Methods Flashcards
shape exp functs
y=e^x close to 0 for neg values, (0,1) then to inf
y=e^-x reflection in y axis
y=-e^x reflection in y
y=-e^-x reflection in x and y
shape of function
f(w) = 1-w -0.99exp(-2w)
f(0) = 1-0-0.9=0.01
between 0 and 1 hits (0.8,0)
n shape
ln(ab)
lna + lnb
ln(a/b)
lna - lnb
ln0
undef
e^0
1
ln1
0
variation of constants
y’+p(t)y=0
y’+p(t)y=0
1) Divide by y
2) integrate
3)take exponential
y=Cexp(- integral p(s).ds)
dont forget negative
alogb
log b^a
log graph
variation of constants
PGF of a Z_n from Z_n-1
Composition of PDF for Z_N -1 on Z_1
E(T) = E(X_1 + X_2 +…)
The sum of the expectations for each one
E(x_1) + E(x_2) +
The Pgf of a sum

Is the product of the Pgfs
Death process
Given death rate, μ
d p_k /dt =μ(k+1) p_{k+1} -μk p_k
Probability of earliest extinction time = P(T less than or equal to t) =Probability, all n cells have died at time. T
= (1-exp(−μt))^n
P(exactly one cell dies)
= N* (1-exp(−μt))* exp(-μt(n-1))
Remember N ways of choosing which one dies
Probability of certain number being alive uses ways of choosing that many
Birth and death, process rates and probability
Z_ Δt
Z_o= n with Δt converging to 0
{n with probability 1-(λ_n +μ_n)Δt
{n-1 with probability μ_nΔt
{n+1 with probability λ_n Δt
Birth and death process when
μ=1
We have partial equation for PGF
Partial derivative with respect to time of G_1 =
−(G₁ −phi(G₁))
With G₁(z,o) =z
Integral over [o,L]of cos(nπx/L) cos(mπx/L)
L/2 for m=n
Integral from 0 to infinity of cos(KX) .dx
πδ(k)
Integral from 0 to infinity, of cosine times an exponential 
Convert the cosine to an exponential using exp(ikx) use the fact that exponential for negative integrates the same and double the limits, while being able to halve the value as well as having two exponentials added together
Limits change to negative infinity to infinity
The integral from negative infinity to infinity of
Exp(ik (x±x_o) * exp(-k²/D. t) .dk
With respect to K
√(π/Dt) * exp( (x±x_o)²/4Dt)
Links to fundamental solution of 1D, Brownian motion, not quite though 
