Arula Ratnakar

@arula-ratnakar

Computational Neuroscience PhD student at Boston University (advisors Cynthia Bradham & Gabe Ocker). Mathematically modeling embryonic neurodevelopment. I also write trippy scifi & mathfiction. Ignyte Award Finalist 2025 6 stories in Clarkesworld

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Arula Ratnakar@arula-ratnakar· 3d

I take a lot of notes, I have an 80 page single spaced document of math notes from my PhD so far 😅:

Arula Ratnakar@arula-ratnakar· 4d

Studying for my oral quals, reviewing Turing stability analysis and reaction diffusion systems today, next I’ll review MSRDJ path integrals and compare them to a kind im learning now that uses creation and annihilation operators called Doi Peliti.

Selfie, tired lol. Wearing black t shirt

Arula Ratnakar@arula-ratnakar· 10d

I used to make so many paintings and drawings by hand in architecture school. It’s been so many years. Now I’m doing something I don’t even know if I’ll be good at yet.

Drawing in colored pencil of a community kitchen

Drawing in colored pencil of a wisteria tunnel

Painting in watercolor and poster paints of a straw roof hut by a dirt path in lush greenery

Arula Ratnakar@arula-ratnakar· 10d

im actually really excited for this QFT class. It's going to be so strange learning about all of this stuff in a non-statistical-field-theory context.

field theory with applications to high-energy and condensed matter physics. Topics include field equations and quantization of many-body systems; Greens function and linear response theory, S-matrix and scattering theory, path integration, perturbation expansions and the Feynman rules, renormalization and effective field theories, epsilon expansion and critical exponents.

Arula Ratnakar@arula-ratnakar· 22d

All of the TAs have to do this but my path is actually so wild. this is one of my first slides lol

Slide showing a drawing with text High School to College: failed 9th grade biology, spent all my time making this 10ft drawing so my gpa sucked but I got good math grades and drew buildings well, so I got into Carnegie Mellon architecture school.

Arula Ratnakar@arula-ratnakar· 25d

Gosh so much is piling up lol My quals are august 18 💀

Arula Ratnakar@arula-ratnakar· 25d

One of my favorite parts of writing is designing my story buildings' architecture. I did, in fact, enjoy imagining and designing buildings when I was in architecture school haha. I can still totally do that in my scifi. I'm designing a really cool building for this story.

I turn on the light in the stairwell as we descend. It bounces off red bricks, the warm wooden paneling on the stairs. Art-glass windows providing glimpses of the living room make the descent colorful and kaleidoscopic, as carefully placed light sources in the living room shine through the stained glass panes onto us.

Arula Ratnakar@arula-ratnakar· 26d

last year at this time I was intimidated by a simple n-dimensional gaussian integral w/ a source term and now I know all of this. wild how much you can learn in a year. (I put all of my notes of what i've been studying wrt math into one document. ik i need to organize the "preliminaries" better)

toc of what is in all of my notes, 83 pages written so far!

Arula Ratnakar@arula-ratnakar· Jul 12

Got this at Readercon. It’s PKD’s first novel, wasn’t published until after his death and it isn’t scifi. Very intrigued to read it. his work is a huge influence on my writing. A Scanner Darkly changed my philosophy for why I write, who I‘m writing for & what I should capture about humanity.

Arula Ratnakar@arula-ratnakar· Jul 8

So, overall, we have our final mean field equations after taking both the N to infinity limit to get sparse connectivity, and taking the K to infinity limit to get a large amount of inputs to the neurons. That's all for today, hope you enjoyed! ☺️ (19/19)

$\begin{align*} \text{population activity rate equations:}\\ \text{General equation: $\boxed{\tau_k \frac{d}{dt}m_k(t) = - m_k(t) + H\left(\frac{-u_k}{\sqrt{\alpha_k}}\right)}$}\\ \text{where, again:}\\ H(z) = \int_z^\infty \frac{dx}{\sqrt{2 \pi}}e^{-x^2/2}\\ \boxed{\begin{align*} \tau_E \frac{d}{dt}m_E(t) &= -m_E(t) + H\left(\frac{-u_E}{\sqrt{\alpha_E}}\right)\\ \tau_I \frac{d}{dt}m_I(t) &= -m_I(t) + H\left(\frac{-u_I}{\sqrt{\alpha_I}}\right) \end{align*}}\\ \text{Input drive $u_k$ equations:}\\ \text{General equation: $\boxed{u_k = (J_{k0}m_0 + J_{kE}m_E + J_{kI}m_I) \sqrt K - \theta_k}$}\\ \text{where as they said in the model definition section, E here is the external input }\\ \text{to excitatory neurons and I is the external input to inhibitory neurons. }\\ \text{(Remember they defined $J_{kE}$ (so $J_{EE}$ and $J_{IE}$) as fixed as 1 in the model }\\ \text{definition section and they renamed the remaining $J_{kI}$ as $-J_{EI} = J_{E}$ and $-J_{II} = J_I$}\\ \boxed{\begin{align*} u_E &= (Em_0 + m_E - J_Em_I)\sqrt{K} - \theta_E\\ u_I &= (Im_0 + m_E - J_Im_I)\sqrt{K} - \theta_I\\ \end{align*}}\\ \text{Variance $\alpha_k$ equations:}\\ \text{General equation: $\boxed{\alpha_k = (J_{kE})^2m_E + (J_{kI})^2m_I}$}\\ \boxed{\begin{align*} \alpha_E &= m_E + J_E^2m_I\\ \alpha_I &= m_E + J_I^2m_I\\ \end{align*}}\\ \end{align*}$

Arula Ratnakar@arula-ratnakar· Jul 8

We can just solve for the equilibrium value for this now, which will become really relevant for other sections in the paper. (18/n)

$Note that at steady state, when $\frac{dm_k}{dt} = 0$ you can solve to get the steady state equilibrium: $$0 = - m_{k_{ss}} + H\left(\frac{-u_k}{\sqrt{\alpha_k}}\right) \rightarrow \boxed{m_{k_{ss}} = H\left(\frac{-u_k}{\sqrt{\alpha_k}}\right)}$$$

Arula Ratnakar@arula-ratnakar· Jul 8

This update to F gives us, now, the mean field equation for the population activity rates: (17/n)

$\begin{align*} \tau_k \frac{d}{dt}m_k(t) &= -m_k(t) + F_k(m_E(t), m_I(t))\\ &\text{so replacing $F_k$ with what we just derived}\\ \Aboxed{\tau_k \frac{d}{dt}m_k(t) &= - m_k(t) + H\left(\frac{-u_k}{\sqrt{\alpha_k}}\right)}\\ &\text{where, again:}\\ H(z) &= \int_z^\infty \frac{dx}{\sqrt{2 \pi}}e^{-x^2/2} \end{align*}$

Arula Ratnakar@arula-ratnakar· Jul 8

We can define a function H(z) which will allow us to simplify the expression and get the F from the paper. (16/n)

$H(z) = P(X \ge z) = \int_z^\infty \frac{1}{\sqrt{2\pi}}e^{-t^2/2}dt$$ The neuron will fire whenever $u_k + \sqrt{\alpha_k}x \ge 0$ Isolating x algebraically gives us: $$x \ge \frac{-u_k}{\sqrt{\alpha_k}}$$ The integral over Dx is sweeping over the Gaussian random noise x, and we have our boundary for what x needs to be greater than, so we can replace the entirety of that integral with $H\left(\frac{-u_k}{\sqrt{\alpha_k}}\right)$. This gives us: $$\boxed{F_k(m_E, m_I) = \int Dx \Theta(u_k + \sqrt{\alpha_k}x) = H\left(\frac{-u_k}{\sqrt{\alpha_k}}\right)}$

Arula Ratnakar@arula-ratnakar· Jul 8

Again, as the K to infinity limit occurs, the probabilities became Gaussian instead of Poisson, and the input into the Heaviside step function becomes that mean u_k equation plus the fluctuations around the mean. (15/n)

$$$\boxed{F_k(m_E, m_I) = \int Dx \Theta(u_k + \sqrt{\alpha_k}x)}$$ Where $Dx = dx \exp (-x^2/2) / \sqrt{2 \pi}$\\$

Arula Ratnakar@arula-ratnakar· Jul 8

Before taking the K to infinity limit, we had the following: (14/n)

$F_k(m_E, m_I) = \sum_{n_1, n_2 = 0}^\infty p_1 (n_1)p_2(n_2) \Theta \left(\sqrt K J_{k0}m_0 + \sum_l \frac{J_{kl}}{\sqrt K} n_l - \theta_k \right)$$ $$p_l(n) = \frac{(m_lK)^n}{n!}e^{-m_lK}$

Arula Ratnakar@arula-ratnakar· Jul 8

So we have our alpha equation now too. (13/n)

$So the term representing the fluctuations around the mean, $\alpha_k$ is: $$\boxed{\alpha_k = (J_{kE})^2m_E + (J_{kI})^2m_I}$$$

Arula Ratnakar@arula-ratnakar· Jul 8

Here we can utilize that property when we derive the terms of our alpha equation. Remember, alpha represents the fluctuations around the mean input. (12/n)

$The weight of the synapse is $\frac{J_{kl}}{\sqrt{K}}$\\ The variance of the input as we know is:\\ $\text{Var}(n_l) = Km_l$\\ So all together: $$\text{Var}\left(\frac{J_{kl}}{\sqrt{K}}n_l\right) = $$ $$\left(\frac{J_{kl}}{\sqrt{K}}\right)^2 \text{Var}(n_l) = \frac{J_{kl}^2}{K}Km_l = J_{kl}^2m_l$$$

Arula Ratnakar@arula-ratnakar· Jul 8

Now this represents the average input, but there is also going to be a variance of that input determining fluctuation around the mean. Let's think about that now. Note the property in statistics where if you scale a random variable X by some constant C, the variance scales like so: (11/n)

$\text{Var}(CX) = C^2\text{Var}(X)$

Arula Ratnakar@arula-ratnakar· Jul 8

This is how that turns into the K to infinity version: (10/n)

$As K goes to infinity, the average number of active inputs $<n_E> = Km_E$ and $<n_I> = Km_I$ So we can rewrite the equation as: $$<u_k^i(t)> = \sqrt{K}J_{k0}m_0 + \frac{J_{kE}}{\sqrt{K}}Km_E + \frac{J_{kI}}{\sqrt{K}}Km_I - \theta_k$$ This simplifies to the equation: $$\boxed{u_k = (J_{k0}m_0 + J_{kE}m_E + J_{kI}m_I) \sqrt K - \theta_k}$$$

Arula Ratnakar@arula-ratnakar· Jul 8

Remember the microscopic equation describing what's entering neuron i in population k: (9/n)

$u_k^i(t) = \sqrt K J_{k0}m_0 + \frac{J_{kE}}{\sqrt K} n_E(t) + \frac{J_{kI}}{\sqrt K} n_I(t) - \theta_k$

Arula Ratnakar@arula-ratnakar· Jul 8

This is what they say we get from taking the K to infinity limit. Let's go through this explicitly and derive it. (8/n)

$So in the limit $K \rightarrow \infty$ we are given that: $$F_k(m_E, m_I) = \int Dx \Theta(u_k + \sqrt{\alpha_k}x) = H\left(\frac{-u_k}{\sqrt{\alpha_k}}\right)$$ Where $Dx = dx \exp (-x^2/2) / \sqrt{2 \pi}$\\ From the above statistics of $n_l$ one obtains that the average input, relative to threshold, $u_k$ into a cell of population k given by: $$u_k = (J_{k0}m_0 + J_{kE}m_E + J_{kI}m_I) \sqrt K - \theta_k$$ and standard deviation of the input $\alpha_k$\\ $$\alpha_k = (J_{kE})^2m_E + (J_{kI})^2m_I$$$

Arula Ratnakar@arula-ratnakar· Jul 8

So even though these neurons' activities are either 0 or 1, the sum of tons of them loses the binary characteristic). The mean and variance of this Gaussian distribution are going to be the mean and variance of the poisson from before: (7/n)

$$$<n_k> = \text{Var}(n_k) = Km_k$$ (they use this notation for variance: $< (\delta n_k)^2>$ but I prefer Var).\\$

Arula Ratnakar@arula-ratnakar· Jul 8

Here are the standard deviations of n_E and n_I. I think there is a typo in the paper, so I corrected the typo and explained below in this pic: (4/n)

$(I think there is a typo in the text and that it should be $\text{std} (n_k) = \sqrt{m_k K}$ (they wrote $\text{std}(n_k) = m_kK$ in the text) because since the inputs are Poisson distributed to be $p_l(n) = \frac{(m_lK)^n}{n!}e^{-m_l K}$ we would expect the mean to be equal to the variance, and the mean is $m_kK$.\\ So the CORRECT standard deviations of $n_E$ and $n_I$ are given by: $$\text{std}(n_k) = \sqrt{m_k K}$$$

Arula Ratnakar@arula-ratnakar· Jul 6

So after taking the N to infinity limit, we now have the following mean field equations for the population activity levels for finite K. Now the next thing to do is take the limit as K goes to infinity, which I'll go over tomorrow. That's all for today, hope you enjoyed! ☺️ (34/n)

$\boxed{\tau_k \frac{d}{dt}m_k(t) = -m_k(t) + F_k(m_E(t), m_I(t))}$$ $$\boxed{F_k(m_E, m_I) = \sum_{n_1, n_2 = 0}^\infty p_1 (n_1)p_2(n_2) \Theta \left(\sqrt K J_{k0}m_0 + \sum_l \frac{J_{kl}}{\sqrt K} n_l - \theta_k \right)}$$ $$\boxed{p_l(n) = \frac{(m_lK)^n}{n!}e^{-m_lK}}$

Arula Ratnakar@arula-ratnakar· Jul 6

Here is why, explicitly: (33/n)

$\begin{align*} p_l(n) &= \sum_{s = n}^{\infty} \frac{K^se^{-K}}{s!}\left(\frac{s!}{n!(s-n)!}\right)m_l^n(1-m_l)^{s-n}\\ \text{the two s! cancel}\\ p_l(n) &= \sum_{s = n}^{\infty} \frac{K^s}{n!(s-n)!}e^{-K}m_l^n(1-m_l)^{s-n}\\ &= \frac{m_l^n}{n!}e^{-K}\sum_{s=n}^\infty \frac{K^s}{(s-n)!}(1-m_l)^{s-n}\\ &=\frac{m_l^n}{n!}e^{-K}\sum_{s=n}^\infty \frac{K^n K^{s-n}}{(s-n)!}(1-m_l)^{s-n}\\ &= \frac{(m_lK)^n}{n!}e^{-K}\sum_{s = n}^\infty \frac{[K(1-m_l)]^{s-n}}{(s-n)!}\\ &\text{Let q = s-n so at lower bound s=n, q = 0 so we can rewrite the sum as:}\\ &\sum_{q=0}^\infty \frac{[K(1-m_l)]^q}{q!}\\ &\text{now we just use the property $\sum_{q=0}^\infty \frac{x^q}{q!}=e^x$}\\ &\sum_{q=0}^\infty \frac{[k(1-m_l)]^q}{q!}= e^{K(1-m_l)}\\ &\text{putting everything together}\\ p_l(n)&=\frac{(m_lK)^n}{n!}e^{-K}e^{K(1-m_l)} = \frac{(m_lK)^n}{n!}e^{-K}e^{K}e^{-m_lK}\\ \Aboxed{p_l(n) &= \frac{(m_lK)^n}{n!}e^{-m_lK}} \end{align*}$

Arula Ratnakar@arula-ratnakar· Jul 6

Again, the average number of connections a cell receives from a population is K. On average each of these synapses has a probability m_l to be active, so (again, here n represents the number of active inputs a neuron is receiving from presynaptic population l): (32/n)

$p_l(n) = \sum_{s = n}^ \infty \frac{K^s}{s!}e^{-K}\binom{s}{n} m_l^n(1 - m_l)^{s-n} = \frac{(m_lK)^n}{n!}e^{-m_lK}$

Arula Ratnakar@arula-ratnakar· Jul 6

When you take the limit of the binomial distribution under these conditions the binomial distribution becomes a Poisson Distribution: (31/n)

$When you take the limit of the binomial distribution under the conditions $N_l \rightarrow \infty$ and $p \rightarrow 0$ while $N_l \cdot p = K$ the binomial distribution becomes a Poisson Distribution:\\ $$\lim_{N_l \rightarrow \infty} \binom{N_l}{s}\left(\frac{K}{N_l}\right)^s\left(1 - \frac{K}{N_l}\right)^{N_l - s} = \frac{K^se^{-K}}{s!}$$ (Here's how) \begin{align*} P(s) &= \binom{N_l}{s}\left(\frac{K}{N_l}\right)^s\left(1 - \frac{K}{N_l}\right)^{N_l - s}\\ &= \frac{N_l!}{s!(N_l - s)!}\left(\frac{K}{N_l}\right)^s\left(1 - \frac{K}{N_l}\right)^{N_l}\left(1 - \frac{K}{N_l}\right)^{ - s}\\ &=\frac{N_l(N_l-1)(N_l-2)\cdot \cdot \cdot(N_l-(s - 1)}{s!}\frac{K^s}{N_l^s}\left(1 - \frac{K}{N_l}\right)^{N_l}\left(1 - \frac{K}{N_l}\right)^{ - s}\\ &= \frac{K^s}{s!}\left(\frac{N_l(N_l - 1)(N_l-2) \cdot \cdot \cdot (N_l - (s-1)}{N_l^s}\right)\left(1 - \frac{K}{N_l}\right)^{N_l}\left(1 - \frac{K}{N_l}\right)^{ - s}\\ &=\frac{K^s}{s!}\left(\frac{N_l}{N_l}\frac{(N_l - 1)}{N_l}\frac{(N_l-2)}{N_l} \cdot \cdot \cdot \frac{(N_l - (s-1)}{N_l}\right)\left(1 - \frac{K}{N_l}\right)^{N_l}\left(1 - \frac{K}{N_l}\right)^{ - s}\\ \text{Taking the limit as N goes to infinity for each term}\\ \lim_{N_l \rightarrow \infty}\left(\frac{N_l}{N_l}\frac{(N_l - 1)}{N_l}\frac{(N_l-2)}{N_l} \cdot \cdot \cdot \frac{(N_l - (s-1)}{N_l}\right) &= \left(1\cdot 1 \cdot 1 \cdot \cdot \cdot 1\right) = 1\\ \lim_{N_l \rightarrow \infty}\left(1 - \frac{K}{N_l}\right)^{N_l} &= e^{-K}\\ \lim_{N_l \rightarrow \infty}\left(1 - \frac{K}{N_l}\right)^{-s} &= (1 - 0)^{-s} = 1\\ \text{All together this leaves us with}\\ \Aboxed{\lim_{N_l \rightarrow \infty} P(s) &= \frac{K^se^{-K}}{s!}} \end{align*}$

Arula Ratnakar@arula-ratnakar· Jul 6

if p is the probability of a cell connecting to a target neuron, 1-p is the probability of no connection. Since we do this for each neuron up to the number of neurons in the presynaptic population, aka N_l times, the probability of ending up with s synapses follows the Binomial distribution (29/n)

$\text{Prob(s)} = \binom{N_l}{s}p^s(1-p)^{N_l - s}$

Arula Ratnakar@arula-ratnakar· Jul 6

This can also be written in the language the paper uses, where they say that the subscript E is interchangeable with 1 and I (uppercase i) is interchangeable with 2, and using their summation notation for the sum. This is how the paper writes what I put above: (26/n)

$\boxed{F_k(m_E, m_I) = \sum_{n_1, n_2 = 0}^\infty p_1 (n_1)p_2(n_2) \Theta \left(\sqrt K J_{k0}m_0 + \sum_l \frac{J_{kl}}{\sqrt K} n_l - \theta_k \right)}$

Arula Ratnakar@arula-ratnakar· Jul 6

F_k denotes the probability the updating cell at time t will be in an updated *active* state, which is given by: (25/n)

$F_k(m_E, m_I) = \sum_{n_E, n_I = 0}^\infty p_E (n_E)p_I(n_I) \Theta \left(u_k^i(t)\right)$$ Remember we derived what $u_k^i(t)$ was just a bit above? Let's plug it in. $$F_k(m_E, m_I) = \sum_{n_E, n_I = 0}^\infty p_E (n_E)p_I(n_I) \Theta \left(\sqrt K J_{k0}m_0 + \frac{J_{kE}}{\sqrt K} n_E(t) + \frac{J_{kI}}{\sqrt K} n_I(t) - \theta_k \right)$