robjohn

West Hills, CA

none

Age: 55

the mean square The Mean Square
(with one standard deviation and several unusual ones)

aka Rob Johnson

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1d
answered Proving that $\sum_{(m,n)\in \Bbb Z \times \Bbb Z}\frac{1}{m^2+n^2+1}$ diverges.
1d
awarded Nice Answer
1d
revised Prove limit of $\displaystyle \frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{n+n}$ exists and lies between $0$ and $1$.
revisit the answer with more detail
2d
answered What is the area of $[r = \frac{4}{2 - \cos \theta}]$?
2d
answered Evaluating $ \sum_{n=1}^\infty \frac{1}{n^2 2^n} $
Mar
26
comment Evaluating the integral of a Fourier Series
I believe that the last integral should be $\frac{1-(-1)^n}n$, which is $\frac2n[n\text{ is odd}]$ where the brackets are Iverson Brackets.
Mar
26
comment A vector $\vec v = p_2 - p_1$ is pointing from $p_2$ to $p_1$ or $p_1$ to $p_2$
You can replace the vectors with real numbers (which are vectors in $\mathbb{R}^1$), and note that $3-1$ points in the direction of $3$ from $1$ (i.e. the positive direction).
Mar
25
revised Volume with a Cubic
mention the disc method
Mar
25
comment Mathjax's DNS provider attack: why do formulas render on my phone, not on my laptop?
Perhaps the DNS cache on your phone is more persistent than that on your laptop.
Mar
25
comment Volume with a Cubic
(+1) a different approach (shells) is always good.
Mar
25
comment Volume with a Cubic
I don't know if the downvote was because I gave a hint, but since there are now two complete answers, I might as well complete the answer to demonstrate the whole idea.
Mar
25
revised Volume with a Cubic
remove the hint and give a complete answer since there are two already posted
Mar
25
revised Volume with a Cubic
give a simpler hint
Mar
25
revised Volume with a Cubic
explain what to do
Mar
25
revised Volume with a Cubic
remove unneeded work
Mar
25
revised Volume with a Cubic
expand the integration to demonstrate why it simplifies
Mar
25
comment Volume with a Cubic
There is no reason to swap the $x$ and $y$. It has only served to confuse things. Upon changing the orientation, the area is given by $\pi y^2=\pi(x-1)^{2/3}$ and the integral should be $$\int_{\color{#C00000}{x}=1}^9\pi(\color{#C00000}{x}-1)^{2/3}\,\mathrm{d}x$$
Mar
25
answered Volume with a Cubic
Mar
25
revised Calculation of limit without stirling approximation
note assumed value
Mar
25
answered Calculation of limit without stirling approximation
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