I'm still pretty lost on the above, but thinking it's a bandlimited gaussian problem which means maybe should be in the frequency domain? Need to look up fourier transfers, haven't done those in 5 years.

Maybe it's just me, but I have no idea what half the terms in your formula mean. What is the range for the summation variables? What do we know about q, p, rho, N, P, W? It would be helpful if you stated what area this problem is from.

OK sorry I should have been more clear. This is related to coding/communications and calculating error bounds when transmitting a message.

The problem involves a system with bandwidth W and a band-limited Gaussian channel with q(x) = N(0,P/(2W)) and p(y|x) = N(x,N_o/2).

q is an arbitrary input weighting distribution
p(y|x) is the memoryless channel statistic
I think P is power and N_o is noise and rho is just a constant

I'm not sure how to deal with the range of summation- they are over all x and all y. I guess over all x is just within the bandwidth W and over all y is from 0 to the height (maybe P?).

It should be continuous so integrals instead of summations.



We know that if rho=1, the solution is = (1/2)ln(1+ P/(2*N_o*W)) and now need to solve for rho in general, but would be happy to understand how to find the solution given rho=1 and then could try to advance from there.

The constants are red herrings and can be pulled out. Looks like you can do basic substitutions to boil this down to basically e^-u^2 (e^-v^2)^(1/(1+p)) and then p is a constant and (x^a)^b is just x^ab so get rid of the 1/1+p term with a substitution too. I think this can all be evaluated directly but I'm on my phone without pen and paper, so, erm... Exercise for the reader?

Isn't p(y|x) = N(x,N_o/2) = just 1/sqrt(pi*N_o) because the Gaussian is

1/sqrt(pi*N_o)*e^([-(x-x)^2]/N_o) = 1/sqrt(pi*N_o)

Y-x not x-x

Been trying to evaluate for the case where rho=1. Here's my work. The N_o cancels early which is definitely wrong.

Hey I have an easy question.

X^5 = 0.7

Can someone show me how to solve this?



Zy

Sure. Rather than just tell you an answer though, when you ask for help solving this problem, are you asking something similar to "what is x?" ? Part of solving it is understanding what AN answer would look like and I just want to make sure we are on the same page.

I'll pm you it's confusing to me tbh.



Zy

OK I finally figured this out! One bad problem with the above is that the q(x) should have an x in the exponent, not a y (ugh). So the integral in terms of x is the most annoying and has to be complete-the-squared. I let a=P/W and b=N_o(1+rho) then combined the exponents and grinded through the x integral.

After that the y integral wasn't so bad since it was just one y^2 term. Of course made some small mistakes that took a while to catch like having an extra 2 in the denominator at the end. Eventually worked out and 3 pages of scratch paper + 2 real pages later got the answer! Thanks for the tips.

I need some help with the following sum. It appears to be geometric in nature and as such I'm trying to take it in the good ol a/(1-r) route. Problem is that I don't know how to work with the factorial term.

None of what you did is correct, mathematically. You can't pull the 2n! outside the sum, since the 2n term is not independent of the index n.

Your hint is to view this as the Taylor series of a trig function, evaluated at a particular value of x.

That explains a lot. I'm prepping for a midterm tomorrow by going through the chapter review, but we're one chapter short of Taylor/maclaurin series so I don't think I've seen this yet.

So I dug around a bit and see that that is the maclaurin expansion of Cosine evaluated at sqrt(pi)/2. Is that the whole name of the game in one of these series/sequence early calc courses? We just learn to recognize known series and manipulate the given series until it takes one of the known forms?

Edit: By the looks of the rest of the chapter there seems to be a bit more going on than just that.

cos[sqrt(pi)/3]. And no, you will learn how to represent functions as a Taylor series which is very important. The Maclaurin series is a special case of Taylor series. It is a power series in x^n, and each term is the nth derivative of f(x) evaluated at 0, divided by n!, times x^n. The first term uses the zeroth derivative or just f(0).

f(x) = f(0)/0! * x^0 + f'(0)/1! *x^1 + f''(0)/2! * x^2 + f'''(0)/3! * x^3 + ...

= f(0) + f'(0) * x + f''(0)/2! * x^2 + f'''(0)/3! * x^3 + ...

There's also an error term so that you know how close it is to f(x) after so many terms. The Taylor series is more general in that you have powers of (x-a), and you evaluate the derivatives at a. You had a Maclaurin series for cos[sqrt(pi)/3)]. For the cosine, the odd terms are zero because the derivatives all give sin(0) or -sin(0) = 0, but the even terms (starting from 0) are either cos(0) or -cos(0), hence the (-1)^n. Since the odd terms are zero, we have only even powers 2n and even values of the factorial (2n)!. Since you had pi^n/3^(2n), this was actually [sqrt(3)/3]^(2n), so x = sqrt(3)/3.

You can use this as a method of computing pi. Do a MacLaurin series for 4*arctan(x) and evaluate it at x=1. You get

pi = 4*(1 - 1/3 + 1/5 - 1/7 + ...)

It converges very slowly, so it's not a good way to compute pi.

1000 terms: 3.13959
10000 terms: 3.141393
100000 terms: 3.141573
1 million terms: 3.141591
exact: 3.141592653589793.....

There are much faster methods based on the arctan.

Haha i would solve the series using differential equations instead, trying to find a more general series with a parameter x that obviously creates then a function f(x) instead of say (Pi)^(1/2)/3 for that particular value of x, that i could then take derivatives of a couple of time to recreate itself (after keeping track of some terms) and find that the solution of that second order linear diff. eq. without first derivative is a trigonometric function (a*sinx +b*cosx+C) (whatever turns out to be here based on boundary conditions).

That way i wouldnt need to remember expansion of particular functions in order to pick the pattern and guess them like Taylor etc.

The advantage of this method of course is that it may solve more complex and more general problems when it works. (but of course it requires to have had diff.equations already, likely not the case here, although probably in their books they have had the Taylor expansions recently enough to be able to pick them up instantly as that cos whatever which is the easiest way if its that straightforward as here)

sqrt(pi)/3, not sqrt(3)/3.

Not to suggest that there's anything wrong with using the Taylor series for the arctan to compute pi. In fact, the guys that computed pi to a record of over a trillion decimal places used the Taylor series for the arctan, but they evaluated several arctans using formulas like these which converge very quickly rather than just arctan(1).

ok this question is making me feel pretty dumb. it should be pretty easy,but for some reason i cannot reason it out:

how many unique customers would i have per month (on average) in a gym class where 26 people attend each class, and this class happens 5 times per day (given that people on average attend 3 classes per week).

thanks in advance!

You get 26*5 people per day = 130 ppl per day = 7*130 ppl per week, but each person was counted 3 times so you get 130*7/3 distinct people per week, or ~303 distinct people per week.

Convert to ppl per month by whatever way makes you comfortable (I'd multiply by 52 and divide by 12).

Why would the people in Week 1 be different from the people in Week 2?

Lol, yeah good point.

for electrical interactions in chemical systems, how can you show that the following are equal:

E= (q+ * q-)/(4*pi*episilon*r)

and E = 332 kcal/mol * (q+ * q-)/r

i'm not really sure what to do here

If a pot is 120k, and a bet is 60k.

IF you can only chop the pot, at best. how do you figure out what % of time you need to be chopping to make the call correct?

Sorry, i've had this kinda theory stuff explained to me a lot and read ToP, but still lack basic understand.