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Theoretical Neuroscience

Chapter 5 Study Guide

Problem 1: Analytical Derivation of the Membrane Potential Update Rule
Problem Statement
Using the differential equation governing the membrane potential:
τV

dV
= V∞ − V,
dt

derive the discrete-time update rule for V (t + ∆t), assuming the input current Ie remains
constant during each time interval ∆t
...

Additionally, establish the numerical stability of this update method by evaluating fixed
points and boundedness as ∆t → ∞
...

τV

Integrating both sides over the interval from t to t + ∆t:
Z V (t+∆t)
Z t+∆t
dV
∆t
dt
V (t+∆t)
=
=
⇒ [− ln |V∞ − V |]V (t)

...

Rewriting this, we obtain the closed-form update rule:
V (t + ∆t) = V∞ + (V (t) − V∞ ) e−∆t/τV
...


Then,
εn+1 = e−∆t/τV εn ,

⇒

|εn+1 | = e−∆t/τV |εn |
...

• Linear Factor: Since e−∆t/τV ∈ (0, 1) for ∆t > 0, it follows that εn → 0 monotonically
...

• Boundedness: The sequence {Vn } remains bounded, as:
sup |Vn | ≤ max {|V∞ |, |V0 |} ,

∀ ∆t > 0
...


Problem 2: Conductance-Based Parameter Synthesis
Problem Statement
Given the definitions:

P
V∞ =

i

gi Ei +
P
i gi

Ie
A

,

cm
τV = P ,
i gi

assume the following conductance and biophysical parameters:
g1 = 2 µS/mm2 ,
cm = 1 nF/mm2 ,

g2 = 3 µS/mm2 , E1 = −70 mV,
A = 0
...


E2 = 50 mV,

Compute:
1
...
The membrane time constant τV
3
...
Steady-State Potential V∞
First compute the weighted sum of reversal potentials:
X
gi Ei = g1 E1 + g2 E2 = (2 µS/mm2 )(−70 mV) + (3 µS/mm2 )(50 mV) = 10 nA/mm2
...

A
0
...

i

Thus, we compute:
P
V∞ =

i

gi Ei +
P
i gi

Ie
A

=

110 nA/mm2
= 22 mV
...
Membrane Time Constant τV
Using the definition:
1 nF/mm2
cm
1 × 10−9 F
=
τV = P
=
= 2
...
2 ms
...
Dimensional Consistency of τV
Dimensional analysis yields:


F/mm2
cm
C/V/mm2
C
=
=
= = s
...
Both computed
quantities are valid in terms of units and physical interpretation
...
Then, demonstrate analytically that shunting reduces both the neuronal gain and the integration time
constant, and discuss the physiological implications for synaptic modulation
...

κ

3

Theoretical Neuroscience

Chapter 5 Study Guide

The governing equation simplifies:
EL + αEs + Rm Ie
τm dV
= −V +

...

dt
τm
τm
Rewriting in canonical linear form:
dV
V − V∗
=−
,
dt
τ∗

where

τ∗ =

τm
,
1+α

V∗ =

EL + αEs + Rm Ie

...

• Gain to Injected Current:
Rm
∂V∗
=
< Rm ,
∂Ie
1+α
hence the voltage response per unit current is divisively attenuated by the shunt
...


Physiological Interpretation
The term gs Ps models an activity-dependent synaptic conductance
...
Consequently, the neuron:
• integrates inputs over a shorter temporal interval (τ∗ ↓),
• produces smaller voltage excursions for identical levels of current or synaptic drive
(gain suppression)
...

4

Theoretical Neuroscience

Chapter 5 Study Guide

Problem 4: Average Release Probability under Facilitation
Problem Statement
Given the expression:
⟨Prel ⟩ =

P0 + fF rτP
,
1 + rfF τP

perform the following:
1
...

2
...
1,
fF = 0
...

3
...


1
...

Given a Poisson spike train of rate r, the mean-field equation for the expected release
probability is:
⟨P ⟩ − P0
d⟨P ⟩
=−
+ rfF (1 − ⟨P ⟩)
...

1 + rfF τP

2
...
1,

fF = 0
...
05 s,

the expression evaluates numerically over the range r ∈ [0, 100] Hz
...
1 toward 1, with diminishing marginal gains at
high frequency
...
Asymptotic Analysis and Convergence Bound
To prove convergence and establish a bound, rewrite:
1 − ⟨Prel ⟩ =

1 − P0
1 − P0
<

...

r
fF τ P

This establishes monotonic, asymptotically fast convergence toward full facilitation
...
That ⟨Prel ⟩ ∼

1
as r → ∞,
r

2
...
That this behavior contrasts with facilitation and limits the output at high presynaptic
rates
...
High-Rate Asymptotic Behavior
From the model:
⟨Prel ⟩ =

P0
= P0 [(1 − fD )rτP ]−1 [1 + o(1)]
1 + (1 − fD )rτP

as r → ∞
...

(1 − fD )τP r

2
...

1 + (1 − fD )rτP
(1 − fD )τP 1 + (1 − fD )rτP

In the limit r → ∞, this approaches:
lim r⟨Prel ⟩ =

r→∞

P0

...


3
...
Specifically, total output saturates at:
P0 [(1 − fD )τP ]−1 ,
a maximum rate determined solely by synaptic recovery dynamics
...
Thus,
facilitation amplifies synaptic throughput, whereas depression enforces output constancy—a
distinction central to neural gain control and synaptic adaptation
...
, and z is
updated at ∆t/2, 3∆t/2,
...


1
...

−(z − z∞ )/τz

Then the system is:
y˙ = (A + B)(y),

with exact flow E(t) = et(A+B)
...
Two Splitting Schemes
Scheme
Simultaneous (Lie)
Alternating (Strang)

Flow over ∆t
L(∆t) = e∆tA e∆tB
∆t
∆t
S(∆t) = e 2 A e∆tB e 2 A

BCH Expansion
2
L − E = ∆t2 [A, B] + O(∆t3 )
3
S − E = ∆t
([A, [A, B]] + 2[B, [A, B]]) + O(∆t4 )
12

Hence, the staggered ABA update is second-order accurate, one order higher than the
simultaneous AB step
...
Scalar Local-Truncation Check
Consider single-step updates for V :
Lie scheme:
Vn+1 = V∞ + (Vn − V∞ )e−∆t/τV
...

2

Using Taylor expansion:
e−x = 1 − x +

x2 x3
−
+ O(x4 ),
2
6

we find:
• Lie error:

∆t2
(Vn
2τV2

• Strang error:

∆t3
24τV2

− V∞ ) +
...


Identical conclusions follow for z, verifying that LTE matches the theoretical BCH expansion
...
Conclusion
1

The alternating scheme, structured as V n → V n+ 2 → z n+1 → V n+1 , cancels the dominant
∆t2 commutator error
...
This establishes second-order accuracy over the first-order Lie
method

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