Mathematics
WannaBeRealAnalysist's user avatar
WannaBeRealAnalysist's user avatar
WannaBeRealAnalysist's user avatar
WannaBeRealAnalysist
  • Member for 7 years, 7 months
  • Last seen this week
  • Canada
Profile Activity
Stats
451
reputation
13k
reached
10
answers
28
questions
Communities
About

I'm just a lonely person trying to learn some university level math but I'm not very smart. I've taken some courses in math in university but don't really belong there. I'm usually the worst performer in the class (whether it's a math class or otherwise).

I do love working on and solving problems but I don't have any friends or teachers. So I have no one to ask for help if I have trouble or have a question.

:(

Badges
View all badges
This user doesn’t have any gold badges yet.
2
silver badges
13
bronze badges
Top tags
View all tags
6
Score
24
Posts
63
Posts %
2
Score
6
Posts
16
Posts %
1
Score
22
Posts
58
Posts %
1
Score
6
Posts
16
Posts %
0
Score
6
Posts
16
Posts %
0
Score
6
Posts
16
Posts %
Top posts
View all questions and answers
All Questions Answers
Score Newest
question
9
Let f be a continuous real valued function on the compact interval [a,b]. Given ϵ>0, show that there is a polynomial p such that: |p(x)−f(x)|<ϵ
Feb 19, 2022
question
7
Is f(x,y) integrable? Question 3-7 from Spivak's Calculus on Manifolds
Nov 23, 2017
question
6
For what values of $(a,b,c)$ is $M$ a manifold?
Feb 17, 2018
answer
5
The limit of $f_n = \frac {e^x \sin(x) \sin(2x) \cdots \sin(nx)}{\sqrt n}$
Nov 5, 2017
question
2
What is a compact 2-D submanifold?
Nov 3, 2017
question
2
Prove that a homogeneous function $f(x_1,x_2,...,x_n)$ of degree $k$ which is $k$ times differentiable at the origin is a polynomial.
Nov 23, 2017
question
2
Is there a proof of the stronger version of the Implicit Function Theorem?
Nov 23, 2017
answer
2
Is the given vector in span(v1,v2)?
Feb 22, 2018
question
2
Prove that $(\mathbb{Q}, +)$ is not isomorphic to $(\mathbb{Q}^n, +)$
Oct 31, 2022
question
2
Prove that [O(2), O(2)] = SO(2). The commutator subgroup of the ortogonal group is equal to the special orthogonal group.
Nov 9, 2022

Your privacy

By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.