Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -




Not registered yet?

Post a reply

Go back

Write your message and submit
:) :| :( :D :o ;) :/ :P :lol: :mad: :rolleyes: :cool: | :dizzy :eek :kiss :roflol :rolleyes :shame :down :up :touched :sleep :wave :swear :tongue :what :faint :dunno

Go back

Topic review (newest first)

2005-11-03 07:32:39

What a great review! Maybe we could adapt some of those ideas to the internet.

2005-11-03 06:19:27

The author of my mathbook, like many great mathematiciains he was a genius, and he's dead. :-/

Well actually Saxon wasn't a 100% mathematician, I think he was an engineer and computer scientist, and a very good teacher. He wrote mathbooks for grades 1 though 12, including algebra 1, 2, trig and calculus. He also wrote books on other subjects like phonics. (yuck!) Saxon's philosophy was that math is not difficult, math is differant, and that people often call things that are unfamiliar to them, difficult, and things that are familiar easy. Time and practice is required for things that are differant to become things that are familiar and thus, easy. This philosophy is the basis for saxons teaching. Where some mathbooks would try to teach all the aspects of a topic in one lesson, saxon presents each topic little by little. Begining with the simplelest problems of that type and saving the more advanced problems for a later lesson. Each short lesson has 30 practice problems to do before moving on. This ensures you become familiar with each concept making comprehension of the more advanced problems easy since you have a solid foundation on the previous topics. "An Incremental Development" is written under the title of each saxon book. I taught myself algebra 1, 2 trigonometry with these books and am now working on calculus. Very good books, I highly reccomend them to anyone.

2005-11-02 09:53:21

Sequences are iffy, because you can take as many sets of differences as you want until eventually all the numbers in a set of differences will be the same because there's only one of them. And then you can just work back to find all the other parts as I have done above. So technically every set of numbers is an arithmetic progression, even if they're just completely random.

If you replaced your '1, 1, 2, 4, 7, 11' with any 6 numbers, I could crank out an nth term that would most probably involve n^5 and lots of big fractions, but the progression would nonetheless exist. Who's Saxon?

2005-11-02 09:46:17

I see. Well my idiot mathbook never taught me the concept of a second differance. Perhaps there was a revised edition of a previous book in the series that I didn't have. If Saxon were still alive this would not have happened. >:-(

2005-11-02 09:17:54

1, 1, 2, 4, 7, 11
  0, 1, 2, 3, 4    <-- Take differences
    1, 1, 1, 1     <-- Take 2nd differences

The 2nd differences are all the same, so put 1/2! = 1/2 as the n term. Then subtract the values that you get from doing n/2 and do it again.

1-0.5, 1-2, 2-4.5, 4-8, 7-12.5, 11-18
0.5, -1, -2.5, -4, -5.5, -7
-1.5, -1.5, -1.5, -1.5, -1.5   <-- Take differences

The differences are all the same, so put -1.5/1! = -1.5 as the n term. Then subtract the values that you get from doing n/2 - 1.5n and do it again.

0.5+1.5, -1+3, -2.5+4.5, -4+6, -5.5+7.5, -7+9
     2    ,    2   ,      2      ,    2  ,      2      ,    2

The values are all 2, so that's the last part of your sequence.

Your overall sequence is therefore n/2 - 3n/2 + 2.

To find the next term, substitute n = 7 into the sequence to get 49/2 - 21/2 + 2 = 16.

You could have got that by observation and guessing anyway, but that's a proper answer with working and stuff. big_smile

2005-11-02 07:05:16

every term in this sequence appears to be the sum of 1 and a term in another sequnce.

2005-11-02 07:03:29

begining my calculus book, some problem types were taught in previous books in the series, so if you don't remember how to do it, you go back and review. Well I have all the previous books but I can't find any problems of this type:

Find the next term in a sequence who's first six terms are 1,1,2,4,7,11.

At first I thought it was an arithmatic sequance, but 1 + d = 1,   1 + 2d = 2,    1 + 3d = 4 etc all yield differant values of d so there is no common differance.

So I thought perhaps it was a geometric progression. If that were the case, then dividing any term by the previous term would give us the common ratio. But again, it doesn't work. 1/1 = 1,   2/1   = 2,    etc.

After some guess work, I noticed the pattern 1,  (1 + 0),  (1 + 1),  (1 + 3),  (1 + 6), (1 + 10)
if you look at the numbers 0 ,1, 3, 6, 10.

1 is one greater then 0. 3 is two greater then 1.  6 is three greater then 3.   10 is four greater then 6. The answer to the problem is consistant with this sequence, but I've never seen a sequence like this. It does not appear to be arithmatic as it does not have a common differance, and it does not appear to be geometric as there is no common ratio.

So what is it and how would I solve it without guessing?

Board footer

Powered by FluxBB