[RAILhead]

Okay, I'm somewhat embarrassed to be asking this (not really), but I'm trying to figure something out. Here's the scenario...

1 person donates $1. Within 24 hours, they get 5 more people to donate $1. The next 24 hours, those 5 new people each get 5 more people to donate $1, etc., etc.

How many days will it take to raise 1 million dollars?

[Ghoser777]

That looks like a geometric series question.

Day 0: $1
Day 1: $5
Day 2: $25
Day n: $5^n

So want you want is the sum of n terms in this sequence exceeding one million dollars. The formula for a geometric series is:

sum = (a(1-r^n)/(1-r)

where a is the first term, n is the number of terms, and r is the rate.

So therefore

1000000 = (1-5)^n / (1 - 5)
1000000 = (1-5^n) / (-4)
-4000000 = 1 - 5^n
4000001 = 5^n
log(4000001) = nlog(5)
n = log(4000001) / log(5)
n = 9.44

So therefore, 10 days.

[Railroader]

Day 9 or 10

1. $1.00
2. $5.00
3. $25.00
4. $125.00
5. $625.00
6. $3,125.00
7. $15,625.00
8. $78,125.00
9. $390,625.00
10. $1,953,125.00

This is all wrong!

[Railroader]

... or day 11 as the mathmatician-like guy above me says. I used MS Excel.

[nonhuman]

Well, it's just a simple function of y = 5^x, so it will take log(1e6)/log(5) days.

Edit: Google tells me that's 8.58405935 days.

Edit 2: It's worth mentioning that my day scale is zero-indexed, so 8.5 days duration lands you approximately halfway through day 10.

[Railroader]

It's not called a pyramid scheme for nuttin'.

1. $1.00
2. $5.00
3. $25.00
4. $125.00
5. $625.00
6. $3,125.00
7. $15,625.00
8. $78,125.00
9. $390,625.00
10. $1,953,125.00
11. $9,765,625.00
12. $48,828,125.00
13. $244,140,625.00
14. $1,220,703,125.00
15. $6,103,515,625.00
16. $30,517,578,125.00
17. $152,587,890,625.00
18. $762,939,453,125.00
19. $3,814,697,265,625.00
20. $19,073,486,328,125.00
21. $95,367,431,640,625.00
22. $476,837,158,203,125.00
23. $2,384,185,791,015,620.00
24. $11,920,928,955,078,100.00
25. $59,604,644,775,390,600.00
26. $298,023,223,876,953,000.00
27. $1,490,116,119,384,770,000.00
28. $7,450,580,596,923,830,000.00
29. $37,252,902,984,619,100,000.00
30. $186,264,514,923,096,000,000.00
31. $931,322,574,615,478,000,000.00
32. $4,656,612,873,077,390,000,000.00
33. $23,283,064,365,387,000,000,000.00
34. $116,415,321,826,935,000,000,000.00

As well as this.

[Railroader]

OK, here's my final set of (possibly) wrong numbers:

1. $1.00
2. $6.00
3. $31.00
4. $156.00
5. $781.00
6. $3,906.00
7. $19,531.00
8. $97,656.00
9. $488,281.00
10. $2,441,406.00

[RAILhead]

Thanks Ghoser.

[brassplayersrock²]

yes, thank you ghoser.

[RAILhead]

How do you set this up in Excel?

[ghporter]

Originally Posted by Railroader 

OK, here's my final set of (possibly) wrong numbers:

1. $1.00
2. $6.00
3. $31.00
4. $156.00
5. $781.00
6. $3,906.00
7. $19,531.00
8. $97,656.00
9. $488,281.00
10. $2,441,406.00

Yep. Ghoser left out adding the previous day's sums.

[CharlesS]

Uh, Ghoser is actually the only one in this thread who didn't forget to add the previous day's sums.

[Ghoser777]

Originally Posted by ghporter 

Yep. Ghoser left out adding the previous day's sums.

No I didn't. I used the geometric series formula that adds all the terms from the beginning to the end. If you're not convinced, you can try my formula for a couple days:

1(1 - 5^1)/ (1-5) = $1
1(1 - 5^2)/ (1-5) = $6 (+5)
1(1 - 5^3) / (1-5) = $31 (+25)

http://en.wikipedia.org/wiki/Geometric_series#Formula

[ghporter]

Originally Posted by Ghoser777 

No I didn't. I used the geometric series formula that adds all the terms from the beginning to the end. If you're not convinced, you can try my formula for a couple days:

1(1 - 5^1)/ (1-5) = $1
1(1 - 5^2)/ (1-5) = $6 (+5)
1(1 - 5^3) / (1-5) = $31 (+25)

http://en.wikipedia.org/wiki/Geometric_series#Formula

Your first post shows 1, 5, 25..., which does not match the geometric series. That's what I was referring to.

[nonhuman]

Originally Posted by CharlesS 

Uh, Ghoser is actually the only one in this thread who didn't forget to add the previous day's sums.

Yeah, but those have no substantive effect on the answer.

[CharlesS]

Originally Posted by ghporter 

Your first post shows 1, 5, 25..., which does not match the geometric series. That's what I was referring to.

His first post also immediately goes on to bust out the formula that adds up all the previous days.

1(1-5^10) / (1-5) = -9765624 / -4 = 2,441,406

Sheesh, I didn't find it that difficult to understand.

[Ghoser777]

Originally Posted by ghporter 

Your first post shows 1, 5, 25..., which does not match the geometric series. That's what I was referring to.

I thought I was being clear, but I guess not. I should have said that those were the new donations for each day, not the cumulative total.

[ghporter]

The first post starts out wrong. I didn't go past that. None of my professors would bother to look past a prima facie error of that magnitude. It should have said something like:

Day 0: $1
Day 1: $5 + $1 from Day 0 = $6
Day 2: $25 + $6 from Day 1 = 31
Day n: $5^n + sum from Day n-1

Looking back, sure the whole geometric series is there. But the first thing Ghoser did was say it looked like a geometric series and then present something OTHER THAN a gemetric series.

Maybe I'm being all curmudgeonly, but it wasn't right from the start, and that's what stands out.

[Chuckit]

It wasn't wrong — it was a totally accurate tally of each day's take. It just wasn't counting what you wanted it to count.

[RAILhead]

I still don't get it. I wish I understood this stuff.

[Ghoser777]

Well, I'm a Math teacher - and I'd accept my own work so =P.

Oh, and I was demonstrating the terms of a geometric series (a geometric sequence to be exactly precise). Which is exactly what I say in my post that I'm doing (summing up the terms in the sequence).

[Ghoser777]

Originally Posted by RAILhead 

I still don't get it. I wish I understood this stuff.

I can explain as much or as little as you want. What level math class is this for?

[RAILhead]

Let's just say...Level None.

[Ghoser777]

Well... then doing manually is probably the best way (with excel). I know how to do formulas in excel, but I don't know a great way to explain it without a video.

[RAILhead]

Okay, I figured it out. The Excel formula is:

=A1+(A1*5)

[CharlesS]

Originally Posted by ghporter 

The first post starts out wrong. I didn't go past that. None of my professors would bother to look past a prima facie error of that magnitude.

It would only be wrong if he claimed that was the answer (in which case it would have been at the end of the post, not the beginning ). What he actually said was:

Originally Posted by Ghoser777 

So want you want is the sum of n terms in this sequence exceeding one million dollars.

If he had posted the list you gave, then that would have been wrong, because then when he added the terms of the sequence, he would have added all the previous days twice.

[Doofy]

Originally Posted by RAILhead 

Okay, I figured it out. The Excel formula is:

=A1+(A1*5)

That sounds about right.

I usually do this kind of thing in columns:

A - Day : 1, =A2+1, =A3+1, etc..
B - People: 1, =B2*5, =B3*5, etc..
C - Donations per day: $5, =C2, =C3, etc..
D - Total for Day: =B2*C2, =B3*C3, etc..
E - Running total: =D2, =D2+D3, etc..

If you do it properly you can pop in a cell for how many people each peeps tells per day. Say G1.

B - People: 1, =B2*$G$1, =B3*$G$1, etc..

Also bring in columns for percentage of people told who sign up, stuff like that.

It might not be the ubermath way of doing things, but it's quick and it works.

[OreoCookie]

Goser's initial answer is almost the correct one, the exponent should be n+1 and not n. But other than that, there is nothing to add to that

\sum_{k = 0}^n x^n = (1 - x^{n+1})/(1-x)

Day 0 corresponds to n+1 = 1
Day 1 corresponds to n+1 = 2
Day 2 corresponds to n+1 = 3
Day 3 corresponds to n+1 = 4
etc.

So the solution is 9 days, not 10. (This also corresponds to Railroader's list, he starts numbering with 1 instead of 0 (1. $1 instead of Day 0: $1).

It's also not `easier' if you want to do it by hand.

[Andrew Stephens]

OK the question we are all asking is how do we buy into Railheads scheme and how long before I'm rich?

[Ghoser777]

Originally Posted by OreoCookie 

Goser's initial answer is almost the correct one, the exponent should be n+1 and not n. But other than that, there is nothing to add to that

\sum_{k = 0}^n x^n = (1 - x^{n+1})/(1-x)

Day 0 corresponds to n+1 = 1
Day 1 corresponds to n+1 = 2
Day 2 corresponds to n+1 = 3
Day 3 corresponds to n+1 = 4
etc.

So the solution is 9 days, not 10. (This also corresponds to Railroader's list, he starts numbering with 1 instead of 0 (1. $1 instead of Day 0: $1).

It's also not `easier' if you want to do it by hand.

It's only n+1 if you say your first day is Day 1. I set my first day as Day 0 (I'm a computer scientist at heart, so I prefer zero-indexing everything), so my formula is also correct. In either formula, n is the number of days that passed.

Also, if you go only 9 days, you're up to only about half a million dollars. You technically have to go 9.44ish days to reach 1 million... but that's kind of an odd answer as there's no indication of at what rate donations are retrieved over the course of a day.

I'll put it to you this way: if you get paid $5 an hour, how long do you have to work to make $23? It depends on if you can get paid every minute or not (highly unlikely). Most people would say that it takes exactly 5 hours to reach that condition even though that would exceed $23.

[brassplayersrock²]

saying that a math teacher is wrong when giving help to others when his solution is correct.... good job people.......

[OreoCookie]

I'm not sure I understand here. (Please no heuristic arguments, I need a math argument )
If you sum up to n-1, then you get (1 - x^n)/(1-x), if you sum up the first n days, starting from 0, of course, the sum has n+1 terms. So the result for Day 1 is (1 - 5^{0+1})/(1-5) = + 4/4 = 1. For Day 1, you get (1-5^{1+1})/(1-5) = +24/4 = 6, etc. If you start counting days at 0, then the exponent is higher. You wrote it yourself: on Day n, you add 5^n dollars to the money earned previously and hence the last term in the sum is 5^n. Hence, the correct exponent is n+1, not n, where n indexes the day and starts at 0.

If you don't believe me, look at the last of Railroaders tables (for simplicity, I've added the day):

Originally Posted by Railroader 

OK, here's my final set of (possibly) wrong numbers:

1. Day 0: $1.00
2. Day 1: $6.00
3. Day 2: $31.00
4. Day 3: $156.00
5. Day 4: $781.00
6. Day 5: $3,906.00
7. Day 6: $19,531.00
8. Day 7: $97,656.00
9. Day 8: $488,281.00
10. Day 9: $2,441,406.00

[ghporter]

Originally Posted by Chuckit 

It wasn't wrong — it was a totally accurate tally of each day's take. It just wasn't counting what you wanted it to count.

True, but when one says "this is a geometric progression" and then leaves out the accumulation portion, it makes it look like that's what he's saying the progression is. And it was late (for me).

[RAILhead]

Man, little did I know that my post would start such a pocket-protector fight!!!

Thanks for the info, all!

[Ghoser777]

Originally Posted by OreoCookie 

I'm not sure I understand here. (Please no heuristic arguments, I need a math argument )
If you sum up to n-1, then you get (1 - x^n)/(1-x), if you sum up the first n days, starting from 0, of course, the sum has n+1 terms. So the result for Day 1 is (1 - 5^{0+1})/(1-5) = + 4/4 = 1. For Day 1, you get (1-5^{1+1})/(1-5) = +24/4 = 6, etc. If you start counting days at 0, then the exponent is higher. You wrote it yourself: on Day n, you add 5^n dollars to the money earned previously and hence the last term in the sum is 5^n. Hence, the correct exponent is n+1, not n, where n indexes the day and starts at 0.

If you don't believe me, look at the last of Railroaders tables (for simplicity, I've added the day):

The question wasn't what day you exceeded 1 million (which depends on whether you say that the first day is day 0 or day 1) - the question was how many days does it take to exceed 1 million.

This formula:

sum = a(1 - r^(n+1))/(1 -r)

holds true if n+1 is the number of terms added in the series. Even look at your example:

Day 1 is (1 - 5^{0+1})/(1-5) = + 4/4 = 1

There n+1 = 1 (as there is 1 day in the sum).

While this formula:

sum = a(1 - r^n)/(1-r)

holds true where n represents the number of terms in the sequence.

[CharlesS]

Originally Posted by ghporter 

True, but when one says "this is a geometric progression" and then leaves out the accumulation portion, it makes it look like that's what he's saying the progression is. And it was late (for me).

Well, that's why it's sometimes a good idea to read the post before hitting the attack button.

[goMac]

This thread reminds me why I'm happy to be done with Calculus.

I know there isn't any Calculus here, I just feel like the Calculus could start at any time...

(If someone asks for the rate of change of the amount of money I will smack them.)

[brassplayersrock²]

<~~~~~ Hasn't been smacked in a while. (I secretly miss it.)

What's the rate of change of the amount of money?

[OreoCookie]

Originally Posted by Ghoser777 

The question wasn't what day you exceeded 1 million (which depends on whether you say that the first day is day 0 or day 1) - the question was how many days does it take to exceed 1 million.

Oh, I thought it was something rather trivial. You're right, I thought the question was `on which day'. My bad

[nonhuman]

Originally Posted by goMac 

This thread reminds me why I'm happy to be done with Calculus.

I know there isn't any Calculus here, I just feel like the Calculus could start at any time...

(If someone asks for the rate of change of the amount of money I will smack them.)

Calculus makes everyone's life better.

[ghporter]

Originally Posted by CharlesS 

Well, that's why it's sometimes a good idea to read the post before hitting the attack button.

Not attack, just observation. But poorly worded, I'll admit.

[CharlesS]

Originally Posted by goMac 

(If someone asks for the rate of change of the amount of money I will smack them.)

Of course, due to the ambiguity of language, you could either mean that you will hit them or that you will kiss them.