I'm having trouble understanding how ML estimates were used to find the upper bound of the Contour Integral defined in $(2.)$. My understanding of the example detailing the integral in question can be followed through from $(1).$
$(0.0)$
If $f$ is a complex-valued, continuous function on the co...
room topic changed to This is the Realm of Simply Beautiful Art: Room for totally random people to hang. Open discussions. [abstract-algebra] [big-numbers] [geometry] [logic] [recreational-mathematics]
I'm having trouble understanding how ML estimates were used to find the upper bound of the Contour Integral defined in $(2.)$. My understanding of the example detailing the integral in question can be followed through from $(1).$
$(0.0)$
If $f$ is a complex-valued, continuous function on the co...
Mmm well I just don't know what is a free resolution xd. I mean, I have a vague idea of what it is, but I haven't studied properly homological algebra :(
Learning about inaccessible cardinals and half wanting to get a girl friend, half thinking the latter would be a bad idea and the former idea much better, then thinking that this thinking would probably make me a horrible boyfriend. Fun high school years. amirite?
@SimplyBeautifulArt sure, I can tell you that when I studied real analysis I really liked the topic, but then I studied abstract algebra and everything changed xd
It's like how I find beauty in using the squeeze theorem for horribly hard limits: https://math.stackexchange.com/questions/2259517/evaluate-riemann-sum/2259586#2259586
I am wondering if there is a way to solve this definite integral depicted below:
$$\int_0^1 \log(-\log(x)\sin(x))\text{d}x.$$
Mathematica numerically evaluates this integral to $-1.63394$. I am not experienced with these types of integrals and have tried approaches in related questions with n...
If $\displaystyle A = \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\cdots\cdots\cdots+\frac{1}{\sqrt{999}}+\frac{1}{\sqrt{1000}}.$
Then $\lfloor A \rfloor$ is, where $\lfloor A\rfloor = A-\{A\}.$
$\bf{My\; Try::}$ Using $$\left(\sqrt{k}+\sqrt{k-1}\right)<2\sqrt{k}<\left(\sqrt{k+1}+\sqrt{k}\ri...
@TheGreatDuck: Bad news: you can only apply Darboux's theorem when the function is differentiable on the interval. So I accidentally assumed the conclusion, derp.
Yes if you refer to periodic motion uppercase "T" is indeed the period and lowercase "t" is time but uppercase "T" may be temperature if you are talking about thermodynamics.
Physics often runs out of letters, even Greek ones!
So you may have cooked your "ham" up to temperature "T" in time "t"...
But what if temperature also is periodic in time? then we're screwed with our T's...
I'm having trouble understanding how ML estimates were used to find the upper bound of the Contour Integral defined in $(2.)$. My understanding of the example detailing the integral in question can be followed through from $(1).$
$(0.0)$
If $f$ is a complex-valued, continuous function on the co...
Notice that:
$$60.41<60.5<61.24$$
but,
$$\lfloor60.5\rfloor=60\ne61$$
so your logic is wrong. To correct it, simply note that:
$$\sum_{k=2}^{1000}\frac1{\sqrt k}=\sum_{a=2}^{10}\frac1{\sqrt a}+\sum_{b=11}^{1000}\frac1{\sqrt b}$$
Likewise, you will find that
$$\sum_{a=2}^{10}\frac1{\sqrt a...
@Simply in the example i'm using looks like I haven't worked thorugh all of it last part uses the Residue Theorm to calculate singularities inside $\gamma_{r}$
the fundamental theorem of calculus then allows you to reduce integrals down to two points if you know enough about the function you are integrating.
Likewise, the residue theorem let's you reduce complex integrals down to discrete points rather than having to Riemann sum along the entire bloody path.