hopefully i will have one member of my family back tomorrow night. my 19 yo son is flying back from croatia via frankfurt & boston on his own, his first international alone. i hope it all works out.
In Spivak chap. 14, problem 1 which consists of finding derivatives of integrals using FTOC Part I and other tricks, why don't we first need to check that the integrand is continuous before applying FTOC?
For example, find the derivative of $F(x) = \int_a^{x^3} \sin^3 t ~ dt$
We let $f(x) = \int_a^x \sin^3 t ~ dt$, and then since $\sin^3 t$ is clearly continuous on the reals, we can apply FTOC to $f$.
sure, yes. it's not so much that you don't 'need' to check any hypothesis, it's just easy to check the hypothesis and so maybe not the focus of attention.
My point is, when given exercises like this am I supposed to check that it satisfies FTOC, even though it's clear that we need to use FTOC to solve the exercise
Or is it just an exercise in applying FTOC without worrying about if we can even apply it
i'd assume it is mostly an exercise in symbolic manipulation, with a background understanding that because the inputs are not 'ugly' the theorems will apply to them. but i've never taught out of that book.
I would assume so as well (and so I didn't bother proving the integrands are continuous), however the dilemma I keep thinking of is that when you do apply FTOC 'in the real world' i.e. not computing symbolic manipulation out of a chapter in a textbook, you will not always know if the integrand is continuous
So should I be preparing for such a scenario by making sure that the integrand is indeed continuous or... what exactly?
there's something beyond this, polite proofs, which is, the FTC holds under very general and subtle conditions, if you replace the riemann integral with another integral more suited to the purpose. the conditions in spivak are sufficient for FTC to apply but not necessary.
but in practice, the functions people run into 'in the real world' do not test these extremes.
@TedShifrin So for example, to prove we can apply FTOC I to find $F'$, where $F(x) = \sin \left( \int_0^x \sin \left( \int_0^y \sin^3 t ~ dt \right) ~ dy \right)$, we would go something like: Define $g(x)$ by $\int_0^x \sin(f(y)) ~ dy$ where $f(y)$ is defined as $\int_0^y \sin^3 t ~ dt$. Then $F(x) = \sin(g(x))$. $\sin$ is continuous, so we need to check if $g$ is continuous.
Then I suppose we need to check if $\sin(f(y))$ is integrable
$a \mid b$ means $a$ divides $b$, or into other words there some integer which we can call $c$ such that $\frac{b}{a} = c$, so multiplying by $a$ we get $b = ac$.
it's genuinely confusing, especially at first. particularly as computer languages have a "mod 3" function and saying a = b mod 3 might actually refer to computing a value of that function in such languages.
whereas in the pure math world it's just a relation, __ = __ mod __ having a meaning without __ mod __ necessarily being given a meaning.
Yeah. But when talking about infinity and finding a pattern that dis includes 1/3 of all infinity when that third is a multiple of 3…wait it might be a 1/6 of infinity…
If $F(x) = \int_{-\cos x}^{\sin x} \frac{1}{\sqrt{1 - t^2}} ~ dt$ for $x \in (0,\pi/2)$, one can check to find that $F'(x) = 0$. So I suppose you could argue that $F(x)$ is constant. But how do you then argue that the value of $F(x)$ does not depend on the value of $x$?
Maybe I'm being dumb, but could it be a different constant... depending on the $x$?
if we find the integral, it comes out to be $\frac{\pi}{2}$ for all $x \in (0,\pi/2)$ so it is not dependent on $x$. Also it has to be same constant for all $x$, if it will be a different constant for say $x=\frac{\pi}{6}$ and $x=\frac{\pi}{3}$ then it will become a discontinuous function and it will not be differentiable for all values of $x$, which is not the case. Please correct me if i am wrong!
In geometry, the folium of Descartes is an algebraic curve defined by the equation
x
3
+
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3
−
3
a
x
y
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{\displaystyle x^{3}+y^{3}-3axy=0\,}
.The name comes from the Latin word folium which means "leaf".
== History ==
The curve was first proposed and studied by René Descartes in 1638. Its claim to fame lies in an incident in the development of calculus. Descartes...
Question, does none degenerate bilinear form imply positive definite, or does positive definite imply none degenerate ?
What do you mean? ZFC proves that for every cardinal $\kappa$ there is a cardinal $\lambda$ such that $\lambda$ does not inject in $\kappa$ (this is already provable in ZF, but then you need AC to prove that cardinals are linearly ordered and so $\lambda>\kappa$)
I'm making this more complicated than it needs to be, just by Cantor's theorem $2^\kappa>\kappa$ for any cardinal
I'm reading "Olver - Asymptotics and Special Functions" and, at some point, for $u\to\infty$ the author says that $u-1+2\exp[-2u+2+O(e^{-2u})]+O(e^{-4u})=u-1+2\exp(-2u+2)+O(e^{-4u})$. He is ignoring the term $2\exp[O(e^{-2u})]$, in the sense that (it seems to me) he puts it equal to $1$ in the multiplication. I know that $aO(f(x))=O(f(x))$ for $a$ constant, but I don't understand why that equality holds.
@runway44 Thank you for your help. I am still getting used to this big O notation, especially when used in identities mixed with non-big O notation terms. Now is almost all clear, only one last thing: when you write $\exp(x)=1+O(x)$ as $x \to 0$ you mean that there exists a constant $K$ such that $\exp(x) \le 1+K|x|$ in a neighborhood of $0$?
Hello, we know that a polar decomoposition of an endomorphism that is in the general linear group (IE it is invertible ) is possible, however, i am asked to find a polar decomposition of an endomorphism that is skew hermetian, without the mention of it being invertible, how to do that?
furthermore i know that i* the endomorphism is hermetian
okay i just found out, that the polar decomposition exists even if it is not invertiable, but then it is not unique, how can i find such decomposition? one matrix is easy, but i am not sure how to find the orthogonal one
mad: how you implement it very much depends on what tools you use. conceptually you want to find an isometry V from the range of |T| to the range of T with T = V |T|. that's uniquely defined on the range of |T|. you then extend that arbitrarily on the orthocomplement if you want a unitary.
@SHASHAANKB.H. To get good at Calculus you need to do problems. I would work on easy problems and hard problems. Also make sure your answers are correct.
it's fun to do integrals. Well, it was fun to do integrals when I was still in high school at least
@SHASHAANKB.H. are you talking about all of those integral problems? Yeah, you have to "master the craft" so to speak to do what those people do with integrals. It's admirable
@user4539917 Unless you want to become a hereditary monarch, in which case all you need to do is be born into the family of a hereditary monarch. That seems like a pretty royal road to me!
I guess I was speaking a little bit in general because that's how I see math. Also, sometimes you just have to do the dirty work, don't forget about it
Say Omega = sum of F_i, we can assume F_i is increasing and then cross it with [-i, i]^d where d is the dimension, thus obtaining exhaustion by compact sets of Omega
I asked someone few hours ago and he said that a compact set K is not support then on K^c , T doesn't vanish so there exists a phi in D(omaga) s.t. T(phi) is non zero
@Jakobian Yes
But I think he made a mistake, K is not compact then K^c is not the largest open set on which T vanishes
I used to say that phrase too, with it being late. But then I asked myself, does it really make a difference. Maybe I am even thinking somewhat more clearly?
Consider the $n!$ points whose coordinates are permutations of $(1,2,\dots,n)\in\Bbb R^n$. Their convex hull is known as the permutahedron. Turns out its ($n-1$-dimensional) volume is $n^{n-2}$
Any way to see this?
($n-1$-dimensional as it lies in the hyperplane $\sum x_i=\sum i$)
Wait that can't be true
$n^{n-2}$ times the volume of the lattice $\{(x_1,\dots,x_n)\in\Bbb Z^n:\sum x_i=1\}$
which works out to $(n-1)!n^{n-2}=n!n^{n-1}$ times the volume of the standard simplex (of points with nonnegative coordinates whose sum is 1)
I think my question isn't necessarily true, and it's not even what I should be looking at. I'm trying to prove a homogenous Markov chain with finite amount of states has a recurring state, and for this I'm trying to prove that taking an essential state, it also needs to be recurrent
@Jakobian Sorry, I was gone for a bit. Write $\Bbb P(X_{n+1} = a|X_n \in \{b,c\}) = \Bbb P(X_{n+1} = a, X_n \in \{b, c\})/\Bbb P(X_n \in \{b, c\})$. Can you compute the probabilities in the numerator and denominator in terms of the transition matrix, lets call it $P$?
For me, Markov chains usually have a completely deterministic initial state.
What can you say in that case?
The probability distribution, written as a vector, of $X_n$ on the whole state-space is $P^n \mathbf{v}$ where $\mathbf{v}$ is the probability-vector on the initial state, yes?
The expected numbers of visits of a Markov chain to a state $b$, starting at state $a$, is something like $P(a, b)/(1 - P(b, b))$, not?
Geometric series
If $X_{a, b}$ is the number of visits to $b$, starting at $a$, then having $P(b, b) < 1$ for all $b$ in a finite state chain would imply $\sum \Bbb E[X_{a, b}] < \infty$. But this is also $\sum_{n = 1}^\infty \sum_b P^n(a, b)$ which is obviously divergent because $\sum_b P^n(a, b) = 1$
why the second order derivative of $f(x,y)$ general is $ \partial ^2f\left( x,y \right) /\partial x^2=f\left( x+\text{1,}y \right) -2f\left( x,y \right) +f\left( x-\text{1,}y \right) $?