something new on the "pi and e" debate... (1 Viewer)

[no_arg]

Well I agree with that completely! It is totally consistent with what I've been saying.

 

[who_loves_maths]

Originally Posted by brett86
i hope people who read no_args post were able to see the real reason he made those comments because in many of my past posts i've encouraged people to think creatively when solving problems in maths.

^ completely agree! like that intriguing integration question you posted a few days back where you tried to get us to come up with creative solutions - that was a good one.

Thanks for encouraging me, and others on BOS, to be more creative brett :uhhuh:

Challenging Integration Question

Originally Posted by brett86
Idyll and Trefoil, if u want to thank people for contributing to this forum, u should thank Slide Rule, FinalFantasy, who_loves_maths, KFunk, McLake, acmilan, turtle_2468...

correction: i think you forgot your own name in that list

 

[no_arg]

I'm afraid it does!
If a=b then a=b to 4 decimal places!

 

[no_arg]

But Gardner does not say a is not equal to b!
He says they are equal to 4 decimal places and thus may be equal

 

[brett86]

Yes they are all exactly correct....unfortunately they are not exactly Pi....you cannot put the current value of Pi into a computer...it's irrational!! for the time being. You are simply proving results about numbers near Pi as it stands

i think uve misunderstood what i was refering to:

π < 3.14159266 & 2.718281828 < e.

So

π^4+π^5

< 3.14159266^4+3.14159266^5

= 403.4287797363725532846657585016733756588576 (exactly)

< 403.42879308396500147612667658990386685186886530139770 4704

= 2.718281828^6 (exactly)

< e^6

buchanan said π < 3.14159266 & 2.718281828 < e, he never said they were equal!

because i calculated 2.71828182^6 on paper and not 718281828^6, here is a modified version:

π < 3.14159266

e > 2.71828182

π⁴ + π⁵ < 3.14159266⁴ + 3.141592665⁵

3.14159266⁴ = 97.40909182902901737436094415467536 (exactly)

3.14159266⁵ = 306.0196879073435359103048143469980156588576 (exactly)

3.14159266⁴ + 3.14159266⁵ = 403.4287797363725532846657585016733756588576 (exactly)

.: π⁴ + π⁵ < 403.4287797363725532846657585016733756588576

Now,

e⁶ > 2.71828182⁶

2.71828182⁶ = 403.428785960133422981798452810123327605718818306624 (exactly)

e⁶ > 403.428785960133422981798452810123327605718818306624 > 403.4287797363725532846657585016733756588576 > π⁴ + π⁵

 

[acmilan]

Well after searching through the usyd library, the only reference on the topic i found was in this book that had puzzles about pi. The exact words of the question was:

eCCeNTRIC π: What is eccentric about (π⁴ + π⁵)^1/6

Unfortunately there were no answers.

However all was not lost, I did find a number of proofs that π is irrational. Here is one (taken word for word) that was titled 'Simple proof that π is irrational' (note: for those that dont know, f⁽ⁿ⁾(x) means the nth derivative of f(x), eg. f⁽²⁾ is the 2nd derivate of f(x))

Let π = a/b, the quotient of positive integers. We define polynomials

the positive integer n being specified later. Since n!f(x) has integeral coefficients and terms in x of degree not less than n, f(x) and its derivatives f⁽ⁱ⁾(x) have integral values for x = 0; also for x = π = a/b, since f(x) = f(a/b - x). By elementary calculus we have:

Now F(π) + F(0) is an integer since f⁽ⁱ⁾(π) and f⁽ⁱ⁾(0) are integers. But for 0 < x < π, 0 < f(x)sinx <

so that the integral in (1) is positive, but arbitrarily small for n sufficiently large. Thus (1) is false and so is our assumption that π is rational.

 

[凍鴛鴦]

How about this? =P

pi^4.0000001584961854419016065242344862178145279880029574746032812229828713251230353378935060344863712520462367633546938829227705917747439148261884526503886984959611414499536285970087596304422991542688619506660619234030430454848993499572971099695325634828463115362988225978145350237706453727820415461088287200169428024732357730310000702501875200104860537432633735115162728913111406395851378617344551943071664503904996285834327519330367747644383544449163405463919285992641672273454420261670564336535159065187835251385

+

pi^5

=

e^6???

Let's see you prove/disprove that =D

 

[FinalFantasy]

wow, first time I see someone here with Chinese username!
I'm gonna try make one too!

 

[Trev]

wow, first time I see someone here with Chinese username!
I'm gonna try make one too!

What's it say? lol...

 

[凍鴛鴦]

Oh yeh BTW, because of the fact that the above quoted number is a closer approximation to e^6 than is pi^4 + pi^5, it proves that pi^4 + pi^5 does not equal to e^6

=)

 

[who_loves_maths]

lol, hongkong is not a very innovative name FF

 

[凍鴛鴦]

Haha... you stole my idea!!! ><

 

[FinalFantasy]

lol, hongkong is not a very innovative name FF

hahaha but i like Hong Kong!!!

and so ur chinese eh?
can read my HK

 

[FinalFantasy]

Haha... you stole my idea!!! ><

it's too good to not steal

 

[who_loves_maths]

Originally Posted by FinalFantasy
hahaha but i like Hong Kong!!!

and so ur chinese eh?
can read my HK

hehe... just cause i can read chinese characters does not mean i have to be chinese.

 

[FinalFantasy]

^that's true..

 

[maniacguy]

I suspect that no_arg's initial point was that some of the so-called proofs* that were presented depended on the fact that there was divergence in some decimal place. Such a proof relies on the assumption that the calculator is sufficiently accurate, which is not always a wise assumption to make.

In short, so the decimal places disagree. Who cares? That could just indicate a defective calculator.

Now, the counterarguement to this (which people seem to be trying to get at, but which has not - as far as I can tell - been articulated yet), is simply that the calculator may not be accurate, however it is possible to bound the error in the approximations used. Having done this, then one can determine an interval for the value of exp(6), and an interval for the value of pi^4 + pi^5. Since the two intervals are disjoint, then the two values are equal.

no_arg, hopefully that settles your objections to the proof by calculator - disagreement in decimal places is considered sufficient because it is taken as given that the error implicit in the calculator's computation is sufficiently small as to be irrelevant in this instance.

(I think Proof 3 on buchanan's list should be acceptable to you anyway, since you only need verify the accuracy of the approximations e > 2.718281828 and pi < 3.14159266, and hopefully you can be convinced of those!)


*This is not meaning to cast aspersions on all of them, however one or two of them I would find extremely dubious if I were to run into them on a dark night. Mostly because I have an inherent dislike for the calculator - I think life would be much better if the current HSC were to ban use of calculators and demand exact computation everywhere.

In the words of Terry Gagen (Sydney University) - "Calculators are the tools of the devil!!"

 

[香港!]

hey buchanan, is that "Toby Buchanan" related to u?

http://community.boredofstudies.org/showthread.php?t=81361&page=1&pp=15

 

[who_loves_maths]

^ why are you obsessed with that 'Toby Buchanan' guy? note that he changed his name to "Tobes" now?

 

[香港!]

^hahaha i'm bored #_#
ya i realised he changed the name, was wondering why the number of posts were diff.