Math Is Fun Forum

Hm. For some reason when I am dividing I am getting a very complicated result, and when I multiply what you gave by the denominator to check it, I also get an arduous result.

Is it not true that (x^2-2x+k)(k^2+(2k-9)x-8k+10) expands out to k^3+k^2 x^2-8 k^2+2 k x^3-12 k x^2+7 k x+10 k-9 x^3+28 x^2-20 x, which is not at all what the numerator was?
I am quite flabbergasted about it

Hi!

The trouble I'm having with the sums is that it's difficult to predict which odd numbers will pop up next. We know how many even and odd numbers come up, and the evens are relatively simple because they are the result of doubling... but odd numbers begin to be skipped in increasing amounts. For example, 15 does not appear in S5, but 11, 13 and 17 do. In S6, 15 appears, but 23, 27, 29 and 31 don't even though 19, 21, 25 and 33 do (and it gets much more complicated later).

The clear response is that the missing odd numbers are those that are 1 more than double a term in the previous set (or another way to put it; one more than another term in the same set). But how to account for this systematically?

If you just wanted the sum of S10, or even all the sums up to S10, that would be a simple matter, but if someone asked for the sum of S72 I could not say at the moment. A very rough approximation after S2 seems to be about 19 * (2.7)^(n-2), i.e. the sums appear to increase by a factor of roughly 2.7 (although that figure may be gradually decreasing so as to be erroneous after S15 or so).

Edit: Oh! The odd numbers that get moved up a step are those that are one more than double all of the ODD numbers in the previous step. Still don't have a formula though.

Hate Number Theory,
In case it was just the sum of S10 you were after, using Excel to speed things along I obtained the following:

Hi!

Hey!
Took me way too long to confirm, but here it is

bobbym,
Your answer makes x a complex number. Is that intended?

If I have interpreted the problem correctly, I think the answer is sought in terms of x, because there are heaps of possible solutions, even over the integers

Hi!

Ho ho ho

Hello!

The formulas are equivalent for SP #172

Well, I suppose you could get the sd from the mean date and see things like how the birthdays are distributed in the year (more in Summer or Winter, for instance). You just model the dates as numbers, e.g. January 1 is 1, February 1 is 32, etc.

But I actually meant the sd of the ages in years (maybe the age that everyone turned this year).

Edit: I don't understand how you know to just add 1. What would you do with 365.2425?

Okay, while we're at it, how about this:

I'd be interested to see how knowledge of the standard deviation of the age of the group can affect the probability, because of the likelihood of leap year birthdays.

I have an odd fascination for factors that have only a slight effect.

I gave up on this one, but your formula checks out! How did you solve this?