Ok, being completely pissed off because of the lack of education received in my math class this year, I have a question for all you Math loving SDMB-ers: What is the formula (or formulae) or method of solving for infinite series. For example, if a man is on a a building 100 feet high, and he drops a ball which bounces half the distance of the drop each time (50 ft, 25 ft, 12.5 ft, etc) how far will the ball have travelled in all?
Well, this specific answer is 300 ft.
There is no formula for finding the sum of an infinite series in general. The example you gave, however, is a special kind called a geometric series, and there is a formula for this. The geometric series
a + ar + ar[sup]2[/sup] + ar[sup]3[/sup] + …
has sum a/(1-r) (r must be strictly between 1 and -1).
Your specific series would be:
100 + (50+50) + (25+25) + (12.5+12.5) + …
Discounting the first term (100), this is a geometric series with a = 50+50 = 100, r = 1/2.
The sum is 100 + a/(1-r) = 100 + 100/(1/2) = 300 feet.
Actually, there is a general method for summing infinite series whenever the terms can be mapped one-to-one with the set of integers, which is often the case.
Howwever, it’s extremely dificult to describe here, involves some pretty advanced mathematics, and it’s been 25 years since I’ve done it. Roughly, you define a complex function of Z = x + i*y (x and y are Cartesian coordinates, and i is the square root of -1) so that the function’s value is infinite at y = 0 and x = 1, 2, 3, … and the first term in the Taylor series of the function expanded around N + 0i is the Nth term of the series that you want to sum. Then all you need is to contour integrate F(Z) around any closed contour that includes all the integers inside, and Bob’s your uncle: the value of the integral is the sum of the series. It’s a “trivial” result from residue analysis. Weird as all get-out, too.
For some reason, few people do this for fun or idle curiosity.
You can also just mess with it until you can derive a closed form solution. For your particular problem, letting S be the sum:
S = 100 + 100 + 50 + 25 + 12.5 + …
2S = 200 + 200 + 100 + 50 + 25 + …
Subtracting the first equation from the second:
2S - S = 200 + 200 + 100 + 50 + … - 100 - 100 - 50 - 25 …
The left side is just S, and on the right side there is a negative term matching each positive term except for the 200 + 200 and one of the - 100’s. So all the terms on the right except those three cancel out, and
S = 200 + 200 - 100 = 300
JonF’s method works pretty well, provided that the contour integral you end up trying to evaluate is analytical-- A great many are not. Typically, the way that you determine that an integral is non-analytical, is to spend a couple of hours or so trying to solve it and/or find it or a close variant in a really big table of integrals. If you don’t succeed in this, well, then, obviously you made a bad choice for your complex function, so you should just start all over again. Simple, really, and not too far above the level of calculus that cogitoergosum knows… Heck, I was doing 'em already in my sophomore year of college! With a general solution that easy, who needs closed-form solutions?
With computers we all can get Ds in math. There’s a simple way to get the distance tne ball will travel after 200 bounces.
Private Sub Command1_Click()
Dim longDrop As Double Dim longTotalDrop As Double Dim intCount As Integer longDrop = 100 longTotalDrop = 100 Do While intCount = 200 longDrop = longDrop / 2 longTotalDrop = longTotalDrop + (longDrop * 2) intCount = intCount + 1 Loop
Hey, alright. I was beginning to think that I was the only person on the Internet who liked BASIC. Granted, QuickBASIC is my program of choice, but Visual will get you points in my book too. Welcome, ScratchDog. (I’m old enough to welcome people now, am I not?) Incidentally, though, you’ll either have to change that Do statement to something like this:
Do While intCount < 200
or just take the easy way, and make it a For loop:
For intCount = 0 to 200 longDrop = longDrop / 2 longTotalDrop = longTotalDrop + (longDrop * 2) Next intCount
I like this method the best, but the values should be:
S = 100 + 50 + 50 + 25 + 25 + 12.5 + 12.5 …
2S = 200 + 100 + 100 + 50 + 50 + 25 + 25 …
The ball falls 100 ft., bounces up 50, falls back the 50 ft., bounces up 25, and so on. It’s the same answer though: 300 ft total.
Both are correct. I just did some partial sums; the second and third terms in your series are the second term in my series, the fourth and fifth terms in your series are the third term in my series; and so on.
It’s a good method, but it often requires some insight or luck to obrtain a form in which two infinite portions cancel each other.
OK, let’s see how well we’ve absorbed the lessons of this thread.
You and I play a game. You flip a fair coin until you throw heads. Say this occurs on roll N. I then pay you N[sup]2[/sup] dollars.
How much should you pay me, to the nearest penny, for the game to be fair?
Please indicate how you arrived at your answer.