Methods

Question 1

shape exp functs

Answer 1

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

Question 2

shape of function
f(w) = 1-w -0.99exp(-2w)

Answer 2

f(0) = 1-0-0.9=0.01
between 0 and 1 hits (0.8,0)
n shape

Question 3

ln(ab)

Answer 3

lna + lnb

Question 4

ln(a/b)

Answer 4

lna - lnb

Question 5

ln0

Answer 5

undef

Question 6

e^0

Answer 6

1

Question 7

ln1

Answer 7

0

Question 8

variation of constants
y’+p(t)y=0

Answer 8

y’+p(t)y=0

1) Divide by y
2) integrate
3)take exponential

y=Cexp(- integral p(s).ds)

dont forget negative

Question 9

alogb

Answer 9

log b^a

Question 10

log graph

Question 11

variation of constants

Question 12

PGF of a Z_n from Z_n-1

Answer 12

Composition of PDF for Z_N -1 on Z_1

Question 13

E(T) = E(X_1 + X_2 +…)

Answer 13

The sum of the expectations for each one
E(x_1) + E(x_2) +

Question 14

The Pgf of a sum


Answer 14

Is the product of the Pgfs

Question 15

Death process

Answer 15

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

Question 16

Birth and death, process rates and probability

Answer 16

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

Question 17

Birth and death process when
μ=1

Answer 17

We have partial equation for PGF
Partial derivative with respect to time of G_1 =
−(G₁ −phi(G₁))

With G₁(z,o) =z

Question 18

Integral over [o,L]of cos(nπx/L) cos(mπx/L)

Answer 18

L/2 for m=n

Question 19

Integral from 0 to infinity of cos(KX) .dx

Answer 19

πδ(k)

Question 20

Integral from 0 to infinity, of cosine times an exponential 

Answer 20

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

Question 21

The integral from negative infinity to infinity of

Exp(ik (x±x_o) * exp(-k²/D. t) .dk

Answer 21

With respect to K

√(π/Dt) * exp( (x±x_o)²/4Dt)

Links to fundamental solution of 1D, Brownian motion, not quite though