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Topic: Number "e" and Straightlinecurves; derivative & integral of y=1/x #13
Uni-textbook 6th ed.:TRUE CALCULUS

Replies: 19   Last Post: Jul 2, 2013 1:43 AM

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 plutonium.archimedes@gmail.com Posts: 18,572 Registered: 3/31/08
Re: 13.2 Re: Number "e" and Straightlinecurves; derivative & integral
of y=1/x #13 Uni-textbook 6th ed.:TRUE CALCULUS

Posted: Jun 30, 2013 12:56 PM

On Sunday, June 30, 2013 10:27:13 AM UTC-5, konyberg wrote:
> On Sunday, June 30, 2013 1:20:38 PM UTC+2, konyberg wrote:
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> > On Sunday, June 30, 2013 10:36:22 AM UTC+2, Archimedes Plutonium wrote:
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> > > On Sunday, June 30, 2013 2:48:17 AM UTC-5, Archimedes Plutonium wrote:
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> > > > I am trying to think of a easy fast trick to determine if Floor pi *10^603 as integer 314159..32000 is a member of the sequence 2, 4, 8, 16, 32, 64, . . .
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> > > > AP
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> > > Rather a silly question on my part, and my only excuse is it is late at night.
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> > > That sequence cannot get away with the last digit being either a 2, 4, 6 or 8, so Floor pi * 10^603 is not a member. So I wonder if Floor pi*10^600 as integer 31415..32 is a member?
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> > > The reason this is bugging me, is that macroinfinity produces microinfinity, so that 1*10^603 produces 1*10^-603 so the first infinity number is 1*10^603 + 1*10^-603. And if that division into 1 is messy, the entire algebra is messy.
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> > > So far, I have explained this away by saying that pi digits determine the exponent in which we find the infinity borderline. Once we determine that it is 10^603, then we claim that 1*10^603 is the borderline and not Floor pi. When we work with 1 and then 603 zeros following, the division into 1 comes out perfectly even and not messy.
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> > > For example, in the 10 Grid system, as we pretend 10 is the borderline and last and largest finite number then 0.1 is the smallest nonzero finite number and the smallest infinity number is 10.1. So everything is smooth fresh and clean. But suppose instead we pretend 13 is the borderline, then 1/13 is messy with 0.076 and we cannot then say 13.076 is the smallest infinity.
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> > > Calculus is cleaning up the infinity borderline numbers.
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> > > AP
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> > Hi. I dont think even if you cut the three zeroes in pi the number will fit the sequence. Try this site and test it: http://apfloat.appspot.com/
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> > 3141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067982148086513282306647093844609550582231725359408128481117450284102701938521105559644622948954930381964428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273724587006606315588174881520920962829254091715364367892590360011330530548820466521384146951941511609433057270365759591953092186117381932611793105118548074462379962749567351885752724891227938183011949129833673362440656643086021394946395224737190702179860943702770539217176293176752384674818467669405132
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> And since the number end in "2" it must be of the form 2^(4n+1), n = 0, 1, 2, 3, ....
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> No more clues. :)
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> KON

And what clue was that.

Now pi and "e" in the 10^604 and 10^605 region look like this:

Floor-pi*10^605

314159265358979323846264338327950288419716939937510582097494459230781640628 620\ ?899862803482534211706798214808651328230664709384460955058223172535940812848 111\ ?745028410270193852110555964462294895493038196442881097566593344612847564823 378\ ?678316527120190914564856692346034861045432664821339360726024914127372458700 660\ ?631558817488152092096282925409171536436789259036001133053054882046652138414 695\ ?194151160943305727036575959195309218611738193261179310511854807446237996274 956\ ?735188575272489122793818301194912983367336244065664308602139494639522473719 070\ ?217986094370277053921717629317675238467481846766940513200056,\ ?

Floor-e*10^605

271828182845904523536028747135266249775724709369995957496696762772407663035 354\ ?759457138217852516642742746639193200305992181741359662904357290033429526059 563\ ?073813232862794349076323382988075319525101901157383418793070215408914993488 416\ ?750924476146066808226480016847741185374234544243710753907774499206955170276 183\ ?860626133138458300075204493382656029760673711320070932870912744374704723069 697\ ?720931014169283681902551510865746377211125238978442505695369677078544996996 794\ ?686445490598793163688923009879312773617821542499922957635148220826989519366 803\ ?318252886939849646510582093923982948879332036250944311730123}

My motivation is to preserve Algebra, for if the last and largest finite number is used to produce the smallest nonzero finite number, it must be a clean number as in say 10 Grid 1/10 and where the first infinity number is 10.1. Or in say 100 Grid, 1/100 produces the first infinite number as 100.01

If we have a sloppy last finite number then divide it into 1 to produce the first infinite number, we tend to have a messed up algebra.

If the last and final finite number is 1 with a lot of zeroes, makes the algebra clean and sterile and furnishes us with what I think of as a **secondary unit** to that of 1 is a unit for integers and 0.001 is a unit for fractions between 0 and 1 if 100 is the last and final finite number.

So I want to preserve the cleanly and sterile algebra, not a messy algebra.

So pi gives us the exponent that is infinity borderline, but does pi give us this clean algebra? The answer is apparently not.

So let me look at "e" and focus on Floor-e*10^604 as the integer 271828..73012.

Is that integer a member of the sequence 2, 4, 8, 16, 32, 64, . . .

Maybe I have to be satisfied that pi at exponent 603 delivers to us the infinity borderline and that out of sheer algebra necessity we take 1*10^603 as the last and largest finite integer and not Floor pi*10^603 for it does not preserve Algebra.

I was hoping that Mathematics, and not the human mind needs a clean sterile and pure borderline of 1 with a lot of zeroes.

AP