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H: Sum of the number of unique factors of numbers from 1 to 100? I want to compute $\sum_{i\; =\; 1}^{i\; =\; 100}{d\left( i \right)}$ where d(i) is the number of factors of i. For example, for 1 it is 1, for 2 is is 2, for 3 it is 2, for 12 it is 6, and so on. AI: HINT: $1$ is a divisor of every number in the set $[100]=\{1,\ldots,100\}$, so it contributes $100$ to the total count of divisors. $2$ is a divisor of the even numbers in $[100]$; there are $50$ of those, so $2$ contributes $50$ to the total count of divisors. Keep going: if $d$ is a positive integer, how many numbers in $[100]$ have $d$ as a divisor? Clearly you need only look at positive integers $d\le 100$, and you can deal with sum of them in batches rather than individually. For instance, how much does each $d$ in $\{51,52,\ldots,100\}$ contribute to the total?
H: A question about rolls A dodecagon labelled by 12 months at each edge is rolled in a game. One “turn” of the game is to roll it until one April appears, then the number of the rolls is recorded. What is the probability to have five consecutive turns with rolls no greater than 10? Could you help me with this stuck? AI: What is the chance you don't get April on one roll? The chance you don't get April on one turn is the tenth power of this. The chance you get $10$ or less is one monus this. the chance you get five in a row $10$ or less is the fifth power of that. I get somewhat less than $7\%$
H: The king comes from a family of 2 children. What is the probability that the other child is his sister? I've found this question on a book and I'd like a review in my answer. The king comes from a family of 2 children. What is the probability that the other child is his sister? AI: First of all, take note about @Micah comment and suppose the male-preference primogeniture is observed. This is a Conditional Probability question. We'll use the followin notation: $P(A\|B)$ means the probability of event $A$ occurs given that event $B$ has occurred. By definition on page 59 of Sheldon Ross book, we know that $ P(A\|B) = \dfrac{P(AB)}{P(B)}$, where $AB = A\cap B $. Another notation is $\|X\|$ that means the number of elements in set $X$. As the king comes from a family of two children, we are given two tips: (1) This family has a boy, the king. (2) The king has a sibling. What we want to know is the probability this sibling be a girl. In other words, what's the probability of the two children be each one of each gender. Let B the event of possible children where at least one is a boy, the king. So $B=\{(b,b), (b,g), (g,b)\}$, where $(x,y)$ means the gender of each child and the possible values are $b$ for boy and $g$ for girl. Then $A$ is the event that the king's sibling is a girl, $A=\{(b,g),(g,b)\}$. The sample space $S$ contains all possible outcomes, $S=\{(b,b),(g,b),(b,g),(g,g)\}$. It follows that $AB=A$ and $\|AB\|=\|A\|=2$, $\|B\|=3$ and $\|S\|=4$. So, we have: $$P(A\|B) = \dfrac{P(AB)}{P(B)}=\dfrac{\dfrac{\|AB\|}{\|S\|}}{\dfrac{\|B\|}{\|S\|}}=\dfrac{\|AB\|}{\|B\|} = \dfrac{2}{3}$$ The end.
H: How to get the inverse of this function? I have the function f(x) = (1+8x) / (3-3x). I have been stuck on trying to get the inverse of this function by isolating for y. I ended up with: x = (1+8y) / (3-3y) but I am not quite sure where to go from there. AI: $$ y = \frac{1+8x}{3-3x}$$ $$(3-3x)y = 1+8x$$ (multiplying each side by $3-3x$) $$8x + 3xy = 3y -1$$ (multiplying out the brackets, adding $3xy$ to each side and subtracting $1$ from each side) $$x(8+3y) = 3y - 1$$ (taking out a factor of $x$ from LHS) $$x = \frac{3y-1}{8+3y}$$ (dividing each side by $8+3y$)
H: Prove that if $\operatorname{rank}(T) = \operatorname{rank}(T^2)$ then $R(T) \cap N(T) = \{0\}$ Let $V$ be a finite-dimensional vector space and let $T:V\to V$ be linear. Prove that if $\operatorname{rank}(T) = \operatorname{rank}(T^2)$, then $R(T) \cap N(T) = \{0\}$. I don't see this implication, at all. Please give hints and explain conceptually. AI: You have $T^2(V)\subseteq T(V)$, and the dimension of these two subspaces is equal so $T(V) = T^2(V)$. Hence, $T$ is 1-1 on $T(V)$.
H: Hiding Eggs Combinatorics How can I figure out if I have 6 red eggs, 3 blue eggs, 1 green egg, and 2 yellow eggs. Aside from their color, they are identical. I have 12 different hiding spots, each big enough for 1 egg. How many ways are there to hide the eggs? AI: HINT: How many ways are there to choose $6$ of the $12$ hiding places for the red eggs? Once you’ve done that, how many ways are there to choose $3$ of the remaining hiding places for the blue eggs? Keep going in this fashion and combine the answers correctly, and you’ll have your answer.
H: Simplifying euler exponent? How would I go about simplifying and finding the exact value for this question: $e^{6\ln(4)}$ I know that $e ^{\ln x} = x$ but how does the $6$ affect this answer? AI: $$a^{bc} = \left(a^c\right)^b$$ If $a=e$ and $c=\ln(4)$, then $a^c = 4$. You should be able to do the rest yourself.
H: Instantaneous rate of change Let $f(x) = ax + b$. Find the instantaneous rate of change of $f(x)$ at the following points: $1, 2, 4\text{ and }8$. Is the instantaneous rate of change for all the points $a$, because the derivative for $f'(x)= a$? AI: Yes, the instantaneous rate of change is $4$. To see this (even) more explicitly, remember that the derivative $f'(x)$ is defined as $\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$, so $$\frac{d}{dx} (ax+b) = \lim_{x\to 0}\frac{(a(x+h) + b) - (ax+b)}{h} = \lim_{h\to0} \frac{ah}{h} = a$$
H: Rate of change of distance from particle (on a curve) to origin A particle is moving along the curve $y = 2\sqrt{4x + 9}$ As the particle passes through the point (4, 10) its x-coordinate increases at a rate of 3 units per second. Find the rate of change of the distance from the particle to the origin at this instant. Okay. Rate of change of the distance from the particle to the origin. So the origin is going to be the point (0,y). So: $y = 2\sqrt{4(0)+9} = 6$. The point (0,6) is the origin, then. Now the problem is asking us for the rate of change of the distance between these two points, we recall that: $D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ So, now we have to differentiate: $$d' = \frac{1}{2}\cdot ((x_2 - x_1)^2 + (y_2 - y_1)^2)^\frac{-1}{2} \cdot \left[ 2(x_2'-x_1') + 2(y_2'-y_1') \right]$$ Then we have to substitute. But I don't know if I'm even correct thus so far. Can someone help? AI: The distance to the origin when the particle is at $(x,y)$ is given by $D(x,y)=\sqrt{x^2+y^2}$. We want $\frac{dD}{dt}$ at a certain instant. I prefer to work with $D^2$. So we have $$D^2=x^2+y^2.$$ Differentiate, using the Chain Rule. We have $$2D\frac{dD}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}.\tag{1}.$$ We know that $y=2\sqrt{4x+9}$. So $$\frac{dy}{dt}=\frac{4}{\sqrt{4x+9}}\frac{dx}{dt}.\tag{2}$$ Now "freeze" things at the instant when $x=4$. We know $\frac{dx}{dt}$ at this instant. We also know $\frac{dy}{dt}$, by (2). We also know $y$ and therefore $D$. Now we can use (1) to find $\frac{dD}{dt}$ at this instant.
H: Using Logarithms \begin{align} -2^{n-1} \ln2 &= -100 \ln 10\\ &\\ -100 \ln 10 &= -230\\ &\\ \dfrac{-230}{\ln (2)} &= -333\\ &\\ -2^{n-1} &> -333\\ &\\ (n-1) \ln(-2) &> \ln(-333) \end{align} Here is where I am stuck. I am not sure if this part is correct: $-n-1=8$. Then solving we would get $-9$. AI: Log is an increasing function. So you can take log in both side of an inequality and that is fine. But you can't take log of negative values like $-2$ The mistake is at $$-2^{n-1}=-333.$$ You missed negative sign at $333$. Then if you correct the next steps it's fine. Also when you multiply both sides of an inequality by negative the sign changes. $$-2^{n-1} ln2 < -100ln10$$ $$-2^{n-1}<\cfrac{-100ln10}{ln2}$$ $$2^{n-1}>\cfrac{100ln10}{ln2}$$ $$n-1 ln(2) >ln(\cfrac{100ln10}{ln2}).$$ And this gives $n>\cfrac{ln(\cfrac{100ln10}{ln2})}{ln2}+1$
H: Making first order logic statements I'm working on an assignment that deals with predicate calculus, and I'm trying to put sentences into first order logic statements. I've got the hang of most of them, but I'm not quite sure how to do a couple of them: Every X except Y likes Z, $$\forall m: \neg Y(m) \rightarrow \mathit{likes}(X(m), Z)$$ X likes all Y who are not Z, $$\mathit{likes}(X, Y(\neg Z))$$ AI: In the first sentence $Y$ and $Z$ must be understood as referring to specific individuals, while $X$ is a dummy name that does not refer to any specific individual. In more technical terms, $Y$ and $Z$ are constants, and $X$ is a variable. If $\mbox{Likes}(x,y)$ is a two-place predicate meaning ‘$x$ likes $y$’, then you want $$\forall X\Big(X\ne Y\to\mbox{Likes}(X,Z)\Big)\;.\tag{1}$$ Read fairly literally, that’s for all X, if X is not Y, then X likes Z, which is pretty clearly equivalent to the desired statement. There is another possible interpretation of the original sentence. it certainly says that everyone who is not $Y$ likes $Z$. In everyday usage everyone except A does X generally also implies that $X$ does not do $X$, though it doesn’t explicitly say so. If you want to include that implication, then add $$\neg\mbox{Likes}(Y,Z)\land$$ at the front of $(1)$. In the second sentence $X$ is a constant: it apparently refers to a particular individual, as does $Z$; this time it’s $Y$ that’s the variable ranging over all possible people. This statement doesn’t say anything one way or the other about whether $X$ likes $Z$, so we have $$\forall Y\Big(Y\ne Z\to\mbox{Likes}(X,Y)\Big)\;:$$ for every Y, if Y is not Z, then X likes Y.
H: For which intervals $[a,b]$ in $\mathbb{R}$ is the intersection $[a,b]\cap Q$ a clopen subset of the metric space $\mathbb{Q}$? For which intervals $[a,b]$ in $\mathbb{R}$ is the intersection $[a,b]\cap Q$ a clopen subset of the metric space $\mathbb{Q}$? My answer is : $[a,b]\cap\mathbb{Q}$ is a clopen subset iff $a,b \in (\mathbb{R}\backslash \mathbb{Q})$, since if $a,b \in \mathbb{Q}$ then $[a,b]\cap \mathbb{Q}$ won't be open. I got this wrong on a p set. Can someone correct what I have done wrong? AI: Your answer is correct if you consider the expression $[a,b]$ ill-formed when $a>b$, but there’s a small flaw in your justification: you’ve said that if $a$ and $b$ are both rational, then $[a,b]\cap\Bbb Q$ is not open (in $\Bbb Q$), which is correct, but to justify your answer you also have to point out that $[a,b]\cap\Bbb Q$ isn’t open in $\Bbb Q$ if even one of $a$ and $b$ is rationa. If you allow intervals $[a,b]$ with $a>b$, then you must include them as clopen irrespective of whether $a$ or $b$ is rational, since they’re all empty.
H: Does the structure group of S^n homotopic to O(n+1)? It is easy to show that Diff(S^1) is homotopic to O(2), but in the case of n bigger than 1 things become really complicated, I cannot see the conclusion directly. AI: The proof for $S^2$ is in lecture notes by Lurie. The proof for $S^3$ is due to Hatcher [added by studiosus]. What is the homotopy type of the group of diffeomorphisms of the 4-sphere? is a problem in the Open Problem Garden.
H: How to sample N points between 0 and R if they are exponentially distributed? The density of my points x $\in$ [0,R] is exponential: $\rho(x) \approx e^x$ How can I sample N points from there? Thanks, AI: This is an important class of problems, with many different sorts of solutions. We give a procedure that is general, but has important drawbacks. A random variable $X$ with density function a constant times $e^x$ on $[0,R]$ and $0$ elsewhere is unlikely to come up in applications. In order for the integral over $[0,R]$ to be $1$, the density function of $X$ will be given by $$f(x)=\frac{e^x}{e^R-1}$$ on $[0,R]$, and $0$ elsewhere. On the interval $[0,R]$, the cumulative distribution function $F(x)$ is given by $$F(x)=\frac{e^x-1}{e^R-1}.$$ To simulate, we use independent uniforms $U$ on the interval $[0,1]$, and solve the equation $$\frac{e^X-1}{e^R-1}=U.$$ We get $$X=\ln(1+(e^R-1)U).$$
H: Finding a counterexample in model theory I'm currently reading about models of Set Theory, and I'm working on exercises to better understand the concepts. In Kunen's most recent Set Theory text, he mentions that if we have a transitive model $M$ and if $\cap^M$ is defined, then $\cap$ is absolute. Then, he gives the following exercise: Describe a two-element non-transitive $M$ that is isomorphic to $\{0,1\}$, such that $\cap^M$ is defined, but $\cap$ is not absolute for $M$, and such that $\subset$ is not absolute for $M$. I'm having trouble coming up with such a two-element $M$ since I haven't seen many examples of models being used outside the theory. Any help would be greatly appreciated! AI: Let $M$ consist of two sets $\{A,B\}$, such that $A\not\subset B$, but also $A\in B$ and $B\notin A$. For example, one can take $A=\{\emptyset\}$ and $B=\{A\}$. It follows that $M$ thinks $A$ has no elements (since it has no elements in $M$, and therefore $M\models A\subset B$, even though this isn't true externally to $M$. Similarly, $M$ thinks $A\cap B=A$, since there are no elements in $M$ that are in both $A$ and $B$, and so this intersection has the same elements in $M$ as $A$ has in $M$. But externally, we can see that $A\cap B\neq A$. So $M$ thinks $A$ is empty and $B$ is singleton $A$, so $M$ is isomorphic to $\{0,1\}$.
H: Parametrize the intersection of 2 planes. Parametrize the intersection of $\frac {x^2} {3}+y^2+\frac {z^2} {10} = 1$ with $z=2$ (level curve) plane. Here is what I did. Plugged in $z=2$ into the plane $\frac {x^2} {3}+y^2+\frac 25=1$. I got $y^2=9-5x^2$ Then I substituted $y^2$ into the plane $\frac {x^2} {3}+9-5x^2+\frac25=1$ to solve for $x^2$. I got $x^2=1.8$ and then got $y=0$. How should I parametrize the intersection from here onward? $\mathbf Error$ in my calculation, $y^2=\frac 35-\frac{x^2}{3}$ AI: You started out correctly, by plugging $z = 2$ into the equation and finding $\frac{x^2}{3} + y^2 + \frac{2}{5} = 1$, or $$\frac{x^2}{3} + y^2 = \frac{3}{5}.$$ The next step is to put the equation in the standard form for an ellipse. (Recall that the standard form is $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$.) \begin{align} \frac{5}{3} \cdot \frac{x^2}{3} + \frac{5}{3} \cdot y^2 &= \frac{5}{3} \cdot \frac{3}{5} \\ \frac{x^2}{\frac{9}{5}} + \frac{y^2}{\frac{3}{5}} &= 1 \end{align} Here we have $h = k = 0$, $a = \sqrt{\frac{9}{5}} = \frac{3}{\sqrt{5}}$, and $b = \sqrt{\frac{3}{5}}$. The next step is to parametrize the ellipse, and recall that the parametrization for the $z$ coordinate is $z(t) = 2$.
H: If $A\subset B\subset C$, $A$ is dense in $B$, and $B$ is dense in $C$ prove that $A$ is dense in $C$. If $A\subset B\subset C$, $A$ is dense in $B$, and $B$ is dense in $C$ prove that $A$ is dense in $C$. Here's my answer: $A$ is dense in $B$: $\bar{A}=B$ B is dense in $C$: $\bar{B}=C$ However $\bar{B}$ is the smallest closed set containing $B$. Therefore $\bar{A}=\bar{B}$ and since $\bar{A}=C$, $A$ is dense in $C$. My instructor gave me a 2/5, what did I did wrong? AI: First, it is kind of bad form to answer with $X,Y,Z$ if the question was phrased in terms of $A,B,C$. But I wouldn't take you 60% off because of that. Regarding the question, note that when you say $\overline X=Y$, you are implying that $Y$ is closed, which is not necessarily the case. To give an intuitive example, you could have $X=\mathbb Q$, $Y=\mathbb R\setminus\{\sqrt2\}$, $Z=\mathbb R$. You could use your reasoning if you think in terms of relative topologies. But in that case you must indicate so when you answer the question.
H: Find the probability function of random variable Y Two balls are drawn at random form a box containing ten balls numbered 0,1,...9.Let random variable Y be their total. Tabulate the probability function of Y if the sampling is without replacement. I dont know how to solve this question.Please help. Thank you very much. AI: There are $\binom{10}{2}=45$ equally likely pairs of numbers. The sum $Y$ must be $\ge 1$ and $\le 17$. For each value of $k$, where $1\le k\le 17$, we need to compute $\Pr(Y=k)$. We do a couple of calculations, and let you deal with the rest. Let $k=1$. Our sum is $1$ precisely if we picked the numbers $0$ and $1$. Thus $\Pr(Y=1)=\frac{1}{45}$. Let $k=5$. Our sum is $5$ if we picked the numbers $0$ and $5$, or $1$ and $4$, or $2$ and $3$. Each of these has probability $\frac{1}{45}$, so $\Pr(Y=5)=\frac{3}{45}$. Deal separately with the possibilities $k=1$ to $17$. When you are finished, check that your numbers add up to $1$. Symmetry will help shorten the calculations.
H: Why is the $\operatorname{diam}{(\mathbb Q)}$ infinite? I was trying to find a counterexample to show that the $\operatorname{diam}{(A)}$ and the $\operatorname{diam}{(Int(A))} $ may not be the same, where $A$ is the subset of the metric space $X$. I chose the metric space $X$ = $\mathbb R$ , and chose $A$ = $\mathbb Q$. So, since $Int(\mathbb Q)$ is empty , i concluded that the $\operatorname{diam}{(Int(\mathbb Q))}$ = $0$. Then, i checked up online, to find that the $\operatorname{diam}{(\mathbb Q)}$ is infinity. So, i think that my counterexample would probably work. I do understand intuitively, why the $\operatorname{diam}{(\mathbb Q)}$ is infinity, What i wanted to know is that if we were to prove this then, we would probably have to show that $a-b$ > $M$ where $a,b \in \mathbb Q$ and $M$ is an arbitrarily large positive real number. Am i correct ? It would be nice if someone could shed some light on this ! AI: It is probably better to make an example that doesn't involve infinity or the empty set. You could take $$ A=(\mathbb Q\cap[0,1])\cup[0,1/2]. $$ Then $$ \mbox{diam}(A)=1,\ \ \mbox{diam}(\mbox{int}(A))=1/2. $$ And of course, you can take $d$ to be any other number in $(0,1)$. As for your question, think of the diameter of the integers (which are part of the rationals).
H: Sigma algebras... "Let $\mathcal{C}$ = {A,B,C} be a partition of E. List the elements of the smallest sigma algebra containing $\mathcal{C}$." Let me see if I got this right...because $\mathcal{C}$ is a partition of E, the smallest sigma algebra is the trivial one {$\emptyset$,E}? What if $\mathcal{C}$ were not a partition? AI: $\{\varnothing,E\}$ does not contain $\mathscr{C}$, so it clearly cannot be the smallest $\sigma$-algebra containing $\mathscr{C}$. That $\sigma$-algebra must contain $A,B,$ and $C$. It must also contain all countable unions of $A,B$, and $C$, so it must include $A\cup B$, $A\cup C$, $B\cup C$, and $A\cup B\cup C=E$, and it must include all of the complements $X\setminus Y$ with $X,Y\in\{A,B,C\}$, so in particular it must include $A\setminus A=\varnothing$. The list so far includes the following sets: $$\varnothing,A,B,C,A\cup B,A\cup C,B\cup C,E\.$$ Is that collection of sets a $\sigma$-algebra? If so, it’s the smallest one containing $\mathscr{C}$, because (as we’ve just seen) every one of those sets must be in any $\sigma$-algebra containing $\mathscr{C}$.
H: Solving system of multivariable 2nd-degree polynomials How would you go about solving a problem such as: \begin{matrix} { x }^{ 2 }+3xy-9=0 \quad(1)\\ 2{ y }^{ 2 }-4xy+5=0 \quad(2) \end{matrix} where $(x,y)\in\mathbb{C}^{2}$. More generally, how would you solve any set of equations of the form: \begin{matrix} { ax }^{ 2 }+bxy+c=0 \\ d{ y }^{ 2 }+exy+f=0 \end{matrix} where $a, b, c, d, e, f \in \mathbb{Q}$ and $(x,y)\in\mathbb{C}^{2}$. I know that there are four complex solutions to a system of equations in this form, but don't know how one would solve for them. AI: Multiply first equation by $5$. Multiply second equation by $9$. Add both Equations. Divide this equation by $y^2$. Let $t={x\over y}$. You get a quadratic in $t$. The steps are[Generalization]: Reduce the two equations to one $${ ax }^{ 2 }+bxy+{cy}^2=0$$ Then divide by $y^2$ $${ a{x^2 \over y^2} }+b{x\over y}+{c}=0$$ Replace ${x\over y}=t$ $${ a{t} }^{ 2 }+b{t}+{c}=0$$
H: Finding Missing Observed Scores given standard deviation and mean The question that i'm being asked it: the mean of 10 observed scores was 20 and the standard deviation if 6.0. the observed scores are {16,11,20,24,29,24,16,20,,} what are the two missing scores I have tried to figure this out and i keep ending up needing to use the quadratic formula and getting two numbers that equate to a different standard deviation. I really do not know what more i can do. please help me AI: Let the missing scores be $a$ and $b$. The non-missing scores have mean $20$, so the missing scores must add up to $40$. The standard deviation, if we use "$n-1$", not $n$, is equal to $$\sqrt{\frac{1}{9} (226+(a-20)^2+(b-20)^2)}.$$ This is equal to $6$, so some manipulation gives $$324=226+(a-20)^2+(b-20)^2.$$ Let $x=a-20$ and $y=b-20$. We get $x+y=0$ and $x^2+y^2=98$. Substitute. We get $x^2=49$. So $x=7$ and $y=-7$, or the other way around. Our missing values are $13$ and $27$.
H: Prove the probability is 0 Let E and F be independent with E = AUB and F=AB. Prove that either P(AB)=0 or else P(not A and not B)=0. I dont know how to solve it. Please help. Thank you very much. AI: Since $E,F$ are independent, you have $P(E) P(F) = P(E \cap F)$. Note that $E \cap F = F$. Hence we have $P(E) P(F) = P(E)$, or $P(E)(1-P(F)) = 0$. This gives the required result.
H: A couple has 2 children. What is the probability that both are girls if the eldest is a girl? This is another question like this one. And by the same reason, the book only has the final answer, I'd like to check if my reasoning is right. A couple has 2 children. What is the probability that both are girls if the eldest is a girl? AI: The sample space is $S = \{(g,g),(g,b),(b,g),(b,b)\}$, where $b$ is for boy, $g$ is for girl the first element of the tuple is the eldest. Let $B$ the event the eldest is a girl, so $B=\{(g,b),(g,g)\}$. $A$ is the event where the two children are girls. $A = \{(g,g)\}$. Then: $$ P(A\|B)=\frac{P(A\cap B)}{P(B)}=\frac{\dfrac{\|A\cap B\|}{\|S\|}}{\dfrac{\|B\|}{\|S\|}}=\frac{1}{2}$$. The end.
H: If $x \in \overline{A}$ and $A \subset X$, $X$ first-countable, then there's a sequence of points in $A$ converging to $x$. Supposedly this relies on first-countability of $X$. Let $x \in \overline{A}$, then by definition there's a neighborhood $U_1$ of $x$ that contains some $x_1 \in A$. If this is the only neighborhood of $x$ then we're done since we can let the sequence be $x_1, \dots , x_1, \dots$ and for all neighborhoods $U \ni x$, there's natural $N$ such that $n \gt N \implies x_n \in U$, where $x_n$ is the sequence of $x_1$'s. That is the topological definition of a sequence converging to a point. So assume there's another neighborhood $U_2'$ of $x$... This is where I'm lost. AI: You haven’t used first countability. Start by using it to say that there is a fmaily $\{U_n:n\in\Bbb N\}$ of open nbhds of $x$ such that if $V$ is any open nbhd of $x$, then there is an $n\in\Bbb N$ such that $U_n\subseteq U$. In other words, $\{U_n:n\in\Bbb N\}$ is a countable local base at $x$. That still isn’t quite good enough, though: you’ll want a nested local base at $x$. For $n\in\Bbb N$ let $B_n=\bigcap_{k\le n}U_k$, and let $\mathscr{B}=\{B_n:n\in\Bbb N\}$. Show that $B_n\supseteq B_{n+1}$ for each $n\in\Bbb N$. Show that $\mathscr{B}$ is a countable local base at $x$. For each $n\in\Bbb N$ choose $x_n\in A\cap B_n$ (why is this possible?), and show that $\langle x_n:n\in\Bbb N\rangle$ converges to $x$.
H: will the rank of a projector matrix be equal to the dimension of vector space it projects to? Let the projector be the $N \times N$ matrix $A$. Let its rank be $r$. Let the dimension of the space for which it is the projector be $m$. Is $m==r$? AI: Yes. The rank of matrix is the dimension of the image, and the image is precisely the "space for which it is the projector", in your language.
H: Counting edges in a specially defined graph $G$ has vertices $\{0, \ldots, pq - 1\}$, where $p, q$ are different primes. There is an edge between $x$ and $y$ if $p \mid x - y$ or $q \mid x - y$. How many edges does $G$ have? I looked at the cases where $p=2$ and $q=3, 5, 7, 11, 13, 17, 23$ and the formula I got was $\|E\|=q^2$, but it doesn't apply to $p=3$. For $p = 3, q = 5$, there are $45$ edges. How do I derive a closed form formula for this? AI: In the set $\{0,\dots,pq-1\}$, there are $p$ elements in each equivalence class modulo $q$, and $q$ in each class modulo $p$. Since there is an edge between $x$ and $y$ iff $x\cong y \pmod p$ of $x\cong y \pmod q$, each of the $p$ equivalence classes modulo $p$ contributes $\frac{q(q-1)}2$ edges, and each of the q equivalence classes modulo $q$ contributes $\frac{p(p-1)}2$ edges. Therefore, overall, there are $\frac{pq(q-1) + pq(p-1)}2$ edges. In the case $p=2$, this does in fact simplify to $q^2$, as you found.
H: Simplifying an inverse trig function? I am trying to figure out how to simplify this expression but I am not quite sure how these inverses work. What sort of approach should I take for this equation? $$\tan\left(2\cos^{-1}\left(\dfrac{x}{5}\right)\right)$$ AI: You will need two things. One is the identity for $tan(2x)$ in terms of $tan(x)$. You can look this up the purpose is to get rid of the $2$ in your problem. The other thing is to know what $\tan(\cos^{-1}(x/5))$ is. You can figure out by realizing that the angle $\cos^{-1}(x/5)$ corresponds to a right triangle with an adjacent side $x$ and a hypotenuse $5$. With this information you should be able to solve the right triangle completely and determine what the tangent of that angle is. The key to handling inverse trig functions is to realize that they represent some angle in a right triangle. For instance $\theta = \tan^{-1}(x+2)$ is an angle whose opposite side is $(x+2)$ and whose adjacent side is $1$. The hypotenuse of this hypothetical triangle is $\sqrt{(x+2)^2+1}$. Therefore we can conclude that $\sin(\tan^{-1}(x+2)) = \frac{x+2}{\sqrt{(x+2)^2+1}}$
H: Does two isomorphic lattices always have the same Hasse diagram? I have done few examples of proving that lattices are isomorphic. I just draw their Hasse diagrams and if they are the same I say its isomorphic. Is there any case where two isomorphic latices don't have the same Hasse diagram? AI: Two lattices (or, more generally, posets) are isomorphic if and only if their order-relation is 'the same'. That is, if $(P, \le_P)$ and $(Q, \le_Q)$ are posets and $f : P \to Q$ is an isomorphism then $a \le_P b$ if and only if $f(a) \le_Q f(b)$. Thus they have the same Hasse diagram.
H: Showing that a function is not a homeomorphism Consider the function $f: [0,1) \rightarrow \mathbb{C}$ given by $f(t) = e^{2\pi i t}$. I must show that the function $f^: [0,1) \rightarrow \mathrm{im}(f)$ is not a homeomorphism, given the standard topologies on both sets, but I am not sure how to proceed. Some work: I am given that $f$ is injective (from which it follows that $f^$ is surjective, and thus bijective and invertible). I also know that $f^$ is continuous. Thus I need to show that the inverse of $f^$ is not continuous. I'm unsure of how to compute the inverse directly (does this require any complex analysis? I have no background in it!), so I am trying to show that $f^$ is not an open map by looking for an open set in $[0,1)$ that does not get mapped to an open set in $\mathrm{im}(f) = S^1$, with respect to the subspace topology induced by $\mathbb{C} \cong \mathbb{R}^2$. Any help is appreciated! AI: Let $f^:[0,1) \to \mathbb{S}^1$ be the map $f^(t)=e^{2 i \pi t}$. Now define the the function $g:\mathbb{S} \to [0,1)$ as the mapping with the rule: $$ e^{2\pi i t } \mapsto t $$ It will be easy for you to show that this $g$ is the inverse of $f^$. Now you should know that a function is continuous iff the inverse image of every open set is open. So consider the inverse image of $[0,\epsilon)$ under $g$ where $\epsilon<1$. The set $[0,\epsilon)$ is an open subset in the relative topology of $[0,1)$. But the inverse image is $$ g^{-1}([0,\epsilon))=\left\{ e^{2\pi i t} \mid t\in [0,\epsilon) \right\} $$ which is not open in $\mathbb{S}^1$. This is easy to verify since there is no open set containing $1=e^{2 i \pi \cdot 0}\in \mathbb{S}^1$ entirely contained in $g^{-1}([0,\epsilon))$. Every such open set will contain an element $e^{2 i \pi s}$ with $s<0$.
H: Keeping Focused: School Mathematics is such a huge field, and I am just wondering how does a mathematician keep focused? For example to learn Stochastic Calculus you need some Analysis, Measure Theory, Probability, Set Theory, and all of those topics usually need some other branch of mathematics. So, how do you know when to say enough with Measure Theory? or enough with .....? And stay on topic, say learning Stochastic Calculus? Also, I find that there is usually one course that just captivates me (Real Analysis) more than another and I often find myself neglecting the others (DFQ, Probability). Since, many here have already finished graduate school, I figured perhaps they might have some good advice for maintaining a healthy and productive balance. AI: How terrific that at least one course bowls you over. The reason you keep discovering you need tools from one "branch" of mathematics to proceed in another "branch" is that there really are no branches. There is just mathematics and it is all thoroughly intertwined. We split things up so that we can teach them, or focus on them, but any split we make is at least partly artificial. So just about the time we think we've learned geometry, someone comes along and solves our hardest problem with algebra. Or we're working away at number theory, and someone brings our attention to elliptic functions. The mathematicians who accomplish the most are the ones who know the most, and the amount, depth and breadth of what they know is amazing. So if you want to be really good, the answer is you just keep on learning, as much as you can in as many areas as you can. However, you must also survive graduate school. To deal with the courses you are less interested in I suggest (after years of experience doing things I don't like) that you attend to what needs to be done for those courses first. Do what you have to do, but not more. Then plunge into what you like, and do more than you have to do. Discuss the interesting stuff with others who are also interested. That is a huge motivator. Now -- do you want more of a life than mathematics, or not? The best mathematicians spend huge amounts of time on the subject. They work all day at it, then at night they go home and for recreation they -- do more mathematics. Is that the lifestyle you want? (I'm not knocking it, it's either for you or it isn't.) If the answer is yes, you will have plenty of time do learn tons of math. If the answer is no, you have to budget your math time so it doesn't take over your life. And that definitely means concentrating on what fascinates you. The good news is that what fascinates you is entrenched in mathematics, and as you concentrate on it a lot of peripheral knowledge will just naturally come to you. Best wishes.
H: What condition can make this event independent? A population consists of F females and M males;the population includes f female smokers and m male smokers. If A is the event that the individual is female and B is the event he or she is a smoker , find the condition on f, m, F and M so that A and B are independent events? The answer is f/F=m/M I dont know how to get that answer. Please help. Thank you. AI: The events $A$ and $B$ are independent precisely if $$\Pr(A\cap B)=\Pr(A)\Pr(B).\tag{1}$$ If $F+M\ne 0$, we have $$\Pr(A)=\frac{F}{F+M},$$ and $$\Pr(B)=\frac{f+m}{F+M}.$$ Also, $$\Pr(A\cap B)=\frac{f}{F+M}.$$ Substituting in (1) we get after some minor cancellation that $A$ and $B$ are independent precisely if $$f=\frac{F(f+m)}{F+M}$$ or equivalently if $$f(F+M)=Ff +mF.$$ Cancel the $fF$. We get $fM=mF$. Reamrks: $1.$ The suggested answer $\frac{f}{F}=\frac{m}{M}$ is almost but not quite equivalent to $fM=mF$. For example, if there are no males, then $\frac{m}{M}$ does not make sense, but we still have independence. In our calculation we assumed $F+M\ne 0$. If $F+M=0$, we have independence. So in fact the given condition is not equivalent to independence. But the non-equivalence is for trivial cases the author presumably did not consider, namely $F=0$ or $M=0$. $2.$ We did a formal calculations, since sometimes one has independence even when "intuition" might suggest we do not. But here the intuition is reasonably clear. The events $A$ and $B$ are independent if $\Pr(B\|A)=\Pr(B)$. That means that given the information that a person is female gives no information about whether or not she is a smoker. That will be the case if the proportion of smokers among females is the same as it is among males, that is, if $\frac{f}{F}=\frac{m}{M}$.
H: Let $M$ be a metric space with the discrete metric or more generally a homeomorph of $M$. Let $M$ be a metric space with the discrete metric or more generally a homeomorph of $M$. I've proven that every subset of $M$ is clopen and every function defined on $M$ is continuous. For the question, I answered: Let $(x_m)_{m\in\mathbb{R}}$ be a sequence of points defined in metric space $M$, converges to $x\in M$, denoted by that $\lim_{m\to\infty}x_m=x$. Then, the sequences that converge in $M$ would be the ones that take the same value for all large value $m$. However he said that is wrong, I think this is definitely correct. Anyone please help? AI: I'm currently grading for a class like this. You have to realize that it is completely possible to have exactly the right answer in your head, yet write a proof that is either wrong or so incomplete that we have to give it no points. You are referencing a good proof. That is, I can take your idea, and make a good proof from it, like so: Let $X$ be any metric space, and $f: M\to X$ any function. Continuity is equivalent to sequential continuity, so it suffices to show that $f$ is sequentially continuous. Let $(a_n)$ be a convergent sequence in $M$ with limit $a$. Then, for some $N$, we have $\|a_n -a\| <1$ for all $n\leq N$. But $M$ has the discrete metric, so $\|a_n -a\| <1 \implies a_n = a$, and it follows that $(a_n)$ is eventually constant. Finally, we have $\lim_{n\to \infty} f(a_n) = \lim_{n\to \infty} f(a) = f(a)$. So the sequence $(f(a_n))$ is also convergent, and we conclude that $f$ is sequentially continuous. But is the proof you actually wrote, actually correct as written? Let's compare: Let $(x_m)_{m\in\mathbb{R}}$ be a sequence of points defined in metric space $M$, converges to $x\in M$, denoted by that $\lim_{m\to\infty}x_m=x$. Then, the sequences that converge in $M$ would be the ones that take the same value for all large value $m$. Okay, your proof is shorter (though it's not that much shorter). But it fails to really explain what it's doing, it never mentions continuity or functions at all, it names one sequence then suddenly switches to talking about "the sequences", it uses the vague word "large", and it incorrectly uses $\mathbb{R}$ as an indexing set instead of $\mathbb{N}$ for further confusion. Finally, it ends with a conclusion about sequences. Sequences! There were no sequences in the problem, so ending your proof with a statement about sequences is confusing even to somebody who understands what you're saying, let alone a person trying to read it for the first time, in the middle of a stack of eighty other problem sets, late at night when his kids want him to read them a story. The point is that your proof is most certainly not "definitely correct". There's a good idea, but it's just not executed very well, both in technical terms and in your awareness of the reader. The art of proof is the communication of mathematical ideas to other humans. You must put yourself in their shoes, and understand that there will always be room for improvement in the quality of your mathematical writing. If an instructor does not understand your proof, that is on you, not them.
H: Finding the eqn. of a plane that passes through the line of intersection of 2 planes and is perpendicular to another plane. Find the equation of the plane that passes through the line of intersection of the planes $4x - 2y + z - 3 = 0$ and $2x - y + 3z + 1 = 0$, and that is perpendicular to the plane $3x + y - z + 7 = 0$. I have attached a picture and that is what I got. Can someone please tell me if my answer is right because the answer at the back of the textbook is different but I am pretty confident with my solution. Thanks! AI: The intersection of the two planes gives a solution $(x,2(x-1),-1)$ with $x \in \mathbb{R}$. If $n$ is the normal for our plane, then we need $n \bot (3,1,-1)$ (and $n \neq 0$). This gives $3 n_1+n_2=n_3 = 0$. The plane will have the form $\langle n, (x,y,z) \rangle = \alpha$. Since the line is on the plane, we have $\langle n, (x,2(x-1),-1) \rangle = \alpha$ for all $x$. Differentiating gives $n_1+2 n_2 = 0$. Solving gives $n = t(-2,1,-5)$ for some $t \neq 0$. I will pick $t=1$ to get $n=(-2,1,-5)$. Setting $x=0$ gives a point on the line as $(0,-2,-1)$ from which we get $\alpha = 3$. Hence one equation of the plane is $-2x+y-5z = 3$.
H: Which are tangents We are asked to see which are tangents and which aren't. I think B3, bottom left and bottom middle are not tangents AI: A tangent to a curve is a straight line that just touches the curve, such that the slope of the tangent is exactly that as the slope of the curve. By that definition, all except for B5 display tangents. EDIT: The picture is unclear, so it is difficult for me to tell in the cases of B3 and B4. A larger picture would help, or you could judge this yourself using the definition I provided above.
H: If $\|z\| = 2$. Then Locus of $z$ Representing The Complex no. $-1+5z,$ is If $\|z\| = 2$. Then Locus of $z$ Representing The Complex no. $-1+5z,$ is $\underline{\bf{My\;\; Try::}}$ Let $z^{'} = -1+5z$, Where $z^{'} = x^{'}+iy^{'}$. So put $5z = z^{'}+1$ in $\|z\| = 2\Leftrightarrow \|5z\| = 10$ $\|z^{'}-(-1+0\cdot i)\| = 10$, means Locus of Conplex no. $z^{'}$ is a Circle with Center at $(-1,0)$ and Radious $ = 10$ Is it Right OR Not, If Not . The please explain me Righta answer. AI: Well done! Your reasoning is spot on!
H: Partial derivative paradox Okay, perhaps not a paradox, but somewhat of a lack of understanding on my part. Let $z$ equal some function of $x$ and $y$, i.e. $z = f(x, y)$ and take partial derivatives $\frac{\partial z}{\partial x} = f_x$ and $\frac{\partial z}{\partial y} = f_y$ all and good. But now say I do partial differentiation with respect to z. $\frac{\partial z}{\partial z} = f_z$ which equals $0$ because $f(x, y)$ is a function of $x$ and $y$, but not $z$, so all $x$ and $y$ are held constant and the derivative of a constant is zero. But that's not the right answer is it? $\frac{\partial z}{\partial z}$ should be equal to 1. Where did my logic go wrong? AI: The problem is that $f(x,y)$ is indeed a function of $z$, you have written it yourself: $f(x,y)=z$, when taking a partial derivative you must always look at the dependance of the function with respect to the variable.
H: The rate of change of the distance from the plane to the radar station Problem statement: A plane flying with constant speed of 4 km/min passes over a ground radar station at an altitude of 6 km and climbs at an angle of 35 degrees. At what rate, in km/min, is the distance from the plane to the radar station increasing 6 minutes later? Hint: The law of cosines for a triangle is $c^2=a^2+b^2− 2ab \cos(\theta)$ where $\theta$ is the angle between the sides of length $a$ and $b$. Okay, so following the typical related rates algorithm: Identify the problem at hand by drawing a picture. I did that. It appears to me that after the plane passes over the ground station, we get an angle between the measures a and b of 90 + 35 degrees. Awesome. Now, let's build a table of values: $$ a = 6\,\mathrm{km}\\ a' = 0 \quad \text{(duh)}\\ b = b'\cdot t = 4\cdot 6 = 24\\ b' = 4\,\mathrm{km}/\mathrm{min}\\ t' = 6min\\ c = \sqrt{a^2 + b^2} = \sqrt{36+24} = 2\sqrt{15} \approx 2.78316\\ c' = ? $$ Okay, let's differentiate the law of cosines, which is given to us in form of a hint: $2cc' = 2aa' + 2bb' - \frac{du}{dt}$ $ u = 2ab\cos(\theta)$ $\frac{du}{dt} = \frac{ds}{dt} \cdot \cos(\theta) - \sin(\theta)\cdot(2ab)$ $\frac{ds}{dt} = 0\cdot ab + (a'b + b'a)$ And so now we rewind: $\frac{du}{dt} = \left[ \left(0 \cdot ab + \left(a'b + b'a \right) \right) \right] \cdot \cos(\theta) - \sin(\theta) \cdot 2ab $ And so: $2cc' = 2aa' + 2bb' - \left[ \left(0 \cdot ab + \left(a'b + b'a \right) \right) \right] \cdot \cos(\theta) - \sin(\theta) \cdot 2ab $ Boom, boom. Let's substutute what we know $2\left(2 \sqrt{15} \right)(c') = 2(6)(0) + 2(24)(4) - \left[ \left(0\cdot (6)(24) + \left((0)(24) + (4)(6) \right) \right) \right] \cdot \cos(125^\circ) - \sin(125^\circ) \cdot 2(4)(24)$ Now what? Well, evaluate. Not sure if that's going to yield the right answer. But I don't think that it will. Please help! I don't know how to solve this kind of problem! It seems my solution/process is incorrect! AI: We use your notation. Let $c=c(t)$ be the distance from the plane to the radar station at time $t$. Let $b$ be the distance from the plane to the point where the climb began. Then $$c^2=36+b^2-12b\cos \theta=36+b^2+12b\sin\phi,\tag{1}$$ where $\theta=125^\circ$ and $\phi=35^\circ$. Differentiating, we get $$2c\frac{dc}{dt}=2b\frac{db}{dt}+12\sin\phi\frac{db}{dt}.\tag{2}$$ "Freeze" at time $6$ minutes after the climb began. We know everything on the right-hand side of (2), since at that instant $b=(6)(4)$. We need to calculate $c$ at time $t=6$. Equation (1) does that, with $b=24$.
H: polynomial with integer coefficients Question: Let $\Pi_{j=1}^n (z-z_j)$ be a polynomial with integer coefficients. Is also $\Pi_{j=1}^n (z-z_j^k)$ for $k=1,2,3,\dots$ a polynomial with integer coefficients? In fact, this is a question that someone asked 3 days ago, but the answer is not clear (I think) For this, I think that it is enough to show that $z_J$ and $z_j^k$ have same minimal polynomial, since there is a $\mathbb{Z}$-automorphism($\mathbb{Z}$ left fixed) which map an element into its conjugate. But this is impossible to show, really. How can I solve this problem? Separately, Who knows why I can't write a comment in some questions?? AI: Yes, they are integers. Note that the $z_j^k$ are algebraic integers, since the $z_j$ are and thus so are the coefficients of $\displaystyle \prod_j(z-z_j^k)$ it suffices to prove that the coefficients are rational. To do this, let $K=\mathbb{Q}(\{z_j\})$. Note that $K$ is the splitting field of $\displaystyle \prod_j(z-z_j)$ a polynomial with $\mathbb{Q}$-coefficients, and thus $K/\mathbb{Q}$ is Galois. Let $\sigma\in\text{Gal}(K/\mathbb{Q})$ and consider any of the coefficients of $\displaystyle \prod_j (z-z_j^k)$ which are elementary symmetric polynomials $e_i(z_1^k,\ldots,z_n^k)$. Then, since $\sigma$ permutes the set $\{z_1,\ldots,z_n\}$ we have that $$\sigma(e_i(z_1^k,\ldots,z_n^k))=e_1(\sigma(z_1)^k,\ldots,\sigma(z_n)^k)=e_i(z_1^k,\ldots,z_n^k)$$ where we used the fact that $e_i$ is the elementary symmetric polynomial. Thus, since $K/\mathbb{Q}$ is Galois, we have that $e_i(z_1^k,\ldots,z_n^k)\in K^{\text{Gal}(K/\mathbb{Q})}=\mathbb{Q}$. Thus, from previous comment it follows that $e_i(z_1^k,\ldots,z_n^k)\in\mathbb{Z}$ as desired.
H: adjustment of Prof.,s in Round Table In How many ways can $5$ Professors of Physics Including Prof. Hardy and $3$ Three Professors of Chemistry Including Prof. Julian be seated on a Round table, If Prof. hardy and Prof. Julian are not adjacent. $\underline{\bf{My\;\;Try:}}$ First we will adjust $4$ Prof. of Physics and $2$ Prof. of Chemistry, which can be done in $6!$ ways. Now we will adjust Hardy and Julian so that they are not adjacent. I Did not Not Understand How can I adjust These two Professors. Help Required. Thanks AI: Since rotating the people, by convention, yields the "same" arrangement, we can assume that Hardy, who just won the Nobel Prize, sits on the one throne among the chairs. Then Julian selects one of the chairs away from Hardy. She has $5$ choices. For each of these $5$ choices, the remaining $6$ people can be arranged in the remaining $6$ chairs in $6!$ ways, for a total of $(5)(6!)$.
H: Question about property of circle We know that equal chords are equidistant from the center. However, I was curious if the lengths involved are proportional as well since the circle is a pretty symmetrical shape. Here's what I mean: Let there be two chords,C1 and C2, in a circle at distances D1 and D2 from the center. Will L1 and L2 be proportional to the distances D1 and D2? Is $ \frac{L1}{L2} = \frac{D1}{D2} $ true? AI: rUse pythagoras with $R$ as the radius of the circle $R^2=D_1^2+\left(\cfrac{L_1}2\right)^2=D_2^2+\left(\cfrac{L_2}2\right)^2$ or you can use trigonometric functions of half the angle subtended by the chord at the centre of the circle, which amounts to the same thing. So it isn't quite as neat as you conjectured, Note that the chord gets shorted as the distance from the centre increases. It is zero when the distance is equal to the radius.
H: How to write $\frac{k}{k}$ using $\sum$ notation? I want to use $\sum$ notation for this:$$\underbrace{\frac{1}{k}+\frac{1}{k}+\ldots +\frac{1}{k}+\frac{1}{k}}_{k\text{ times}}$$ I guessed$$\sum_1^k\frac{1}{k} ,$$but it equals $$1+\frac{1}{2}+\dots+\frac{1}{k-1}+\frac{1}{k}.$$ AI: You are getting confused because you are not being explicit about the index of the summation. The summation $$\sum_{i=1}^k\frac{1}{k}$$ does represent $$\underbrace{\frac{1}{k}+\frac{1}{k}+\cdots+\frac{1}{k}}_{k\text{ times}}.$$ You're confusing this with the summation $$\sum_{i=1}^k\frac{1}{i}$$ which represents $$\frac{1}{1}+\frac{1}{2}+\cdots+\frac{1}{k}.$$
H: Equivalent definitions of algebra homomorphisms I'm studying Atiyah-Macdonald's commutative algebra book and I'm trying to prove this equivalence: One implication If $h\circ f=g$. I can prove that $h(ax)=ah(x)$ but I have failed to prove that $f(x_1+x_2)=f(x_1)+f(x_2)$ (in order to complete the proof of A-module homomorphism) and $f(x_1x_2)=f(x_1)f(x_2)$ (to complete the prove of ring homomorphism). what I proved was $h(f(x_1)+f(x_2))=h(f(x_1))+h(f(x_2))$ and $hf(x_1)hf(x_2)=h(f(x_1))h(f(x_2))$ since $f$ is not necessarily surjective, we don't have that $h$ is a A-algebra homomorphism. Another implication I need hints, I have no idea how to proceed. I need help. Thanks a lot AI: For the implication shown, it is required that $h$ is a homomorphism of rings, the cited statement is to read as follows: For ring homomorphisms $f \colon A \to B$ and $g \colon A \to C$ a homomorphism of rings $h \colon B \to C$ is a homormorphism of $A$-algebras iff $g = h \circ f$. For the missing implication, note that the $A$-module structures are given by $f$ and $g$ respectively, we have for $a \in A$ if $h$ is an $A$-module homomorphism: \begin{align} (h\circ f)(a) &= h\bigl(f(a)\bigr)\\ &= h(a \cdot 1_B)\\ &= a \cdot h(1_B)\\ &= a \cdot 1_C\\ &= g(a) \end{align} So $g = h \circ f$.
H: Permutations with repetition for some elements Suppose we have $N$ slots, each of which can be filled with $X$ options, but $2$ of these slots can only be filled in $1$ way (out of $X$ ways), then what is the number of permutations possible ? For example $N=4$,$\ X=\{a,b,c\}$ and one element must be $a$ and one must be $c$ then the number of permutation in this case I've calculated as follows: $\{Possible\ configuration\} \implies Permutations\ for\ this\ configuration $ $\{a,a,a,c\} \implies 4 $ $\{a,a,b,c\} \implies 12$ $\{a,a,c,c\} \implies 6$ $\{a,b,b,c\} \implies 12$ $\{a,b,c,c\} \implies 12$ $\{a,c,c,c\} \implies 4$ Thus Total = $50$ I am looking for a generalized formula for this (possibly with even 2 replaced with a parameter) TIA AI: We do the count for your case in another way. There are $3^4$ strings with no restriction. We count the bad strings, that are missing at least one of $a$ and $c$. There are $2^4$ strings with no $a$, and $2^4$ with no $c$. But we have double-counted the one string with no $a$ and no $c$. Thus there are $31$ bad strings, and therefore $50$ good. Exactly the same idea works if we have $n$ slots, and $k$ letters that include $a$ and $c$, and we want to have at least one $a$ and at least one $c$. There are $k^n$ strings. The number of bad strings is $(k-1)^n+(k-1)^n -(k-2)^n$. Subtract. The Inclusion/Exclusion technique should be useful for variants of your problem.
H: Why are these two statements about vector products equivalent? Let $w_1, \dots, w_m \in \mathbb{C}^d$. Condition (1) is: $\sum_i \|\langle v, w_i \rangle \|^2 = \eta$ whenever $\\|v\\| = 1$. Condition (2) is: $\sum_i u_i u_i^* = I^d$, where $u_i = w_i / \sqrt{\eta}$ This paper claims that the two conditions are equivalent (top of page 3, just beneath the statement of Corollary 1.3). I can't figure out why that would be. Can you help point me in the right direction? AI: Note that for $\def\norm#1{\left\\|#1\right\\|}\def\C{\mathbb C}$$v \in \C^d$ with $\norm v = 1$, that $\def\abs#1{\left\|#1\right\|}\def\sp#1{\left<#1\right>}$ \begin{align} \sum_i \abs{\sp{v,w_i}}^2 &= \sum_i \overline{\sp{v,w_i}}\\\sp{w_i,v} &= \sum_i \sp{w_i,v}\sp{v,w_i}\\ &= \sum_i v^w_i w_i^v\\ &= v^ \sum_i w_i w_i^* \cdot v\\ &= \eta \cdot v^\sum_i u_i u_i^ \cdot v\\ \end{align} So the sum is equal to $\eta$ for all such $v$ iff the quadratic form represented by $A = \sum_i u_i u_i^$ is equal to $v \mapsto \norm v^2$, that is iff $A$ is the identity.
H: Expected Value of Flips Until HT Consecutively Suppose you ﬂip a fair coin repeatedly until you see a Heads followed by a Tails. What is the expected number of coin ﬂips you have to ﬂip? By manipulating an equation based on the result of the first flip, shown at this link: http://www.codechef.com/wiki/tutorial-expectation the answer is 6. This also makes sense intuitively since the expected value of the number flips until HH or TT is 3. But is there a way to tackle this problem by summing a series of probabilities multiplied by the values? Thank you! AI: Consider the string of tosses that precede the first HT. If there is ever an H in that string, every toss after that (but within the string) must also be H. Thus, if there are $m$ tosses before the first HT, they must take the form $T^kH^{m-k}$ for some $k\in\{0,\ldots,m\}$. There are therefore exactly $m+1$ possibilities for this initial string of length $m$. If the first HT is completed on toss $n$, then $m=n-2$, and there are $n-1$ possibilities for the initial string, each occurring with probability $\left(\frac12\right)^{n-2}$, assuming a fair coin. The total probability of completing the first HT on toss $n$ is therefore $(n-1)\left(\frac12\right)^n$, and you want $$\sum_{n\ge 2}n(n-1)\left(\frac12\right)^n\;.$$ If $f(x)=\frac1{1-x}$, for $\|x\|<1$ we have $f(x)=\sum_{n\ge 0}x^n$, so $$\frac1{(1-x)^2}=f\,'(x)=\sum_{n\ge 1}nx^{n-1}\;.$$ Now differentiate again and multiply by $x^2$: $$\frac{2x^2}{(1-x)^3}=x^2f''(x)=\sum_{n\ge 2}n(n-1)x^n\;.$$ Note that the answer is not $6$; you must have misapplied the method from the link. Let $x$ be the expected number of tosses, and let $y$ be the expected additional number of tosses if you have just thrown H. Then $$x=\frac12(x+1)+\frac12(y+1)\;,$$ and $$y=\frac12+\frac12(y+1)\;.$$ Then $y=2$, so $x=\frac12(x+1)+\frac12(3)=\frac{x}2+2$, and $x=4$.
H: How to solve $x+(x\cdot\frac{20}{100})=600$? I was solving a puzzle, and finally I got this equation as the result but I couldn't solve it. $$x+\left(x\times\left(\frac{20}{100}\right)\right)=600$$ How to solve this equation? Please provide a step by step solution. AI: $$x+\left(x\times\left(\frac{20}{100}\right)\right)=600$$ $$x+\left(x\times\left(\frac{1}{5}\right)\right)=600$$ $$x+ \frac{x}{5}=600$$ $$5x+ {x}=600 * 5$$ $$6x=3000$$ $$x=500$$
H: How many solutions are there to this equation involving the floor function: $(n+1)x-\lfloor nx \rfloor = c$? How many solutions are there for this equation: $(n+1)x-\lfloor nx \rfloor = c$ I can prove some basic properties of floors and ceiling, but here I'm stumped. AI: Let $y_1(x) = (n+1)x$ and $y_2(x) = c + \lfloor nx \rfloor$. You want to find $x^$ such that $y_1(x^) = y_2(x^)$. Let's look now to $y_2(x)$: $$y_2(x) = k + c ~~ \forall x \in \left[\frac{k}{n}, \frac{k+1}{n}\right), k \in \mathbb{Z}$$ So you want to solve, for each $k$, the following equation: $$(n+1)x = k + c ~~ \forall x \in \left[\frac{k}{n}, \frac{k+1}{n}\right)$$ The solution is: $$x^ = \frac{k+c}{n+1}$$ This solution is feaseble if $x^*$ is in the set $\left[\frac{k}{n}, \frac{k+1}{n}\right)$. Then: $$\frac{k+c}{n+1} \geq \frac{k}{n} \wedge \frac{k+c}{n+1} < \frac{k+1}{n}$$ Assuming that $n \in \mathbb{N}$, then: $$n(k+c) \geq k(n+1) \wedge n(k+c) < (k+1)(n+1)$$ $$k \leq nc \wedge k > nc-n-1$$ So you have solutions for each $k$ integer in the set $\left(nc-n-1, nc\right]$. The length of this set is $n+1$, so you have $n +1$ solutions.
H: Regarding continuity We know that $\tau=\{G\subset \mathbb{N}: n\in G, m\mid n\Rightarrow m\in G\}$ is a topology on $\mathbb{N}$. I have shown that if $m\mid n\Rightarrow f(m)\mid f(n)$, then $f:(\mathbb{N},\tau)\to (\mathbb{N},\tau)$ is continuos. But I couldn't show the converse. Let $f:(\mathbb{N},\tau)\to (\mathbb{N},\tau)$ be continuous and let $m\mid n$. How to show that $f(m)\mid f(n)$? Please give a hint. AI: Define, for all $n \in {\mathbb N}$, $U_n = \{ m \in {\mathbb N} : m \| n\}$. First show that $U_n$ is open. Then, given a continuous $f \colon {\mathbb N} \to {\mathbb N}$ and $m \| n$, look at $f^{-1}(U_{f(n)})$, which is open because $f$ is continuous. Expand what this means and use $m \| n$ to conclude that $f(m) \| f(n)$.
H: Proof of the derivative of $x^n$ Can someone please explain why $(x^n)'=n\cdot x^{n-1}$? Sorry for not writing it in math characters, I'm new here. AI: Note that if $x>0$, then $\ln(x^n)=n\ln(x)$. Taking the derivative of each side, using $(\ln(t))'=\dfrac{1}{t}$, and the chain rule on the left side, $$\dfrac{1}{x^n}\cdot(x^n)' = n\cdot\dfrac{1}{x}.$$ Multiplying by $x^n$ on both sides yields $(x^n)'=nx^{n-1}$.
H: 82% of an event in a persons life, what is the daily chance? If there is an 82% chance that within your average lifetime, lets assume 70 years (25567 days), that "E" event will happen: what is the percent chance that it will happen on any given day? I'm assuming that it'll be something like .02347% or something small. My problem is I've only taken up through Advanced Algebra and that was six years ago. I would have searched for this on google to solve it myself, but I'm not even sure what to call this beyond "math I don't know". So if you could help me identify the solution, how to solve it, and what this is (i.e. I imagine terms like 'graphing polonomials' 'deductive statistics' or some such name that would give me an idea of what exactly it is that I should be learning to handle similar questions on my own). Thanks for the help in advance. AI: If the probability that it happens in a specific day is $p$, then the probability that it does not happen is $(1-p)$ (complimentary events). Thus the probability that it will not happen in $25567$ days is $(1-p)^{25567}$ (product rule, independent events). Therefore the probability that it will happen in a lifetime is $1 - (1 - p)^{25567}$ (again, complimentary events), and that is your $82\%$. So, we have that \begin{align} 0.82 &= 1 - (1-p)^{25567} \\ 0.18 &= (1-p)^{25567}\\ \sqrt[25567]{0.18} &= 1 - p\\ 0.999933 &= 1-p\\ 0.000067 &= p \end{align} so the probability of it happening in a given day (assuming it doesn't change with day of the week, time of the year, age, and so on) is $0.0067\%$. This is a bit backwards from the standard textbook example (it would give you the daily probability and ask you to calculate the lifespan probability), but it's more or less elementary probability, as they should teach in any high school. The two main concepts used in setting up the first equation was complimentary probability and the product rule. Once the equation was set up, however, we basically forget all about the probability, and focus on the algebra of isolating and calculating $p$. When we have that, we go back to the probability interpretation, and say what it means.
H: Generalization of Fatou's lemma for nonpositive but bounded measurable functions. Let $(f_n)^{\infty}_{n=1}$ be a sequence of measurable (not-necessarily $\ge 0$). Let $g \gt 0$ be a measurable function with $\int g d\mu < \infty$ (integrable) such that $f_n\ge -g$ a.e. relative to $\mu$ in $E\in S$ ($S$ $\sigma$-algebra). I want to show that $$\int_E \liminf f_n d\mu \le \liminf \int_E f_n d\mu.$$ My attempt: Let $N_n=\{x\in E; f_n(x) < g(x)\}$. Then $\mu(N_n)=0$ for all $n \in \mathbb{N}$ by hypothesis. Let $N= \cup N_n$, it follows that $N$ is also $\mu$-null. With $N$ as above, one can see that $$(f_n+g)\chi_{E \setminus N}\ge 0,$$ everywhere, so applying Fatou's Lemma one gets: $$ \int_{E\setminus N} \liminf (f_n+g) d\mu \le \liminf \int_{E\setminus N} (f_n +g) d\mu.$$ Since $\liminf g =g, \liminf \int g= \int g,$ and $\int g < \infty$, if I could "open" both integrals as the sum of the integrals of each function, I could cancel out the terms involving $g$, yet I can't seem to make it work. I have a semi-linearity theorem for positive measurable functions and a linearity theorem for integrable functions, but the $f_n$'s are neither positive nor integrable. I was told that I should prove a linearity theorem concerning non positive measurable functions whose integrals can be extended real numbers, i.e., we might define $$\int f = \int f^+ -\int f^-,$$ given that said difference is well-defined (we do not get $\infty - \infty$). I can't seem so see the light in the proof of said linearity. Any insight on the issue would be greatly appreciated. AI: You can apply Fatou's lemma to $h_n = f_n + g \geq 0$. Since $\liminf h_n = \liminf f_n + g$, it yields, $$ \int (\liminf f_n + g)\,d\mu \leq \liminf \int (f_n + g)\,d\mu $$ Conclude using the condition $\int g\,d\mu < \infty$
H: Find the values of $a,b$ so that the given limit equals $2$ $$\lim_{x\rightarrow 0}{(a\sin^2x)(b\log\cos x)\over x^4}={1\over 2}$$ My initial thought was to apply L'Hopital Rule so that at some stage a condition appears where i will have to set some value for $a,b$ to get the limit. This doesn't work though as the numerator keeps expanding. Need some hint on this AI: HINT: $$\lim_{x\to0}\frac{\sin x}x=1$$ and $$\frac{\ln(\cos x)}{x^2}=\frac{\ln(\cos^2x)}{2x^2}=\frac{\ln(1-\sin^2x)}{-\sin^2x}\left(\frac{\sin x}x\right)^2\left(-\frac12\right)$$ We know $$\lim_{h\to0}\frac{\ln(1+h)}h=1$$
H: Some questions about mathematics This question is a soft one. Well, So far I have noticed stuff that is nice in math, particularly in algebra, topology and analysis. For instance, in algebra, there is theorem that says that we can think of groups just as some set of permutations. So, in other words, can we say all groups are just permutations? Also in topology we classify surfaces. In fact, we have that every compact connected surface is either a sphere, an n-torus of $n-$ projective planes. Is there any similarity in measure theory? It seems like math is just like comparing things. Is this true? Also, one last question, What is the big picture that everyone talks about? thanks AI: I was beginning to think that this question was getting a little too broad, then I got to the end and you pretty much pulled in all of mathematics! All groups are permutation groups, in the following sense: Any group $G$ acts on itself by left multiplication. In other words, for any $g\in G$, the map $h\mapsto gh$ is a permutation of $S=\|G\|$, the underlying set of $G$. This gives an injective homomorphism $G\to Sym(S)$. For this reason, we can think of any group as a permutation group (though this is not necessarily a useful way to think about all groups). There are many classification theorems in measure theory (and in all areas of mathematics). One simple one is the correspondence between Borel measures on $\mathbb{R}$ and increasing, right-continuous functions $\mathbb{R}\to\mathbb{R}$. Another is the uniqueness of Haar measure. A lot of math is comparing things, yes. But usually not as an end in itself. Mathematics has many diverse goals. My personal approach to mathematics is to seek out language that broadens the scope of mathematical discussion. One example of what I'm talking about is Emmy Noether's "noetherian hypothesis" that transformed the study of polynomial rings into modern ring theory, by observing that a single abstract property could replace lots of inelegant, technical computation, while simultaneously bringing many different kinds of objects under the same umbrella. Anybody who claims that they see the big picture is compensating for insecurity about their mathematical ability. When you have learned a thousand or ten thousand amazing things in mathematics, the picture starts to feel quite big, but it is never finished. First you learn to count, then you learn to do arithmetic, then you learn that the integers are a set, then you learn that they are a ring, then you learn that they are a principal ideal domain, then you learn that they are the regular functions on an affine scheme whose closed points are the primes, then you learn that their unique factorization is related to the triviality of that scheme's line bundles, then you learn that their algebraic extensions are related to the scheme's covering spaces, and then you learn that our inability to prove the Riemann hypothesis (and claim its $1 million prize) is related to our inability to define an "absolute point" that joins the primes together into a curve. And that is just a single mathematical object! It goes on and on like that in all directions.
H: What factor has to be applied to $\phi(ab)\propto\phi(a)\phi(b)$ for non-coprime $a,b$? For $a,b$ coprime, it is known that $\phi(ab)=\phi(a)\phi(b)$. But is there a connection between $\phi(ab)$ and $\phi(a),\phi(b)$ if they are not coprime? AI: Note that Euler's product formula $$\phi(n) = n\prod_{p\mid n}\left(1-\frac1p\right)$$ implies $$\begin{align} \phi(a\cdot b) &= ab\prod_{p\mid ab}\left(1-\frac1p\right) \\ &= ab\frac{\prod_{p\mid a}\left(1-\frac1p\right)\prod_{p\mid b}\left(1-\frac1p\right)}{\prod_{p\mid\gcd(a,b)}\left(1-\frac1p\right)} \cdot\frac{\gcd(a,b)}{\gcd(a,b)} \\ &= \phi(a)\phi(b)\frac{\gcd(a,b)}{\phi(\gcd(a,b))} \end{align}$$ Or to obtain a more symmetrical expression: $$\phi(ab)\phi(\gcd(a,b)) = \phi(a)\phi(b)\gcd(a,b)$$ or even more symmetrical (courtesy of lhf): $$\frac{\phi(ab)}{ab}\cdot\frac{\phi(d)}{d} = \frac{\phi(a)}a\cdot\frac{\phi(b)}b\quad\text{where}\ d = \gcd(a,b)$$
H: closure and interior of subsets of a metric space Suppose $X$ is a metric space and $S$ and $A$ are subsets of $X$. If $S \subset A \subset Cl(S)$ , then $Cl(A) = Cl(S)$. Also if $Int(S) \subset A \subset S$, then $Int(A) =Int(S)$. What if, $Int(S) \subset A \subset Cl(S)$ , then would $Cl(A) = Cl(S)$ and $Int(A) =Int(S)$ ? AI: What if $X=\Bbb R$, $S=(0,1)\cup\{2\}$, and $A=(0,1)$? Or $S=(0,1)\cup(1,2)$ and $A=(0,2)$? If $S$ is nice enough so that $\operatorname{cl}\operatorname{int}S=\operatorname{cl}S$, you can get one of the equalities that you want, and if $\operatorname{int}\operatorname{cl}S=\operatorname{int}S$, you can get the other; see if you can see which is which and then prove the results.
H: partial derivatives continuous $\implies$ differentiability in Euclidean space I am given this theorem: If $f \in C^1(A,\mathbb R^m)$, i.e. every partial derivative of $f$ is continuous on $A$, and $A$ is open in $\mathbb R^n$, then $f$ is differentiable on $A$. Is the following stronger assertion also true? If every partial derivative of $f:A\to\mathbb R^m$ is continuous at $c\in A$, and $A$ is open in $\mathbb R^n$, then $f$ is differentiable at $c$. Also, is the requirement that $A$ be an open set really necessary in the above statements? EDIT: After doing a little more research, here is the strongest version of the theorem I've come up with (there is an even stronger version, but it isn't quite as neat or succinct): If every partial derivative of $f:A\subset\mathbb R^n\to\mathbb R^m$ exists, and is continuous, at $c\in $int$(A)$, then $f$ is differentiable at $c$. AI: $A$ needn't be an open set in the "stronger" assertion, but $c$ should be an interior point of $A$ for the definition of the derivative to work correctly. A general version is (see Tao's Analysis II for a proof) Let $E\subseteq \mathbf R^n$, $f:E\to\mathbf R^m$, $F\subseteq E$ and $x_0$ an interior point of $F$. If all partial derivatives of $f$ exist on $F$ and are continuous at $x_0$, then $f$ is differentiable at $x_0$ with $\forall \mathbf v=(v_1,\ldots,v_n)\in\mathbf R^n$ $$f'(x_0)\mathbf v=\sum\limits_{i=1}^n\frac{\partial f}{\partial x_i}(x_0)v_i$$
H: Vanishing of a multivariable polynomial on a lattice Let be $p(x_1,...,x_n) \in K[x_1,...,x_n]$ be a polynomial of degree $d$. Suppose there is a $n$-dimensional hyperbox $B = I \times \stackrel{n}{...} \times I = I^n$. Divide $I$ to $d$ segements by $d+1$ points. This creates a lattice of $(d+1)^n$ point on $B$. Suppose that $p$ vanishes on the lattice (that is, for any point $(v_1,...,v_n)$ in the lattice, $p(v_1,...,v_n)=0$). Then we want to show that $p \equiv 0$ is the zero polynomial. Is this true for $K = \mathbb{R}$? Is it true when $K$ is algebraically closed field (e.g. $K = \mathbb{C}$)? If it is, is there a reference for that proposition? AI: I think you can do this inductively, as follows. First, look at a 1-dimensional edge $E$ of your box: this contains $d+1$ lattice points, and $p$ is zero at each of them, so $p$ must be zero along the edge. Now look at a 2-dimensional face $F$ containing $E$; it has $d+1$ lines $E_i$ parallel to $E$ (including $E$ itself), and the same argument shows that $p$ must be zero along each of these. But now any line $L$ joining $E \, (=E_1)$ to the opposite edge $E_{d+1}$ intersects all the $E_i$, so $p$ has at least $d+1$ zeros along $L$, hence must be identically zero along $L$. Since this is true for any such $L$, in fact $p$ must be zero along the whole face $F$. Now keep going in the same way.
H: Members of equivalence classes in $Z(p^\infty)$ Let $p$ be a prime and let $Z(p^\infty)$ be the following subset of the group $\mathbb{Q}/\mathbb{Z}$: $$Z(p^\infty)=\{\overline{a/b}\in\mathbb{Q}/\mathbb{Z}\mid a,b \in \mathbb{Z} \text{ and } b=p^i \text{ for some } i\geq0\}.$$ I am not sure if this is worth my time trying to show but I was wondering how I can be sure that if I pick some $\frac{c}{d} \in \overline{\frac{a}{p^i}}$ that $\frac{c}{d}=\frac{m}{p^k}$ for $k\geq 0$. Pick $\overline{\frac{a}{p^i}} \in Z(p^\infty)$ and assume that $\frac{a}{p^i}$ is in reduced form. Let $\frac{c}{d} \in \overline{\frac{a}{p^i}}$ with reduced form. I want $d=p^k$ for some $k\geq0$. If $i=0$, it is easy. Suppose $i>0$. As $\frac{c}{d} \in \overline{\frac{a}{p^i}}$, $$\frac{c}{d} \sim \frac{a}{p^i} \Leftrightarrow \frac{a}{p^i}-\frac{c}{d} \in \mathbb{Z} \Rightarrow p^i \mid (ad-cp^i).$$ $$\frac{a}{p^i}-\frac{c}{d}=\frac{ad-cp^i}{dp^i}$$ and so $p^i\mid (ad-cp^i)$ which means $\mathbb{Z} \ni \frac{ad-cp^i}{p^i}=\frac{ad}{p^i}-c \Rightarrow p^i \mid ad$ but as $\frac{a}{p^i}$ is in reduced form, $\gcd(a,p^i)=1$ and so $p^i \mid d$. At this point, I tried to do something with $\mathrm{lcm}(c,d)=cd$ and show $d \mid p^i$ but I'm not having any success there. Any advice at this point or alternate solutions? Thanks very much. AI: I suggest a different approch for this last part of the solution. You have proved that $p^i \mid d$. Since $\frac{ad-cp^i}{p^id} \in \mathbb Z$ you also have that $d \mid ad - cp^i$ and so $d \mid cp^i$. By hypothesis $\gcd(d,c)=1$ (you have taken $\frac{c}{d}$ in reduced form) this implies that $d \mid p^i$. From what just proved we have to conclude that $d=p^i$.
H: Binomial Type sequence If $p_0(n),p_1(n)\ldots$ is a sequence of polynomials satisfying $$\sum_{k \geq 0}p_k(n) \frac {x^k}{k!}= \left ( \sum_{ k \geq 0}p_k(1) \frac {x^k}{k!} \right )^n$$ then $$p_k(m+n)= \sum_{i=0}^{k} {k\choose i}p_i(m)p_{k-i}(n), \; k \geq 0$$ Any help would be appreciated. AI: Hint : Try to write $$\sum_{k\geq 0}p_{k}(n+m)\frac{x^{k}}{k!}$$ as the product of 2 exponential generating functions and use the Cauchy product (binomial convolution) of 2 exponential generating functions.
H: Compound quadratic problem The first issue I have is that I am not sure why this is called a 'compound quadratic problem', but anyway to proceed: Suppose that $x-y=14$ and $$(x+y)(x^2+y^2)(x^4+y^4)=a(x^b-y^b)$$ where $a$ and $b$ are constants. Find $a$ and $b$. I am stuck as to how to approach this problem. I have tried factoring (which becomes imaginary as it is the sum of squares, at which point I have no idea what to do), and expanding both sides of the equation. I had an idea that using the difference of two squares on the right hand side might get me somewhere but there is still always a term with a negative coefficient (to demonstrate what I mean): $$(x^b-y^b)=(x^\frac{b}{2}+y^\frac{b}{2})(x^\frac{b}{2}-y^\frac{b}{2})=(x^\frac{b}{2}+y^\frac{b}{2})(x^\frac{b}{4}+y^\frac{b}{4})(x^\frac{b}{4}-y^\frac{b}{4})$$ Any help appreciated. AI: You can use this also to understand: $$x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$$ Daniel Fischer's method is brilliant of course.
H: Understanding cohomology with compact support I am trying to understand the definition of (singular) cohomology with compact supports. My understanding of singular cohomology goes like this. Let $X$ be a topological space. Define the singular chain group $C_n(X)$ to be the free abelian group generated by singular n-simplicies, which are functions $\sigma : \Delta^n \to X.$ The homology of the complex $C_{\bullet}(X)$ is singular homology. If we define $C^n(X):= \operatorname{Hom}(C_n(X), \mathbb{Z})$ to be the dual of the singular cochain group we get singular co-chains. If we now take the cohomology of this new complex, we get singular cohomology. Now I am trying to understand cohomology with compact supports, which this source begins to define (on page 7 of the pdf) as such: Given a (singular, simplicial, cellular) cochain complex $C^{\bullet}$ on a space $X$ , consider the subcomplex $C^{\bullet}_c$ of cochains which are compactly supported: each cochain is zero outside some compact subset of $X$ . What does this last statement mean? Isn't a cochain a map from $C_n(X)$ to $\mathbb{Z}$? How can take have a support inside $X?$ Please help me clear my deep misunderstanding. Thank you. AI: The idea is that a cochain $\varphi \in C^n(X)$ is compactly supported if there's a $K \subseteq X$ compact subset of $X$ such that $\varphi\|_{C_n(X \setminus K)} = 0$. Edit a little remark: for every $K$ compact subset of $X$ there's an embedding $i \colon X \setminus K \hookrightarrow X$ which give rise to a injective embedding of chain complexes $i_* \colon C_\bullet(X \setminus K) \to C_\bullet (X)$, so we can think of $C_n(X \setminus K)$ as being a submodule of $C_n(X)$ and to be exact what I meant above by $\varphi\|_{C_(X \setminus K)}$ should be written more formally as $\varphi\|_{i_*(C_n(X \setminus K))}$. So compactly supported co-chain of $X$ are those co-chains in $C^\bullet(X)$ that vanish on all the simplexes that have image contained in a subspace $X \setminus K$ (for some $K$ compact subset of $X$), i.e. those simplexes $\sigma \colon \Delta^n \to X$ that factors through the inclusion map $i \colon X \setminus K \to X$. You can find out more about this in Hatcher's book Algebraic Topology.
H: Fourier series of a square wave signal with a bias Given a $f(t)$ of the kind: $$f(t)=1, \{kt_0\le t\le kt_0+\tau\}$$ $$f(t)=a,\{kt_0+\tau\le t\le (k+1)t_0\}$$ with $a\lt 1$ what is the Fouries series development of f(t)? Thanks AI: $f$ is the sum of a constant and a multiple of a characteristic function of an interval, $$f(t) = a + (1-a)\cdot \chi_{[0,\tau]}(t).$$ So the Fourier series of $f$ is the sum of the Fourier series of the two parts. The Fourier series of the constant is trivial if the basis contains a constant function, as is the case for the most common bases $\{e^{2\pi int/t_0}\}_{n\in\mathbb{Z}}$ and $\{\cos (2\pi nt/t_0)\}_{n\in\mathbb{N}} \cup \{\sin (2\pi nt/t_0)\}_{n\in\mathbb{Z}^+}$, both systems scaled to obtain an orthonormal basis. So it remains to find the coefficients of the Fourier series of the characteristic function. For the exponential basis, $$\begin{align} \int_0^{t_0} \chi_{[0,\tau]}(t) e^{-2\pi int/t_0}\,dt &= \int_0^\tau e^{-2\pi int/t_0}\,dt\\ &= \begin{cases} \qquad \tau &, n = 0\\ \frac{t_0}{2\pi in}\left(1-e^{-2\pi in\tau/t_0}\right) &, n \neq 0. \end{cases} \end{align}$$ For the trigonometric basis, we get $$\begin{align} \int_0^{t_0} \chi_{[0,\tau]}(t)\sin (2\pi nt/t_0)\,dt &= \int_0^\tau \sin(2\pi nt/t_0)\,dt\\ &= \frac{t_0}{2\pi n}\left(1 - \cos (2\pi n\tau/t_0)\right), \end{align}$$ and $$\begin{align} \int_0^{t_0} \chi_{[0,\tau]}(t)\cos (2\pi nt/t_0)\,dt &= \int_0^\tau \cos(2\pi nt/t_0)\,dt\\ &= \begin{cases} \qquad\tau &, n = 0\\ \frac{t_0}{2\pi n}\sin (2\pi n\tau/t_0) &, n \neq 0. \end{cases} \end{align}$$
H: Centre of a simple algebra is a field How can one show that the centre of simple algebra is a field? I have tried it and proved that the inverse exists for every element of centre but cannot prove that inverse of every element belongs to centre. Please help me. AI: If $ab=ba$ and $a^{-1}$ exists, then $a^{-1}b=a^{-1}(ba)a^{-1}=a^{-1}(ab)a^{-1}=ba^{-1}$.
H: Countability over indexed families I'm strugling with countability over indexed sets compared to ordinary sets. Basically we say that any set $A$ is countable if there is a bijective function $f$ such that $f:\mathbb{N}\longrightarrow A$. Here my problem is that I don't know if there is a similar standard definition when we talk about families of the form$\{a_i\mid i\in I\}$. Essentially what I'm looking for is something like: "A family of sets is countable if and only if..." Intuitively I'd say that a family of sets $\{a_i\mid i\in I\}$ is countable if there is bijection between $I$ and $\mathbb{N}$, but here I don't know if this is a consequence of the definition or this can be admited as such. Any comments are appreciated. Thanks. AI: The answer depends on one's definition of "family of sets". There are basically two possibilities: A family of sets $(a_i)_{i \in I}$ is a mapping $f:I \to A$; A family of sets $\{a_i\}_{i \in I}$ is the image of a mapping $f: I \to A$. (Remark: Both notations (round vs. curly braces) are seen for both concepts; I distinguish them notationally for clarity.) The difference is in whether we consider $a_i$ to uniquely determine $i$, when a set occurs multiple times in the family. In the first case, it's easy: set-theoretically, we have $f = \{(i, a_i): i \in I\}$, and it's clearly bijective with $I$ (projection to first coordinate). So in this case, $(a_i)_{i \in I}$ is countable precisely when $I$ is. In the second case, it's more difficult. Countability of $I$ means that $\{a_i\}_{i \in I}$ is finite or countable. But even if $I$ is uncountable, it may be that $\{a_i\}_{i \in I}$ is countable. So in either case, because of our definition of "family of sets" as a set, we can apply the definition of countability for sets. But in only one of these cases this can be translated to a definition in terms of $I$.
H: Relation between quotients and subalgebras If I have two algebras $A,B$, and one is the quotient of the other, i.e. there exists a surjective morphism $\phi : A \to B$. Then is $B$ isomorphic to some subalgebra of $A$? I think so, because I just need to select for each equivalence class $\phi^{-1}(b)$ one element $a \in \phi^{-1}(b)$, and furthermore if I had selected $a_1 \in \phi^{-1}(b_1), a_2 \in \phi^{-1}(b_2)$, and $a_3 \in \phi^{-1}(b_1 \cdot b_2)$, then $a_3 = a_1 \cdot a_2$ must hold. Which could be realized I think. On the other hand, if I have a subalgebra $A'$ of $A$, could it always be realisied as a quotient, i.e. is there some surjective $\psi : A \to A'$. I think this should be possible. But then if quotients and subalgebras are so related, then they are in some sense the same concepts, but as I see it they are threated quite differently in textbooks and theorems, so they are not the same, which make me doubt my constructions. Could someone please clarify? AI: The answer is NO of course. Take for example any algebra $A$ with has no 1-element subalgera and take their 1-element quotient $B$.
H: How to solve matrix equation $AXH+AHX−BH=0$ How to solve matrix equation $AXH+AHX−BH=0$? All matrices are square, $A$, $B$ known constant matrices and invertible, $H$ can take any value, $X$ represent the solution to be found. I have seen about the Sylvester Equation like in this post Solving a matrix equation $AX=XB$ in a CAS, but I'm not sure how to apply it because of the presence of matrix H. AI: The equation is still linear in $X$, that means that if you consider $X$ a vector, call it $\Xi$ (made up of all the entries of the matrix $X$ in a definite order) you can write it in the form $$ Q\Xi = W $$ where $W$ is the vector likewise made of the entries of the matrix $BH$. Now the problem reduces to finding $Q$ and its inverse (you should check if $Q$ is invertible, it is not obvious to me that it will be). The matrix $Q$ represents the action of $A$ and $H$ over $X$ sometimes acting on the left and sometimes on the right. I you call the right-acting operator $^R$ and the left one $^L$ then in fact you have $Q = A^L \otimes H^R + (AH)^L \otimes {\bf 1}^R$ and you can write the matrix as a tensor product, and then invert it. To simplify matters you could also start multiplying your equation on the left by $A^{-1}$, but the method to solve remains the same. Your answer is then $\Xi = Q^{-1} W$ which you can map back to $X$ in matrix form.
H: Automorphisms of $\mathbb{G}_m$. Let $\mathbb{G}_m$ be the multiplication group whose underlying set is $k^$, where $k$ is a field. How to show that as an algebraic group there are only two automorphisms of $\mathbb{G}_m$? How many automorphisms are there of $\mathbb{G}_m$? I think that if $\phi: \mathbb{G}_m \to \mathbb{G}_m$ is an automorphism, then the induced map $\phi^: k[t, t^{-1}] \to k[t, t^{-1}]$ is also an automorphism. Thank you very much. AI: The character lattice of $\mathbb G_m$ is the ring of endomorphisms of $\mathbb G_m$, and it isomorphic to $\mathbb Z$; the element $n \in \mathbb Z$ corresponds to the character $t \mapsto t^n$. The only units in $\mathbb Z$ are $\pm 1$, corresponding to the identity, and to the involution $t \mapsto t^{-1}$, so these are the only automorphisms of $\mathbb G_m$. Note that in terms of rings, the induced map $\phi^*: k[t,t^{-1}] \to k[t,t^{-1}]$ has not only to be an automorphism of $k$-algebras, but it has to respect the Hopf algebra structure (so that it corresponds to an automorphism of $\mathbb G_m$ as an algebraic group, and not just as a variety).
H: Uniform continuity? Suppose that $f$ is continuous on the interval $I$, and $\|f(x)\|$ is uniformly continuous. Can we have the conclusion that $\sin^3f(x)$ is uniform continuous? My intuition is that it is not. But I could not construct a counterexample. Would you help me out? Thank you. AI: We have the conclusion - assuming real values, and not complex ones. The composition of uniformly continuous functions is uniformly continuous - for $f\circ g$ and an arbitrary $\epsilon > 0$, there is a $\delta_1 > 0$ with $\lvert x-y\rvert \leqslant \delta_1 \Rightarrow \lvert f(x) - f(x)\rvert \leqslant \epsilon$ by the uniform continuity of $f$, and by the uniform continuity of $g$, there is a $\delta_2 > 0$ with $\lvert \xi-\eta\rvert \leqslant \delta_2 \Rightarrow \lvert g(\xi) - g(\eta)\rvert \leqslant \delta_1$; so for $\lvert\xi-\eta\rvert \leqslant \delta_2$ we have $\lvert f(g(\xi)) - f(g(\eta))\rvert \leqslant \epsilon$ then. And for continuous (real-valued) $f$ (on an interval), the uniform continuity of $\lvert f\rvert$ implies uniform continuity of $f$. Let $\epsilon > 0$ be given. By the uniform continuity of $\lvert f\rvert$, there is a $\delta > 0$ such that $$\lvert x - y\rvert \leqslant \delta \Rightarrow \bigl\lvert \lvert f(x)\rvert - \lvert f(y)\rvert \bigr\rvert \leqslant \epsilon/2.$$ Then we have $\lvert x-y\rvert \leqslant\delta \Rightarrow \lvert f(x)-f(y)\rvert \leqslant \epsilon$. For if $f(x)$ and $f(y)$ have different sign - and for real valued functions, that is the only way to have close absolute modulus without being close - then by the intermediate value theorem, there must be a $z$ between $x$ and $y$ with $f(z) = 0$. But if $z$ is between $x$ and $y$, then $\lvert z - x\rvert \leqslant \lvert x-y\rvert \leqslant \delta \Rightarrow \lvert f(x)\rvert \leqslant \epsilon/2$; similarly, $\lvert f(y)\rvert \leqslant \epsilon/2$. So If $\lvert x -y \rvert \leqslant \delta$, either the two values $f(x)$ and $f(y)$ have the same sign (or at least one of them is $0$), in which case $\epsilon/2 \geqslant \bigl\lvert \lvert f(x)\rvert - \lvert f(y)\rvert\bigr\rvert = \lvert f(x)-f(y)\rvert$, or we have $\lvert f(x) - f(y)\rvert \leqslant \lvert f(x)\rvert + \lvert f(y)\rvert \leqslant \epsilon/2 + \epsilon/2 = \epsilon$. Since $\varphi \mapsto \sin^3 \varphi$ is uniformly continuous (bounded derivative) on $\mathbb{R}$, the uniform continuity of $\sin^3 f(x)$ follows.
H: Is there a continuous function on R such that $f(f(x))=e^{-x}$? Is there a continuous function on R such that $f(f(x))=e^{-x}$? I have tried to take derivative of the two sides,but I can't get anything I want.what can I do? AI: No. Hint: An injective continuous function is monotonic and for any monotonic $f(x)$ the function $f(f(x))$ should be increasing.
H: Represent $\frac{1}{3} \ln(x+2)^3 + \frac{1}{2}[\ln x - \ln (x^2+3x+2)^2]$ as a single logarithm I am having some trouble trying to find the single logarithm for the following: $$\frac{1}{3} \ln(x+2)^3 + \frac{1}{2}[\ln x - \ln (x^2+3x+2)^2]$$ I understand that I have to use the addition and subtraction rules but the coefficients are confusing me. These are some of the steps I took: $$\begin{align} \frac{1}{3} \ln(x+2)^3 + \frac{1}{2} [\ln \frac{x}{(x^2+3x+2)^2}] \end{align}$$ Now from here, if I used the addition rule then what would happen to the coefficients? and also, is there a way for me to simplify the powers $2$ and $3$? Hints would be appreciated! Thanks! AI: $$\frac{1}{3} \ln(x+2)^3 + \frac{1}{2}[\ln x - \ln (x^2+3x+2)^2]$$ $$=\ln(x+2)+\frac{1}{2} \ln(x)-\ln(x^2+3x+2)$$ $$=\ln\frac{(x+2)(\sqrt x)}{x^2+3x+2}$$ $$=\ln \frac{\sqrt{x}}{x+1}$$ Properties of $\ln(x)$ that have been used : $$a \ln(b)= \ln (b^a), ln(a)+ln(b)=\ln(ab)$$
H: Ackermann function and primitive recursiveness If we define $b_n(m) := a(n,m)$ for all $n$ and $m \in \mathbb{N}$. For which $n$ is the function $b_n$ primitive recursive and for which $n$ it is not a primitive recursive function? Can anyone please help me out with this? AI: I assume that (as the title suggests) $$ a(n,m) = \begin{cases} m +1 & n= 0\\ a(n-1, 1) & n > 0, m=0\\ a\bigl(n-1, a(n,m-1)\bigr) \end{cases} $$ denotes the ackermann function. Rewriting this in terms of $b_n$, we have $b_0 = \cdot + 1$ is the successor function and $$ b_n(m) = \begin{cases} b_{n-1}(1) & m = 0 \\ b_{n-1}\bigl(b_n(m-1)\bigr)\end{cases} $$ If $b_{n-1}$ is primitive recursive, this defines a primitive recursive function $b_n$. As $\cdot + 1$ is primitive recursive, by induction, all $b_n$ are.
H: A chess problem in Kanamori's "The Higher Infinite"? Just after Corollary 21.17 (on p289) of Kanamori's The Higher Infinite, he outlines the direction in which he wants to take his discussion of iterated ultrapowers. However, immediately after he presents a chess problem. The passage reads as follows: Having developed the analogue of the $0^\sharp$ theory in §9, we now proceed to derive more information with iterated ultrapowers that sharpens the focus. But first, a respite from the rigors: Instead of yet another recipe, we offer the following chess problem (M. Henneberger, first and second prize, “Revista de Sah” 1928): White: King on b1, Rooks on b7 and c7, and Bishop on b5. Black: King on a8, Rook on a3, and Pawn on f2. White to play and win. Send complete solutions to the author for a small prize. My question is simply the following; is this just a chess problem, or is there some joke about large cardinals that I am missing? AI: There's a review of the book here, and it mentions the problem, and gives no indication that it's at all related to the mathematics. I think it's safe to assume you're not missing a joke.
H: Volume of a tetragonal pyramid Express the volume $V$ of a regular tetragonal pyramid as a function of its altitude $x$ and the edge of a lateral face (lateral edge) $y$ The answer given by the book is $\frac{2}{3} (y^2 - x^2) x $. But,I've found the lateral edge is $2 \sqrt{y^2 - x^2 } $ and I thought that the area of the basis is $ 4(y^2 - x^2)$. What am I doing wrong? Thanks for your help! AI: Denote by $d$ the diagonal of the base. There is a orthogonal triangle with edges $d/2$, $x$ and $y$, so by the Pythagorean theorem $$ y^2 = x^2 + \frac 14d^2 $$ that is $$ d^2 = 4(y^2 - x^2) \iff d = 2\sqrt{y^2 - x^2}. $$ Now the sidelength $a$ of the base is given by $a = d/\sqrt 2$ (again by Pythagoras), so $$ a = \frac d{\sqrt 2} = \sqrt 2 \cdot \sqrt{y^2 - x^2} $$ So the area $A$ of the base is $$ A = a^2 = 2 \cdot (y^2 - x^2) $$ and hence the volume $$ V = \frac 13 A \cdot x = \frac 23 \cdot (y^2 - x^2). $$
H: Combinatorial Algebra with Variables The problem is ${m+1} \choose {m-1}$. The answer is $\frac{m(m+1)}{2}$. I am stuck on solving this algebraically. If someone could tell me where the two comes from I would be helped, because I know how to get to m(m+1) just not two as the divisor. AI: Here you go: According to the definition of $n \choose m$, it $\frac{n!}{m!(n-m)!}$ So let's take your "choose" and apply the formula: ${m+1 \choose m-1} = \frac{(m+1)!}{(m-1)!2!}$ = $\frac{(m+1)(m)(m-1)!}{2(m-1)!}$ = $\frac{m(m+1)}{2}$ Here it is...
H: Solve $4^{9x-4} = 3^{9x-4}$ I am having some trouble trying to solve $$4^{9x-4} = 3^{9x-4}$$ I tried to make each the same base but then I'm becoming confused as to what to do next. These are the steps I took: $$\begin{align} 4^{9x-4} &= 3^{9x-4} \\ \log_4(4^{9x-4}) &= \log_4(3^{9x-4}) \\ \end{align}$$ Where do I go from there? Thanks! AI: Magic property of logarithms: $$\log_b(x^y) = y\log_b(x)$$ So here, $$\log_4(4^{9x - 4}) = (9x-4)\log_4(4)$$ and $$\log_4(3^{9x-4}) = (9x-4)\log_4(3)$$ so that $$(9x-4)\log_4(4) = (9x-4)\log_4(3).$$ Now can you solve it?
H: Help with proof of injection and surjection For the record, I am sorry, I haven't yet learnt how to use LaTeX I have a function $f(x) = 2x^3 - 1$ My proof of injection is as follows: $f$ is one to one for all $x_1,x_2$ element $X$, if $f(x_1) = f(x_2)$ then $x_1 = x_2$ Proof $f(x_1) = f(x_2)\\ 2x_1^3 - 1 = 2x_2^3 - 1\\ 2x_1^3 = 2x_2^3\\ x_1^3 = x_2^3$ Therefore $x_1 = x_2$ so $f(x)$ is one to one by direct proof - contraposition of 'if $x_1\neq x_2$, then $f(x_1)\neq f(x_2)$. I am unsure how to approach the problem of surjection. I understand the concept, and I can show that it has a domain and a range which is an element of the real numbers, so it is definitely onto, but I don't know how to prove it. AI: You need to take $y\in\Bbb R,$ and show that there is some $x\in\Bbb R$ such that $f(x)=y$. In other words, you should show that $$y=2x^3-1$$ has a real solution $x$.
H: Induced subgraphs (graph theory) I have the following graph theory question that I am stuck on: Prove or disprove: For every graph G and every integer $r \geq \text{max} \{\text{deg}v: v \in V(G) \}$ , there is an r-regular graph H containing G as an induced subgraph. Thanks for any help. AI: Can you construct an r-regular graph with an arbitrary number of vertices as long as the number of vertices exceeds a given quantity? If so, then you may construct H from G as follows: Let be the degree of the nth vertex of G. For every vertex in G with degree less than r, add a vertex to the graph and an edge connecting that vertex to and repeat this process until the degree of is r. Then you will have a graph in which the degree of the vertices originally in G is r, and the degree of the vertices not originally in G is 1. If you cannot construct an (r-1)-regular graph with the number of additional vertices (the ones not originally in G), then add two vertices and connect them with an edge. Repeat this process until you can construct an (r-1)-regular graph with the number of additional vertices. Now, for the vertices whose degree is 1, add the edges of the (r-1)-regular graph you constructed. This will complete the creation of H, and G will be an induced subgraph of H, which is seen easily by choosing the vertices originally in G, to which no edges were added in the creation of H.
H: Finding critical points of $x^{(2/3)}(5-x)$ So I tried this out and got stuck with this: $$0 = 3x^{(7/6)} + 2x - 10$$ I didn't think I could use a quadratic for this since its to the power of $7/6$ Here is the working I did: We know its a critical point when f'(a) = 0 So I found the derivative of f(x) which is $$2(5-x)/3x^{1/2} - x ^{2/3}$$ So I set this equal to 0 $$2(5-x)/3x^{1/2} - x ^{2/3} = 0$$ $$2(5-x)/3x^{1/2}=x ^{2/3}$$ $$2(5-x)=x ^{2/3}\times3x^{1/2}$$ $$10-2x=3x ^{2/3 +1/2}$$ $$10=3x ^{7/6} + 2x$$ But this would be such a messy answer, so I think I have done something wrong with my working. Do you have any ideas? AI: If $$f(x) = x^{2/3}(5-x) =5x^{2/3}-x^{5/3},$$ then $$f'(x) = \frac{10}{3}x^{-1/3} - \frac{5}{3}x^{2/3}$$ Of course, to find critical points, we need to solve for $$f'(x) = 0 \iff \frac{10}{3}x^{-1/3} = \frac{5}{3}x^{2/3}$$ Assuming if $x\neq 0$, multiply both sides of the equation by $\;\frac 35 x^{1/3}\;$ to find the solution. Note: Do check what happens at $x = 0$, where the function is defined, but not the derivative.
H: Regression analysis on temperature/sensor data Looking for a solution to what I thought should be an easy problem, but has me running in circles somehow... I'm working with two sets of data: he first set is raw values from a sensor (accelerometer), and the second set is the temperatures at which these values were read. The values from the sensor are meaningless (as in, they're in sensor ticks) until run through the formula $Value = (Ticks - OFFSET) * SCALE + Temp * CONST$ which gives me a $Value$ in useful units (in this case, $m/s^2$). The $OFFSET$ and $SCALE$ values are known; I'm trying to find a $CONST$ such that $Value = 0 m/s^2$ across all $Temp$. The story here is that the sensor output changes over temperature even if its acceleration isn't changing due to the thermal characteristics of the sensor. I had a similar problem with another sensor (granted, a different formula) and I was able to use simple linear regression to find the $CONST$, but for some reason it doesn't work for this data... One possible theory I've come up with is that the output being dependent on the temperature violates the independence requirement of linear regression, but if that is the case, what should I use instead to calculate a single $CONST$ that will result in the equation being zero (or near-zero) for all temperatures and tick values? Thanks! AI: You are assuming that OFFSET is linear over temperature and that SCALE doesn't change with temperature. If those assumptions were true, a linear regression would solve your problem. You can make more complicated functions for OFFSET and SCALE-make each quadratic in temperature, for example. Say your model is $OFFSET=A_0+A_1t+A_2t^2, SCALE=B_0+B_1t+B_2t^2$ (the $CONST$ will get absorbed in these) and feed your data to a multidimensional minimizer. Programs are available in any numerical analysis text or in data analysis packages.
H: Standard notation for the set of integers $\{0,1,...,N-1\}$? I was wondering if there exist a standard notation for the set of integers $\{0,1,...,N-1\}$. I know for example $[N]$ could stand for the set $\{1,2,...,N\}$ but what about the former, i.e. $\{0,1,...,N-1\}$? AI: It depends on the context, but sometimes $[N]$ denotes the set $\{0,1,\ldots,N-1\}$. In modern set theory the integers are represented using sets so $N$ is actually the set $\{0,1,\ldots,N-1\}$ (and $0=\varnothing$, of course). I am unaware of a particular "standard" for this set, in some places you can see $\Bbb N^{<N}$, but in other places that notation would denote all the sequences of length $<N$. And as always I give the following advice. You can introduce a notation, just be clear about it: We shall denote by $S(N)$ the set $\{0,\ldots,N-1\}$... That's a perfectly legitimate notation (if it doesn't clash with other notations using the letter $S$ of course).
H: Using a function in Matlab Very new to MATLAB and Im trying to use the FFT function. I got a video which showed me that a function and a normal m file is needed. Created that but now dont know how to call the function from the m file. Here's is the function: function [ X,freq ] = positiveFFT( x,Fs ) %This function takes data,and discards negatives only taking the %positives,getting rid of complex values %adopted from blinkdagger.com N=length(x); k=0:N-1; T=N/Fs; freq=k/T; %create the frequency range X=fft(x)/N; %normalise the data cutOff=ceil(N/2); %cut off frequency X=X(1:cutOff); freq=freq(1:cutOff); end and the m-file:fo=4; %freqency of the sine wave(4cycles,4Hz) Fs=100; %sampling rate Ts=1/Fs; %sampling time interval t=0:Ts:1-Ts; n=length(t); %number of samples y=2sin(2pifot); [YfreqD,freqRng]=positiveFFT(y,Fs); stem(freqRng,abs(YFreqD)); AI: I think you should change stem(freqRng,abs(YFreqD)); as stem(freqRng,abs(YfreqD));
H: How many ways can you distribute 3 types of candies to 8 children? I have a big bag of candy. Peppermints, Chocolates, and Caramels. There are eight sweet children who deserve candy. One each, they are not that sweet. So I give each child a candy. How many ways are there for the candies to be distributed? AI: I guess there will be a standard answer to this question somewhere and wouldn't be surprised if this is a duplicate. However, assuming you've got enough of each sort of candy (which 'big bag of' suggests), each child can have either of the three types of sweets. So, child one has 3 possible candies, child two has 3 possible candies and so on. For the first two children there are $3^2=9$ options this way (here you can still write down the options, if you like) and each successive child adds a factor $3$. Can you see why? The result then would be $3^8$ ways to distribute candies (which equals $6561$, if I am to believe google). The questoin would be a lot harder if the amount of candies of each type would be restricted.
H: Hamiltonian Graphs and connected graphs Prove or disprove: There exists an integer k such that every k-connected graph is hamiltonian. AI: Attempting to do these questions serve as good practice if you are doing a graph theory course. An idea, I think look at $M_{2k+1}$ for all $k \geq 3$.
H: Estimating the average number of passengers in cars in a parking lot. All the workers at a certain company drive to work and park in the company’s lot. The company is interested in estimating the average number of workers in a car. Which of the following methods will enable the company to estimate this quantity? Randomly choose $n$ workers, find out how many were in the cars in which they were driven, and take the average of the $n$ values. Randomly choose $n$ cars in the lot, find out how many were driven in those cars, and take the average of the $n$ values. My intuition goes for number 2, but I'm not able to justify it formally. AI: In the first case, you introduce "inflation" in the sense of over-counting: any two randomly chosen workers may have very well driven in the same car. The more passengers in a car, say k of them in car $C_i$, ($1 \leq i\leq n$) the greater the chance that two or more workers will each, separately, report "k" passengers for the same car, overcounting the number of passengers in car $C_i$. That is, two or more of the selected workers may belong to the same "set." In the second case, since the passengers in each car belong to disjoint sets, randomly selecting the cars, and then determining the number of passengers in the selected cars, ensures that the number of passengers in one car, if counted once, will not be counted again.
H: Solve using Pigeonhole principle There are 45 candidates appear in an examination. prove that there are at-least two candidates in class whose roll numbers differ by a multiple of 44. How can I prove this using pigeonhole principle? AI: Consider the possible remainders mod $44$ as the boxes. There are $44$ of these. Now there are $45$ roll numbers in total. Place each in its box corresponding to its remainder mod $44$. The pidgeonhole principle says that there will be at least two roll numbers $a,b$ such that they lie in the same box, i.e. $a\equiv b \bmod 44$. But then $44\|(a-b)$.
H: $L_2$ is of first category in $L_1$ (Rudin Excercise 2.4b) We mean here $L_2$, and $L_1$ the usual Lebesgue spaces on the unit-interval. It is excercise 2.4 from Rudin. There's several ways to show that $L_2$ is nowhere dense in $L_1$. But in (b) they ask to show that $$\Lambda_n(f)=\int fg_n \to 0 $$ where $g_n = n$ on $[0,n^{-3}]$ and 0 otherwise, holds for $L_2$ but not for all $L_1$. Apparantly this implies that $L_2$ is of the first Category, but I dont know how. Second, I can show this holds for $L_2$ but I cant find a counterexample in $L_1$. Theorem 2.7 in Rudin says: Let $\Lambda_n:X\to Y$ a sequence of continuous linear mappings ($X,Y$ topological vector spaces) If $C$ is the set of all $x\in X$ for which $\{\Lambda_n x\}$ is Cauchy in $Y$, and if $C$ is of the second Category, then $C=X$. So if we find a $f\in L_1$ such that $\Lambda_n(f)$ is not Cauchy, then we proved that $L_2\subset C \subset L_1$ is of the first category. However I dont see why showing that $\Lambda_n(f)$ does not converge to 0 for some $f\in L_1$ is enough here. Am I missing something? AI: The simplest functions in $L_1 \setminus L_2$ are $f_\alpha \colon x \mapsto x^\alpha$ with $-1 < \alpha \leqslant -\frac12$. Computing $\int fg_n$ for such an $f_\alpha$ yields $$\begin{align} \int f_\alpha g_n &= n\int_0^{n^{-3}} x^\alpha\,dx \\ &= \frac{n}{1+\alpha}n^{-3(1+\alpha)}\\ &= \frac{n^{-2-3\alpha}}{1+\alpha}. \end{align}$$ We see that the sequence of integrals does not converge to $0$ iff $-2-3\alpha \geqslant 0 \iff \alpha \leqslant -\frac23$. Regarding the second part, choosing $-1 < \alpha < -\frac23$ gives an $f\in L_1$ with $\Lambda_n(f) \to \infty$, so $\Lambda_n(f)$ certainly is not a Cauchy sequence. Choosing $\alpha = -\frac23$ gives an $f\in L_1$ such that $\Lambda_n(f)$ is constant, hence a Cauchy sequence, but does not converge to $0$. Now, if $L_2$ were of the second category in $L_1$, then the fact that $\Lambda_n(f) \to 0$ for all $f\in L_2$ would imply that $\Lambda_n(f) \to 0$ for all $f\in L_1$, by part $(b)$ of theorem 2.7. But picking $\alpha < -\frac23$ to get an $f\in L_1$ such that $\Lambda_n(f)$ is not a Cauchy sequence seems preferable, since it's more direct.
H: What is the dot product between a vector of matrices? There is a notation used in many sources (e.g. Wikipedia: http://en.wikipedia.org/wiki/Exponential_family) for the natural parameters of exponential family distributions which I do not understand, and I cannot find a description of. With vector parameters and variables, the exponential family form has the dot product between the vector natural parameter, ${\boldsymbol\eta}({\boldsymbol\theta})$ and the vector sufficient statistic, ${\mathbf{T}}({\mathbf{x}})$, in the exponent. i.e. $e^{{\boldsymbol\eta}({\boldsymbol\theta})^{\top}{\mathbf{T}}({\mathbf{x}})}$. However, many examples of these parameters for different distributions are vectors composed of matrices & vectors. E.g. the multivariate Normal distribution has parameter $[\Sigma^{-1}\mu\space\space-\frac{1}{2}\Sigma^{-1}]$ and sufficient statistic $[\mathbf{x}\space\space\mathbf{xx^{\top}}]$. So what are these "vectors" and moreover, how is the dot product between them defined? Does this notation have a name? AI: I believe you're supposed to "vectorize" the matrix, i.e. rearrange into a $n^2 \times 1$ vector. Equivalently, you can take $A\cdot B = \textrm{tr}(A^TB)$ as the definition. EDIT - Example: If $A = \left(\begin{array}{cc}a & b \\ c & d \end{array}\right)$, $B = \left(\begin{array}{cc}e & f \\ g & h \end{array}\right)$, then $$ A^TB = \left(\begin{array}{cc}ae + cg & af + ch \\ be + dg & bf + dh \end{array}\right) $$ and the trace is $ae + bf + cg + dh$. Likewise, if we first vectorize the matrices $$ \widetilde{A} = \left(\begin{array}{c}a & b & c & d \end{array}\right)^T\\ \widetilde{B} = \left(\begin{array}{c}e & f & g & h \end{array}\right)^T\\ $$ it's straightforward to see $\widetilde{A}\cdot\widetilde{B} = \textrm{tr}(A^TB)$.
H: Differentiability of the function $f(z)=\|z\|^p z$($p>0$). Suppose $p>0$ in a real number, is the function $f (z) =\|z\|^p z$ a differentiable function? Moreover, if $f (z) =\|z\|^p z$ is differentiable, dose $f$ belong to the space $C^{\left\lfloor p \right\rfloor,p-\left\lfloor p \right\rfloor}(\mathbb{C})$? P.S. $\left\lfloor p \right\rfloor$ denotes the integer part of $p$. AI: I assume it's real-differentiability; otherwise see DF's comment. When $p$ is an even integer, $f$ is a polynomial, hence infinitely differentiable. From now on I assume $p$ is not an even integer. At $z\ne 0$, the function is infinitely smooth (chain rule encounters no obstacle). Observe the homogeneity: $f(tz)=t^{p+1}f(z)$ for $t>0$, and use it to conclude that the total derivative $Df$ is homogeneous of degree $(p+1)-1 = p$. The $k$ order derivative is homogeneous of degree $p+1-k$. A function that is homogeneous of degree $d$ is: differentiable at $0$ (with zero derivative) if $d>1$ not differentiable at $0$ if $d<1$. not differentiable at $0$ if $d=1$, unless it is a linear function. (The latter won't happen for us since $f$ is not a polynomial.) All of the above are not hard to prove from the definition of derivative. Since we begin with homogeneity $p+1$, where $p>0$, the derivative can be taken $\lceil p \rceil $ times before its degree of homogeneity becomes $\le 1$ and the music stops. So, $f\in C^{\lceil p \rceil, p+1-\lceil p \rceil}$.
H: How to find support of functions $\textbf{Support}$:$f$ is real valued function with domain $E^n$ the support of $f$ is the smallest closed set $K$ such that $f(x)=0$ for all $x$ is not in $K$ Find the support $(1) f(x)=x-\|x\|$ $\displaystyle(2) f(x,y)=\frac{x}{e^{x^2+y^2}} $ $(3) f(x,y)=1$ if either x or y is a rational number,$f(x,y)=0$ if both are irrationals. $(4)f(x,y)=(x-y)\|x+y\|-(x+y)\|x-y\|$ if $\|x\|+\|y\|<1$ otherwise $f(x,y)=0$ if $\|x\|+\|y\|\geq1$ I tried the first one$(1)$ it that function assumes non-zero values in the interval$(-\infty,0)$and the smallest closed set containing this interval is $(-\infty,0]$ In the second example,$f(x,y)=0$ iff $x=0$ therefore the set in which function assumes non-zero values is $E^2-\{0\}$ the smallest closed set is $E^2$ please help me to solve $(3)$ and $(4)$.thanks in advance. AI: Your solutions to (1) and (2) are correct. In (3), the function $f$ is nonzero at least on $\mathbb Q^2$ (in fact on the bigger set $\mathbb Q\times\mathbb R\cup \mathbb R\cup \mathbb Q$). Since the rationals are dense in the reals, the closure of $\mathbb Q^2$ is already all of $\mathbb R^2$. For (4), it may be advisable to sketch the area in question. We are given that $f$ is zero when $\|x\|+\|y\|\ge 1$; this is (the outside of) a diamond shaped figure. The expression $(x-y)\|x+y\|-(x+y)\|x-y\|$ is zero if $x-y$ and $x+y$ have the same sign, that is either $x+y\le 0$ and $x\le y$ (this is a rotated quadrant of the plane) or $x+y\ge 0 $ and $x\ge y$ (another rotated quadrant). On the other hand, if $x-y$ and $x+y$ have opposite signs, then $$(x-y)\|x+y\|-(x+y)\|x-y\| = \pm((x-y)(x+y)+(x+y)(x-y)) =\pm2(x^2-y^2),$$ which is zero only on the diagonals (we already had that and it doesn't matter anyway when taking the closure). Now look at your sketch. It should look like two small diamonds.
H: distribution of the product of a poisson and a bolzmann What is the distribution of the product of two variables for which each of them has its own distribution(specifically one poisson and one bolzmann)? I found on wikipedia that for the sum of the two you should use a convolution, but how is this for the product? The situation for which I need this: i want to know the distribution of the total amount of energy of particles in a given time interval when the number of particles is poisson distributed and the energy of each particle is bolzmann distributed. AI: The situation you describe is not the product of a Poisson and a Boltzmann, because different particles have different energies. Instead, it's a compound Poisson distribution. If $X_i$ are independent random variables with your Boltzmann distribution, and $Y$ is independent of the $X_i$ and has a Poisson distribution, your random variable is $$Z = \sum_{i=1}^Y X_i$$ Note that $E[e^{sZ} \| Y ] = \left(E[e^{sX}]\right)^Y$, so if $g_X(s) = E[e^{sX}]$ is the moment generating function of the $X_i$ and $g_Y(s) = E[e^{sY}]$ is the moment generating function of $Y$, the moment generating function of $Z$ is $$ E[e^{sZ}] = E \left[\left(E[e^{sX}]\right)^Y\right] = g_Y(\log g_X(s)) = e^{\lambda (g_X(s)-1)}$$
H: Given $ h(x)=f(x)+O(g(x)) $ estimate using asymptotic notation $\frac{1}{h(x)}$ Given $ h(x)=f(x)+O(g(x)) $ and knowing that $ \lim_{x \to \infty}=\frac{g(x)}{f(x)}=0$ (int other words $f(x)=o(g(x))$) find such F(x) and G(x), $\frac{1}{h(x)}=F(x)+O(G(x)) $. Because $ f(x)=o(g(x)) $ that for any $e>0$ we can find $n_{0}$ such for every $n>n_{0}$ $g(n)<e\cdot f(n)$ . And knowing that $ h(x)=f(x)+O(g(x)) \leq f(x)+ c \cdot g(x) $ for some $ c>0 $ for all but finite amount of $x$'s we take $ e=\frac{1}{c} $ from the limit and have that $h(x)=f(x)+O(g(x)) \leq f(x)+f(x) \Rightarrow \frac{1}{h(x)}>= \frac{1}{2 \cdot f(x)}$. But this has gotten me completely nowhere and I don't know that to do... AI: You have $\frac{h(x)}{f(x)}=1+O(\frac{g(x)}{f(x)})$. Now note that $\lim \frac{g(x)}{f(x)}=0$ implies that the reciprocal is also $\frac{f(x)}{h(x)}=1+O(\frac{g(x)}{f(x)})$, hence $$\frac{1}{h(x)} = \frac1{f(x)}+O\left(\frac{g(x)}{f^2(x)}\right).$$ In case you didn't know the helpful result I used: Lemma. If $F(x)=1+O(G(x))$ with $G(x)\to 0$, then $\frac1{F(x)} = 1+O(G(x))$. Proof: As $F(x)=1+O(G(x))$, there is some $x_0$ such that $\|F(x)-1\|<cG(x)$ for $x>x_0$. And by the second condition, there exists $x_1$ such that $\|G(x)\|<\frac1{2c}$ for $x>x_1$. Hence $\|F(x)-1\|<\frac12$ and then $F(x)>\frac12$ for $x>x_2:=\max\{x_0,x_1\}$. Therefore $$\left\|\frac1{F(x)}-1\right\|=\frac{\|1-F(x)\|}{\|F(x)\|}<2cG(x)$$ for all $x>x_2$. $_\square$ Note that we are not "wasting" anything in the conclusion because when starting with $\frac1{F(x)}=1+O(G(x))$ and applying the lemma again, we obtain our original $F(x)=1+O(G(x))$ back.
H: Coequalizer in $\mathsf{Sets}$ Assume $f,g:X \to Y$ are arrows in $\mathsf{Sets}$. Then the coequalizer is given by $c:Y \rightarrow Y/R$ where $R \subseteq Y\times Y$ is the smalles equivalence relation on $Y$ s.t. $\forall x \in X: (f(x),g(x)) \in R$. Given any $h:Y \rightarrow Z$ there is a unique $\overline h: Y/R \to Z$ s.t. $\overline h \circ c = h$. I know that $\overline h ([y]) = h(y)$. My question: How can I prove that $[y]=[y'] \Rightarrow h(y) = h(y')$ ? AI: Given any $h:Y\to Z$ such that $hf=hg$ ... Now, "have the same image under $h$" determines an equivalence relation $H$, and since $(f(x),g(x))\in H$ it follows that $R\subseteq H$.
H: Integral of the function $S(x)=\ln\left(1-\frac{x}{\exp(x)}\right)$ I have to check if the following series: $$S(x)=\sum_{k=1}^{\infty}\frac{x^k}{k\exp(kx)}$$ gives a function of $x$ $$S(x)=-\ln\left(1-\frac{x}{\exp(x)}\right)$$ for which: $$J=\left\|\int_{0}^{+\infty}S(x)dx\right\|\lt\infty$$ I used Maple and Mathematica to solve the integral without any result. Does anyone have an idea how to calculate $J$? Thanks AI: For $x \geqslant 0$, all terms in $$S(x) = \sum_{k=1}^\infty \frac{x^k}{x\exp(kx)}$$ are non-negative. Hence we have $$\begin{align} J &= \int_0^\infty S(x)\,dx\\ &= \sum_{k=1}^\infty \frac{1}{k}\underbrace{\int_0^\infty x^ke^{-kx}\,dx}_{t = kx}\\ &= \sum_{k=1}^\infty \frac1k \int_0^\infty \left(\frac tk\right)^ke^{-t}\,d\left(\frac tk\right)\\ &= \sum_{k=1}^\infty \frac{1}{k^{k+2}} \int_0^\infty t^ke^{-t}\,dt\\ &= \sum_{k=1}^\infty \frac{k!}{k^{k+2}}, \end{align}$$ using $\int_0^\infty t^{\alpha-1}e^{-t}\,dt = \Gamma(\alpha)$ for $\Re \alpha > 0$. By Stirling's approximation, $$k! \sim \sqrt{2\pi k}k^ke^{-k} \Rightarrow \frac{k!}{k^{k+2}} \sim \frac{\sqrt{2\pi}}{k^{3/2}e^k},$$ so the sum converges. For a decent approximation of the value, one needs not compute many terms, but I don't know if there is a known closed form for the sum.
H: Change of coordinates with Jacobian We know that the change of variable in $\mathbb{R^n}$ with a $T: V \to U$ is a diffeomorphism of open sets in $\mathbb{R^n}$ and $f$ is an integrable function on $U$. Then $$\int_U f dx_1 \cdots dx_k = \int_V (f \circ T) \|\det(dT)\|dy_1 \cdots dy_k.$$ How can i prove this in Manifolds : $ dT_1 \wedge … \wedge dT_n = J(T) dy_1 \wedge … \wedge dy_n$ where $J(T)$ is the Jacobian determinant of T. Ok this is my new answer but please check it! σ $T^∗dx_1···dT_n = dT_1 \wedge … \wedge dT_n = (\sum_i^n\frac{∂T_1}{∂y_{i_1}} dy_{i_1})\wedge ···\wedge (\sum_i^n\frac{∂T_n}{∂y_{i_n}} dy_{i_n}) = $ $\sum_{1=i_1,…,1_n}^n\prod_{j=1}^n (\frac{∂T_j}{∂y_{i_1}}) dy_{i_1}\wedge ···\wedge dy_{i_n} = $ $\sum_{σ \in S_n} \prod_{j=1}^n (\frac{∂T_j}{∂y_{σ(j)}}) dy_{σ(1)}\wedge ···\wedge dy_{σ(n)} = $ $\sum_{σ \in S_n} \epsilon(σ) \prod_{j=1}^n (\frac{∂T_j}{∂y_{σ(j)}}) dy_{1}\wedge ···\wedge dy_{n} =$ $ det (\frac{∂T_j}{∂y_{j}})_{i,j} dy_{1}\wedge ···\wedge dy_{n} =$ $ J(T) dy_{1}\wedge ···\wedge dy_{n}$ Is it correct? Thanks for your comments please check it!! AI: The relationship between $dx$ and $dy$ is via the pullback action of $T$. See wikipedia, for example. The Jacobian arises from writing the pullback action explicitly and using the definition of a determinant.
H: convergence in $L^p$ implies convergence of $p$-th moments is there a simple way to prove that $X_n \rightarrow_{L^p} X$ implies that $\mathrm{E}(X^p_n) \rightarrow \mathrm{E}(X^p)$? the proof for $p=1$ is easy. but what about the case $p>1$? I would appreciate any comments. many thanks! AI: If $0<a<b$ then, by the Mean Value Theorem, there exists $c\in[a,b]$ such that $$ \|a^p-b^p\|=p\,c^{p-1}\,\|a-b\|\implies \|a^p-b^p\|\le p\,b^{p-1}\,\|a-b\|\le p\,(a^{p-1}+b^{p-1})\,\|a-b\|. $$ Then $$ \|E(X_n^p)-E(X^p)\|\le p\,E((X_n^{p-1}+X^{p-1})\|X_n-X\|). $$ Now use Hölder's inequality an the boundedness of $E(X_n^p)$ to deduce that $E(X_n^p)\to E(X^p)$.
H: basic linear system of equations I am stuck with the following linear algebra problem: given a basis $ \{e_{1} ... e_{n}\} $, I can define products $ \{p_{1} ... p_{m} \} $ as linear combination of the basis and the products itself: $$ p_{i} = \sum_{j=1}^m u_{ij} p_{j} + \sum_{j=1}^n a_{ij}e_{ij} $$ My goal is to represent each product only as a combination of the basis. $$ p_{i} = \sum_{j=1}^n s_{ij} e_{j} $$ any hint for a solution? AI: Your matrix $P=UP+AE$. This means that $P=(1-U)^{-1}AE$. This will break down if $1-U$ is degenerate, e.g., if $U=1$, and your representation is trivial: $u_{ij}=\delta_{ij}$ and $a_{ij}=0$.
H: Iteration of an operator Let $f_0(x)$ be integrable on $[0,1]$, and $f_0(x)>0$. We define $f_n$ iteratively by $$f_n(x)=\sqrt{\int_0^x f_{n-1}(t)dt}$$ The question is, what is $\lim_{n\to\infty} f_n(x)$? The fix point for operator $\sqrt{\int_0^x\cdot dt}$ is $f(x)=\frac{x}{2}$. But it's a bit hard to prove this result. I have tried approximate $f(x)$ by polynomials, but it's hard to compute $f_n$ when $f_0(x)=x^n$ since the coefficient is quite sophisticated. Thanks! AI: 1) If $f_0(x)\equiv 1$, It is easy to check $f(x)=\frac x2$. 2) If there exists $m,M >0$ such that $m<f_0(x)<M$, then $$\sqrt{mx}=\sqrt{\int_0^xmdt} \leq f_1(x)\leq \sqrt{\int_0^xMdt}=\sqrt{Mx}$$ and $$ \sqrt{\int_0^x\sqrt{mt}dt} \leq f_2(x)\leq \sqrt{\int_0^x\sqrt{Mt}dt} $$ and so on Thanks to 1), it follows that $f(x)=\frac x2$. 3). If $\inf\{f_0(x)\}=0$, then Approximation ! choose $\epsilon >0$, think interval $[\epsilon,1]$, begin with $f_1(x)$, not $f_0(x)$ there exists $m,M >0$ such that $m<f_1(x)<M, x\in[\epsilon,1]$, $$\sqrt{m(x-\epsilon)}=\sqrt{\int_\epsilon^xmdt} \leq f_2(x)\leq \sqrt{\int_0^xMdt}=\sqrt{Mx}, x\in [\epsilon,1]$$ and so on, we get that $$\frac{x-\epsilon}2\le f(x)\le\frac x2$$ Let $\epsilon \to 0$, it follows that $f(x)=\frac x2$.
H: Fibonacci Sequence Exercise I need some help checking the following solution. The Fib sequence is defined by $a_1 = 1, a_2 = 1$ and for all $n\geq 2$, $a_{n+1} = a_n + a_{n-1}$. Thus, the sequence begins: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55,... Prove that for all $n\geq 1$, $a_n <\left(\frac{5}{3}\right)^n$. So far I have: by induction $a_{n+1} = \left(\frac{5}{3}\right)^n+1 = a_n+a_{n-1} < \frac{5}{3}\cdot\left(\frac{5}{3}\right)^n $ from the definition and factoring. $\frac{5}{3}\cdot\left(\frac{5}{3}\right)^n > a_n+a_{n-1} $ $\frac{5}{3}\cdot\left(\frac{5}{3}\right)^n > a_n\cdot \frac{5}{3}$ from above $a_n\cdot \frac{5}{3} > a_n + a_{n-1}$ This is where I get stuck, is this a complete solution? Or is there further computation needed? AI: We need Strong induction here Let $a_n<\left(\frac53\right)^n$ for integer $1\le n\le m$ $\implies \displaystyle a_{m+1}=a_m+a_{m-1}<\left(\frac53\right)^m+\left(\frac53\right)^{m-1}=\left(\frac53\right)^{m-1}\left(\frac53+1\right)$ which will be $<\left(\frac53\right)^{m+1}$ if $\displaystyle \left(\frac53+1\right)<\left(\frac53\right)^2\iff 8\cdot3^2<3\cdot5^2$
H: Difference between Probability and Probability Density This question is from DeGroot's "Probability and Statistics" : Unbounded p.d.f.’s. Since a value of a p.d.f.(probability density function) is a probability density, rather than a probability, such a value can be larger than $1$. In fact, the values of the following p.d.f. are unbounded in the neighborhood of $x = 0$:$$f(x) = \begin{cases} \frac{2}{3}x^{-\frac{1}{3}} & \text{for 0<$x$<1,} \\ 0 & \text{otherwise.} \\ \end{cases}$$ Now, I don't know how the p.d.f. can take value larger than $1$.Please let me know the difference between the probability and probability density. AI: Simply put: $\rho(x) \delta x$ is the probability of measuring $X$ in $[x,x+\delta x]$. With $\rho(x):=$ probability density. $\delta x:=$ interval length. A probability will be obtained by computing the integral of $ \rho(x) $ over a given interval (i.e. the probability of getting $X\in [a,b] $ is $\int_a^b \rho(x) dx$. While $\rho(x)$ can diverge, the integral itself will not, and this is due to the fact that we ask that $\int_\mathbb{R}\rho(x) dx=1$, which means that the probability of measuring any outcome is 1 (we are sure that we will observe something). If the integral over the whole range gives 1, the integral over a smaller portion will give less than 1, because p.d.f. can't be negative (a negative probability is meaningless).

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