24
denote by £ the set of leaf indices, where a leaf is a branch with no children. We
suppose that each leaf is sealed at its distal end, i.e.,
∂xvb(0,t) = 0, be C. (2.24)
Initially the neuron is at rest, implying that ∂tvb{x, 0) = 0. We solve for the rest state
and denote it by υb{x), and similarly for the gating variables, which yields the initial
conditions
'(⅛(.τ, 0) = ⅝(χ∙) (2.25)
wbcf(x, 0) = wbcf (æ) ≡ wcftθθ (⅝(x)). (2.26)
With the rest state defined, it is easy to modify (2.17) to use current injection instead
of synaptic conductance. If we substitute the rest state t⅛(rr) for the Vb(x) in the
synaptic input term, then this is equivalent to directly injecting current into the cell,
which yields
ɪ sb
ʃɪɪŋ,b(*) ≡ τr Σ2 ‰(i)<5(≈ - Xbs)(υb(x) - Ebs). (2.27)
Z7Γ z—'
s=l
2.3.2 Linearizing the Cable Equation
The linearization process follows the same technique as that described in §2.2.
Consider (2.17) for branch b at its resting potential t⅛. If gbs = gbs + εgbs(‰ t), where
More intriguing information
1. A Rational Analysis of Alternating Search and Reflection Strategies in Problem Solving2. European Integration: Some stylised facts
3. Comparative study of hatching rates of African catfish (Clarias gariepinus Burchell 1822) eggs on different substrates
4. Density Estimation and Combination under Model Ambiguity
5. Estimating the Technology of Cognitive and Noncognitive Skill Formation
6. A multistate demographic model for firms in the province of Gelderland
7. The name is absent
8. Climate Policy under Sustainable Discounted Utilitarianism
9. 101 Proposals to reform the Stability and Growth Pact. Why so many? A Survey
10. The name is absent