Are there any significant differences between studying functional analysis from a normed space perspective versus a metric space perspective?

Does it matter if functional analysis was introduced from a normed space versus a metric space formulation? Are all major theorems from functional analysis (such as Banach contraction mapping, Hahn Banach theorem…) directly transferable if we exchanged every instance of norm with metric? I am asking because strictly speaking a metric is different than a … Read more

If B(t)B(t) is Brownian motion then prove W(t)W(t) defined as follows is also Brownian motion

Let B(t) be standard Brownian motion on [0.1] Define W(t) as follows W(t)=B(t)−∫t0B(1)−B(s)1−sds Prove W(t) is also Brownian motion So I’m not sure how to deal with the integral here. In order to show it, too, is Brownian motion I think I would need to Make an argument that the transformation is linear and hence … Read more

Finding an approximation of a function’s root

I have the polynomial function f(x)=x5+2×2+1. I am trying to find an approximation to its root in [−2,−1], with the precision of 0.1, and with a minimal number of steps. The answer I was given was −17/16. I find it incorrect, and I wish to ask for your assistance. I have calculated f(−2) and f(−1), … Read more

Expectation in kernel density estimate

Let $X_1,\ldots,X_n$ be i.i.d. random variables with common density $f$. Let $K(\cdot)$ be a probability density function defined on the real line. Then for a nonstochastic $h$: $$E[\hat{f}]=\frac{1}{nh}\sum_{i=1}^n E\left[K\left(\frac{x-X_i}{h}\right)\right]$$ $$=\frac{1}{h}E\left[K\left(\frac{x-X_i}{h}\right)\right]=\frac{1}{h}\int K\left(\frac{x-u}{h}\right)f(u)du$$ $$=\int K(y)f(x+hy)dy$$ I’m having trouble to understand how they get the two last equalities. Since $K$ is a probability density function $$\frac{1}{h}E[K\left(\frac{x-X_i}{h}\right)]=\frac{1}{h}\int K\left(\frac{x-X_i}{h}\right)d\left(\frac{x-X_i}{h}\right)$$ but … Read more

How do we know 3 is a divisor?

The following is a supposedly true claim, and I came across it in the Coursera course Introduction to Mathematical Thinking (Assignment 7/7) as part of an explanation by the instructor to the proof why \sqrt{3} was irrational. The line that perplexes me goes like this: If 3 divides the square of some x ∈ ℕ … Read more

Integration problem solving without contour integration

Can the following question be solved without using contour integration. F:(0,∞)×(0,∞)→R be given by F(α,β)=∫∞0cos(αx)x4+β4dx Show that F(α,β)F(β,α)=α3β3 as long as there is no positive integer n such that α=(4n−1)π√24β Answer Switching to integration parameter t=αβx gives F(α,β)=∫∞0cos(βt)(βαt)4+β4βαdt=α3β3F(β,α) AttributionSource : Link , Question Author : Dhamnekar Winod , Answer Author : random

How can I deepen my knowledge in Mathematics without getting a degree in Mathematics?

I’m an undergraduate student, I’m pursuing a Bachelor Degree in Computer Engineering. I had some exams about Mathematics (calculus 1/2/3, probability and statistics, algorithms and data structures, complex functional analysis). While I don’t want to become a researcher/professor (I would like to be a Software Engineer) I also love Mathematics, and I feel like I … Read more

Finding the area enclosed by the locus of centroids of equilateral triangles inscribed in an ellipse.

Let a and b be the lengths of the semi-major and semi-minor axes of an ellipse, respectively. How to find the area enclosed by the locus of the centroids of equilateral triangles inscribed in the ellipse. Answer. How to answer this question using calculus,trigonometry? Answer Since a circle of center $G(x_G,\,y_G)$ and radius $R > … Read more

Equivalence Relations: Understanding Compatibility

Synopsis My textbook, near the end of the section on equivalence relations, mentions the problem of “defining functions on a quotient set”. Specifically, assume that R is an equivalence relation on A and that F:A→A. We ask if there exists a corresponding function ˆF:A/R→A/R such that for all x∈A, ˆF([x]R)= [F(x)]R. After introducing this notion, he … Read more