad \qquad \quad =\quad \frac { 1 }{ 2 } +\frac { 1 }{ 3 } \\ \qquad \qquad \qquad \qquad \qquad \quad =\quad \frac { 5 }{ 6 }
$$
Identities used:
*
*$a^2-b^2 =(a-b)(a+b) $
*$a^3-b^3 = (a-b)(a^2+ab+b^2)$
|
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|
some basic cardinal arithmetic on $\text{cf}(\aleph_{\omega_1})$ I'm reading The Joy of Sets by K. Devlin, by self-study. I've just seen a statement $\text{cf}(\aleph_{\omega_1})=\omega_1$ without proof, but I think this is slightly harder to prove than more obvious one, $\text{cf}(\aleph_{\omega})=\omega$.
Concretely, $\text{cf}(\aleph_{\omega_1})\le \omega_1$ is trivial, but.. how about the other direction? I've tried and I guess $\aleph_{\omega_1}^{\aleph_0}=\aleph_{\omega_1} $ (as cardinal exp.) implies the result desired. How can I prove the last identity, or is there a simpler proof on the cofinality?
|
This is a consequence of a more general fact:
Suppose $\gamma$ is a limit ordinal and I have a sequence of ordinals $\langle \alpha_{\eta}:\eta<\theta\rangle$, with each $\alpha_\eta<\omega_\gamma$. Then this induces a sequence of ordinals $\langle \beta_\eta: \eta<\theta\rangle$, each $<\gamma$, as follows: $\beta_\eta$ is the least $\delta$ such that $\alpha_\eta<\omega_\delta$.
The fact: if $\langle \alpha_\eta: \eta<\theta\rangle$ is cofinal in $\omega_\gamma$, then $\langle \beta_\eta: \eta<\theta\rangle$ is cofinal in $\gamma$.
Setting $\gamma=\omega_1$ addresses your specific question.
|
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|
Cohomology of Grassmannian of 2-planes in $\mathbb C^4$ The cohomology of the Grassmannian of 2-planes in $\mathbb C^4$ can be deduced from computations in section 4.D of Hatcher's book, using the Leray-Hirsch theorem for fiber bundles.
However, I was told this specific case is not that hard, because this Grassmannian can be given a simple CW structure, and then one can proceed directly (by computing cellular homology and then using universal coefficients, etc).
However, my geometric intuition is terrible, and I do not see it. How can one give $G(2,4)$ a nice CW structure?
|
This is a slight adaptation of the construction for real Grassmannians in chapter 6 of Milnor and Stasheff's Characteristic Classes. In particular, the cohomology is trivial to compute, since cells appear only in even dimensions.
|
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|
Paul Garret's Proof of the Cayley-Hamilton Theorem I am trying to understand the proof of Cayley-Hamilton Theorem given in Paul Garrett's notes.
We have a finite dimensional vector space $V$ over a field $k$ and are given a linear operator $T\in \mathcal L(V)$.
The proof on pg 431 in the above link starts out as:
The module $V\otimes_k k[x]$ is free of rank $\dim_k(V)$ over $k[x]$. Also, $V$ is also a $k[T]$ module by the action $v\mapsto Tv$. (I understand this much).
Now the next line reads:
So $V\otimes_k k[x]$ is a $k[T]\otimes_k k[x]$ module.
This I do not understand.
What is the general fact at play here?
EDIT: To expand on my (complete) lack of understanding, I do not see how are we giving a $k[T]\otimes_k k[x]$ module structure to $V\otimes_k k[x]$.
And of course, I am looking for a 'general principle' at work here.
For example, when we said that $V\otimes_k k[x]$ is a $k[x]$-module of rank $\dim_k(V)$, what we are using is the following: We have a natural injection $i:k\to k[x]$. So we can extend the scalars on the $k$-module $V$ and get a $k[x]$-module $V\otimes_k k[x]$.
Since $V\otimes_k k[x]\cong k[x]^n$, where $n=\dim_k(V)$, we also know that the rank of $V\otimes_k k[x]$ as a $k[x]$-module is same as $\dim_k(V)$.
|
$V$ is a $k[T]$-module.
$V \otimes $ Something is a ($k[T] \otimes$ Something) - module.
|
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|
What are some usual norms for matrices? I am familiar with norms on vectors and functions, but do there exist norms for spaces of matrices i.e. $A$ some $n \times m$ matrix?
If so, that would that imply matrices also form some sort of vector space?
|
Let $A=(a_{i,j})\in\mathcal M_{n,m}(\Bbb C)$. These some norms on $\mathcal M_{n,m}(\Bbb C)$ which are equivalent
*
*$$\|A\|=\sum_{i,j}|a_{i,j}|$$
*$$\|A\|^2=\sum_{i,j}|a_{i,j}|^2$$
*$$\|A\|=\max_{i}\sum_{j}|a_{i,j}|$$
*$$\|A\|=\max_{j}\sum_{i}|a_{i,j}|$$
and of course we can see an analogy with the norms defined on $\Bbb C^{nm}$.
|
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Is the chance of a variable also a parameter for a probability distribution? I'm new to statistics and I'm a bit confused about the concepts of 'chance of a variable' and 'parameters of a probability distribition'.
Is chance also a parameter? And if so: can computing the chance of a variable be considered as an estimation for the parameters?
|
Maybe you are new enough to statistics that you've only thought about normal and binomial distributions.
BINOMIAL. A binomial random variable $X$ counts the number of Successes in a simple experiment. Perhaps $X$ is the number of times you get a 6 when you roll a die twice. There are two parameters of the distribution of this
random variable. The first is $n = 2$, the number of independent 'trials'
(here rolls of the die). The second is the probability $p = 1/6,$ which is the probability (or 'chance') that you get a six on any one roll. But it is not ALWAYS the case that a parameter is
the 'chance' of anything.
From this information we can derive the distribution of $X$, which
gives probabilities $P(X = 0) = (5/6)^2 = 25/36,\;
P(X = 1) = 2(1/6)(5/6) = 10/36$ and $P(X = 2) = (1/6)^2 = 1/36.$
Notice that summing the probabilities of each of the possible values of $X$ must always give 1. Here, we have $10/36 + 25/36 + 1/36 = 1.$
One can prove that the mean $\mu$ of this distribution is $1/3$ and that the standard deviation is $\sigma = 0.6202.$
NORMAL. A normal random variable $Y$ also has two parameters, the mean $\mu$
(or center) of the distribution, and the standard deviation $\sigma$, which says how spread out the values are likely to be.
This is a continuous random variable, so it is not possible to list the values it will take. Instead of a list of individual probabilities, we give
a density function. The total area under the density function is
is 1. For example, if $\mu = 100$ and $\sigma = 15$ then it turns
out that $P(Y > 100) = 1/2 = 50\%$; half of the area under the density
curve is to the right of 1/2. Also, $P(Y > 130) = 0.02275$ (just under $2.5\%$). (Perhaps this random variable models IQ scores;
there are not many people with IQs above 130.)
In this case the mean and standard deviation
are the same as the parameters, but neither of them has any
direct interpretation as the 'chance' of anything.
MORE GENERAL. As you see additional random variables and their distributions,
it is natural to try to see if the parameters have any
intuitive interpretation. Often the answer is YES for the most
commonly-used distributions, and it is worthwhile to
try to understand the intuitive connection. It may have
to do with a 'chance', a 'rate', something to do with the
shape of the distribution, or something else. There is no
general rule for this. (Also, sometimes
the parameters are very abstract and any kind of intuitive
interpretation is difficult.)
Below are plots of the two distributions mentioned as examples
above. A vertical red line shows the location of the mean in
each case.
|
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|
Solve the equation $x(\log \log k - \log x) = \log k$ I want to solve this equation by expressing $x$ in function of $k$. Is it possible?
Thanks.
|
differentiating both sides with respect to x and taking $$loglogk = p$$
$$d(px)/dx - d(xlogx)/dx = 0$$
$$p - x*d(logx)/dx - logx = 0$$
$$p - 1 - logx = 0$$
$$logx = p - 1$$
$$x = e^p/e$$
$$x = (logk)/e$$
|
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|
AP in Chessboard The natural numbers $1,2,...,n^2$ are arranged in a $nXn$ chessboard. In how many ways can we arrange the numbers such that the numbers on every row and every column are in arithmetic progression?
I know that $1$ has to be put in one of the corners. I have also observed that the given criteria is maintained if the numbers are arranged one after another like $1^{st}$ row- $1,2,...,n$. $2^{nd}$ row- $n+1....2n$ and so on. In this way I get $4$ different ways(by rotating the board). Trying to prove that these are the only possible ways. Am I on the right path? Please explain.
|
If $n \ge 2$ the $8$ are the only possibilities. First prove that the condition of arithmetic progressions means that if cell $(0,0)$ is the top-left, which must contain $1$, and cell $(1,0)$ contains $1+a$ and cell $(0,1)$ contains $1+b$ then cell $(x,y)$ must contain $1+ax+by$ for any positive integers $x,y < n$. Clearly $1 \le a,b \le n$ and $a \ne b$. It suffices to consider the case where $a < b$, since the other case is just a reflection across the diagonal. If $b < n$, cell $(0,a)$ and cell $(b,0)$ both contain $1+ab$, which is not allowed. Therefore $b = n$. Thus $a = 1$ otherwise no cell contains $2$. [Prove it!]
|
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|
How do I compute the gradient vector of pixels in an image? I'm trying to find the curvature of the features in an image and I was advised to calculate the gradient vector of pixels. So if the matrix below are the values from a grayscale image, how would I go about calculating the gradient vector for the pixel with the value '99'?
21 20 22 24 18 11 23
21 20 22 24 18 11 23
21 20 22 24 18 11 23
21 20 22 99 18 11 23
21 20 22 24 18 11 23
21 20 22 24 18 11 23
21 20 22 24 18 11 23
Apologies for asking such an open ended question, I've never done much maths and am not sure how to start tackling this.
|
Suppose the image is continuous and differentiable in $x$ and $y$. Then $I(x,y)$ is the value of the pixel at each $(x,y)$, i.e. $I: \mathbb{R}^2 \mapsto \mathbb{R}$. Recall that the gradient at a point $(u,v)$ is:
$$
\nabla I(u,v) = \begin{bmatrix} \frac{\partial I}{\partial x}(u,v) \\ \frac{\partial I}{\partial y}(u,v) \end{bmatrix}
$$
Given a discrete grid, you should approximate the partial derivative in $x$ and $y$ directions using finite difference approximations at the point of interest.
Assume your function $I$ is sampled over points $\{1, \ldots, 7 \} \times \{1, \ldots, 7 \}$ in image-coordinates, i.e. $I(1,1) = 21$, $I(1,7) = 23$, etc... So you're looking for the gradient at $(4,4)$. If you assume the resolution between points is 1, then the forward difference approximation in the $x$ direction gives:
$$
\frac{\partial I}{\partial x}(4,4) \approx I(5,4) - I(4,4) = 24 - 99
$$
Do the same in $y$ to obtain the full gradient at the point.
|
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|
Alternative area of a triangle formula The problem is as follows:
There is a triangle $ABC$ and I need to show that it's area is: $$\frac{1}{2} c^2 \frac{\sin A \sin B}{\sin (A+B)}$$
Since there is a half in front I decided that base*height is equivalent to $c^2 \frac{\sin A \sin B}{\sin (A+B)}$. So I made an assumption that base is $c$ and went on to prove that height is $c \frac{\sin A \sin B}{\sin (A+B)}$. But I end up expressing height in terms of itself.. i.e. $h \equiv \frac{ch}{a\cos B + b \cos A}$. How do I prove this alternative area of triangle formula?
|
$$\triangle =\dfrac12bc\sin A=\dfrac12(2R\sin B)c\sin A\cdot\dfrac c{2R\sin C}$$
Now $A+B=\pi-C\implies\sin(A+B)=\sin(\pi-C)=?$
|
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|
How can I solve this integral equation by converting it to a differential equation Let we have the following integral equation :$$y(x)=e^{-x}cos(x)-\int_{0}^{x}e^{-x+t}cos(x)y(t)dt$$
How can I solve this integral equation by converting it to a differential equation
|
Given $\displaystyle y(x)=e^{-x}\cos(x)-\cos(x) \cdot \int_{0}^{x}e^{-x+t}y(t)dt...........(1)$
Using Differentitation Under Integral Sign.
Given $\displaystyle \frac{d}{dx}\left(y(x)\right) = \frac{d}{dx}\left[e^{-x}\cdot \cos x\right] - \frac{d}{dx}\left[\cos x \cdot \int_{0}^{x}e^{-x+t} \cdot y(t)dt\right]$
We Get $\displaystyle y'(x) = -e^{-x}\cdot \sin x+\cos x\cdot e^{-x}-\cos x\cdot y(x)+\int_{0}^{x}e^{-x+t}\cdot y(t)dt\cdot \sin x $
Now Put Value of $\int_{0}^{x}e^{-x+t}\cdot y(t)dt$ from $\bf{1^{st}}$ equation.
$\displaystyle y'(x) = -e^{-x}\cdot \sin x+\cos x\cdot e^{-x}-\cos x\cdot y(x)+\left[\frac{e^{-x}\cdot \cos x-y(x)}{\cos x}\right]\cdot \sin x$
You get an Differential equation
|
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Integral of $\frac{1}{x^2+x+1}$ and$\frac{1}{x^2-x+1}$ How to integrate two very similar integrals. I am looking for the simplest approach to that, it cannot be sophisticated too much as level of the textbook this was taken from is not very high. $$\int \frac{1}{x^2+x+1} dx$$ and$$\int \frac{1}{x^2-x+1} dx$$
|
For the first one write $x^2 + x + 1 = (x + 1/2)^2 + 3/4$ and the second one $x^2 - x + 1 = (x - 1/2)^2 + 3/4$ these both are arctan
|
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Which numbers are square modulo 9? How can I prove that n = $1,4,7,9$ for every integer k such that $k^2 = n$ (mod9)?
|
You would better state it using $0$ instead of $9$, i.e., $\forall{k}:k^2\equiv0,1,4,7\pmod9$.
There you go:
*
*$k\equiv0\pmod9 \implies k^2\equiv0^2\equiv0\pmod9$
*$k\equiv1\pmod9 \implies k^2\equiv1^2\equiv1\pmod9$
*$k\equiv2\pmod9 \implies k^2\equiv2^2\equiv4\pmod9$
*$k\equiv3\pmod9 \implies k^2\equiv3^2\equiv9\equiv0\pmod9$
*$k\equiv4\pmod9 \implies k^2\equiv4^2\equiv16\equiv7\pmod9$
*$k\equiv5\pmod9 \implies k^2\equiv5^2\equiv25\equiv7\pmod9$
*$k\equiv6\pmod9 \implies k^2\equiv6^2\equiv36\equiv0\pmod9$
*$k\equiv7\pmod9 \implies k^2\equiv7^2\equiv49\equiv4\pmod9$
*$k\equiv8\pmod9 \implies k^2\equiv8^2\equiv64\equiv1\pmod9$
|
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Coin - Probability explanation Let's say we flip a fair coin $1$ time. The probability of obtaining at least one heads is $50\%$.
However let's say we flip the coin $2$ times. The probability of obtaining at least one heads becomes $75\%$.
I can't seem to wrap my head around why our probability of obtaining at least one heads increases as we increase the number of times we flip the coin since each flip of the coin is an independent event and each time we have a $50\%$ of getting heads so shouldn't the probability of obtaining at least one heads be $50\%$?
|
If you are going to flip the coin twice, you know that if do fail to obtain a head on the first try, there will still be another chance, so how can the probability of obtaining at least one head in two flips be any thing but greater than the probability of obtaining one head on the first flip?
$$\begin{align} & \; \mathsf P(\text{at least one head in two flips}) \\[1ex] = & \;\mathsf P(\text{head first try}) + \mathsf P(\text{tail first try})\,\mathsf P(\text{head second try}) \\[1ex] = & \quad \frac 1 2 + \frac 1 2\times \frac 1 2 \\[1ex] = & \qquad \frac 3 4\end{align}$$
|
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Finding roots of the polynomial $x^4+x^3+x^2+x+1$ In general, how could one find the roots of a polynomial like $x^4+x^3+x^2+x^1+1$? I need to find the complex roots of this polynomial and show that $\mathbb{Q (\omega)}$ is its splitting field, but I have no idea of how to proceed in this question. Thanks in advance.
|
Since $(x-1)(x^4+x^3+x^2+x+1)=x^5-1$, the roots are the fifth roots of $1$, excluding $1$. Note that the set of fifth roots of $1$ is a group of prime order, so it is cyclic and any element is a generator. Thus, if $\omega$ is any of the roots of the polynomials, the full set of roots is given by $\omega,\omega^2,\omega^3,\omega^4$. In particular all roots belong to $\mathbb{Q}[\omega]$.
The roots are
$$
\cos\frac{2\pi}{5}+i\sin\frac{2\pi}{5}\\
\cos\frac{4\pi}{5}+i\sin\frac{4\pi}{5}\\
\cos\frac{6\pi}{5}+i\sin\frac{6\pi}{5}\\
\cos\frac{8\pi}{5}+i\sin\frac{8\pi}{5}
$$
However, you can also determine them “more explicitly”. Rewrite the equation as
$$
x^2+\frac{1}{x^2}+x+\frac{1}{x}+1=0
$$
and recall that
$$
x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2
$$
so, setting $t=x+\dfrac{1}{x}$, the equation is
$$
t^2+t-1=0
$$
Find the two roots $t_1$ and $t_2$; next solve the equations
$$
x+\dfrac{1}{x}=t_1,\qquad x+\dfrac{1}{x}=t_2
$$
|
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|
Prove that a sequence whose second difference is a nonzero constant is quadratic. For example, if {$a_0, a_1, a_2, a_3, ...$} is the sequence,
the first difference is {$a_1-a_0, a_2-a_1, a_3-a_2, ...$},
and the second difference is {$(a_2-a_1)-(a_1-a_0), (a_3-a_2)-(a_2-a_1), ...$}.
I think that using facts from up to Calculus, perhaps derivatives, should be enough. I find myself going in circles and don't know how to approach this.
|
Let $d$ denote the constant second difference. Moreover, let
\begin{align*}
c\equiv&\,a_1-a_0-\frac{d}{2},\\
\end{align*}
Claim: $a_n=(d/2)n^2+cn+a_0$ for all $n\in\{0,1,2\ldots\}$.
Proof: The claim is obviously true for $n=0$. For $n=1$, $$\frac{d}{2}\times n^2+cn+a_0=\frac{d}{2}+\left(a_1-a_0-\frac{d}{2}\right)+a_0=a_1.$$
Proceed by induction: suppose that the claim is true for $0,1,\ldots,n$ for some integer $n\geq1$. The task is to prove that it is true for $n+1$. Now:
\begin{align*}
d=&\,(a_{n+1}-a_n)-(a_n-a_{n-1})=a_{n+1}-2a_n+a_{n-1}\\
=&\,a_{n+1}-2\left[\frac{d}{2}\times n^2+cn+a_0\right]+\left[\frac{d}{2}\times (n-1)^2+c(n-1)+a_0\right],
\end{align*}
where the first equality comes from the definition of $d$, and the third one is due to the induction hypothesis. Now, rearrange for $a_{n+1}$:
\begin{align*}
a_{n+1}=&\,d+2\left[\frac{d}{2}\times n^2+cn+a_0\right]-\left[\frac{d}{2}\times (n-1)^2+c(n-1)+a_0\right]\\
=&\,d+dn^2+2cn+2a_0-\frac{d}{2}\times(n^2-2n+1)-c(n-1)-a_0\\
=&\,\underbrace{d}_{\spadesuit}+\underbrace{dn^2}_{\heartsuit}+\underbrace{2cn}_{\clubsuit}+\underbrace{2a_0}_{\diamondsuit}-\underbrace{\frac{d}{2}\times n^2}_{\heartsuit}+\underbrace{dn}_{\star}-\underbrace{\frac{d}{2}}_{\spadesuit}-\underbrace{cn}_{\clubsuit}+\underbrace{c}_{\clubsuit}-\underbrace{a_0}_{\diamondsuit}\\
=&\,\underbrace{\frac{d}{2}}_{\spadesuit}+\underbrace{\frac{d}{2}\times n^2}_{\heartsuit}+\underbrace{c(n+1)}_{\clubsuit}+\underbrace{a_0}_{\diamondsuit}+\underbrace{dn}_{\star}\\
=&\,\frac{d}{2}\times n^2+dn+\frac{d}{2}+c(n+1)+a_0\\
=&\,\frac{d}{2}\times(n+1)^2+c(n+1)+a_0.
\end{align*}
The proof is complete. $\quad\blacksquare$
|
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Closed plus finite dimensional in a TVS If $E$ is a topological vector space (TVS), $F_1$ a closed subspace of $E$, and $F_2$ a finite dimensional subspace of $E$, such that $F_1 \cap F_2=\{0\}$, is $F_1+F_2$ necessarily closed? If yes, are the projection from $F_1+F_2$ onto $F_1$ and $F_2$ respectively, continuous?
|
Yes If we suppose that $E$ is a normed space , the sum of a closed subspace $Y$ and a finite dimensional space $F$ is closed, in fact :
suppose that a sequence $(y_n+f_n) \subset Y+F$ converge vers $x\in X$, Let $P$ the (continuous Why? ) projection from $Y+F$ to $F$, the sequence $(y_n+f_n)$ is bounded (because it converge to $x$ so it exist a subsequence $(f_{n_k})$ that converge to $f\in F$ (because F is finite dimensional vector space so every bounded closed set is compact) and so $y_{n_k}$ converge to $y\in Y$ as difference of two convergent sequences, and so $x=y+f\in Y+F$
To complete the proof we need to proof that the projection from $Y+F$ to $F$ is continuous, so for that we put
$$
\delta=\min\{d(f,Y) : \|f\|=1 \}
$$
$\delta>0$ because the set $S_F=\{ f\in F ; \|f\|=1\}$ is compact, and the function $d(.,Y)$ is a continuous function so it exist $f'\in S_F$ such that $\delta=d(f',Y)$, then if $\delta=0$ this implies that $f'\in Y\cap F=\{0\}$ but $\|f'\|=1\neq 0$, absurde.
so
$$
\|y+f\|\geq \delta\|f\|
$$
so the projection from $Y+F$ to $F$ is of norme $\leq \delta^{-1}$
and the projection from $Y+F$ to $Y$ is of norme $\leq 1+\delta^{-1}$
|
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Is the standard definition of vector wrong? The definition of a vector is usually something like "a quantity that has both a magnitude and a direction". But, in the context of, say, economics rather than physics, does this definition make sense? Let's say we are working in 4 dimensional space? Lastly, how does this definition make sense in matrix algebra when referring to a column vector, say, taken from a list of the number of vehicles sold by brand?
|
That is not a very general definition of a vector, no. But it is hardly "wrong" for that reason.
Direction and magnitude are intimately tied to spatial intuition.
The magnitude referred to is usually a "length" measured by a real number, but there are vector spaces where that idea makes no sense.
The same goes for "direction" too. You can, at best, only get a very primitive notion of direction, as discussed here
The best definition of a vector is probably "element of a vector space," with all the axiomatic behaviors implied. That does not mean it must be particularly useful for developing intuition about what a vector "is."
The direction-magnitude is a particularly apt way of describing vectors in 2 and 3 dimensional geometry, so it has its uses. After convincing yourself about higher dimensional real spaces, and the spaces over other fields, you will probably find yourself abandoning direction and magnitude in favor of the most general definition.
|
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$\int\limits_{0}^{\pi/2}\frac{1+2\cos x}{(2+\cos x)^2}dx$ $\int\limits_{0}^{\pi/2}\frac{1+2\cos x}{(2+\cos x)^2}dx$
I tried to solve it.
$\int\limits_{0}^{\pi/2}\frac{1+2\cos x}{(2+\cos x)^2}dx=\int\limits_{0}^{\pi/2}\frac{4+2\cos x}{(2+\cos x)^2}-\frac{3}{(2+\cos x)^2}dx=\int\limits_{0}^{\pi/2}\frac{2}{2+\cos x}-\frac{3}{(2+\cos x)^2}dx$
But i could not solve further.Please help me in completing.
|
You may observe that: $$\frac{2 \cos (x)+1}{(\cos (x)+2)^2}=\frac{\cos (x)}{\cos (x)+2}+\frac{\sin ^2(x)}{(\cos (x)+2)^2}=\frac{\frac{d}{dx}\sin(x)}{\cos (x)+2}-\frac{\sin(x)\frac{d}{dx}(\cos x +2)}{(\cos (x)+2)^2}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Show that $W$ is a symplectic matrix i.e. $W^T J W=J$ Good day,
This is my first question, I hope all information is given. If not, feel free to ask.
Currently I am reading "New Trends for Hamiltonian Systems and Celestial Mechanics" by E. A. Lacomba & J. Libre, 1994. On page 325 it says the following:
$\rightarrow$ Let $I$ be the identity matrix and $J=\begin{pmatrix} 0 & I \\ -I & 0 \end{pmatrix}$.
Further we have
$W=\frac{1}{\sqrt{N}} \left[ \omega^{i \cdot j} \right]_{0 \leq i,j \leq N-1} =\frac{1}{\sqrt{N}}\begin{pmatrix}
1 & 1 & \dots & 1 \\
1 & \omega & \dots & \omega^{N-1} \\
\vdots & \vdots & \ddots & \vdots \\
1 & \omega^{N-1} & \dots & \omega^{(N-1)^2} \end{pmatrix} $ with $ \omega=e^{2 \pi i / N} $ the nth root of unitary.
Using the complex quantity $ \omega $ makes it easier to write down and perform this transformation (here not necessary), but one should not forget that in the real space $\omega$ corresponds to a 2 $ \times $ 2 submatrix of the form
$ \omega \leftrightarrow \begin{pmatrix}
\cos 2 \pi /N & -\sin 2 \pi /N \\
\sin 2 \pi /N & \cos 2 \pi /N \end{pmatrix} $
Keeping this relationship in mind one sees that for $W^T$ we have to use the conjugate transposed matrix, i.e. $W^T=\overline{W}^t.$ Thus we have a unitary matrix with
$ W^{-1}=W^T $ (...) Due to the structure of $W$ we have in $ \mathbb{R}^{2N} $
$$W^T J W = J$$ $\leftarrow$
Okay, now my question: How do I get $W^T J W = J$? There is no proof given, maybe it's trivial.
Here are my thoughts:
*
*Maybe I could prove it by using the proportions $W^T=W^{-1}$ and $J^T=J^{-1}=-J$ but I couldn't get it, I think I have to multi plicate it formally.
*Okay, it's mentioned it should be in $\mathbb{R}^{2N}$, but W itself is in $\mathbb{C}^{N \times N}$, so I should get it in its real form. But how?
a. By using the correspondence of $\omega$ to its rotating matrix. Like that:
$$ \frac{1}{\sqrt{N}} \begin{pmatrix}
1 & 0 & 1 & 0 & \dots & 1 & 0 \\
0 & 1 & 0 & 1 & \dots & 0 & 1 \\
1 & 0 & \cos\left( \frac{2 \pi}{N} \right) & -\sin\left( \frac{2 \pi}{N}\right) & \dots & \cos\left( \frac{2 \pi (N-1)}{N} \right) & -\sin\left( \frac{2 \pi (N-1)}{N} \right) \\
0 & 1 & \sin\left( \frac{2 \pi}{N} \right) & \cos\left( \frac{2 \pi}{N} \right) & \dots & \sin\left( \frac{2 \pi (N-1)}{N} \right) & \cos\left( \frac{2 \pi (N-1)}{N} \right) \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
1 & 0 & \cos\left( \frac{2 \pi (N-1)}{N} \right) & -\sin\left( \frac{2 \pi (N-1)}{N} \right) & \dots & \cos\left( \frac{2 \pi (N-1)^2}{N} \right) & -\sin\left( \frac{2 \pi (N-1)^2}{N} \right) \\
0 & 1 & \sin\left( \frac{2 \pi (N-1)}{N} \right) & \cos\left( \frac{2 \pi (N-1)}{N} \right) & \dots & \sin\left( \frac{2 \pi (N-1)^2}{N} \right) & \cos\left( \frac{2 \pi (N-1)^2}{N} \right) \end{pmatrix} $$
It would be unitary. But I believe this approach is wrong, not sure, but I tried to compute the case $N=4$ to get $W^T J W = J$ but without success. This form can be simplifed by using $\omega^N=1$ e.g. $\omega^{N-1}=\omega^{-1}$ and so on.
b. Compute the eigenvalues and try to compute a real Jordan Normal Form, but this could be tiring and at $N=4$ this lead me to failure.
Okay. These were my approaches. I'm thankful for every help, let it be another approach/proof or a corrections to mine.
Thanks a lot,
Marvin
|
Preliminaries: Let us see how the product of two complex $n\times n$ matrices $A$ and $B$ looks like in terms of their real and imaginary parts
$$
A=A_r+iA_i,\quad B=B_r+iB_i\quad\Rightarrow\quad AB=A_rB_r-A_iB_i+i(A_rB_i+A_iB_r).
$$
It means that the complex multiplication for matrices works the same as for numbers as long as one respects the order.
Similar to numbers one can construct the $2n\times 2n$ real matrices
$$
A\sim\left[\matrix{A_r & A_i\\-A_i & A_r}\right], \quad
B\sim\left[\matrix{B_r & B_i\\-B_i & B_r}\right]\quad\Rightarrow\quad
$$
$$
\quad\Rightarrow\quad AB\sim \left[\matrix{A_rB_r-A_iB_i & A_rB_i+A_iB_r\\-(A_rB_i+A_iB_r) & A_rB_r-A_iB_i}\right]=\left[\matrix{A_r & A_i\\-A_i & A_r}\right]\left[\matrix{B_r & B_i\\-B_i & B_r}\right].
$$
So the operations of matrix multiplication and taking the real counterpart are perfectly commuting.
Answer: The matrix $J=\left[\matrix{0 & I\\-I & 0}\right]$ is the real counterpart of the complex $n\times n$ matrix $iI$. The complex matrix $W$ is unitary, i.e. $W^*W=I$ (or $W^TW=I$ in your notations, but I prefer to keep ${^T}$ for real transpose to avoid confusions). It means that
$$
iI=W^*W(iI)=W^*(iI)W=(W_r^T-iW_i^T)(iI)(W_r+iW_i).
$$
Note here that $iI$ is a scalar multiple of identity, thus commutes with any matrix. Now taking the real counterparts for each matrix on both sides gives
$$
J=\left[\matrix{W_r^T & -W_i^T\\W_i^T & W_r^T}\right]J
\left[\matrix{W_r & W_i\\-W_i & W_r}\right]=
\left[\matrix{W_r & W_i\\-W_i & W_r}\right]^TJ\left[\matrix{W_r & W_i\\-W_i & W_r}\right].
$$
|
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|
Does $\cos (\pi/5)$ belong to $\mathbb{Q} (\sin(\pi/5))$? I need to know if $$\cos(\pi/5) \in \mathbb{Q} (\sin(\pi/5))?$$
I can compute explicitly such $\cos$ and $\sin$, but I have some difficulties how to deduce from this an answer.
|
Hint : show that $c=\frac{3}{2}-2s^2$, where $c=\cos(\frac{\pi}{5})$
and $s=\sin(\frac{\pi}{5})$. You can check it using Euler's identities for example.
|
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|
Given any base for a second countable space, is every open set the countable union of basic open sets? Second countability implies the existence of a countable topological base.
Does it also imply that any open set in a second countable space itself is a countable union of basic open sets?
I mean, in a second countable space together with a (not necessarily countable) topological base, can every open set be written as a countable union of basic open sets?
|
Given any second-countable space $X$ and any base $\mathcal B$ for $X$, every open $U \subseteq X$ is the union of a countable subfamily of $\mathcal B$.
First fix a countable base $\mathcal D$ for $X$, and let $U \subseteq X$ be open.
Note that there is a $\mathcal B_U \subseteq \mathcal B$ such that $U = \bigcup \mathcal B_U$. Now for each $V \in \mathcal B_U$ and $x \in V$ there is a $W_{x,V} \in \mathcal D$ such that $x \in W_{x,V} \subseteq V$. It follows that $\mathcal D_U = \{ W_{x,V} : V \in \mathcal B_U, x \in V \} \subseteq \mathcal D$, and is therefore countable. Also, $$
U = \bigcup_{V \in \mathcal B_U} V = \bigcup_{V \in \mathcal B_U} \bigcup_{x \in V} \{ x \} \subseteq \bigcup_{V \in \mathcal B_U} \bigcup_{x \in V} W_{x,V} \subseteq U,$$ so $\bigcup \mathcal D_U = U$.
For each $W \in \mathcal D_U$ pick $V_W \in \mathcal B$ such that $W \subseteq V_W \subseteq U$ (e.g., pick any $V$ for which $W = W_{x,V}$ for some $x \in V$). Then $\mathcal B'_U = \{ V_W : W \in \mathcal D_U \}$ is a countable subfamily of $\mathcal B$. Note, too, that
$$
U = \bigcup_{W \in \mathcal D_U} W \subseteq \bigcup_{W \in \mathcal D_U} V_W \subseteq U,$$ and so $\bigcup \mathcal B'_U = \bigcup_{W \in \mathcal D_U} V_W = U$.
(In fact, some countable subfamily of $\mathcal B$ is even a base for $X$: just take $\mathcal B' = \bigcup \{ \mathcal B'_U : U \in \mathcal D \}$ where $\mathcal B'_U \subseteq \mathcal B$ is countable with $\bigcup \mathcal B'_U = U$ for each $U \in \mathcal D$.)
|
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|
Area of shaded region circle help
Find the area of the shaded region
Area of the sector is $240^\circ$ or $\frac{4\pi}{3}$
Next find $\frac{b\cdot h}{2}$ which is $\frac{2\cdot2}{2}$ which is $2$.
Then subtract the former from the latter: $\frac{4\pi}{3} - 2$
Therefore the answer is $~2.189$?
Is this correct?
|
Here, aperture angle $\theta=120^\circ=\frac{2\pi}{3}$
Area of shaded portion $$=\text{(area of the sector)}-\text{(area of isosceles triangle)}$$
$$=\frac{1}{2}(\theta)(r^2)-\frac{1}{2}(r^2)\sin\theta$$
$$=\frac{1}{2}\frac{2\pi}{3}(2)^2-\frac{1}{2}(2)^2\sin\frac{2\pi}{3}$$
$$=2.456739397$$
Edit:
Note: Area of an isosceles triangle having each of the equal sides $a$ & an angle $\theta$ included between them then the area of the triangle $$=\frac{1}{2}(a)(a)\sin\theta=\frac{1}{2}a^2\sin\theta$$
|
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|
What function looks like $\overbrace{}$? Here's my awesome drawing:
Basically it's a function that takes a high (or infinite) value at $0$, then falls off logarithmically for a while before falling off exponentially.
It doesn't need to be symmetric, it would be ok if the negative $x$ values were
reversed in sign or something like that. I really only care about $x\ge0$.
Edit: I would prefer if the function is not periodic.
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How about $\frac{x^2-1}{x^2(x-2)(x+2)}$
|
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|
Step functions on $[a,b]$ are dense in $\mathcal C^0([a,b])$.
Let $\|f \|=\sup_{[a,b]}|f|$. We consider ($\mathcal C^0([a,b]),\|\cdot \|)$ and $(\mathcal E([a,b]),\|\cdot \|)$ where $\mathcal E([a,b])$ is a set of the step functions on $[a,b]$. I have to show that $\mathcal E([a,b])$ is dense in $\mathcal C^0([a,b])$.
There is my proof:
Let $f\in\mathcal C^0([a,b])$ and $\varepsilon>0$. Since $f$ is continuous on $[a,b]$ it's also uniformly continuous and thus, there is a $\delta>0$ s.t. $|f(x)-f(y)|<\varepsilon$ for all $x,y\in[a,b]$ s.t. $|x-y|<\delta$. Let $a=x_0<x_1<...<x_n=b$ such that $x_{i+1}-x_i<\delta$ for all $i$. We set $\varphi(x)=f(x_i)$ for all $i=0,...,n-1$ and $\varphi(x_n)=f(x_n)$. Therefore, if $x\in[a,b[$, there is a $i$ such that $x_i\leq x<x_{i+1}$ and thus $$|f(x)-\varphi(x)|=|f(x)-\varphi(x_i)|<\varepsilon$$ since $|x-x_i|<\delta$. Finally, since $f(b)=\varphi(b)$ we get that
$$\forall x\in[a,b], |f(x)-\varphi(x)|<\varepsilon$$
and thus $$\|f-\varphi\|=\sup_{[a,b]}|f-\varphi|<\varepsilon$$
what prove the claim.
Q1) Is it correct ?
Q2) Something is strange to me. If $A$ is dense in $B$, in particular $A\subset B$, but here, how can $\mathcal E([a,b])\subset \mathcal C^0([a,b])$ since an element of $\mathcal E([a,b])$ is not necessarily continuous ?
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Yes. Uniform continuity means that you can use a finite number of balls to cover the domain, and in each of those balls, the values of the functions are $\varepsilon$-close. Hence, you can build a step function that is $\varepsilon$-close to the given function.
|
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|
Integration of $\sin(\theta)$ I hope I'm not asking a silly question.
We can integrate $\sin(\theta)$ simply by the following identity:
$$\int_0^\frac{\pi}{2} \sin\theta\ \mathsf d\theta = \left[-\cos\theta \vphantom{\frac 1 1} \right]_0^\frac{\pi}{2}=1.$$
But how can we do this by summation ?
For example, $$\int_0^{100} x\ \mathsf dx = \left[\frac{x^2}{2}\right]_0^{100} = 5000 \approx \sum_{x=1}^{100}x= 1+2+\cdots+100 = 5050.$$
How can we do the same for initially mentioned problem ?
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$$\cdots\approx\frac{\pi}{ 200}\sum_{k=1}^{100}\sin\left(\frac{k}{100}\right)$$
In general:
if $b-a$ small,
$$\int_a^b f(x)dx\approx\frac{b-a}{100}\sum_{k=1}^{100}f\left(a+k\frac{(b-a)}{100}\right)$$
since $$\frac{b-a}{n}\sum_{k=1}^nf\left(a+k\frac{b-a}{n}\right)\underset{n\to\infty }{\longrightarrow} \int_a^b f(x)dx.$$
|
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|
Composition of homotopy classes with self-maps of spheres Are there some general rules/formulas on the relation between the homotopy class $[f]\in \pi_i(S^n)$ and the homotopy class of the
composition
$S^i\stackrel{a}{\to} S^i\stackrel{f}{\to}S^n\stackrel{b}{\to}S^n$
where $a,b$ are maps of degree $d_a,d_b$ respectively? I think that composition with $a$ always multiplies $[f]$ by $d_a$, but composition with $b$ seems to be harder. Is $[b\circ f]$ always a multiple of $[f]$?
Apparently, if $f: S^3\to S^2$ is the Hopf map, the homotopy class of the composition is $(d_a \times d_b^2) [f]$ (see this book, p. 205).
In the stable dimension range, however, composition with $b$ seems only to multiply $[f]$ by $d_b$, if I understand this wikipedia paragraph correctly (supercommutativity).
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(this is not a full answer, but for comments it's too long)
For given (smooth) map $f:S^3\to S^2$ if you take two non-critical values $p,q\in S^2$, then linking number of $f^{-1}(p)$ with $f^{-1}(q)$ equals to degree of $f$. Maybe, something similar occurs in case $i=2n-1$ for all $n$.
And when you take a suspension of the diagrams
$S^i\stackrel{a}{\to} S^i\stackrel{f}{\to}S^n\stackrel{b}{\to}S^n$,
for stable dimensions $i,n$ answer will be the same, because of map $\pi_i(S^n)\to\pi_{i+1}(S^{n+1})$ being surjective and equalities $d_{\Sigma a}=d_a$, $d_{\Sigma b}=d_b$.
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|
Correctly calculating permutations and combinations without duplicate patterns Given 16 balls each numbered 1 through 16, and 5 glass tubes numbered 1 through 5; how many ways are there to slot all 16 balls into the glass tubes, selected one at a time, with the only condition that each slot should always have at least 1 ball? The ball and glass numbers matter.
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We assume that order of balls in the tubes matters.
Line up the $16$ balls in some order. There are $16!$ ways to do this. There are $15$ interball gaps. We choose $4$ of them to place a separator into in the usual Stars and Bars style. This can be done in $\binom{15}{4}$ ways.
Thus the total number of ways is $16!\binom{15}{4}$. The idea generalizes.
Remark: Numerically, this gives the same result as the recursion of dREaM.
|
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|
Fraction field of $F[X,Y](f)$ isomorphic to $F(X)[Y]/(f)$ Assume $F$ is a field and $f$ is an irreducible polynomial in $F[X,Y]$ which involves the variable $Y$.
Then, by Gauss's lemma, $f$ is irreducible also in $F(X)[Y]$ so that $F(X)[Y]/(f)$ is a field (where $(f)$ is the ideal generated by $f$).
I'm looking for a simple way to see that the fraction field of the integral domain $F[X,Y]/(f)$ is isomorphic to $F(X)[Y]/(f)$.
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Let $A$ be a UFD, and $f\in A[Y]$ irreducible with $\deg f\ge 1$. Then the field of fractions of $A[Y]/(f)$ is $K[Y]/(f)$, where $K$ is the field of fractions of $A$.
Set $S=A-\{0\}$. Then $K[Y]/(f)=S^{-1}A[Y]/S^{-1}(f)\simeq S^{-1}(A[Y]/(f))$, so $K[Y]/(f)$ is a ring of fractions of the integral domain $A[Y]/(f)$ and moreover it is a field, so necessarily $K[Y]/(f)$ is the field of fractions of $A[Y]/(f)$.
|
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|
Angle between segments resting against a circle Motivation:
A couple of days ago, when I was solving this question, I had to consider a configuration like this
Now, I didn't intentionally make those two yellow bars stand at what appears to be a $90^{\circ}$-angle but it struck me as an interesting situation, so much so that I thought the following question might be an interesting one to solve.
The Question:
Given a circle of radius $r$, a horizontal line a distance $c>r$ from the circle's centre, and two points $A$ and $B$ on that line located as indicated in the picture below, find the angle $\Theta$ as a function of the parameters given ($a,b,c,r$). The blue and red lines passing through $M$ are tangents to the circle, at $P$ and $Q$ respectively.
|
Set Cartesian coordinates on the plane so that the center of the circle
is $(0,0)$ and the black line $\overline{AB}$ is parallel to the $x$-axis.
The coordinates of $A$ are $(-(a - b), c)$ and the distance $OA$ is
$\sqrt{(a - b)^2 + c^2}$. Therefore
\begin{align}
\angle OAB & = \arcsin\left( \frac{\sqrt{(a - b)^2 + c^2}}{c} \right)
= \arctan \left( \frac{c}{a-b} \right) & \text{and} \\
\angle OAQ & = \arcsin\left( \frac{\sqrt{(a - b)^2 + c^2}}{r} \right).
\end{align}
The coordinates of $B$ are $(b, c)$ and the distance $OB$ is
$\sqrt{b^2 + c^2}$, so
\begin{align}
\angle OBA & = \arcsin\left( \frac{\sqrt{b^2 + c^2}}{c} \right)
= \arctan \left( \frac cb \right) & \text{and} \\
\angle OBP & = \arcsin\left( \frac{\sqrt{b^2 + c^2}}{r} \right).
\end{align}
Since $\angle BAM = \angle OAB + \angle OAQ$
and $\angle ABM = \angle OBA - \angle OBP$,
and $\theta = \pi - \angle BAM - \angle ABM$,
\begin{align}
\theta & = \pi - (\angle OAB + \angle OAQ) - (\angle OBA - \angle OBP) \\
& = \pi - \arctan \frac{c}{a-b}
- \arcsin \frac{\sqrt{(a - b)^2 + c^2}}{r}
- \arctan \frac cb
+ \arcsin \frac{\sqrt{b^2 + c^2}}{r} .
\end{align}
Another approach: consider the figure below, which shows line $\overline{AB}$
and segments $\overline{OP}$ and $\overline{BP}$. It also shows the
perpendicular from $O$ to $\overline{AB}$, which intersects $\overline{AB}$
at $C$ and $\overline{BP}$ at $R$.
From the original problem statement, we have $OP = r$, $BC = |b|$, and $OC = c$.
Let $OR = |u|$, with $u$ positive if $R$ is between $O$ and $C$ as shown. Then $CR = |c - u|$ and $PR = \sqrt{u^2 - r^2}$, and by similar triangles,
$$ \frac{\sqrt{u^2 - r^2}}{r} = \pm\frac{c - u}{b}.$$
There are actually three cases represented here:
*
*$b > 0$, shown in the figure;
*$-r < b < 0$, in which case $R$ is on the extension of $\overline{OC}$ beyond $C$, $u > c$, and $\triangle BCR$ has (positive) leg lengths $-b$ and $u - c$; and
*$b < -r$, in which case $R$ is on the extension of $\overline{OC}$ beyond $O$, $u < -r$, and $\triangle BCR$ has (positive) leg lengths $-b$ and $c - u = c + |u|$. This is the case that requires the "$-$" option of the $\pm$ sign.
In all three cases I assume $c > r$.
Squaring both sides of this equation and rearranging terms appropriately,
we get:
$$ (b^2 - r^2)u^2 + 2cr^2 u - (b^2 + c^2)r^2 = 0. \tag 1$$
If $b^2 \neq r^2$ this is a quadratic equation in $u$, and it has roots
$$ u = \frac{-cr^2 \pm br \sqrt{b^2 + c^2 - r^2}}{b^2 - r^2}.$$
We want the positive root if $b > -r$ and the negative root if $b < -r$, so
$$ \angle CBP = \arccos \frac ru
=
\begin{cases}
\arccos \dfrac{b^2 - r^2}{-cr + b \sqrt{b^2 + c^2 - r^2}}
& \text{if $b > r$} \\
\arccos \dfrac{b^2 - r^2}{-cr - b \sqrt{b^2 + c^2 - r^2}}
& \text{if $b < r$ and $b \neq -r$.}
\end{cases}$$
But if $b = r$, then
$ u = \dfrac{c^2+r^2}{2 c} $
and
$$ \angle CBP = \arccos \frac ru
= \arccos \frac{2cr}{c^2+r^2} ,$$
whereas if $b = -r$ then $\angle CBP = \frac\pi2$.
And oh, look, all of these formulas apply to $\angle BAQ$ in the original figure
if we substitute $b - a$ for $b$
(and $b - a < -r$ provided that $\angle BAQ$ is acute, as shown), so
if we assume $b > -r$,
\begin{align}
\theta & = \pi - \angle BAQ - \angle CBP \\
& \begin{aligned}
= \pi & - \arccos \dfrac{(a - b)^2 - r^2}
{-cr + (a - b) \sqrt{(a - b)^2 + c^2 - r^2}} \\
& - \begin{cases}
\arccos \dfrac{b^2 - r^2}{-cr + b \sqrt{b^2 + c^2 - r^2}}
& \text{if $b > r$} \\
\arccos \dfrac{b^2 - r^2}{-cr - b \sqrt{b^2 + c^2 - r^2}}
& \text{if $-r < b < r$} \\
\arccos \dfrac{2cr}{c^2+r^2} & \text{if $b = r$}
\end{cases}
\end{aligned} \\
& = \arcsin \dfrac{(a - b)^2 - r^2}
{-cr + (a - b) \sqrt{(a - b)^2 + c^2 - r^2}} \\
& \qquad\qquad +
\begin{cases}
\arcsin \dfrac{b^2 - r^2}{-cr + b \sqrt{b^2 + c^2 - r^2}}
& \text{if $b > r$} \\
\arcsin \dfrac{b^2 - r^2}{-cr - b \sqrt{b^2 + c^2 - r^2}}
& \text{if $-r < b < r$} \\
\arcsin \dfrac{2cr}{c^2+r^2} & \text{if $b = r$}
\end{cases}
\end{align}
So that's just two trig functions, though there are three cases depending
on the value of $b$.
As suggested in a comment, we could get this down to one trig function if
we could find the three sides of $\triangle ABM$ without using trigonometry;
but I think this would involve saying something about the triangles
$\triangle OPM$ and $\triangle OQM$, and I do not yet see how to do it.
|
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two variable nonhomogeneous inequality Let $$x\ge 0,y\ge 0,x\neq 1,y \neq 1$$Prove the inequality
$$\dfrac {x}{(y-1)^2} +\dfrac {y}{(x-1)^2} \ge \dfrac {x+y-1}{(x-1)(y-1)} $$
|
hint: The endpoints case you can handle with ease, for more general case that: $x, y > 1\to x(x-1)^2 +y(y-1)^2 \geq (x+y-1)(xy-(x+y-1))\iff x(x^2-2x+1)+y(y^2-2y+1)\geq xy(x+y-1)-(x+y-1)^2\iff (x^3+y^3)-2(x^2+y^2)+(x+y)\geq xy(x+y-1)-(x^2+y^2+1+2xy-2x-2y)\iff f(x,y)=x^3+y^3-(x^2+y^2)+1+3xy-(x+y) -xy(x+y)\geq 0$. Taking partial derivatives:
$f_x=3x^2-2x+3y-1-2xy-y^2 = 0=f_y=3y^2-2y+3x-1-x^2-2xy\to f_x-f_y = 0\to (x-y)(4(x+y)-5)=0\to x=y , x+y = \dfrac{5}{4}$. Consider each case separately will yield desire result.
Note: If you want to avoid using calculus, there is another way ( I let you test it to see if it works) that you can prove this at this point:
You prove : $f(x,y) \geq f(y,y)$, and $f(y,y) \geq 0$. The latter inequality is $(y-1)^2 \geq 0$.
Another approach is let $a = x-1,b = y-1$, and assume $a,b > 0\to \dfrac{a+1}{b^2}+\dfrac{b+1}{a^2}\geq \dfrac{a+b+1}{ab}$.But this is quite simple...because $LHS =\dfrac{a}{b^2}+\dfrac{b}{a^2}+\dfrac{1}{a^2}+\dfrac{1}{b^2}\geq \dfrac{a^3+b^3}{a^2b^2}+\dfrac{2}{ab}\geq \dfrac{ab(a+b)}{a^2b^2}+\dfrac{2}{ab}=\dfrac{a+b+2}{ab} > RHS$
|
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Compute $\int_M \omega$
Let $M=\{(x,y,z): z=x^2+y^2, z<1\}$ be a smooth 2-manifold in $\Bbb{R}^3$. Let $\omega=xdy\wedge dz+ydz\wedge dx+zdx\wedge dy\in \Omega^2(\Bbb{R}^3)$. Compute $$\int_M \omega.$$
I parametrised $M$ (up to a null-set) with $g(r,\theta)=(r\cos \theta, r\sin \theta,r)$ where $(r,\theta)\in (0,1)\times(0,2\pi)$, then computed $g^*\omega$ and found that it is zero, hence $$\int_M \omega = \int_{(0,1)\times(0,2\pi)} g^*\omega=0.$$
Is my result correct?
I feel like there might be an easier way to see that the integral is zero, avoiding finding an explicit parametrisation. Is that true? If so, how could one have argued?
|
If $f(x,y)=(x,y,x^2+y^2)$ then $$ \int_M \omega = \int_{D} f^\ast \omega $$ where $D$ is a unit disk. Hence $$ \int_D (-x^2-y^2)dxdy = \int_D (-r^2) rdrd\theta = 2\pi \frac{-r^4}{4}\bigg|_0^1 = -\frac{\pi}{2} $$
|
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|
Get the right "upper" Sequence I want to find out the limit for the sequence $\frac{n!}{n^n}$
using the squeeze theorem.
My idea was:
$\frac{n}{n^n} \leq \frac{n!}{n^n} \leq \frac{n!+1}{n^n} $
So the limits of the smaller sequence and the bigger sequence are 0, thus
the limit of the original sequence is 0.
But is the bigger sequence $\frac{n!+1}{n^n}$ a good choice? Or is there
a more obvious sequence to choose?
|
$$\frac {n!}{n^n} =\frac {n \times (n-1)\times \dots \times 2 \times 1}{n^n}\le \frac {\overbrace {n\times n \times \dots \times n}^{{n-1}\text {times}}\times 1}{n^n}=\frac {n^{n-1}}{n^n}=\frac1n$$
|
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|
Using the $\ln(\cdot)$ for $(1-e^{-x})$ The given function:
$$B= A(1-(e^{-x}))$$
Now, I want to 'destroy' the e-function by taking the logarithm of it.
First, since
$\ln(ab) = \ln(a) + \ln(b)$ we get that $\ln(b) = \ln(a) + \ln(1-e^{-x})$.
Using $\ln(1) =0$, $\ln(b) = \ln(A) +x$.
Is this the right way to handle the function?
Thank you.
|
No it is not true that $\ln(a+b)=\ln a+\ln b$ which you used to split up $\ln(1-e^{-x})$
A better approach is to first solve for the $e^{-x}$
$$e^{-x} = 1-\frac{B}{A}$$
Now take the logarithm.
|
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|
If $t=\tanh\frac{x}{2}$, prove that $\sinh x = \frac{2t}{1-t^2}$ and $\cosh x = \frac{1+t^2}{1-t^2}$.
If $t=\tanh\frac{x}{2}$, prove that $\sinh x = \frac{2t}{1-t^2}$ and $\cosh x = \frac{1+t^2}{1-t^2}$. Hence solve the equation $7\sinh x + 20 \cosh x = 24$.
I have tried starting by writing out $\tanh\frac{x}{2}$ in exponential form and then squaring it but I can't make any progress from this.
|
Note that
(i) tanh($x$) = sinh(x) / cosh($x$)
(ii) 1 - tanh$^2(x$) = 1/ cosh$^2(x$)
(iii) sinh($x + y$) = sinh($x$)cosh($y$) + sinh($y$)cosh($x$)
and apply these to the term
$\frac{2t}{1 - t^2}$ with $t =$ tan($x/2$) to directly obtain your result for sinh($x$). The formula for cosh$(x)$ follows then directly by applying (i) again.
|
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Proofs that Dirichlet's function is not differentiable Define $f: (0,1) \to (0,1)$ by
$f(x)=
\begin{cases}
\frac{1}{q}, & \text{if $x=\frac{p}{q}$ in lowest terms with $p,q \in \mathbb{N}$} \\
0, & \text{if $x$ is irrational}
\end{cases}
$
The version of this proof I found from Spivak's Calculus is, for irrational $a$, say $a=m.a_1a_2a_3\dots$ is the decimal expansion of $a$, consider $[f(a+h)-f(a)]/h$ for $h$ rational, and also for $$h=-0.00\dots0a_{n+1}a_{n+2}\dots$$
Now I tried to prove this using the sequential criterion for limits.
Let $x_n=\frac{p_n}{q_n}$ be in reduced form, a sequence in $(0,1)$ converging to $a$. Then we can write $y_n=(\frac{1}{q_n}-0)/(\frac{p_n}{q_n}-a)=1/[q_n(\frac{p_n}{q_n}-a)]$. I first thought that since $(\frac{p_n}{q_n}-a)$ converges to $0$, $y_n$ would diverge to infinity, but then I realized that $q_n$ may also diverge to infinity, so I cannot guarantee this.
I'm wondering if there's another way using sequential criterion to prove this. Also what other solutions are there? I would greatly appreciate any solution.
|
Fix an irrational $x\in(0,1)$. Suppose $f'(x)$ exists; then $f'(x)=0$. For each prime $q$, pick $r_q$ to be a multiple of $1/q$ satisfying $|x-r_q|\leq 1/q$. Then $|f(x)-f(r_q)|/|x-r_q|\geq 1$. So $|f'(x)|\geq 1$,a contradiction.
|
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|
Integration by Parts Question: Integrate $x^3e^x$ Evaluate $$\int x^3e^x \mathrm{d}x$$
I tried to use integration by parts to do this and I let $u = x^3$ and $\mathrm{d}v = e^x \mathrm{d}x$. So I get $$\int x^3e^x \mathrm{d}x= \int x^3e^x \mathrm{d}x- \int e^x \cdot 3x^2\mathrm{d}x$$
How do I do this if my answer has the original integral in it?
|
$$x^n\int e^x\ dx\ne\int x^ne^x\ dx$$
$$I_n=\int x^ne^x\ dx=x^n\int e^x\ dx-\int\left[\dfrac{d(x^n)}{dx}\int e^x\ dx\right]dx$$
$$\implies I_n=x^ne^x-nI_{n-1}$$
|
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Characterization of uniform integrability of random variables Let $\{X_n \}$ be a sequence of random variables on a probability space $(\Omega,\mathcal{F},P)$.
Then, $\{X_n\}$ is uniformly integrable if $$\lim_{M \to \infty} \sup_n \int_{|X_n| > M } |X_n| = 0 \tag{1}$$
From this, I know that $$\displaystyle \sup_n E(|X_n|) < \infty \tag{2}$$
But, then I am wondering whether (2) implies (1). If not, can you give me a counterexample?
|
No, the converse is, in general, not true. Just consider $((0,1),\mathcal{B}(0,1))$ endowed with Lebesgue measure and the sequence of random variables $$X_n(x) := 2n \cdot 1_{(0,1/n)}(x).$$ Then $$\mathbb{E}(|X_n|) = 2$$ for all $n \in \mathbb{N}$, but $$\limsup_{M \to \infty} \sup_{n \in \mathbb{N}} \int_{|X_n|>M} |X_n| \, d\mathbb{P} \geq \int_{|X_M|>M} |X_M| \, d\mathbb{P} = 2.$$
Remark: If the sequence $(X_n)_{n \in \mathbb{N}}$ is bounded in $L^p$ for some $p>1$, i.e. $$\sup_{n \in \mathbb{N}} \mathbb{E}(|X_n|^p)<\infty,$$ then $(X_n)_{n \in \mathbb{N}}$ is uniformly integrable.
|
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solution of variable coefficient equation Consider the equation $$u_x + yu_y = 0$$
and I know that this PDE has solution $u(x,y) = f(e^{-x}y)$
Can someone help me to derive this PDE to get the solution? Thank you
|
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dx}{dt}=1$ , letting $x(0)=0$ , we have $x=t$
$\dfrac{dy}{dt}=y$ , letting $y(0)=y_0$ , we have $y=y_0e^t=y_0e^x$
$\dfrac{du}{ds}=0$ , letting $u(0)=f(y_0)$ , we have $u(x,y)=f(y_0)=f(e^{-x}y)$
|
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Find the number of real roots of $1+x/1!+x^2/2!+x^3/3! + \ldots + x^6/6! =0$.
Find the number of real roots of $1+x/1!+x^2/2!+x^3/3! + \ldots + x^6/6! =0$.
Attempts so far:
Used Descartes signs stuff so possible number of real roots is $6,4,2,0$
tried differentiating the equation $4$ times and got an equation with no roots hence proving that above polynomial has $4$ real roots.
But using online calculators I get zero real roots. Where am I wrong?
|
I think the notion that the fourth derivative having no real roots proves that the polynomial itself has four real roots is your problem. Can you explain your reasoning a bit more? I mean to say, $x^6+x^4+1$ clearly has no real roots (it is everywhere positive), but its fourth derivative $360x^2+24$ has no real roots either (it is likewise everywhere positive).
The polynomial in your problem does indeed have no real roots.
|
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|
showing an inequality not using stirling formula I don't know how to show that
$$
\frac{k^k}{k!}\leq e^{k}
$$
without using Stirling's approximation. I want to show it directly. I guess I need some inequality to achieve this but I don't know.
|
A somewhat different approach:
In the pmf for a Poisson$(k)$ $P(X=k) = \frac{e^{-k} k^k}{k!}$ but $P(X=k)<1$ (since it's a probability).
The result follows by multiplying by $e^k$.
|
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Division in Summations Suppose $a_n=\dfrac{2^n}{n(n+2)}$ and $b_n=\dfrac{3^n}{5n+18}$.
I need to find the value of:
$$\displaystyle\sum_{n=1}^{\infty}\dfrac{a_n}{b_n}$$
I think this problem is meant for me to compute each sum differently and then divide. Is this some property of summations that we need to utilize here?
|
$$\begin{equation}
\begin{split}
\sum_{n=1}^\infty \frac{a_n}{b_n}
&
= \sum_{n=1}^\infty \frac{2^n(5n+18)}{3^nn(n+2)} \\
&
= \sum_{n=1}^\infty \left[ \left(\frac{2}{3}\right)^n \left(\frac{5}{n+2} + 9\left(\frac{1}{n} - \frac{1}{n+2} \right) \right) \right] \\
&
= \sum_{n=1}^\infty \left[ \left(\frac{2}{3}\right)^n \left(\frac{9}{n} - \frac{4}{n+2}\right) \right] \\
&
= \sum_{n=1}^\infty \frac{2^n}{3^{n-2}n} - \sum_{n=1}^\infty \frac{2^{n+2}}{3^n(n+2)} \\
&
= \sum_{n=1}^\infty \frac{2^n}{3^{n-2}n} - \sum_{n=3}^\infty \frac{2^{n}}{3^{n-2}n} \\
&
= \sum_{n=1}^2 \frac{2^n}{3^{n-2}n} = \frac{2}{3^{-1}} + \frac{2^2}{3^{0}2} \\
&
= 6 + 2 \\
&
= 8 \end{split} \end{equation} $$
|
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Jacobson radical of $\mathbb{F}_{2}(t)[x]/(x^4-t^2)$ Let $\mathbb{F}_{2}$ be the field of two elements. Let $R=\mathbb{F}_{2}(t)[x]/(x^4-t^2)$.
Why is $R/J(R)$ equal to $\mathbb{F}_{2}(t)[x]/(t-x^2)$? here $J(R)$ denotes the Jacobson radical of $R$.
|
The Jacobson radical of $R$ is, by definition, the intersection of maximal ideals of $R$. In our case $R$ is local with the maximal ideal $(x^2+t)/(x^2+t)^2$. If we want to see this as an $R$-module, then we can conclude that it is isomorphic to $R/(x^2+t)$, but I can't see how is this isomorphic to $\mathbb F_2(t)[x]/(x^2+t)$ (which is actually isomorphic to $R/J(R)$).
|
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Why does basic algebra provide one value for $x$ when there should be two? I have the equation $x^2=x$.
If I divide $x$ from both sides I get $x=1$.
Yet clearly $x$ can also equal $0$.
What step in this process is wrong? It seems to me that there's only one step. And isn't dividing the same thing from both sides a valid step?
I hope this isn't a stupid question because I feel dumb asking about something so basic.
EDIT: To clarify, what I'm looking for is not only an explanation for why my methodology is wrong, but also a better methodology that will keep me from missing possible solutions in the future.
|
The given quadratic equation $x^2=x$ will have two real roots given as follows $$x^2-x=0$$ $$x(x-1)=0$$ $$x=0\ \ \ \ \textrm{or}\ \ \ x-1=0$$
$$x=0\ \ \ \ \textrm{or}\ \ \ x=1$$ Hence, we get $x=0$ or $x=1$
|
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Prove that if $A$, $B$ are countable, then $A \times B$ is countable? Is $A\times B$ referring to the axis here? So an $X$ and $Y$ coordinate plane?
$A$ is countable, therefore a bijection occurs from $A \rightarrow \mathbb{N}$.
$B$ is countable, therefore a bijection occurs from $B \rightarrow \mathbb{N}$.
If these statements hold true then any elements belonging to the sets can be linked and ordered one-by-one from any natural number, thus any coordinate $(a,b)$ would be countable as the elements it represents ($x$ and $y $ coordinates) are countable. Together they preserve their nature of cardinality.
Am I wrong?
|
Let $f:A \to \mathbb{N}$ be a bijection from $A$ to $\mathbb{N}$
Let $g:B \to \mathbb{N}$ be a bijection from $B$ to $\mathbb{N}$
Define $h:A \times B \to \mathbb{N}$ as follows:
$h(x,y)=2^{f(x)}\cdot3^{g(y)}$
|
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Is this proof about the countability of $\Bbb Q \times \Bbb Q \times \cdots \times \Bbb Q$ sound? If $\Bbb{Q}$ is countable, prove that the set $\Bbb{Q}^n$ for $n = 2,3,...$ is countable.
Base case: $n = 2 \rightarrow \Bbb{Q}^2 = \Bbb{Q}\times\Bbb{Q}$ which, by Proposition 4.5 (see bottom of question), is countable.
Assume the statement is true for some $k \in \Bbb{Z}, k \ge 2,$ s.t., $\Bbb{Q}^{k}$ is countable.
Then $\Bbb{Q}^{k+1} = \Bbb{Q}^{k}\Bbb{Q}^{1}$ which is a relation $\Bbb{Q}^{k}\times\Bbb{Q}^{1}$.
$\Bbb{Q}^{k}$ is countable by the inductive hypothesis and $\Bbb{Q}^{1}$ was introduced as a countable set. By Proposition 4.5, $\Bbb{Q}^{k+1}$ is a countable set.
By the PMI, $\Bbb{Q}^n$ is countable for $n = 2,3,...$
[Proposition 4.5: Let D be a countable set. Then D$\times$D (the set of all pairs $(x,y), x,y \in D$) is countable.]
|
As you wrote in a comment, the proposition you are trying to use doesn't apply directly to $\Bbb Q^k\times \Bbb Q$.
However, if we know that $D$ is countable and $E$ is some other countable set, then there is a bijection $f: D \to E$, and the map from $D \times D$ to $D \times E$ given by $(x,y) \mapsto (x,f(y))$ is a bijection. Since by the proposition $D \times D$ is countable, we have shown that $D \times E$ is countable too.
This one you can apply to your proof, and the rest works as you wrote it.
Assume the statement is true for some $k \in \Bbb{Z},$ s.t., $\Bbb{Q}^{k}$
Are you missing something at the end of this sentence? I think it should read either "Assume the statement is true for some $k \in \Bbb{Z}$" or "Assume that for some $k \in \Bbb{Z},$ s.t., $\Bbb{Q}^{k}$ is countable".
|
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Group $G$ acting on $\Omega$ such that each $\alpha \in \Omega$ has unique $p$-element fixing $\alpha$.
Let $G$ be a group acting on a set $\Omega$ and let $p$ be a prime. Suppose that for each $\alpha \in\Omega$ there is a $p$-element $x \in G$ such that $\alpha$ is the only point fixed by $x$. If $\Omega$ is finite, show that $G$ is transitive on $\Omega$; and if $\Omega$ is infinite, show that $G$ has no finite orbit on $\Omega$.
Any hints how to solve this problem?
EDIT: My initial thoughts were all wrong as I am assumed uniqueness of the $p$-element there.
|
First suppose that $\Omega$ is finite, and so $G/K$ is finite, where $K$ is the kernel of the action, so we can assume that $K=1$ and $G$ is finite.
For any $\alpha \in \Omega$, there is a nontrivial $p$-subgroup $Q(\alpha)$ of $G$ with the unique fixed point $\alpha$. Let $P(\alpha)$ be a Sylow $p$-subgroup of $G$ containing $Q(\alpha)$. Since the orbits of $P$ are unions of orbits of $Q$ and have length a power of $p$, $P(\alpha)$ must fix $\alpha$, which is its unique fixed point.
Now for $\alpha,\beta \in \Omega$, an element $g \in G$ conjugating $P(\alpha)$ to $P(\beta)$ must map $\alpha$ to $\beta$, so $G$ is transitive.
Now suppose that $\Omega$ is infinite and has a finite orbit $\Delta$. Then $|\Delta| \equiv 1 \bmod p$. Let $\alpha \in \Omega \setminus \Delta$. Any $p$-element fixing $\alpha$ must also fix some point in $\Delta$, contrary to assumption.
|
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computing the cubed root of a complex number... I do know how to calculate the cubed root of a complex number....like if I'm given that $x^3=p$, where $p$ is a complex number, then $$x= r^{1/3}\left(\cos\left(\frac{2k\pi+m}{3}\right) + i\sin \left(\frac{2k\pi+m}{3}\right)\right)$$ where $p$ is $r\left(\cos m +i\sin m\right)$ and $k=0,1,2$
But can I write it in this way??
$$x^3=p \implies x=p^{1/3},\,p^{1/3}w , \,p^{1/3} w^2, \dots$$
where $w$ is the cubed root of unity?
I thought that if I write the roots in this way and put any value of $p^{1/3}$, I get all the three values of $x$ what I got using the first method...thanks
|
HINT: use that $$x^3-a=(x-a^{1/3})(x^2+xa^{1/3}+a^{2/3})$$
|
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"source": "stackexchange",
"question_score": "2",
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|
Prove that $\sum\frac{(\log n)^2}{n^3}$ converges This question is from Serge Lang's textbook, in a chapter that comes before the ratio and integral tests are introduced, so those can't be used. I've already proved that $\sum\frac{\log n}{n^3}$ converges and have an inkling that this result may be useful, but I can't figure out how.
|
Well then use $\ln(n)^2 = 4\ln(\sqrt{n})^2 < 4\sqrt{n}^2$.
|
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|
Find a basis for $U\cap V$
Let $$a = (0,2,3,-1)^T \quad b=(0,2,7,-2)^T \quad c = (0,-2,1,0)^T \quad u = (1,2,0,1)^T\quad v = (2,2,1,2)^T$$
Let $U= \langle a,b,c \rangle, V = \langle u,v\rangle$
Then a) find a basis for $U$ and $V$, b) find a basis for $U\cap V$.
I have solved a) and found a basis $\{a,b\}$ for $U$ and $\{u,v\}$ for V. But How can I combine this information to find a basis for $U\cap V$?
I have tried:
Let $x \in U\cap V$, then $x = \lambda_1 a+\lambda_2 b = \mu_1u+\mu_2 v$.
Solving $\lambda_1 a + \lambda_2 b - \mu_1 u - \mu_2 v = 0$ results in
$$\lambda_1 = -2r\quad \lambda_2 = r\quad \mu_1 = -2r\quad \mu_2 = r\quad (\forall r \in \mathbb{Q})$$
This means $a,b,u, v$ are lineair dependent. But now what? I know I can remove one basisvector and this would result in set of independent basis vectors, but doesn't that break the spanning property?
I'm somewhat confused, could someone clarify, no solutions please.
|
Now you have two ways of expressing your vector $x$ in terms of a single arbitrary value $r$. Since $x$ was arbitrary any element of the intersection has this form. So use one of the two ways to find a fixed vector $w$ such that your arbitrary $x = rw$ for some $r$. As a check to your work so far, you should get the same $w$ from using either $a, b$ or from $u, v$, (up to a constant multiplier). Now can you figure out what the basis of $U \cup V$ is?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1398300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Order of a corrector-predictor method Given an explicit method:
$$ x_{i+1} = x_i+ h \Phi(t_i,x_i,h) $$
as predictor method and an implicit method:
$$ x_{i+1} = x_i + h \Psi(t_i,x_i,x_{i+1},h) $$
as corrector method, it follows that
$$
x_{i+1} = x_i + h \Psi(t_i,x_i,x_{i+1}^*, h), \quad
(x^*_{i+1} = x_i+\Phi(t_i,x_i,h))
$$
is an explicit predictor-corrector method. If $\tau_p(h)$, $\tau_c(h)$ and $\tau_{pc}(h)$ denote respectively the local truncation error of predictor, corrector and predictor-corrector method, then
$$ \tau_{pc}(h) = \tau_c(h) + \mathcal O(h \tau_p(h)) $$
(which is what I haven't proved).
From the definition of local truncation error, we have:
$$
\begin{array}{rlrl}
x^{(*)}(t_{i+1}) &= x(t_i) + h \Phi(t_i, x(t_i), h) + h \tau_p(t_i,h),& \tau_p(h) &= \max_i \tau_p(t_i,h)\\
x(t_{i+1}) &= x(t_i) + h \Psi(t_i, x(t_i), x(t_{i}+h), h) + h \tau_c(t_i,h),& \tau_c(h) &= \max_i \tau_c(t_i,h)\\
x(t_{i+1}) &= x(t_i) + h \Psi(t_i, x(t_i), x^*(t_i+h), h) + h \tau_{pc}(t_i,h),& \tau_{pc}(h) &= \max_i \tau_{pc}(t_i,h)\\
\end{array}
$$
So, substracting from the third equation the second ecuation, and since $h > 0$:
$$ \tau_{pc}(t_i,h) = \tau_c(t_i,h) - [ \Psi(t_i,x(t_i),x(t_{i+1}),h)-\Psi(t_i,x(t_i),x^*(t_{i+1}),h)] $$
And only is needed to prove that the last substraction is $\mathcal O(h \tau_p(h))$. So,
$$
\|\Psi(t_i,x(t_i),x(t_{i+1}),h)-\Psi(t_i,x(t_i),x^*(t_{i+1}),h)\|\leq
L\|x(t_{i+1})-x^*(t_{i+1})\|
$$
$$
\leq L\|x(t_{i+1})-x(t_i)-h \Phi(t_i,x(t_i),h)-h\tau_p(t_i,h)\|\leq L\|...\| + Lh\|\tau_p(h)\|
$$
where $L$ is a Lipschitz constant.
How could I justify this? Any hint? Thanks in advance.
|
Pure predictor and pure corrector schemes
$$
\begin{array}{rlrlc}
x(t_{i+1}) &= x(t_i) + h \Phi(t_i, x(t_i), h) + h \tau_p(t_i,h),& \tau_p(h) &= \max_i |\tau_p(t_i,h)| & (1)\\
x(t_{i+1}) &= x(t_i) + h \Psi(t_i, x(t_i), x(t_{i+1}), h) + h \tau_c(t_i,h),& \tau_c(h) &= \max_i |\tau_c(t_i,h)| & (2)
\end{array}
$$
And the predictor-corrector scheme
$$
\begin{array}{rlrlc}
x^* &= x(t_i) + h \Phi(t_i, x(t_i), h) & & & (3)\\
x(t_{i+1}) &= x(t_i) + h \Psi(t_i, x(t_i), x^*, h) + h \tau_{pc}(t_i,h),& \tau_{pc}(h) &= \max_i |\tau_{pc}(t_i,h)| & (4)
\end{array}
$$
Subtracting $(2)$ from $(4)$:
$$
\tau_{pc}(t_i, h) = \tau_{c}(t_i, h) + \Psi(t_i, x(t_i), x(t_{i+1}), h) - \Psi(t_i, x(t_i), x^*, h)
$$
Thus
$$
\tau_{pc}(h) \leq \tau_{c}(h) + L ||x(t_{i+1}) - x^*||
$$
Now subtract $(1)$ from $(3)$ and get the desired result.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Determining if a map on a space is continuous by checking on a dense subset I recently read a proof of something my calculus teacher had told me, namely that the set of continuous maps $f: \mathbb{R} \to \mathbb{R}$ had the cardinality of $\mathbb{R}$. The proof was simple enough: There must be at least that many, as it includes all constant maps, but can be no more because we can injectively map the continuous maps on $\mathbb{R}$ to $\mathbb{R}^{\mathbb{Q}}$ by $\phi : f \mapsto (f(q))_{q \in \mathbb{Q}}$, and $\mathbb{R}^{\mathbb{Q}}$ has the same cardinality as $\mathbb{R}$, as this is enough to determine the value of $f$ on any real value by taking $f(x) = \lim_{q \to x, q \in \mathbb{Q}} f(q)$; that is, assuming the map is continuous.
But there's also the following: Let $g: \mathbb{Q} \to \mathbb{R}$ be given by
\begin{align*}
g(x) & = \begin{cases}
0, & x < \sqrt{2} \\
1, & x > \sqrt{2} .
\end{cases}
\end{align*}
Then $g$ is continuous on $\mathbb{Q}$, but does not extend to a continuous map on $\mathbb{R}$. So my question is: Is there a way to determine if a map $g : \mathbb{Q} \to \mathbb{R}$ extends to a continuous map $f: \mathbb{R} \to \mathbb{R}$?
Thanks.
|
It is necessary and sufficient that for any Cauchy sequence $\{x_n\}$ of rational numbers, the sequence of $\{f(x_n)\}$ is also Cauchy.
Since the real numbers are the completion of the rational numbers, if a sequence in the rational numbers converges (i.e., is Cauchy), then it converges to a real number, and you need the values of $f$ to converge as well (so it must also be Cauchy).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1398601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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|
Derivative of integral $\int_0^{\infty} e^{-x \cosh t} dt$ I am given the integral $y = \int_0^{\infty}e^{-x\cosh t} dt$ and wish to show that this integral solves the modified Bessell equation: $x^2y'' + xy' -x^2y = 0.$
To do this I need to calculate the derivatives $y'$ and $y''$. I suspect integration by parts is necessary, but do not know how to continue. I have tried substituting $e^-x\cosh t$ and $\cosh t$ by their Taylor series and integrating term by term, but this does not seem to simplify things.
My question is: How do I calculating $y'$?
|
You can check the equation as follows:
\begin{align*}
x^2(y'' - y)
&= x \int_{0}^{\infty} x (\cosh^2 t - 1) e^{-x \cosh t} \, dt \\
&= x \int_{0}^{\infty} x \sinh^2 t \, e^{-x \cosh t} \, dt \\
&= x \left[ -\sinh t \, e^{-x \cosh t} \right]_{0}^{\infty} + x \int_{0}^{\infty} \cosh t \, e^{-x\cosh t} \, dt \\
&= -xy'.
\end{align*}
@anthus, I agree that Leibniz integral rule (interchanging of integration and differentiation) is often not obvious, and is simply false in some cleverly designed examples.
In this case, however, we have a simple idea to circumvent such complication: think everything backward!
For example, suppose that you want to know that $y$ is differentiable with
$$ y'(x) = -\int_{0}^{\infty} \cosh t \, e^{-x \cosh t} \, dt. \tag{*} $$
To this end, let us work backward: That is, we will show that $y$ is an antiderivative of the right-hand side.
Denote the RHS of (*) as $y_1 (x)$. Clearly $y_1 (x)$ is a well-defined function. Now for any $0 < a < b$, Tonelli's theorem guarantees that we can interchange the order of integration and we have
\begin{align*}
\int_{a}^{b} y_1(x) \, dx
&= \int_{0}^{\infty} \int_{a}^{b} (-\cosh t) e^{-x \cosh t} \, dt dx \\
&= \int_{0}^{\infty} (e^{-b \cosh t} - e^{-a \cosh t}) \, dt \\
&= y(b) - y(a).
\end{align*}
This means that $y$ is an antiderivative of $y_1$, hence $y$ is differentiable and $y' = y_1$.
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
$\int_{0}^{\infty}\frac{\ln x dx}{x^2+2x+2}$ $$\int_{0}^{\infty}\frac{\ln x .dx}{x^2+2x+2}$$
$$\int_{0}^{\infty}\frac{\ln x .dx}{x^2+2x+2}=\int_{0}^{\infty}\frac{\ln x .dx}{(x+1)^2+1}\\
=\ln x\int_{0}^{\infty}\frac{1}{(x+1)^2+1}-\int_{0}^{\infty}\frac{1}{x}\frac{1}{(x+1)^2+1}dx$$ and then lost track,answer is $\frac{\pi \ln 2}{8}$.
Any hint will solve my problem.
|
$\bf{Another\; Solution::}$ Let $$\displaystyle I = \int_{0}^{\infty}\frac{\ln x}{x^2+2x+2}dx.$$
Put $\displaystyle x = \frac{2}{t}\;,$ Then $\displaystyle dx = -\frac{2}{t^2}dt$ and Changing Limits, We get
$$\displaystyle I = \int_{\infty}^{0}\frac{\ln\left(\frac{2}{t}\right)}{\frac{4}{t^2}+\frac{4}{t}+2}\cdot -\frac{2}{t^2}dt = \int_{0}^{\infty}\frac{\left(\ln 2-\ln t\right)}{t^2+2t+2}dt$$
Now Using $$\displaystyle \bullet \int_{a}^{b}f(x)dx = -\int_{b}^{a}f(x)dx$$ and $$\displaystyle \bullet \int_{a}^{b}f(t)dt = \int_{a}^{b}f(x)dx$$
So $$\displaystyle I = \ln 2\int_{0}^{\infty}\frac{1}{x^2+2x+2}dx - \int_{0}^{\infty}\frac{\ln x}{x^2+2x+2}dx = \ln 2\int_{0}^{\infty}\frac{1}{(x+1)^2+1^2}dx-I$$
So we get $$\displaystyle I = \ln 2\times \left[\tan^{-1}\left(x+1\right)\right]_{0}^{\infty}-I$$
So we get $$\displaystyle I = \frac{\ln 2}{2} \times \left[\frac{\pi}{2}-\frac{\pi}{4}\right] = \frac{\pi}{8}\cdot \ln 2$$
|
{
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"url": "https://math.stackexchange.com/questions/1398875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Evaluate $\int e^x \sin^2 x \mathrm{d}x$ Is the following evaluation of correct?
\begin{align*} \int e^x \sin^2 x \mathrm{d}x &= e^x \sin^2 x -2\int e^x \sin x \cos x \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int e^x (\cos^2 x - \sin^2x) \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int e^x (1 - 2\sin^2x) \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int -2 e^x \sin^2x \mathrm{d}x + 2 e^x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 e^x -4 \int e^x \sin^2x \mathrm{d}x \end{align*}
First two steps use integration by parts. In the first step we differentiate $\sin^2 x$. In the second step we differentiate $\sin x \cos x$.
Using this, we reach $$5\int e^x \sin^2 x \mathrm{d}x = e^x \sin^2 x -2e^x \sin x \cos x + 2 e^x$$
$$\int e^x \sin^2 x \mathrm{d}x = \frac{e^x \sin^2 x -2e^x \sin x \cos x + 2 e^x}{5}+C$$
I can't reach the form that most integral calculators give, which has terms $\cos(2x)$ and $\sin(2x)$ by just using trig identities, so I wonder whether the result is correct. I would also be interested in a method that immediately gives the form $$-\frac{e^x[2 \sin(2x)+ \cos(2x)-5]}{10}+C$$
|
$$\int \left(e^x\sin^2(x)\right)\text{d}x =$$
$$\int \left(e^x\left(\frac{1}{2}(1-\cos(2x))\right)\right)\text{d}x =$$
$$\frac{1}{2}\int \left(e^x-e^x\cos(2x)\right)\text{d}x =$$
$$\frac{1}{2} \left(\int \left(e^x\right) \text{d}x-\int \left(e^x\cos(2x)\right) \text{d}x\right) =$$
$$\frac{1}{2} \left(\int e^x \text{d}x-\int e^x\cos(2x) \text{d}x\right) =$$
$$\frac{1}{2} \left(e^x-\int e^x\cos(2x) \text{d}x\right) =$$
For the integrand $e^x\cos(2x)$, use the formula:
$$\int\exp(\alpha x)\cos(\beta x)\text{d}x=\frac{\exp(\alpha x)(\alpha \cos(\beta x))+\beta\sin(\beta x)}{\alpha^2+\beta^2}$$
$$\frac{1}{2} \left(e^x-\frac{e^x(2\sin(2x)+\cos(2x))}{5}\right) + C =$$
$$-\frac{e^x(2\sin(2x)+\cos(2x)-5)}{10} + C $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1398965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Left cosets of $A_6$ in $S_6$ Which may be the all of left cosets of $A_6$ in $S_6$?
$\{A_6,(156)A_6\},\{A_6,(34)A_6\},\{A_6,(42)(35)A_6\}, \{A_6,(46523)A_6\}$ or $\{A_6\}$
I dont understand why the answer is $\{A_6,(34)A_6\}$ .I know if we multiply $A_6$ elements in $S_6$ with $A_6$ we get $A_6$ itself. why multiplying $S_6-A_6$ element with $A_6$ is equal to $(36)A_6$?
|
$A_{6}$ is the group of all even permutations of $6$ elements. If you multiply that set by an even permutation, you get $A_{6}$ back, but if you multiply by any odd permutation you get every odd permutation (you need to prove that every distinct even permutation is sent to a distinct odd permutation). So the problem is just: which of those 5 permutations is odd? I leave that to you.
|
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"timestamp": "2023-03-29T00:00:00",
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|
For which $n \in \mathbb{N}$ $f(x) = x^{2n}+x^n+1$ is irreducible in $\mathbb{F}_2[x]$? I have $$f(x) = x^{2n}+x^n+1 \in \mathbb{F}_2[x].$$ When is this polynomial irreducible? It is obvious that for even $n$ this polynomial is reducible.
But I don't have any idea about odd $n$.
|
Before we delve into $\Bbb{F}_2[x]$ let's consider the irreducibility of $f(x)$ in $\Bbb{Z}[x]$.
Because
$$
f(x)(x^n-1)=x^{3n}-1,
$$
the zeros of $f$ are roots of unity of order dividing $3n$.This makes us consider the cyclotomic polynomial $\Phi_{3n}(x)\in \Bbb{Z}[x]$. Its degree is given by the Euler (totient) function $\phi(3n)$. If $n$ is divisible by an odd prime $p$ other than $3$, then
$$
\phi(3n)\le \frac{3-1}3\cdot\frac{p-1}p\cdot3n<2n.
$$
This means that for $f(x)$ to be irreducible in $\Bbb{Z}[x]$ it is necessary that $n$ is a power of three. The usual business with cyclotomic polynomials shows that this condition is also sufficient for irreducibility in $\Bbb{Z}[x]$.
A less trivial fact is that when $n=3^k$ the polynomial $f(x)$ remains irreducible in $\Bbb{F}_2[x]$ as well. This is a consequence of the fact that $2$ is a primitive root modulo $3^{k+1}$. I don't want to repeat the argument here. Instead, I refer you to an earlier answer of mine. See the addendum for the gory details.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Understanding the Group Structure A Group is an algebraic structure consisting of a set of elements together with an operation that combines any two elements to form a third element. The operation satisfies four conditions
*
*Closure
*Associativity
*Existence of Identity
*Existence of Inverse
Intuitively I understand the purpose of Closure and Associativity Property. But I'm not getting the intuition behind Identity and Inverse. Whats the purpose of having these elements in a group.
I searched everywhere and find only common definition which are straight forward. But nobody discuss why it's important. Can anybody explain it to me.
|
Groups abstract symmetries of a set, that is, bijections $X \to X$ of a set $X$:
*
*Function composition is associative.
*Every bijection has an inverse, which is also a bijection.
*The identity map is a bijection.
Closure just allows us to consider sets of bijections that are smaller than the full group of all bijections $X \to X$.
All groups are groups of symmetries of a set: that's Cayley's theorem.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1399268",
"timestamp": "2023-03-29T00:00:00",
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|
Can we make a subgroup of a group by selecting exactly one element from each distinct left cosets of a subgroup of the given group? Let $G$ be a group and $H$ be a subgroup of $G$ ; can we select exactly one element from each distinct left coset of $H$ such that the set of all those elements form a subgroup of $G$ ? How do we characterize those groups and for given groups , those subgroups , for which it is possible ?
|
If $H$ is a retract of $G$ or a kernel of a retraction of $G$ then it is possible (in other words: if $H$ is a semidirect factor of $G$). But I am not sure if that exhausts all possibilities.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1399352",
"timestamp": "2023-03-29T00:00:00",
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|
Theorem 3.22 from baby Rudin $\sum a_n$ converges if and only if for every $\varepsilon >0$ there is an integer $N$ such that $$\left|\sum_{k=n}^{m}a_k\right|\leqslant \varepsilon$$ if $m\geqslant n\geqslant N$.
In particular, by taking $m=n$ above inequality becomes $$|a_n|\leqslant \varepsilon \quad(n\geqslant N).$$
In other words: If $\sum a_n$ converges, then $\lim_{n\to \infty}a_n=0$
The condition $a_n\to 0$ is not, however, sufficient to ensure convergence of $\sum a_n$. For instance. the series $\sum \frac{1}{n}$ diverges.
Reading all this I have one question.
If for every $\varepsilon >0$ there is an integer $N$ such that $\left|\sum_{k=n}^{m}a_k\right|\leqslant \varepsilon$ if $m\geqslant n\geqslant N$ then $\sum a_n$ converges.
If we put here $m=n$ why can not conclude that $\sum a_n$ converges? Where is the mistake?
|
The mistake is that
$$
\sum_{k=m}^m a_k = a_m
$$
and you are saying nothing about the series $\sum_k a_k$. You are only looking at the single term $a_m$.
|
{
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|
Find the derivative of $f(x) = \int_{-\infty}^\infty \frac{e^{-xy^2}}{1+y^2}\ dy.$ Problem statement: Find the derivative of
$$f(x) = \int_{-\infty}^\infty \frac{e^{-xy^2}}{1+y^2}\ dy$$
and find an ordinary differential equation that $f$ solves. Find the solution to this ordinary differential equation to determine an explicit value for $f$.
My attempt: Normally, to find $f'(x)$ in this situation, if this was a calculus problem, I would write:
$$\frac{d}{dx} f(x) = \frac{d}{dx}\int_{-\infty}^\infty \frac{e^{-xy^2}}{1+y^2}\ dy = \frac{d}{dx}\left(\int_{-\infty}^0\frac{e^{-xy^2}}{1+y^2}\ dy + \int_0^\infty \frac{e^{-xy^2}}{1+y^2}\ dy\right)$$
and then I would differentiate under the integral sign:
$$f'(x) = \int_{-\infty}^0\frac{\partial}{\partial x}\frac{e^{-xy^2}}{1+y^2}\ dy + \int_0^\infty \frac{\partial}{\partial x}\frac{e^{-xy^2}}{1+y^2}\ dy,$$
which leaves me with
$$f'(x) = \int_{-\infty}^0 \frac{-y^2e^{-x(y^2+1)}}{1+y^2}\ dy + \int_{0}^\infty \frac{-y^2e^{-x(y^2+1)}}{1+y^2}\ dy.$$
However, this is actually an analysis problem, and as such I am having a really hard time justifying all of these steps. I have already proved that if $F(x) = \int_a^x f(y) \ dy$, then $F$ is absolutely continuous, and therefore the derivative exists a.e. Now, I also know from the FCT that if $f$ is integrable, $F'(x) = f(y)$ a.e. But I still can't quite figure out how to justify moving the derivative in the integral sign, particularly when I have infinite bounds, which I do. Any help would be much appreciated here!
|
Using Differentiation Under the Integral Sign, we get
$$
\begin{align}
f(x)&=\int_{-\infty}^\infty\frac{e^{-xy^2}}{1+y^2}\,\mathrm{d}y\tag{1}\\
f'(x)&=\int_{-\infty}^\infty\frac{-y^2e^{-xy^2}}{1+y^2}\,\mathrm{d}y\tag{2}\\
f(x)-f'(x)&=\int_{-\infty}^\infty e^{-xy^2}\,\mathrm{d}y\tag{3}\\
&=\sqrt{\frac\pi{x}}\tag{4}
\end{align}
$$
Explanation:
$(1)$: Given
$(2)$: Differentiation Under the Integral Sign. Alternatively, use Fubini's Theorem to show
$$\begin{align}
f(x)
&=\int_{-\infty}^\infty\frac{e^{-xy^2}}{1+y^2}\,\mathrm{d}y\\
&=\int_{-\infty}^\infty\int_x^\infty\frac{y^2e^{-ty^2}}{1+y^2}\,\mathrm{d}t\,\mathrm{d}y\\
&=\int_x^\infty\int_{-\infty}^\infty\frac{y^2e^{-ty^2}}{1+y^2}\,\mathrm{d}y\,\mathrm{d}t\\
\end{align}$$
Then the FTC to show that
$$f'(x)=\int_{-\infty}^\infty\frac{-y^2e^{-xy^2}}{1+y^2}\,\mathrm{d}y
$$
$(3)$: subtract $(2)$ from $(1)$
$(4)$: evaluate the integral in $(3)$
|
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|
Trig limit in Spivak's Calculus $$\lim_{x\rightarrow 1} (x-1)^3 \sin\frac{1}{(1-x)^3} = 0$$
To prove that this is true, the chapter on limits has things like $\lim_{x\rightarrow a}(f\cdot g)(x) = \lim_{x\rightarrow a}f(x)\cdot \lim_{x\rightarrow a} g(x)$, when both limits exist. Now, in the case of the problem, $\lim_{x\rightarrow 1} (x-1)^3 = 0$, but $\lim_{x\rightarrow 1} \sin\frac{1}{(1-x)^3}$ doesn't exist. However, it is bounded, and because the first limit tends to zero, it is clear that the multiplication tends to zero. But since there is a discontinuity where the limit doesn't exist, I can't find a way to express this formally in the terms of the theorems that I should know at this point.
So there's my doubt, I guess. How should I proceed?
|
$$
\underbrace{-|x-1|^3}_A \quad \le \quad \underbrace{|x-1|^3\cdot\sin(\cdots)}_B \quad \le \quad \underbrace{|x-1|^3}_C.
$$
You don't need to know the nature of the function of $x$ inside the sine function in order to deduce the inequalities $A\le B\le C$. If $A$ and $C$ both approach $0$, then so does $B$. Most calculus textbooks have a theorem that states this explicitly.
|
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|
Why do we not have to prove definitions? I am a beginning level math student and I read recently (in a book written by a Ph. D in Mathematical Education) that mathematical definitions do not get "proven." As in they can't be proven. Why not? It seems like some definitions should have a foundation based on proof. How simple (or intuitive) does something have to be to become a definition? I mean to ask this and get a clear answer. Hopefully this is not an opinion-based question, and if it is will someone please provide the answer: "opinion based question."
|
Frequently, a definition is given, and then an example or proof follows to show that whatever has been defined actually exists. Some authors will also attempt to motivate a definition before they give it: for example, by studying the symmetries of triangles and squares and how those symmetries are related to each other before going on to define a general group.
A definition is distinguished from a theorem or proposition or lemma in that a definition does not declare some fact to be true, it merely assigns meaning to some group of words or symbols. The statement of a theorem says that "such-and-such" thing is true, and then must back up the claim with a proof.
|
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|
Find all integers such that $2 < x < 2014$ and $2015|(x^2-x)$ Find all integers, $x$, such that $2 < x < 2014$ and $2015|(x^2-x)$.
I factored it and now I know that $x > 45$ and I have found one solution so far: $(156)(155)= (2015)(12)$. It's just that I don't think I'm approaching it the right way.
I'm new to this kind of stuff and I'm just doing this for fun so any help please?
|
You need a couple of pieces of information to complete this.
First of all, a property of prime numbers: If $p$ is a prime number such that $p ~|~ ab$, then $p~|~a$ or $p~|~b$. Also $2015=5 \cdot 13\cdot 31$.
Now, if $2015~|~N$, then $5~|~N$. Since $2015~|~x(x-1)$, $5~|~x$ or $5~|~(x-1)$; thus $x\equiv0\pmod5$ or $x\equiv1\pmod5$.
Continue with the other two factors: $13~|~x(x-1)$, so $x\equiv0\pmod{13}$ or $x\equiv1\pmod{13}$. There are now four possibilities for $x$:
*
*$x\equiv0\pmod5$ and $x\equiv0\pmod{13}$
*$x\equiv0\pmod5$ and $x\equiv1\pmod{13}$
*$x\equiv1\pmod5$ and $x\equiv0\pmod{13}$
*$x\equiv1\pmod5$ and $x\equiv1\pmod{13}$
When you continue with the 31 factor, you will get eight possibilities.
Here's the second piece of information you need: The Chinese Remainder Theorem: https://en.wikipedia.org/wiki/Chinese_remainder_theorem
Now, use the Chinese Remainder Theorem to find $x\mod (5\cdot13\cdot31)$ for each of the 8 cases, and find the solutions which are in the interval $[2,2013]$. (There are six of them, one of which is 156.)
|
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|
A k-lipschitz function Let $f:M\to \mathbb{R}$ a k-lipschitz function, i.e, $\vert f(x)-f(y)\vert\leq kd(x,y)$, for any $x,y\in M$. Show that $f(x)=\displaystyle \inf_{y\in M}[f(y)+kd(x,y)]=\displaystyle\sup_{y\in M}[f(y)-k d(x,y)]$, for all $x\in M$.
Any hint pls!. Regards
|
Hint: what is $f\left(y\right)+kd\left(x,y\right)$ at $y=x$?
|
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|
Dimension of quotient ring
What is the dimension of the following quotient ring, $\mathbb{Z}[x,y,z]/\langle xy+2, z+4 \rangle$, where $\mathbb{Z}$ is the ring of integers?
I realized this is isomorphic to $\mathbb{Z}[x,-2/x]$. How does $-2/x$ affect the dimension since the ring is $\mathbb{Z}$.
|
Let $R=\mathbb Z[X,Y]/(XY+2)$. We have $\dim R\le2$. Furthermore, since $x$ is a non-zero divisor on $R$ we have $\dim R\ge\dim R/(x)+1=2$. (Note that $R/(x)\simeq(\mathbb Z/2\mathbb Z)[Y]$.)
|
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Calculate $\iint{f d\mu dv}$ and $\iint{f dv d\mu}$ The purpose of this problem is to show that in Fubini-Tonelli theorem, the condition $f \in L^{+}(X \times Y)$ or $f \in L^1$ is necessary. Here is the problem:
Let $X = Y = \mathbb{N}$, $\mathcal{M} = \mathcal{N} = \mathcal{P}(\mathbb{N})$, $\mu = v =$ counting measure. Define $f(m, n) = 1$ if $m = n$, $f(m, n) = -1$ if $m = n + 1$ and $f(m, n) = 0$ otherwise. Then $\int{|f|}d(\mu \times v) = \infty$ and $\iint{fd\mu dv}$ and $\iint{fdvd\mu}$ exist but are not equal.
I can prove that $\int{|f|}d(\mu \times v) = \infty$, but I don't know how to calculate $\iint{fd\mu dv}$ and $\iint{fdvd\mu}$. Anyone can help me. I really appreciate.
|
We have that $\iint{fd\mu dv}=\int\left(\int f d\mu\right) dv$. And for each fixed $n$, $f(m,n)$ has only two nonzero values which are of equal measure. Thus for each fixed $n\in Y$:
$$\int_{m\in X} f(m,n) d\mu=f(n,n)+f(n+1,n)+0=1-1=0.$$
So we get $\iint{fd\mu dv}=0$. And for each fixed $m\in X$ for $m\neq 1$:
$$\int_{n\in Y} f(m,n) dv=f(m,m)+f(m,m-1)+0=1-1=0.$$
Thus
$$\iint{f dvd\mu}=\int_{n\in Y}f(1,n)dv+\int_{m\in X,m\neq1}\left(\int_{n\in Y}f dv \right)d\mu =1+0.$$
So we get that $0=\iint{fd\mu dv}\neq\iint{f dvd\mu}=1$ and the iterated integrals are not equal.
|
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|
Confusion with the eccentricity of ellipse Confusion with the eccentricity of ellipse. On wikipedia I got the following in the directrix section of ellipse.
Each focus F of the ellipse is associated with a line parallel to the minor axis called a directrix. Refer to the illustration on the right, in which the ellipse is centered at the origin. The distance from any point P on the ellipse to the focus F is a constant fraction of that point's perpendicular distance to the directrix, resulting in the equality e = PF/PD. The ratio of these two distances is the eccentricity of the ellipse. This property (which can be proved using the Dandelin spheres) can be taken as another definition of the ellipse.
Besides the well-known ratio e = f/a, where f is the distance from the center to the focus and a is the distance from the center to the farthest vertices (most sharply curved points of the ellipse), it is also true that e = a/d, where d is the distance from the center to the directrix.
It is given that $e=\frac fa=\frac ad$
In my book it was only given that $e=f/a$ (in my book there is nothing given about directrix of an ellipse).
My question
Knowing that $e=f/a$ how can I get $e=a/d$ and $e=PF/PD$?
|
Hint: If the eccentricity $e$ & the major axis $2a$ of an ellipse are known then we have the following
*
*Distance of each focus from the center of ellipse
$$=\text{(semi-major axis)}\times \text{(eccentricity of ellipse)}=\color{red}{ae}$$
*Distance of each directrix from the center of ellipse
$$=\frac{\text{semi-major axis} }{\text{eccentricity of ellipse}}=\color{red}{\frac ae}$$
|
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|
Showing 2 vector spaces are isomorphic. I am trying to understand how to show two vector spaces are isomorphic. You do this by showing there is an isomorphism that can be mapped between the two spaces.
What I don't understand is my lecturer's way of showing that the isomorphism is one-to-one?
It seems to me that he only showed that f is one-to one for the case of $w=0$. This, doesn't generalize to other points does it?
|
In general, showing a map $f:A \to B$ is one-to-one involves showing that for any $x,y \in A$ such that $x \ne y$, we have $f(x) \ne f(y)$. However, an isomorphism between vector spaces is by definition a linear map, and your lecturer should have shown (maybe much earlier) that a linear map is one-to-one if and only if the kernel of $f$ is $\{0\}$, i.e., $f$ maps nothing to zero except zero. Indeed, this follows from the fact that $f(x)-f(y)=f(x-y)$ if $f$ is linear.
Proof of last claim:
Let $f$ be linear and one-to-one. Suppose for sake of contradiction that the kernel of $f$ is nontrivial, i.e., there exists some $x \ne 0$ such that $f(x)=0$. Then for any $y$ we have $f(y+x)=f(y)+f(x)=f(y)$ but $y+x \ne y$, contradicting the fact that $f$ is one-to-one.
For the converse, let $f$ be linear and have kernel $\{0\}$. Let $a \ne b$. Then $f(b)-f(a) = f(b-a) \ne 0$ because $b-a \ne 0$; thus $f$ is one-to-one.
|
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|
Show that Cauchy's function is infinitely differentiable Show that
$$f(x)=
\begin{cases}
exp(-\frac{1}{x^2}), & \text{if $x\gt 0$} \\[2ex]
0, & \text{if $x\le 0$ }
\end{cases}$$
is infinitely differentiable.
Clearly $f^{(n)}(x)=0$ for all $x\lt 0$ and $f_{-}^{(n)}(0)=0$. Also since the derivatives of $exp(-1/x^2)$ produce $exp(-1/x^2)$ and a polynomial in $1/x$, using the rules for differentiation we can evaluate $f^{(n)}$ if $x\gt 0$ with any $n$. It remains to show that $f_{+}^{(n)}(0)=0$.
Now this final part is where I'm struggling. How can I show this part? I would greatly appreciate any help.
|
Hint: Use induction and l'Hopital. Consider $\lim_{x\rightarrow 0^+}\frac{e^{-1/x^2}}{x^2}$. Let $y=1/x$, so this becomes $\lim_{y\rightarrow\infty}\frac{e^{-y^2}}{y^{-2}}=\lim_{y\rightarrow\infty}\frac{y^2}{e^{y^2}}$. After one step of l'Hopital, you have your answer. This approach can be generalized for any $n\geq 0$.
|
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|
What is a usual order relation? I've just started learning about relations and now I'm at partial order relations and total order relations; essentially, I'm trying to convey that I'm very much a beginner to this relations stuff.
My textbook includes the following remark:
The usual order relation on the real line $ \Bbb R $ is a total order relation.
A little later in my textbook is the following exercise (a portion of it, anyways):
Let $A$={1,2}.
List all the partial order relations on A. (The usual order relation on A is $R$={$\mathsf (1,2) \cup E)$}, where $E$={$(1,1),(2,2)$} is the relation of equality on A. ...)
The textbook also refers to a usual order relation again, somewhere later, so I would like to know what is meant by the term. I tried to infer something from the exercise whose excerpt I included, but I still don't understand. I went to mathworld.com and I didn't find anything there.
|
The usual ordering on $\Bbb{R}$ can be uniquely defined as the total order $\lt$ on $\Bbb{R}$ that satisfies the following two properties:
(Let $a,b,c \in \Bbb{R}$)
1) $c \gt 0 \iff c$ is positive.
2) $a \lt b, \ \ 0 \lt c \implies ac \lt bc$
Proof that this defines uniquely $\lt$: Suppose that $\lt'$ also satisfies these properties and is a total order on $\Bbb{R}$. Then if $a \lt b$, assume that $b \lt' a$. Then we have $0 \lt' a-b$, or by (1) that $a-b$ is positive. Then let $c = a-b \gt' 0$. This connects us to $\lt$ by (1) since we can also then write $c = a-b \gt 0$. Now we need one more defining property to complete the proof:
3) $a \lt b \implies a + c \lt b + c$
$\square$.
|
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|
Given a graph G, if X(G) = k, and G is not complete, must we have a k-colouring with two vertices distance 2 that have the same colour? As the title asks, If given a graph, $G$, with chromatic number $k$, and $G$ is not complete must there exist a $k$-colouring of $G$, $f$, where there are two vertices $x,y$ such that $d(x,y) =2$ and $f(x) = f(y)$.
Note that not all $k$-colourings of $G$ have this property. Just consider the $9$-cycle with $k=3$ and colour it $a-b-c-a-b-c-a-b-c$, but the $9$-cycle has a $3$-colouring $a-b-a-b-a-b-a-b-c$ which does.
In the case where $k =3$, start by noticing that $G$ has a cycle C, where $|V(C)|>=3$. Furthermore you can find a chordless cycle with this property.
If possible, identify vertices $v,w$ on $C$ which are distance $2$ apart which have the same colour and repeat until no longer possible. Now given any three adjacent vertices on the cycle, $x,y,z$ where the edges are $xy$, $yz$, if there is a vertex $p$ adjacent to $y$, the colour of $p$ is either the same as $x$ or the same as $z$ and then we identify whichever happens to work. Continue this procedure until you are left with just a cycle, at which point you can unidentify vertices and give them the same colour as which they originated. The last two vertices which are unidentified are the desired vertices.
Thanks for any help :)
Edit: $G$ is a simple connected graph
|
I'll give an alternative proof to Leen's.
Suppose we have a graph $G$ where $\chi(G) =k$ and $G$ is not complete. We will proceed by showing that using kempe chains we can always get two vertices distance $2$ with the same colour.
Now take any $k$-colouring of $G$, $\alpha$. Since $G$ is not complete, there is at least one colour class $C$ of $G$ which has more than one vertex in it. If there are two vertices of $C$ which are distance $2$ apart, then we are done. Otherwise let $x,y \in V(C)$. Since $G$ is connected, there is a shortest path, $P$ between $x$ and $y$. Then there is a vertex $z \in N(x)$ which lies on $P$. Apply a kempe chain operation on the $x$ and $z$. This results in a new colouring $\alpha'$, and now we have reduced the distance between colours in $C$, as $d(x,y) = d(y,z) +1$. Now we can apply this same technique on $z$ and $y$, continually reducing the distance until we have a $k$-colouring where there are two vertices which have the same colour and are distance $2$ apart.
|
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|
What method of numerical integration is this? I am trying to update some old code that finds the area under a curve from $17$ evenly spaced discrete data points. I'd like to update it to calculate from $65$ data points. I'd like to use the same methodology, so I'm trying to determine the method used to approximate it. The code looks like this:
$$
\begin{align}
Area = \frac{20\cdot 10^{-9}}{3}\cdot &(Data(0)+Data(16) + Data(1)\cdot2 + Data(2)\cdot 4\\ &+ Data(3)\cdot 2 +Data(4)\cdot4+ \cdots +Data(15)\cdot2)
\end{align}
$$
The $ 20\cdot 10^{-9} $ comes from the $x$-axis spacing between points, but I'm having trouble understanding why they alternate between multiplying by $2$ and $4$ in the summation. Does anyone have an idea what type of approximation follows a pattern like this?
|
This appears to be the composite Simpson rule: https://en.wikipedia.org/wiki/Simpson%27s_rule#Composite_Simpson.27s_rule
A summary: start with a partition of $[a,b]$ into $N$ subintervals of equal length, and add a point in the middle of each of the subintervals. Now you have $2N+1$ evaluation points. (Note that this is always odd, regardless of whether $N$ was even or odd.)
On each of the original $N$ subintervals, you take a quadratic interpolant through the three evaluation points, integrate that interpolant. Then you sum over the intervals. On each subinterval, the weights arising from integrating the interpolant turn out to be $1,4,1$. (There is nothing special going on here, it's just a calculation.) But the endpoints of the subintervals other than the first and last one get used twice (as the left endpoint of one interval and the right endpoint of another), so they get double counted, which is where the $2$s come from.
|
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