Oops. (2 Viewers)

[buchanan]

I was trying to edit a post in the <a href="http://community.boredofstudies.org/showthread.php?t=81019">pi and e thread</a> when McLake closed it. The - should have been a + in my last post in that thread at the bottom. So here it is again.

Here's another proof using only integers.

e*10⁹ > 2718281828 and &pi;*10⁸ < 314159266

Hence

(e⁶-&pi;⁴-&pi;⁵)*10⁵⁴

> 2718281828⁶-314159266⁴*10²²-314159266⁵*10¹⁴

= 13347592448191460918088230491193011265301397704704

> 0

Hence &pi;⁴+&pi;⁵&ne;e⁶

 

[no_arg]

And how pray tell does a computer raise a large integer to a power?
These are just calculations they prove nothing.
You may as well try to prove the result using a toaster!
Goldbach conjecture?? No worries I'll use the hair-dryer!
Poincaré Conjecture...that one's easy I'll use the electric drill..the cord is a bit dodgy but I'm sure it'll be OK.
Riemann hypothesis?? I used the distributer cap from my Volvo and the electric motor from the windscreen wipers.
Wow maths has become so easy! I don't even need to think any more!

Appalling is the word!
Shame Shame Shame

 

[yrtherenonames]

Don't you think this argument has become just a little sad?

 

[no_arg]

It was sad from the start

 

[buchanan]

Actually Pi^4+Pi^5=e^6 is an exact formula

I just think it's sad that no_arg still thinks &pi;⁴+&pi;⁵=e⁶ exactly. It doesn't.

&pi;⁴+&pi;⁵=403.42877...

and e⁶=403.42879...

Hence &pi;⁴+&pi;⁵&ne;e⁶

It's quite simple, really.

 

[yrtherenonames]

Ok then we're agreed its sad. Next case please!
And Mr. Buchanan, stop ruining our dreams of flukey transcendental numbers coinciding.

 

[no_arg]

1/(√(10001)-100)=200.00499987500624960940234489433

√(10001)+100=200.00499987500624960940234169938

Therefore 1/(√(10001)-100) is not equal to √(10001)+100

It's simple really ....but hang on?? They are equal!


You cannot trust your machines...that is a simple truth.

What could possibly guide you to accept the calculations on the one hand and then reject them on the other?

It should also be pointed out that your spurious argument above involving powers of 10 could just as easily be used to prove that these two equal irrationals are unequal.

 

[brett86]

Pi^4+Pi^5=e^6 is an exact formula

no_arg, everyone knows that u know π⁴+π⁵ ≠ e⁶



but just incase a hsc-er comes and takes ur comments seriously:

this was on:

http://mathworld.wolfram.com/eApproximations.html

a site from Wolfram Research

hence, π⁴+π⁵ is an approximation to e⁶

 

[brett86]

π < 3.14159266

e > 2.71828182

π⁴ + π⁵ < 3.14159266⁴ + 3.14159266⁵

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

hence, π⁴ + π⁵ ≠ e⁶

no calculator needed, u can check for urself no_arg

now that everyone knows no_arg is an idiot, can u close this thread again McLake so he cant respond?

now im goin to sleep

EDIT: thankyou so much for closing the other one, no_arg was really pissing everyone off

 

[tywebb]

well maybe no_arg has a point.

consider an easier problem, like e < 3.

one might naively "prove" it by saying e=2.7... and therefore e < 3. but that's not really a proof.

a better way is to consider the graph of y=e^-x and the chord joining (0,1) and (1, e^-1) and comparing areas.

the area of the trapezium

= (1/2)(1+e^-1)

> area under curve

= 1-e^-1

so e+1 > 2e-2 and e < 3.

maybe we can do something similar to prove &pi;⁴+&pi;⁵ < e⁶


ok?

 

[buchanan]

one might naively "prove" it by saying e=2.7... and therefore e < 3. but that's not really a proof.

I disagree.

"e=2.7... . Therefore e < 3" is a perfectly valid proof.

 

[no_arg]

3.141592664^4 = 97.40909182902901737436094415467536 (maybe)

As has already been pointed out simple division is questionable on any machine let alone exponentiation! I have already displayed how two identical numbers can appear differently on a computer.

I agree wholeheartedly with Buchanan's better half Tywebb.....a proof is required not a calculation

I believe a starting point is to consider the function y=Pi^x+Pi^(x+1)-e^(x+2)....induction is also a possibility!


I must admit to being very surprised at Brett86's intolerence to differing points of view! e and Pi are already related at the complex level with e^(iPi)=-1 so a relation at the real level is certainly on the cards. You have to accept that even in the mathematical world a difference of opinion will sometimes arise! This is no reason to play the censorship card. I myself am open minded on this issue and would gladly consider any mathematical proof regarding this problem.

 

[buchanan]

I tried it a similar way to brett86 and got

&pi; < 3.14159266 & 2.718281828 < e.

So

&pi;⁴+&pi;⁵

< 3.14159266⁴+3.14159266⁵

= 403.4287797363725532846657585016733756588576 (exactly)

< 403.42879308396500147612667658990386685186886530139770 4704

= 2.718281828⁶ (exactly)

< e⁶

Hence, using only rational numbers, I've proved that &pi;⁴+&pi;⁵&ne;e⁶.

I believe a starting point is to consider the function y=Pi^x+Pi^(x+1)-e^(x+2)....induction is also a possibility!

I don't think induction would do it because it isn't generally true that eⁿ⁺² > &pi;ⁿ⁺¹+&pi;ⁿ for positive integers n, e.g., e⁷-&pi;⁶-&pi;⁵=-170.7... . I think the reason for this is basically that &pi;^x dominates e^x for large positive x, simply because &pi; > e.

Maybe we can look at y=e^x+2-&pi;^x+1-&pi;^x, as per no_arg's suggestion:

Looking at that graph shows clearly that there is a positive number &alpha; such that e^&alpha;+2-&pi;^&alpha;+1-&pi;^&alpha;=0. Thanks to who_loves_maths, &alpha;=(ln(&pi; +1) -2)/(1 - ln(&pi;))=4.000000303... > 4 (by calculator), and so &pi;⁴+&pi;⁵ < e⁶.

You might like prove other things about the shape of this graph, like:

* The x-axis is an asymptote as x&rarr;-&infin;

* It is concave up till it gets to an inflection (a,b) where 0 < a < &alpha;, and is thereafter concave down

* It is continuous and smooth everywhere.

* There is a stationary point (c,d) where a < c < &alpha;

* y > 0 for all x < &alpha;

* y < 0 for all x > &alpha;

If we want to avoid calculators altogether, then again thanks to who_loves_maths, maybe if we can show &alpha;=(ln(&pi; +1) -2)/(1 - ln(&pi;)) is irrational, then it couldn't be 4 and hence &pi;⁴+&pi;⁵&ne;e⁶.

 

[no_arg]

Oh I get it now
Together you are saying that
403.428785960133422981798452810123327605718818306624 (exactly)
=403.42879308396500147612667658990386685186886530139770 4704 (exactly)

Exactly what form of exactly are you exactly using?
This is my point exactly
The only way of dealing with exact quantities is to do the math!
Thank you for displaying the problem so clearly!

 

[no_arg]

Here's a nice argument using the above spurious logic:

Let x=1/(sqrt(10001)-100

and y=sqrt(10001)+100

Then correct to 12 significant figures:

x= 200.005000125 and y=200.004999875

Thus x>200.005 and y<200.0049999

Therefore x>y implying 1/(sqrt(10001)-100>sqrt(10001)+100.

Hence 1>1 and thus 1 is not equal to 1.

 

[who_loves_maths]

Originally Posted by no_arg
Actually Pi^4+Pi^5=e^6 is an exact formula
Any discrepancies you get when calculating the quantities stem directly from errors in the way the calculator handles Pi and e. This also applies of course to "advanced" packages such as MAPLE MATLAB etc. Calculators and software cannot deal directly with irrational quantities hence the apparent "error" in the tail of the decimal places.

i tried to read through all of the "pi and e" thread but got tired mid way through. but i did read up to this ^(above) comment and i think it's quite obvious that (pi^4 + pi^5) = e^6 is only an approximation, and not exact:

To determine whether or not this is exact:

Let: pi^x + pi^(x+1) = e^6 ; where 'x' is a variable.

taking the logarithm of both sides:

ln(pi^x + pi^(x+1)) = 6 -----> ln[(pi^x)(pi + 1)] = 6

Log rules: 6 = ln(pi^x) + ln(pi +1) = xln(pi) + ln(pi + 1)

-----> x = [6 - ln(pi + 1)]/ln(pi)


from here, i don't think one would need a calculator to see, with the 'pi's and 'ln(pi)'s, that [6 - ln(pi + 1)]/ln(pi) is not an integer. Moreover, it's not equal to 4.

it follows that the original "equation" (pi^4 + pi^5) = e^6 is not exact, only an approximation.


P.S. since i didn't read through all of the other thread, i apologise if someone has already posted this sort of method up before.

also: [6 - ln(pi + 1)]/ln(pi) = [approx.] 4.000000038


Edit: upon reading some of the past long posts in this thread again, i see that no_arg had previously suggested to consider the function: y = pi^(x) + pi^(x+1) - e^(x+2). Sorry no_arg, i hadn't seen your suggestion before posting this (similar) method here.

 

[David_O]

Nobody here believes no_arg's statement.
But let us suppose it was true, then none of us could ever use it anyway because it would the expression is both utterly useless and unusable without a proof.

I await the closure of this thread too.

 

[KFunk]

-----> x = [6 - ln(pi + 1)]/ln(pi)

also: [6 - ln(pi + 1)]/ln(pi) = [approx.] 4.000000038

Nice work, I think that's the closest someone has come to a proof which might settle this argument.

 

[who_loves_maths]

Originally Posted by buchanan
Looking at that graph shows clearly that there is a positive number α such that e^(α+2)-π^(α+1)-π^α=0. So if we can show that α > 4, this also will prove π^4+π^5 < e^6.

Maybe you could find α with Newton's Method to any desired degree of accuracy, but of course that would require the use of a computer or calculator! Anyway, this should show α > 4.

...

In fact α=4.000000303.... > 4. Hence π^4+π^5 < e^6.

This is not my preferred way of doing it, and in any case, I had to use a calculator to get the value of α using Newton's method anyway.

Buchanan, you don't need to using approximation techinques such as Newton's Method to solve the simple equation: e^(a+2) - pi^(a+1) - pi^a = 0 ; you can find the solution in exact form quite easily:

---> e^(a+2) = pi^(a+1) + pi^a

take log of both sides: a+2 = ln[(pi^a)(pi + 1)] = aln(pi) + ln(pi +1)

---> a(1 - ln(pi)) = ln(pi +1) -2

hence, a = [ln(pi +1) -2]/(1 - ln(pi))

ie. no calculators, nor Newton's Method, required to find the root


P.S. like you suggested however, a = [ln(pi +1) -2]/(1 - ln(pi)) = [approx.] 4.000000303...

 

[buchanan]

Good. Thanks for that.

So &alpha;=(ln(&pi; +1) -2)/(1 - ln(&pi;))=4.000000303.... > 4 (by calculator) and therefore &pi;⁴+&pi;⁵ < e⁶.

But if the idea is to avoid calculators altogether, won't you need to prove &alpha;=(ln(&pi; +1) -2)/(1 - ln(&pi;)) is irrational? Then it couldn't be 4 and hence &pi;⁴+&pi;⁵&ne;e⁶.

Perhaps all that is now required is to in fact prove (ln(&pi; +1) -2)/(1 - ln(&pi;)) is irrational.