A bunch of matricies could be sparse; for fluid dynamics, for instance, has a \(10^{6} \times 10^{6}\) matrix, but may only have \(7 \times 10^{6}\) non-zero entries; but the inverse could be fully dense!

In these cases, we almost never want to form a in inverse if needed.

If we really need to invert this, performing a LU-Factorization is going to be a very good idea.

• take audio

• calculate Mel Scale representation

• apply a series of Filter Banks which attenuates the input to highlight groups of frequencies

  

• we then run a discrete-cosine transform to obtain MFCCs, because much of the output results will still correlate with each other

Group Meetings

A spinal tap is a medical procedure whereby cerebralspinal fluid is collected by puncturing the lumbar; used to diagnose problems where biomakers from the brain are needed.

A stationary point of an ODE is considered “stable” if, at the stationary point \(y=c\), the function with initial condition.

If you start near a stationary point, the function will either diverge \(t\to \infty\) to that stationary point, or converge to a stationary point. Whether the functions done that makes it “stable”/“unstable”.

For an autonomous ODEs \(y’(t) = f(y(t))\), suppose \(y(t) = c\) is a stationary solutiona:

• \(c\) is stable (i.e. \(t\to \infty, y \to c\) for \(y_0 \approx c\)) if the graph of \(f\) near \(c\) crosses from positive to negative; that is, when \(f’( c) < 0\)
• \(c\) is unstable (i.e. \(t\to -\infty, y \to c\) for \(y_0 \approx c\)) if the graph of \(f\) near \(c\) crosses from negative to positive; that is, when \(f’(t) > 0\)
• \(c\) is semi-stable (i.e. stable on one side, unstable on the other) if the graph of \(f\) near \(c\) has the same sign on both sides; meaning \(f’( c) = 0\) and \(f’’( c)\neq 0\)
• if \(f’( c) = 0\) and \(f’’( c) \neq 0\), we are sad and should investigate more

away from zeros, the concavity of \(y(t)\) could be checked for \(f f’\). when its positive, \(y(t)\) is concave up; when its negative \(y(t)\) is concave down.