Aspyn wishes to invest an unknown sum in an account where interest is compounded continuously. Assuming Aspyn makes no additional deposits or withdrawals, what annual interest rate will allow her investment to double in exactly five years?
I interpret this as the ODE $P'(t) = rP(t)$ with the initial condition $P(5) = 2P(0)$, where $t$ is the number of years the sum has been in the account, $P(t)$ is account balance at $t$, and $r$ is a constant interest rate to be computed. I found the general solution $P(t) = Ce^{rt}$, then tried $2P(0) = Ce^{r5}$, leading to $C = 2e^{-5r}P(0)$ and the particular solution $P(t) = 2e^{rt - 5r}P(0)$.
I can solve this for $r$, but I'm not sure how to get rid of the $P(0)$ without reintroducing $C$. Am I doing this right?
Not sure not really my area of focus economics. But I can introduce her well actually I think he is frightened of me now well and truly but she should try to get in touch with me ol mate Anthony, he had a lot of coin, seemed a bit creepy but yeah Anthony buck. He'll be ready to
@user10478 You can't get rid of P(0), no, but you don't need to: It's 5 years to double any initial investment.
So once you solve for $r$ you're done.
To put it another way: your formula works if P(0) is one dollar or one million dollars. after 5 years, it'll have doubled---from what to what hardly matters re: the problem (though it'd make a lot of practical difference if you were Aspyn, of course!)
Show that the tractrix has the following property: for each point on the curve, the line segment of the tangent line from the curve point to the $y-$ axis has length 1.
My attempt:- Let the given Tractrix can be parametrised by the formula, $$c:(0,\pi/2)\to \mathbb R^2,$$ $$c(t)=\sin(t) \hat{i}...
So, what kind of object would the set of all semigroup homomorphisms under function composition be? It's certainly not a magma, because there exist elements which cannot be combined with each other, which I believe disqualifies it, correct?
I've got a set-theoretic question: if we assume ZF + every essentially surjective fully faithful functor between small categories is an equivalence, can we prove AC?
essentially surjective means that every object in the target category is isomorphic to some object in the image. fully faithful means the map on Hom-sets is bijective for every pair of objects. $F$ is an equivalence if there's a functor $G$ that goes in the opposite direction such that $GF$ and $GF$ are isomorphic to the respective identity functors
I think I have a proof: if $(X_i)_{i \in I}$ is a non-empty family of non-empty sets, then consider the following categories: (1) $I$ as a discrete category (2) Consider the disjoint union $X$ of the sets $X_i$, each as a codiscrete category (i.e. there's a morphism between each pair of elements in the same set), then we have a functor $F:X_i \to I$ by sending each element $x_i \in X_i$ to $i$. This is essentially surjective and fully faithful, having a quasi-inverse $G: I \to X$ is the same as a choice function
So essentially surjective + fully faithful => equivalence of categories for functors between small categories is equivalent to choice
@MatheinBoulomenos I'm confused, pick a surjective map $f\colon X\to Y$ and think about $X$ and $Y$ as trivial groupoids (the only morphisms are the identity morphisms), $f$ is now an (essentially) surjective fully faithul functor, right?
I think the same proof should also show that essentially surjective + fully faithful => equivalence of categories is equivalent to choice for classes over something like NBG set theory as a meta-theory
yeah, exactly: there's an adjunction between the slice category $\mathbf{Top}/X$ and the category of presheaves on $X$ $\mathbf{PSh}(X)$, for one direction, you take a topological space $E \to X$ to the presheaf defined by taking preimages, for the other direction you take the espace etale. This restricts to an equivalence of categories between etale spaces over $X$ (with local homeomorphisms as structure maps) and sheaves over $X$ and sheaves over $X$ and to an equivalence between covering spaces and locally constant sheaves
So, a question: Given any function with a semigroup as it's domain, can you evaluate whether it's a semigroup homomorphism just by looking at the definition of the function?
@Mathein you use the fact that 1, $2^{4n}$, ..., $(4n-1)^{4n}$ are all 1 mod p and then take the first difference of that sequence. Those are all divisible by p, and all factor as a difference of squares $(a^{2n} + b^{2n})(a^{2n} - b^{2n})$. Then by primality p has to divide one of those factors. There’s a Lemma that says that if (a,b)=1 then every factor of the sum of the squares of a and b is a sum of two squares, so all you need to do is show that p divides the first factor of
The difference of squares in at least one case and the Lemma gives you the proof
Not yet, I’ve been quite busy at work so I’ve been getting home and going straight to bed lol, I have a day off today which I’m spending in a different city reading Edwards‘ „genetic intro to ANT“
@Mathein showing that the first factor is divisible by p in at least one case just comes from the fact that the nth difference in the sequence $\{ x^n\}$ is n!
maybe I’ll try a couple of exercises from those sheets today :)
It drives me crazy that people say "localization at $\mathfrak{p}$" when it should be "localization away from $\mathfrak{p}$"
I mean, I get it, it's the stalk over $\mathfrak{p}$, so "at" makes sense. But geometrically the picture is you're inverting elements in $A \setminus \mathfrak{p}$
@ÍgjøgnumMeg That's because the contravariance of the ring <-> affine scheme duality, though, innit. $A_\mathfrak{p}$ tells you stuff about the structure of $\text{Spec}\, A$ near $\mathfrak{p}$
Think about localizing at elements, or multiplicative subsets generated by them, for a change. $\Bbb Z$ localized at the multiplicative subset generated by $p$, which is $\Bbb Z[1/p]$
Tensoring with this guy, or extending scalars from $\Bbb Z$ to $\Bbb Z[1/p]$, kills $p$-torsions of abelian groups
That's why that should be "localization at $p$"
Tensoring with $\Bbb Z_{(p)}$ however kills everything but $p$-torsions :3
I think I said at the end of the project that $\Bbb Z[\zeta_p]$ is the ring of integers of $\Bbb Q_p$ (which was almost definitely a typo but I'm sad that I missed it before my hand-in hahaha)
Problem: Show that any group $G$ of order 108...By Sylow's theorem, I know that the number of sylow $3$-subgroups is either $1$ or $4$. But I don't know how to rule out the latter possibility. I could use a hint.
Okay, so the action is not faithful (just a rephrasing of non-injectivity). It looks like the action is transitive, so the orbit stabilizer theorem says $N_G(P) = P$ for every Sylow 2-subgroup $P$...Is this heading in the right direction, or am I getting sidetracked ?
If $G$ is a finite group it always acts transitively on it's Sylow $p$-subgroups, so you get a permutation representation $G \to S_n$ where $n$ is the number of Sylow $p$-subgroups. This is surjective by transitivity, and it's kernel is the normalizer of a Sylow $p$-subgroup. This in particular means the normalizer has index $n$.
you can generalize and slightly strengthen this approach to show that if $G$ has a subgroup of index $n$ where $G$ does not divide $\frac{n!}{2}$, then $G$ is not simple (this applies to the normalizer of a 3-Sylow in this example which is equivalent to the given solution)
You have proven that the normalizer is normal here, which is a subgroup of index 4 in a group of order 108, i.e., a subgroup of order 27.
The philosophy is if a large order group has too few Sylow p-subgroups for some prime p, then it's not going to be simple.
Or as Mathein says it's really a divisibility thing than order-comparison because if it's simple it has to embed in $S_n$ where $n$ is the number of Sylow $p$-subgroups.
It's just bloating up the usual four 3-Sylows in A_4 intersecting trivially by a Z_9, doesn't matter how you do it
Easiest to take the trivial semidirect product.
Sounds hard to classify all such extensions, though. You'd need to calculate H^2(A_4, Z_9) and H^2(A_4, Z_3 x Z_3) for all twisted coefficients, no idea how one does that.
If $R \setminus S = \bigcup \mathfrak{p}$ for some primes $\mathfrak{p}$ of $R$ then $S = \bigcap R\setminus \mathfrak{p}$, now if $xy \in S$ then $xy \in R\setminus \mathfrak{p}$ for at least one prime, so that $xy \neq 0$ in the quotient, meaning that $x \in R\setminus \mathfrak{p} \subset S$ so $S$ is saturated
pls confirm
hahaha
nah I want $xy \in R\setminus \mathfrak{p}$ for aaaall the $\mathfrak{p}$
but that doesn't change anything I just used the wrong words
So a primitive of $\frac{1}{x\ln{x}}$ is $\ln(\ln{x})$, but while trying to evaluate the integral within 0 and 1 of the former we get an indeterminate since the latter doesn't exist there.
I'd like to show that operations on irrational numbers produce (most often) another irrational number. The approach I have in mind is to show that rational numbers make a cut $(A,B)$ and have $max(A)$ or $min(B)$, which is untrue for irrationals. And then from the bounds of these numbers reconsturct the resulting one. Is it worthwhile?
You can actually cook up a crude bound for the degree of the minimal polynomial without Cayley-Hamilton by noting that the space of linear operators is n^2 dimensional, where n is the dimension of the vector space
Problem: Determine up to similarity all matrices of finite order in $GL_2(\Bbb{Q})$...If $A \in GL_2(\Bbb{Q})$ has finite order, then $A^k = I$ for some $k \in \Bbb{N}$, so its minimal polynomial $m_A$ divides $t^k - 1$. It's either a first or second degree polynomial, but I'm not sure how to proceed.
So the minimal polynomial is an atmost 2 degree polynomial which divides $t^k - 1$. What's special about these polynomials over $\Bbb Q$? What does a completely factorizable minimal polynomial say about the matrix?
Hmm...okay. I'm still hung up on why $m_A$ equals either $t-1$ or $t^2-1$. I mean, it's intuitively clear that if it divides $t^k-1$, then should have one of the two forms; but I don't see the proof. Is it just an annoying calculation? If so, I can figure that out later.
I was recently going through this notes. However I didn't understand the proof of Proposition 1.11 (page 8). Specifically the following, "But then using a compatibility of $e_1$ and $ε_1$ we get that the composite natural transformation is $e_2$ as required." I don't understand how we can get $e_2$. Since I haven't understood this part I didn't go further. Can anyone help me here?
The given answer is correct. I do not understand your comment.
More precisely, $1$ is a factor of any number (obviously), whereas $x$ is not a factor (but this also isn't implied by what the answer says). Saying $x=1$ is a factor doesn't quite make sense.
Diving multivariable polynomials is hard. The factor theorem in the form that is used still holds though. You can consider $K[X_1,...,X_n]$ as $K[X_1,...,X_{n-1}][X]$ and apply the usual factor theorem.
I'm trying to show that any two fields with the same order are isomorphic. If I could construct a non-zero ring homomorphism, I'd be done...But this doesn't seem possible. I could use a hint.
So, if $E$ and $F$ are finite fields of the same order, are $f(x) = x^n - 1_F$ and $g(x) = x^n-1_E$ considered the same polynomial, where $n = |F^\times|$?
If not, I don't see how they could be splitting fields of a common polynomial.
Speaking of suspicious, I think it's important to erase some words from the English dictionary, even if it means using additional characters, the meaning is conveyed more precisely. For example, one should no longer use the word "uniform" but rather refer to that word's meaning by using "respect costume"
Oh no you are right it's about pseudo respect, not actual respect, that's why costume is important, so people know it's not actual authority the individual possesses
@fargle banach algebras are vector spaces, derivatives are well defined over complete vector spaces ($d_x f$ is linear part of the best affine approximation of $f$ at $x$)
:-) oh well we all have questions unanswered who knows you may get lucky and find someone cares for yours, banach algebras are just vector spaces, this is another good example of my original remark this morning actually
We consider the following linear $(n,m)$-code $C$ :
I want to determine the canonical generator matrix and the parity-check matrix.
$$$$
Is the generator matrix the matrix with rows the above codes? I mean the following:
\begin{equation*}G=\begin{pmatrix}0& 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 &...
I need a book called "A=B" someone buy it for me, im holding my breath until I get it, and if I die it will be all your fault and the story will go viral and the whole world will know what cruel monsters you are
Show that the tractrix has the following property: for each point on the curve, the line segment of the tangent line from the curve point to the $y-$ axis has length 1.
My attempt:- Let the given Tractrix can be parametrised by the formula, $$c:(0,\pi/2)\to \mathbb R^2,$$ $$c(t)=\sin(t) \hat{i}...
We consider the following linear $(n,m)$-code $C$ :
I want to determine the canonical generator matrix and the parity-check matrix.
$$$$
Is the generator matrix the matrix with rows the above codes? I mean the following:
\begin{equation*}G=\begin{pmatrix}0& 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 &...
