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seen Dec 18 at 12:57

Oct
24
awarded  Popular Question
Sep
30
awarded  Explainer
Sep
7
revised Mathematics - The big picture
edited body
Sep
7
answered Mathematics - The big picture
Sep
3
comment Does anyone recognise this recursion sastisfied by the Bell numbers?
no problem, I can't think of better way to use these points :). Thanks again for the neat reply.
Sep
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comment Does anyone recognise this recursion sastisfied by the Bell numbers?
Great! Thank you, I'll award the bounty as soon as the site lets me (20 hours).
Sep
2
accepted Does anyone recognise this recursion sastisfied by the Bell numbers?
Sep
2
revised Does anyone recognise this recursion sastisfied by the Bell numbers?
added 158 characters in body
Sep
1
comment Does anyone recognise this recursion sastisfied by the Bell numbers?
Thank you for the reply, just to check, you mean you computed the above numerically for different $n$s and for each $n$ you found $R_n$ to be zero?
Aug
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revised Does anyone recognise this recursion sastisfied by the Bell numbers?
deleted 5 characters in body
Aug
30
asked Does anyone recognise this recursion sastisfied by the Bell numbers?
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awarded  Yearling
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12
awarded  Good Question
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awarded  Curious
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awarded  Inquisitive
Apr
7
accepted Is $|x-y|^n\leq 2^n(|x|^n+|y|^n)$?
Apr
7
comment Is $|x-y|^n\leq 2^n(|x|^n+|y|^n)$?
@lyj Oh yes of course, thank you.
Apr
7
asked Is $|x-y|^n\leq 2^n(|x|^n+|y|^n)$?
Apr
1
comment Approximating $C^2$ functions with compactly supported $C^2$ functions
@Igor, thank you for the comment. I just needed pointwise convergence, and cheap effective diet pills's suggestion seems to do the trick. Sorry I'm not sure how to follow through with your suggestion (not familiar with how to "blend spines together with the partition of the unity bump function").
Apr
1
comment Approximating $C^2$ functions with compactly supported $C^2$ functions
@cheapeffectivedietpills, thanks this works perfectly (for example, with $\phi(x)=\exp(1-1/(1-x^2))$ if $|x|<1$ and zero otherwise). If you copy paste your comment into an answer box, I'll accept and we can close the question.