who_loves_maths
I wanna be a nebula too!!
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- HSC
- 2005
sorry for making this thread on "pi and e" again, but i had no choice on account of the last one being closed.
Anyhow, just found out something interesting.
Wolfram - Pi
specifically, it was this line:
the fact that 'ln(pi)' is not known in the maths community to be irrational or rational yet can mean that proving either 1) [ln(pi +1) -2]/(1 - ln(pi)), or 2) [6 - ln(pi +1)]/ln(pi) to be irrational is not yet even possible (certainly beyond the reach of HSCers on this forum i think).
so buchanan, i guess you can stop trying that 'dead-end' now
the only way to use [ln(pi +1) -2]/(1 - ln(pi)), or [6 - ln(pi +1)]/ln(pi) to prove the case against no_arg is then to show that they are, as i said before, non-integral - which will suffice as a proof for this particular problem of approximation.
alternatively, if the "equality" of [ln(pi +1) -2]/(1 - ln(pi)) = [6 - ln(pi +1)]/ln(pi) can be proven false without using a calculator, then it also follows that the original "equality" of pi^4 + pi^5 = e^6 is a mere approximation...
Anyhow, just found out something interesting.
Originally Posted by buchanan
Perhaps all that is now required is to in fact prove (ln(π +1) -2)/(1 - ln(π)) is irrational.
Originally Posted by no_arg
all you've done is rewrite the question!
Clearly [ln(pi +1) -2]/(1 - ln(pi))=4
It is perfectly possible for the ln of an irrational to be rational!...
Just went to look for some info on 'pi' on Wolfram, and was surprised at a line on the website:Originally Posted by no_arg
Does it really matter whether or not I actually believe that [ln(pi +1) -2]/(1 - ln(pi))=4?
Wolfram - Pi
specifically, it was this line:
this might mean that the problem of showing pi^4 + pi^5 = e^6 is an approximation isn't as simple to prove (ie. without calculator) as showing that [ln(pi +1) -2]/(1 - ln(pi)) is irrational...It is not known if (pi+e), pi/e, or ln(pi) are irrational.
the fact that 'ln(pi)' is not known in the maths community to be irrational or rational yet can mean that proving either 1) [ln(pi +1) -2]/(1 - ln(pi)), or 2) [6 - ln(pi +1)]/ln(pi) to be irrational is not yet even possible (certainly beyond the reach of HSCers on this forum i think).
so buchanan, i guess you can stop trying that 'dead-end' now
the only way to use [ln(pi +1) -2]/(1 - ln(pi)), or [6 - ln(pi +1)]/ln(pi) to prove the case against no_arg is then to show that they are, as i said before, non-integral - which will suffice as a proof for this particular problem of approximation.
alternatively, if the "equality" of [ln(pi +1) -2]/(1 - ln(pi)) = [6 - ln(pi +1)]/ln(pi) can be proven false without using a calculator, then it also follows that the original "equality" of pi^4 + pi^5 = e^6 is a mere approximation...