'http://www.BillHowell.ca/Neural nets/Paper reviews/230120 journal paper review - math discarded scribblings.txt' 02Feb2023 initial ************************** ************************** Other stuff (not used) +--> (notation example : r = reviewer in (r2.4)) : (r2.4) d[dt: V(u)] <= + d[dt: M(t) *V(u) - p1*V(u)^k1 - p2*V(u)^k2] <= + d[dt: M(t) *V(u)] - d[dt: p1*V(u)^k1] - d[dt: p2*V(u)^k2] <= + M(t) *d[dt: V(u)] + d[dt: M(t)]*V(u) - p1*k1*d[dt: V(u)^(k1 - 1)] - p2*k2*d[dt: V(u)^(k2 - 1)] It does NOT help to do this derivative! Authors' approach is to use bounded integrals, which makes more sense. <--+ +--+ looking at : d[dt: e^{ -(1 - k2) *∫[ds, t0 to t: M_hat(s)] } ] = e^{ -(1 - k2) *∫[ds, t0 to t: M_hat(s)] } *d[dt: -(1 - k2) *∫[ds, t0 to t: M_hat(s)] ] = e^{ -(1 - k2) *∫[ds, t0 to t: M_hat(s)] } * -(1 - k2)*d[dt: ∫[ds, t0 to t: M_hat(s)] ] using equivalence of dt = ds = e^{ -(1 - k2) *∫[ds, t0 to t: M_hat(s)] } * -(1 - k2) *(M_hat(t) - M_hat(t0)) |---> *********************** so ASSUMING M_hat(t0) = 0 (!!??!!) - this must be stated (r2.9.2) d[dt: eTrm(t)] = -(1 - k2)*M_hat(t)*eTrm(t) +--+ looking at : d[dt: e^{ -(1 - k1) *∫[ds, t_bar to t: M(s)] } ] = e^{ -(1 - k1) *∫[ds, t_bar to t: M(s)] } *d[dt: -(1 - k1) *∫[ds, t_bar to t: M(s)] ] = e^{ -(1 - k1) *∫[ds, t_bar to t: M(s)] } * -(1 - k1) *d[dt: ∫[ds, t_bar to t: M(s)] ] using equivalence of dt = ds = e^{ -(1 - k1) *∫[ds, t_bar to t: M(s)] } * -(1 - k1) * M(s) ??? so (r2.10.2) d[dt: eTrm2(t)] = -(1 - k1)*M_hat(s)*eTrm2(t) +-----+ 03Feb2023 start by looking at d[dt: X(t)] ≤ (1 - k1)*[ M(t)*X(t) - p1 ] (r2.11.10) ∫[ds, t_bar to t: d[dt: X(t)*e^{ -(1 - k1)*∫[ds, t_bar to t: M(s)] } ] / (1 - k1)] <= ∫[ds, t_bar to t: -p1 *e^{ (k1 - 1)*∫[ds, t_bar to t: M(s)] } ] <= -p1 *∫[ds, t_bar to t: e^{ (k1 - 1)*∫[ds, t_bar to t: M(s)] } ] p3c2h0.9 this is the same as the authors' LHS ????????????????????? ( X(t) *e^{ -(1 - k1)*∫[dv, t_bar to s: M(v)] } - X(t_bar) ) / (1 - k1) (r2.11.16) ∫[ds, t_bar to t: -p1* e^{ (k1 - 1)*∫[ds, t_bar to t: M(s)] } ] <= -p1 *e^{(k1 - 1)*α1} * ∫[ds, t_bar to t: 1 ] <= -p1 *e^{(k1 - 1)*α1} *(t - t_bar) p3c2h0.33 this is the same as the authors' RHS <= -p1 *e^{(k1 - 1)*α1} *(t - t_bar) eTrm2(t) = e^{ -(1 - k1)*∫[ds, t0 to t: M(s) ] } WRONG!!! but integrand is a constant <= -p1 *e^{(k1 - 1)*(−μ*(t − t_bar) + α2)} *∫[ds, t_bar to t: 1 ] <= -p1 *e^{(k1 - 1)*(−μ*(t − t_bar) + α2)} *(t - t_bar) reducing negatives <= -p1 *e^{(1 - k1)*( μ*(t − t_bar) + α2)} *(t - t_bar) +-----+ 04Feb2023 +--+ p3c2h0.4 Thus, we further get Note that the RHS of p3c2h0.45 has not changed, and we are looking for a more compact version of the LHS. (r2.10.20) ( X(t)*eTrm2(t) - X(t_bar) ) / (1 - k1) <= -p1 *e^{(k1 - 1)*α1} *(t - t_bar) LHS ( X(t)*eTrm2(t) - X(t_bar) ) / (1 - k1) = ( X(t)*e^{ -(1 - k1)*∫[dv, t_bar to s: M(v)] } - X(t_bar)) / (1 - k1) this is the same as the 3rd line (2nd inequality) p3c2h0.5 Reversing (r2.10.16) step-by-step -p1 *e^{(k1 - 1)*α1} *(t - t_bar) >= ∫[ds, t_bar to t: -p1 *e^{ (k1 - 1)*∫[ds, t_bar to t: M(s)] } ] divide each side by -p1 (constant - can come out of integral) so must reverse >= (this is an easy change to miss) e^{(k1 - 1)*α1} *(t - t_bar) <= ∫[ds, t_bar to t: e^{ (k1 - 1)*∫[ds, t_bar to t: M(s)] } ] ????????????? p3c2h0.4 X(t)*e^{(k1 - 1)*α1} <= -p1 *e^{(k1 - 1)*α1} *(t - t_bar) divide both sides by e^{(k1 - 1)*α1} (r2.10.25) X(t) <= -p1*(t - t_bar) +--+ p3c2h0.6 By using t = t_bar + 1 / (p1*(k1 - 1)) so p1 = 1 / {(t - t_bar) +-----+ 05Feb2023 rearrange (r2.12.10) (r2.12.11) -1 + p1/μ*e^{ (1 - k1)*( μ*(t − t_bar) - α2)} - p1/μ*e^{-(1 - k1)*α2} <= 0 (r2.12.12) p1/μ*e^{ (1 - k1)*( μ*(t − t_bar) - α2)} <= 1 + p1/μ*e^{-(1 - k1)*α2} drop the term 1 (NOT a legitimate action!), take ln of each side (r2.12.13) ln(( p1/μ *e^{ (1 - k1)*( μ*(t − t_bar) - α2)} )) <= ln(( p1/μ *e^{-(1 - k1)*α2} (r2.12.14) ln(p1/μ) + ln(( e^{ (1 - k1)*( μ*(t − t_bar) - α2)} )) <= ln(p1/μ) + ln(( e^{-(1 - k1)*α2} )) subtract ln(p1/μ) from both sides, take ln (r2.12.15) (1 - k1)*( μ*(t − t_bar) - α2) <= -(1 - k1)*α2 simplify (r2.12.16) μ*(t − t_bar) - α2 <= -α2 (r2.12.17) μ*(t − t_bar) <= 0 (r2.12.17) t <= t_bar WRONG!! but this was not a legitimate deduction anyways Given assumptions : p3c1h0.8 V(u(t)) ≤ 1 p3c1h0.85 t_bar <= t0 + 1/(p2*(k2 - 1)) ~= T1 p3c2h0.70 X(t) = V(u(t))^(1 - k1) p3c2h0.75 d[dt: V(u)] <= M(t)*V(u) - p1*V(u)^k1 (given for Part 2) |---> *********************** p3c2h0.25 Integrate (2.9) from t 0 to T 1 and by (2.5), it obtains ... sub (2.5) into (r2.9.12), with ASSUMPTION as integral bounds should be 0 to +∞ instead of t0 to T1!!!! ??? It may be possible that integral of (t0 to t) <= (0 to +∞), but that needs to be proven. <---| *********************** p3c2h0.33 unproven ASSUMPTION : see the step-by-step math check of Lemma 4 at http://www.billhowell.ca/Neural%20nets/Paper%20reviews/230120%20journal%20paper%20review%20-%20math%20only.txt looking at RHS of (r2.9.10) : (r2.9.15) ∫[ds, t0 to T1: -p2* e^{(k2 - 1)*∫[ds, t0 to t: M_hat(s)] } ] = -p2* ∫[ds, t0 to T1: e^{(k2 - 1)*∫[ds, t0 to t: M_hat(s)] } ] looking at (2.5) ∫[dv: 0 to +∞: M_hat(v)*dv] < α1 where M_hat(v) = max{M(v), 0} (constant >= 0 over interval) sub into (r2.9.12), with ASSUMPTION as integral should be 0 to +∞ !!!! It may be possible that integral of (t0 to t) <= (0 to +∞), but that needs to be proven. ... However, with that assumption, the next steps follow... ∫[ds, t0 to T1: -p2* e^{ (k2 - 1)*∫[ds, t0 to t: M_hat(s)] } ] <= -p2* ∫[ds, t0 to T1: e^{ (k2 - 1)*α1} ] but e^{ (k2 - 1)*α1} is a constant, therefore (r2.9.16) ∫[ds, t0 to T1: -p2* e^{ (k2 - 1)*∫[ds, t0 to t: M_hat(s)] } ] <= -p2*e^{(k2 - 1)*α1} * ∫[ds, t0 to T1: 1 ] <= -p2*e^{(k2 - 1)*α1} * (T1 - t0) p3c2h0.33 this is the same as the authors' RHS <= -p2*e^{(k2 - 1)*α1 }* (T1 - t0) ... Note that use of (2.6 avoids this issue, but would yield a slightly different form. I do not feel that this point, even if I am correct, undermines my confidence in the authors' paper. Reversing (r2.9.16) step-by-step -p2 *e^{(k2 - 1)*α1} *(T1 - t0) >= ∫[ds, t0 to T1: -p2 *e^{ (k2 - 1)*∫[ds, t0 to t: M_hat(s)] } ] divide each side by -p2 (constant - can come out of integral) so must reverse >= (this is an easy change to miss) e^{(k2 - 1)*α1} *(T1 - t0) <= ∫[ds, t0 to T1: e^{ (k2 - 1)*∫[ds, t0 to t: M_hat(s)] } ] ????????????? p3c2h0.4 Y(T1)*e^{(k2 - 1)*α1} <= -p2 *e^{(k2 - 1)*α1} *(T1 - t0) divide both sides by e^{(k2 - 1)*α1} (r2.9.25) Y(T1) <= -p2*(T1 - t0) +--+ p3c2h0.6 By using T1 = t0 + 1 / (p2*(k2 - 1)) so p2 = 1 / {(T1 - t0) #] 06Feb2023 p3c2h0.9 Integrating (2.11) from t_bar to t, using t_bar <= t0 + 1/(p2*(k2 - 1)) +--+ p3c2h0.9 Integrating (2.11) from t_bar to t looking at LHS of (2.11) : (r2.11.11) ∫[ds, t_bar to t: d[dt: X(t)*e^{ -(1 - k1)*∫[ds, t_bar to t: M(s)] } ] / (1 - k1)] = ( X(t) *e^{ -(1 - k1)*∫[ds, t_bar to t : M(s)] } - X(t_bar)*e^{ -(1 - k1)*∫[ds, t_bar to t_bar: M(s)] } ) / (1 - k1) but ∫[ds, t_bar to t_bar: M(s)] -> is zero as bounds (upper=lower) so 1 = e^{ -(1 - k1)*∫[ds, t_bar to t_bar: M(s)] } (r2.11.12) ∫[ds, t_bar to t: d[dt: X(t)*e^{ -(1 - k1)*∫[ds, t_bar to t: M(s)] } ] / (1 - k1)] = ( X(t) *e^{ -(1 - k1)*∫[ds, t_bar to t: M(s)] } - X(t_bar) ) / (1 - k1) simplified ( X(t)*eTrm2(t) - X(t_bar)) / (1 - k1) looking at RHS of (2.11) : (r2.11.15) ∫[ds, t_bar to t: -p1* e^{(k1 - 1)*∫[ds, t_bar to t: M(s)] } ] = -p1 *∫[ds, t_bar to t: e^{(k1 - 1)*∫[ds, t_bar to t: M(s)] } ] looking at givens (2.6) ∫[ds, t0 to t: e^{(k1 - 1)*∫[ds, t_bar to t: M(s)] } ] <= - μ*(t - t0) + α2 where t > t0 p3c1h0.85 t_bar <= t0 + 1/(p2*(k2 - 1)) = T1` (r2.11.16) t0 >= t_bar - 1/(p2*(k2 - 1)) sub (r2.11.16) into (2.6) (r2.11.17) ∫[ds, t0 to t: e^{(k1 - 1)*∫[ds, t_bar to t: M(s)] } ] <= - μ*(t - (t_bar - 1/(p2*(k2 - 1)) )) + α2 <= - μ*(t - t_bar + 1/(p2*(k2 - 1)) ) + α2 <= - μ*(t - t_bar) + μ/(p2*(k2 - 1)) + α2 unfortunately, the inequality <= is no longer certain, as (r2.11.16) throws it into doubt. sub (r2.11.17) into (r2.11.15) (r2.11.18) ∫[ds, t_bar to t: -p1* e^{(k1 - 1)*∫[ds, t_bar to t: M(s)] } ] <= -p1 *((- μ*(t - t_bar) + μ/(p2*(k2 - 1)) + α2 )) Let (reversing signs) (r2.11.20) eTrm3(t) = e^{(1 - k1) *(( μ*(t - t_bar) - μ/(p2*(k2 - 1)) - α2 ))} |--> take derivative to see what happens (r2.11.25) d[dt: eTrm3(t)] = d[dt: e^{(1 - k1) *(( μ*(t - t_bar) - μ/(p2*(k2 - 1)) - α2 ))} ] = e^{(1 - k1) * ( μ*(t - t_bar) - μ/(p2*(k2 - 1)) - α2 ) } * d[dt: (1 - k1) * ( μ*(t_bar- t_bar) - μ/(p2*(k2 - 1)) - α2 ) ] = e^{(1 - k1) * ( μ*(t - t_bar) - μ/(p2*(k2 - 1)) - α2 ) } * (( + (1 - k1) * ( - μ/(p2*(k2 - 1)) - α2 ) + (1 - k1) * d[dt: μ*(t_bar- t_bar)] but d[dt: μ*(t_bar- t_bar)] = 0, so NUTS!! this is WRONG! (r2.11.26) d[dt: eTrm3(t)] = e^{(1 - k1) * ( μ*(t - t_bar) - μ/(p2*(k2 - 1)) - α2 ) } ????????????????????????????????? switch signs within exponential term = e^{(1 - k1) * ( μ*(t - t_bar) - μ/(p2*(k2 - 1)) - α2) } - e^{(1 - k1) * ( - μ/(p2*(k2 - 1)) - α2) } >> this is close to the correct functional form NOTE : In an earlier derivation, I directly substituted t_bar for t), which meant that the term μ/(p2*(k2 - 1)) did not appear. I obtained : d[dt: eTrm3(t)] = e^{ (1 - k1)* ( μ*(t - t_bar) - α2) } - e^{-(1 - k1) * α2 } This is essentially the same as the authors' p4c1h0.12 e^{ (1 - k1)* ( μ*(t - t_bar) - α2) } - e^{-(1 - k1) * α2 } but I am missing the factor p1/μ in the full expression, instead having only p1 also - this doesn't directly address what is needed for <--| |--> integrate eTrm3(t) from (r2.11.20) generic d[dt: e^x(t) ] = e^x(t) *d[dt: x(t) ] therefore ∫[dt, ti to tf: d[ds: e^x(s)] ] = ∫[dt, ti to tf: e^x(t) *d[dt: x(t) ] ] apply this to eTrm3(t) (r2.11.30) ∫[dt, t_bar to t: d[ds: eTrm3(s)] ] = ∫[dt, t_bar to t: eTrm3(t) *d[dt: eTrm3(t)] ] consider LHS of (r2.11.30) (r2.11.35) ∫[dt, t_bar to t: d[ds: eTrm3(s)] ] = eTrm3(t) - eTrm3(t_bar) = ∫[ds, t_bar to t: e^{(k1 - 1) *((- μ*(t - t_bar) + μ/(p2*(k2 - 1)) + α2 ))} ] - ∫[ds, t_bar to t_bar: e^{(k1 - 1) *((- μ*(t - t_bar) + μ/(p2*(k2 - 1)) + α2 ))} ] consider RHS of (r2.11.30) ∫[dt, t_bar to t: eTrm3(t) *d[dt: eTrm3(t)] ] from (r2.11.25) sub d[dt: eTrm3(t)] = ∫[dt, t_bar to t: eTrm3(t) *(( e^{(1 - k1) * ( μ*(t - t_bar) - μ/(p2*(k2 - 1)) - α2) } - e^{(1 - k1) * ( - μ/(p2*(k2 - 1)) - α2) } )) ] <--| Combine LHS (r2.11.12) & RHS (r2.11.16) to get full form (r2.11.40) ( X(t)*eTrm2(t) - X(t_bar) ) / (1 - k1) <= -p1 *∫[ds, t_bar to t: e^{(k1 - 1)*(-μ*(t - t_bar) + α2)} ] re-arrange, subbing eTrm2(t) (r2.11.41) X(t) *e^{ -(1 - k1)*∫[ds, t_bar to t: M(s)] } <= X(t_bar) + (1 - k1)*( -p1 *∫[ds, t_bar to t: e^{(k1 - 1)*(-μ*(t - t_bar) + α2)} ] ) ???????????????? This is the same as the authors' result : p3c2h0.9 (r2.11.25) X(t) *e^{ -(1 - k1)*∫[ds, t_bar to t: M(s)] } <= X(t_bar) - p1/μ*(( e^{(1 - k1)*( μ*(t - t_bar) - α2)} - e^{-(1 - k1)*α2} )) >> so p3c2h0.33 is confirmed?????????????? >> result is very similar to Part 1 with symbol changes p4c1h0.12 the last expression should be <= rather than = X(t) *e^{ -(1 - k1)*∫[ds, t_bar to t: M(s)] } <= X(t_bar) - p1/μ*(( e^{(1 - k1)*( μ*(t - t_bar) - α2)} - e^{-(1 - k1)*α2} )) >> Actually, the derivative expression (r2.11.17) seems to have been used, rather than the integral eTrm3(t)? ???????????????? 06Feb2023 = e^{(1 - k1) *( μ*(t - t_bar) - α2 ) } * (( + (1 - k1) *( - α2 ) + (1 - k1) *d[dt: μ*(t_bar- t_bar)] NOTE : In this derivation, I directly substituted t_bar for t0, which meant that the term μ/(p2*(k2 - 1)) did not appear. I obtained : d[dt: eTrm3(t)] = e^{ (1 - k1)* ( μ*(t - t_bar) - α2) } - e^{-(1 - k1) * α2 } This is essentially the same as the authors' p4c1h0.12 e^{ (1 - k1)* ( μ*(t - t_bar) - α2) } - e^{-(1 - k1) * α2 } but I am missing the factor p1/μ in the full expression, instead having only p1 also - this doesn't directly address what is needed for from (r2.11.25) sub d[dt: eTrm3(t)] = ∫[dt, t_bar to t: eTrm3(t) *(( e^{(1 - k1) * ( μ*(t - t_bar) - μ/(p2*(k2 - 1)) - α2) } - e^{(1 - k1) * ( - μ/(p2*(k2 - 1)) - α2) } )) ] # enddoc