7,431 reputation
720
bio website linkedin.com/in/gt6989b
location New York, NY
age 35
visits member for 2 years, 7 months
seen 2 days ago

Apr
13
comment Joint Probability Distribution Function
@Did is that integral not the probability that X=Y.
Apr
10
answered Sets of feasible directions
Apr
10
revised Sets of feasible directions
edited tags
Apr
10
revised Evaluating the surface integral?
deleted 5 characters in body; edited tags
Apr
9
answered Comparing two quantities
Apr
9
answered Testing series for absolute convergence
Apr
9
revised Linear algebra [Intermediate level]
added 13 characters in body
Apr
9
reviewed Approve suggested edit on Show that $\frac{1}{x^4 \sin^2 (x) +1} \in L^1([0, \infty))$
Apr
9
answered I'm not quite sure I understand my book's reasoning for the answer
Apr
9
revised Show that $\frac{1}{x^4 \sin^2 (x) +1} \in L^1([0, \infty))$
edited title
Apr
9
comment Prove that the function $f(x)=1/2^n$ is integrable on $[0,1]$
You need to show that there is some partition, any refinement of which only yields smaller differences between $L$ and $U$
Apr
9
comment Logarithmic Equations and solving for the variable
@ajotatxe I'm sorry, I revised a couple of times, not sure if your remark is still relevant?
Apr
9
comment Logarithmic Equations and solving for the variable
+1 for thinking about the problem before posting here.
Apr
9
reviewed Approve suggested edit on Logarithmic Equations and solving for the variable
Apr
9
answered Logarithmic Equations and solving for the variable
Apr
9
revised Logarithmic Equations and solving for the variable
added 17 characters in body; edited tags
Apr
9
comment Prove that the function $f(x)=1/2^n$ is integrable on $[0,1]$
$$L\left(2^{-n}, \pi\right) \le \int_0^1 2^{-x} dx \le U\left(2^{-n}, \pi\right)$$ and you need to show that the difference between the RHS and the LHS is converging to $0$ in the limit as $n \to \infty$.
Apr
9
comment Prove that the function $f(x)=1/2^n$ is integrable on $[0,1]$
You can use $f(x) = 2^{-n}$, not $2^n$.
Apr
9
comment Prove that the function $f(x)=1/2^n$ is integrable on $[0,1]$
Yes, you need to exhibit the partition and show that the difference between the upper and the lower bound is bounded by $\epsilon$ as you indicated (i.e. goes to $0$ as $n \to \infty$).
Apr
9
comment Prove that the function $f(x)=1/2^n$ is integrable on $[0,1]$
Any thoughts on how to apply the theorem? We will be happy to both comment and hint once you exhibit some work on the problem.