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Multipliers for series for manipulating signs of the terms 
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#1
Jan1014, 07:31 PM

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There are multipliers that can be used when building infinite series that can create several different orders for the signs of consecutive terms by, for example, (1)^n to get,
 +  +  +... but I have been having difficulty figuring out any beyond the following, +  +  + ... + +   + +... +    +   ... and,  + + +  + + +... What else do we have? 


#2
Jan1114, 01:13 AM

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eg:  +  +  + is (1)^n +  +  +  = 1x( +  +  +) is (1)^n = (1)^(n+1) = (1)^(n1) 


#3
Jan1114, 09:55 AM

P: 553

The first set can obviously be done with the old standby (1)^n to get,  +  +  +... and (1)^(n+1) or (1)^n to get the opposite set, +  +  + ... as you had suggested. For the   + +   + +... I came up with this, i^(n(n+1)) The negative of course will give + +   + +  ... For those last two in my post it took a bit of work to derive but the final form is, +/ (ni^(n(n+1))+n(1)^n+1) / ((ni^(n(n+1))+n(1)^n+1)^2)^(1/2) to get +    +    +... and the  + + +  + + + ... depending on the sign That was a fun solution to get as I used whole numbers to get the sign pattern and then simply divided by the square root of that quantity squared for each term to get back to '1' or '1'. The '+1' in each term was to eliminate the '0's' The problem I face now is there are no combinations of these forms that will yield something different. If I can get to say + + +    + + +... then that will give some room for manipulation. I have also been working on +    +    + by trying to eliminate the middle positive so we would instead have +        +... (the negative being just the opposite sign). What do we currently have in mathematics? Or do you have any techniques of your own? 


#4
Jan1114, 06:02 PM

P: 553

Multipliers for series for manipulating signs of the terms
I just finished deriving a 'general solution' for any combination of 'pairs' of positive and negative terms. For example the solution will give,
++++++... ++++++++... ++++++++++... ++++++++++++... etc. etc. How exciting! What does mathematics have available today? It would be fun to compare notes! 


#5
Jan1114, 10:44 PM

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You want to find a way to label the combinations so you can summarize the realtionships.
The ones you've found are periodic functions ... you can also get interference i.e.: (++++++++++...)x(++++++++)=(++++++...) 


#6
Jan1214, 07:52 AM

P: 553

I have been playing with these quite a bit, there is a wealth of interference patterns that can be produced (infinite) much like your example above but given enough time they all eventually repeat their patterns. The last thing to find (for now) is a pattern that does not repeat but grows at a steady rate with time, e.g. ++++++++++++... or ++++++++++++++++++++... With this last tool building infinite series will become substantially easier (albeit they are still tricky buggers!). These are periodic but change with time, what would the proper 'label' for functions of this type? Periodic expansive functions? I haven't even began to think about how to do this and school starts tomorrow :P ...although it's been a wonderfully productive break! 


#7
Jan1214, 06:48 PM

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You've noticed that ##\cos(n\pi)## gives alternating 1 and 1 and built from there.
You may also like to look at binary interference ... so instead of + and  you have 0 and 1. In a way you just need to look for functions with the pattern of zeros you need. i.e. treat either the +'s pr the 's as a background that you change individual bits of. It's a very big subject .... i.e. try for this one: ++++++++++++++++++++... or: +++++++++++++... or: +++++++++++... ... do you recognize the patterns? They may be easier to see as strings of 1's and 0's. 


#8
Jan1314, 01:55 PM

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#9
Jan1314, 03:34 PM

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How about: ++++++++++++++++++++++++++... ... repeats after 90 terms. +++++++++++... ... repeats after 60 but: +++++++++++++++... ... does not repeat. Working out generating functions for sequences, even ones with easy algorithms, is pretty tough. But if you enjoy this sort of thing, you'll probably enjoy cryptography. 


#10
Jan1314, 09:38 PM

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#11
Jan1314, 09:57 PM

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(Not providing that information also tells you something about the pattern  it's metadata.) Even with that information there is more than one generator. I'll leave you to it. 


#12
Jan1314, 10:00 PM

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