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How I feel about Heidi Robert Richardson February 7, 2009 So if Heidi asked me how I felt about her this is what I would say. I would say 1. The derivative of the natural log of the homogenous solution to the differential equation y 00 +y. 2. The tennis term for the tangential component of acceleration (aT ) for a particle moving with the vector function r(t) = h2t + 1, 7π, 4 − 3ti at t = 2. 3. The maximmum likelihood estimation n independent observations of a poisson distribution with parameter λ. Let’s figure out what I just said!

Part 1 y 00 + y

=

0

To solve let’s try setting y equal to ert ert

y

=

y0

= rert

y

00

y 00 + y r2 ert + ert 2

r +1

= r2 ert = r2 ert + ert =

0

=

0

r2

= −1

r

= ±i

Therefore eit is a solution to my differential equation. The log of this is it and the derivative of that is simply i. So the answer to the first part of the riddle is i.

Part 2 The equation for the tangential component (aT ) of acceleration is v = r0 (t) and a = r00 (t) r(t)

= h2t + 1, 7π, 4 − 3ti 1

v·a kvk

where

r0 (t)

= h2, 0, 3i p 2 |r (t)| = (2 + 02 + 32 ) p (13) = 0

r00 (t) aT

= h0, 0, 0i h2, 0, 3i · h0, 0, 0i p = (13) = 0+0+0 = 0

The tennis term for this is of course love. So the solution to the second part of the riddle is love.

Part 3 Assuming independence, the joint pdf is f (x1 , x2 , ..., xn , λ)

=

n Y

f (xi , λ)

i=1

= = lnf (x1 , x2 , ..., xn , λ)

e−λ λxi i = 1n xi ! Pn xi −λ e λ i=1 Qn i=1 xi

Y

= −(nλ)lne +

n X

xi (lnλ) − ln

i=1

n Y

! xi

i=1

= So to find the M LE we take the derivative of the log of the function, set it equal to zero and solve for λ.

l0 (λ) n λ

Pn xi = −n + i=1 λ Pn i=1 xi = Pnλ i=1 xi = n

Which is simply the mean of the sample distribution which is an estimator of µ which is the greek letter for u. So in summary if we combine all the parts we get what I would say to Heidi if she asked me how I felt about her: i love you!!!!!!!!!!

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