zetafunction
Mar25-10, 05:52 AM
using an exponential regulator exp(-\epsilon n) the sum
1+2+3+4+5+6+7+............= -1/12+ 1/\epsilon^{2}
and for Casimir effect 1+8+27+64+125+............= -1/120+ 1/\epsilon^{4}
can i simply remove in the calculations of divergent series 1+2+3+4+5.. and similar the epsilon terms imposing renormalization conditions ??
how about for the rest of sums 1+2^{m}+3^{m}+..........= \zeta (-m) + 1/\epsilon ^{m+1}
if i introducte a power regulator n^{-s} in the limit s-->0+ i would get
\zeta(s-m)=\zeta(-m) but i am not sure, why this work
for example in the definition of a functional determinant (in differential geommetry )
\prod_{i} \lambda_{i} apparently there is no divergent term proportional to log(\epsilon) as one would expect since the product is divergent
1+2+3+4+5+6+7+............= -1/12+ 1/\epsilon^{2}
and for Casimir effect 1+8+27+64+125+............= -1/120+ 1/\epsilon^{4}
can i simply remove in the calculations of divergent series 1+2+3+4+5.. and similar the epsilon terms imposing renormalization conditions ??
how about for the rest of sums 1+2^{m}+3^{m}+..........= \zeta (-m) + 1/\epsilon ^{m+1}
if i introducte a power regulator n^{-s} in the limit s-->0+ i would get
\zeta(s-m)=\zeta(-m) but i am not sure, why this work
for example in the definition of a functional determinant (in differential geommetry )
\prod_{i} \lambda_{i} apparently there is no divergent term proportional to log(\epsilon) as one would expect since the product is divergent