A little math for ATOT!

[TecHNooB]

Sum from k = 0 to k = infinity of k*(0.5)^(k+1)

I haven't seen my calc book in quite a few semesters

 

[Tim]

Do your homework on your own.
(I don't know the answer)

 

[TecHNooB]

Originally posted by: theplaidfad
Do your homework on your own.
(I don't know the answer)

I'm stuck on this problem If only I was as smart as you

 

[Legendary]

My intuition says 1

 

[xanis]

about teefiddy

 

[Ticky]

1) Go to used bookstore
2) Buy Calc book
3) ....
4) Profit!

 

[PlasmaBomb]

As k gets very large 0.5^k+1 gets very small...

 

[ConstipatedVigilante]

This sounds like a limit problem...

 

[blinky8225]

It's 1. I'll edit it with a solution in a bit.

So the series is:
Let the lim(n) approach infinity
0 + (.5)^2 + 2 (.5)^3 + ... + n(.5)^(n+1) = (0.5)^2 * [1 + 2 (.5) + ... + n(.5)^(n-1)]

Let.....S = [1 + 2 (.5) + 3 (.5)^2 ... + n(.5)^(n-1)]
then 0.5 * S = [1 (.5) + 2 (.5)^2+..+(n-1)(0.5)^(n-1) + n(.5)^(n)]

See how they line up beautifully?

So S - 0.5*S = 0.5*S = 1 + (0.5) + (0.5)^2 + ... + (0.5)^(n-1) - n(0.5)^n

We can drop the last term because it equals 0 as n approaches infinity
So let 0.5 * S = T = 1 + (0.5) + (0.5)^2 + ... + (0.5)^(n-1)

This is just a geometric series.
So T * 0.5 = (0.5) + (0.5)^2 + ... + (0.5)^(n)
Now all the terms cancel out except for the first and last
So T - T * 0.5 = 0.5 * T = 1 - (0.5)^n.

Again we can drop the last term and solve for T.
T = 2. Thus, from there we find S = 4.

And finally going back, remember the complete sum is S * (0.5)^2 == 1. Yay, isn't math fun?!

 

[drinkmorejava]

Lol, why was I thinking this was an integral problem.

 

[Atomic Playboy]

It's a trick question; k is a letter, not a number. You can't calculate letters.

 

[blinky8225]

What class is this anyway? Probability? It looks a little like calculating the expectation of a geometric series.

 

[TecHNooB]

Originally posted by: blinky8225
It's 1. I'll edit it with a solution in a bit.

So the series is:
Let the lim(n) approach infinity
0 + (.5)^2 + 2 (.5)^3 + ... + n(.5)^(n+1) = (0.5)^2 * [1 + 2 (.5) + ... + n(.5)^(n-1)]

Let.....S = [1 + 2 (.5) + 3 (.5)^2 ... + n(.5)^(n-1)]
then 0.5 * S = [1 (.5) + 2 (.5)^2+..+(n-1)(0.5)^(n-1) + n(.5)^(n)]

See how they line up beautifully?

So S - 0.5*S = 0.5*S = 1 + (0.5) + (0.5)^2 + ... + (0.5)^(n-1) - n(0.5)^n

We can drop the last term because it equals 0 as n approaches infinity
So let 0.5 * S = T = 1 + (0.5) + (0.5)^2 + ... + (0.5)^(n-1)

This is just a geometric series.
So T * 0.5 = (0.5) + (0.5)^2 + ... + (0.5)^(n)
Now all the terms cancel out except for the first and last
So T - T * 0.5 = 0.5 * T = 1 - (0.5)^n.

Again we can drop the last term and solve for T.
T = 2. Thus, from there we find S = 4.

And finally going back, remember the complete sum is S * (0.5)^2 == 1. Yay, isn't math fun?!

That's so unintuitive and roundabout. I would never make those arbitrary decisions

O well, thanks

Class~ Probabilistic Methods in ECE

 

[pontifex]

not a clue...

 

[Fenixgoon]

here's my math problem:

show that (A x B) dot (B x C) x (C x A) = (A dot (B x C))^2

x = cross product
dot = dot product

using indicial notation, permutation tensor, Kronecker delta, and probably the delta-epsilon identity somewhere

 

[Q]

3

 

[blinky8225]

Originally posted by: TecHNooB

Originally posted by: blinky8225
It's 1. I'll edit it with a solution in a bit.

So the series is:
Let the lim(n) approach infinity
0 + (.5)^2 + 2 (.5)^3 + ... + n(.5)^(n+1) = (0.5)^2 * [1 + 2 (.5) + ... + n(.5)^(n-1)]

Let.....S = [1 + 2 (.5) + 3 (.5)^2 ... + n(.5)^(n-1)]
then 0.5 * S = [1 (.5) + 2 (.5)^2+..+(n-1)(0.5)^(n-1) + n(.5)^(n)]

See how they line up beautifully?

So S - 0.5*S = 0.5*S = 1 + (0.5) + (0.5)^2 + ... + (0.5)^(n-1) - n(0.5)^n

We can drop the last term because it equals 0 as n approaches infinity
So let 0.5 * S = T = 1 + (0.5) + (0.5)^2 + ... + (0.5)^(n-1)

This is just a geometric series.
So T * 0.5 = (0.5) + (0.5)^2 + ... + (0.5)^(n)
Now all the terms cancel out except for the first and last
So T - T * 0.5 = 0.5 * T = 1 - (0.5)^n.

Again we can drop the last term and solve for T.
T = 2. Thus, from there we find S = 4.

And finally going back, remember the complete sum is S * (0.5)^2 == 1. Yay, isn't math fun?!

That's so unintuitive and roundabout. I would never make those arbitrary decisions

O well, thanks

Class~ Probabilistic Methods in ECE

Well, just do it with variables by letting a1 = (0.5)^2 and r = (0.5). Then, you find a fairly simple formula, you can memorize and never do this again.

 

[TecHNooB]

Originally posted by: blinky8225

Originally posted by: TecHNooB

Originally posted by: blinky8225
It's 1. I'll edit it with a solution in a bit.

So the series is:
Let the lim(n) approach infinity
0 + (.5)^2 + 2 (.5)^3 + ... + n(.5)^(n+1) = (0.5)^2 * [1 + 2 (.5) + ... + n(.5)^(n-1)]

Let.....S = [1 + 2 (.5) + 3 (.5)^2 ... + n(.5)^(n-1)]
then 0.5 * S = [1 (.5) + 2 (.5)^2+..+(n-1)(0.5)^(n-1) + n(.5)^(n)]

See how they line up beautifully?

So S - 0.5*S = 0.5*S = 1 + (0.5) + (0.5)^2 + ... + (0.5)^(n-1) - n(0.5)^n

We can drop the last term because it equals 0 as n approaches infinity
So let 0.5 * S = T = 1 + (0.5) + (0.5)^2 + ... + (0.5)^(n-1)

This is just a geometric series.
So T * 0.5 = (0.5) + (0.5)^2 + ... + (0.5)^(n)
Now all the terms cancel out except for the first and last
So T - T * 0.5 = 0.5 * T = 1 - (0.5)^n.

Again we can drop the last term and solve for T.
T = 2. Thus, from there we find S = 4.

And finally going back, remember the complete sum is S * (0.5)^2 == 1. Yay, isn't math fun?!

That's so unintuitive and roundabout. I would never make those arbitrary decisions

O well, thanks

Class~ Probabilistic Methods in ECE

Well, just do it with variables by letting a1 = (0.5)^2 and r = (0.5). Then, you find a fairly simple formula, you can memorize and never do this again.

Which formula are you referring to? I enjoyed seeing the full solution btw

 

[maziwanka]

blinky's solution isn't unintuitive. you just needed to get that method into your "toolbag"

 

[MrPickins]

Originally posted by: blinky8225
It's 1. I'll edit it with a solution in a bit.

So the series is:
Let the lim(n) approach infinity
0 + (.5)^2 + 2 (.5)^3 + ... + n(.5)^(n+1) = (0.5)^2 * [1 + 2 (.5) + ... + n(.5)^(n-1)]

Let.....S = [1 + 2 (.5) + 3 (.5)^2 ... + n(.5)^(n-1)]
then 0.5 * S = [1 (.5) + 2 (.5)^2+..+(n-1)(0.5)^(n-1) + n(.5)^(n)]

See how they line up beautifully?

So S - 0.5*S = 0.5*S = 1 + (0.5) + (0.5)^2 + ... + (0.5)^(n-1) - n(0.5)^n

We can drop the last term because it equals 0 as n approaches infinity
So let 0.5 * S = T = 1 + (0.5) + (0.5)^2 + ... + (0.5)^(n-1)

This is just a geometric series.
So T * 0.5 = (0.5) + (0.5)^2 + ... + (0.5)^(n)
Now all the terms cancel out except for the first and last
So T - T * 0.5 = 0.5 * T = 1 - (0.5)^n.

Again we can drop the last term and solve for T.
T = 2. Thus, from there we find S = 4.

And finally going back, remember the complete sum is S * (0.5)^2 == 1. Yay, isn't math fun?!

Reminds me of some stuff we did in Discrete Math.

 

[blinky8225]

Originally posted by: TecHNooB

Originally posted by: blinky8225

Originally posted by: TecHNooB

Originally posted by: blinky8225
It's 1. I'll edit it with a solution in a bit.

So the series is:
Let the lim(n) approach infinity
0 + (.5)^2 + 2 (.5)^3 + ... + n(.5)^(n+1) = (0.5)^2 * [1 + 2 (.5) + ... + n(.5)^(n-1)]

Let.....S = [1 + 2 (.5) + 3 (.5)^2 ... + n(.5)^(n-1)]
then 0.5 * S = [1 (.5) + 2 (.5)^2+..+(n-1)(0.5)^(n-1) + n(.5)^(n)]

See how they line up beautifully?

So S - 0.5*S = 0.5*S = 1 + (0.5) + (0.5)^2 + ... + (0.5)^(n-1) - n(0.5)^n

We can drop the last term because it equals 0 as n approaches infinity
So let 0.5 * S = T = 1 + (0.5) + (0.5)^2 + ... + (0.5)^(n-1)

This is just a geometric series.
So T * 0.5 = (0.5) + (0.5)^2 + ... + (0.5)^(n)
Now all the terms cancel out except for the first and last
So T - T * 0.5 = 0.5 * T = 1 - (0.5)^n.

Again we can drop the last term and solve for T.
T = 2. Thus, from there we find S = 4.

And finally going back, remember the complete sum is S * (0.5)^2 == 1. Yay, isn't math fun?!

That's so unintuitive and roundabout. I would never make those arbitrary decisions

O well, thanks

Class~ Probabilistic Methods in ECE

Well, just do it with variables by letting a1 = (0.5)^2 and r = (0.5). Then, you find a fairly simple formula, you can memorize and never do this again.

Which formula are you referring to? I enjoyed seeing the full solution btw

Well for any sequence, where you have a1[1+ 2r + 3r^2 + 4r^3+...+(infinity)r^(infinity)], you will find that the it sums to a1 / (1-r)^2 if you solve in terms of variables. Only ff 0<r<1 of course.

a1 is the first term of the series. In this case it was (1/2)^2. r is the ratio which you multiply each term by, which in this case was (1/2). You don't have to actually factor out a1 like I did. For me, I just find it easier on the eyes.

Furthermore since you're taking probability, let r = q. a1 = pq. p^2 = (1 - r)^2. Then, you just have q/p if you're counting failures of course. It looks a little different if you're counting number of trials. In that case, it's just 1/p, even easier.