pi and e. (1 Viewer)

[haboozin]

just out of interest

i think 2001 HSC Q8 b iii was:
prove e is irrational.

 

[lucifel]

yeh but it was a lead on question. It's not too hard if you followed all the steps.
Oh and knew what, 'proof by contradiction' meant. or whatever the other ways of sayign it is.

 

[MAICHI]

i think 2001 HSC Q8 b iii was:
prove e is irrational.

I think our lecturer at UNSW set that question. I remember him talking about it in the first session.

 

[acmilan]

source: http://www.mathpages.com/

 

[mojako]

(pi^4 + pi^5)^1/6 is e or is approximately e. ?

if (pi^4 + pi^5)^1/6 was e then people wouldn't have invented a new symbol "e"
so its just an approximation

 

[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.

 

[brett86]

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.

e = (π⁴ + π⁵)^¹/₆ is just an approximation

there is no proof that suggests that e is exactly equal to (π⁴ + π⁵)^¹/₆

 

[no_arg]

I think you meant to say that there is no proof that

Pi^4+Pi^5 is not equal to e^6

 

[no_arg]

You can't prove claims about irrational numbers using a calculator!

For example 1/(sqrt(10001)-100) is exactly equal to sqrt(10001)+100
THis is easily verified by rationalising the denominator.

But when you evalaute on a calculator

1/(sqrt(10001)-100)=200.0050001 and

sqrt(10001)+100=200.0049999

even though they are identical quantities!

It's the same above!

Pi^4+Pi^5 i= e^6

Discrepancies when evaluating on a calculator or in MAPLE are simply due to round off error...they don't actually exist!

 

[buchanan]

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

No it isn't.

π⁴+π⁵≠e⁶

π⁴+π⁵=403.42877...
and e⁶=403.42879...

Hence π⁴+π⁵≠e⁶

It's only an approximation.

Sometimes roundoff errors do occur however, like 2x2²⁴⁹-2²⁵⁰.
Some calculators don't give 0.

Reference:

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

 

[no_arg]

You must therefore also conclude that

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


You cannot prove facts about irrationals using a calculator. Calculators only deal with rationals!

 

[babydoll_]

I conclude that you should all shut the fuck up

 

[buchanan]

I conclude that you should all shut the f**k up

no_arg and I have tried for a long time to shut each other up, and we have both failed.

Anyway, I have both proved π⁴+π⁵≠e⁶ and provided a reference to support my claim.

no_arg is claiming that π⁴+π⁵=e⁶ exactly and has neither proved this claim nor provided a reference to back up his claim.

 

[babydoll_]

babydoll_ likes using obscene concepts, although most people don't..

 

[buchanan]

Nobody cares

(pi^4 + pi^5)^1/6 is e or is approximately e. ?

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

 

[babydoll_]

This is what I think of you and no_arg

babydoll_ likes obscene pictures... but most people don't appreciate them

 

[Xayma]

You must therefore also conclude that

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


You cannot prove facts about irrationals using a calculator. Calculators only deal with rationals!

Calculators only deal in rationals true.

You cannot prove facts about irrationals using a calculator, not necessairly true.

You could never prove that is equal using a calculator but you can prove it is unequal.

Lets take two numbers

1.0000000±.0000001
2.0000000±.0000001

Now this means that the multiplication of them is between 1.0000001*2.0000001 and 0.9999999*1.9999999

Now we know π and e to more then 7 decimal places yet the difference doesn't occur till the 7th decimal place meaning we know it is accurate to 5 (rounding may take out the 6th) that is it is equal to 2.00000±1

π and e are known more then accurate enough to know that a difference of 2 in the 5th decimal place isn't accounted for in error given that calculators use 12 significant figures this occurs at the 8th. It is possible that it could be an error (due to the large number of multiples) however, with larger numbers of significant figures used it would be possible to exclude this error (since buchanan had a refrence to a mathematics journal one can assume that a larger number had been used).

 

[no_arg]

Surely you are not suggesting that putting numbers into a calculator could ever
constitute a proof? Pleeeeassse

It should be also pointed out that being approximately equal does not preclude things from also being equal! Many approximations in the literature are later refined to equality. The references given are unfortunately very dated and no longer reflect the state of play on these issues.

 

[buchanan]

Surely you are not suggesting that putting numbers into a calculator could ever constitute a proof? Pleeeeassse

It seems you haven't seen the 1989 Four unit HSC Question 8b. Such things seem OK to the HSC exam committee. So they should seem OK to HSC students.

And I bet you don't think the 4-colour problem has been solved yet either, hey? (It was done by computer!)

Many approximations in the literature are later refined to equality.

Well π⁴+π⁵≈e⁶ hasn't been because they are not equal!

The references given are unfortunately very dated and no longer reflect the state of play on these issues.

And I suppose you still think π⁴+π⁵=e⁶? Prove it or provide a more up to date reference supporting your claim. Alternatively, accept that you are wrong and I am right. Oh, but you couldn't do that, could you no_arg?

 

[buchanan]

There is nothing wrong with my proof and there is nothing wrong with my references.

"References a little dated"?

Well in maths that's a bit like saying:

"Euclid's proof that there are infinitely many primes is wrong because we've all moved on from there. We now know better. Especially the world renowned mathematical experts at UNSW! They're so brilliant, you know. They know Euclid's proof is no longer valid because it's so like really really old. The UNSW experts are more up to date. Their maths is so really really trendy, that the new breakthrough is that π⁴+π⁵=e⁶ exactly! WOW! Gee. I didn't know that. We can all live in peace and harmony knowing we have such brilliant mathematicians at UNSW. Well done guys. Tell the world of your new insights. You'll be the envy of the whole mathematical world with that one I bet."

I bet. ....

... hmmmm ... Yeah. That's right no_arg. I'll put money on it. I bet you $50 you're wrong! Or couple of cartons of VB. Do you accept?