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H: Is this a multiplication?
6x-13+4(-3)x=9+2x
I'm like, really dumb, I can't tell if if the 4(-3) bit is a multiplication
AI: $6x-13+4(-3)x=9+2x$
is the same as the equation $6*x-13+4*(-3)*x=9+2*x$
which is the same as $(6*x)-13+(4*(-3)*x)=9+(2*x)$
which is the same as $(6*x)-13+((-12)*x)=9+(2*x)$
which is the same as $(6*x)-13-(12*x)=9+(2*x)$
What happens is something similar to what happens when you add an x into the equation. Just like $6x$ is the same as $6*x$ the same happens with parenthesis. so $4(3)$ is the same as $4*3$ |
H: What would be the right domain for a function that takes time as a parameter?
I want to define a function that takes a value which represents time and returns an integer.
So when using it, the function would look something like $f(t)$
What is the right (or commonly used) domain for such a function?
$f:T \rightarrow Z$
What would $T$ actually be? (Note I don't want to discretize it in timesteps or anything)
AI: $T$ would be the set of times that you were considering. If you were dealing with a set of times with a discrete timestep, then it would be something along the lines of $\mathbb{N}$. On the other hand, the time could start at some initial time $a$ go continuously to some time $b$, in which case you'd have a domain of $[a,b]\subset \mathbb{R}$.
In general, however, your domain can be anything you want, and just depends upon the context. While it wouldn't be very clear what was meant, it could be represented by complex numbers. Usually, however, it is a subset of the real line. |
H: Solution Verification: Functions/Sets Question
Which of the following is a true statement? (Assume a finite domain.)
a. If a function is not a one-to-one correspondence, its domain must
contain more points than its image.
b. If a function is one-one, its domain and range are the same set.
c. If the domain of a function has more points than its image, it is
not surjective.
d. If a function is onto, its domain and range are the same set.
B, C, D all seem false to me, but I am still unsure that the domain must contain more points than its image to satisfy not one-to-one correspondence.
B seems false because the sets need not be the same but just need the same number of points. Same with D.
C is false because even if the domain of a function has more points than its image, it can still be surjective.
This leaves me with A.
AI: Apparently you’re to understand one-to-one correspondence as simply a one-to-one (injective) function, i.e., a one-to-one correspondence between its domain and its range, irrespective of whether the range is all of the codomain. On that understanding (a) is indeed true: the function can fail to be one-to-one only by sending two points of the domain to the same point of the range, and since the domain is finite, that means that the range must be smaller than the domain.
As you say, the other three are obviously false in general. |
H: Factorial lower bound: $n! \ge {\left(\frac n2\right)}^{\frac n2}$
A professor in class gave the following lower bound for the factorial
$$
n! \ge {\left(\frac n2\right)}^{\frac n2}
$$
but I don't know how he came up with this formula. The upper bound of $n^n$ was quite easy to understand. It makes sense. Can anyone explain why the formula above is the lower bound?
Any help is appreciated.
AI: Suppose first that $n$ is even, say $n=2m$. Then
$$n!=\underbrace{(2m)(2m-1)\ldots(m+1)}_{m\text{ factors}}m!\ge(2m)(2m-1)\ldots(m+1)>m^m=\left(\frac{n}2\right)^{n/2}\;.$$
Now suppose that $n=2m+1$. Then
$$n!=\underbrace{(2m+1)(2m)\ldots(m+1)}_{m+1\text{ factors}}m!\ge(m+1)^{m+1}>\left(\frac{n}2\right)^{n/2}\;.$$ |
H: Equation with fractions
If $P=\frac{h}{1-h}$ then $h$ is equal to?
Answer is: $\frac{P}{1+P}$
I understand that $\frac{P}{1+P}$ is the right answer for when I replace $\frac{P}{1+P}$ for h the answer solves the equation, but what I can't do is find the answer by myself, how do I get to $\frac{P}{1+P}$?
AI: $$P = \frac{h}{1-h}$$
Cross-multiplying you get
$$(1-h) \cdot P = h$$
After this, simply work to get the expression in terms of h.
$$P - P\cdot h = h$$
$$P = h + P \cdot h$$
$$P = (1 + P) \cdot h \implies h = \frac{P}{1 + P}$$ |
H: Prove that every unitary matrix $U$ is unitarily diagonalizable
I just can't show that a unitary matrix $U$ is unitarily diagonizable. I know I need to show that $U$ is unitarily similar to a diagonal matrix, and this result is presumably a consequence of the spectral theorem.
EDIT: I was reading this wrong, and I am supposed to be proving this result without the use of the spectral theorem. I have written a proof using the notion of block matrices which I believe is correct. Can anyone help me prove this result without the spectral theorem, using inner products? I suppose the question reduces to:
Prove that a normal matrix is unitarily diagonalizable (using inner products).
AI: We shall show that unitary matrices are normal, from which the Spectral theorem shall directly apply. The defining property of a normal matrix is $TT^\ast=T^\ast T$. Since $TT^\ast=I$ for $T$ unitary, clearly $T^\ast T=I$ as well, and $TT^\ast=I=T^\ast T$.
If you are using the inner product definition: Recall that $T$ is unitary iff $\langle Tv,Tw\rangle=\langle v,w\rangle$ for every $v,w$ in your inner product space $V$, as in general, $\langle Tv,Tw\rangle=\langle v,T^\ast Tw\rangle=\langle v,w\rangle$ if $TT^\ast=I=T^\ast T$. $T$ normal iff $\langle Tv,Tw\rangle=\langle T^\ast v,T^\ast w\rangle$, and $T$ unitary clearly has this property. |
H: $H
Does anyone know of a counter example or a proof of the following proposition? If it doesn't hold in general are there any classes of groups for which it holds?
Let $G$ be a non-abelian finite group and let $H<K<G$ with $H$ a maximal subgroup of $K.$ If $gHg^{-1}<K$ then $g^{-1}Hg<K.$
Thank you
AI: Consider $G=S_4$ and put $H=\{(1),(12)\}$ and $K=S_3$. Then $H$ is maximal in $K$ because of their orders being $2,6$ respectively, and $2|x|6$ has no nontrivial solutions.
Now let $g=(1432)$. Then [I'm doing compositions left-to-right] $gHg^{-1}=\{(1),(23)\}<K$, however $g^{-1}Hg=\{(1),(14)\}$ is not a subgroup (or even a subset) of $K.$
Note this could be switched around if one wants to do composition in the opposite right-to-left order. |
H: Multiplication, What is It?
What is multiplication? Upon review logarithms, and square roots, I realized that I have no intuitive grasp of multiplication-well no more so than I have for addition. Is it simply another thing we need to memorize? I understand that things like $\sqrt2$ could just be memorize as the thing, that when applied to itself, gives two... But this makes a lot less sense to me then thinking about things like $2+2=4$. Logarithms are a function that express a number as a power of some base, but when you get fractions for the logarithms, I don't really know what that means-What is $2^{1.2}$, or any decimal power for that matter.
Upon rereading the link posted below, I think I can repose my question a bit. If these operations are axiomatic, as I felt they were, the idea of a square root is the inverse of exponentiation. How do I grasp exponentiation intuitively? Multiplication and addition have natural roots in our minds, if exponentiation is also fundamental, what is it? Also when is exponentiation not just repeated multiplication?
AI: First, you shouldn't find expressions like $\sqrt{2}$ weird. Saying that $\sqrt[a]{b}$, where $a$ and $b$ are positive integers, is the number whose $a^\text{th}$ power is $b$ is perfectly fine. Just think about the graph of $x^a$. This is fudging a little bit, but we know $x^a$ is continuous, meaning it's a solid line, and we know it goes up to $\infty$ in the positive direction. So we know that the lines $y = x^a$ and $y = b$ intersect at some point. When you write down $\sqrt[a]{b}$ that's just notation for the $x$-coordinate of that point of intersection.
Now if you're fine with $\sqrt[a]{b}$ then you should be fine with raising a number to a fraction. When we write $x^\frac{a}{b}$ we just mean $\sqrt[b]{(x^a)}$ or $(\sqrt[b]{x})^a$ (they're the same).
After you're fine with raising numbers to fractions you might ask what does $x^a$ mean if $a$ is $\pi$ or some other number that's not a fraction. This is maybe farther than you want to go at the moment, so I won't give you a detailed answer (I'm sure this questions exists elsewhere on the site). I'll just say that $x^a$ is the number that $x^b$ gets closer to as you choose fractions $b$ that get closer and closer to $a$. |
H: is a plane smooth surface?
let f(u,v)=a + u.p + v.q : the equation of the plane
where p,q are unit vectors perpendicular to each other. a a point on the plane.
I do not understand how f can have partial derivatives of all orders, since derivative of wrt. u and v are p and q, respectively.
after this, aren't the derivatives zero?
AI: The second order and higher partial derivatives of $f$ are indeed $0$, which is not a problem.
The zero function has partial derivatives of all orders. |
H: weave of two sequences
Let x, y, z be three sequences of real numbers. z is said to be a weave (not "the weave", because weaves are not unique) of x and y iff x and y are disjoint subsequences of z that span all of z. For example, (0,0,1,0,1,0,0,1,0,1,...etc) is a weave of (0,0,0,...) and (1,1,1,..). My question is this: If x and y both converge to L, does any weave of them also converge to L? If so, give me a proof. If not, give me a counterexample.
AI: Suppose that $\mathbf{x}$ and $\mathbf{y}$ converge to $L$. Then $\mathbf{x}$ and $\mathbf{y}$ are both Cauchy, and for each $\epsilon>0$ there exists $N$ such that $n,m>N$ implies that $|x_n-x_m|<\epsilon$. A similar statement holds for $\mathbf{y}$. Take any weave $\mathbf{z}$ of $\mathbf{x}$ and $\mathbf{y}$, and choose $\epsilon>0$. Then choose $N_x'$ such that $|x_n-x_m|<\epsilon/2$ for $n,m>N_x'$, choose $N_x''$ such that $|x_n-L|<\epsilon$ for $n>N_x''$, define $N_x=\max\{N_x',N_x''\}$, and similarly for $\mathbf{y}$, giving $N_y$. Now let $N$ be a natural number such that if $i$ is the index of $\mathbf{z}$ such that the $N_x$th entry of $\mathbf{x}$ appears and $j$ the index of $\mathbf{z}$ such that the $N_y$th entry of $\mathbf{y}$ appears, then $N>\max\{i,j\}$. Now let $n,m>N$. Then $|z_n-z_m|$ is either $|x_n-x_m|$, $|y_n-y_m|$, or $|x_n-y_m|$. In the former two cases, these are clearly less than $\epsilon$. In the latter case, we use the triangle inequality to give $|x_n-L+L-y_m|\leq |x_n-L|+|y_m-L|<\epsilon$. Thus, $\mathbf{z}$ is Cauchy. Since a Cauchy sequence converges to the same limit that any subsequence converges to, it converges to $L$. |
H: Calculate modulo large number
How do I calculate 4^23 mod 31? I think it can be done using Euler's Totient Function, but I don't know how.
AI: This should be a relatively easy example. $4^{23}=2^{46}$. Now, since $2^5=32\equiv1\pmod{31}$,
$$2^{46}=(2^5)^9\times2=32^9\times2\equiv1\times2\equiv2\pmod{31}$$ |
H: How to integrate $\int e^{-x}\arctan(e^x) \, dx$
After trying this multiple ways, I give up. Here's the integral:
$$\int e^{-x}\arctan(e^x)\,dx$$
I have set
$u=\arctan(e^x)$ and $dv=e^{-x}d\,x$
and have obtained $du=\dfrac{e^x \, dx}{1+e^{2x}}$ and $v=-e^{-x}$
Using the integration by parts formula,
$$\int u\, dv = uv - \int v\,du$$
I got $-e^{-x}\arctan(e^x)+\int \dfrac{1}{1+e^{2x}}\,dx$
How would I solve $\int \dfrac{1}{1+e^{2x}}\,dx$ ? That's the part I'm stuck on.
AI: $$\int \frac{dx}{1+e^{2 x}} = \int dx \frac{e^{-x}}{e^x+e^{-x}} = - \int du \frac{1}{u+u^{-1}}$$
which is
$$-\int du \frac{u}{1+u^2} = -\frac12 \log{(1+u^2)} + C = -\frac12 \log{\left (1+e^{-2 x}\right)} + C$$ |
H: Function with a continuous domain but a discrete range
Does it makes sense for a function to have a discrete range even though the range is continuous? If yes how is it defined, and is it called something specific?
To explain what I mean if one had to model time against whether the light is on or off (to indicate when light goes on and light goes off). The range will just be 0 and 1, nothing in between, while the domain is a continuous value, time.
AI: It makes sense and your example is a good one. A function is required to return a single value for each element of the domain, but doesn't have to be continuous. A couple other example of functions on $\Bbb R$ are $\lfloor x \rfloor$ and the function that is $1$ if $x$ is rational and $0$ otherwise. |
H: Laplace transform of $t^2e^{at}$??
I'm trying to prove that $$\mathcal{L}\{t^2e^{at}\} = \frac{2}{(s-a)^3}.$$
I've gotten to the last integration by parts where
$$
\lim_{n\to\infty}\int_0^n\frac{1}{(a-s)^22e^{(a-s)t}}dt = \left. \lim_{n\to\infty}\frac{2}{(a-s)^3}e^{(a-s)t} \right|_0^n.
$$
Now what do I do? I can't find a way to make that last term converge?
AI: Evaluating the integral for a fixed $n$ gives
$$\frac{2}{(a - s)^3} e^{(a - s)n} - \frac{2}{(a - s)^3} e^0$$
Assume that $s > a$ and let $n$ go to infinity. Then since $(a - s) n \to -\infty$, the first term disappears and the limit is
$$-\frac{2}{(a - s)^3} e^0 = -\frac{2}{(-1)^3 (s - a)^3} = \frac{2}{(s - a)^3}$$ |
H: Confused with finding C in economic integral
Question:
An automobile company is ready to introduce a new line of cars. They project that the sales will increase by:
$P'(t)=10-10e^{-0.1t}, 0\leq t\leq 24$
in t months after the campaign has started.
(A) - What will be the total sales S(t) t months after the beginning of the campaign if we assume no sales at the beginning of the campaign?
So I integrate the marginal function and get:
$10t-100e^{-0.1t}+C$ but my book gets the same answer except that C is +100. How does my calculus text book author arrive at this conclusion?
AI: Hint: How many cars are sold at time $t=0$ months? Use that condition (also known as initial condition) in your expression for $P(t)$ and solve for $C$. |
H: Linear Functional on $V$
Need some help understanding step two.
Suppose $\varphi$ is a linear functional on $V$. Then there is a unique vector $v \in V$ such that $\varphi (u) = \left \langle u, v \right \rangle$ for every $u \in V$
We start off with $\varphi (u) = \varphi (\left \langle u, e_1 \right \rangle e_1 + \cdot \cdot \cdot + \left \langle u , e_n \right \rangle e_n)$ where $(e_1 .... e_n ) $ is an orthogonal basis of $V$. Next $\varphi = \left \langle u, e_1 \right \rangle \varphi (e_1) +\cdot \cdot \cdot + \left \langle u,e_n \right \rangle \varphi (e_n) $, where did $\varphi (e_n ) $ come from?
$=\left \langle u, \varphi (e_1 )^* e_1 + \cdot \cdot \cdot + \varphi (e_n)^* e_n \right \rangle$
AI: $\varphi$ is a linear functional: $\varphi(av) = a\varphi(v)$ for every scalar $a$ and every vector $v$. Note that $\langle u,e_j\rangle$ is a scalar, and you have it. |
H: Conditional distribution of binomial random variables is hypergeometric
Let's say $X$ and $Y$ are binomial random variables with parameters $n$ and $p$ and $X+Y=m$. I want to show that the conditional distribution of $X$ if $X+Y=m$ is a hypergeometric distribution.
I'm thinking about putting these in terms of coin flips with the $m^\text{th}$ head occurring after $2n$ tosses. $X$ and $Y$ have $n$ tosses with probability of heads $p$. I'm not sure how to relate this to the hypergeometric distribution however.
AI: We prove a somewhat more general result. Let $X$ and $Y$ be independent binomially distributed random variables with the same probability $p$ of "success" and number of trials equal to $a$ and $b$ respectively. (In your problem, we have $a=b=n$.)
Let us find the probability that $X=k$ given that $X+Y=m$. By the usual formula for the conditional probability, this is the probability that $X=k$ and $Y=m-k$ divided by the probability that $X+Y=m$.
The probability that $X=k$ and $Y=m-k$ is $\binom{a}{k}p^k(1-p)^{a-k}\binom{b}{m-k}p^{m-k} (1-p)^{b-(m-k)}$. This is $\binom{a}{k}\binom{b}{m-k}p^m(1-p)^{a+b-m}$.
The probability that $X+Y=m$ is $\binom{a+b}{m} p^m(1-p)^{a+b-m}$.
Divide. We get
$$\frac{\binom{a}{k}\binom{b}{m-k}}{\binom{a+b}{m}}.\tag{1}$$
This is precisely the probability of getting $k$ red balls if we choose, without replacement, $m$ balls from $a+b$ balls $a$ of which are red and $b$ of which are blue. Formula (1) is a version of the familiar hypergeometric distribution function. |
H: Seating $2n$ people around a table - Why $(2n - 1)!$ and not $(2n)!$?
There've been numerous questions about this so please let me know if this is a duplicate.
Page 12 in http://www.am.qub.ac.uk/users/g.gribakin/sor/Chap1a.pdf says:
Let $A(i, r) =$ couple $i_r$ sit next to each other.
To compute the generic term $P[ \, A(i, 1) \cap A(i, 2) \cap \ldots \cap A(i, r) \, ]$,
we proceed as follows.
There are $(2n - 1)!$ ways of seating $2n$ people at a roundtable. Why? Put the first person on some seat, then arrange the other $(2n - 1)$ around them.
My first thought was that seating $2n$ people around a roundtable means selecting $2n$ seats without replacement and with ordering. There'd be $(2n)$! ways of doing so.
How and why is this wrong? Shouldn't both ways work? It seems more guileful to "put the first person ... $(2n - 1)$ around them"?
AI: The thing you're missing here is that in these sorts of problems, we generally consider two arrangements "the same" if one is a rotation of the other. So, for instance, the arrangements (read, say, clockwise around the table) ABCD and BCDA are generally considered the same... each person has the same neighbors to their left and right.
So, we think of this differently: once we've placed one person, the rest of the people flow from there in a permutation. This is where $(2n-1)!$ comes from: that first person's seat doesn't matter, but once they're in place any arrangement of the other $2n-1$ is distinct.
Another way to think of this is as follows: if we count up all the permutations of $\{1,2,\ldots,2n\}$, then you can note that there are $2n$ rotations that all lead to the same round-table configuration. Thus there are
$$
\frac{(2n)!}{2n}=(2n-1)!$$
ways to arrange them around the table.
If we care which particular chair each person lands in, rather than just what their arrangement is, then you are absolutely right that it would be $(2n)!$ instead of $(2n-1)!$. |
H: Did I do this Laplace transform correctly?
1) $w'' + w = t^2 + 2$; $w(0) = 1$, $w'(0) = -1$
2) $s^2W(s) - sw(0) - w'(0) = \frac{2 + 2s^2}{s^3}$
3) $s^5W(s) - s^4w(0) - s^3w'(0) = 2 + 2s^2$
4) $ W(s) = \frac{2 + 2s^2 - s^3 + s^4}{s^5}$
5) $W(s) = 2\left(\frac{1}{s^5}\right) + 2\left(\frac{1}{s^3}\right) - \frac{1}{s^2} + \frac{1}{s}$
6) $w(t) = \frac{2}{4!}t^4 + t^2 - t + 1$
Anyone?
AI: It seems that there is a defect in your solution. In fact, we have: $$s^2W(s)+1-s+W(s) = 2/s^3+2/s$$ instead which leads us to have: $$W(s)=\frac{2+2s^2-s^3+s^4}{s^3(s^2+1)}$$ |
H: A convergent-everywhere expression for $\zeta(s)$ for all $1\ne s\in\Bbb C$ with an accessible proof
I'm looking for a way to define the Riemann zeta function $\zeta(s)=\sum_{n\in\Bbb N_0}n^{-s}$ on the whole complex plane, without having to use analytic continuation, or perhaps more accurately, in a way which can be written down as a single formula. (In principle, an analytic continuation can be written down as a formula, but no one ever seems to do it this way, and always end up appealing to drawings in the end.) Any formula will do, even a piecewise one, as long as there is a relatively elementary proof that the expression is equal to $\sum_{n\in\Bbb N_0}n^{-s}$ when $\Re[s]>1$ (and to constrain piecewise solutions which just use the standard definition in this region, you should also need to be able to show that it is a reasonable extension in some way, i.e. continuity or analyticity).
My current best bet is the Laruent expansion about the pole at $s=1$ (in terms of Stieltjes constants):
$$\zeta(s)=\frac1{s-1}+\sum_{n=0}^\infty\frac{(-1)^n(s-1)^n}{n!}\!\!\lim_{m\to\infty}\left[\sum_{k=1}^m\frac{(\ln k)^n}k-\frac{(\ln m)^{n+1}}{n+1}\right]$$
However, I was unable to find a proof online of the equality of this expression to the standard one, and I expect the proof to be rather involved. Does anyone have any suggestions for more workable formulas for $\zeta(s)$?
AI: Considering the accepted answer for this related question and following Gerry Myerson's answer elsewhere, I found Sondow's paper. |
H: Definition of $H_\lambda$ (hereditary cardinality)
It seems to me that the definition of $H_\lambda$ (the set of sets of hereditary cardinality less than $\lambda$) on the web page at Cantor's Attic is not quite correct. From the page:
$H_\lambda=\{x: |\operatorname{trcl}(x)|<\lambda\}$ where $\operatorname{trcl}(x)$ denotes the transitive closure of $x$.
This is correct when $\lambda$ is a regular cardinal, but if $\lambda$ is singular then the definition should include any cofinal subset of $\lambda$ with size less than $\lambda$, which this definition doesn't. Indeed, if $\lambda$ is a singular cardinal, $A \subset \lambda$ is cofinal, and $|A| < \lambda$, then because $A$, its elements, their elements, and so on all have cardinality less than $\lambda$ then we should say that $A$ has hereditary cardinality less than $\lambda$ if we want the terminology to make any sense.
I think the closest definition of $H_\lambda$ that would be correct is
$H_\lambda=\{x: \forall y \in \operatorname{trcl}(\{x\})\, (|y| <\lambda)\}$.
My question for Math.SE is whether I'm correct that there is a mistake, and if so, what the best definition would be (so that I can fix the page.) In my experience $H_\lambda$ is only ever used when $\lambda$ is regular, so the existing definition could be fixed by restricting it to this case, which is what Jech does in the exercises to Chapter 12. On the other hand, the second definition proposed above has the feature that it generalizes better to the case of "hereditarily in $P$" where $P$ is some class other than $\{y : |y| < \lambda\}$, I think. There might be a third definition that is even better. What is standard? What is best?
AI: You are correct. For the reasons you discuss, the appropriate definition of $H_\lambda$ for singular $\lambda$ is the set of $x$ such that $x$ and all sets in its transitive closure have size below $\lambda$. (Though I could not instantly find a reference, this is actually standard among those that consider the set at all, which are not too many, and its use is somewhat limited. For example, I would imagine one may want to consider the class in settings where choice fails, but I have yet to see an application in that context.)
On the other hand, it is also standard (and far more common) to only define $H_\lambda$ for $\lambda$ regular.
Mostly as a curiosity: Some people (Forster, for example), present $H_\kappa$ as $\bigcap\{y\mid \mathcal P_\kappa(y)\subseteq y\}$. This presentation generalizes: We can define $$\mathcal P_\phi(x)=\{y\subseteq x\mid \phi(y)\},$$ so $\mathcal P_\kappa(x)$ is $\mathcal P_{|y|<\kappa}(x)$. One can then define $H_\phi$ as $\bigcap\{y\mid\mathcal P_\phi(y)\subseteq y\}$. This notation seems to go back to
Maurice Boffa. Sur l'ensemble des ensembles héréditairement de puissance inférieure à un cardinal infini donné, Bull. Soc. Math. Belg., 22, (1970), 115–118. MR0309732 (46 #8837).
Note this version coincides with the standard one for $\kappa$ regular, but is the problematic one for $\kappa$ singular. |
H: Data preprocessing
How would you preprocess 2 dimensional data to have 0 mean? Say you have a matrix $M $ that is $p \times q $. Would you calculate the mean of each row, get a vector of length $q $ and subtract each element of the vector from the corresponding column?
AI: If your data are 2-dimensional, then the matrix $M$ would have the size $p \times 2$ or $2 \times p$. In either case, you need to take the mean along the $p$ dimension. This will give you a 2-element mean vector $m$. Then you have to subtract it from every row (or column) of $M$. |
H: Prove using Rolle's Theorem that an equation has exactly one real solution.
Prove that the equation $x^7+x^5+x^3+1=0$ has exactly one real solution. You should use Rolle’s Theorem at some point in the proof.
Since $f(x) = x^7+x^5+x^3+1$ is a polynomial then it is continuous over all the real numbers, $(-\infty,\infty)$. Since $f(x)$ is continuous on $(-\infty,\infty)$ then it is certainly continuous on $[-1,0]$ and the Intermediate Value Theorem (IVT) applies. The IVT states that since $f(x)$ is continuous on $[-1,0]$ we can let $C$ be any number between $f(-1)=-2$ and $f(0)=1$, namely $C=0$, then there exists a number $c$ with $-1 < c < 0$ such that $f(c)=C=0$. Since $f(x)$ is continuous on $[-1,0]$ and differentiable on $(-1,0)$ then Rolle’s Theorem applies. Rolle’s Theorem states that if a function $f\colon [a,b] \to \mathbf{R}$ is continuous on $[a,b]$ and differentiable on $(a,b)$ then if $f(a)=f(b)$, there exists a point $c \in (a,b)$ such that $f'(c)=0$. We assume that there is more than one real solution for this equation, namely $f(a)=0=f(b)$. If there exists more than one real solution for $f(x)=0$ then $f(a)=0=f(b)\implies a=b$, and thus there is only one real solution to the equation, as desired.
But I feel I am missing something, and I'm not sure how to show that the equation is differentiable on the interval and that Rolle's Theorem applies.
AI: Let $y = x^7+x^5+x^3+1$
$y(0) = 1, y(-1)=-2$
By IVT, there exists at least one real root in $x\in(-1,0)$ such that $y(x)=0$.
Now I claim that there is EXACTLY one such real root, by using the method of contradiction.
Suppose not, there exists at least $2$ real roots $x_1,x_2$ such that $y(x_1)=0,y(x_2)=0$
Since $y$ is differentiable, by Rolle's theorem, there exists a number $a\in(x_1,x_2)$ such that $y'(a)=0$. However, $y'(x)=7x^6+5x^4+3x^2>0$ for all $x\ne0$.
... |
H: counting and probability question - help needed
I am stuck on how to start this exercise. Any help is welcome.
An instructor gives an exam with 14 questions. Students are allowed to choose any 10 to answer. Suppose the exam instructions specify that at most 1 of questions 1 and 2 may be included among the 10. How many different choices of 10 questions are there?
I have made a few attempts so far with no luck. Here is some of what I have tried.
1. $\binom{14}8 = 3003$ This is incorrect
2. $\binom{14}9 = 2002$ This is incorrect
3. $\binom{10}8 = 45 $ This is incorrect
I have tried others with similar results, so any help is appreciated. I just don't know how to get this started.
Thanks,
Tony
AI: Divide into cases:
(i) We use neither Q1 nor Q2. Then we must choose $10$ from the remaining $12$.
(ii) We use one of Q1 or Q2. We must choose $1$ from these $2$, and $9$ from the remaining $12$.
Add the answers to (i) and (ii).
Another way: Alternately, find the total number of choices, with no restriction. Then subtract the number of bad choices, where we picked both Q1 and Q2. There are $\binom{12}{8}$ bad choices. |
H: Infinite series for Euler-Mascheroni constant
Problem
Show that
$$
\gamma = \tfrac 12 \log 2 + {1 \over \log 2} \sum_{n=2}^\infty (-1)^n {\log n \over n}.
$$
Progress
I tried writing the terms $1/k$ of the harmonic sum in the definition of $\gamma$ as $\int_0^1 x^{k-1} dx$, and interchanging the order of summation and integration. This gives
$$
\gamma = \lim_{n \to \infty} \int_0^1 {1 - x^n - \log n + x \log n \over 1 - x} dx,
$$
which might just be a dead end. I also know the alternate form
$$
\gamma = \lim_{s \to 1} {\zeta'(s) \over \zeta(s)} + {1 \over s-1} = \lim_{s \to 1} -\sum_{n=1}^\infty \Lambda(n) n^{-s} + {1 \over s-1},
$$
but I don't see how it could help.
Notes: $\log x$ is the natural logarithm, $\Lambda(n)$ is the von Mangoldt function.
AI: this problem is from analysis book: Problems In Mathematical Analysis I Page (70) |
H: How to find the following integral? $\int\tfrac{x}{\sqrt{1+3x^2}}\mathrm dx$
Find: $$\int\dfrac{x}{\sqrt{1+3x^2}}\,\mathrm dx$$
I can't fully integrate this, I get $1/x+\sqrt3 x$ and then I don't know what to do, not sure if I even started it correctly, thanks in advance.
AI: Check first (chain rule) that
$$\int\frac{f'(x)}{\sqrt{f(x)}}dx=2\sqrt{f(x)}+C$$
and now observe that
$$\frac x{\sqrt{1+3x^2}}=\frac16\frac{(1+3x^2)'}{\sqrt{1+3x^2}}$$ |
H: What does the notation $G /^r H$ mean?
$G /^r H$
I saw this notation in an answer to a question and am not sure what it means.
The exact context is as follows:
$G=Sym(5)$ acts on the set $G/^r H$ of all right cosets of $H$ in $G$.
AI: Space of right cosets. When $H$ is normal, you'd just interpret $G/H$ as the quotient space. But when it's just a subgroup, $G/H$ represents a coset space. But there are two distinct ones, so you need a way to distinguish them. That answer used a superscript $r$ to denote right cosets. I've also seen $H\backslash G$ for right cosets (and then $G/H$ means left cosets). |
H: Evaluating the integral, $\int_{-\infty}^{\infty} e^{-x^2/a}\ln\left(1 + be^{-cx^2}\right)dx$
I recently got stuck on evaluating the following integral, $$\int_{-\infty}^{\infty} e^{-x^2/a}\ln\left(1 + be^{-cx^2}\right)dx$$ where $a>0$, $b>0$. I don't know if there is an effective substitution to use.
AI: First assume that $c > 0$. Simply plugging the series expansion
$$\log(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} x^{n}, \quad |x| < 1$$
and integrating term by term, we have
$$ \int_{-\infty}^{\infty} e^{-x^{2}/a} \log ( 1 + b e^{-c x^{2}} ) \, dx
= \sqrt{a \pi} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n\sqrt{acn + 1}} b^{n}, $$
which is valid for $a, c > 0$ and $|b| < 1$. Of course, the left-hand side extends holomorphically for any complex $b$ avoiding the cut $(-\infty, -1)$.
For $c < 0$, the behavior changes completely. For $b > 1$, we write
$$ \int_{-\infty}^{\infty} e^{-x^{2}/a} \log ( 1 + b e^{-cx^{2}} ) \, dx
= \int_{-\infty}^{\infty} e^{-x^{2}/a} \left\{ |c|x^{2} + \log b + \log ( b^{-1} e^{-|c|x^{2}} + 1 ) \right\} \, dx. $$
The applying a similar technique,
$$ \int_{-\infty}^{\infty} e^{-x^{2}/a} \log ( 1 + b e^{-cx^{2}} ) \, dx
= \sqrt{a\pi} \left( \frac{a|c|}{2} + \log b + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \sqrt{a|c|n + 1}} b^{-n} \right). $$
Edit. Actually we may assume $a = 1$ upon applying the substitution $x \mapsto \sqrt{a} x$ and absorbing $\sqrt{a}$ to $c$. Then we have the following formal series expansion with respect to $c$:
$$ \int_{-\infty}^{\infty} e^{-x^{2}} \log ( 1 + b e^{-cx^{2}} ) \, dx = -\pi \sum_{k=0}^{\infty} \frac{\operatorname{Li}_{1-k}(-b)}{k!\Gamma\left(\frac{1}{2}-k\right)} c^{k} \quad \text{as } c \to 0. $$
This series asymptotic in the sense that it diverges for any $c \neq 0$, but for any $n$ its first finite $n$ terms gives an approximation of the integral with an error $O(c^{n+1})$. |
H: Number of strings consisting of k ones and n zeros such that no two ones are adjacent
What is number of strings consisting of k ones and n zeros such that no two ones are adjacent?
I already know the answer from wikipedia $\binom{n+1}{k}$ but i want to know its source.
AI: Write down the $n$ $0$'s in a row, with a little gap between them, like this:
$$0\quad0\quad0\quad0\quad0\quad0\quad0\quad0\quad0\quad0\quad0\quad0\quad0\quad0\quad0\quad0\quad0\quad0$$
This determines $n+1$ "gaps," $n-1$ of them real gaps between consecutive $0$'s, and the two "endgaps." We must choose $k$ of these to slip a $1$ into.
Remark: This also works if say $n=5$ and $k=8$, if we use the convention that for non-negative integers $a$ and $b$, we have $\binom{a}{b}=0$ if $a\lt b$. |
H: How to approach proving $\lim\limits_{n \to \infty}(\int^{b}_{a}(f(x))^ndx)^{\frac{1}{n}}=\max\{f(x):x\in[a,b]\}$?
Let $f:[a,b] \to \mathbb{R}$ be positive and continuous. Let $M = \max\{f(x):x \in [a,b]\}.$
Could anyone advise me on how to establish the following result:
$$\lim_{n \to \infty}(\int^{b}_{a}(f(x))^ndx)^{\frac{1}{n}}=M.$$
Fix $n \in \mathbb{N}.$ By Mean Value theorem, $(\int^{b}_{a}(f(x))^ndx)^{\frac{1}{n}}=f(c)(b-a)^{\frac{1}{n}},$ for some $c \in [a,b]?$
Thank you.
AI: Note that $\lim_{n\to\infty} \left(\sum a_i^n\right)^{1/n}=\max(a_i)$, since $\lim_{n\to\infty} \frac{\max(a_i)^n}{a_j^n}=\infty$ if $a_j\neq\max(a_i)$. Therefore we have
$$\begin{align} \lim_{n\to\infty} \left(\int_{a}^b f(x)^n\mathrm dx\right)^{1/n}&=\lim_{n\to\infty}\left(\lim_{k\to\infty}\sum_{i=ak}^{bk} f\left(\frac{i}{k}\right)^n\cdot k^{-1}\right)^{1/n}\\
&=\lim_{k\to\infty} \max\left(\left\{f\left(\frac{i}{k}\right)\mid ak\leq i\leq bk\right\}\right) \\
&=\max(\{f(x)\mid a\leq x\leq b\})\end{align}$$ |
H: How to find this Differential Equation ?
Find differential equation of a circle in $XY-Plane$ such that it passes through $(-1,1)$ and $(1,1)$
I need some clue please. I am not able to find the general equation. Is it a trick problem ?
AI: Hint: Find the equation of the circle. The equation of a circle with centre (g,h) is
$$(x-g)^2+(y-h)^2=r^2$$
Substitute x=1,y=1 and x=-1,y=1 to calculate g and h in terms of r. Differentiate both sides w.r.t x. r will be a constant. |
H: Can any one tell me the books for power series?
Can any one tell me the books for power series? I want to find the power series for sqrt(x). I surf on the internet but there is no success.
So please tell me the name of the book where I can find the power series of sqrt(x)/
AI: A few sites where you can find useful information
http://hardycalculus.com/calcindex/IE_powerseriestables.htm
http://en.wikipedia.org/wiki/Taylor_series
https://www.projectrhea.org/rhea/index.php/PowerSeriesFormulas
http://www.reading.ac.uk/AcaDepts/sp/PPLATO/imp/h-tutorials/series_02_power.pdf |
H: how to find accumulation point of $z_n=e^{in}$
How can I find accumulation point of
a) $z_n=e^{in} $
b)$z_n=i^n$
c)$z_n=(1-\frac{1}{n})+(-1)^ne^{\frac{1}{n}}i$
I tried at b) $\lim_{n\to\infty}(i^2)^{\frac{n}{2}}=(-1)^{\frac{n}{2}}=-1,+1$
at c) $\lim_{n\to\infty}(1-\frac{1}{n})+(-1)^ne^{\frac{1}{n}}i=lim_{n\to\infty}(1)+(-1)^ni=1+i,
1-i$
Can you help me out?
AI: a) Hint: $|e^{in}|=1$, so we have a bounded sequence and ther must be at least one acccumulation point; they form a nonempty closed subset of $S^1$. If $z$ is an accumulation point then so is $e^iz$.
In b) you must not replace $i^n$ with $(-1)^{\frac n2}$. Just have a look at the first few terms of the sequence
Your result for c) is correct. |
H: Prove that if $F : A \rightarrow B$ and $F^{-1}$ is a function, then $F$ is Injective
Statement: if $F : A \rightarrow B$ and $F^{-1}$ is a function, then $F$ is $1-1$
Proof: If $F$ is not $1-1$, then there exist $x_{1}, x_{2} \in A$ where $x_{1} \neq x_{2}$ and $F(x_{1}) = F(x_{2})$. Therefore, $F^{-1}(y) \neq F^{-1}(y)$, so $F^{-1}$ is not a function. By contraposition, if $F : A \rightarrow B$ and $F^{-1}$ is a function, then $F$ is $1-1$. $\square$
Is my proof valid?
AI: Your proof has the right idea (it is 'valid'), but the phrasing could be better, and if I were grading this, I would not give you full marks for "Therefore, $F^{-1}(y) \neq F^{-1}(y)$". Here's how I would do it:
Assume that $F$ is not injective. Then there exist $x_1, x_2 \in A$, such that $x_1 \neq x_2$ and $F(x_1) = F(x_2) = y \in B$. Thus, we would have $F^{-1}(y) = x_1, F^{-1}(y) = x_2$. As a function cannot map an element of its domain to more than one element of its range, this means that $F^{-1}$ would not be a function. But $F^{-1}$ is a function, so our assumption that $F$ is not injective must be false. Therefore, $F$ is injective. |
H: A question of straight lines
If the straight lines $x+y-2=0$, $2x-y+1=0$ and $px+qy-r=0$ are concurrent, then what is the slope of the member of family of lines $2px+3qy+4r=0$ which is farthest from origin?
I wrote the coefficients of the variables of the given lines in a determinant, equated it to $0$ and got $p+5q-3r=0$. Then, I substituted $r$ in the equation of the family of lines. What do I do next? How to find the line farthest from the origin?
AI: Apply the distance formula for the distance between a line and a plane to get
$$d=\frac{r}{\sqrt{p^2+q^2}}$$
Clearly r=$\frac{p+5q}{3}$
Substitute in the above equation to get
$$d=\frac{\frac{p+5q}{3}}{\sqrt{p^2+q^2}}$$
Dividing numerator and denominator by p,we get d in terms of q/p=$\lambda$. Now q/p is the slope of the line. Differentiate d w.r.t $\lambda$ and equate it to 0 to find the maximum value of the slope. |
H: Finding the limit $\lim_{n \to \infty}{\frac{\Sigma_{0}^{n}(1/n)}{\ln(n)}}$
Let
$$
\lim_{n \to \infty}{\frac{
\sum_{1}^{n}(\frac{1}{n})}{\ln(n)}}
$$
Please provide some hint or a solution.
Thanks!
AI: Using Stolz–Cesàro theorem you get that
$$\lim\limits_{n\to\infty} \frac{\sum_{k=1}^n\frac1k}{\ln n}
=\lim\limits_{n\to\infty} \frac{\frac1{n+1}}{\ln (n+1)-\ln n}
=\lim\limits_{n\to\infty} \frac{\frac1{n+1}}{\ln \frac{n+1}n}
=\lim\limits_{n\to\infty} \frac{\frac1{n+1}}{\ln (1+\frac1n)}.$$
(Of course, you should also verify that the assumptions of Stolz-Cesaro theorem are fulfilled.)
Now you can use the well-known limit $\lim\limits_{t\to0} \frac{\ln(1+t)}t=1$.
I will also mention that there are two equivalent formulations of Stolz-Cesaro theorem.
The same problem appeared in the book
Wieslawa J. Kaczor, Maria T. Nowak: Problems in mathematical analysis: Volume 1; Real Numbers, Sequences and Series as Problem 2.3.21. The problem is stated on p.39 and solved on p.186. |
H: Factorise $13$ into a product of irreducibles in $\Bbb Z[i]$
I need to factorise $13$ into a product of irreducibles in $\Bbb Z[i]$ but I'm having trouble factorising it at all.
So far I have
$13=(a+bi)(c+di)$
$13=ac + cbi + adi - bd$
So
$bc = -ad$
$ac-bd=13$
I don't know how to go any further for working out the factors for it. Any direction would be appreciated.
AI: $13=2^2+3^2=(2-3i)(2+3i)$
General result: If $p$ is a prime and
$p \equiv 1$ mod $4$ then $p=(a-bi)(a+bi)$ for some $a,b \in \mathbb{Z}$ and $(a-bi),(a+bi)$ are irreducible in $\mathbb{Z}[i]$ .
You can see the proof of this result in the page 290, ABSTRACT ALGEBRA , authors: David S.Dummit and Richard M.Foot. |
H: Find all matrices that satisfy $\mathrm B \mathrm A = \mathrm I_2$
Given the matrix $$A=\begin{pmatrix}1&8\\3&5\\2&2\\ \end{pmatrix}$$ find all $2 \times 3$ matrices in $B \in M_{2 \times 3}(\mathbb R)$ with $BA=I_2$.
Here's what I did:
$$\begin{pmatrix}a&b&c\\d&e&f\\ \end{pmatrix} \begin{pmatrix}1&8\\3&5\\2&2\\ \end{pmatrix} = \begin{pmatrix}1&0\\0&1\\ \end{pmatrix}$$
and then multiplying things out:
$$\begin{pmatrix} {a+3b+2c}&{8a+5b+2c}\\{d+3e+2f}&{8d+5e+2f}\\ \end{pmatrix} = \begin{pmatrix}1&0\\0&1\\ \end{pmatrix}$$
So would I just set
$$a+3b+2c=1$$
$$8a+5b+2c=0$$
$$d+3e+2f=0$$
$$8d+5e+2f=1$$
But then this gives $4$ equations in $6$ unknowns, so wouldn't there be infinitely many solutions? Did I do this correctly? Matrices aren't my strong point...
Thanks.
AI: Since the second column is not a multiple of the first column, the matrix $\pmatrix{1 & 8\cr 3 & 5\cr 2 & 2\cr}$ has rank $2$. That implies that a solution exists. One way to get a solution is to take any two rows of $A$ (check that they are linearly independent) and use the inverse of that $2 \times 2$ matrix for the corresponding columns of $B$, and $0$'s in the other column. Thus if you take the first two rows of $A$, the inverse of $\pmatrix{1 & 8\cr 3 & 5\cr}$ is $\pmatrix{-5/19 & 8/19\cr 3/19 & -1/19\cr}$, corresponding to
$B = \pmatrix{-5/19 & 8/19 & 0\cr 3/19 & -1/19 & 0\cr}$.
For the general solution, write the system using block matrices as
$$ [ B_1 \ b ] \left[\matrix{A_1\cr a^T\cr} \right] = B_1 A_1 + b a^T = I$$
(where $B_1$ and $A_1$ are $2 \times 2$, $b$ is $2 \times 1$ and $a^T$ is $1 \times 2$). We can solve for $B_1$ in terms of $b$: $B_1 = (I - b a^T) A_1^{-1}$. We already have $A_1^{-1} = \pmatrix{-5/19 & 8/19 \cr 3/19 & -1/19 \cr}$, so with $b = \pmatrix{b_1\cr b_2}$ and $a^T = (2, 2)$ we get
$$ B_1 = \pmatrix{1 - 2 b_1 & - 2 b_1\cr -2 b_2 & 1 - 2 b_2\cr} \pmatrix{-5/19 & 8/19 \cr 3/19 & -1/19 \cr} = \pmatrix{ (-5 + 4 b_1)/19 & (8 - 14 b_1)/19\cr
(3 + 4 b_2)/19 & (-1 - 14 b_2)/19\cr}$$
i.e.
$$ B = \pmatrix{ (-5 + 4 b_1)/19 & (8 - 14 b_1)/19 & b_1\cr
(3 + 4 b_2)/19 & (-1 - 14 b_2)/19 & b_2\cr}$$
where $b_1$ and $b_2$ are arbitrary. |
H: Computing $[T]_\beta$
Let $V=\mathbb R^2$, $T(a,b)=\begin{pmatrix} 10a-6b\\17a-10b\\ \end{pmatrix}$ and $\beta$={$(1,2),(2,3)$}.
Where $T$ is a linear operator on $V$ and $\beta$ is an ordered basis. I have to compute $[T]_\beta$ and determine whether $\beta$ is a basis consisting of eigenvectors for $T$.
I know this is probably very simple, but by book's explanation remains unclear. I just need to know how to compute $[T]_\beta$, I can figure out the eigenvectors part.
Thanks
AI: $$T(1,2)=(10-12, 17-20)=(-2,-3)=a(1,2)+b(2,3)$$ and $$T(2,3)=(20-18, 34-30)=(2,4)=c(1,2)+d(2,3)$$ for some values $a,b,c,d$. Now set a system of achieved equations and then try to find these unknown values. Indeed, $$[T]_{\beta}=\begin{pmatrix} a &c\\b &d\\ \end{pmatrix}$$ |
H: Evaluating series by contour integration, the residue theorem, and cotangent
I'm trying to understand this section in Tristan Needham's book Visual Complex Analysis about what he says is a standard method for evaluating series via a contour integral. My specific question is about the computation of the residues of $f(z) \cot \pi z$ at $z = n$. Is it in general true that $\operatorname{Res} (f \cdot g) = f \cdot \operatorname{Res} (g)$, provided that $f$ is analytic at the point of interest?
From Section 9.III.5, page 441:
AI: Consider the definition
$$\operatorname*{Res}_{z=z_0} f(z) g(z) = \frac{1}{i 2 \pi} \oint_{C} dz \, f(z) g(z)$$
where $C$ is a simple closed contour enclosing the pole of $g$ at $z=z_0$, while $f$ is analytic at $z=z_0$. We may then write
$$f(z) = a_0 + a_1 (z-z_0) + a_2 (z-z_0)^2 + \cdots$$
$$g(z) = \cdots + b_{-2} (z-z_0)^{-2} + b_{-1} (z-z_0)^{-1} + b_0 + b_1 (z-z_0) + \cdots$$
Then
$$\operatorname*{Res}_{z=z_0} f(z) g(z) = a_0 b_{-1} + a_1 b_{-2} + a_2 b_{-3} + \cdots$$
If $z_0$ is a simple pole, then only $b_{-1}$ is nonzero, and the assertion is true. If, however, $z_0$ is a multiple pole, then the residue also depends on the values of the derivatives of $f$ at the pole as well.
The above formula for the sum in terms of the residues works because $z=n$ is a simple pole of $\cot{\pi z}$. However, if $f$ has a multiple pole at a non-integer value, then the residue of $f(z) \cot{\pi z}$ must be evaluated in terms of derivatives of $\cot{\pi z}$. |
H: calculus antiderivative of the function. Find the position of the particle
A particle is moving with the given data:
$$a(t)=\cos{t}+\sin{t},\, s(0)=8,\, v(0)=5.$$
Find the position $s(t)$ of the particle.
I don't understand the problem. What do those symbols stand for?
AI: If we assume $s(t)$ to be position function then, $a(t)=s''(t),~v(t)=s'(t)$ according to the definition. Here, we have $$s''(t)=\cos(t)+\sin(t)$$ so by solving this ODE, we get $v(t)=s'(t)=\sin(t)-\cos(t)+C_1$ and by taking an integration again we get: $$s(t)=-\cos(t)-\sin(t)+C_1t+C_2$$ in which $C_1,C_2$ are some constants. Now use the initial values you 're given to find proper constants $C_1,C_2$. |
H: Let $a_1 > 1$, and for $n \in N$, define $a_{n+1} = 2 - 1/a_n$.
Let $a_1 > 1$, and for $n \in N$, define $a_{n+1} = 2 - {1\over a_n}$.
Show that the sequence $a_n$ is monotone and bounded. Find $lim_{n\to\infty} a_n$
So far I have it set up like this (not sure if it's the right way)
monotonic: $a_n < a_{n + 1}$ (increasing) or $a_n > a_{n+1}$ (decreasing)
base case: $n = 1$
we know that $a_1 > 1$ and $a_{n+1} = 2 - {1\over a_n}$
$a_{1+1} = 2 - {1\over a_1}$
I have no idea how to show that it's decreasing as n goes up with the base case being 1. Would I have to show the next one ($a_3$)?
AI: Start with an experiment. Say let $a_1=2$ and see what is happening.
It appears as if the terms are getting smaller and this is your proposition.
The sequence is monotone decreasing
$$a_{n+1}-a_n=1-\frac{2}{a_n}-a_n=\frac{2a_n-1-a_n^2}{a_n}=\frac{-a_n^2+2a_n-1}{a_n}.$$
Can you say anything about the quadratic $-x^2+2x-1$? The proof of this proposition is in there.
Further experimentation suggests that the limit is one and indeed note that if $a_n\approx1$ then
$$a_{n+1}= 2-\frac{1}{a_n}\approx 2-1=1,$$
so $a_{n+1}\approx a_n$. Perhaps $a_n>1$? In fact this follows inductively... if $a_n>1$ then so is $a_{n+1}$. Can you show this?
Now we think that $a_n\rightarrow 1$. We know the limit exists because the sequence is bounded and monotone. What could you say if the limit were $1+\varepsilon$? |
H: composition of two uniformly continuous functions.
Let $f : \mathbb{R} \rightarrow \mathbb{R}$ and $g : \mathbb{R} \rightarrow \mathbb{R}$ are two uniform continuous functions. Which of the following options are correct and why?
$f(g(x))$ is uniformly continuous.
$f(g(x))$ is continuous but not uniformly continuous.
$f(g(x))$ is continuous and bounded.
My attempt:
Every uniformly continuous function maps a Cauchy sequence to a Cauchy sequence. (Here I have a doubt, as the converse may not be true). So if $\{x_n\}$ be a Cauchy sequence, $\{f(x_n)\}$ and $\{g(f(x_n))\}$ both will be Cauchy sequence. So $g(f(x))$ will be uniformly continuous, i.e. 1 is true.
Composite function of two continuous functions will be continuous. As 1 is true, 2 is false.
$f(x) = x $ is uniformly continuous. $g(x) = \log(x)$ is uniformly continuous in $[1,\infty)$. So $g(f(x)) = \log(x)$ is uniformly continuous in $[1, \infty)$, but not in $\mathbb{R}$.
Thank you for your help.
AI: Your suspicion is justified. The argument you give for the validity of 1 is faulty since it only shows the composite preserves Cauchy sequences, but it does not show that the composite is uniformly continuous. Try working directly with the definition of uniform continuity instead to give a direct proof.
You can make things easier with the counter example to 3: Take $f(x)=g(x)=x$ on $\mathbb R$. |
H: Relatively Prime problem
If $a$ and $b$ are relatively prime integers then $b$ and $a$ plus some multiple of $b$ are also relatively prime. I can see how it works for concrete examples but can't prove it. i.e. $(a,b)=1$ implies $(b,a+kb)=1$.
AI: If $d$ divides $b$ and $a+kb$
$d$ will divide $a+kb-k\cdot b=a$ as $k$ is an integer
$\displaystyle \implies d$ will divide $(a,b)=1$ |
H: Help with simple limit calculation
I have tried searching the site for an answer but I couldnt find any even though it's a simple calculation.
$$ \lim_{n \rightarrow \infty} \left(\frac{4^n + 7^n}{4^{n-1} + 7^{n-1}}\right) $$
Thanks , Danny.
AI: $$\lim_{n \rightarrow \infty} \left(\frac{4^n + 7^n}{4^{n-1} + 7^{n-1}}\right)=\lim_{n \rightarrow \infty}\frac{7^n}{7^{n-1}} \left( \frac{(\frac{4}{7})^n+1}{(\frac{4}{7})^{n-1}+1} \right)=7 \frac{\lim_{n \rightarrow \infty} \left( (\frac{4}{7})^n+1 \right)}{\lim_{n \rightarrow \infty} \left( (\frac{4}{7})^{n-1}+1 \right)}$$
$$\lim_{n \rightarrow \infty}(\frac{4}{7})^n=0, \quad \lim_{n \rightarrow \infty}(\frac{4}{7})^{n-1}=0$$
So
$$7 \frac{\lim_{n \rightarrow \infty} \left( (\frac{4}{7})^n+1 \right)}{\lim_{n \rightarrow \infty} \left( (\frac{4}{7})^{n-1}+1 \right)}=7.\frac{(0+1)}{(0+1)}=7$$ |
H: Are all continuous bijective translations isometries?
Let $(M, d)$ be a metric space. I define a translation on $M$ to be a function $f$ from $M$ to $M$ such that $d(x,f(x))=d(y,f(y))$ for all $x$ and $y$ in $M$. In a previous question, I asked if every translation was an isometry in the post Are all metric translations isometries. The answer was no, even with continuity, and even with bijectiveness. But what if we have both bijectiveness AND continuity? Is that sufficient to force the translation to be an isometry?
AI: Let $M = \{-1, 0, 1, 2\}$ with the standard distance (in $\mathbb R$). Let
$$f(-1) =0\ , f(0) = -1\ , f(1) = 2\ , f(2) =1$$
Then $d(x, f(x)) = 1$ for all $x$ and $4= d(f(0), f(1)) \neq d(0, 1) = 1$ |
H: Volume of a solid with base of circular disk, parallel crosssections perpendicular to base are squares.
Working on a problem of volume using integration:
The problem is this:
The base of is a circular disk with radius . Parallel crosssections
perpendicular to the base are squares.
I already have an idea how the solid figure would look like but I am lost at finding a way to define my variables.
Anyone here who can give me, at least a hint?
Thanks
AI: Consider the circle $y=\pm \sqrt{r^2-x^2}$, where $r$ is the radius of the base. Then the cross-sectional area of the volume is $A(x) = (2 y)^2= 4 (r^2-x^2)$. The volume is
$$\int_{-r}^r dx \, A(x) = 8 \int_0^r dx \, (r^2-x^2)$$
I assume you got it from here. |
H: Missing step in rearrangement
Can someone explain the missing step in the following rearrangement ?
From $ \displaystyle a^2 + b^2 + \frac {a^2y}{x} + \frac {b^2x}{y} \geq (a+b)^2 $ to $ \displaystyle \frac{(a+b)^2}{x+y} \leq \frac{a^2}{x} + \frac{b^2}{y} $ ?
AI: We have
$$ \left(\frac{a^2}{x} + \frac{b^2}{y}\right)(x+y)
=\frac{a^2}{x}x +\frac{a^2}{x}y+ \frac{b^2}{y}x +\frac{b^2}{y}y
=a^2 +\frac{a^2}{x}y+ \frac{b^2}{y}x +b^2y $$
and therefore
$$ \left(\frac{a^2}{x} + \frac{b^2}{y}\right)(x+y) = a^2 + b^2 + \frac {a^2y}{x} + \frac {b^2x}{y} \geq (a+b)^2. $$
And this gives, assuming $x+y>0$
$$ \frac{(a+b)^2}{x+y} \leq \frac{a^2}{x} + \frac{b^2}{y}.$$
If $x+y<0$, the inequality would be reversed! |
H: Which one is bigger? $e^{\pi} $ or $\pi^e$
$e^{\pi}$ or $\pi^e$,
Can we find which one is bigger by using calculus?
Thanks.
AI: $$e^\pi>\pi^e\iff\pi>e\log\pi\iff\frac{\log e}e=\frac1e>\frac{\log\pi}\pi$$
Now look at the function
$$f(x):=\frac{\log x}x\;,\;\;x\ge e\implies f'(x)=\frac{1-\log x}{x^2}\le0\implies f(x)$$
is monotone descending, and thus
$$e<\pi\implies \frac1e=\frac{\log e}e=f(e)>f(\pi)=\frac{\log\pi}\pi$$ |
H: How do I simulate a simple pendulum?
I have the equation of motion of a simple pendulum as
$$\frac{d^2\theta}{dt^2} + \frac{g}{l}\sin \theta = 0$$
It's a second order equation.
I am trying to simulate it using a SDL library in C++. I know how to solve first order differential equation using Runge-Kutta method. But I can't combine all these. Can anybody help me to solve the differential equation to get the correct simulation?
Update
BTW, I am not sure whether it should be posted here or in any stack-exchange forums. But If you could help me or redirect me to any other forums, that would be helpful.
I have simulated it using Runge-Kutta method and I am adding my code below.
//simplified equations 1
double thetadot(double u)
{
return u;
}
//simplified equations 2
double udot(double theta)
{
return (-g / l) * sin(theta);
}
int main(int argc, char *argv[])
{
double theta, thetanext, u, unext, ku1, ku2, ku3, ku4, kt1, kt2, kt3, kt4;
if (SDL_Init(SDL_INIT_VIDEO) != 0)
return 1;
atexit(SDL_Quit);
SDL_Surface *screen = SDL_SetVideoMode(width, height, 0, SDL_DOUBLEBUF);
if (screen == NULL)
return 2;
//putting inital values to the function
u = u0;
theta = theta0;
while(true)
{
SDL_Event event;
while(SDL_PollEvent(&event))
{
if(event.type == SDL_QUIT)
return 0;
}
double x = xoffset + l * sin(theta);
double y = yoffset + l * cos(theta);
SDL_LockSurface(screen);
//string hanging position
draw_circle(screen, xoffset, yoffset, 10, 0x0000ff00);
fill_circle(screen, xoffset, yoffset, 10, 0x0000ff00);
//draw string
draw_line(screen, xoffset, yoffset, x, y, 0xff3366ff);
//draw bob's current position
fill_circle(screen, (int)x, (int)y, r, 0xff004400);
draw_circle(screen, (int)x, (int)y, r, 0xff3366ff);
SDL_Delay(300);
//SDL_FreeSurface(screen);
SDL_Flip(screen);
//Numerical integration of equation 1
kt1 = thetadot(u);
kt2 = thetadot(u) + 0.5 * h * kt1;
kt3 = thetadot(u) + 0.5 * h * kt2;
kt4 = thetadot(u) + h * kt3;
thetanext = thetadot(u) + (h / 6) * (kt1 + 2 * kt2 + 2 * kt3 + kt4);
//Numerical integration of equation 2
ku1 = udot(theta);
ku2 = udot(theta) + 0.5 * h * ku1;
ku3 = udot(theta) + 0.5 * h * ku2;
ku4 = udot(theta) + h * ku3;
unext = udot(theta) + (h / 6) * (ku1 + 2 * ku2 + 2 * ku3 + ku4);
//updating values
u = unext;
theta = thetanext;
}
return 0;
}
And the output is coming as follows
Can anybody let me know where it went wrong?
AI: You can convert your equation to a first order system by introducing a new function $u = \theta'$.
The system satisfied by $u$ and $\theta$ is:
\begin{align*}
\theta' &= u \\
u' &= - \frac{g}{\ell} \sin \theta.
\end{align*}
You can solve this first order system using Runge-Kutta. |
H: Cartesian Product - compute the number of unique combinations
given the number of sets
and given the number of elements in each set
What mathematical equation efficiently computes the number of unique combinations
Example:
lets say there are 3 sets {1,2,3},{1},{1,2}
and we know that the 3 sets have 3,1,2 elements respectively
The Cartesian product would be {(1,1,1),(1,1,2),(2,1,1),(2,1,2),(3,1,1),(3,1,2)} and thus the count of these 'unique combinations' would be 6.
I'm looking for the math, that would allow me to find the count without calculating the Cartesian product first
AI: If I understand your question correctly, it is just the product of the cardinalities of each set.
As an example, let $A$ and $B$ be sets of cardinality $a$ and $b$ respectively. For each of the $a$ elements of $A$ we have a different combination for every element of $B$, so in total we have $a\cdot b$ possible combinations.
Starting from this you can easily prove my claim by induction. |
H: Let $x > 0 \in \mathbb R$. Find sequences $\{a_n\}_{n=0}^{\infty}$, $\{b_n\}_{n=0}^{\infty}$ such that $a_n + b_n = n$ and $a_n/b_n \rightarrow x$
Let $x > 0 \in \mathbb R$. Find sequences $\{a_n\}_{n=0}^{\infty}$, $\{b_n\}_{n=0}^{\infty}$ of natural numbers ($a_n, b_n \in \mathbb N$) such that $a_n + b_n = n$ and $a_n/b_n \rightarrow x$ for $n \rightarrow \infty$
I've proved using the squeeze theorem for sequences that $\frac 1 n\lfloor x n\rfloor \rightarrow x $ for $n \rightarrow \infty$. However I don't know how to apply this result to find the sequences in question.
Can anyone help me out ?
AI: For each $n \geq 0$, pick $a_n$ and $b_n$ to satisfy $a_n + b_n = n$, $a_n/b_n = x$.
Substitution gives us
\begin{align*}
& b_n x + b_n = n \\
\implies & (x + 1) b_n = n \\
\implies & b_n = \frac{n}{x+1}.
\end{align*}
And $a_n = n - b_n = n - \frac{n}{x+1} = \frac{nx}{x+1}$.
Edit: to get natural number sequences, I think we can take the floor of $a_n$ and the ceil of $b_n$, but we need to prove the limit of the ratio is unchanged.
Define sequences $q$ and $r$ by $\lfloor a_n \rfloor = a_n + q_n$, $\lceil b_n \rceil = b_n + r_n$.
Then
\begin{align*}
\frac{a_n +q_n}{b_n + r_n} &= \frac{a_n/b_n + q_n/b_n}{1 + r_n/b_n} \\
& \to x \qquad \text{as } n \to \infty.
\end{align*}
This proves that
\begin{equation*}
\frac{\lfloor a_n \rfloor}{\lceil b_n \rceil} \to x
\end{equation*}
as $n \to \infty$. |
H: No. of $t,s \in [0,1]$ such that $f(t)=f(s)$ for a continuous function $f:[0,1]\rightarrow [0,1]$
Question is :
Let $f:[0,1]\rightarrow [0,1]$ be a continuous function such that $f(0)=f(1)$. Let
$$A=\{ (t,s) \in [0,1]\times [0,1] : t\neq s ; f(t)=f(s)\}$$
The number of elements in $A$ is...... ?
As $f(0)=f(1)$ and i should take care of continuity of $f$ I some how feel that the figure should be like :
which is telling me that there are infinitely many points $(t,s)$ such that $t\neq s$ but $f(t)=f(s)$
I am not sure if this should be the case and if it is so, I am not sure how to write this precisely.
I would be thankful if some one can help me to clear this problem..
Thank you
P.S : This is the first time (after so many years) I am using Paint. So, the figure looks very awkward but please adjust with that.
AI: Firstly, assume $f\not\equiv C$, since then this is trivial ($A = [0,1]^2$)
Then use the extreme value theorem together with $f(0) = f(1)$ to get at least one extremum within $(0,1)$ (if both were on the border, we'd be in the first case).
Call this extremum $(\xi, f(\xi))$ and assume it is a maximum (else consider $-f$), then by intermediate value theorem we have
$$\forall \eta \in (f(0), f(\xi)) \quad \exists
\begin{cases}
\alpha\in (0,\xi) & f(\alpha) = \eta \\
\beta\in (\xi, 1) & f(\beta) = \eta
\end{cases}$$
And thus
$$\forall \eta \in (f(0), f(\xi)) \qquad \exists \alpha \neq \beta: (\alpha, \beta) \in A$$
Since $(f(0), f(\xi))$ is infinite because $f\not\equiv C$, $A$ is also infinite for this case. |
H: Is there a proof that $n^xm^x = (n^x)^{(\log(mn)/\log(n))}$?
This isn't a homework question, just something I'm curious about, but you can treat it that way if you like.
So the other day I was playing with my calculator and I noticed that
$$
2^x10^x = (2^x)^{(\log(20)/\log(2))}
$$
I tried it out with some other numbers and came to the conclusion that
$$
n^xm^x = (n^x)^{(\log(nm)/\log(n))}
$$
So I wanted to see if there is a way to prove that.
I already know that $m = n^{(\log(m)/\log(n))}$ and I figured that there must be a relation. So from that I can see that $mn = n^{(\log(nm)/\log(n))}$. However I don't understand why that would mean that $(nm)^x = (n^x)^{\log(nm)/\log(n)}$.
Is what I say actually true? How do the powers fit into the proof?
AI: It's true, and you're almost done. Probably recalling $$a^{x\cdot y} = (a^x)^y = (a^y)^x$$ is all you need. |
H: Sum to $n$ terms.
Evaluate the following expression:
$\sum_{i=1}^n \frac{i \cdot 2^{i}}{(i+2)!}$
I can't find a way to use telescoping sums here. What else could be done?
AI: $$\frac{i\cdot 2^i}{(i+2)!}=\frac{(i+2-2)\cdot 2^i}{(i+2)!}=\frac{2^i}{(i+1)!}-\frac{2^{i+1}}{(i+2)!}$$
Observe that if we set $\displaystyle \frac{2^i}{(i+1)!}=f(i),$
$\displaystyle\frac{2^{i+1}}{(i+2)!}=f(i+1)$ |
H: Which sentences survive the passage from $X$ to the set of all functions $I \rightarrow X$?
Suppose $X$ is a mathematical structure with a single underlying set which we will also denote $X$, equipped with some functions and relations. Letting $I$ denote an arbitrary non-empty set, we see that the set of all functions $I \rightarrow X$ can be made into a mathematical structure $Y$ in an obvious way. For example, if $X$ equipped with a binary operation $+$, then for all $f,g \in Y$ we define $(f+g)(i)=f(i)+g(i).$ Similarly, if $X$ is equipped with a binary relation symbol $\leq$, then for all $f,g \in Y$ we define $$f \leq g \leftrightarrow \forall i \in I(f(i) \leq g(i)).$$
Now in the passage from $X$ to $Y$, certain sentences are preserved. For example, if $+$ is commutative on $X$, well this is the sentence $$\forall x,y \in X(x+y=y+x).$$ Its easy to see that this is preserved, in the sense that $$\forall x,y \in X(x+y=y+x) \rightarrow \forall f,g \in Y(f+g=g+f).$$
In general, it seems obvious that universally quantified equalities survive the passage; but, that's not all.
Is there a nice characterization of those sentences which, if satisfied by $X$, are also satisfied by $Y$?
Discussion. The three sentences for a partial order are all preserved, namely
$\forall x(x \leq x)$
$\forall x,y,z(x \leq y \wedge y \leq z \rightarrow x \leq z)$
$\forall x,y(x \leq y \wedge y \leq x \rightarrow x = y)$
However, linearity is not; and, this is of the form
$\forall x(x \leq y \vee y \leq x).$
So, this suggests that $\vee$ is maybe "suspect." However, sometimes its okay: for example, if we have three constant symbols $0,1$ and $2$ and we know that $X$ satisfies $(0 = 1) \vee (1 = 2),$ well then so to does $Y$.
AI: Let me start with some definitions:
A positive literal is an atomic formula and a negative literal is the negation of an atomic formula.
A clause is a disjunction of literals.
A Horn clause is a clause in which no more than one literal is positive.
A sentence is universal Horn is if it is the universal closure of a Horn clause.
It is not hard to see that universal Horn sentences that hold in $X$ are also satisfied in the $Y$. For example, the axioms for rings or partial orders can be written as universal Horn sentences but being a linear order or a field cannot be done since this are not preserved under products as you noticed. |
H: Finding where f is increasing/decreasing
$$f'(x)= (x+1)^2(x-4)^5(x-2)^4 $$
For the critical points, I got $x = -1,2,$ and $4$.
According to my professor's answer key, the interval is increasing on $(4, \infty)$ and decreasing on $(-\infty,4)$. I got increasing on $(-1,2)$ and $(4,\infty)$ and decreasing on $(-\infty,1)$ and $(2,4)$. Why are my answers wrong and why doesn't my professor consider the $-1$ and $2$ when determining where the function increases/decreases?
AI: Observe that $$(x+1)^2(x-2)^4(x-4)^4\ge0$$ for real $x$
So, the sign of $f'(x)$ will be dictated by that of $x-4$ unless $(x+1)^2(x-2)^4\ne0$ |
H: Conjecture similar to Fermat's Theorem.
I was wondering about a problem which i could reduce to asking the following
Does there exist a set $a,b,c$ of prime numbers such that $$a^a+b^b=c^c$$
Is it really a tough problem or do you think it can be solved with some amount of work .
Thank you for your ideas and help .
AI: Assume $a, b, c$ are primes with $a<b<c$. Then $b$ and $c$ are both odd, so that $a^a$ must be even, so that $a=2$.
Then we have $4+b^b=c^c$, but with $c$ being at least 2 more than $b$, this is impossible, so there are no solutions with all primes.
Edit:
A slightly different argument shows that there are no solutions in natural numbers even if we do not insist that $a, b, c$ be prime:
$$(n+1)^{n+1} > n^{n+1} = n\times n^n$$
so that for $n > 2$ we have
$$(n+1)^{n+1} > 2n^n \geq a^a + b^b$$
for any $a, b \leq n$, so it is impossible to add 2 terms of the sequence $1^1, 2^2, 3^3, \ldots$ to get another term of the sequence. |
H: Is the compact interval $[0,1]$ in the usual topology compact in this new topology?
Let $\mathbb{R}$ be a topological space with topology consisting of the sets $A \cup B$, where $A$ is open in the usual topology, and $B \subseteq \mathbb{R} \setminus \mathbb{Q}$. Is the interval $[0,1]$ compact in this topology?
I think that it might be, as if we have an open cover of $[0,1]$ in this topology, the cover consists of open sets of the form $A \cup B$. $B$ is a set of irrational numbers in the interval $[0,1]$. We can always find an open interval in $[0,1]$ to cover each irrational number, since for any irrational $p \in [0,1]$ there are $a,b \in \mathbb{Q} \cap [0,1]$ such that $a < p < b$, and then $p \in (a,b)$. Therefore now we have an open cover entirely composed of sets from $A$. But we know closed intervals in the usual topology are compact, so then $[0,1]$ is compact in the $A \cup B$ topology.
Could you please tell me whether my argument is correct or not?
AI: It is not a compact space.
Let $\pi_n$ be a positive increasing succession converging to an irrational number $p<1$. You had that $\{p\}=\emptyset \cup \{p\}$ is open in $[0,1]$.
So in particular $X=\cup_{n>0} \{(0,\pi_{n})\} \cup \{[0,\pi_1),\{p\},(p,1]\}$ is an open cover, from where you cannot extract a finite one. |
H: Prove that $\{(x,y) \in \mathbb R^2 | y = x^2 \}$ is not compact
I know I need to choose an open cover and then show it has no finite subcover.
If I use $((-n,-n^2),(n,n^2)) \forall n \in \mathbb N$ does this work?
AI: Hint : prove that the given set is not bounded. |
H: Laplace transform of the following function
find the laplace transform of the function :
$$f(t) =\begin{cases}
t^2, & 0<t<1 \\
2\cos t+2, & t>1 \\
\end{cases}$$
My attempt:
$$L\{f(t)\}=\int_{0}^{1}e^{-st} \ t^2 \ \text{d}t+\int_{1}^{\infty}e^{-st} \ (2\cos t+2) \ \text{d}t$$
Now,
$$\int_{0}^{1}e^{-st} \ t^2 \ \text{d}t=\frac{-1}{s}e^{-s}-\frac{2}{s}te^{-s}-\frac{2}{s^3}e^{-s}+\frac{2}{s^3}$$
And
$$\int_{1}^{\infty}e^{-st} \ (2\cos t+2) \ \text{d}t$$
But the integration is not stopping.
AI: $$\begin{align} \int_1^{\infty} dt \, e^{-s t} \cos{t} &= \Re{\left [\int_1^{\infty} dt \, e^{-(s-i) t} \right ]}\\ &= \Re{\left [\frac{e^{-(s-i)}}{s-i} \right ]}\\ &= e^{-s} \Re{\left [(\cos{1}+i \sin{1}) \frac{s+i}{s^2+1}\right ]} \\ &= \frac{s \cos{1}-\sin{1}}{s^2+1} e^{-s} \end{align}$$
$$ \int_1^{\infty} dt \, e^{-s t} = \frac{e^{-s}}{s}$$
Multiply by $2$, add, done. |
H: How to evaluate $\displaystyle \lim_{n\to \infty}\left[a^n+b^n+c^n\right]^{1/n}$
What is the $\displaystyle \lim_{n\to \infty}\left[a^n+b^n+c^n\right]^{1/n}$, assuming that $0<a<b<c$?
I think that, as 1/n tends to zero, the limit 1. Is this correct?
AI: Hint:
Assuming that $0\le a\le b\le c$, then
$$(c^n)^{\frac{1}{n}}\le(a^n+b^n+c^n)^{\frac{1}{n}}\le(3c^n)^{\frac{1}{n}}$$
Now let $n\to\infty$,... |
H: How find this equation solution $2\sqrt[3]{2y-1}=y^3+1$
find this equation roots:
$$2\sqrt[3]{2y-1}=y^3+1$$
My try: since
$$8(2y-1)=(y^3+1)^3=y^9+1+3y^3(y^3+1)$$
then
$$y^9+3y^6+3y^3-16y+9=0$$
Then I can't.Thank you someone can take hand find the equation roots.
AI: Let $f(y) = \frac12 (y^3+1)$, we have
$$2\sqrt[3]{2y-1} = y^3+1\quad\iff\quad f^{-1}(y) = f(y) \quad\implies\quad y = f(f(y))$$
Since $f(y)$ is a strictly increasing function in $y$, we have
If $f(x) > x$, then $f(f(x)) > x$.
If $f(x) < x$, then $f(f(x)) < x$.
This means
$$\begin{align}
y = f(f(y)) \implies & y = f(y)\\
\iff & y^3+1 - 2y = 0\\
\iff & (y^2 + y -1 )(y-1) = 0\\
\implies & y = 1\text{ or } \frac{-1\pm \sqrt{5}}{2}
\end{align}$$ |
H: Drawing phase planes
I have to draw the phase plane of Y'=$
\begin{pmatrix}
-3 & 1 \\
1 & -3 \\
\end{pmatrix}
$Y
The general solution of the system was,
Y(t)=C$_1$$
\begin{pmatrix}
1 \\
1 \\
\end{pmatrix}
$e$^{-2t}$+C$_2$$
\begin{pmatrix}
1 \\
-1 \\
\end{pmatrix}
$e$^{-4t}$
I know how to get the direction of the trajectories(plug a point to the equation y= Y'=AY).But I don't understand to which direction these trajectories bend.This gives a improper node.
In the class the lecturer considered separate cases when c$_1$=0 ,c$_2$=+/-1 and c$_2$=0,c$_1$=+/-1.
I don't understand how the direction to which these trajectories bend was decided with this .Any help on how I can decide on the direction.
AI: You can look at the eigenvalues.
You can also look at the derivative in the different quadrants where
we have:
$$\dfrac{dy}{dx} = \dfrac{-3x+y}{x-3y}$$
You can draw the solution curves using various initial conditions. In your approach, we have $c_1 = 0, c_2 = \pm 1$. We could, for example, parametrically plot $x(t) = e^{-4t}, ~y(t) = -e^{-4t}$ to see how the trajectory looks for various values of $t$. We see (the direction arrow is pointing in toward the origin):
You can also see: Difference between improper node and proper node for phase portrait
Since both eigenvalues are negative and real, this is an asymptotically stable node. We have the critical point at $(0,0)$.
This leads to the phase portrait: |
H: Positively non-positive (from Brilliant.org) Whats wrong with my method?
(taken from brilliant.org)
For how many positive integers $N$ between $3$ and $1000$ (inclusive) is the following statement true:
If $\{a_i\}^N_{i=1}$ is a set of $N$ (not necessarily distinct) real numbers such that
$a_1+a_2+…+a_N=0$,
then we must (always) have
$a_1a_2+a_2a_3+…+a_{N−1}a_N+a_Na_1≤0$?
I think that this is true for all $N$, because we can have all $a_{2k}$ to be a positive number and all $a_{2k+1}$ to be a negative number, then if $N$ is odd then $a_1$ is $0$, this way positive numbers and negative numbers are alternating so the product of them is negative so the sum of the products will be negative, so the answer is $998$, but when I entered it on Brilliant it says my answer is wrong, what is wrong with my method
AI: In order for that statement to be true for a given $N$, the conclusion must hold true for all possible choices of $a_1,a_2,\ldots,a_N$ whose sum is $0$.
In other words, you can not decide that a particular $N$ works by specifying that the $a_{2k}$'s are nonnegative and the $a_{2k+1}$'s are negative, since, while the conclusion of the statement might hold for that choice, it might not hold for other choices. For example, your method doesn't seem to cover the possibility that $a_1=1,a_2=1,\ldots,a_{N-1}=1$ and $a_N=1-N$. |
H: Indefinite intergral of $\int { \sqrt{ x^2-a^2} \over x } \mathrm dx $
I need to integrate $$\int { \sqrt{x^2-a^2} \over x } \mathrm dx $$ using substition, and show it equals $$ \sqrt{x^2 - a^2} - a(\operatorname{arcsec} ({x\over a })) +c $$
I've tried $x=a\sin t$ but I couldn't finish it out.
Thank in advance for any help.
AI: Use the substitution:
$$x = a \sec(\theta)\implies dx = a \sec\theta\tan\theta\,d\theta$$
And use the identity
$$\sec^2(\theta)-1 = \tan^2(\theta).$$
$\require{cancel}$
Then your integral becomes: $$\begin{align}\int { \sqrt{x^2-a^2} \over x } dx & = \int \dfrac{\sqrt{a^2\sec^2 \theta - a^2}(\cancel{a \sec\theta}\tan\theta \,d\theta) }{\cancel{a\sec\theta}}\, \\ \\ & = a\int \sqrt{\tan^2\theta}\, \tan\theta\,d\theta \\ \\ &= a\int \tan^2 \theta \,d\theta \\ \\ & = a\int (\sec^2 \theta - 1)\,d\theta \\ \\ & = a \int \sec^2 \theta \,d\theta - a\int d\theta\\ \\ & = a \tan\theta - a\theta + c\end{align}$$
Now, back substitute to get $$\sqrt{x^2 - a^2} - a\left(\operatorname{arcsec} \left({x\over a }\right)\right) +c$$ |
H: estimate on $| \nabla (u |u|^2) - \nabla(w|w|^2)|$ for $u,w \in H^1$
suppose $u, w \in H^1 (R^2)$. I'd like to know where does the following inequality come from (it appears in a proof I've been reading and I can't figure it out)
$$ | \nabla (u |u|^2) - \nabla(w|w|^2)| \leq C | \nabla(u-w)| \cdot(|u|^2 + |w|^2)
+ C|u-w|\cdot(|\nabla u| + |\nabla w|)(|u|+|w|)$$
$u$ and $w$ are complex valued. Can anyone provide some suggestions? C is some constant, its value isn't importnat.
Perhaps I should mention that I'm going to use it with integrals ($L^2$), but I'm pretty sure it should be a pointwise inequality
AI: Let $f:\mathbb{C}\to\mathbb{C}$ be the function $f(x,y) = (x+iy) (x^2 + y^2)$. Its derivative is
$$ \mathrm{d}f = (x^2 + y^2) \mathrm{d}(x + i y) + 2(x + i y) (x \mathrm{d}x + y \mathrm{d}y) $$
which satisfies
$$ |\mathrm{d}f| \leq C(x^2 + y^2) $$
Integrating along the straight line joining $v,w\in \mathbb{C}$ gives
$$ |f(w) - f(v) | \leq C \max(|v|^2,|w|^2) |v-w| \leq C(|v|^2 + |w|^2) |v-w|$$
Now let $w,v$ be differentiable $\mathbb{C}$ valued functions.
Since you are applying to functions in $L^2$ this is okay by density.
Since there seems to be some confusion, let me do this in a bit more detail.
Let $u_s = (1-s) v + s w$ so that $u_0 = v$ and $u_1 = w$ be a one parameter family of functions. What we should do is to compute
$$ \nabla f(v) - \nabla f(w) = \nabla f(u_0) - \nabla f(u_1) = \int_1^0 \partial_s \nabla f(u_s) \mathrm{d}s $$
The term (writing $u = u_s$ and $\dot{u} = \partial_s u_s$)
$$ \partial_s \nabla f(u) = \nabla \left[ 2\bar{u} u \dot{u} + u^2 \dot{\bar{u}}\right] $$
Putting in the $\nabla$ you see that by the product rule you have
$$ \partial_s \nabla f(u) = 2 \nabla \bar{u} u \dot{u} + 2 \bar{u} \nabla u \dot{u} + 2 \bar{u} u \nabla \dot{u} + 2 u \nabla u \dot{\bar{u}} + u^2 \nabla \dot{\bar{u}} $$
Now, since $u$ is linear in $s$, for each component of the $\nabla$ the term with $\dot{u}$ is signed. So putting in absolute value signs and integrating both sides give you the desired inequality. |
H: Correct interpretation of Kleene (Intro to Metamathematics) symbol $\vdash^x$ in Predicate Calculus
Rif. S.C.Kleene, IM (1952) : which is the correct interpretation (or the "modern equivalent") of the "x" used as exponent of the "turnstile" as in:
$$A(x) \vdash^x \forall xA(x)$$
[see Derived Rules (Th.2, pag.98)] ?
Is it similar to Principia Mathematica usage of → (original "horseshoe") with "x" subscripted to indicate universally quantified variables ?
If so, the "implicit" universal quantifier has as his scope the Assumption only or all the context of the proof ?
There are some related examples at pag.149 [Examples 2 to 4].
I think that they are Kleene's version of similar examples of wrong application of quantifier rules in Nat Deduction.
$$x=0$$ (Assumption)
$$\forall xx=0$$ ($\forall$-Intro, illegal)
$$x=0 → \forall xx=0$$ (→Intro, discharging the Assumption)
$$\forall x(x=0 → \forall xx=0)$$ ($\forall$-Intro: at this step the Ass has been discharged)
$$0=0 → \forall xx=0 $$ ($\forall$-Elim).
The first occurrence of forall-Intro is illegal, because the assumption x=0 depends on itself.
AI: There's some text on it, right above, and in, Theorem 2:
The variable "x" written as superscript on the symbol "$\vdash$" in two of the rules is to mark the application of Rule 9 or 12 with respect to x in constructing the resulting deduction.
For the predicate calculus (or full number-theoretic system), all the rules hold, provided that in each subsidiary deduction the free variables are held constant for the assumption formula to be discharged.
Where "Rule 9 and 12" come from p.82:
$$\begin{align}
9.&&\frac{C\supset A(x)}{C\supset \forall x A(x).}\\
12.&&\frac{A(x) \supset C}{\exists x A(x) \supset C.}
\end{align}$$
Now, what does this mean? The $\vdash^x$ notation basically conveys that $x$ is no longer allowed to occur in any "discharged assumption" (introduced on p.94). As such, it is not but the counterpart of the "freeness" condition that is present in rules 9 and 12.
It would therefore seem that the scope of the implicit universal quantification is the collection of all assumptions done in the proof (in that every assumption is implicitly quantified over $x$, whether or not this makes any difference or not).
(But the language is convoluted, so I'm not entirely certain.) |
H: How to prove(or disprove) $\begin{vmatrix} A&B\\ B&A \end{vmatrix}=|A^2-B^2|$
Let $A$ and $B$ be square matrices of the same size.
(1) If $f$ is not invertible and $AB=BA$, show that
$$\begin{vmatrix}
A&B\\
B&A
\end{vmatrix}=|A^2-B^2|.$$
(2) If $A$ is invertible and $AB\neq BA$, then do we have
$$\begin{vmatrix}
A&B\\
B&A
\end{vmatrix}=|A^2-B^2|?$$
My try: I have only proved this: if $A$ is invertible and $AB=BA$, then
$$\begin{vmatrix}
A&B\\
B&A
\end{vmatrix}=|A^2-B^2|.$$
For $(1)$ and $(2)$, I can't prove. Thank you.
AI: Let
$$
J=\left(\begin{array}{cc}I&0\\0&-I\end{array}\right)
$$
and
$$
M(A,B)=\left(\begin{array}{cc}A&B\\B&A\end{array}\right).
$$
Then $\det J^2=\det I=1$, so $\det(JM(A,B)J)=\det M(A,B)$. But (check this)
$$JM(A,B)J=M(A,-B),$$
so $\det M(A,B)=\det M(A,-B)$.
On the other hand using $AB=BA$ we get
$$
M(A,B)M(A,-B)=\left(\begin{array}{cc}A^2-B^2&-AB+BA\\BA-AB&A^2-B^2\end{array}\right)=\left(\begin{array}{cc}A^2-B^2&0\\0&A^2-B^2\end{array}\right).
$$
Putting these together we get
$$
\det M(A,B)^2=\left|\begin{array}{cc}A^2-B^2&0\\0&A^2-B^2\end{array}\right|.
$$
Can you take it from here? |
H: Limit with Stolz-Cesàro theorem: $\lim\limits_{n\to \infty} \frac{1+2\sqrt2+3\sqrt3+\ldots+n\sqrt n}{n^2 \sqrt{n}}$
$$\lim_{n\to \infty} \frac{1+2\sqrt2+3\sqrt3+\ldots+n\sqrt n}{n^2 \sqrt{n}}= \text{?}$$
Book has no answers.
It's on Stolz-Cesàro theorem lesson, if that helps. Can't find a solution.
AI: This is the direct application of the Stolz-Cesàro theorem
$$
\begin{align}
&\lim_{n\to \infty} \frac{1+2\sqrt{2}+...+n\sqrt{n}}{n^2 \sqrt{n}}\\
=&\lim_{n\to \infty} \frac{(1+2\sqrt{2}+...+(n+1)\sqrt{n+1})-(1+2\sqrt{2}+...+n\sqrt{n})}{(n+1)^2 \sqrt{n+1}-n^2\sqrt{n}}\\
=&\lim_{n\to \infty} \frac{(n+1)\sqrt{n+1}}{(n+1)^2 \sqrt{n+1}-n^2\sqrt{n}}\\
=&\lim_{n\to \infty} \frac{(n+1)\sqrt{n+1}((n+1)^2 \sqrt{n+1}+n^2\sqrt{n})}{((n+1)^2 \sqrt{n+1}-n^2\sqrt{n})((n+1)^2 \sqrt{n+1}+n^2\sqrt{n})}\\
=&\lim_{n\to \infty} \frac{(n+1)\sqrt{n+1}((n+1)^2 \sqrt{n+1}+n^2\sqrt{n})}{(n+1)^5-n^5}\\
=&\lim_{n\to \infty} \frac{(n+1)^4+(n+1)n^2\sqrt{n(n+1)}}{5n^4+10n^3+10n^2+5n+1}\\
=&\lim_{n\to \infty} \frac{(1+n^{-1})^4+(1+n^{-1})\sqrt{1+n^{-1}}}{5+10n^{-1}+10n^{-2}+5n^{-3}+n^{-4}}\\
=&\frac{2}{5}
\end{align}
$$ |
H: What is the closure of $ C^\infty_c(\mathbb{R}^n\setminus\{0\})$ in Sobolev $ W^{1,p} $ norm?
For $1 \leq p < \infty, n\geq 1 $ my guess of the answer was $ W^{1,p}(\mathbb{R}^n)$ but I can't prove the inclusion $ \overline{C^\infty_c(\mathbb{R}^n\setminus\{0\})} \subseteq W^{1,p}(\mathbb{R}^n) $.
Any hints or partial solutions( like for $ p = 2$) would be extremely helpful. Thanks.
AI: In the case $n = 1$ or ($n \ge 2$ and $p> n$), you have the embedding from $W^{1,p}$ to $L^\infty$. In this case $\overline{C_c^\infty}\subsetneq W^{1,p}(\mathbb R^n)$.
Indeed, every limit point of $\overline{C_c^\infty}$ is continuous and zero at $0$.
In the other case, you get $\overline{C_c^\infty} = W^{1,p}(\mathbb R^n)$. This can be seen by taking an arbitrary function in $W^{1,p}(\mathbb R^n)$, mollifying it and truncating it (smoothly) at $0$ and $\infty$.
All steps but the truncation at $0$ are standard. For this truncation, recall, that there is a function $\varphi \in C_c^\infty(\mathbb R^n)$, which is equal to one in a neighbourhood of $0$ and has an arbitrary small $W^{1,p}$-seminorm. Multiplication with $1-\varphi$ will do the job. |
H: Finding $\lim_{x\to+\infty} (3^x-x)^{1/(4x)}$
I have to find the limit of $(3^x-x)^{1/(4x)}$ as $x\to+\infty$ without using de l'Hôpital's method or Taylor series.
I've tried to use some notable limits as $(1+1/t)^t$ or other but the problem is the fact that $x$ goes to infinity, then I tried to use substitution but I can't manage it.
Please can somebody help me in a clear way?
AI: $$f(x)=\left(3^x-x\right)^{1/4x}=\sqrt[4]3\left(1-\frac x{3^x}\right)^{1/4x}\xrightarrow[x\to\infty]{}\sqrt[4]3$$
The above is based on
$$\begin{cases}\lim_{x\to x_0}f(x)=1\\\text{and}\\\lim_{x\to x_0}g(x)=0\end{cases}\;\implies\;\lim_{x\to x_0}f(x)^{g(x)}=1$$ |
H: Conditional Probability coin tossing
I was going through some exercises on probability and came across a question. Two people A and B are tossing a fair coin with A tossing first. The process is repeated till someone gets a heads. What is the probability of A winning.
I came across a solution where it says since A tosses first, probability of A winning in first outcome is 1. If B tosses and wins then it's probability is 1/2. Then it goes on to state that P(B) = 1/2 P(A)
P(B) + P(A) = 1
Therefore P(A) = 2/3 and P(B) = 1/3
I could not grasp how the probability of A tossing first and winning is 1 and not 1/2. Is this solution correct. If it is could anyone please clarify my doubt. Thanks
AI: I suspect that you came across some form of the following argument. Let $p$ be the probability that the first player wins. What is the probability that the second player wins? With probability $\frac12$ the first player tosses tails, and in effect the game starts over with $B$ now becoming the first player and having probability $p$ of winning. Thus, $B$ wins precisely when $A$ tosses tails and $B$ then goes on to win as if he were the first player, so $B$ wins with probability $\frac12p$. Clearly $p+\frac12p=1$, so $p=\frac23$. |
H: Compact Operators and Complete Metrics Spaces
I have a couple of questions about compact operators and compactness in complete metric spaces:
1.I have the following implications: Let $Y$ be a metric space with $A$ a subset of $Y$. $A$ is precompact iff $\bar{A}$ is sequentially compact iff any sequence in $A$ has a convergent subsequence in $Y$.
Do these implications only hold if $Y$ is a complete metric space?
2.The following are characteristics of a bounded compact operator T:
A bounder operator $T$ is compact iff any of the following is true:
Image of the unit ball in $X$ under $T$ is relatively comapact in $Y$.
Image of any bounded set under $T$ is relatively compact in $Y$.
Image of any bounded set under $T$ is totally bounded in $Y$.
For any sequence $(x_{n})_{n}$ from the unit ball in $X$, the sequence$(Tx_{n})_{n}$ contains a Cauchy sequence.
Do these implications depend on whether $X$ and $Y$ are Banach Spaces?
3.Are compact operators always defined on Banach Spaces?
AI: Ad 1., yes. A subset of a metric space can be precompact (totally bounded) without its closure being compact (which for metric spaces coincides with being sequentially compact) if the space is not complete. The equivalence that every sequence in $A$ has a convergent (in $Y$) subsequence if and only if $\overline{A}$ is (sequentially) compact also holds in non-complete metric spaces and of course always implies that $A$ is precompact.
Ad 2., it depends on $Y$ being a Banach space. See first part for why.
Ad 3., no, compact operators are also defined on other topological vector spaces, for example Fréchet spaces. One usually wants the target space to be complete when considering compact operators, but that is not much of a restriction, since every (Hausdorff) topological vector space has a completion. |
H: Given $\begin{pmatrix} a & b\\ c & d\end{pmatrix}∈GL_2^+(R)$ , $\beta(w)=\frac{aw+b}{cw+d},\Im(w)>0$.Is $\beta$ bijective?
Given any matrix $A=\begin{pmatrix} a & b\\ c & d\end{pmatrix}∈GL_2^+(R)$, we can define a function $\beta:H\to{ \mathbb{C} }$ by $$\beta(w)=\frac{aw+b}{cw+d},w∈H$$,where $H$ is the upper complex plane.Is $\beta$ bijective?
I compute the imaginary of $\beta(w)$,$$\Im(\beta(w))=\frac{|A|}{(cw+d)^2}\Im(w)$$
then $\beta(w)∈H$.So $\beta:H\to{H}$. And I think it is a bijection,but I fail to prove that it is an onto. Could you show that $\beta$ is a bijection or not?Thank you in advance!
AI: You can try to compute the inverse of $\beta$. In other word, we solve for $w$ in term of $\beta$, namely, $z=\frac{d\beta-b}{-c\beta +a}$
So for for each $k \in \mathbb{C}$ ,$\exists$ $z=\frac{dk-b}{-ck +a}$, such that $\beta(z)=k$ which gives the surjectivity of $\beta$. |
H: Find the eigenvalues and eigenvectors of A geometrically
I am really confused with this question:
Find the eigenvalues and eigenvectors of A geometrically: $$ A = \begin {pmatrix} 0 & 1 \\ 1 & 0 \end {pmatrix} $$^ reflection in the line $y=x$.
Thanks.
AI: I am trying to understand the question. Forgive me if I am totally off.
But reflection in $y=x$ means geometrically that any vector on the line, stays where it is, which corresponds with an eigenvalue of 1. Any perpendicular vector to the line, is mapped exactly to its "other" side, which corresponds with an eigenvalue of -1. (The entries of the vector are switched signs) |
H: Dual of the space of all convergent sequences
I need to find what it wrong with my logic and Ii will be glad if someone can told me what I do wrong.
Define $C$ be the subspace of $ l^{\infty} $ that consists of convergent sequences and let $C_0$ be the subspace of $C$ that consists of sequences that converge to 0.
I need to show that $C_0^*$ is isometric to $l^1$ .
I already show that $C_0$ is isometric to $l^1$ and I want to use the fact that if $Z = X \oplus Y $ then $Z^* = X^* \oplus Y^* $.
So I claim that $ C \cong C_0 \oplus \mathbb C$ because every converge sequence I can write like this:
let $\{x_n\}_{n=1} ^ \infty$ that converge to $d$ then we can look at the following sequence $y_n =(x_n- d) + d $
by this observation I conclude that $C \cong C_0 \oplus \mathbb C$
because every constant sequence we can id by it first element.
Then by what I write above I get $C^* \cong C_0^* \oplus \mathbb C ^* $.
So I get $C^* \cong l^1 \oplus \mathbb C ^*$
and to get that is isometric to $l^1$ i think that it wrong because I think that there not such a function that will be 1-1 in this case.
So my questions are
1) what I doing wrong, because I need to prove this and not to say that its wrong?
2) If all that I write is correct can I conclude from this that $C_0$ not isometric to $C$ (because $C \cong C_0 \oplus \mathbb C$)?
thank
AI: When you use sums of Banach spaces it is better emphasize the type of norm you are using on this sum. With slight modifications your proof reads as follows
$$
c^*\cong(c_0\oplus_\infty\mathbb{C})^*\cong_1 c_0^*\oplus_1\mathbb{C}^*\cong_1 \ell_1\oplus_1\mathbb{C}
$$
It is remains to show that $\ell_1\oplus_1\mathbb{C}\cong_1 \ell_1$, but it is really easy - just consider isometric bijective linear operator
$$
I:\ell_1\oplus_1\mathbb{C}\to\ell_1:(x_1,x_2,\ldots)\oplus_1 z \mapsto(z,x_1,x_2,\ldots)
$$ |
H: Convert 59 to octal
I'm reading Discrete Mathematics by Kevin Ferland and I'm stuck with exercise 33 of Chapter 0: Write the octal number equivalent of 59.
I'm following this procedure:
59/8 | 3
7/8 | ??
By using an online calculator, the result should be 73, however how can the reminder of 7/8 be 7?
AI: When you divide $7$ by $8$ you get an integer quotient of $0$ and a remainder of $7$: $7=0\cdot8+7$. This is the unique representation of $7$ in the form $8q+r$, where $q$ and $r$ are integers, and $0\le r<8$. |
H: Continuity of a multivariable absolute value function
The function is as follows:
$$f(x,y)=\sqrt{\left |xy \right |} $$ I have to check whether it is continuous, differentiable and has defined partial derivatives at $(0,0)$. My attempt is as follows:
Function is discontinuous at the origin.
Not differentiable at the origin because of the pointy peak (haha); and
Partial derivatives are as follows: $f_x=\frac{\sqrt y}{2\sqrt x}$ and $f_y=\frac{\sqrt x}{2\sqrt y}$ and are not defined at origin.
Is my reasoning correct?
AI: Since $\sqrt{|x|} \to 0$ as $x \to 0$, it follows immediately that
$$
\sqrt{|xy|} = \sqrt{|x|} \sqrt{|y|} \to 0
$$
as $|x|+|y| \to 0$. Hence $f$ is continuous at the origin.
Since $f=0$ on both coordinate axes, the two partial derivatives do exist at the origin, and are both zero. Can you study the issue of differentiability by yourself? Hint: check that the directional derivative along the direction $\mathbf{v}=(1,1)$ does not exist at the origin. |
H: Need help understanding the factorial formula $n!=n(n-1)(n-2)\cdots(3)(2)(1)$
The caption says the following:
If $n$ is an integer such that $n \ge 0$ then $n$ factorial is defined as,
$$n!=n(n-1)(n-2)\cdots(3)(2)(1)$$
if $n \ge 1$ by definition.
I'm really just confused by the $(3)(2)(1)$ in the formula, and if $n!$ is really just (for example): $5! = (5)(4)(3)(2)(1)$ then why wouldn't this just end when it hits $(n-4)$ if $n$ is 5?
I'm trying to understand this since sequence and series requires the cancellation of factorials under certain tests.
AI: It seems that your confusion here is with the presentation of the formula; so, let me give you a slightly different description. Perhaps that description will make the notation you used make more sense.
For $n\in\mathbb{N}$, we define $n!$ to be the product of all natural numbers between $1$ and $n$, inclusive: that is,
$$
n!=\prod_{i=1}^{n}i=1\cdot2\cdot3\cdots\cdot(n-1)\cdot n.
$$
The "$\cdots$" notation takes a bit of getting used to; it is basically intended to mean "and so on" or "continuing in this way". It expresses that there's a pattern at work, and that the pattern continues on in the obvious way. The first few terms are used to establish the pattern, and the final terms describe where the pattern stops. |
H: Expected value and variance on exponential distribution
The length of time Y necessary to complete a key operation in the construction of houses has an exponential distribution with mean 10 hours. The formula C = 100 + 40Y + 3Y^2 relate the cost C of completing this operation to the square of the time to completion. Find the mean and variance of C.
For E(C), I found 1100
For V(C), I found V(100+40Y+3Y^2) = V(100) + V(40Y)+ V(3Y^2)
= 1600V(Y) + 9V(Y^2). However, I do not know how to evaluate 9V(Y^2)?
Could anyone check my answer and help me out to evaluate 9V(Y^2)?
AI: The density function of $Y$ is $\lambda e^{-\lambda y}$ (for $y\gt 0$).
Your answer for $E(C)$ is correct. The proposed procedure for calculating the variance of $C$ is not correct. The formula for the variance of a linear combination of independent random variables was used. But $40Y$ and $3Y^2$ are not independent.
We can simplify the calculation a little by discarding the constant $100$. That does not change the variance. So we want
$$E((40Y+3Y^2)^2 -(E(40Y+3Y^2))^2.$$
To calculate the first term, we need
$$\int_0^\infty (40y+3y^2)^2 \lambda e^{-\lambda y}\,dy.$$
Expand. We can save some integrations by looking up or computing the mgf of our exponential. |
H: Are these disjoint/dependent?
Given $P(A) = 0.7, P(B) = 0.6, P(A^c | B^c) = 0.25$, are:
I) $A$ and $B$ disjoint?
II) $A$ and $B$ dependent?
So, what I said:
$I)$ Since $P(A) + P(B) = 1.3 > 1$ then $P(A \cap B) \neq \emptyset$, thus $A, B$ are non-disjoint.
II) $P(A^c | B^c) = \frac{P(A^c \cap B^c)}{P(B^c)} = 0.25 \rightarrow P(A^c \cap B^c) = 0.1$. But $P(A^c \cap B^c) = 1 - [P(A) + P(B) - P(A \cap B)] \rightarrow P(A \cap B) = 0.55$. From here we check: is $P(A \cap B) = P(A) \cdot P(B)? \rightarrow$ is $0.55 = 0.7 \cdot 0.6?$ NO! thus $A, B$ are dependent.
Is that correct or am I missing something? (new to probability)
AI: You are correct and so are your derivations, except that $P(A\cap B)=.4$ not 0.55.
$P(A^c\cap B^c)=.1=1-P(A\cup B)=1-[P(A)+P(B)-P(A\cap B)] \rightarrow 0.9=.7+.6-P(A\cup B) \rightarrow P(A \cup B) = .7+.6-.9 = .4 \neq .7*.6 = .42$
Disjoint events have additive probabilities for their union and depnendent events are simply not independent, so you verified this witha a counterexample. |
H: Generating points from a standard Gaussian
I'm new to Gaussian distributions and I'm trying to generate say, $ N$ points from a $ M$ dimensional standard gaussian. What does this mean? How would I do this in matlab?
AI: here's how in Matlab. In general, you can generate a uniform random variable on [0,1] then feed it into the inverse CDF (i.e,$\Phi^{-1}$) for the gaussian to get a random variable. |
H: Parametric Equations: Find $\frac{\mathrm d^2y}{\mathrm dx^2}$.
Find $\dfrac{\mathrm d^2y}{\mathrm dx^2}$, as a function of $t$, for the given the parametric equations: $$\begin{align}x&=3-3\cos(t)\\y&=3+\cos^4(t)\end{align}$$
$\displaystyle\dfrac{\mathrm d^2y}{\mathrm dx^2}=\ldots$
I don't really understand this section that I am learning at all, is there any useful website I can look over to help me understand this concept better?
Thanks!
AI: It is known that $$ \dfrac {\mathrm{d}y}{\mathrm{d}x} = \dfrac {\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}}. $$Do you know how to find $y'(t)$ and $x'(t)$? Do this to get $\frac{\mathrm{d}y}{\mathrm{d}x}$ and then differentiate again. I am not finishing the whole thing for you since you haven't shown your work, but I think this hint should help you enough to finish. |
H: How many inverse relations
How many inverse relations are there for an n-element set?
I know that $R \circ R^{-1}=R^{-1} \circ R$ where $R$ is an invertible relation, but that's as far as I can get.
AI: Every relation has an inverse relation, so this question is really asking how many relations there are on some ground set $A$. A relation is a subset of $A\times A$, which has size $|A\times A|=|A|\times |A|=n^2$. There are $$2^{n^2}$$ subsets of $A\times A$, and therefore this quantity of inverse relations. |
H: proving a sequence has exactly four limits points
Given the sequence \begin{align*}
x_n:=\begin{cases}
1,&\textrm{if }n\equiv0 \mod 4\\
2,&\textrm{if }n\equiv 1\mod 4\\
3,&\textrm{if }n\equiv 2\mod 4\\
4,&\textrm{if }n\equiv 3\mod4
\end{cases}
\end{align*}
I want to show that the limit points are exactly $1,2,3,4$
So $1$ is a limit point since $x_{4n}$ converges to $1$.
$2$ is a limit point since $x_{4n+1}$ converges to $2$.
Doing the same for $3,4$ I know $1,2,3,4$ are limit points.
But why are there no other limit points?
Suppose $y$ is another limits, $y\neq1,2,3,4$. So there is a subsequence $x_{n_l}$ with $x_{n_l}\rightarrow y$ for $l\rightarrow \infty$.
Now I am stuck. How can you get a contradiction?
AI: Choose $\epsilon = \frac{1}{2}\min\{|y-1|,|y-2|,|y-3|,|y-4|\}$ (so e.g., if $y = 1.1$, take $\epsilon = 0.05$.)
Then you know that for every $n$, $|x_n - y| > \epsilon$. Do you see how you get a contradiction now?
Full proof:
Suppose $y$ is a limit point of this sequence. Then, for every $\epsilon > 0$ there is a subsequence $\{x_{n_l}\}$ such that there exists $L$ so that whenever $l > L$, we have $|x_{n_l} - y| < \epsilon.$
Now, choose $\epsilon = \frac{1}{2}\min\{|y-1|,|y-2|,|y-3|,|y-4|\}$. If $y$ is not $1,2,3$ or $4$, then $\epsilon > 0$, and we have for every $x_n$ (i.e. not just the subsequence) that $|x_n - y| > \epsilon$. Thus, there can be no such $L$ as described in the first paragraph; i.e. we cannot simulaneously have
$$
|x_{n_l} - y| < \epsilon
$$
as required for convergence as well as
$$
|x_{n_1} - y| > \epsilon
$$
as we just demonstrated. Thus we have a contradiction. |
H: Prove that the symmetric derivative of a function exists whenever the derivative exists.
Let $f$ be a function defined on an interval $(a,b)$ and let $c \in (a,b)$. The symmetric derivative of $f$ at $c$ is defined by $f'_s(c)=lim_{h\to 0} \frac{f(c+h)-f(c-h)}{2h}$ provided that the limit exists. Prove that $f'_s(c)$ exists whenever $f'(c)$ exists, but that it is possible for $f'_s(c)$ to exists even when $f'(c)$ does not exist.
I have already proved it is possible for $f'_s(c)$ to exist when $f'(c)$ DNE, using the function $f(x)=|x|$.
But my struggle is with part 1: I know that we should assume that $f'(c)$ exists, and since $f'(c)$ exists then we know $lim_{x\to c} \frac{f(x)-f(c)}{x-c}$ exists. Now my struggle is getting from this to the definition of symmetric derivative.
AI: \begin{eqnarray}
\frac{f(x+h)-f(x-h)}{2h} &=& \frac{1}{2} \left( \frac{f(x+h)-f(x)+f(x)-f(x-h)}{h} \right) \\
&=& \frac{1}{2} \left( \frac{f(x+h)-f(x)}{h} + \frac{f(x)-f(x-h)}{h} \right) \\
&=& \frac{1}{2} \left( \frac{f(x+h)-f(x)}{h} + \frac{f(x-h)-f(x)}{-h} \right)
\end{eqnarray}
Now take limits.
Addendum: Since $\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h \to 0} \frac{f(x-h)-f(x)}{-h} = f'(x)$, we have that the corresponding limit on the left hand side exists and
$\lim_{h \to 0} \frac{f(x+h)-f(x-h)}{2h} = f'(x)$. |
H: Do densities of invariant distributions satisfy the Fokker Planck equation?
Suppose that $\{X_t\}_{t\in[0,\infty)}$ is a $\mathbb{R}^n$ valued homogenous diffusion process with drift vector $b$ and diffusion matrix $A$. Is it ever true that if the process has an invariant distribution with a density $\pi(x)$, then $\pi$ is a time independent solution of the Fokker Planck equation, that is
$$0=-\sum_i\frac{\partial }{\partial x_i}[b_i(x)\pi(x)]+\sum_{i,j}\frac{\partial^2}{\partial x_i\partial x_j}[A_{i,j}(x)\pi(x)]?\quad\quad(*)$$
I've spend a bit of time scanning through standard references (Freidman, Oskendal, Karatzas-Shreve, Arnold) looking results connecting the densities of invariant distributions to one of the Kolmogorov equations and I've had little success. The closest I've gotten is a couple of results in Section 31 of these notes which establish the above and its converse (under some technical assumptions) if $A=\frac{1}{2}I$ and $b$ is $C^\infty$. However, glancing through the proof it doesn't seem these assumptions play a fundamental role in it (then again, I only have a working knowledge--at best--of this material and I might be missing some subtleties), in fact the author states:
Remark. The only property of the operator L we used was to conclude
q > 0. We may therefore expect a similar result for more general operators.
Does anyone know under what more general conditions the above holds? A reference with this type of results would be great.
Thanks in advance.
Aside: In case anyone is wondering why I even think the above should hold here's my chain of thought: If we know that the process's transition probabilities have a density $p(t,x)$, then $p$ satisfies the forward Kolmogorov equation:
$$\frac{\partial p}{\partial t}(t,x)=-\sum_i\frac{\partial }{\partial x_i}[b_i(x)p(t,x)]+\sum_{i,j}\frac{\partial^2}{\partial x_i\partial x_j}[A_{i,j}(x)p(t,x)].$$
Suppose that we know that $X_t$ converges (in some sufficiently strong fashion) as $t\to\infty$ to some random variable $X_\infty$. Suppose that $X_\infty$'s distribution has a density $\pi(x)$. Then we have that $p(t,x)\to\pi(x)$ as $t\to\infty$. So we should have
$$\frac{\partial p}{\partial t}(t,x)\to 0\text{ as } t\to\infty$$
giving $(*)$. Of course, the above is totally heuristic and (I imagine) much more attention needs to be payed to the technicalities (for example, above I've assumed that the derivative of a sequence of functions is converges to the derivative of the limit of the sequence, which I suspect much depends on what fashion the sequence converges).
AI: You might be interested in this reference by arnold on the work of R. Khasminskii over Lyapunov function method in the context of sde (especially by theroem 2.2).
Best regards |
H: Prove that $\alpha$ is algebraic over $K$.
Let $\alpha$ be a transcendental element over a field $E$, and $F=E(\alpha)$. Prove that for any subfield $K$ of $F$ containing $E$ as a proper subset, $\alpha$ is algebraic over $K$.
Can anyone give me some hint? I have no idea how to start.
AI: Since $K$ is a subset of $E(\alpha)$ properly containing $E$, it must contain as an element some nonconstant rational function of $\alpha$. To get a feel for what's going on, suppose for instance that $\frac{\alpha^2+1}{\alpha} \in K$. |
H: FoxTrot Bill Amend Problems
So I found this on the Wolfram website today:
So I was wondering about how one might be able to (if possible) solve those four problems by hand. Here are the problems, $\LaTeX$ed:
$ \lim_{x \to +\infty} \dfrac {\sqrt{x^3-x^2+3x}}{\sqrt{x^3}-\sqrt{x^2}+\sqrt{3x}} $
$ \displaystyle\sum_{k=1}^{\infty} \dfrac {(-1)^{k+1} k^2}{k^3+1} $
$ \dfrac {\mathrm{d}}{\mathrm{d}u} \left[ \dfrac {u^{n+1}}{(n+1)^2} \cdot \left[ (n+1) \ln u - 1 \right] \right] $
$ \displaystyle\int_0^{2\pi}\displaystyle\int_0^{\frac{\pi}{4}}\displaystyle\int_0^4 \left( \rho \cos \phi \right) \rho^2 \sin \phi \, \mathrm{d}\rho \mathrm{d}\phi \mathrm{d}\theta$
Ideas
The degree of the numerator is $\frac{3}{2}$. The degree of the
denominator is $\frac{3}{2}$. The expression is of an indeterminate
form, namely $\frac{\infty}{\infty}$, so we use l'Hoptial's Rule.
No ideas, really.
Derivatives are always pretty easy, although this one is a bit
bashy. Basically bash Product and Chain Rules, etc.
Integration by Parts bash? The $\mathrm{d}\theta$ part is very
trivial. I mean for the $\rho$ and $\phi$.
AI: I can derive the sum quite easily, using partial fractions and the residue theorem.
Note that $$\frac{k^2}{k^3+1} = \frac13 \left [\frac{1}{k+1} + \frac{2 k-1}{k^2-k+1}\right ]$$
Now, note that
$$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k+1} = 1-\log{2}$$
For the other piece, note the fortuitous coincidence that
$$\sum_{k=1}^{\infty} \frac{(-1)^{k+1} (2 k-1)}{k^2-k+1} = -\frac12 \sum_{k=-\infty}^{\infty} \frac{(-1)^{k} (2 k-1)}{k^2-k+1}$$
(One may see this by showing that the map $k \mapsto -k$ produces the $k+1$th term of the summand.)
Now, the sum on the RHS may be evaluated using residue theory. In general, one may show that
$$\sum_{k=-\infty}^{\infty} (-1)^k f(k) = -\pi \sum_n \operatorname*{Res}_{z=z_n} [\csc{\pi z} \, f(z)]$$
where the $z_n$ are the non-integral poles of $f$. In this case, the poles are at $z=e^{\pm i \pi/3}$, and we are also lucky to have an $f$ of the form $g'/g$, so that the residue at each pole is simply $1$. Therefore, we have
$$\sum_{k=1}^{\infty} \frac{(-1)^{k+1} (2 k-1)}{k^2-k+1} = \frac{\pi}{2} \left ( \csc{\left (\pi e^{i \pi/3}\right )}+ \csc{\left (\pi e^{-i \pi/3}\right )} \right ) = \pi \, \text{sech}{\left ( \frac{\sqrt{3} \pi}{2}\right )}$$
Therefore,
$$\sum_{k=1}^{\infty} \frac{(-1)^{k+1} k^2}{k^3+1} = \frac13 \cdot \left [ 1-\log{2} + \pi \, \text{sech}{\left ( \frac{\sqrt{3} \pi}{2}\right )}\right ]$$ |
H: Use a factorial argument to show that $C(2n,n+1)+C(2n,n)=\frac{1}{2}C(2n+2,n+1)$
I need some help, showing that the left hand side is equivalent to the right hand side. I tried but I get stuck, I am not sure if I am on the right path. Here is my attempt:
$C(2n,n+1) + C(2n,n)$
Equivalent to this
$\frac{(2n)!}{(n-1)!(n+1)!}+\frac{(2n)!}{n!n!}$
Now I add them together when I have a common denominator and I get
$\frac{(2n)!(n+n+1)}{(n)!(n+1)!}\frac{(2n)!(2n+1)}{(n)!(n+1)!} = \frac{(2n+1)!}{n!(n+1)!}$
But I do not know how to go from there. I don't know how to transform it to look like
$\frac{1}{2}C(2n+2,n+1)=\frac{(2n+2)!}{2(n+1)!(n+1)!}$
Any hint, tip, that will help me will be really useful.
AI: Multiply and divide the answer you got by $2(n+1)$. |
H: time taken to complete work
A can do a piece of work four hours faster than B. They worked together for two hours and then the remaining part of the work was done by A in an hour. How many hours would B take to complete the job if he were to work alone?
AI: let t be the time A takes in hours, then A does $\frac{1}{t}$ per hour. B does $\frac{1}{t+4}$ per hour. we know $\frac{3}{t}+\frac{2}{t+4}=1$ $\rightarrow 3+\frac{2t}{t+4}=t\rightarrow 3(t+4)+2t=t^2+4t\rightarrow t^2-t-12=0\rightarrow (t-4)(t+3)=0$ so $t=4$
therefore A takes 4 hours and B takes 8 hours. |
H: Proof for the length of the shortest 4-connected path and 8-connected path on a chessboard
I have a chessboard with a square marked with A as in the following figure:
$$\begin{array}{|c|c|c|} \hline
8&1&2\\ \hline
7&A&3\\ \hline
6&5&4\\ \hline
\end{array}$$
The $4$-connected neighbors squares of A are marked with $1, 3, 5, 7$.
The $8$-connected neighbors squares of A are marked with $1, 2, 3, 4, 5, 6, 7, 8$.
Then I have a chessboard with two marked squares, an origin square and a destination square: a $4$-connected path from the origin to the destination is a sequence of squares that connects the origin to the destination and all the squares in the sequence are $4$-connected neighbors; an $8$-connected path from the origin to the destination is a sequence of squares that connects the origin and the destination and all the squares in the sequence are $8$-connected neighbors. The length of a path is the number of squares in the sequence.
The left part of the following figure shows a $4$-connected path from the origin square marked with $1$ to the destination square marked with $7$; the length of the path is $7$. The right part of the figure shows an $8$-connected path from the origin $1$ to the destination $5$; the length of the path is $5$.
$$\begin{array}{}
\begin{array}{|c|c|c|} \hline
&6&7\\ \hline
&5&\\ \hline
&4&\\ \hline
&3&\\ \hline
1&2&\\ \hline
\end{array}&\qquad&\begin{array}{|c|c|c|} \hline
&&5\\ \hline
&4&\\ \hline
&3&\\ \hline
&2&\\ \hline
1&&\\ \hline
\end{array}
\end{array}$$
I have read in stackoverflow that the length of the shortest $4$-connected path between origin and destination is the Manhattan distance between origin and destination and that the length of the shortest $8$-connected path is the Chebyshev distance between origin and destination.
How to prove it?
AI: Impose a rectangular coordinate system that makes the lower left cell $\langle 0,0\rangle$ and the upper right cell $\langle m,n\rangle$, where $m$ identifies the column and $n$ the row. (In other words, in the example $m=2$ and $n=4$.) Any $4$-connected path from $\langle 0,0\rangle$ to $\langle m,n\rangle$ must take at least $m$ steps to the right and at least $n$ steps up, so it must have length at least $m+n$. On the other hand, it’s clear that any step down or to the left is wasted, since it must be compensated for by an extra step up or to the right, so $m+n$ is the minimum length of a $4$-connected path from $\langle 0,0\rangle$ to $\langle m,n\rangle$. Clearly $m+n=|m-0|+|n-0|$, the Manhattan distance between $\langle 0,0\rangle$ and $\langle m,n\rangle$.
Suppose that $m\le n$. From $\langle 0,0\rangle$ we can take $m$ steps up and to the right, reaching $\langle m,m\rangle$, and another $n-m$ steps up will take us to $\langle m,n\rangle$ in a total of $n$ steps. Similarly, if $m\ge n$ we can take $n$ steps up and to the right to $\langle n,n\rangle$ and then $m-n$ steps to the right, this time reaching $\langle m,n\rangle$ in $m$ steps. Thus, there is always an $8$-connected path of length $\max\{m,n\}$ from $\langle 0,0\rangle$ to $\langle m,n\rangle$. It only remains to show that there is no shorter $8$-connected path from $\langle 0,0\rangle$ to $\langle m,n\rangle$, since $\max\{m,n\}$ is the Chebyshev distance between $\langle 0,0\rangle$ and $\langle m,n\rangle$.
Suppose that $m\le n$, so that $\max\{m,n\}=n$; the other case is completely similar. Each step changes each coordinate by at most $1$, so that if $r$ steps take you to $\langle k,\ell\rangle$, then $k,\ell\le r$. In particular, if $r<n$, then $\ell<n$, and you cannot reach $\langle m,n\rangle$ in $r$ steps. Thus, at least $n$ steps are required. |
H: Beautiful Mathematical Images
My Maths department is re-branding itself, and we've been asked to find suitable images for the departmental sign. Do you have a favourite mathematical image that could be used for the background of an A1-sized sign?
AI: I produced the following images, which I personally like ;-)
A part of the Mandelbrot set:
A projective curve in $\mathbb{P}^2$:
$\mathbb{P}^2$" /> |
H: Probability with 13 trees
A company is planting trees and we know that 90% of the trees survive. What is the probability that from 13 trees:
1, at most $10$ survive
2, at least $10$ survive
3, exactly $10$ trees survive.
AI: Here is a similar problem. Suppose we flip 10 coins. What is the probability that at least 9 are heads?
Case 1: All 10 are heads. This has probability $(1/2)^{10}$.
Case 2: Exactly 9 are heads. If the first 9 are heads, and the last is tails, this has probability $(1/2)^{9}(1/2)^{1}$. However we can choose which 9 are heads in ${10\choose 9}$ ways. Hence the total probability is ${10 \choose 9}(1/2)^{9}(1/2)^{1}$.
Putting it together, we get $(1/2)^{10}+{10\choose 9}(1/2)^9(1/2)^{1}$.
The reason I didn't combine the $(1/2)^9(1/2)^1$ into $(1/2)^{10}$ (you can), is to distinguish the $1/2$ that comes from the probability of heads, and the $1/2$ that comes from the probability of tails. If instead we got heads with probability $p$ and tails with probability $1-p$, the answer would be $$p^{10}+{10\choose 9}p^9(1-p)^1$$ |
H: Notation of double-sided infinite sum
The notation $\sum_{k=1}^\infty a_k$ always means $$\lim_{n\rightarrow\infty}\sum_{k=1}^n a_k.$$
What about $\sum_{k=-\infty}^\infty a_k$, such as in the Laurent series? Does it always means $$\lim_{n\rightarrow\infty}\sum_{k=-n}^n a_k,$$ or does the meaning depend on the context?
AI: The notation must be used only when $\dagger$ is satisfied.
$$\lim_{N_1 \to \infty} \lim_{N_2 \to \infty} \sum_{k=-N_1}^{k=N_2} a_k \text{ and }\lim_{N_2 \to \infty} \lim_{N_1 \to \infty} \sum_{k=-N_1}^{k=N_2} a_k \text{ exists and }$$
$$\lim_{N_1 \to \infty} \lim_{N_2 \to \infty} \sum_{k=-N_1}^{k=N_2} a_k = \lim_{N_2 \to \infty} \lim_{N_1 \to \infty} \sum_{k=-N_1}^{k=N_2} a_k \tag{$\dagger$}$$
If $\dagger$ is not satisfied, the notation doesn't make sense. Sometimes, it can also be used if both $$\lim_{N_1 \to \infty} \lim_{N_2 \to \infty} \sum_{k=-N_1}^{k=N_2} a_k \text{ and }\lim_{N_2 \to \infty} \lim_{N_1 \to \infty} \sum_{k=-N_1}^{k=N_2} a_k$$ are either $+ \infty$ or $- \infty$. |
H: My Odometer, Speedometer, and the Time
I was driving from home to university earlier this week when I glanced at my dashboard just soon enough to notice the odometer tick up while simultaneously registering my speed and the time. I thought to myself,
"How might I find the probability that the odometer will tick up once more before the minute space on the clock increments?"
That is my question. If this belongs in physics, please move it. Assume nonrelativistic velocity (unless, of course, you're into that sort of thing).
AI: I think I see what you are getting at.
Lets say you see your odometer tick up and you note the location of the minute hand and your speed. What is the probability that the odometer will tick again before the minute hand moves to the next minute.
We can model this as follows:
Let $T_1$ be the number of seconds into the given minute that your odometer first ticks up. Let V be the noted speed, in mph. And assume your odometer has increments of $\Delta d$ miles.
Lets assume that your odometer is equally likely to tick up for the first time near the beginning of the minute as towards the end, so that $T_1 \sim Uniform(0,60)$. Now, what is the number of seconds to get another tick given your noted velocity, V? It is: $T_2(V)=\frac{3600\Delta d}{V}$. The probability of getting two ticks in a minute, given your noted speed and the occurrance of a first tick, can be written as:
$P(T_1+T_2\leq 60|V)=P(T_1\leq 60-\frac{3600\Delta d}{V})=\frac{60-\frac{3600\Delta d}{V}}{60}=max\{0,1-\frac{60\Delta d}{V}\}$ |
H: Can you think of a good approximation to this integral?
So, I have
$$\int_{0}^{\arcsin\left(\frac{r}{g}\right)}\left(g\cos\theta-\sqrt{r^{2}-g^{2}\sin^{2}\theta}\right)^{2}d\theta$$
but the integral, which one can evaluate with mathematica, has a singularity at the upper limit to the integral. Can anyone think of a good approximation? You can use g<<1.
AI: Sorry, but I do not see any singularity so long as $r \le g$, which I believe is the case. Expanding the square in the integrand, you get three very doable integrals:
$$g^2 \int_0^{\arcsin{r/g}} d\theta \, \cos^2{\theta} = \frac12 g^2 \arcsin{\frac{r}{g}} + \frac12 r \sqrt{g^2-r^2}$$
$$\int_0^{\arcsin{r/g}} d\theta \, (r^2-g^2 \sin^2{\theta}) = r^2 \arcsin{\frac{r}{g}} - \frac12 g^2 \arcsin{\frac{r}{g}} + \frac12 r \sqrt{g^2-r^2}$$
$$-2 g \int_0^{\arcsin{r/g}} d\theta \, \cos{\theta} \sqrt{r^2-g^2 \sin^2{\theta}} = -\frac{\pi}{2} r^2$$
Adding, I get the integral is, without approximation,
$$ r^2 \arcsin{\frac{r}{g}} + r \sqrt{g^2-r^2} - \frac{\pi}{2} r^2$$
I did all of these out by hand, by the way. If you need to see the steps, I will be happy to oblige. |
H: Conditional expectation by $\sigma (G_n,Y)$ when $Y$ is $G_\infty$-measurable
Let $G_n$ be a filtration (an increasing sequence of sigma-algebras), $Y$ a random variable that is $G_\infty$-measurable, and $X$ a random variable. Is it true that in $L^2$-norm,
$$ \mathbb{E}[X \mid \sigma (G_n,Y)]- \mathbb{E}[X \mid G_n] \stackrel{n \rightarrow + \infty}{\rightarrow} 0 $$
AI: Lemma Let $X \in L^2$ and $(\mathcal{G}_n)_{n \in \mathbb{N}}$ a filtration. Then $$\lim_{n \to \infty} \mathbb{E}(X \mid \mathcal{G}_n) = \mathbb{E}(X \mid \mathcal{G}_{\infty}) \quad \text{in} \, L^2$$ where $\mathcal{G}_{\infty} := \sigma(\mathcal{G}_n; n \in \mathbb{N})$.
Proof: Set $X_n := \mathbb{E}(X \mid \mathcal{G}_n)$. Obviously, $(X_n,\mathcal{G}_n)_{n \in \mathbb{N}}$ is a martingale. Since the martingale is bounded in $L^2$, the martingale convergence theorem implies $X_n \to Y$ in $L^2$ for some random variable $Y \in L^2$. It remains to show that $Y= \mathbb{E}(X \mid \mathcal{G}_{\infty})=:X_{\infty}$. By the tower property, we have $$\mathbb{E}(Y \mid \mathcal{G}_n) = X_n = \mathbb{E}(X \mid \mathcal{G}_n) = \mathbb{E}(X_{\infty} \mid \mathcal{G}_n)$$
Hence, for any $G \in \bigcup_{n \in \mathbb{N}} \mathcal{G}_n$: $$\int_G Y \, d\mathbb{P} = \int_G X_{\infty} \, d\mathbb{P}$$ Since $\bigcup_{n \in \mathbb{N}} \mathcal{G}_n$ is a generator of $\mathcal{G}_{\infty}$ which is stable under intersections, this implies
$$\int_G Y \, d\mathbb{P} = \int_G X_{\infty} \, d\mathbb{P}$$
for any $G \in \mathcal{G}_{\infty}$. By choosing $G := [\pm (X_{\infty}-Y)>0]$, we find $X_{\infty} = Y$ a.s. This finishes the proof.
The lemma shows that $\mathbb{E}(X \mid \mathcal{G}_n) \to \mathbb{E}(X \mid \mathcal{G}_{\infty})$ in $L^2$. Applying the lemma for the filtration $\mathcal{H}_n := \sigma(\mathcal{G}_n,Y)$ yields $$\mathbb{E}(X \mid \sigma(\mathcal{G}_n,Y)) = \mathbb{E}(X \mid \mathcal{H}_n) \to \mathbb{E}(X \mid \mathcal{H}_{\infty}) = \mathbb{E}(X \mid \mathcal{G}_{\infty})\quad \text{in} \, L^2$$ using that $\mathcal{H}_{\infty} = \mathcal{G}_{\infty}$ since $Y$ is $\mathcal{G}_{\infty}$-measurable. Consequently,
$$\mathbb{E}(X \mid \sigma(\mathcal{G}_n,Y))-\mathbb{E}(X \mid \mathcal{G}_n)\to 0 \quad \text{in} \, L^2.$$ |
H: Using partial fraction for $\cot \pi z$ to compute infinite sum
I want to compute the values $\sum_{n=1}^\infty \dfrac{1}{n^2}$ and $\sum_{n=1}^\infty \dfrac{1}{n^4}$ and $\sum_{n=1}^\infty \dfrac{1}{n^6}$ by comparison to the partial fraction development of $\cot \pi z$.
First, I note that $$\pi z\cot(\pi z)=1+2\sum_{n=1}^\infty\dfrac{z^2}{z^2-n^2}=1-2z^2\sum_{n=1}^\infty\dfrac{1}{n^2}\cdot\dfrac{1}{1-\dfrac{z^2}{n^2}}$$
Using the sum of geometric series formula, this equals
$$1-2z^2\sum_{n=1}^\infty\dfrac{1}{n^2}\cdot\left(1+\dfrac{z^2}{n^2}+\dfrac{z^4}{n^4}+\ldots\right)$$
Interchanging the order of summation (since both series are convergent), I get $$1-2z^2\sum_{k=0}^\infty\left(\sum_{n=1}^\infty\dfrac{1}{n^{2(k+1)}}\right)$$
So, I need to find the $z^2,z^4,z^6$ terms of the Taylor expansion of $z\cot(\pi z)$. I can write it as $\dfrac{z\cos(\pi z)}{\sin(\pi z)}$, and both $\cos$ and $\sin$ have Taylor expansions. But how can I get the Taylor expansion for $z\cot(\pi z)$ (in a not-messy way)?
AI: You have lost a few $z^{2k}$ on the way.
$$1-2z^2\sum_{n=1}^\infty\dfrac{1}{n^2}\cdot\left(1+\dfrac{z^2}{n^2}+\dfrac{z^4}{n^4}+\ldots\right) = 1-2z^2\sum_{k=0}^\infty \left(\sum_{n=1}^\infty \frac{1}{n^{2(k+1)}}\right){\color\red{z^{2k}}}.$$
If you only want the very few first terms of the Taylor expansion, dividing the expansions of $\cos$ and $\sin$ is sufficiently non-messy, but going beyond the $z^6$ coefficient starts to be uncomfortable. A technique to make the computation of the Taylor series of $\frac{f(z)}{g(z)}$ less taxing is to make an ansatz
$$\frac{f(z)}{g(z)} = \sum_{n=0}^\infty \frac{c_n}{n!}z^n,$$
and then use the Cauchy product of that with the Taylor series of $g$ to obtain a recursion for the $c_n$.
For the Taylor series of $z\cot z$, we can fiddle a bit to find a representation that yields a nicer recursion,
$$\cot z = \frac{\cos z}{\sin z} = i\frac{e^{iz}+e^{-iz}}{e^{iz}-e^{-iz}} = i + i\frac{2e^{-iz}}{e^{iz}-e^{-iz}} = i + 2i\frac{1}{e^{2iz}-1}.\tag{1}$$
So if we have the Taylor series of $\frac{w}{e^w-1}$, everything follows nicely from that. So we make the ansatz
$$\frac{w}{e^w-1} = \sum_{k=0}^\infty \frac{B_k}{k!}w^k.\tag{2}$$
The $B_k$ are known (famous) as the Bernoulli numbers. From the way we arrived there, it is clear that $\frac{w}{e^w-1}+\frac{w}{2}$ is an even function, hence the only odd term in the series is $-\frac{w}{2}$, and $B_{2k+1} = 0$ for $k > 0$.
Multiplying $(2)$ with the Taylor series of $e^w-1$ produces
$$\begin{align}
w &= \left(\sum_{k=0}^\infty \frac{B_k}{k!}w^k\right)\left(\sum_{m=1}^\infty \frac{w^m}{m!}\right)\\
&= \sum_{n=1}^\infty \left(\sum_{k=0}^{n-1}\frac{B_k}{k!(n-k)!}\right)w^n\\
&= \sum_{n=1}^\infty \left(\sum_{k=0}^{n-1}\binom{n}{k}B_k\right)\frac{w^n}{n!}.
\end{align}$$
Thus we have $B_0 = 1$ and the relation
$$\sum_{k=0}^{n-1}\binom{n}{k}B_k = 0\tag{3}$$
for $n > 1$. It is still not trivial (in terms of amount of computation) to compute the $B_{2k}$ for larger $k$, but it is much easier and less work than dividing the Taylor series of $\cos$ and $\sin$.
I'll leave it to you to connect $(1)$ and $(2)$ to obtain the Taylor series of $\pi z\cot \pi z$ in terms of the Bernoulli numbers, and to compute the first Bernoulli numbers with the relation $(3)$. |
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