Calculating Derivative of $\sigma_N(t)$: A Justification

[Feynman]

:surprise:
Hello,
I've to calculate the derivate of :
$$\displaystyle \sigma_{N}(t):=\sum_{i=1}^{N}g_{i}(t)$$
and $$g_{i}(t)$$ verify the differential equation:

$$\displaystyle \frac{d g_i}{d t}(t) = <br /> \sum_{k=1}^{i-1} \frac{1}{k}K(k,i-k) g_{i-k}(t) g_k(t) - \sum_{l=1}^{\infty} \frac{1}{l}K(l,i) g_i(t)g_l(t)$$.

I've to justify:
$$\displaystyle \partial_{t}\sigma_{N}(t):= - \sum_{i,j\leqN;i+j>N}\frac{1}{j}K(j,i) g_i(t)g_l(t)$$
Thanks.

 

[Wong]

It sounds like just some kind of troublesome computation...You may like to "visualise" the problem by putting things in an array (much the same way like you put entries to the matrix). Note the similarities in the two sums in your differential equation (both include summand of the form (1/m*K(m, n)g_n(t)g_m(t)). Then put the summands in an array. Say for example if you are taking m to be the row index and n to be the column index then the sum SUM(l from 1 to infinity) {1/l*K(l, i)g_i(t)g_l(t)} would occupy the entire ith column of your array. From this you may easily "visualise" that your proposition is true.

 

[M98Ranger]

Hello there,

To justify the calculation of the derivative of $\sigma_N(t)$, we can use the chain rule for derivatives. Since $\sigma_N(t)$ is defined as a sum of functions $g_i(t)$, we can write it as:

$\displaystyle \sigma_N(t) = g_1(t) + g_2(t) + ... + g_N(t)$

Now, using the chain rule, we can calculate the derivative of $\sigma_N(t)$ as:

$\displaystyle \frac{d\sigma_N(t)}{dt} = \frac{d}{dt}(g_1(t) + g_2(t) + ... + g_N(t))$

$\displaystyle = \frac{dg_1(t)}{dt} + \frac{dg_2(t)}{dt} + ... + \frac{dg_N(t)}{dt}$

Substituting the given differential equation for each $g_i(t)$, we get:

$\displaystyle \frac{d\sigma_N(t)}{dt} = \sum_{i=1}^{N}\left(\sum_{k=1}^{i-1} \frac{1}{k}K(k,i-k) g_{i-k}(t) g_k(t) - \sum_{l=1}^{\infty} \frac{1}{l}K(l,i) g_i(t)g_l(t)\right)$

Next, we can rearrange the sums and use the property of summation to write it as:

$\displaystyle \frac{d\sigma_N(t)}{dt} = \sum_{i=1}^{N}\left(\sum_{k=1}^{i-1} \frac{1}{k}K(k,i-k) g_{i-k}(t) g_k(t)\right) - \sum_{l=1}^{\infty}\left(\sum_{i=1}^{N} \frac{1}{l}K(l,i) g_i(t)\right)g_l(t)$

Finally, we can use the definition of $\sigma_N(t)$ to write it as:

$\displaystyle \frac{d\sigma_N(t)}{dt} = \sum_{i=1}^{N}\left(\sum_{k=1}^{i-1} \frac{1}{k}K(k,i-k) g_{i