pi and e. (1 Viewer)

[babydoll_]

Here's an idea

 

[buchanan]

1989 four unit HSC Question 8(b)(i)(α)

"The difference between a real number r and the greatest integer less than or
equal to r is called the fractional part of r, F(r). Thus F(3.45) = 0.45. Note that
for all real numbers r, 0 < F(r) < 1.

Let a = 2136 log₁₀2.
Given that F(a) = 7.0738 · · ·×10⁻⁵
observe that F(2a) = 14.1476 · · ·×10⁻⁵
F(3a) = 21.2214 · · ·×10⁻⁵

Use your calculator to show that log₁₀ 1.989 < F(4223a) < log₁₀1.990."

Well, no_arg, it seems it's OK to use the calculator to prove things, according to the HSC exam committee.

 

[brett86]

in the image below the first input represents the value of π⁴ + π⁵ and the second input represents e⁶

mathematica calculates the sum to a point where it is accurate to the given number of decimal places

as u can see the 2 numbers are different from the 5th decimal place onwards

hence π⁴ + π⁵ ≠ e⁶

 

[acmilan]

^Wouldnt that bring no_arg to the same arguement as before? Its still using a calculator of some sort?

 

[Dumsum]

Let's see the implementation of some mathematical skill and the production of a proof!!!

Where's the proof that it is?

 

[brett86]

all right then, if that is not enough to convince no_arg then look at this site from mathworld:

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

it directly says that π⁴ + π⁵ = e⁶ is an approximation

mathworld also says that (π⁴ + π⁵)^¹/₆ is accurate to 7 decimal places which is consistent with what jago said before:

((pi^4) + (pi^5))^(1 / 6) = 2.7182818109

e^1 = 2.718281828

 

[acmilan]

all right then, if that is not enough to convince no_arg then look at this site from mathworld:

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

it directly says that π⁴ + π⁵ = e⁶ is an approximation

Buchanan posted that already

Evidently, wanton-wonton cares or else he wouldn't have started this thread. I think I've settled the matter. no_arg may not agree. But I don't need him to.

Anyway, here are some more approximations for e:
http://mathworld.wolfram.com/eApproximations.html

 

[brett86]

just wondering, is anyone besides no_arg convinced that π⁴ + π⁵ = e⁶?

 

[brett86]

^Wouldnt that bring no_arg to the same arguement as before? Its still using a calculator of some sort?

it shouldnt bring him to the same argument because mathematica is calculating a sum rather than using a stored constant

 

[brett86]

sorry, no_arg is just really pissing me off

he knows his argument is untrue and is arguing it just to annoy others

a hsc student might read his posts and believe it

i think one of the moderators should edit his posts so people know what hes saying isnt correct

 

[buchanan]

How about this.

&pi; < 3.14159266 & 2.718281828 < e.

So

&pi;⁴+&pi;⁵

< 3.14159266⁴+3.14159266⁵

= 403.4287797363725532846657585016733756588576 (exactly)

< 403.428793083965001476126676589903866851868865301397704704

= 2.718281828⁶ (exactly)

< e⁶

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

RIP no_arg!

 

[no_arg]

The intel chip can't even be trusted to divide rational numbers properly let alone raise to the sixth power!If you want to convince me I'm afraid it will have to be mathematics...remember that stuff?

"The Pentium microprocessor is the CPU (central processing unit) for what are now possibly the widest-selling personal computers. Unlike previous CPUs that Intel made, the 486DX and Pentium chips included a floating-point unit (FPU) also know as a math coprocessor. Previous Intel CPUs did all their arithmetic using integers; programs that used floating-point numbers (non-integers like 2.5 or 3.14) needed to tell the chip how (for example) to divide them using integer arithmetic. The 486DX and Pentium chips have these instructions built into the chip, in their FPUs. This makes them much faster for intense numerical calculations, more complex, and more expensive. The problem for Intel is that all Pentiums manufactured until sometime this fall had errors in the on-chip FPU instructions for division. This caused the Pentium's FPU to incorrectly divide certain floating-point numbers. "

 

[buchanan]

If you want to convince me I'm afraid it will have to be mathematics...remember that stuff?

Well, you're the one claiming &pi;⁴+&pi;⁵=e⁶ exactly, and your invective does not constitute proof.

I've now proved &pi;⁴+&pi;⁵&ne;e⁶ two ways, and several others on this thread have also proved it.

You're still claiming &pi;⁴+&pi;⁵=e⁶ exactly, so I think the burden of proof is upon you, not me anymore.

So prove it, if you want to convince me.

 

[brett86]

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

this was on:

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

a site from Wolfram Research

do u really expect people to believe u over them

 

[lucifel]

by the way he is arguing for the sake of it, (like some prick at my school), the answer is obvious, of course he expects us to take his word over ANY number of experts.

 

[buchanan]

Exactly! And the reference to Castellanos is

Castellanos, D. "The Ubiquitous Pi. Part I." Math. Mag. 61, 67-98, 1988.

no_arg has no publication refuting Castellanos's claim, and if he ever submits such a thing for publication, it would get rejected!

 

[brett86]

i know most people know what no_arg is saying isnt true but it is possible that a hsc student might read what he has wrote and believe it

i really think one of the mods of the 4u forum should edit his posts so people know what hes saying isnt true

or at least make a post saying no_arg is wrong, because people acknowledge that the mods, turtle_2468 and McLake, give reliable information

 

[buchanan]

Unfortunately, they are both offline at the moment, so we'll have to wait till they come back later.

 

[lucifel]

hmm, but i really think if they see EVERYONE making a mockery out of this arrogant fool, it is rather unlikely they can take him seriously. And to make things clear, arrogant fool = no_arg.

 

[no_arg]

You want to censor me just because you can't provide a rigorous proof of a result?? I am amazed! Instead of feverishly punching the buttons on your calculator maybe you should have been considering properties of the function
f(x)=ln(x^4+x^5)?? A little bit of calculus a shift in coordinates and its done.
No need to start burning the books...clearly you can't see beyond your liquid crystal displays. Mathematics is not computing!

Good Bye

Disappointed yet again